robert huben with help from matt mills 2015-2016 · 2019. 11. 8. · algebra 901-902 notes robert...

Algebra 901-902 Notes

Robert HubenWith help from Matt Mills

2015-2016

Contents

1 MATH 901 51 Roots of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Separable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Inseparable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Fundamental Theorem of Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 Norm and Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Solvable Groups and Radical Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248 Algebraic Independence and Transcendental Extensions . . . . . . . . . . . . . . . . . . . . . 319 Introduction to Rings and Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3310 Exact Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3511 Notherian Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3612 Simple Rings and Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3813 Semisimple Rings and Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3914 Filtrations and Length of Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4215 Classification of Semisimple Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4516 Weyl Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5117 k-Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5418 Jacobson Radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5619 Projective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6120 Group Rings are Semisimple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6221 Representaions of Finite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

2 MATH 902 701 Representation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703 Localization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 824 Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 895 Category Theory and Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 936 Projective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1017 Injective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1028 Integral Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1059 Affine Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11210 Algebraic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11411 Invariant Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11612 Extensions Of A Field of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11713 Representation Theory Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12014 Primary Decompositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12415 Associated Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

1

3 Appendix 1341 Projective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1342 Injective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1353 Semisimple Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1354 Semisimple Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1365 Localization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1366 Hom Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1367 Tensors and Flat Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1378 Primary Decompositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1379 Integral Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13810 Techniques For 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

4 Index 139

2

List of Named Theorems

1 Homework Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Eisenstein’s Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Primitive Element Theorem 1 (PET 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 Primitive Element Theorem II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Homework Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1211 Fundamental Theorem of Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1612 Artin’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1616 Fundamental Theorem of Galois Theory, Part 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 193 Homework Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 Homework Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2118 Linearly Independent Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2219 Hilbert’s Satz 90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 Homework Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2733 Example (Don’t got time for this!) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3636 Hilbert Basis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386 Homework Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3817 Lemma (Zassenhaus Lemma) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4240 Schreier Refinement Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4344 Artin-Wedderburn Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4945 Rieffel’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4947 Artin-Wedderburn, Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5049 Example (The Weyl Algebra) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5149 Jacobson Density Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5350 Artin-Wedderburn, Part 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 547 Homework Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5451 Burnside . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5555 Nakayama’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5957 The Splitting Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6060 Maschke’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

69 Krull’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 808 Homework Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8274 Nakayama’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 859 Homework Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9880 Hom−⊗ Adjointness/Adjunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10083 Baer’s Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10286 Lying Over Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10887 Incomparable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10888 Going Up Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10990 Going Down Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11092 Nullstellensatz - Strong Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11393 “Trick of Rabinowitsch” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11395 Nullstellensatz - geometric version . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

3

106 Schur . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122110 First Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129111 Krull’s Intersection Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

150 First Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137151 Second Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137154 Krull’s Intersection Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138155 Lying Over Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138156 Incomparability Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138157 Going Up Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138158 Going Down Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

4

Chapter 1

MATH 901

1 Roots of Unity

1.1 Day 1 - August 24

Here we go!Let n ≥ 1 ∈ N, and let Un = {ω ∈ C : ωn = 1}. Recall that this is a cyclic group generated by e

2πin , or

generated by e2kπin for gcd(k, n) = 1. These generators are called primitive n-th roots of unity .

Remark 1. We will typically remember n and let ω be a primitive n-th root of unity. We will use |ω| todenote the order of ω, i.e. n.

Definition 1. For n ≥ 1, let Φn(x) =∏|ω|=n

(x− ω).

Example 1.

Φ1(x) = x− 1

Φ2(x) = x+ 1

Φ4(x) = x2 + 1

Remark 2. Note that deg(Φn(x) = φ(n).

Note also that xn − 1 =∏

|ω| dividing n

(x− ω) =∏d≥1d|n

∏|ω|=d

(x− ω) =∏d|n

Φd(x).

Lemma 1. Φn(x) ∈ Z[x]

Proof. Induct on n: Base case is true as Φ1 = x− 1.

Inductive case: n > 1. Let f(x) =∏d|nd<n

Φd(x). By the inductive hypothesis, f(x) ∈ Z[x]. Note that

xn − 1 = f(x)Φn(x), and note also that f(x) is monic.Recall that for all rings R, R[x] has a division algorithm for monic polynomials. In this case, there exists

a unique q(x), r(x) ∈ Z[x] such that xn − 1 = f(x)q(x) + r(x) and where deg(r) < deg(f) or r(x) = 0.But this is also true in C[x]. Since the quotient and remainder are unique, and f(x) divides xn − 1 in

C[x], then there is no remainder, and thus xn − 1 = f(x)q(x), so q(x) = Φn(x), but q(x) ∈ Z[x].

Theorem 1. Φn(x) is irreducible in Q[x].

Proof. Assume for sake of contradiction that Φn(x) is reducible in Q[x]. Then by Gauss’ Lemma, we can infact factor it in Z[x]. That is, there exist f(x), g(x) ∈ Z[x] (non-constant), such that Φn(x) = f(x)g(x).

5

Furthermore, we can assume WLOG that f(x) is irreducible in Q[x]. Also, since Φn(x) is monic, thenf(x) and g(x) are (WLOG) monic as well.

Recall that the roots of Φn(x) are precisely the primitive n-th roots of unity. Let ω ∈ C be a root off(x). Then |ω| = n. Let p be any prime such that p6 | n. Then |ωp| = n, so ωp is a root of Φn(x).

We wish to show that f(ωp) = 0. If this is not true, then g(ωp) = 0. Then, ω is a root of g(xp) ∈ Z[x].Note that since f(x) is irreducible, then f(x) = Min(ω,Q) (that is, it is the unique monic polynomial ofleast degree for which ω is a root). Then we know that f(x)|g(xp) (in Q[x]).

Since f(x) is monic, we can argue (as we did in the lemma) that g(xp) = f(x)h(x) for some h(x) ∈ Z[x].Now consider this in Zp[x]. Then g(xp) = f(x)h(x). But g(xp) = g(x)p (basically because of the

Frobenius Automorphism). Let β be a root of f(x) (in some algebraic closure of Zp[x]). Then g(β)p = 0, sog(β) = 0.

Then Φn(x) = f(x)g(x) in Zp[x]. Thus β is a multiple root of Φn(x). However, Φn(x)|xn − 1. Hence,xn − 1 ∈ Zp[x] has a multiple root. But p was chosen so that p 6 |n, so the formal derivative of xn − 1 isnxn−1, which only has roots at x = 0 (and 0 is certainly not a root of xn − 1). This is a contradiction.

Thus f(ωp) = 0. But by the same argument, ωp2

is a root, and so is ωp3

, and so on. But this gives thatevery primitive n-th root of unity is a root of f(x), so deg(f) = deg(Φn), which is a contradiction.


Recall from last class that we proved that each cyclotomic polynomial Φn(x) is irreducible and has integercoefficients.

Example 2. Let’s find Φ12(x). We know it is a divisor of x12 − 1 = (x6 − 1)(x6 + 1). The roots of Φ12(x)have order 12, so none of them are roots of x6−1 (which have order dividing 6). Thus Φ12(x) divides x6 +1.We can factor like so: x6 + 1 = (x2 + 1)(x4 − x2 + 1). Similarly, the roots of x2 + 1 have order 4, so Φ12(x)divides x4 − x2 + 1. But Φ12(x) is monic and has degree 4, so Φ12(x) = x4 − x2 + 1.

Corollary 1. Let E be a splitting field of xn − 1 over Q. Then [E : Q] = φ(n).

Proof. We know that E = Q(ω) (where ω = e2πin ). Then [Q(ω) : Q] = deg(Min(ω,Q)) = deg Φn(x) =

φ(n).

2 Separable Extensions

Definition 2. Let F be a field, and let F be a fixed algebraic closure of F . Let f(x) ∈ F [x] and let α ∈ Fbe a root of f(x). If (x−α)2|f(x) (in F [x]), then we say α is a multiple root of f(x). Otherwise, α is calleda simple root .

Proposition 1. If f(x) is a polynomial, then f(x) has no multiple roots (in F ) if and only if gcd(f, f ′) = 1.

Definition 3. Let f(x) ∈ F [x] be irreducible. Then f(x) is called separable if f(x) has no multiple roots inF .

Proposition 2. Let f(x) ∈ F [x] be irreducible. Then (1) if charF = 0, then f(x) is separable. Also, (2) ifcharF = p, then f(x) has multiple roots if and only if f(x) = g(xp) for some g(x) ∈ F [x].

Definition 4. Let E/F be an algebraic extension. An element α ∈ E is separable over F if Min(α, F ) isseparable. We say E/F is a separable extension if every element of E is separable over F .

Definition 5. A field F is called perfect if every algebraic extension of F is separable. (Equivalently, everyirreducible polynomial in F [x] is separable.)

Theorem 2. Let F be a field. Then (1) if charF = 0, then F is perfect. Also, (2) if charF = p, then F isperfect if and only if F = F p = {αp : α ∈ F}.

Proof. (1) follows immediately from proposition 2.We shall now prove (2). Suppose F is perfect. Let α ∈ F . Consider xp − α ∈ F [x]. Let β be a root of

xp − α in F . Then βp = α.

6

Also, Min(β, F )|xp − α. But xp − α = xp − βp = (x − β)p. Thus Min(β, F ) = (x − β)i for some1 ≤ i ≤ p. But we assumed F is perfect, so β is separable, so its minimal polynomial has no multiple roots.Thus i = 1. But by definition of minimal polynomial, Min(β, F ) ∈ F [x], so x− β ∈ F [x], and thus β ∈ F .Thus α = βp ∈ F p.

Therefore, F ⊂ F p. But the opposite direction, that F p ⊂ F , is easy, so F p = F .Conversely, assume F = F p. Suppose for the sake of contradiction that F is not perfect. Then there

exists a minimal polynomial f(x) ∈ F [x] which has multiple roots. So f(x) = g(xp) for some g(x) ∈ F [x].Then f(x) = bmx

pm + bm−1x0(m−1) + ... + b0 where the bi ∈ F . By hypothesis, for each bi, we can find an

ai ∈ F such that bi = api .Thus f(x) = (amx

m + am−1xm−1 + ...+ a0)p. This contradicts the fact that f(x) is irreducible. Thus F

is perfect.

Recall the following: Let E/F be an algebraic extension, and let σ : F → L be a nonzero field map (recallthat these must be injective, so we call them embeddings) where L is algebraicly closed. Then, there exists

an “extension” or “lifting” of σ to a τ : E → L such that τ∣∣∣F

= σ. (You draw this with a diagram too.)

In this context, let Sσ = {τ : E → L | τ∣∣∣F

= σ}.

Proposition 3. Let E/F be as above, and let σ : F → L1, π : F → L2 be nonzero field maps. Supposealso L1 and L2 are algebraically closed. Then, |Sσ| = |Sπ|.

Proof. “Without loss of generality”, we can assume that L1 and L2 are algebraic closures of σ(F ) and π(F )respectively. We can do this because E/F is algebraic, so τ(E) will be algebraic over σ(F ). Thus τ(E) liesinside the algebraic closure of σ(F ) inside L1. We can make a similar argument for L2.

Consider the following diagram which I can’t draw.

The long story short is that we get a λ1 : L1 → L2 such that λ∣∣∣σ(F )

= πσ−1.

We claim that λ is an isomorphism.This follows because L1 is an algebraic closure of σ(F ). Since λ(L1) is an algebraic closure of π(F )

inside L2, L2/λ(L1) is an algebraic extension. As λ(L1) is algebraicly closed, then L2 = λ(L1). Thus λ issurjective. Also, λ is a field map, so it is automatically injective. Thus λ is an isomorphism.

Let τ ∈ Sσ. Then λτ : E → L1 → L2 and λτ∣∣∣F

= λ(τ∣∣∣F

) = λσ = πσ−1σ = π. Thus λτ ∈ Sπ. This gives

a map λ : Sσ → Sπ by (σ) = λσ. Similarly we can get an λ−1 : Sπ → Sσ by λ−1(σ) = λ−1σ. But λ−1 and

λ are inverses, so |Sσ| = |Sπ|.

Definition 6. Let E/F be algebraic. The separable degree of E/F , denoted [E : F ]S is |Sσ| for anyembedding σ : F → L (where L is algebraically closed).

In particular, fix an algebraic closure E of E. Then [E : F ]S = |{τ : E → E : τ fixes F}|. We coulddenote this as |S1| if we were feeling funny. We will show next time that the separable degree is multiplicative.


Recall from last class that if E/F is an algebraic extension, we use [E : F ]S to denote the separable degree.

Lemma 2. Let E/F be an algebraic extension, and let α ∈ E. Let f(x) = Min(α, F ). Then [F (α) : F ]S =number of distinct roots of f(x) in F . (In particular, if α is separable over F , then [F (α) : F ]S = deg f(x) =[F (α) : F ].

Proof. We know that [F (α) : F ]S = number of distinct embeddings of F (α) → F which fix F . If σ :F (α)→ F fixes F , then σ is determined by σ(α). Note that f(σ(α) = σ(f(α)) = σ(0) = 0.

Thus σ(α) is a root of f(x) in F . So [F (α) : F ]S ≤ # of distinct roots of f(x). If β ∈ F is a root of f(x),then there exists σ : F (α)→ F such that σ(α) = β. Thus, [(α) : F ]S ≥ # of distinct roots of f(x). But f(x)is minimal and separable, so it has deg f(x) distinct roots.

7

Proposition 4. Let K ⊂ L ⊂ E be fields such that [E : K] is finite. Then [E : K]S = [E : L]S [L : K]S .Moreover, [E : K]S ≤ [E : K].

Proof. (Proof of multiplicativity.) Fix an algebraic closure E of E. Then E = L = K. Let S = {π : L →E|π fixes K}. Then |S| = [L : K]S .

Given π ∈ S, let Tπ = {τ : E → E|τ∣∣∣L

= π}. The |Tπ| = [E : L]S . Note also that if π1 6= π2, then Tπ1

and Tπ2are disjoint.

Let T =⋃π∈S Tπ. Note that every σ ∈ T is a function mapping E to E, and also fixes K. Furthermore,

if σ : E → E fixes K, then σ∣∣∣L

: L→ E fixes K, so σ∣∣∣L

= π ∈ S. Thus σ ∈ Tπ ⊂ T . Hence |T | = [E : K]S .

Hence |T | = [E : K]S , but |T | =∑π∈S|Tπ| =

∑π∈S

[E : L]S = [L : K]S [E : L]S .

Proof. (Proof of “moreover”.) Since E/K is finite, E = K(α1, ...αn). We use induction on n to show that[E : K]S ≤ [E : K].

If n = 1, then by the lemma we have have that the [E : K]S is the number of roots of the minimalpolynomial of α1, and [E : K] is the degree of the polynomial of α1. But it is an ultra-classical result thatthe number of roots is less than or equal to the number of degrees.

If n > 1, assume the inductive hypothsis. Then let L = K(α1, ...αn−1). Then E = L(αn). By induction,[L : K]S ≤ [L : K]. By the n = 1 case, [E : L]S ≤ [E : L], so [E : K]S ≤ [E : K].

Example 3. Let p be a prime and Fp be the field of p elements. Let t be an indeterminant. Let E = Fp(t),and K = Fp(tp). Then t is algebraic over K, since t is a root of f(x) = xp − tp ∈ K[x].

We claim that in fact, f(x) = Min(t,K). Let g(x) = Min(t,K). Then g(x)|f(x) in K[x], and hencein E[x]. However, in E[x], xp − tp = (x − t)p. Thus g(x) = (x − t)i for some 1 ≤ i ≤ p. But theng(x) = xi − itxi−1 + ... ∈ K[x]. Thus −it ∈ K. But if i < p, then i is a unit in Fp, so t ∈ K. But this isabsurd, since K = Fp(tp).

Thus [E : K] = p. But there is exactly one root of xp − tp (namely, x = t), so [E : K]S = 1.

Remark 3. Let E/F be algebraic, and let α ∈ E. Then α is separable over F if and only if [F (α) : F ]S =[F (α : F ]. (The proof of this was hidden in the lemma: [F (α) : F ]S is the number of distinct roots, whichequals [F (α : F ] if and only if all the roots are distinct, aka the polynomial is separable.)

Remark 4. Suppose E/F is algebraic and separable, and suppose L is an intermediate field. (It is directlytrue from definitions that L/F is separable.) Then E/L is separable.

Proof. Let α ∈ E. Let f(x) = Min(α, F ), and let g(x) = Min(α,L). Since f(x) ∈ L[x], and f(α) = 0, sog(x)|f(x). But since f(x) is separable, it has no repeated roots, and thus g(x) has no repeated roots. Henceα is separable over L.

Theorem 3. Let E/F be a finite extension. Then E/F is separable if and only if [E : F ]S = [E : F ].

Proof. Suppose E/F is separable. Then we can write E = F (α1, ..αn). We will now induct on n.If n = 1, then the theorem follows from remark 3.If n > 1, then suppose the inductive hypothesis. Let L = F (α1, ...αn−1). Then E = L(αn). As E/L is

separable by remark 4, then by the inductive hypothesis, [L : F ]S = [L : F ]. Furthermore, by the n = 1 case,[E : L]S = [E : L]. Thus because both degree and separable degrees are multiplicative, [E : F ]S = [E : F ].

Conversely, suppose [E : F ]S = [E : F ]. Let α ∈ E. Then [E : F (α)]S [F (α) : F ]S = [EF ]S = [E : F ] =[E : F (α)][F (α) : F ]. But since [E : F (α)]S ≤ [E : F (α)] and [F (α) : F ]S ≤ [F (α) : F ], the the only wayfor us to have equality when we multiply these is to have equality in both of these expressions. That is,[F (α) : F ]S = [F (α) : F ]. By remark 3, we know that α is separable over F . But α was arbitary, so everyelement is separable, so E is separable.

Robert’s alternate proof of the converse: suppose E/F were not separable. Then there would be someα ∈ E such that Min(α, F ) has repeated roots. Then by remark 3, [F (α) : F ]S < [F (α) : F ]. By the“moreover” part of the proposition, [E : F (α)]S ≤ [E : F (α)]. Combining these two inequalities, we have[E : F (α)]S [F (α) : F ]S < [E : F (α)][F (α) : F ]. But since both separable degree and regular degree aremultiplicative, the left hand side is [E : F ]S and the righthand side is [E : F ].

8

Exercise 1. Let K ⊂ L ⊂ E be fields with E/K algebraic. Suppose E/L and L/K are separable. Thenprove E/K is separable.

Corollary 2. Let E = F (α1, ...αn), where each αi is separable and algebraic over F . Then E/F is algebraicand separable.

Proof. Induct on n. If n = 1, then by remark 3, we have that [F (α1) : F ]S = [F (α1) : F ]. But by thetheorem, F (α1)/F is separable.

(Rest of the proof is a sketch.) In the inductive case, we use multiplicativity.


Definition 7. A field F is called separably closed if there does not exist an extension field E ) F such thatE/F is algebraic and separable.

Given a field K, a field L is a separable closure of K if L ⊂ K and if L/K is separable algebraic and Lis separably closed.

Proposition 5. Let F be a field, and let F be a fixed algebraic closure of F . Let F sep = {α ∈ F |α is separable over F}.Then F sep is a field and is a separable closure of F .

Proof. Certainly, F sep contains F . Then it suffices to show that F sep is closed under the four field operations.Let α, β ∈ F sep. Since α, β are separable, then F (α, β)/F is separable. But α+β, α−β, αβ, αβ if (β 6= 0)

∈ F (α, β) ⊂ F sep. Thus F sep is closed under these operations, so it is a field.Suppose E/F sep is an algebraic and separable extension. Without loss of generality, we can assume

E ⊂ F (since we can assume everything is happening within a single algebraic closure). We have thatE/F sep is separable, and F sep/F is separable, so by homework problem 2 [edit: see immediately below],E/F is separable. That is, every element of E is an element of F which is separable over F , so E ⊂ F sep.Thus E = F sep.

Homework Problem 1. Claim: Let K ⊂ F ⊂ E be a tower of fields such that F/K and E/F are algebraicand separable. Then E/K is algebraic and separable.

Definition 8. Let E,F be subfields of a field L. Then the compositum or join of E and F is EF =⋂K subfield of L

E∪F⊂K

K.

Example 4. Let E = K(α1, ...αm) ⊂ L and let F = K(β1, ...βn) ⊂ L. Then EF = K(α1, ...αm, β1, ...βn).

Exercise 2. If E/K and F/K are algebraic then EF/K is algebraic. If E/K and F/K are separable thenEF/K is separable.

Theorem 4 (Eisenstein’s Criterion). Let D be a Unique Factorization Domain, and let F = Q(D) (iethe field of fractions of D). Let f(x) = anx

n + ...a1x + a0 ∈ D[x], and suppose there exists an irreducibleelement π ∈ D such that π6 | an, π|ai for all i < n, and π26 | a0. Then f(x) is irreducible in F [x].

Proof. Suppose for the sake of contradiction that f(x) is reducible in F [x]. By Gauss’s Lemma, f(x) isreducible in D[x], so let f(x) = g(x)h(x) be a non-trivial factorization in D[x].

Let g(x) = bmxm+ ...+b0, and let h(x) = clx

l+ ...c0. In (D/(π))[x], write g(x) = bmxm+ ...bjx

j (bj 6= 0),and similarly write h(x) = clx

l + ...cixi (ci 6= 0)

In (D/(π))[x], we have that

anxn = f(x)

= g(x)h(x)

= (bmxm + ...+ bix

i)(clxl + ...cjx

j)

= bmclxn + ...+ bicjx

i+j

9

If i < m or j < l, then bicj = 0 in D/(π). But D/(π) is a domain since π is prime, so bi = 0 or cj = 0,which contradicts our assumption. Thus i = m and j = l. In particular, b0 = c0 = 0, so π|b0 and π|c0. Buta0 = b0c0, and thus π2|a0, which is a contradiction.

Proposition 6. Let K be a field and u be a transcendental element over K. Let F = K(u). Then xn − uis irreducible in F [x] for all n ≥ 1.

Proof. Let F = Q(D) where D = K[u]. Since u is transcendental over K, then K[u] is a polynomial ring, soit is a UFD. Then u is an irreducible element in D, so by Eisenstein’s Irreducibility Criterion (with π = u),xn − u is irreducible in F [x].

Example 5. Let K be a field of characteristic p > 0. Let u be transcendental over K, and let v betranscendental over K(u).

Let E = K(u, v) and let F = K(up, vp). What is [E : F ]?Let L = K(u, vp). Consider the tower F ⊂ L ⊂ E. We can see that Min(v, L) = xp − vp is irreducible

over K(u, vp) as vp is transcendental over K(u).Also, Min(u, F ) = xp − up, as up is transcendental over K(vp) (take Professor Marley’s word for this).

Thus [E : F ] = [E : L][L : F ] = p2.

Remark 5. In the previous example, note that if g(u, v) ∈ E, then g(u, v)p ∈ Kp(up, vp) ⊂ K(up, vp) = F .Thus Min(g(u, v), F )|xp − g(u, v)p, so [F (g(u, v)) : F ] ≤ p for all g ∈ E. Thus E/F has no primitiveelement.

Recall from 818 the following theorems:

Theorem 5 (Primitive Element Theorem 1 (PET 1)). Let E/F be a finite separable extension. Thenthere exists α ∈ E such that E = F (α).

Theorem 6. (Primitive Element Theorem 2 (PET 2)) Let E/F be a finite field extension. Then there existsα ∈ E such that E = F (α) if and only if there exist only finitely many intermediate fields of E/F .

2.3 Day 5 - September 2

We shall prove the following theorem (which was stated last class):

Theorem 7 (Primitive Element Theorem II). Let E/F be a finite extension. Then E = F (α) for someα ∈ E if and only if there exists only finitely many intermediate fields of E/F (that is, if there exist onlyfinitely many fields L such that F ⊂ L ⊂ E.

Proof. Let S = {intermediate fields of E/F} = {L field|E ⊂ L ⊂ F}.Suppose E = F (α). Let f(x) = Min(α, F ), and let E be a fixed algebraic closure of E.Let T = {monic polynomial factors of f(x) in E[x]}. Note that f(x) = (x−α1)...(x−αn) ∈ E[x], so the

monic factors of f(x) are of the form (x−αi1)...(x−αik) for some i1...ik distinct elements of {1, ...n}. ThusT is finite.

Define λ : S → T by L 7→Min(α,L). (Since Min(α,L)|Min(α, F ), then Min(α,L) ∈ T .)We wish to show that λ is injective. Suppose L1, L2 ∈ S andMin(α,L1) = Min(α,L2) = xn+cn−1x

n−1+...c0 ∈ Li[x] for i = 1, 2. It suffices to show that L1 = F (c0, ..., cn−1) (since by the exact same argument itwould follow that, L2 = F (c0, ..., cn−1).

Let K = F (c0, ...cn−1). Since the ci ∈ L1, then certainly K ⊂ L1. It then suffices to show that[E : L1] = [E : K] (since then [L1 : K] = 1).

10

Let g(x) = Min(α,L1). Note g(x) ∈ K[x] and is irreducible in L1[x], so it is also irreducible in K[x].Thus g(x) = Min(α,K). Note that E = K(α) = L1(α). Thus

[E : K] = deg(Min(α,K))

= deg(g(x))

= deg(Min(α,L1))

= [E : L1]

Following this string of implications all the way back, we can conclude that λ is injective. Since λ injects Sinto T , and T is a finite set, then S is also finite, as desired.

Conversely, suppose there are only finitely many intermediate fields. We have two cases: either F is finiteor infinite.

If F is finite, then E is also finite, so E× is cyclic (since the unit group of any finite field is cyclic). ThusE× =< α > (in the sense of generating a group) for some α, so E = F (α).

If F is infinite, then since the extension is finite, we can write E = F (α1, ...αn) for some α1, ..., αn ∈ E.We then induct on n.

If n = 1, then E = F (α1), as desired.If n > 1, we can really just show that n = 2 case. So suppose E = F (α, β). Then let Λ = {F (α +

cβ)|c ∈ F} ⊂ S. Since S is finite, then Λ is finite, but |F | = ∞, so there exist c1 6= c2 ∈ F such thatF (α+ c1β) = F (α+ c2β). Let us call this field L. Then (c1 − c2)β = (α+ c1β)− (α+ c2β) ∈ L, so β ∈ L.Then α ∈ L as well, so L = F (α, β), and thus F (α, β) is generated by a single element.

3 Inseparable Extensions

Let us turn our attention to inseparability.

Theorem 8. Let F be a field of characteristic p > 0, and let α ∈ F . Then,

1. α is separable over F if and only if F (α) = F (αp)

2. If α is inseparable over F then [F (α) : F (αp)] = p and Min(α, F (αp)) = xp − αp.

3. For all n ≥ 1, [F (αpn−1

) : F (αpn

)]S = 1, so [F (α) : F ]S = [F (αpn

) : F ]S

4. There exists n ∈ N such that αpi

is separable over F for all i ≥ n.

5. Let n be the least exponent such that αpn

is separable over F . Then [F (α) : F ] = pn[F (α) : F ]S .

Proof. (1) Suppose α is separable over F . Then α is separable over F (αp), but we know thatMin(α, F (αp))|xp−αp = (x − α)p. Since Min(α, F (αp)) has distinct roots then it must be that deg(Min(α, F (αp))) = 1, soα ∈ F (αp). Thus F (α) = F (αp).

[Alternative proof of the forward direction: Since α is separable, then [F (α) : F (αp)] = [F (α) : F (αp)]S =# of distinct roots of Min(α, F (αp)) = 1.]

Conversely, suppose F (α) = F (αp). Suppose for the sake of contradiction that α is not separableover F . Let f(x) = Min(α, F ). Then f(x) = g(xp) for some g(x) ∈ F [x]. Thus g(αp) = 0. Then[F (αp) : F ] ≤ deg(g(x)) < deg(g(xp)) = [F (α) : F ]S . This is a contradiction, so F (α) = F (αp).

(2) We know that Min(α, F (αp))|xp−αp. By homework problem 5 [edit: see immediately below], eitherxp−αp is irreducible in F (αp)[x] or it splits completely. If it splits, then α ∈ F (αp), which implies that α isseparable by part (1) of the theorem. But this is a contradiction, since we assumed α is inseparable. Thusxp − αp is irreducible, so it is the minimal polynomial of α as desired.

11

(3) We know that [F (α) : F (αp)]S = # of distinct roots of Min(α, F (αp)) = 1. From part (2), Min(α, F (αp)) =

(x − α)p, so it only has one distinct root. By applying this argument to αpn−1

, we get that [F (αpn−1

) :

F (αpn

)]S = 1 for all n. Thus [F (α) : F (αpn

)]S = [F (α) : F (αp)]S ...[F (αpn−1

) : F (αpn

)]S = 1, so[F (α) : F ]S = [F (α) : F (αp

n

)]S [F (αpn

) : F ]S = [F (αpn

) : F ]S for all n.

(4) Consider the descending chain of F -vector spaces F (α) ⊃ F (αp) ⊃ F (αp2

) ⊃ .... Since dimF (Fα) =

[F (α) : F ] < ∞, then this chain stabilizes. Thus there exists an n ∈ N such that F (αpn

) = F (αpn+1

).

But (αpn

)p = αpn+1

, so by part (1), αpn

is separable over F . Thus F (αpn

) is separable over F but since

αpi ∈ F (αp

n

) whenever i ≥ n, then αpi

is separable for all i ≥ n.

(5) We have the following tower of fields: F (α) ⊃ F (αp) ⊃ F (αp2

)... ⊃ F (αpn

). For each of these, thedegree of the extension is p by part (2), and the separable degree is 1 by part (3). By assumption, αp

n

isseparable, so F (αp

n

) is separable. Thus [F (αpn

) : F ] = [F (αpn

) : F ]S .But

[F (α) : F ] = [F (α) : F (αp)][F (αp) : F (αp2

)]...[F (αpn−1

) : F (αpn

)][F (αpn

) : F ]

= pn[F (αpn

) : F ]

= pn[F (αpn

) : F ]S

= pn1n[F (αpn

) : F ]S

= pn[F (α) : F (αp)]S [F (αp) : F (αp2

)]S ...[F (αpn−1

) : F (αpn

)]S [F (αpn

) : F ]S

= pn[F (α) : F ]S

Homework Problem 2. Let E/F be an extension, and suppose charF = p > 0. Let α ∈ E. Then eitherxp − α is either irreducible over F , or it factorizes completely.

Corollary 3. Let E/F be a finite extension, and let CharF = p. Then [E : F ] = pn[E : F ]S for somen ≥ 0.

Proof. Let E = F (α1, ..., αk) for some α1, ..., αk ∈ E. We induct on k.If k = 1, we have this from part (5) of the previous theorem.If k > 1, then let L = F (α1, ..., αk−1). By the inductive hypothesis, [L : F ] = p`[L : F ]S for some ` ∈ N.

Also, E = L(αk). By the k = 1 case, [E : L] = pt[E : L]S for some t ∈ N. Then we multiply, and get that[E : F ] = [E : L][L : F ] = p`+t[E : L]S [L : F ]S = p`+t[E : F ]S .

Definition 9. If E/F is a finite extension, with [E : F ] = pk[E : F ]S , then the inseparable degree of E/Kis pk.

Theorem 9. (Really easy) The inseparable degree is multiplicative.


Andrew sez:im helping!On homework 1 problem 4, we can assume the degrees are finite.

Definition 10. Let F be a field of characteristic p > 0, and let α ∈ F . Then α is purely inseparable (oftenwritten p.i.) if αp

n ∈ F for some n ≥ 0.This is equivalent to Min(α, F )|xpn − β in F [x].We say E/F is purely inseparable if each α ∈ E is p.i. over F .

Lemma 3. Let α ∈ F . Then the following are equivalent

1. α is p.i. over F .

2. [F (α) : F ]S = 1

3. [F (α) : F ]i = [F (α) : F ]

12

Proof. We note that 2 and 3 are equivalent by the definitions of separable and inseparable degrees.For 1 and 2, note that α is p.i. over F if and only if αp

n ∈ F for some n ≥ 0. But this is the case if andonly if [F (αp

n

) : F ] = 1 for some n ≥ 0. But this is the case if and only if [F (αpn

) : F ]S = 1 for some n ≥ 0(since for some n sufficiently large, αp

n

is separable by a theorem from last class). But this is the case if andonly if [F (α) : F ]S = 1 since [F (α) : F ]S = [F (αp

n

) : F ]S .

Theorem 10. Let E/F be a finite extension. Let E = F (α1, ...αn) for some αi ∈ E. Then the followingare equivalent.

1. E/F is p.i.

2. Each αi is p.i. over F

3. [E : F ]S = 1

4. [E : F ] = [E : F ]i

Proof. Note that (3) and (4) are equivalent by the definition of separable and inseparable degree.Note that (1) implies (2) by definition.Suppose (2) holds. We induct on i to prove that [F (α1, ...αi) : F ]S = 1. The i = 1 case is the lemma.If i > 1, let L = F (α1, ...αi−1). By the inductive hypothesis, [L : F ]S = 1. As αi is p.i. over F , then

[L(αi) : L]S = 1, so (L(αi) : F ]S = 1, but L(αi) = F (α1, ..., αi). Thus (2) implies (3).Suppose 3 is true. Then let β ∈ E. Then [F (β) : F ]S ≤ [E : F ]S = 1, so by the lemma β is p.i. over F .

Thus (3) implies (1), and all are equivalent.

Definition 11. Let E/F be an algebraic extension, and let charF = p. Let L = {α ∈ E|α is p.i. over F}.Then L is a field (proof omitted, but straightforward). Certainly, L is intermediate between E and F . Wecall L the inseparable closure of F inside E.

4 Normal Extensions

Question 1. If E/F is an algebraic extension, is E generated by the separable closure of F and theinseparable closure of F?

Answer: yes, if E/F is normal.

Lemma 4. Let E/F be an algebraic extension. And let σ : E → E be a field map fixing F . Then σ is anautomorphism of E.

Proof. Let β ∈ E. Let f(x) = Min(β, F ), and let S = {all roots of f(x) in E}. Observe that S is anonempty finite set. As f(x) has coefficients in F , and σ fixes F , then σ(β) is a root of f(x). Thus σ mapselements of S to elements of S. But S is finite, and σ is injective (as it is a field map), so in fact σ permutesS. Thus in particular there exists a γ ∈ S such that σ(γ) = β. Thus σ is onto, so it is a bijection.

Proposition 7. Let E/F be an algebraic extension. The following are equivalent.

1. E is a splitting field for some (possibly infinite, but definitely nonempty) set of polynomials in F [x].

2. Any embedding σ : E → E which fixes F is an automorphism of E.

3. Any irreducible polynomial in F [x] which has a root in E splits completely in E[x].

13

Proof. Suppose (1), and let S denote the set of polynomials for which E is the splitting field. Then letT = {α ∈ E : α is a root of some f(x) ∈ S}. Certainly, E = F (T ).

Let σ : E → E and suppose σ fixes F .By the same argument as in the previous lemma, σ is a permutation of T , so namely σ(T ) = T ⊂ E.

Thus σ(T ) ⊂ E. But σ fixes F and sends elements of T to elements of T , and E is generated by F and T ,

so σ∣∣∣E

maps into E. But by the previous lemma, σ(E) = E, and thus σ is an automorphism.

Thus (1) implies (2).Suppose (2). Let f(x) ∈ F [x] be irreducible and let α ∈ E be a root of f(x). Let β ∈ E be a root of

f(x). Then there exists an isomorphism between F (α) and F (β) which fixes F (since both are isomorphicto F [x]/(f(x))) and for which α is mapped to β. Let this automorphism be called τ .

Then τ extends to an embedding σ of E into E, but by (2), then σ is in fact a field automorphism. Thusτ(α) = β ∈ E, so every root of f(x) is in E, so f(x) factors completely. Thus (2) implies (3).

Suppose (3). Let S = {Min(α, F ) : α ∈ E}. By (3), all polynomials in S split completely in E, so E isthe splitting field for S over F . Thus (3) implies (1), and all three criteria are equivalent.

Definition 12. If E/F satisfies any of the equivalent conditions in Proposition 7, then E/F is called anormal extension.

Example 6. If [E : F ] = 2, then E/F is normal.

Remark 6. Let E/F and F/K be field extensions. If E/K is normal, then E/F is normal. This followsdirectly from the criterion 1 in Proposition 7.

However, F/K may not be normal. For instance, let K = Q, F = Q( 3√

2), let ω be a primitive third rootof unity, and let E = Q(ω, 3

√2). Then E is the splitting field of x3 − 2, but F/K is not normal since x3 − 2

has a root but does not factor completely.

Remark 7. If E/F and F/K are normal extensions, then it does NOT follow that E/K is normal.For instance, if K = Q, F = Q(

√2), and E = Q( 4

√2), then E/F is degree 2, hence normal, and similarly

F/K is also degree 2, hence normal.However, x4 − 2 has a root in E but does not factor completely.


Recall from last class that if K ⊂ L ⊂ E are field extensions and E/K is normal, then E/L is normal. If Eis the splitting field for a set S ⊂ K[x], then E is also the splitting field for S as a subset of L[x].

Exercise 3. Let {Eα}α∈Λ be a set of subfields of a field L, and suppose each Eα is normal over a fixed field

F . Then⋂α∈Λ

Eα is normal over F .

Definition 13. Let E/F be an algebraic field extension. Then normal closure of E/F is⋂

E⊂K⊂EK/F normal

K. By

the previous exercise, this is normal, and by construction it is the smallest normal extension of F containingE.

Example 7. Let E = F (α1, ..., αn). Let fi(x) = Min(αi, F ), and let f(x) = f1(x)...fn(x).Then the normal closure of E/F is the splitting field for f(x) over F .

Proposition 8. Let E/F be a normal algebraic extension. Suppose E/F is inseparable. Then there existsα ∈ E \ F such that α is purely inseparable over F .

(Remark: This is false if E/F is not assumed to be normal.)

Proof. Let α ∈ E, such that α is inseparable over F . Let f(x) = Min(α, F ). Since α is inseparable over F ,then its minimal polynomial is of the form f(x) = g(xp) for some g(x) ∈ F [x].

In E[x], we can factorize g(x) = (x− α1)...(x− αn).

14

So f(x) = g(xp) = (xp − α1)...(xp − αn). For each i, let βi ∈ E be a root of xp − αi. Then βpi = αi forall i, so f(x) = (xp − βp1)...(xp − βpn) = ((x− β1)...(x− β2))p.

But each βi is a root of f(x), and f(x) ∈ F [x] is irreducible and has a root α ∈ E. But E/F is normal,so f(x) factors completely in E[x]. Thus each βi ∈ E.

Let `(x) = (x − β1)...(x − βn) ∈ E[x]. Also, `(x)p = f(x) ∈ F [x] (and `(x) 6∈ F [x] since f(x) isirreducible). Therefore, there exists a coefficient in `(x) which is not in F . Let us call that coefficient c. Butthe corresponding coefficient in `(x)p is cp, which is in F . Thus c ∈ E \ F , and c is purely inseparable (bythe definition).

Definition 14. If E/F is a field extension, then we say F insep = {α ∈ E|α is purely inseparable over F}.We call F insep the inseparable closure of F (over E).

Time for Galois Theory!

Definition 15. Let E/F be a field extension. Let Aut(E/F ) = {φ : E → E|φ fixes F} (⊂ {φ : E →E|φ fixes F}. We call Aut(E/F ) the automorphism group of E/F (and yeah, it is a group).

Remark 8. Let E/F be a finite extension. Then

1. |Aut(E/F )| ≤ [E : F ]S with equality if and only if E/F is normal.

2. |Aut(E/F )| ≤ [E : F ] (this follows immediately from the first part), with equality if and only E/F isnormal and separable.

Definition 16. Let E/F be an algebraic extension. If E/F is normal and separable, then we say that E/Fis a Galois extension. In this case, Aut(E/F ) is called the Galois group, and we denote it Gal(E/F ).

(By the previous remark, note that if E/F is finite, then E/F is Galois if and only if |Gal(E/F )| = [E :F ].)

Example 8. Let E be the splitting field of x3 − 2 over Q. The roots of x3 − 2 are 3√

2, ω 3√

2, ω2 3√

2, whereω is a primitive third root of unity.

So E = Q(ω, 3√

2). We can draw a chart of the intermediate fields as Q ⊂ Q( 3√

2) ⊂ E and Q ⊂ Q(ω) ⊂ E(wow, it would really help if I could draw diagrams in LaTeX... I should probably learn). Note that[Q( 3√

2 : Q] = 3 and [Q(ω) : Q] = 2, so 3|[E : Q] and 2|[E : Q]. Thus [E : Q] ≥ 6. But at the same time itis at most 6 (since it can at most permute the roots, and there are only 3! = 6 permutations of 3 roots), so[E : Q] = 6.

Then Gal(E/F ) = {permutations of the roots of x3 − 2}. Alternatively, if σ : E → E by ω 7→ ω2 and3√

2 7→ 3√

2, and τ : E → E by ω 7→ ω and 3√

2 7→ ω 3√

2. Then we can present the group as < σ, τ |σ2 = 1, τ3 =1, τσ = στ2 >.

Remark 9. If you are working over a field of characteristic 0, separability is free. Therefore, Galois-ness isequivalent to normality.

Example 9. Let E be the splitting field for xn − 1 over Q. Then E = Q(ω), where ω = e2πin (a primitive

n-th root of unity).Furthermore, Min(ω,Q) = Φn(x) which has degree φ(n), so [Q(ω) : Q] = φ(n).Define ψi : E → E by ω 7→ ωi. Then Gal(E/Q) = {ψi| gcd(i, n) = 1} (and one should check that ψi = ψj

if and only if i ≡ j mod n). Then Gal(E/Q) ∼= Z×n , so |Gal(E/Q)| = |Z×n | = φn. Thus the extension isGalois.

(But actually, we already knew this: the extension is separable because it is over fields of characteristic0, and it is normal since it is a splitting field. Thus it is Galois.)

5 Fundamental Theorem of Galois Theory


Special thanks to Matt Mills for letting me use his laptop today!

15

Example 10. Let E be the splitting field of x5 − 2 over Q. Then E = Q(ω, 5√

2), where ω is a primitive5-th root of unity.

Then what is the degree of the extension? Well [Q(ω) : Q] = 4 and [Q( 5√

2) : Q] = 5, so the degree ofQ(ω, 5

√2/Q) = 20.

What is the Galois group? Let L1 = Q(ω), L2 = Q( 5√

2), and E = Q(ω, 5√

2). Then we define τ : E → Eby 5√

2 7→ 5√

2ω (and fixing L1). Observe that τ has order 5. We also define σ : E → E by ω 7→ ω2 (andfixing L2). Observe that σ has order 4.

Thus σ and τ generate Gal(E/Q) = G. What is the relation between τ and σ? By Sylow’s theorem, < τ >is normal in G. Thus στσ−1 = τ i for some i between 1 and 4. In particular, σ(τ( 5

√2)) = σ(ω 5

√2) = ω2 5

√2.

Also, τ i(σ( 5√

2)) = τ i( 5√

2) = ωi 5√

2. Thus i = 2, and we can present G by G =< σ, τ |σ4 = 1, τ5 = 1, στ =τ2σ >.

Theorem 11 (Fundamental Theorem of Galois Theory). Let E/F be a finite Galois extension, and letG = Gal(E/F ). Then we have maps from {intermediate fields of E/F} to {subgroups of G} and vice-versa,given by

• If H ≤ G, let EH = {α ∈ E|σ(α) = α for all σ ∈ H}. We call EH the fixed field of H, and it is indeedan intermediate field between E/F . We say φ(H) = EH .

• Let L be an intermediate field of E/F . Then E/L is Galois, and Gal(E/L) ≤ G. We say ψ(L) =Gal(E/L).

Then φ and ψ are inverses. That is L = EGal(E/L), and H = Gal(E/EH).

Proof. We first show that L = EGal(E/L). Recall that EGal(E/L) denotes the set of elements that are fixedby every automorphism which fixes L. Then certainly L ⊂ EGal(E/L).

Instead suppose α ∈ EGal(E/L). Let σ : L(α)→ L = E be any embedding which fixes L. The number ofsuch σs is [L(α) : L]S , but since E/L is Galois, it is separable, so this equals [L(α) : L].

Let τ : E → E be any extension of σ. As τ fixes L, and E/L is normal, we have that τ : E → E, so

τ ∈ Aut(E/L) = Gal(E/L). Since α ∈ EGal(E/L), τ(α) = α. But recall that τ∣∣∣L

= σ, so σ(α) = α, and

since α was arbitrary, then σ = id. Hence, 1 = [L(α) : L]S = [L(α) : L], so α ∈ L. Thus L = EGal(E/L).[Proof that H = Gal(E/EH) comes later.]

Lemma 5. Let E/F be a separable algebraic extenion such that there exists n satisfying [F (α) : F ] ≤ n forall α ∈ E. Then [E : F ] ≤ n.

Proof. Choose α ∈ E such that [F (α) : F ] is as large as possible (we can find such an α because these degreesare bounded above by n).

Suppose for the sake of contradiction that F (α) 6= E. Then there exists some β ∈ E \ F (α). Then[F (α, β) : F ] > [F (α) : F ]. But by the first primitive element theorem (which we did not prove in class),there exists γ ∈ F (α, β) such that F (α, β) = F (γ), so this is a contradiction of the assumption that σ wasmaximal. Thus E = F (α), and [E : F ] = [F (α) : F ] ≤ n.

Theorem 12 (Artin’s Theorem). Let E be a field, and let G be a finite subgroup of Aut(E). LetEG = {α ∈ E|σ(α) = α for all σ ∈ G}. Then E/EG is Galois and finite, and G = Gal(E/EG).

Note that Artin’s Theorem makes it very easy to prove the second part of the Fundamental Theorem ofGalois Theory:

Proof. If H is a finite subgroup of Aut(E) = Gal(E/F ), then by Artin’s Theorem, H = Gal(E/EH).

16


We shall begin by proving Artin’s Theorem, which will finish our proof of The Fundamental Theorem ofGalois Theory.

Theorem 13. Let E be a field, and let G be a finite subgroup of Aut(E). Let EG = {α ∈ E|σ(α) =α for all σ ∈ G}. Then E/EG is Galois, and Gal(E/EG) = G.

Proof. Let α ∈ E. Let φ1(α), ...φr(α) be the distinct maps of α under G (where each φi ∈ G).Let G = |n|. Then r ≤ n.If τ ∈ G, as τ is one-to-one, then τφ1(α), ..., τφr(α) are distinct, so τ is merely permutes the set

{φ1(α), ...φr(α)}. Let f(x) =

r∏i=1

(x − φi(α)). Then for all τ ∈ G, fτ (x) = f(x) (where fτ (x) denotes

the polynomial given by applying τ to the coefficients of f(x)). Thus f(x) ∈ EG[x].Also note that f(α) = 0.So, Min(α,EG)|f(x). Since f(x) has distinct roots, then so does Min(α,EG). Thus α is separable over

EG for all α ∈ E. Thus E/EG is separable.Note that [EG(α) : EG] ≤ degf(x) = r ≤ n for all α ∈ E. Thus [E : EG] ≤ n by the lemma from last

class.Note that for all α ∈ E, Min(α,EG) splits in E[x] (since f(x) does). But if f(x) ∈ E[x] has a root α,

then f(x) = kMin(α,EG), so f(x) splits completely. Thus the extension is normal. Thus E/EG is Galois.Certainly, G ≤ Gal(E/EG). Then |Gal(E/EG)| ≥ |G| = n. But |Gal(E/EG)| = [E : EG] since this is a

Galois extension, but [E : EG] ≤ n, so G = Gal(E/EG).

Remark 10. Let E/F be a finite Galois Extension, and let G = Gal(E/F ). Then,

1. If L is an intermediate field, then |Gal(E/L)| = [E : L] (since E/L is Galois).

2. If H ≤ G, then |G| = [E : EH ]. This follows from part 2 of the FTGT since H = Gal(E/EH).

3. If L1 and L2 are intermediate fields, then L1 ⊃ L2 ⇐⇒ Gal(E/L1) ⊂ Gal(E/L2), and if H1 andH2 are subgroups of G, then H1 ≤ H2 ⇐⇒ EH1

⊃ EH2. The forward directions are easy, and the

backwards directions follow from the forward directions plus the Fundamental Theorem.

Example 11. Let E = Q(ω, 3√

2), the splitting field of x3 − 2 over Q. Let’s find generators of all theintermediate fields of E/Q.

Let G = Gal(E/Q) =< σ, τ > where σ fixes ω and sends 3√

2 7→ ω 3√

2, and τ fixes 3√

2 and sends ω 7→ ω2.Since this group is isomorphic to S3, then every proper subgroup is of order 2 or 3, and is hence cyclic.

In fact, the subgroups are < σ >,< τ >,< στ >,< σ2τ >. Then the subfield lattice is the subgroup latticebut order-reversed.

Since all of our extensions in this case are prime, then we can merely check that we adjoin a single thingsfixed by our generator and which is not rational.

Then the subfields are E<σ> = Q(ω), E<τ> = Q( 3√

2), E<στ> = Q(ω2 2√

3), and E<σ2τ> = Q(ω 3√

2).

Proposition 9. Let E/F be a finite Galois extension. Then E/F is separable, so by the first PrimitiveElement Theorem, E = F (α) for some α ∈ E. Let H ≤ Gal(E/F ) = G. Then

1. Min(α,EH) =∏h∈H

(x− h(α)).

2. If Min(α,EH) = xm + cm−1xm−1 + ....c0, then EH = F (cm−1, ..., c0).

17

Proof. Let E/F be a finite Galois extension, and suppose E = F (α), and let H ≤ Gal(E/F ). Furthermore,

let f(x) =∏h∈H

(x− h(α)).

For all τ ∈ H, we have that fτ (x) = f(x) since τH = H (where fτ (x) denotes τ applied to all thecoefficients of f). Thus f(x) ∈ EH [x].

Note that degf(x) = |H| = [E : EH ] = [EH(α) : EH ]. Since f(α) = 0, then Min(α,EH)|f(x). But sincef(x) and Min(α,EH) have the same degree (and are monic), then f(x) = Min(α,EH), as desired.

Let L = F (c0, ..., cm−1), and note that E = L(α). Since fτ (x) = f(x), then each ci is fixed by eachelement in H, so L ⊂ EH . But f(x) ∈ L[x], so f(x) is irreducible in L[x] (since it is irreducible in EH [x]).Thus f(x) = Min(α,L), so [E : L] = [L(α) : L] = degf(x) = [E : EH ], so L = EH .

Example 12. Let ω be a primative 19th root of unity. Let E = Q(ω), a splitting field for x19− 1 over Q[x].Then Gal(E/Q) ∼= Z×19

∼= C18 (where C18 is the cyclic group of 18 elements).Note that Gal(E/Q) is generated by φ2 : ω 7→ ω2. Thus the subgroups of G are generated by φ18 = φ9

2,φ7 = φ6

2, φ8 = φ32, and φ4 = φ2

2.Let Ei denote the subfield of E fixed by φi. Then the subfield orders are given by Q ⊂ E8 ⊂ E18 ⊂ E,

Q ⊂ E4 ⊂ E7 ⊂ E, and E8 ⊂ E7.How do we write these nicely? Let’s do E7, for which H7 =< ω7 >. Then the corresponding polynomial

is (x − ω)(x − ω7)(x − ω11) = x3 − (ω + ω7 + ω11)x2 + (ω8 + ω12 + ω18)x − 1. Thus E7 = Q(ω8 + ω12 +ω18, ω + ω7 + ω11).


Recall Artin’s Theorem from last class:

Theorem 14. Let E be a field, and let G be a finite subgroup of Aut(E). Let EG = {α ∈ E|σ(α) =α for all σ ∈ G}. Then E/EG is Galois, and Gal(E/EG) = G.

Here’s an application of it:

Let K be a field and let x1, ..., xn be indeterminants over K. Let E = K(x1, ..., xn) = { f(x1,...,xn)g(x1,...,xn) |f, g ∈

K[x1, ...xn]}. Given σ ∈ Sn, define σ : E → E by sending xi to xσ(i).

For example, if σ = (13)(24), then σ(3x2x

33−x

24

x21x2+x3+x4

2) =

3x4x31−x

22

x23x4+x1+x4

4.

Observe that some polynomials are fixed by all σ. For instance, x1 + ...+ xn and x1x2...xn are fixed byall σ.

Let L = ESn (the fixed field under Sn). We call L the field of symmetric rational functions.What are field generators for L/K?By Artin’s theorem, since Sn is a finite subgroup of the automorphism group of E, then E/L is Galois

and Gal(E/L) = Sn. Also, [E : L] = |Sn| = n!.

Let t be another indeterminant over E. Consider f(t) =

n∏i=1

(t − xi) ∈ E[t]. Note that fσ(t) = f(t).

Thus all coefficients of f(t) are in L. That is, f(t) ∈ L[t]. Let si be the coefficients in L such thatf(t) = tn − s1t

n−1 + s2tn−2 + ...+ (−1)nsn. Then, for instance, s1 = x1 + ...+ xn and sn = x1x2...xn.

Theorem 15. If K, L, E, and si are as before, then L = K(s1, ..., sn).

Proof. We’ve already shown that si ∈ L for all i, so K(s1, ..., sn) ⊂ L.But observe that f(t) splits in E = K(s1, ..., sn)(x1, ..., xn), and in fact E is the smallest field that f(t)

splits in (since if it splits, each xi must be in the splitting field). Thus E is actually the splitting field of f(t)over K(s1, ..., sn).

Since degf(t) = n then [E : K(s1, ..., sn)] ≤ n! (by elementary abstract algebra). However, [E : L] = n!by Artin’s theorem. Then

n! · [L : K(s1, ..., sn)] = [E : L][L : K(s1, ..., sn)]

= [E : K(s1, ..., sn)]

≤ n!

18

Thus [L : K(s1, ..., sn)] = 1, and L = K(s1, ..., sn)

Theorem 16 (Fundamental Theorem of Galois Theory, Part 3). Let E/F be a finite Galois extension,let G = Gal(E/F ) and let L be an intermediate field. Let H = Gal(E/L). Then L/F is normal if and onlyif H CG. Furthermore, if L/F is normal, Gal(L/F ) ∼= G/H.

Proof. Suppose L/F is normal. Define φ : G → Gal(L/F ) by σ 7→ σ∣∣∣L

. It is easy to see φ is a group

homomorphism.

What is the kernel of φ? Well, σ ∈ kerφ if and only if σ∣∣∣L

= 1L if and only if σ fixes L, which is the case

if and only if σ ∈ H. Thus H = kerφ. But every kernel is normal in the group, so H CG, as desired.Furthermore, since every τ ∈ Gal(L/F ) can be extended to some σ ∈ Gal(E,F ), then φ is onto.Then we can apply an Isomorphism Theorem (of groups): that G/ker(φ) ∼= im(φ) = Gal(L/F ). Since

ker(φ) = H, then G/H ∼= Gal(L/F ), as desired.Conversely, supposeHCG. We wish to show that L/F is normal using the definition that every embedding

of L into L = F which fixes F in fact is an automorphism of L. To that end, let σ : L→ F be an embeddingwhich fixes F . Let α ∈ L, and let h ∈ H. We can extend σ to τ : E → E. But since E/F is Galois, it isnormal, so since τ fixes F , then in fact τ maps into E. Thus τ ∈ Aut(E/F ) = Gal(E/F ) = G. Since HCG,then there exists some h′ ∈ H such that τ−1hτ = h′. Thus τ−1hτ(α) = h′(α). But H = Gal(E/L), andα ∈ L, so h′(α) = α. Thus by transitivity, τ−1hτ(α) = α, so h(τ(α)) = τ(α).

But h was arbitrary, so τ(α) is fixed by every element of H = Gal(E/L). Thus τ(α) ∈ L. Recall thatσ(α) = τ(α), so σ(α) ∈ L. But α was an arbitrary element of L, so σ maps every element of L to an elementof L. That is, σ is an automorphism of L/F , as desired.

6 Norm and Trace

Definition 17. Let E/F be a finite extension. Let σ1, ..., σr be the distinct embeddings of E → F fixingF . (So r = [E : F ]S by the definition of separable degree.)

Define the norm of E/F as NEF : E → E by α 7→

(r∏i=1

σi(α)

)[E:F ]insep

.

Similarly, we define the trace of E/F as TrEF : E → E by α 7→ [E : F ]insep

(r∑i=1

σi(α)

).

Example 13. Let E = Q(√

2) = {a + b√

2|a, b ∈ Q} (and F = Q, of course). Then there are only two σi:the identity, and σ : a+ b

√2 7→ a− b

√2. Then N(a+ b

√2) = a2 − 2b2 and Tr(a+ b

√2) = 2a.

Example 14. Let E = Q( 3√

2) = {a+ b 3√

2 + c 3√

4|a, b, c ∈ Q} (and F = Q, of course). Then there are nowthree σi: the identity; σ, which is generated by σ : 3

√2 7→ ω 3

√2 (where ω is a primitive 3rd root of unity); and

σ2. ThenN(a+b 3√

2+c 3√

4) = a+b 3√

2+c 3√

4+σ(a+b 3√

2+c 3√

4)+σ2(a+b 3√

2+c 3√

4) = a3+2b3+4c3−6abc ∈ Q.Also, Tr(a+ b 3

√2 + c 3

√4) = 3a (since 1 + ω + ω2 = 0, and a few steps of algebra).

Example 15. (An inseparable example) Let E = Fp(t), and F = Fp(tp) (where t is some indeterminant).Then we have seen that E/F is purely inseparable, with [E : F ]insep = p, and [E : F ]S = 1. Sincethe separable degree is only 1, there is only the identity map, so for all β ∈ E, N(β) = βp ∈ F . Also,Tr(β) = p · β = 0.

Remark 11. The trace always “degenerates” whenever E/F fails to be separable. That is, if [E : F ]insep >1, then Tr(β) = 0 for all β ∈ E.

19


Recall from last class the norm and trace: if E/F is a finite extension, and σ1, ...σr are the distinct embeddings

of F into E, then we say NEF (α) =

(r∏i=1

σi(α)

)[E:F ]insep

, and TrFE(α) is the additive version for all α ∈ E.

Lemma 6. Let E/F be a finite separable extension. Then for all α ∈ E, NEF (α) ∈ F and TrEF (α) ∈ F .

Proof. Let σ1, ...σr : E → F be the distinct embeddings of E into F which fix F . Let L be the normalclosure of E/F . In particular, by the primitive element theorem, E/F is generated by as single element, soE = F (α) for some α ∈ E, and then L is the splitting field of Min(α, F ) over F .

As proved on the homework [edit: see immediately below], L/F is separable, and since it is normal, thenL/F is Galois (and also finite). Let G = Gal(L/F ). For each φ ∈ G, φσ1, ...φσr are distinct embeddingsfrom E → F fixing F . (We can do this because Im(σi) ⊂ L for all i as L/F is normal.)

Thus each φ ∈ G permutes σ1, ...σr.Let α ∈ E. Then

φ(NEF (α)) = φ(σ1(α), ..., σr(α))

= (φσ1)(α)...(φσr)(α)

= σ1(α)...σr(α)

= NFE (α)

Thus NFE is fixed by all φ ∈ G. Thus NE

F (α) ∈ LG = F , as desired.The proof for trace is identical.

Homework Problem 3. Let E/F be a separable algebraic field extension, and let L be the normal closureof E/F . Then L/F is separable.

Remark 12. Recall that if [F (α) : F ]insep = pn, then αpn

is separable over F .

Theorem 17. Let E/F be a finite extension. Then for all α ∈ E, NEF (α) ∈ F and TrEF (α) ∈ F .

Proof. We know that [E : F ]insep = pn. By the lemma, it suffices to consider the case n > 0.

The claim for trace is boring: TrEF (α) = pn(blah) = 0 ∈ F for all α ∈ E.It then suffices to check NE

F (α) ∈ F . Let L be the separable closure of F in E. By a homework problem[edit: see immediately below], E/L is purely inseparable. That is, [E : L]S = 1. Let r = [E : F ]S = [L :F ]S = [L : F ].

Then there are r embeddings of L into F which fix F (by the definition of separable degree). Let us callthem σ1, ..., σr.

Extend each σi to τi : E → F . But each embedding of E → F which fixes F can be restricted back downto some σi, so τ1, ...τr are the distinct embeddings of E → F which fix F .

Let α ∈ E. Then [L(α) : L]insep ≤ [E : F ]insep = pn, so αpn

is separable over L. Thus αpn

is separableover F since L/F is also separable. Therefore αp

n ∈ L.Then

NEF (α) =

(r∏i=1

τi(α)

)[E:F ]insep

=

(r∏i=1

τi(α)

)pn

=

r∏i=1

τi(αpn)

=

r∏i=1

σi(αpn)

= NLF (αp

n

)

20

Since L/F is separable and αpn ∈ L, then by the lemma, NL

F (αpn

) ∈ F . Thus NEF (α) ∈ F as desired.

Homework Problem 4. Let E/F be an algebraic field extension and let K = F sep, the separable closureof F in E. Then E/K is purely inseparable.

Proposition 10. Let E/F be a finite extension. Then

1. For all α, β ∈ E, N(αβ) = N(α)N(β). In particular, NEF : E× → F× is a group homomorphism.

2. For all α, β ∈ E, and c ∈ F , Tr(α + β) = Tr(α) + Tr(β) and Tr(cα) = cTr(α). Thus TrEF : E → F isan F -linear transformation (a “linear functional”).

3. For α ∈ F , NEF (α) = α[E:F ].

4. For α ∈ F , TrEF (α) = [E : F ]α.

5. If K is an intermediate field, then NEF = NK

F ◦ FEK and TrEF = TrKF ◦TrEK .

Proof. Observe that (1) and (2) follow from the fact that the σis are automorphisms, so they respect additionand multiplication.

Observe that (3) and (4) follow from the fact that the number of σi is the separable degree, and that ifα ∈ F , then σi(α) = α for all i.

We will now prove (5). Let φ1, ..., φs be the distinct embeddings E → F fixing K. (So s = [E : K]S).Let σ1, ..., σt : K → F be the distinct embeddings fixing F .Extend each σi to τi : E → F .We claim that the {τiφj} are distinct embeddings of E → F fixing F .Suppose τiφj = τkφl. Then restrict these both to K. But the φs fix K, so they are the identity when

restricted to K, and thus τi

∣∣∣K

= τk

∣∣∣K

. But τi

∣∣∣K

= σi and τk

∣∣∣K

= σk. Thus i = k, and by applying τ−1i , we

get that φj = φl so j = l. Thus the τiφj are distinct.Note that #{τiφj} = st = [E : K]S [K : F ]S = [E : F ]S . But then we have found the right number of

distinct embeddings of E → F fixing F , so this is the correct list of all embeddings.Then,

NKF N

EK(α) = NK

F

∏j

φj(α)[E:K]insep

=

∏i

τi

∏j

φj(α)

[E:K]insep

[K:F ]insep

=

∏i,j

τiφj(α)

[E:F ]insep

= NEF (α)

The proof of trace is similar but with sums instead of products.

Lemma 7. Let σ1, ...σr : E → L be distinct nonzero field homomorphisms. Then {σ1, ...σr} is a linearlyindependent set over L, in the sense that if c1σ1 + ... + crσr = 0 as a linear transformation, where eachci ∈ L, then ci = 0 for all i.

Proof. We induct on r. If r = 1, then if c1σ1 = 0, then c1 = c1σ(1) = 0, so c1 = 0.In the inductive case, suppose r > 1, and assume for the sake of contradiction that c1σ1 + ...crσr = 0,

where ci 6= 0 for some i. By the inductive hypothesis, we can in fact assume that ci 6= 0 for all i. Ourinductive hypothesis also lets us assume that there is no shorter linear dependence among the σis.

21

As σ1 6= σ2, there exists some β ∈ E such that σ1(β) 6= 0 and σ1(β) 6= σ2(β).Then for all α ∈ E, we have that

r∑i=1

ciσi(α) = 0

=

r∑i=1

ciσi(αβ)

=

r∑i=1

ciσi(α)σi(β)

Thus we can divide by σ1(β) to get that c1σ(α1) +

r∑i=2

(ciσi(β)

σ1(β)

)σi(α) = 0 and subtracting from a

previous equation, we get that a non-zero linear combination of σ2, ...σr is 0. Thus we have found a lineardependence among r − 1 of the σs, which contradicts the inductive hypothesis. Thus σ1, ..., σr are linearlyindependent.


The first exam will be coming soon: probably either Wednesday October 7 or Wednesday October 14. Butmaybe we won’t have class on Friday October 16.

Recall the following theorem from character theory:

Theorem 18 (Linearly Independent Characters). Let σ1, ...σr be distinct field embeddings from Einto L. Then σ1, ...σr are linearly independent (as linear transformations) over L.

Corollary 4. If E/F is finite separable, then TrEF 6= 0. If E/F is inseparable, then TrEF = 0.

Proof. If E/F is separable and finite, TrEF = σ1 + ... + σr. This is a non-zero linear combination, so it isnon-zero.

If E/F is inseparable, then recall that the trace is multiplied by the inseparable degree. The inseparabledegree is pn for some n > 1, but p is the characteristic, so TrEF is always zero.

Definition 18. Let E/F be a finite extension. The extension is called cyclic (respectively abelian, solvable,nilpotent, etc.) if E/F is Galois, and Gal(E/F ) is cyclic (respectively abelian, solvable, nilpotent, etc.).

Theorem 19 (Hilbert’s Satz 90). Let E/F be a finite cyclic extension, let σ be a generator for Gal(E/F ),and let β ∈ E. Then N(β) = 1 if and only if β = α

σ(α) for some α ∈ E.

Proof. Suppose β = ασ(α) . Let n be the order of σ. ThenNE

F (β) =

n−1∏i=1

σi(β) = σi(α

σ(α)) =

σ(α)...σn(α)

σ2(α)...σn+1(α)=

1 since σn = id.Suppose instead that N(β) = 1. Then by Theorem 18, {1, σ, ...σn−1} is linearly independent.

Let φ : E → E by φ(x) = β · σ(x). Note that φn(x) =

(n−1∏i=0

σ(β)

)(σn(x)) = N(β)σn(x) = 1 · x = x for

all x ∈ E.

Let g : E → E be given by g(x) =

n−1∑i=0

φi(x). By gathering up all of the σ terms in φi, we can see that g

is a linear combination of σis. But one of its coefficients (namely the one on the 1 term) is nonzero, so sincethe σis are linearly independent, then g 6= 0. Let u ∈ E such that g(u) 6= 0. Let α = g(u). Then

22

φ(α) = φ(g(u))

= φ(

n−1∑i=0

φi(u))

=

n∑i=1

φi(u)

=

n−1∑i=0

φ(u)

= g(u)

= α

Thus βσ(α) = α, so β = ασ(α) as desired.

Remark 13. Let F be a field and let n ≥ 1 such that Char F - n. Then xn − 1 has n distinct roots on F .

The set of roots, called Un is a subgroup of F×

. Furthermore, Un is cyclic. Any cyclic generator of Un iscaled a primitive n-th root of 1 .

Theorem 20. Let E/F be a finite extension. Assume F contains a primitive n-th root of 1. Also assumeChar F - n. Then E/F is cyclic of degree dividing n if and only if there exists some α ∈ E such thatE = F (α) and αn ∈ F .

Proof. Suppose E/F is cyclic of degree dividing n. Let [E : F ] = d (so n|d). Let ζ ∈ F be a primitive n-throot of 1. Then ω = ζ

nd is a primitive d-th root of unity.

Note that NEF (ω−1) = (ω−1)[E:F ] = 1. By Hilbert’s Satz 90, there exists α ∈ E such that ω−1 = α

σ(α) ,

where Gal(E/F ) =< σ >.Then σ(α) = ωα, so σi(α) = ωiα for i = 0, ..., d− 1.Since Char F - d, we can say that α, ωα, ..., ωd−1α are all distinct, so [F (α) : F ]S ≥ d, since α 7→ ωiα

are d distinct embeddings of F (α) into E. But d ≤ [F (α) : F ]S ≤ [F (α) : F ] ≤ [E : F ] = d, so all of theseequal d. Thus E = F (α).

It then suffices to check that αn ∈ F . Since E/F is Galois, it suffices to check that αn is fixed by allelements of Gal(E/F ). Note that σ(αn) = (σ(α))n = ωnαn = αn, so αn is fixed by σ. But Gal(E/F ) isgenerated by σ, so αn is fixed by all automorphisms, and thus αn ∈ F , as desired.

Conversely, suppose there exists an α ∈ E such that E = F (α), and β = αn ∈ F .Then α is a root of xn − β ∈ F [x]. By hypothesis, F contains a primitive n-th root of unity. Let ω be

such a root of unity. Then ωiα (for i = 0, ..., n− 1) are all of the roots of xn − β. Thus F (α) is the splittingfield of xn − β.

So F (α)/F is normal. Also, Min(α, F )|xn − β, so Min(α, F ) has distinct roots. Thus α is separableover F , so F (α)/F is separable. Since it is separable and normal, F (α)/F is Galois.

Let G = Gal(E/F ). If σ ∈ G, then σ(α) = ωiσ (α) for some iσ = 0, ..., n − 1. Define ψ : G →< ω > by

σ → σ(α)α .

We claim ψ is a group homomorphism. Let σ, π ∈ G. Say σ(α) = ωiα and π(α) = ωjα. Thenσπ(α) = (ωi+j)α. Thus ψ(σ)ψ(π) = i+ j = ψ(σπ), so ψ is a homomorphism.

Furthermore, note that σ ∈ kerψ if and only if ψ(α) = 1 if and only if σ(α) = α, which is the caseprecisely when σ = id. Thus ψ is injective, so G is isomorphic to a subgroup of a cyclic group of order n.Thus G is cyclic of order d|n.

Remark 14. This is the key step to showing that not all equations are solvable by radicals.

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The first Exam will be Monday, October 12, from 3-6. The actual exam will be only 2 hours, and we get tochoose when we start.

Recall the following foreshadowing theorem from the last class:

Theorem 21. Let E/F be a finite extension. Assume F contains a primitive n-th root of 1. Also assumeChar F - n. Then E/F is cyclic of degree dividing n if and only if there exists some α ∈ E such thatE = F (α) and αn ∈ F .

But now let’s ignore it and look at something else:

7 Solvable Groups and Radical Extensions

Definition 19. Let G be a group. A commutator is an element of the form xyx−1y−1 for some x, y ∈ G.The commutator subgroup G′ of G is the subgroup of G generated its commuators.

Definition 20. A subgroup H ≤ G is a characteristic subgroup if φ(H) = H for all Y ∈ Aut(G). We usethe notation H Char G.

Remark 15. We have the following results:

1. H Char G ⇐⇒ φ(H) ⊂ H for all φ ∈ Aut(G).

2. H Char G implies H CG.

3. G′ Char G

Let us prove the last part:

Proof. It suffices to show that if φ ∈ Aut(G), then φ(G′) ⊂ G′. It suffices to show that each commutator issent into G′. But φ(xyx−1y−1) = φ(x)φ(y)φ(x)−1φ(y)−1, so G′ Char G.

Remark 16. Since G′ Char G, then G′ CG.

Proposition 11. Let G be a group and let H ≤ G. Then

1. G/G′ is abelian.

2. If H CG and G/H is abelian, then G′ ≤ H.

3. If G′ ≤ H, then H CG and G/H is abelian.

Proof. (Part 1) In G/G′, for all x, y ∈ G, xyx−1y−1 = 1, so xy = yx.(Part 2) Let x, y ∈ G. Then in G/H, xy = yx, so xyx−1y−1 = 1, so xyx−1y−1 ∈ H, so G′ ⊂ H.(Part 3) Note H/G′ is a subgroup of G/G′. Then H/G′CG/G′, as G/G′ is abelian. Thus HCG because

normality lifts.

Definition 21. Let G(0) = G, and for all i, let G(i+1) = (G(i))′.The derived normal series of G is ...CG(2) CG(1) CG(0) = G.A group G is called solvable if G(n) = 1 for some n ≥ 0.

Remark 17. If G is a group, then G(1) = 1 if and only if G is abelian.

Example 16. Let’s compute the derived normal series of S3. Let H =< (123) > CS3. Then S3/H is

abelian, so S(1)3 ⊂ H. Since S

(1)3 6= 1, then H = S

(1)3 .

Then S(2)3 = H ′ = 1 since H is abelian. So S3 is solvable.

24

Lemma 8. Let φ : A→ B be a surjective group homomorphism. Then φ(A(i)) = B(i) for all i ≥ 0.

Proof. We induct on i. If i = 0, then A(i) = A and B(i) = B. But since φ is surjective, then φ(A) = B, asdesired.

If i = 1, then for all x, y ∈ A, φ(xyx−1y−1) = φ(x)φ(y)φ(x)−1φ(y)−1, so φ(A(1)) ≤ B(1). If aba−1b−1 ∈B(1), then there exist x, y ∈ A such that φ(x) = a and φ(y) = b, so then φ(xyx−1y−1) = aba−1b−1. Thusφ(A(1)) = B(1)

Suppose i > 1, and assume φ(A(i−1)) = B(i−1). Then φ∣∣∣A(i−1)

is a surjective homomorphism, so we can

apply the i = 1 case to this, proving the inductive case.

Corollary 5. Suppose H C G and φ : G → G/H is the natural map. Then φ(G(i)) = (G/H)(i). Also,

φ(G(i)) = G(i) = G(i)HH .

Theorem 22. Let G be a group, and let H ≤ G. Then

1. If G is solvable, so is H. If, in addition, H CG, then G/H is solvable.

2. Conversely, if H CG, and both H and G/H are solvable, then so is G.

Proof. If G is solvable, then G(n) = 1 for some n. Since H(i) ⊂ G(i) for all i, then H(n) = 1, so H is solvable.

If H CG, then by the corollary, (G/H)(n) = G(n) = 1, so G/H is solvable.

Conversely, suppose we have G(n)HH = (G/H)(n) = 1 for some n. Hence G(n) ⊂ H, so we have H(m) = 1

for some m, and thus G(m+n) = 1. Thus G is solvable.

Proposition 12. Any p-group is solvable.

Proof. Let G be a p-group, and let |G| = pn. We induct on n. We use the fact that any p-group has anon-trivial center: If n = 1, then G is cyclic, hence abelian, hence solvable. If n > 1, then consider thecenter Z(G). This is certainly normal, and it is abelian, hence solvable. Furthermore, G/Z(G) is smaller,so by the inductive hypothesis it is solvable. Thus by the previous theorem, G is solvable by the previoustheorem.

Exercise 4. Any group of order pq is solvable.Why? If p = q, then we are done. If p < q, then the Sylow q-subgroup Q is normal and solvable. Thus

|G/Q| = p, so G/Q is a Ap-group, hence solvable. Thus G is solvable.

Exercise 5. Any group of order pqr (where p, q, r are primes) is solvable. I should prove this by extensiveapplication of Sylow’s Theorem.


Definition 22. A solvable series for a group G is a normal series {1} = GtC ...CG0 = G such that Gi/Gi+1

is abelian for all i.

Proposition 13. If G is a group, then G is solvable if and only if G has a solvable series.

Proof. Suppose G is solvable. Then its derived series is a solvable series.Suppose instead that G has a solvable series {1} = Gt C ... C G0 = G. We claim that G(i) ≤ Gi for all

i. We prove this by induction on i. If i = 0, this is easy, since G0 = G. Suppose G(i−1) ≤ Gi−1. ThenG(i) = (G(i−1))′ ≤ G′i−1 ≤ Gi (the last containment is because Gi−1/Gi is abelian). Thus G(t) ≤ Gt = {1},so G is solvable.

This proof also implies that the derived series is the shortest possible normal series.

Corollary 6. If G is a group, then G is solvable if and only if G has a normal series where the factor groupsare cyclic of prime order.

25

Proof. Suppose G has a normal series where the factor groups are cyclic of prime order. Then it has a normalseries, so it is solvable by the previous proposition.

Suppose instead that G is solvable. We wish to consider the “longest possible” solvable series of G. First,note that any time a quotient in a solvable series is the identity, we can remove it. But then in each step,the group gets strictly smaller, so the lengths of solvable series are bounded by |G|. Thus there is a solvableseries of maximal length. Let {1} = Gt C ...CG0 = G be such a series.

Then suppose for the sake of contradiction that someGi/Gi+1 is not cyclic of prime order. Since |Gi/Gi+1|is a natural number greater than 1, it must be composite. But Gi/Gi+1 is abelian, so there exists someintermediate group H such that Gi+1 ≤ H ≤ Gi and such that H/Gi+1 is a subgroup of order p. ThenH/Gi+1CGi/Gi+1 as Gi/Gi+1 is abelian. Then HCGi, and Gi+1CHCGi. Furthermore, this containmentsare proper, so we have that this can be inserted into the solvable series. This contradicts the maximality ofthe solvable series, so Gi/Gi+1 is cyclic of prime order for all i.

We will use this exercise for the next theorem:

Exercise 6. Let An denote the alternating group on n elements. Then An is generated by the 3-cycles.

Theorem 23. Let An denote the alternating group on n elements. Then An is not simple (i.e. it containsno proper, nontrivial normal subgroups) for all n ≥ 5.

Proof. Suppose K CAn, and suppose K 6= {1}.We will show that K contains a 3-cycle. Choose σ ∈ K \ {1} such that σ fixes a maximal number of

elements of {1, 2, ..., n}. Assume for the sake of contradiction that σ is not a 3-cycle. Write σ as a productof disjoint cycles. Then there are two possible forms (much WLOGing is going on):

1. σ = (123...)... (i.e. there exists a 3-cycle or longer)

2. σ = (12)(34)... (i.e. there are no 3-cycles, i.e. σ is the product of disjoint transpositions)

If we are in case (1), then note that since σ is not a 3-cycle, then σ must move at least two more elements.Without loss of generality, let those elements be called 4 and 5. Also let β = (354), and let π = βσβ−1.Then in case (1), π = (124...). In case (2), π = (12)(35).... Note that π 6= σ.

Now let τ = πσ−1 = βσβ−1σ−1. Since π 6= σ, then τ ∈ K \ {1}.We now wish to show that τ fixes more elements than σ, which will contradict our choice of σ. Note that

if i > 5, and σ(i) = i, then τ(i) = i.Also note that in case (1), τ(2) = 2, but σ moves 1,2,3,4, and 5. Thus τ fixes more elements than σ.In case (2), τ(1) = 1 and τ(2) = 2, but σ moves 1,2,3, and 4, so τ fixes more elements than σ.Thus in either case, we have a contradiction. Thus σ is a 3-cycle, so K contains at least one 3-cycle.We will now show that σ contains every other 3-cycle. Without loss of generality, let the 3-cycle in K be

(123). Let (ijk) be any other 3-cycle. We construct γ ∈ Sn by γ(1) = i, γ(2) = j, γ(3) = k. Furthermore,we choose two other elements l,m such that γ(4) = l and γ(5) = m, and we fill out the rest of γ any otherway. If γ is even, then γ ∈ An, so γ(123)γ−1 = (ijk) ∈ K since K is normal. If γ is odd, then let π = (lm)γ.Then π ∈ An, so π(123)π−1 = (ijk) ∈ K. Thus K contains every 3-cycle.

Then by the exercise, since An is generated by 3-cycles, K = An. Thus the only normal subgroups of Anare {1} and An.

Corollary 7. If n ≥ 5, then An is not solvable.

Proof. Since An is not abelian, A′n 6= {1}. However, A′n C An, so since An is simple, A′n = An. Thus

A(i)n = An 6= {1} for all i, so it not solvable.

Remark 18. The quadratic formula says that if F is a field with characteristic not equal to 2, and f(x) =

ax2 + bx+ c ∈ F [x] (for a 6= 0), then the roots of f are −b±√b2−4ac

2a .

Remark 19. Suppose F is a field, and x3 + bx2 + cx+ d ∈ F [x]. If CharF 6= 2, 3, then by replacing x byx− b

3 , we get a polynomial of the form f(x) = x3 + px+ q.The following result of Cardano (in the 1500s) found that the roots of f(x) can be found in the field

E = F (ω, δ, y1, y2) where ω3 ∈ F , δ2 ∈ F (ω), y21 ∈ F (ω, δ), y3

2 ∈ F (ω, δ, y1). This is called a root tower .

26


Recall the following theorem from last class:

Theorem 24. Let E/F be a finite extension, and suppose Char F - n. Suppose also that F contains aprimitive nth root of unity. Then E/F is cyclic of degree ni|n if and only if E = F (α) for some α such thatαni ∈ F .

Recall also the remark on the cubic formula from last class:

Remark 20. Suppose F is a field, and x3 + bx2 + cx+ d ∈ F [x]. If CharF 6= 2, 3, then by replacing x byx− b

3 , we get a polynomial of the form f(x) = x3 + px+ q.The following result of Cardano (in the 1500s) found that the roots of f(x) can be found in the field

E = F (ω, δ, y1, y2) where ω3 ∈ F , δ2 ∈ F (ω), y21 ∈ F (ω, δ), y3

2 ∈ F (ω, δ, y1). This is called a root tower .Namely,

ω3 = 1

δ2 = 12p3 − 81q2

y21 =

27

2q +

3

2δ

y32 =

27

2q − 3

2δ

Definition 23. A field extension E/F is called radical if there exists a sequence of fields F = E0 ⊂ E1 ⊂... ⊂ Et = E where for each i, Ei = Ei−1(ui) and umii ∈ Ei−1 for some mi ∈ N.

Definition 24. A polynomial f(x) ∈ F [x] is called solvable by radicals over F if f(x) splits completely insome radical extension of F .

Theorem 25. Let f(x) ∈ F [x] be a separable polynomial, and let E be its splitting field. Suppose also thatChar F - [E : F ]. Then if Gal(E/F ) is solvable, then f(x) is solvable by radicals (over F ).

Proof. Let n = [E : F ]. We wish to reduce to the case where F contains a primitive n-th root of unity.To this end, let ω be a primitive nth root of unity, and let L = F (ω). Note that since Char F - [E : F ],

then there are indeed φ(n) distinct nth roots of unity.Note also that f(x) is solvable by radicals over L if and only if its splitting field lies inside a radical

extension R. This is the case if and only if L = E0 ⊂ ... ⊂ Et = R for some appropriately chosen Ei. LetE−1 = F . Then E0 ⊂ ... ⊂ Et is a radical tower if and only if E−1 ⊂ E0 ⊂ ... ⊂ Et = R is a radical tower.This is the case if and only if f(x) is solvable by radicals over E−1 = F .

Thus it suffices to show f(x) is solvable by radicals over L.By a homework problem [edit: see immediately below], EL/L is Galois and G = Gal(EL/L) is isomorphic

to a subgroup of Gal(E/F ). By assumption Gal(E/F ) is solvable, and a subgroup of a solvable group issolvable. Thus G = Gal(EL/L) is solvable.

That is, there exists a normal series {1} = Gt C ...CG1 CG0 = G. Recall that by a previous result, wecan assume Gi/Gi+1 is cyclic. If ni = #Gi/Gi+1, then certainly, ni|#G = n.

Let Ei be the subfield of EL fixed by Gi. (So Et = EL and E0 = F ). Then Gal(Et/Ei) = Gi CGi−1 =Gal(Et/Ei−1). Since the one Galois group is normal in the other, then Ei/Ei−1 is a normal extension, andGal(Ei/Ei−1) ∼= Gi−1/Gi, which is cyclic.

Thus by Theorem 24 Ei = Ei−1(ui), where unii ∈ Ei. Thus EL/L is a radical extension, so f(x) issolvable by radicals over L. Therefore it is solvable by radicals over F .

Homework Problem 5. Let K ⊂ E ⊂ L and let K ⊂ F ⊂ L be fields. Suppose E/K is finite and Galois.Then EF/F is Galois, and Gal(EF/F ) is isomorphic to a subgroup of Gal(E/K).

We now want to prove some form of the converse of this theorem.

27

Lemma 9. Let E/F be a radical extension, and let L be the normal closure of E/F . Then L/F is a radicalextension.

Proof. Since E/F is radical, E = F (u1, ..., u`), where there exists some m ∈ N such that umi ∈ F (u1, ...ui−1)for all i.

Let σ1, ...σr be the distinct embeddings of E → F fixing F .Let K = F ({σj(ui)}i,j}, and recall that L denotes the normal closure of F in E. We claim that L = K.For each j, let τj be an extension of σj to L → L. Then τj(ui) = σj(ui) ∈ L for all i, j (since L is the

normal closure), so K ⊂ L.Let π : K → F be an embedding fixing F . It suffices to show that this sends elements of K to elements

of K. In other words, it suffices to show that π(σj(ui)) ∈ K for all i, j. Extend π and σj to τ and

τj ∈ Aut(L/F ). Then (πσj)(ui) = (ττj)(ui). But ττj

∣∣∣E

: E → F and fixes E, so ττj

∣∣∣E

= σk for some k.

Thus (πσj)(ui) = (ττj)(ui) = σk(ui) ∈ K. Thus K/F is normal, so K ⊂ L. Thus K = L.It then suffices to show that K/F is radical. To do this, you adjoin the roots in a moderately clever way.

First you adjoin σi(u1) for all i, since for all i, σi(u1)n1 = σi(un11 ), and since E/F is radical, un1

1 ∈ F . Also,σi fixes F , so σi(u1)n1 ∈ F .

Then you adjoin σi(u2) for all i, and then continue through all the ujs. Finally, you have adjoined allσj(ui), so K/F is radical. Thus L/F is radical.


Recall from last class the following lemma (which we proved):

Lemma 10. Let E/F be a radical extension, and let L be the normal closure of E/F . Then L/F is aradical extension.

We will now prove another lemma:

Lemma 11. Let L/K be a Galois radical extension. Then the Galois group is solvable.

Proof. Since L/K is radical, there is a root tower K = K0 ⊂ K1 ⊂ ... ⊂ Kt = K, where Ki = Ki−1(ui)where umii ∈ Ki−1 for all i.

We wish to show that we can in fact choose mi such that Char K - mi for all i. Suppose for the sakeof contradiction that this is not the case. Then Char K = p, a prime, and let mi = lpt where p - l. Then

(uli)pt ∈ Ki−1, so uli is purely inseparable over Ki−1. But L/Ki−1 is separable, so uli ∈ Ki−1. Thus we can

replace mi with l, and p - l.Let m = m1...mt. Then Char K - m, and umi ∈ Ki−1 for all i.Let ω be a primitive mth root of unity. We will now show that, without loss of generality, we can assume

ω ∈ K. That is, we will show that Gal(L/K) is solvable if Gal(L(ω)/K(ω)) is solvable.Consider the five extensions in K ⊂ K(ω) ⊂ L(ω) and K ⊂ L ⊂ L(ω) (namely K(ω)/K), L(ω)/K(ω),

L/K, L(ω)/L, and L(ω)/K). We claim that all of these extensions are Galois and radical.In no particular order, we can see that K(ω)/K and L(ω)/L are the splitting fields of xm − 1, so they

are Galois and radical. Also, L/K is radical and Galois by our hypothesis. Then since L(ω)/L and L/K areradical and Galois, then L(ω)/K is radical and Galois since “radicalness” and “radicalness” are transitive.It then suffices to show that L(ω)/K(ω) is radical and Galois. This extension must be radical because L/Kis radical, and you can take the same radical tower with the same exponents. Furthermore, L(ω) = LK(ω),and by Homework Problem 5, LK(ω)/K(ω) is Galois.

Therefore all of these extensions are radical and Galois.Let L′ = L(ω) and K ′ = K(ω). Let K ′i = Ki(ω).Then, let Hi = Gal(L′/K ′i). By the theorem on m-cyclic extensions, K ′i/K

′i−1 is cyclic. Thus Hi CHi−1

and Hi−1/Hi is cyclic. Then H0 = Gal(L′/K ′) and Ht = Gal(L′/L′) = {1}, so Gal(L′/K ′) is solvable.That is, Gal(L(ω)/K(ω)) is solvable.

Also, Gal(K(ω)/K) is isomorphic to a subgroup of Z×n , which is abelian. Thus Gal(K(ω)/K) is solvable.But Gal(K(ω)/K) ∼= Gal(L(ω)/K)/Gal(L(ω)/K(ω)), and by a previous theorem, since Gal(K(ω)/K)

and Gal(L(ω)/K(ω)) are both solvable, then so is Gal(L(ω)/K).

28

But Gal(L/K) ∼= Gal(L(ω)/K)/Gal(L(ω)/L), and the quotient of a solvable group is solvable. ThusL/K is solvable.

Theorem 26. Let F be a field, and let f(x) ∈ F [x] be a separable polynomial which is solvable by radicalsover F . Let E be the splitting field of f(x) over F . Then Gal(E/F ) is solvable.

Proof. Since f(x) is solvable by radicals, there exists some radical extension L/F such that E ⊂ L.By the first lemma (Lemma 10), we can assume L/F is normal (if not, replace it by its normal closure).

Let G = Aut(L/F ). Note that |G| = [L : F ]S ≤ [L : F ] <∞. Define φ : G→ Gal(E/F ) by σ 7→ σ∣∣∣E

. Note

that φ does indeed map into Gal(E/F ) since E/F is normal.Furthermore, any τ ∈ Gal(E/F ) can be extended to some σ ∈ Aut(L/F ) since L/E is algebraic, and

L/F is normal. Thus φ is surjective.Thus to show that Gal(E/F ) is solvable, it suffices to show that G = Aut(L/F ) is solvable.Let LG be the fixed field of G. By Artin’s Theorem, we know that L/LG is Galois, and G = Gal(L/LG).

Also, L/LG is radical as F ⊂ LG and L/F is radical.But by the second lemma (Lemma 11), G is solvable, as desired.

Recall that if u1, ..., un, x are indeterminants over F , then let g(x) =

n∏i=1

(x − ui) ∈ F [u1, ..., un, x] ⊂

F (u1, ...un)[x]. Let si be the correct polynomial in the uis such that g(x) = xn − s1xn−1 + ... + (−1)nsn.

Then the si are elements of F [u1, ..., un].We showed that F (u1, ..., un)/F (s1, ..., sn) is a Galois extension, and the Galois group is Sn. Furthermore,

we showed that F (u1, ..., un) is the splitting field for g(x) over F (s1, ..., sn).

Definition 25. Let F be a field. The general equation of degree n over F is xn − t1xn−1 + ... + (−1)ntnwhere t1, ..., tn are indeterminants over F .

Theorem 27. Let F be a field and let t1, ..., tn be indeterminants. Furthermore, let L = F (t1, ..., tn), andlet f(x) = xn − t1xn−1 + ... + (−1)ntn ∈ L[x]. Let E be the splitting field of f(x) over L. Then E/L isGalois and Gal(E/L) ∼= Sn.

(A proof will be given next class.)

Remark 21. For n ≤ 4, Sn is solvable. Therefore the general equation of degree n ≤ 4 is solvable byradicals, so there exists a single formula for all functions of degree n. That is, there exists a quadratic, cubic,and quartic formula.

7.4 Day 17 - October 2

We will now prove this statement, which we said last class:

Theorem 28. Let F be a field and let t1, ..., tn be indeterminants. Furthermore, let L = F (t1, ..., tn), andlet f(x) = xn − t1xn−1 + ... + (−1)ntn ∈ L[x]. Let E be the splitting field of f(x) over L. Then E/L isGalois and Gal(E/L) ∼= Sn.

Proof. Let E = L(y1, ..., yn) where y1, ..., yn are the n roots of f(x) in L. Then f(x) =

n∏i=1

(x− yi) in E[x].

Then by definition of the elementary symmetric functions, ti = si(y1, ..., yn) for each i.Then E = L(y1, ..., yn) = F (y1, ...yn) since L = F (t1, ..., tn) and each ti ∈ F (y1, ..yn).Consider the diagram [which I can’t draw lol] of F [y1, ...yn] ← F [u1, ..., un], where this arrow is φ given

by plugging in yi for ui. Below this should be F [t1, ..., tn] → F [s1, ..., sn] with containment arrows goingup. Here the function is ψ given by sending ti to si. Since the tis and uis are indeterminants, these aresurjections.

Note that φψ(ti) = φ(si) = φ(si(u1, ..., un)) = si(y1, ..., sn) = ti. Thus φψ is the identity on F [t1, ..., tn].Thus ψ is injective and thus is an isomorphism.

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Now consider a diagram of the field of fractions:We have L = F (t1, ..., tn) ⊂ F (y1, ..., yn) = E, and B = F (s1, ..., sn) ⊂ F (u1, ..., un) = A (and these are

the respective splitting fields). But we also have σ : F (t1, ..., tn)→ F (s1, ..., sn) by σ : a(t)b(t) 7→

a(s)b(s) .

But f(x) ∈ L[x] and fσ(x) = xn − s1xn−1 + ...+ (−1)nsn =

n∏i=1

(x− ui). Thus A is the splitting field for

fσ over B. As E is the splitting field for f(x) over L, we get an isomorphism τ : E → A such that τ∣∣∣L

= σ.

Thus we’ve proven A/B is Galois, and Gal(A/B) ∼= Sn. But because we have isomorphisms, E/L is Galoisand Gal(E/L) ∼= Sn.

Corollary 8. If n ≤ 4, and Char F - n!, then the general equation of degree n over F is solvable byradicals.

Proof. This follows from the fact that Sn is solvable for n ≤ 4.

Corollary 9. (Abel’s Theorem)If n ≥ 5, the general equation of degree n is not solvable by radicals (over any field).

Proof. This follows from the fact that Sn is not solvable for n ≥ 5.

Exercise 7. Let p be prime. Then Sn is generated by any p-cycle and any transposition.

Proposition 14. Let f(x) ∈ Q[x] be any irreducible polynomial of prime degree with precisely two non-realroots. Let E be the splitting field of f(x). Then G = Gal(E/Q) ∼= Sp.

Proof. As degf(x) = p, and is separable, then every element of G is a permutation of the roots of f(x).Thus G ≤ Sp.

Let α ∈ E be a root of f(x). Then [F (α) : F ] = p, since f(x) is the minimal polynomial (assuming,without loss of generality, that f(x) is monic). Thus p

∣∣|G|, so G contains an element of order p. But theonly elements of order p in Sp are p-cycles, so G contains a p-cycle. Also, if τ denotes complex conjugactionrestricted to E, then τ ∈ Gal(E/F ), but τ is a transposition. Thus by the exercise, G = Sp.

Example 17. Let f(x) = x5 − 2x3 − 8x− 2 ∈ Q[x]. By Eisenstein, f(x) is irreducible.We now use calculus to show that there are exactly three real roots. Observe that f ′(x) = 5x4−6x2−8 =

(5x2 + 4)(x2− 2). The only zeroes of f ′(x) are ±√

2, so it has at most 3 real roots (recall from calculus thatbetween any two zeroes of a function is a zero of its derivative).

Note also that f(−3) = −167, f(−1) = 7, f(0) = −2, and f(3) = 163, so by the intermediate valuetheorem, f(x) has three real roots. Thus by the proposition, this is a polynomial whose roots do not live inroot towers!

Remark 22. RIP Quintic Formula

Question 2. (Inverse Galois Problem) Is every finite group the Galois group of a finite extension of Q?This is an open problem.


Lemma 12. Let m,n be distinct positive integers, such that m|n. Let p be a prime such that p - n. ThenΦn(x) and xm − 1 have no common root mod p.

Proof. We know that xn − 1 =∏d|n

Φd(x) = Φn(x) ·

∏d|nd<n

Φd(x)

. Let f(x) =∏d|nd<n

Φd(x). Then xn − 1 =

Φn(x) · f(x). Note that xm − 1 divides f(x).Suppose for the sake of contradiction that a ∈ Z is a common root of xm − 1 and Φn(x) mod p. Then a

is a double root of xn − 1. But p - n, so xn − 1 has distinct roots mod p. This is a contradiction.

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Theorem 29. Let n > 1 be an integer. Then there exists infinitely many primes congruent to 1 mod n.

Proof. Suppose for the sake of contradiction that there were only finitely many primes congruent to 1 modn. Let p1, ..., pk be the complete list of them.

We know that Φn(x) is monic, so Φn(s) > 1 for all s sufficiently large.Choose l sufficiently large such that Φn(lnp1...pk) ≥ 2. Note that the constant term of Φn(x) = ±1. Let

a = lnp1...pk.As l, n, and each pi divides each term of Φ(a) except the constant term, then we can see that n - Φ(a)

and pi - Φ(a).Since Φ(a) ≥ 2, there exists a prime p dividing it. Then Φn(a) ≡ 0 (mod p), and since Φn(x)|xn − 1,

then an − 1 ≡ 0 (mod p). Thus an ≡ 1 (mod p).Also, p - n, (since if p|n, then p|a, so p - Φn(a), a contradiction). By the lemma, am 6≡ 1 (mod p) for all

m|n and m < n, so the order of a in Z×p is n.But by Fermat’s little theorem, ap−1 ≡ 1 (mod p), so n|p − 1 so p ≡ 1 (mod n). But p is not any pi

since no pi divides Φ(a). This is a contradiction of our assumption that p1, ...pk is a complete list of primescongruent to 1 mod n.

Theorem 30. Let G be a finite abelian group. Then there exists a root of unity ω ∈ C and a field E ⊂ Q(ω)such that F/Q is Galois and Gal(F/Q) ∼= G.

Proof. Recall that by the classification of finite abelian groups, G ∼= Cn1× ... × Cnk where each Cni is a

cyclic group of order ni.Let p1, ...pk be distinct primes such that pi ∼= 1 mod ni. (Remark: in order to do the pathological case

n1 = n2 = ... = nk, we need the fact that there are infinitely many primes congruent to 1 mod ni.)Let m = p1...pk. Then Zm ∼= Zp1 × ...× Zpk (as rings).So, Z×m ∼= Z×p1 × ...× Z×pk as abelian groups.But for all primes pi, Z×pi is cyclic, and has pi − 1 elements, so Z×m ∼= Cp1−1 × ...× Cpk−1. As ni|pi − 1,

there exists Hi ≤ Cpi−1 of order pi−1ni

. Let H = Hi × ...×Hk. Then Z×m/H∼= Cn1

× ...× Cnk .

Let ω be a primitive mth root of unity, and let E = Q(ω). We’ve seen Gal(E/Q) = Z×m. Let F = EH .As Z×m is abelian, F/Q is Galois and Gal(F/Q) ∼= Z×m/H ∼= G.

Stuff before this point: · will be on the exam. Stuff after it will not!

8 Algebraic Independence and Transcendental Extensions

Now let’s start on algebraically independent things.

Definition 26. Let E/F be a field extension, and let S ⊂ E. We say S is algebraically dependent over F ifthere exist s1, ..., sn ∈ S and f(x1, ..., xn) ∈ F [x1, ..., xn] such that f is nonzero, but f(s1, ..., sn) = 0.

We say S is algebraically independent over F if it is not algebraically dependent.

Proposition 15. If S is the singleton {s}, then S is algebraically dependent over F if and only if s isalgebraic. (Equivalently, S is algebraically independent over F if and only if s is transcendental.)

Example 18. Let E = F (x, y) (the field of rational functions in x and y). Let S = {x2, xy, y2}. Then iff(u, v, w) = uw−v2, then f(x2, xy, y2) = 0 and f is not the zero polynomial, so S is algebraically dependentover F .

Remark 23. Recall the concept from linear algebra that if S is linearly independent and v 6∈ S, then S∪{v}is linearly independent if and only if v 6∈ Span(S).

We now write a version of this for algebraically independent sets over fields.

Lemma 13. Let E/F be a field extension, and let S ⊂ E be an algebraically independent set over F . Letu ∈ E. Then S ∪ {u} is algebraically independent if and only if u is transcendental over F (S).

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Proof. Suppose u is not transcendental over F (S).Then it is algebraic. That is, there exist f(x) ∈ F (S)[x]such that f(u) = 0. Since f(x) has coefficients in F (S), there exists s1, ..., sn ∈ S such that each coefficient isa polynomial in the si. Then f can be thought of as a root of the nonzero polynomial in n+ 1 variables.

Suppose instead that u is transcendental over F (S). Suppose that g is a polynomial in x1, ..., xn suchthat g(s1, ..., sn−1, u) = 0. Then this can be interpreted as a polynomial in one variable and s1, ..., sn−1, soit is a polynomial in F (S)[x] with u as a root. Thus it is the zero polynomial.


Exam will be next week. The room was decided, but will be announced via email. We need to answer 4− 5problems, but there will be more questions than that on the exam. Also, one of the problems will be ahomework problem. Also, Friday will be review.

Recall the following Lemma

Lemma 14. Let E/F be a field extension and S ⊂ E an algebraically independent set over F , and letu ∈ E. Then U ∪ {u} is algebraically independent if and only if u is transcendental over F (S).

Proof. We proved the “only if” direction last time, so we will prove the “if” direction now.By way of contradiction, suppose u is algebraic over F (S). Then there exist s1, ..., sn) ∈ S such that u

is algebraic over F (s1, ..., sn). That is, there exists a nonzero polynomial f(x) ∈ F (s1, ..., sn)[x] such that

f(u) = 0. The coefficients of f(x) are of the form ai(s1,...,sn)bi(s1,...,sn) . If we clear all denominators, we can find f1(x)

with coefficients in F [s1, ..., sn]. But all denominators are non-zero, so the leading coefficient is non-zero.Also, f1(u) = 0 still. Then, thinking of this as a polnyomial in x1, ..., xn, x, we have found u as a root.But {s1, ..., sn, u} is algebraically independent over F , so this polynomial cannot evaluate to 0. This is acontradiction, and we are done.

Definition 27. Let E/F be a field extension. A set S ⊂ E is called a transcendence base for E/F if S isalgebraically independent over F and E/F (S) is algebraic.

Example 19. Let X and Y be indeterminants over F , and let E = F (X,Y ). Then {x, y} is a transcendencebase for E/F . But so is {X2, Y 5}.

Theorem 31. Let E/F be a field extension. Then if L is an algebraically independent set, there existsa transcendence base of E/F containing L. Since ∅ is algebraically independent, it can be extended to atranscendence base, so there always exists a transcendence base.

Proof. We use Zorn’s Lemma. Let Λ = {T | L ⊂ T ⊂ E, T is alg. ind. overF}. This is partially ordered bycontainment. Also, L ∈ Λ, so Λ is nonempty.

Suppose C is a chain of Λ. Let C =⋃T∈C

T . Then TC is algebraically independent, since any finite subset

of TC is contained in some T ∈ C.Thus TC is an upper bound for C.Thus by Zorn’s Lemma, Λ has a maximal element, say S. Since S ∈ Λ, then S is algebraically independent.

If E/F (S) is not algebraic, it has a nonalgebraic element u. Then S ∪{u} is algebrically independent, whichcontradicts the maximality of S. Thus S is a transcendence base for E/F .

Theorem 32. Let E/F be a field extension, and let S, T be two transcendence bases for E/F . If |S| <∞,then |S| = |T |.

Proof. Let S = {s1, ...sn}. Suppose for the sake of contradiction that for all t ∈ T , {t, s2, ..., sn} is notalgebraically independent. Then F (T ) is algebraic over F (s2, ..., sn). Since E/F (T ) is algebraic, thenE/F (s2, ..., sn) is algebraic. Thus s1 is algebraic over F (s2, ..., sn), so S was not algebraically independent.This is a contradiction, so there exists some t1 ∈ T such that {t1, s2, ..., sn} is algebraically independent.

Suppose for the sake of contradiction that E/F (t1, s2, ..., sn) were not algebraic. Since E/F (s1, ..., sn)is algebraic, then E/F (s1, ..., sn, t) is algebraic. Thus F (s1, ..., sn, t)/F (s2, ..., sn, t) is not algebraic, so s1 is

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transcendental over F (s2, ..., sn, t), so {s1, ..., sn, t} is algebraically independent by the lemma. But t1 is alge-braic over F (s1, ..., sn), and this contradicts the lemma. Thus E/F (t1, s2, .., sn) is algebraic, so {t1, s2, ..., sn}is a transcendence base.

By repeated application of this, we can replace all the elements of S with elements of T . But the elementsof a transcendence base are always distinct, so in particular these elements of T must be distinct, so |S| ≤ |T |.One could also do this process going the other way, giving that |T | ≤ |S|, so |T | = |S| as desired.

Definition 28. The transcendence degree of E/F is the number of elements in any transcendence base forE/F .

Proposition 16. If E/F is a field extension, then E/F is algebraic if and only if Tr .deg(E/F ) = 0.

Theorem 33. Let E/F/L be a tower of fields. Then Tr .deg(E/L) = Tr .deg(E/F ) + Tr .deg(F/L).

Proof. Let S ⊂ E be a transcendence base for E/F and let T ⊂ F be a transcendence base for F/L. Notethat S ∩ F = ∅, so S ∩ T = ∅. Thus |S ∪ T | = |S|+ |T | = Tr .deg(E/F ) + Tr .deg(F/L).

Thus it suffices to show that S ∪ T is a transcendence base of E/L.We first show that E/L(S ∪T ) is algebraic. We know that F is algebraic over L(T ), so F (S) is algebraic

over L(T )(S) = L(S ∪ T ). As E/F (S) is algebraic, E/L(S ∪ T ) is algebraic.We now need to show S∪T is algebraically independent over L¿ Let f(x1, ..., xm, y1, ...yn) ∈ L[x1, ..., xm, y1, ...yn]

and suppose f(s1, ..., sm, t1, ...tn) = 0 for some si ∈ S and ti ∈ T . We need to show f = 0.

Write f =∑j

gj(y1, ..., yn)hj(x1, ..., xm). Then the hjs are distinct monomials in x1, ..., xm. Let

l(x1, ..., xm) = f(x1, ..., xm, t1, ..., tn) ∈ L(T )[x1, ..., xm] ⊂ F [x1, ..., xm]. But l(s1, ..., sm) = 0, and S is

algebraically independent over F . Thus l(x1, ..., xm) = 0. Then∑

gj(t1, ...tn)hj(x1, ..., xm) = 0. As the hjs

are linearly independent over F , then gj(t1, ..., tn) = 0 for all i. As {t1, ..., tn} is algebraically independentover L, gj(y1, ..., yn) = 0 for all i. Thus f = 0, so S ∪ T is algebraically independent.

Thus S ∪ T is a transcendence base, so Tr .deg(E/L) = |S ∪ T | = Tr .deg(E/F ) + Tr .deg(F/L) asdesired.

9 Introduction to Rings and Modules


Definition 29. A ring has a multiplicative identity, but may not be commutative. You know the rest.

Example 20. (Matrix Rings) If R is a ring, then let Mn(R) denote the set of n× n matrices with entriesfrom R. Then Mn(R) is also a ring. Furthermore, if n ≥ 2, this ring is noncommutative (except maybe forR = 0).

Example 21. (Group Rings) Let R be a ring (usually commutative), and let G be a group. Let R[G]be a free R-module with basis G (i.e. it is the set of R-linear combinations of elements of G). ThenR[G] =

⊕g∈GRg. We also give R[G] a multiplication operation by defining (r1g1)(r2g2) = r1r2(g1g2).

For a particular example, considerR = Z andG = C3, the cyclic group of 3 elements. ThenR[G] = Z(C3).If a is a generator of C3, then (2 + 3a)(1− a2) = 2− 2a2 + 3a− 3 = −1 + 3a− 2a2.

In general, R[G] is commutative if and only if G is abelian.

Example 22. (Skew Polynomial Rings) Let R be a commutative ring, and σ : R → R be a ring homo-morphism. Let R[x, σ] = R[x] as a left R-module, but define multiplication in R[x, σ] by (axm)(bxn) =aσn(b)xm+n.

In particular, if Char R = p, then f : R → rp is a ring homomorphism. Then R[x, f ] is given byaxmbn = abp

m

xm+n.

Example 23. (Real Quaternions) Considering the following matrices in M2(C): 1 = id, i = [i, 0, 0,−i],j = [0, 1,−1, 0], and k = [0, i, i, 0]. We let H = R1 ⊕ Ri ⊕ Rj ⊕ Rk. In fact, H is a ring with the usual

33

multiplication (it suffices to check that multiplication works out). One can check that i2

= j2

= k2

= ijk =−1, so it is a ring.

In fact, H is a division ring. Since if α = r01 + r1i + r2j + r3k, set α = r01 − r1i − r2j − r3k. Thenαα = (r2

0 + r21 + r2

2 + r23)1, so α−1 = α

|α|2 .

Definition 30. Let R be a ring. A left R-module is an abelian group (M,+) and an operation · : R×M →Msuch that the usual axioms hold. That is,

1. (r + s)m = rm+ sm

2. r(m+ n) = rm+ rn

3. r(s(m)) = (rs)m

4. 1 ·m = m

[Remark: some people do without the fourth property. To make it clear that we have the fourth property,one might say it is a unital module.]

A right R-module is the same, but the operation is on M ×R and obeys (m(s))r = m(sr) instead of thethird property.

Definition 31. If R is a ring, a left (respectively, right) ideal of R is just a left (respectively, right) R-submodule of R. An ideal of R is a left ideal which is also a right ideal (a “2-sided ideal”).

Definition 32. Let R be a ring, and let M be a left R-module. M is called (left) Noetherian or (respectively,(left) Artinian) if M satisfies the Ascending Chain Condition (respectively, the Descending Chain Condition)on (left) submodules.

Recall that the ascending chain condition is that if N1 ⊂ N2, ..., is a chain of submodules of M , thenthere exists n such that Nn = Nn+1 = ... (i.e. the chain stabilizes). [Remark: the notation implies that thechain is countable, but countable chains stabilizing is equivalent to any chain stabilizing.] The decreasingchain condition is the same but for decreasing chains of submodules.

A ring R is left or right Noetherian or Artinian if and only if it is that adjective as an R-module.

Definition 33. Let R and S be rings. We say M is an R − S bimodule if M is a left R-module, a rightS-module, and (rm)s = r(ms) for all r ∈ R, m ∈M , and s ∈ S.

Example 24. Let R and S be rings, and let M be an R − S bimodule. Let A denote the set of uppertriangular matrices (r,m, 0, s) where r ∈ R, m ∈ M , and s ∈ S. Then this is indeed a ring and all of thestuff holds.

Exercise 8. Show that A is left Noetherian if and only if R and S are left noetherian and M is leftNoetherian.

Also show that the same statement holds if “left” is replaced with right, and/or “Noetherian” is replacedby “Artinian”.

Example 25. Let A = (Q,Q, 0,Z). Recall that Q is a field, so it is Noetherian (as a ring), and Z is aPrincipal Ideal Domain, so it is Noetherian (as a ring). But Q is not Noetherian as a (right) Z-module. ThusA is left Noetherian, but not right Noetherian.

If instead A = (R,R, 0,Q), then this is left Artinian but not right Artinian.


Review day!This was a theorem from 818:

Theorem 34. Let E/F be algebraic, and let σ : E → E be a field embedding which fixes F . Then σ is infact onto. That is, σ is an automorphism.

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Proof. One proof uses vector space dimension: E is an F -vector space, so if E/F is finite, then σ is anF -linear transformation, so since σ is injective, it is also surjective. The infinite dimensional case can bereduced to the finite dimensional case since E/F is algebraic.

Alternatively, let α ∈ E, and let fα(x) = Min(α, F ). Let β1, ..., βn be the roots of fα ∈ E. For each i,σ(βi) = βj for some j, so if you restrict σ to {β1, ..., βn}, then it is still injective. Therefore it is a permutationof the βis, so it is surjective. Since α is some βi, and α was arbitrary, then σ is surjective.

10 Exact Sequences


Back to rings:Exact sequences!

Definition 34. A sequence of R-linear maps of left R-modules Ni with fi : Ni → Ni+1 (where Ni is called“degree i”) is said to be exact in degree i+ 1 if imfi = kerfi+1. We say the entire sequence is exact if it isexact at in all of its degrees.

We often write our exact sequence as

...→ Nifi→ Ni+1

fi+1→ Ni+1 → ...

Example 26. We have that 0→ Af→ B is exact if and only if kerf = 0 if and only if f is injective.

Example 27. Similarly, Ag → B → 0 is exact if and only if g is surjective.

Example 28. By combining these, 0→ A→ 0 is exact if and only if A = 0.

Example 29. Lastly, 0→ Af→ B → 0 is exact if and only if f is an isomorphism.

Definition 35. An exact sequence of the form 0 → Af→ B

g→ C → 0 is called a short exact sequence. Ashort exact sequence has the property that f is injective, im f = ker g, and g is onto.

Example 30. Let N be a submodule of M . Then 0 → Ninclusion→ M

projection→ M/N → 0 is a short exactsequence.

Example 31. Let M1 and M2 be modules. Then 0→M1f→M1⊕M2

g→M2 → 0 is a short exact sequencewhen f : u 7→ (u, 0) and g : (u, v) 7→ v. This short exact sequence is called a split short exact sequence sincethere are maps f ′ : M1 ⊕M2 →M1 and g′ : M2 →M1 ⊕M2 such that f ◦ f ′ = idM1 and g′ ◦ g = imM2 .

Example 32. The following short exact sequence is not a split exact sequence: 0 → Z f→ Z g→ Z/2Z → 0,where f(n) = 2n and g(m) = m.

Remark 24. We have been leaving off the word “left” a lot, but it should probably be around. Also, mosttheorems give another theorem if you find/replace the word “left” with the work “right”.

Theorem 35. [Two theorems in one!] Let R be a ring and let 0 → Lf→ M

g→ N → 0 be a short exactsequence of left R-modules. Then M is Noetherian (respectively, Artinian) if and only if both L and N areNoetherian (respectively, Artinian).

Proof. We know that L ∼= f(L), and N ∼= M/ker g = M/f(L). Thus without loss of generality, we canassume L ⊂M and N ⊂M/L.

If M is Noetherian (respectively, Artinian), ideals in L are ideals in M , so any descending (respectively,ascending) chain stabilizes in L. Thus L is Noetherian (respectively, Artinian). Also, a chain in N = M/Lcan be lifted to a chain in M , which must stabilize, and when we project back down, the result must stabilize.Thus N = M/L is Noetherian (respectively, Artinian), as desired.

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Conversely, suppose both L and N are Noetherian (Artinian case will be identical, but not proven).Let A ⊃ A2 ⊃ ... be a descending chain of left R-submodule of M . Then A ∩ L1 ⊃ A ∩ L2 ⊂ ... is adescending chain of L-modules. Since L is Noetherian, this chain stabilizes. Thus there exists a k such thatAk ∩ L = Ak+1 ∩ L for all i ≥ 0.

Also, A1 +L ⊃ A2 +L ⊃ ..., so A1+LL ⊃ A2+L

L ⊃ ... is a descending chain in N = M/L. As M/L has the

descending chain condition, there exists an l such that Al+LL = Al+1+L

L for all i ≥ 0. But this is the case ifand only if Al + L = Al+i + L.

Let j = max(k, l). We wish to show that Aj = Aj+i for all i ≥ 0. Fix some i ≥ 0. Let u ∈ Aj ⊂ Aj +L =Aj+i+L. Then u = aj+i+ lj+i for some aj+i ∈ Aj+i and l ∈ L. Then u−aj+i = lj+i ∈ Aj ∩L = Aj+i∩L ⊂Aj+i. Thus u ∈ Aj+i, so Aj ⊂ Aj+i, so Aj = Aj+i. Thus the chain has stabilized.

As remarked, the Artinian proof is similar.

Corollary 10. If M,N R-modules, then both M and N are Noetherian (respectively, Artinian) if and onlyif M ⊕N is Noetherian (respectively, Artinian).

Proof. Apply the previous theorem to 0→M →M ⊕N → N → 0, which is a short exact sequence.

Corollary 11. A module M is Noetherian (respectively, Artinian) if and only if Mn =⊕n

i=1M is Noethe-rian (respectively, Artinian).

Proof. Apply the previous corollary over and over, and induct on n.

Corollary 12. Let R be a left Noethereian (respectively, Artinian) ring, and let M be a finitely generatedleft R-module. Then M is left Noetherian (respectively, Artinian).

Proof. Since M is finitely generated, then M = Rx1 + ... + Rxn for some x1, ..., xn ∈ M . Then define andR-linear map φ : Rn →M by (r1, ..., rn) 7→

∑rixi. Note that φ is onto.

Since R is left Noetherian (respectively, left Artinian), then Rn is left Noetherian (respectively, leftArtinian), so M = Rn/ker φ is Noetherian.

Remark 25. Let φ : R → S be a ring homomorphism (including the restriction that φ(1) = 1). Then φdefines a left- and right-R-module structure on S. In particular, for r ∈ R and s ∈ S, define r · s = φ(r)sand s · r = sφ(r). If im φ ⊂ Z(S) = {t ∈ S|at = ta for all a ∈ S} (the so-called “center” of S), then S iscalled an R-algebra.

Remark 26. Let φ : R→ S be a ring homomorphism. Suppose R is left Noetherian (respectively, Artinian)and suppose S is finitely generated as a left R-module. Then S is left Noetherian (respectively, Artinian),as a ring.

Proof. By Corollary 12, S is left Noetherian (respectively, Artinian) as an R-module. But every left Smodule is a left R-module because of φ. Thus S satisfies the ascending chain condition on left ideals, sincesuch left ideals are left R-modules.

Example 33 (Don’t got time for this!).

Remember, no class on Friday!

11 Notherian Rings


More rings!

36

Example 34. Recall that if R is a commutative ring, and G is a finite group, then R[G] is the group ringgenerated by R and G. As a module, it is a free module over R with basis G. We define multiplication thelogical way to make it a ring.

Define φ : R→ R[G] by r 7→ r · 1. Then φ is a ring homomorphism and imφ ⊂ Z(R[G]).Therefore, R[G] is an R-algebra. Note that R[G] is a finitely generated left and right R-module.Therefore, if R is Noetherian, R[G] is (left and right) Noetherian as a ring. Similarly, if R is Artinian,

R[G] is left and right Artinian.

Proposition 17. Let R be a ring, and let M be a left R-module. Then the following are equivalent

1. M is (left) Noetherian.

2. Every (left) R-submodule of M is finitely generated.

3. Every nonempty subset of (left) R-submodules of M has a maximal element. (I.e. there exists anA ∈ Λ such that if A ≤ B and B ∈ Λ, then A = B.

Proof. (1 implies 2) Let N be a submodule of N . Choose x1 ∈ N , and let L1 = Rx1 ⊂ N . If N 6= L1, choosean x2 ∈ N \ L1, and let L2 = L1 +Rx2. Note L1 ( L2 ⊂ N .

We repeat this process: if Ln 6= N , then take some xn+1 ∈ N \ Ln, and let Ln+1 = Ln + Rxn+1. Thisgives us L1 ( L2 ( L3 ( .... If for all n, Ln 6= N , this would give us an infinite strictly increasing chain,contradicting Noetherian-ness. Thus N = Ln for some n, but N = Rx1 + ...+Rxn, so N is finitely generated.Thus every submodule is finitely generated, as desired.

(2 implies 3; actually, not 3 implies not 2) Suppose Λ is a nonempty collection of submodules with nomaximal element. Then there exists an infinite an infinite ascending chain of submodules N1 ( N2 ( N3 ( ...

For all i ≥ 2, choose some xi ∈ Ni \ Ni−1, and let N denote the module generated by x2, x3,... Note

that N ⊂∞⋃i=1

Ni. If N were finitely generated, then all of its generators would come from some Nk. Thus

xi ∈ Nk for all i, so in particular xk+1 ∈ Nk, which is a contradiction. Thus N is not finitely generated, asdesired.

(3 implies 1) If L1 ⊂ L2 ⊂ ..., then let Λ = {Li|i ∈ N}. Then Λ has a maximal element Lk. That is, thechain stabilizes at Lk.

Remark 27. If a ring is Noetherian, condition (3) from this proposition means we don’t have to worryabout citing Zorn’s lemma! Condition (3) is really powerful!

Example 35. Let F be a field, and let σ : F → F be a nonzero field homomorphism which is not surjective.(For example, it might be that F = Q(t), where t is an indeterminant, and we define σ : F → F byf(t)g(t) 7→

f(t2)g(t2) .)

Then we have F [x;σ] is left Noetherian, but not right Noetherian. (We will prove this.)(Hold up, what the heck is F [x;σ]? As an F -vector space, F [x;σ] = F [x]. We give it a different

multiplication though: (anxn)(bmx

m) = anσn(bm)xn+m.)

Before we prove the claim from the example, we prove a lemma.

Lemma 15. Let f(x), g(x) ∈ F [x;σ], with g(x) 6= 0. Then there exists q(x), r(x) ∈ F [x;σ] such thatf = qg + r, and deg r < deg g.

Proof. Let f(x) = amxm + ...+ a0, and let g(x) = bnx

n + ...b0. If m < n, we are already done, since we canlet q = 0, and r = f .

But if m ≥ n, then consider f − am(σm−n(bn))−1xm−n. Observe that this has no degree m term, so itsdegree is strictly less than m. Then proceed by induction to strip away all the degrees.

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Proof. (Of the claim in the example)Let R = F [x;σ].We first show that every left ideal of F [x;σ] is principal. Let I 6= 0 be a left ideal of F [x;σ]. Choose

g ∈ I \ {0} of minimal degree. Then certainly Rg ⊂ I. If f ∈ I, then by the lemma, there exists a quotientand remainder such that f = qg + r, so r = f − qg. Since I is a left ideal, then r ∈ I, but r has degreestrictly less than the degree of g, so r = 0. Thus f = qg, so f ∈ Rg. Thus Rg = I, so all left ideals areprincipal.

Thus in particular, all left ideals are finitely generated, so F [x;σ] is left Noetherian.We now wish to show that R is not right Noetherian. We shall do so by constructing an infinite ascending

chain of right ideals. Since σ is no surjective, choose some b ∈ F \ σ(F ). Let M0 = bxR, and for all For alli > 0, let Mi = xibxR+Mi−1. Certainly, xibx ∈Mi. We wish to show that xnbx 6∈Mn−1.

Suppose for the sake of contradiction that xnbx ∈ Mn−1. Then xnbx = xn−1bxfn−1(x) + ... + bxf0(x)for some fn−1, ..., f0 ∈ R. Then bxf0(x) = xg(x) for some g(x) ∈ R. Let r = deg f0 = deg g.

If f0 = arxr + ... + a0, and g1 = crx

r + ... + c0, then bσ(ar) = σ(cr). Thus b = σ( crar ) ∈ σ(F ), which isa contradiction. Thus M0 ( M1 ( M2 ( ... is a strictly ascending chain of right modules, so R is not rightNoetherian.

Recall the Hilbert Basis Theorem, in all of its glorious generality:

Theorem 36 (Hilbert Basis Theorem). Let R be a left Notherian ring, and let X be a variable. ThenR[x] is left Noetherian.

By induction, R[x1, ..., xn] is left Noetherian for commuting variables x1, ..., xn.

Remark 28. Subrings of Noetherian (or Artinian) rings are not necessarily Noetherian or Artinian.

Example 36. Note Z ⊂ Q, and since Q is a field, it is Artinian. But Z is not Artinian.

12 Simple Rings and Modules


Definition 36. Let R be a ring. An ideal in R is a left ideal that happens to be a right ideal as well. Thatis, I is an ideal if (I,+) is a subgroup of (R,+), and rI ⊂ I and Ir ⊂ I for all r ∈ R.

Definition 37. The two trivial ideals of a ring R are (0) and R.A ring R is simple if it has no nontrivial ideals.

Example 37. Let R = M2(Q). Then I = {(a 0b 0

) ∣∣∣a, b ∈ Q} is a left ideal but not a right ideal.

Another example is R = M2(Z). Let I = {2A|A ∈ R} = M2(2Z). Then I is a nontrivial ideal.

Remark 29. Any division ring is simple. By a homework problem [edit: see immediately below], Mn(D) issimple whenever n ≥ 1 and D is a division ring.

Homework Problem 6. Let n ∈ N, let R be a ring, and let S = Mn(R). Then there is a natural bijectionbetween the two-sided ideals in R and the two-sided ideals in S given by I ↔Mn(I).

Example 38. If φ : R→ S is a ring homomorphism, then ker φ is an ideal of R.If R is simple, every such φ : R→ S is either 0 or injective.

Example 39. Let M be a (left) R-module, and let AnnR M = {r ∈ R|rM = 0}. Then AnnR M is a(two-sided) ideal of R. This is the case because if r ∈ AnnR M , and s ∈ R, then (sr)M = s(rM) = s ·0 = 0,and (rs)M = r(sM) ⊂ rM = 0.

Aside 1. Let x ∈M . Then AnnR x = {r ∈ R|rx = 0}. Then AnnR x is a left ideal, but not necessarily aright ideal. However, AnnR Rx is an ideal.

Definition 38. Let R be a ring. An R-module M is called simple if M 6= 0, and M has no non-trivialsubmodules.

38

Remark 30. A simple ring is not necessarily simple as a left R-module. For instance, take M2(Q). Weshowed that this is simple, but it had a nontrivial left ideal. Therefore it is not simple as a left R-module.

Proposition 18. An R-module M is simple if and only if M ∼= R/I for some maximal left ideal I.

Proof. (⇐) Suppose M ∼= R/I for some maximal ideal I. Recall that the submodules of R/I are in one-to-onecorrespondence with the submodules of R containing I. As I is maximal, the only submodules containingI are R and I. So the only submodules of R/I are R/I and I/I = 0. Thus M = R/I is simple (as aR-module), as desired.

(⇒) Let x ∈M , x 6= 0. Then Rx is a submodule of M , but it is nonzero, so Rx = M since M is simple.Define an R-homomorphism φ : R → Rx = M by r 7→ rx. Then by an isomorphism theorem, M ∼= R/I,where I = ker φ. Since R/I is simple, I is maximal as a left ideal.

Remark 31. If R is commutative, then left ideals are precisely two-sided ideals, so M is simple if and onlyif M = R/m where m is a maximal ideal. But R/m is a field, so the only simple R-modules are fields.

If R is a local ring (i.e. there is a unique maximal ideal), then this implies that for all simple R-modulesare isomorphic.

Example 40. For the complex numbers, things are nicely behaved by Hilbert’s Nullstellensatz.For a not-algebraically-closed field, things need not be very nice. For instance, let R = R[x]. Let

m1 = (x). Then R/m1∼= R. Let m2 = (x2 + 1). Then R/m2 ∼= R(i) = C. Thus two simple modules have

led to different fields.

Example 41. Let D be a division ring, and let R = Mn(D), where n ≥ 1. Consider M = Dn =

{

a1

...an

∣∣∣ai ∈ D}. Then M is a left R-module by matrix multiplication.

We claim that Dn is a simple R-module. It suffices to show that Ru = Dn for all u ∈ Dn \{0}. Note thatDn is generated by the vector (1, 0, 0, ..., 0). This vector is generated by any vector of the form (a1, 0, ..., 0)(where a1 6= 0) since we can divide by a1. But this is generated by any nonzero vector by multiplying bythe right matrix which kills all-but-one term, and also the right row swapping matrix.

Example 42. Let R = S3, let R = C[S3] (or any algebraically closed field of characteristic not equal to 2or 3). Recall that R = C1⊕ C(12)⊕ ...⊕ C(132).

Let I1 = C(1 +(12) + (13) + (23)+ (123) + (132). This is a C-vector space, at least. But also, multiplyingby any element of S3 will only permute the terms, so it is a left R-module.

Also, as a vector space it is dimension 1, so there are no nontrivial submodules (which would be subspaces).Thus I1 is a simple left R-module.

Let u1 = (1) + (12) and let u2 = (23) + (132) and let u3 = (13) + (123). Let v1 = u1 − u2 and letv2 = u1 − u3.

Let I2 = Cv1 + Cv2. We can check that v1 and v2 are linearly independent, so this has dimension 2, asa C-vector space. Also, by checking left multiplication, I2 is a left ideal.

Finally, by a complicated argument that will remain a secret, but apparently involving character theory,I2 has no nontrivial submodules. Thus I2 is is simple as a left module.

13 Semisimple Rings and Modules


No homework was posted yet, since Tom was at a conference!(Recall that an ring is simple if it only has two (two-sided) ideals. A module is called simple if it is

nonzero and has no proper submodules.)

Definition 39. An R-module M is called semisimple if every submodule N of M is a direct summand ofM (i.e. there exists N ′ ⊂M such that M = N ⊕N ′ (i.e.i.e. M = N +N ′ and N ∩N ′ = 0)).

39

Aside 2. Let {Nλ}λ∈I be a collection of submodules ofM . Then∑λ∈I

Nλ = {finite sums of elements of Nλs}.

This is a submodule of M , and is the smallest such submodule containing all of the Nλs.

We write∑λ∈I

Nλ =⊕λ∈I

Nλ if for all x ∈ N , there exist unique aλ ∈ Nλ for each λ ∈ I such that x =∑λ∈I

aλ

(that is, each element can be written uniquely as a sum). This is called the internal direct sum.The external direct sum is the set of ordered I-tuples such that all but finitely many entries are nonzero,

with termwise addition.Exercise: The sum is direct if and only if the sum is isomorphic to the external direct product.

Exercise: The sum is direct if and only if for all λ ∈ I, Nλ ∩ (∑δ 6=λ

Nδ) = (0).

Remark 32. A simple module is always semisimple. A simple ring is not always semisimple (as a module).

Claim 1. Let D be a division ring. Then any D-module is semisimple.

Proof. Let M be a D-module, and let N be a submodule. Let β be a basis for N , and extend to a basis β′

of M . Let N ′ = SpanD(β′ \ β). hen M = N ⊕N ′.

Lemma 16. Any submodule of a semisimple module is semisimple.

Proof. Let M be semisimple and let A be a submodule of M . Let B be a submodule of A. As B is asubmodule of M , M = B ⊕N for some N .

We wish to show that A = B ⊕ (N ∩ A). Certainly, B + (N ∩ A) ⊂ A. Let a ∈ A. Then a ∈ X, soa = b + n, for some b ∈ B, n ∈ N . But n = a − b ∈ A. Thus n ∈ N ∩ A. Thus A ⊂ B + (N ∩ A), soA = B + (N ∩A).

Finally, we wish to show that this sum is direct. Note that B∩ (N ∩A) ⊂ B∩N = (0), so B and (N ∩A)are disjoint. Thus the sum is direct, from the second criterion in the aside.

Remark 33. If M is semisimple, then so is M/N . This is the case because M/N = (N + N ′)/N ∼= N ′,a submodule of M . Thus M/N is isomorphic to a submodule of M , so it is semisimple by the previouslemma.

Proposition 19. Every nonzero semisimple module contains a simple submodule.

Proof. Let M be a nonzero semisimple R-module. Let x ∈M be a nonzero element. Then Rx is semisimplesince it is a submodule of a semisimple module. We then wish to show that Rx contains a simple submodule.

Let Λ = {A|A ⊂ Rx, x 6∈ A}. Note that Λ 6= ∅ since (0) ∈ Λ. Also, for any chain C ⊂ Λ, by taking theunion of the elements of C, we get a new submodule of Rx, and since each term in C does not contain x,the union does not contain x. Thus the union is in Λ, so every chain in Λ has an upper bound in Λ.

Thus by Zorn’s Lemma, there exists a maximal element of Λ, which we shall call B. Since Rx issemisimple, Rx = B ⊕B′.

Suppose for the sake of contradiction that B′ is not simple. Then there exists L ( B′ such that L 6= 0.Note x 6∈ B + L, or else B + L = Rx. Let b′ ∈ B′. Then b′ = b + l, for some b ∈ B, l ∈ L. Thenb = b′ − l ∈ B′ ∩B. Thus b′ = l, so B′ = L. Thus B + L ∈ Λ, so B ( B + L, since L 6= 0. This contradictsthe maximality of B, so this is a contradiction.

(Alternatively: If B′ is not simple, it is still semisimple, so B′ = L ⊕ L′. Thus R = B ⊕ (L ⊕ L′) =(B ⊕ L)⊕ L′. Then B ⊕ L is strictly larger, contradicting maximality of B.)

Thus B′ ⊂ Rx ⊂M is simple.

Theorem 37. Let M be an R-module. The following are equivalent:

1. M is semisimple.

2. M is the sum of a simple submodule.

3. M is the direct sum of simple submodules.

40

(Proof is long, and will be done on Wednesday.)

Definition 40. A ring R is called left semisimple if R is semisimple as a left R-module.Similarly, it is called right semisimple if R is semisimple as a right R-module.

Remark 34. Much later on, we will see that a ring is left semisimple if and only if it is right semisimple.In this case, we just call the ring semisimple.

Claim 2. A ring R is left semisimple if and only if it is the (direct, or not direct) sum of finitely manysimple left ideals.

There is an analogous statement for right semisimple.

Proof. (⇐) If R is the sum or direct sum of finitely many simple left modules, then by the theorem, R issemisimple.

(⇒) By the theorem, R =∑λ∈J

Iλ where Iλ are simple left ideals. Write 1 = aλ1+...+aλk for λ1, ..., λk ∈ J ,

aλj ∈ Iλj . Since R = R · 1, then R = Raλ1 + ...+Raλk ⊂ Iλ1 + ...+ Iλk ⊂ R.Thus R = Iλ1 + ...+ Iλk .

Exercise 9. Generalize the previous claim to finitely generated modules.Show that any division ring is semisimple.Let D be a division ring, and let R = Mn(D). For k = 1, ..., n, let Ik denote the set of matrices which are

zero except in the k-th column. Then each Ik ∼= Dn, which is a simple left R-module. Also, R = I1⊕ ...⊕ In,so R is left semisimple.


Recall the following theorem from last class. We will prove it now:

Theorem 38. Let M be an R-module. The following are equivalent:

1. M is semisimple.

2. M is the direct sum of a simple submodule.


Proof. If M = 0, these are all true. Suppose for the rest of the proof that M is nonzero.(1 ⇒ 2) Suppose M is semisimple. Let T = {E|E ⊂ M,E is simple}. Since M 6= 0, then by a theorem

from the previous class, M contains a simple submodule. Thus T 6= ∅.Let Λ = {J ⊂ T |

∑E∈J E = ⊕E∈J}. Certainly, for each t ∈ T , {t} ∈ Λ, so Λ 6= 0. Note that the union

of a chain is still in Λ for the following reason: the union would fail to be a direct sum if some element couldbe written two different ways. But this would only involve finitely many elements, so some element of thechain would not be in Λ, a contradiction.

Thus we can apply Zorn’s Lemma. Let J be a maximal element of Λ. Then∑E∈J

E =⊕E∈J

E. Let

N =∑E∈J

E.We wish to show that M = N . Certainly, N ⊂ M , so since M is semisimple, M = N ⊕N ′ for

some submodule N ′ of M . Since N ′ is a submodule of a semisimple module, it is semisimple.Therefore, if N ′ is nonzero it contains a simple submodule E′. Note that since E′∩N = (0), then E′ 6∈ J .

But J ∪ {E′} ∈ Λ for the same reason. This contradicts the maximality of of J .Thus N ′ = 0, so M = N .(2⇒ 3) Suppose M is the direct sum of simple submodules. Certainly, every direct sum is a sum, so M

is the sum of simple submodules.

41

(3 ⇒ 1) Suppose M is the sum of simple submodules. We have M =∑E∈T

E for some collection T of

simple submodules.Let A be a submodule of M . We wish to find a submodule A′ such that M = A⊕A′.Let Λ = {J ⊂ T |(

∑E∈J

E) ∩ A = 0}. Note that ∅ ∈ Λ, so Λ 6= ∅. Also, the union of chains in Λ are in

Λ for the following reason: a union would fail to be in Λ if some nonzero finite sum of elements were in A.Then this finite sum would be in some element of the chain, a contradiction.

Thus we can apply Zorn’s Lemma. Let J be maximal in Λ, and let A′ =∑E∈J

E. Certainly, A′ ∩A = (0).

Thus the sum A+A′ is direct. Let N = A+A′. We wish to show that N = M .If N 6= M , then there exists a simple submodule E′ of M such that E′ 6⊂ N . Since E′ is simple, then

E′ ∩ N = (0), so E′ + N is direct. Suppose a = e + e′ for some a ∈ A, e ∈ N , and e′ ∈ E′. Thene′ = a − e ∈ N ∩ E′ = (0). Thus e′ = 0, so a = e ∈ A ∩ A′ = (0), so a = e = 0. Thus J ∪ {E′} ∈ Λ. sinceE′ 6∈ J , then this contradicts the maximality of J . Thus N = M , so M = A ⊕ A′. Thus M is semisimple,as desired.

14 Filtrations and Length of Modules

Let’s move on to a new topic!

Definition 41. Let M be an R-module. A series for M is a finite chain (also known as a filtration) ofsubmodules of M such that 0 = Mn ⊂Mn−1 ⊂ ... ⊂M0 = M .

The factor modules of this series are Mi/Mi+1|i = 0, ..., n− 1} and we count them with multiplicity.The length of a series is the number of non-zero factor modules (which is equal to the number of strict

inequalities).A refinement of a series for M is another series for M which “contains” the original series as a subseries.

That is, if the original series is 0 = Mn ⊂ .... ⊂ M0 = M , and 0 = M ′k ⊂ ... ⊂ M ′0 = M is another series,then the new series is a refinement of the original one if for each i there exists a j such that Mi = M ′j .

A proper refinement is a refinement which strictly increases the length.We say two series for a module are equivalent if there exists a bijection between the two sets of nonzero

factor modules, such that the corresponding factor modules are isomorphic.

Example 43. Let R = Z and let M = Z. Then (0) ⊂ 18Z ⊂ 18Z ⊂ 3Z ⊂ Z. Then the factor modules are18Z, 0,Z6,Z3.

A refinement of this series might be (0) ⊂ 72Z ⊂ 18Z ⊂ 18Z ⊂ 9Z ⊂ 3Z ⊂ 3Z ⊂ Z. Then the factormodules are 72Z,Z4, 0,Z2,Z3, 0,Z3.

Another series might be 0 ⊂ 72Z ⊂ 24Z ⊂ 12Z ⊂ 4Z ⊂ Z. Then the factor modules are 72Z,Z3,Z2,Z3,Z4.Thus the third series is equivalent to the second. Note that the third series is not a refinement of the first.

Theorem 39. (Schreier Refinement Theorem) Let M be an R-module. Then any two series for M haverefinements which are equivalent.

We will prove this next class. For now, we prove a lemma.

Lemma 17 (Zassenhaus Lemma). Let M be an R-module, and let A ⊂ A′ and B ⊂ B′ be R-submodules

of M . ThenA+ (A′ ∩B′)A+ (A′ ∩B)

∼=B + (A′ ∩B′)B + (A ∩B′)

.

Proof. We will first show both modules are isomorphic toA′ ∩B′

A′ ∩B +A ∩B′. By symmetry, it suffices to show

this is true for just one of the modules. That is, it suffices to show thatA+ (A′ ∩B′)A+ (A′ ∩B)

∼=A′ ∩B′

A′ ∩B +A ∩B′.

LetA′ ∩B′

A′ ∩B +A ∩B′= L

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Define φ : A+(A′∩B′)→ L by a+u 7→ u. We need to show this is well-defined. Suppose a1+u1 = a2+u2.Then u1 − u2 = a2 − a1 ∈ A ∩A′ ∩B′ ⊂ A ∩B′. Thus u1 = u2, so the function is well-defined.

We then need to check two things: first, that this function is surjective. This is straightforward.Also, we need to check that kerφ = A+ (A′ ∩B). This is left as an exercise.


Homework will be due next Monday (November 9) instead of Friday!Recall the following lemma from last class:

Lemma 18. (Zassenhaus Lemma) Let M be an R-module, and let A ⊂ A′ and B ⊂ B′ be R-submodules

of M . ThenA+ (A′ ∩B′)A+ (A′ ∩B)

∼=B + (A′ ∩B′)B + (A ∩B′)

.

We will use it to prove this important theorem:

Theorem 40 (Schreier Refinement Theorem). Let M be an R-module. Then any two series for Mhave refinements which are equivalent.

Proof. Let 0 = Mn ⊂ ... ⊂M0 = M and 0 = Nt ⊂ ... ⊂ N0 = M be two series for M .Fix i ∈ {0, 1, ..., n}. Let Mi,j = Mi+1 +Mi ∩Nj for j = 0, ..., t.(What does this look like? Well, Mi,0 = Mi+1 +Mi ∩N0 = Mi+1 +Mi = Mi. Also, Mi,t = Mi+1. Also,

Mi,j ⊂Mi,j+1 so we have “series” such that Mi+1 = Mi,t ⊂Mi,t−1 ⊂ ... ⊂Mi,0 = Mi.)Then we get a refinement of the first series by taking 0 = Mn,t ⊂Mn,t−1 ⊂ ... ⊂M0,1 ⊂M0,0 = M .Similarly, we define Ni,j = Ni+1 + Mj ∩ Ni for j ∈ {0, ..., n}. Thus we get a refinement of the second

series by taking 0 = Nt,n ⊂ ...N0,1 ⊂ N0,0 = M .We now wish to show that these series are equivalent.First we will show that that Ni,j/Ni,j+1

∼= Mj,i/Mj,i+1. This follows from the Zassenhaus Lemma bysetting Ni+1 = A, Ni = A′, Mj = B, and Mj+1 = B′.

There are also a few other links in the refined series, where something like Mi+1,t ⊂ Mi,t but this is infact equivalence, since these both equal Mi+1, so the factor modules are 0, so we don’t need to pay attentionto them.

Thus the two series are equivalent, as desired.

Definition 42. Let M be an R module. A composition series for M is a series in which all the nonzerofactor modules are simple modules.

Remark 35. A series is a composition series if and only if it has no proper refinement. (Proper, in thiscase, means that we have inserted a strictly contained module).

A series is a composition series if and only if the series is equivalent to all of its refinements.

Proposition 20. Let M be an R-module. Then any two composition series of M are equivalent.

Proof. Take any two composition series of M . By the Schreier Refinement Theorem, these can be refined toequivalent series. But they are composition series, so they are equivalent to their own refinements. Thus bytransitivity they are equivalent.

Definition 43. Let M be an R-module. If M has a composition series, the length of M , denoted λR(M),is the length of any composition series for M . (Note that this is well defined by the previous proposition.)If M does not have a composition series, then we define λR(M) =∞.

If λR(M) <∞, then we say M is a finite length module.

Example 44. Let R = Z, and let M = Z24. Then recall that simple Z-modules are simple abelian groups,i.e. cyclic groups of order p. Then a composition series for Z24 could be

0 ((12)

(24)(

(6)

(24)(

(2)

(24)(

Z(24)

43

Then the factor groups are Z2, Z2, Z2, and Z3. Thus λR(M) = 4.

Example 45. Let D be a division ring. Then every D-module M has a basis β, so M ∼=⊕α∈β

D.

Then D is a simple (left) D-module. Suppose M = Dn (i.e. dimD(M) = n, i.e. |β| = n).Then let Mi = D ⊕D ⊕ ... ⊕D ⊕ 0 ⊕ 0 ⊕ ... ⊕ 0 ⊂ Dn, where there are n − i copes of D, and i copies

of 0. Then 0 = Mn ( ... ( M0 = M , and Mi/Mi+1∼= D (which is simple as a left D-module). Thus

λD(M) = n = dimD(M).

Remark 36. We use RM to denote M thought of a left R-module, and we use MR to denote M thoughtof as a right R-module.

Example 46. Let D be a division ring, and let R = Mn(D). For k =∈ {1, ..., n}, we let Ik be the subset withzero entries off of the kth column. Then recall that each Ik is a simple left ideal of R, and R = I1 ⊕ ...⊕ In.By doing the same thing as before, we can see that λR(RR) = n.

By symmetry (exchanging rows with columns), λR(RR) = n as well. And since it is a D-vector space,λD(DR) = n2.

Proposition 21. Let M be an R module. Then λR(M) < ∞ if and only if M is both Noetherian andArtinian.

Proof. (⇒) Suppose λR(M) < ∞. Then there is a composition series of length λR(M) = r. Consider anyfinite chain of submodules of M : 0 = Mn ⊂ ... ⊂M0 = M .

By the Schreier Refinement Theorem, we can refine this chain of submodules into a composition series.The refined chain will have r proper containments.

Note that a refinement can introduce new proper containments, but can’t destroy them. Thus everychain of submodules has at most r proper containments. That is, any ascending and descending chain ofsubmodules of M stabilizes after at most r proper containments.

(⇐) Suppose M is both Noetherian and Artinian. If M = 0, then we are done. If M 6= 0, then becauseM is Artinian, we can choose a minimal nonzero submodule of M , which we shall denote N1. Note that N1

is simple, since if it had a proper submodule, it would not be minimal.Now we repeat this process: if N1 6= M , choose a submodule that is minimal among all submodules

properly containing N1, which we shall denote N2. Then N2/N1 is simple as before.Thus by repitition, we get 0 = N0 ( N1 ( N2 ( ...Since M is Noetherian, this chain cannot go on forever. The only way it can stop is if some Nk = M .

Thus we have a composition series 0 = N0 ⊂ N1 ⊂ ... ⊂ Nk = M . Thus M has a composition series.

14.2 Day 28 - November 2

Recall that homework will be due next Monday (November 9) instead of Friday!

Proposition 22. Claim: Let R be a ring, and let 0 → A → B → C → 0 be a short exact sequence ofR-modules. Then, λR(B) = λR(A) + λR(C).

Proof. Without loss of generality, we can assume A is a submodule of B and C = B/A.First we consider the case where there are infinite lengths. Recall that a module has finite length if and

only if it is Noetherian and Artinian. But also, we have proven that a quotient B is Noetherian (respec-tively, Artinian) if and only if both its submodule A and its “supermodule” C are Noetherian (respectively,Artinian). Thus λR(B) =∞ if and only λR(A) =∞ or λR(C) =∞. Thus if one of λR(A),λR(B), or λR(C)is infinite, then the claim holds.

Suppose instead that all the lengths are finite. Let n = λR(A) and let m = λR(C). Then there existscomposition series 0 = An ( ... ( A0 = A and 0 = Bm/A ( ... ( B0/A = B/A = C for A and C,respectively. Since these are composition series, Aj/Aj+1 and Bi/Bi+1 is simple for all appropriate j and i.

Then 0 = An ( ... ( A0 = Bm ( ... ( B0 = B is a series for B. Note that Bi/Bi+1∼= Bi/A

Bi+1/Ais simple

for all i. Thus this series is in fact a composition series. Thus λR(B) = m+n = λR(A) +λR(C), as desired.Thus in all cases, λR(B) = λR(A) + λR(C).

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Remark 37. Recall that if A1 and A2 are modules, then 0 → A1 → A1 ⊕ A2 → A2 → 0 is a short exactsequence. Thus λ(A1 ⊕A2) = λ(A1) + λ(A2).

By induction, we have that λ(

n⊕i=1

Ai) =

n∑i=1

λ(Ai).

Example 47. Let K be a field, let R = K[x], and let M = R/(x2). We wish to show that 0 ( Rx ( M is

a composition series for M . Note that M/Rx = R/(x2)(x)/(x2)

∼= R/(x) ∼= K, which is simple.

We can also define f : R → Rx by r → rx, which is a module homomorphism. Note that rx = 0 ifand only if rx = 0 which is the case if and only if rx ∈ (x2) which is the case if and only if r ∈ (x). Thusker f = (x), so Rx/(0) ∼= Rx ∼= R/(x) which is simple. Thus λR(M) = 2.

Also, R is a K-algebra, and λK(M) = dimK(M) = 2, since {1, x} is a K-basis for M .

Example 48. Let R = R[x]. Let M = R/(x2 + 1). Note that x2 + 1 is irreducible as a polynomial,so (x2 + 1) is maximal as an ideal. Thus this quotient is a simple R-module. Thus λR(M) = 1, butλR(M) = dimR(M) = 2.

In general, length does not have to equal vector space dimension.

15 Classification of Semisimple Modules

Proposition 23. Let R be a ring. The following are equivalent

1. R is semisimple.

2. RR is a direct sum of finitely many simple left ideals.

3. RR is a sum of finitely many simple left ideals.

Proof. (1⇒ 2) Suppose R is semisimple. We know from the previous characterization of semisimple modules

that R is the direct sum of simple submodules, i.e. ideals. That R =⊕α∈Λ

Iα for some simple left ideals Iα.

Then we can write the multiplicative identity as 1 = iα1+ ... + iαn where each iαj ∈ Iαj . Then for all

r ∈ R, r = riα1+ ...+ riαn ∈ Iα1

+ ...+ Iαn . Thus R = Iα1+ ...+ Iαn . But R was the direct sum of the Iαs,

so R =

n⊕i=1

Iαn . Thus R is the direct sum of finitely many simple left ideals, as desired.

(2→ 3) Suppose RR is the direct sum of finitely many simple left ideals. Then since every direct sum isa sum, RR is the sum of finitely many simple left ideals.

(3→ 1) Suppose RR is the sum of finitely many simple left ideals. Then by the previous characterizationof semisimple modules, we know that RR is semisimple. That is, R is semisimple.

Remark 38. If R is left semisimple, then RR = I1 ⊕ ... ⊕ In, where each Ij is simple. Thus for each j,

λR(Ij) = 1, so λR(R) =

n∑j=1

λR(Ij) = n. Thus R is both left Noetherian and left Artinian.

Exercise 10. Generalize the previous proposition to any finitely generated module. Use this to show thatany finitely generated semisimple module has finite length.

Proposition 24. Suppose R is left semisimple. Then every (left) R-module is semisimple.

Proof. Let M be an R-module. Let {xα}α ∈ Λ be a generating set. Then M =∑α∈Λ

Rxα. Let F =⊕α∈Λ

R,

which is semisimple, since R is semisimple.

45

Define φ : F →M by∑rαeα 7→

∑rαxα. Note that φ is surjective homomorphism. Thus M ∼= F/ kerφ.

That is, M is the quotient of a semisimple module. By a previous theorem, the quotient of semisimplemodules is semisimple, so M is semisimple.

Remark 39. (Prelude to the Artin-Wedderburn Theorem)Recall that if D is a division ring and n is a positive integer, then Mn(D) is left (and right) semisimple.Recall also that R1, ..., Rl are left semisimple, then so is R1 × ..×Rl.From these two facts, it follows that Mn1(D1)× ...×Mnl(Dl) is semisimple for division rings D1, ...Dl.

Next time, we will prove the Artin-Wedderburn Theorem!


Let R be a ring, let I be a left ideal, and let M be a left R-module. Then IM = {∑finite

aiui|ai ∈ I, ui ∈M}.

Then IM is a left submodule of M .In this case IR = I and RI = I. Also, for x ∈M , IRx = Ix = {ix|i ∈ I}, and this is a submodule of M .

Lemma 19. Let R be a ring, and I be a simple left ideal and M a simple module. If I 6∼= M then, IM = 0.

Proof. We prove the contrapositive. Suppose IM 6= 0. Then there exists some i ∈ I and u ∈ M such thatiu 6= 0.

Define φ : I → M by a 7→ au. Then we can check that φ is an R-module homomorphism. Note thatimφ 6= 0 since φ(i) = iu 6= 0. Also, φ(I) is a submodule of M , so since M is simple, imφ = M . That is, φ issurjecive.

Also, kerφ 6= I since φ(i) 6= 0. But kerφ is a subideal of I, and since I is simple then kerφ = 0. Thus φis injective, so φ is an R-module isomorphism.

Proposition 25. Let R be a left semisimple ring. Then

1. Every simple R module is isomorphic to a simple left ideal.

2. There are only finitely many (up to isomorphism) simple R-modules.

Proof. (Proof of 1) Let M be a simple R-module. Choose u ∈M \ {0}. Define φ : R→M by r 7→ ru. Notethat Ru 6= 0, and Ru is a submodule of M , so since M is simple, then Ru = M . Thus φ is surjective. As Ris left semisimple, R = kerφ ⊕ I for some ideal I. Then for r ∈ R, there exist unique a ∈ kerφ and b ∈ Isuch that r = a + b. But φ(r) = φ(a + b) = (a + b)u = bu. Then φ

∣∣∣I

: I → M is an isomorphism. Thus

M ∼= I.Since M is isomorphic to I, and M is simple, then I is simple, as desired.(Proof of 2) Since R is semisimple, we can write it as the finite sum of simple left ideals. That is,

R = I1 + ...+ Ik for some simple left ideals Ijs. Let J be a simple left ideal of R. It suffices to show that Jis isomorphic to some Ij .

Suppose J were not isomorphic to any Ij . Then by the lemma, JIj = 0 for all j. But J = JR =J(I1 + ...Ik) = JI1 + ...JIk = 0. This is a contradiction, so J must be isomorphic to some Ij .

Theorem 41. Let R be a left semisimple ring. Let I1, ..., Ik be representatives of the distinct isomorphismclasses of simple left ideals (that is, every simple left ideal of R is isomorphic to exactly one Ij). For each i,

let Ri =∑

L left idealL∼=Ii

L. Then

1. For all i, Ri is a ring with identity.

2. For all i, Ri is a left semisimple with a unique (up to isomorphism) simple left ideal.

3. For all i, Ri is simple.

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4. Finally, R ∼= R1 × ...×Rk (as rings).

Proof. (Proof of 1) As R is semisimple, R =∑

L simple left ideal

L = R1 + ...+Rk.

Each Ri is a left ideal of R, so Ri is closed under + and ·. We can write 1 = e1 + ...+ek for some ei ∈ Ri.We wish to show that ei is the multiplicative identity for Ri.

First observe that for i 6= j, RiRj = (∑Lα∼=Ii

Lα)(∑Lβ∼=Ij

Lβ) =∑α,β

LαLβ =∑

0 = 0 (since non-isomorphic

simple ideals multiply to 0).Let x ∈ R. We can write x = x1 + ... + xk for some xi ∈ Ri. Then xi = xi × 1 = xi(e1 + ... + ek) =

xiei = (xi + ...xk)ei = xei. Thus the xis are uniquely determined for all i. Thus R = R1 ⊕ ...⊕ Rk (as leftR-modules).

Observe that xi = xi · 1 = xi(e1 + ... + ek) = xiei, and similarly xi = 1 · xi = (e1 + ... + ek)xi = eixi.Thus ei is a multiplicative identity for Ri. Thus each Ri is a ring with identity.

(Proof of 4) Define ψ : R→ R1×...×Rk by x 7→ (xe1, ...xek) = x(e1, ..., ek). Certainly, ψ is R-linear. Notealso that ψ(xy) = (xye1, ..., xyek) = x(e1, ..., ek)y(e1, ..., ek) = ψ(x)ψ(y). Thus ψ is a ring homomorphism.

Also, suppose ψ(x) = 0. Then xe1 = 0 for all i, and x = x(e1 + ...+ ek) = 0. Thus ψ is injective.Finally, suppose (y1, ..., yk ∈ R1 × ... × Rk. Then let x = y1 + ... + yk. Then ψ(x) = (xe1, ..., xek) =

(y1, ..., yk), so ψ is surjective. Thus ψ is a ring isomorphism. Thus R ∼= R1 × ...×Rk.(Proof of 2) Fix an i. Then Ri is a sum of left ideals, so Ri is semisimple. Then by something Zorn-ish,

it contains a simple left ideal.Suppose J is a simple left ideal of Ri. We wish to show that J ∼= Ii.Let K be a left ideal of Ri. Note that RK = (R1 + ...+Rk)K = RiK ⊂ J . Hence K is a left ideal of R.

Also, if K is a left ideal of R contained in Ri, then certainly K is a left ideal of Ri. Therefore the left idealsof Ri are precisely the left ideals of R contained in Ri.

Thus since J is a simple left ideal of Ri, then J is a simple left ideal of Ri. Therefore J ∼= Ij for some j.But J = RiJ ∼= RiIj = 0 for all i 6= j, so i = j and J ∼= Ii. Thus there is a unique simple left ideal of Ri,up to isomorphism.

(Proof of 3) Let J 6= 0 be a two-sided ideal of Ri. Then J contains some simple left ideal of Ri, whichwe can denote L. Since Ri is semisimple, Ri = L⊕ L′ where L′ is another left ideal of Ri. Then 1 = e+ e′,where e ∈ L and e′ ∈ L′. Thus e = e2e+ ee′, so ee′ = e− e2 ∈ L′ ∩ L = (0). Thus e = e2.

Hence e2 = e ∈ Le 6= 0 (as e 6= 0). Since L is simple Le = L. Let K be a simple left ideal of Ri. ThusK ∼= Ii ∼= L. Let ψ : L → K be an isomorphism. Then K = ψ(L) = ψ(Le) = Lψ(e) ⊂ Jψ(e) ⊂ J (since J

is a right ideal). Thus J = Ri =∑K∼=Ii

K.

Next time, we will prove the Artin-Wedderburn Theorem!


Next exam will be take-home.Recall from last class the following powerful theorem:

Theorem 42. Let R be a left semisimple ring. Let I1, ..., Ik be representatives of the distinct isomorphismclasses of simple left ideals (that is, every simple left ideal of R is isomorphic to exactly one Ij). For each i,

let Ri =∑

L left idealL∼=Ii

L. Then

1. For all i, Ri is a ring with identity.

2. For all i, Ri is a left semisimple with a unique (up to isomorphism) simple left ideal.

3. For all i, Ri is simple.

47

4. Finally, R ∼= R1 × ...×Rk (as rings).

This leads to the following corollary:

Corollary 13. Let R be a left semisimple ring. Then R is simple if and only if R has a unique (up toisomorphism) simple left ideal.

Proof. We write R = R1 × · · · ×Rk by the previous theorem.Suppose R has a unique (up to isomorphism) simple left ideal. Then k = 1, so R = R1, and by the

previous theorem, R1 is simple. Thus R is simple.Suppose instead that R is simple. Recall that each Ri can be interpreted as residing inside R, and that

they are ideals in R (since RiRj = 0 for i 6= j). In particular, R1 is a two-sided ideal in R. Since R is simple,and R1 6= 0, then R = R1. Thus R has a unique simple left ideal.

Definition 44. Let R be a ring, and let Ro = {ro|r ∈ R}. We define + on Ro by ro + so = (r + s)o, Wedefine · on Ro by roso = (sr)o. We say that Ro with these operations is the opposite ring of R.

Remark 40. Observe that R is commutative if and only if the map r 7→ ro is a ring isomorphism of R andRo.

Note that I is a left ideal in R if and only if Io = {io|i ∈ I} is a right ideal in Ro. Because of this, R isleft Noetherian (respectively, Artinian, semisimple, etc) if and only if Ro is right Noetherian (respectively,Artinian, semisimple, etc).

Finally, (Ro)o ∼= R.

Definition 45. LetM be anR-module. Define EndR(M) = {φ : M →M | φ is a R−module homomorphism}.Then EndRM is a ring under + and · (which in this case is composition). It has a multiplicative identityelement (namely, the identity function). We can also verify distributivity (but I won’t).

Proposition 26. Let R be a ring. Then EndR(R) (where we think of R as a left R-module) is isomorphicto Ro.

Proof. Define f : EndR(R)→ Ro by φ 7→ φ(1)o. Let us now check that this is an R-module homomorphism.It is easy to verify that f(φ+ ψ) = f(φ) + f(ψ). Also,

f(ψφ) = (ψφ)(1)o

= (ψ(φ(1)))o

= (ψ(φ(1) · 1))o

= (φ(1)ψ(1))o

= φ(1)oψ(1)o

= f(ψ)f(φ)

Note that the fourth equality is because ψ is R-linear. Thus f is an R-module homomorphism. It is easy tocheck that f is injective. For surjectivity, we can see that for ro ∈ Ro, we can define g : R→ R by x 7→ xr.Then g ∈ EndR(R), and f(g) = ro. Thus f is surjective, so it is bijective. Thus f is isomorphic to Ro.

Exercise 11. Let M be an R-module. Then M is simple if and only if EndR(M) is a division ring.

Remark 41. Let M be an R-module, and let Mn = M ⊕ · · · ⊕M (where we have n copies).For i = 1, ..., n, let fi : M → Mn by u 7→ (0, . . . , u, . . . , 0) where the u is in the ith component. Also

define πi : Mn 7→M by (u1, ..., un) 7→ ui. Note that πifi = 1M for all i, and πjfi = 0 for all i 6= 0.Let ψ ∈ EndR(Mn). Define ψi,j = πjψfi. Note that this is a function M → Mn → Mn → M . Thus

ψi,j ∈ EndR(M) for all i, j. Define [ψ] := [ψi,j ] ∈Mn(EndR(M)).

48

Claim 3. Let f : EndR(Mn)→Mn(EndR(M)) by ψ 7→ [ψ]. Then f is a ring isomorphism.

Proof. This is left as an exercise for the reader.

Corollary 14. If R is a ring, then EndR(Rn) ∼= Mn(EndR(R)) ∼= Mn(Ro).

Remark 42. Let M be an R module. We use R′ or R′(M) to denote EndR(M). Then M is an R′-module,by the action φu = φ(u) for φ ∈ R′ and u ∈M (one can verify that this is indeed an R′-module).

Also, R′′ = R′′(M) = End′R(M). For a ∈ R, let rMa : M → M by u 7→ ua. Then rMa ∈ R′(M). LetlMa : M →M by u 7→ au. Then lMa 6∈ R′(M) in general.

However, we can show that lMa ∈ R′′(M). Let us do so. Note that lMa (u+v) = a(u+v) = au+av = la(u)+la(v). Thus lMa is additive. Also, for φ ∈ R′, we can see that lMa (φu) = lMa (φ(u)) = aφ(u) = φ(au) = φ·lMa (u).

Let M be an R-module. Then there is a ring homomorphism λ : R→ R′′(M) by a 7→ lMa . One can againverify additivity, and it is boring. Also, (lMa l

Mb )(u) = la(bu) = abu = lMabu.

Next time, we will prove the following theorem due to Rieffel:

Theorem 43. Let R be a simple ring and let I be a left ideal. Then λ : R→ R′′(I) is an isomorphism.

Note that we still have not proven the Artin-Wedderburn Theorem.

Theorem 44 (Artin-Wedderburn Theorem). Let R be a left semisimple ring. Then there exists uniquepositive integers l, n1, ..., nl and unique division rings D1, . . . , Dl such that R ∼= Mn1

(D1)×· · ·×Mnl(Dl).


Definition 46. Let R be a ring, and let S, T be nonempty subsets of R. Define ST = {∑finite

siti|si ∈ S, ti ∈

T}. This is the product of S and T .

Remark 43. We have two elementary properties of these products: first, that (ST )U = S(TU), and thatif f : R → R′ is a ring homomorphism, then f(ST ) = f(S)f(T ). These are easy to prove, but we will skipthat.

Last time, we stated the following theorem:

Theorem 45 (Rieffel’s Theorem). Let R be a simple ring and let I be a nonzero left ideal of R. Thenλ : R→ R′′(I) given by a 7→ l1 is an isomorphism.

Let’s prove it now!

Proof. Since λ is R-linear (on both sides), then kerλ is a two-sided ideal of R. Then λ(1) = l1 6= 0 sinceI 6= 0. Thus kerλ 6= R, so since R is simple, then kerλ = 0. That is, λ is injective.

Note that IR is a two-sided ideal of R. Therefore IR = R.We now wish to show that λ(I) is a left ideal of R′′. Certainly, it is an additive subgroup. It then suffices

to show that if f ∈ R′′ and la ∈ λ(I), then fla ∈ λ(I). Fix some x ∈ I, let rx denote the operation of rightmultiplying by x. Then rx ∈ R′ = EndR(I). Therefore

(fla)(x) = f(la(x))

= f(ax)

= f(rx(a))

= rx(f(a))

= f(a)x

= lf(a)(x)

49

Therefore, λ(I) is a left ideal of R′′, so R′′λ(I) = λ(I). Then we have that

R′′ = R′′λ(R)

= R′′λ(IR)

= (R′′λ(I))λ(R)

= λ(I)λ(R)

= λ(IR)

= λ(R)

Therefore λ is surjective, so λ is a bijection. Thus λ is an isomorphism.

Theorem 46. Let R be a simple ring. Then the following are equivalent:

1. R is left semisimple.

2. R is left Artinian.

3. R ∼= MN (D) for some n ≥ 1 and some division ring D.

Proof. We have already show that (3) implies (1), and that (1) implies (2).It then suffices to show that (2) implies (3). To that end, suppose R is left Artinian. Then R contains a

simple left ideal I. By Rieffel’s Theorem, R ∼= R′′(I) = EndR′(I). Since I is simple, then R′ = EndR(I) = Dis a division ring.

Hence, R ∼= EndD(I).We now wish to show that I is a finite dimensional D-vector space. Suppose for the sake of contradiction

that it is not. Then it would be infinite dimensional, so let {e1, e2, ...} be a countably infinite linearlyindependent set in I. For each n ≥ 1, let Jn = {f ∈ EndR(I)|f(ei) = 0whenever1 ≤ i ≤ n}. Observe thateach Jn is a left ideal of EndD(I). Additionally, note that J1 ⊃ J2 ⊃ ....

Finally, for each n, we can construct gn : I → I with gn ∈ EndD(I) such that g(en) 6= 0 but g(ei) = 0for i 6= n (you can do this since we are working in a D-vector space). Then gn ∈ Jn−1 \ Jn, so we have thatJ1 ) J2 ( .... Therefore EndD(I) is not left Artinian.

But R ∼= EndD(I) is left Artinian, so this is a contradiction. Thus dimD(I) = l < ∞. That is, I ∼= Dl.Therefore R ∼= EndD(Dl) ∼= Ml(D

o), as desired.

Theorem 47 (Artin-Wedderburn, Part I). Let R be a ring. Then the following are equivalent:

1. R is left semisimple.

2. R is right semisimple.

3. R ∼= Mn1(D1)× ...×Mnk(Dk) for some positive integers n1, ..., nk and division rings D1, ..., dk).

Proof. (1)⇒ (3) Suppose R is left semisimple. We previously proved that R ∼= R1 × ...×Rk where each Riis a simple left semisimple ring. Then by the previous theorem, Ri ∼= Mni(Di), as desired.

(3)⇒ 2) Suppose R is a product of matrices of division rings. We have shown that matrices of divisionrings are semisimple, so their product is semisimple.

(2)⇒ (1). Suppose R is right semisimple. Then Ro is left semisimple, so since (1)⇒ (2)⇒ (3), then Ro

is right semisimple. Then R ∼= (Ro)o is left semisimple.

Remark 44. Due to this theorem, a ring is left semisimple if and only if it is right semisimple. Thus werefer to this as simply “semisimple”.

If R is a semisimple ring, then λR(RR) < ∞ and λR(RR) < ∞, so R is both left and right Noetherianand Artinian.

50

Exercise 12. Let R ∼= Mn1(D1) × ... × Mnk(Dk), where each Di is a division ring. Then λR(RR) =λR(RR) = n1 + ...+ nk.

Remark 45. History time! Wedderburn came before Emmy Noether (he publish the result in 1907), so hedidn’t know about chain conditions. He proved the theorem for the case of rings that contain a field and arefinite dimensional over this field.

Artin came after Emmy Noether (he published in the 1920s) and generalized the result to rings satisfyingthe descending chain condition.

Note that we still have not proven the Artin-Wedderburn Theorem.

Theorem 48. (Artin-Wedderburn Theorem) Let R be a left semisimple ring. Then there exists uniquepositive integers l, n1, ..., nl and unique division rings D1, . . . , Dl such that R ∼= Mn1

(D1)×· · ·×Mnl(Dl).

16 Weyl Algebra


Let’s look at an example of a ring that is simple, but not semisimple.

Example 49 (The Weyl Algebra). Let F be a field, and let x be a variable. Then F [x] is a polynomialring. Note that an F -basis for F [x] is {1, x, x2, ...}.

Let R = EndF (F [x]). Let f(x) ∈ F [x], and let µf : F [x] → F [x] by g 7→ fg. Then µf ∈ R for allf ∈ F [x]. In fact, the map ρ : F [x]→ R by f 7→ µf is an injective ring homomorphism.

Identify F [x] with image(ρ), so we can assume F [x] ⊆ R.Define also d : F [x]→ F [x] by d(xj) = jxj−1 for all j = 0, 1, 2, ... (and extending linearly to all of F [x].

Note that d is the derivative operator: d(f(x)) = f ′(x). Therefore d ∈ R.Finally, the (first) Weyl algebra, denoted A1(F ), is the subring of R generated by F [x], and d.Because we are lazy, we will write A for A1(F ).

Proposition 27. In the Weyl algebra, for any j ≥ 1, dxj − xjd = jxj−1.

Proof. To check two operators are equal, we need only to check that they behave the same on all basiselements. Therefore consider xl for some l = 0, 1, 2, ....

Then

(dxj − xjd)xl = (dxj)xl − (xjd)xl

= d(xj+l − xj(lxl−1)

= (j + l)xj+l−1 − lxj+l−1

= jxj+l−1

= (jxj−1)(xl)

for all l. Thus dxj − xjd = jxj−1.

Proposition 28. The set B = {xidj |i, j ≥ 0} is an F -basis for A, if char F = 0.

Proof. Because of the previous proposition, for any expression involving xs and ds, we can move all the dsto the right in each term. Hence B is at least a spanning set for A.

It then suffices to show that B is linearly independent.Let φ : A → A be given by φ(f) = fx − xf . Observe that φ(f + g) = (f + g)x − x(f + g) =

fx−xf + gx−xg = φ(f) +φ(g). Thus φ is additive. Since it also respects scalar multiplication by elementsof F , then φ is F -linear.

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Also,

φ(fd) = (fd)x− x(fd)

= f(xd+ 1)− xfd= fxd+ f − xfd= f + (fx− xf)d

= f + φ(f)d

That is, φ(fd) = f + φ(f)d.By using this result and induction, we will show that φ(xidj) = jxidj−1. If j = 0, then φ(xi) =

xi+1 − xi+1 = 0, as desired. If j = 1, then φ(xid) = xi + φ(xi)d = xi. If j > 1, then

φ(xidj) = φ(xidj−1d)

= xidj−1 + φ(xidj−1)d

= jxidj−1

This completes the induction. By linearity, we can extend this to all polynomials f(x). That is, φ(f(x)dj) =jf(x)dj−1 for all polynomials f(x).

Finally we can show that B is an F -basis for A. Suppose∑finite

ai,jxidj = 0.

We can write this equation as fn(x)dn + fn−1(x)dn−1 + ... + f0(x) for some polynomials fi. Thenby applying φn, we get that φn(fi(x)di) = 0 for i ≤ n − 1, and φn(fn(x)dn) = n!fn(x). Thus 0 =φ(fn(x)dn + ... + f0(x)) = n!fn(x). Since our field is characteristic 0, then n! 6= 0, so fn(x) = 0. Thereforewe can remove this term from our linear combination, and repeat until all terms are zero. Thus B is linearlyindependent, so B is a basis.

Proposition 29. If F is a field of characteristic 0, then A1(F ) is a simple ring.

Proof. Let I 6= 0 be a (two-sided) ideal of A. If I 6= 0, then choose some nonzero g ∈ I. Observe that for allh ∈ I, then φ(h) = hx− xh ∈ I.

We write g = fn(x)dn + ... + f0(x), where fn(x) 6= 0. By applying φn to g, we get that phin(g) =n!fn(x) ∈ I. Therefore I contains some f(x) ∈ F [x] \ {0}.

Now we repeat this process with d: df(x)− f(x)d = f ′(x) ∈ I. Continuing, we will get that I contains anonzero constant. Therefore, I contains a nonzero constant. Since F is a field, then I = A1(F ). Thereforethe only two-sided ideals in A are A and 0, so A is simple.

We now have a few exercises that are useful.

Exercise 13. A is neither left nor right Artinian.

Exercise 14. A is a domain.

Exercise 15. A is not semisimple. In fact, A does not contain a simple left ideal.

Now let’s move on to the Jacobson density theorem.

Remark 46. Let M and N be R-modules, and let f : M → N be an R-module homomorphism.For n ≥ 1, define f (n) : Mn → Nn by f (n)(u1, ..., un) = (f(u1), ..., f(un)). Then f (n) is an R-module

homomorphism.Let E be an R-module. Let f ∈ R′′(E) = EndR′(E). That is, f : E → E such that f is additive, and

fφ = φf for all φ ∈ R′ = EndR(E). Then f (n) : En → En is also also R′(E)-linear.We claim that f (n) is R′(En)-linear. That is, f (n) ∈ R′′(En).This is because R′(En) = EndR(En) = Mn(EndR(E)) = Mn(R′). We need to show that f (n)φ = φf (n)

for all φ ∈ R′(En) = Mn(R′).

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We represent φ and f (n) as matrices. Then f (n) = fIEn and φ = φi,j with each φi,j ∈ R′(E). Thenf (n)φ = fIEn ◦ φi,j = [fφi,j ] = [φi,jf ] since f is R′-linear. This in turn equals [φi,j ] · fIEn . Thus f (n) ∈R′′(En) as desired.


Recall the following: suppose E is an R-module, R′ = EndR(E) and R′′ = EndR′ E, and f : E → E is inR′′(E). Then f (n) : En → En and f ∈ R′′(En).

Lemma 20. Let E be a semisimple R-module, and let f ∈ R′′(E). Let x ∈ E. Then there exists a ∈ Rsuch that f(x) = ax.

Proof. Certainly, Rx is a submodule of E. Then since E is semisimple, E = Rx⊕N for some other submoduleN . Define π : E → E by rx+n 7→ rx. Then π is R-linear, so π ∈ R′(E). Also, π(x) = x, so f(x) = f(π(x)).Since f is R′-linear, then f(π(x)) = π(f(x)) = ax for some a ∈ R. Thus by transitivity, f(x) = ax for somea ∈ R.

Theorem 49 (Jacobson Density Theorem). Let R be a ring, let E be a semisimple R-module, and letf ∈ R′′(E). Let x1, ..., xn ∈ E. Then there exists an a ∈ R such that f(xi) = axi for all i.

Proof. We will use λa to denote the map x 7→ ax for some a ∈ R.We have that f (n) ∈ R′′(En) by the result from last class. Also, En is semisimple since E is semisim-

ple. Let u = (x1, ..., xn) ∈ En. By the lemma there exists a ∈ R such that f (n)(u) = au. That is,(f(x1), ..., f(xn)) = (ax1, ..., axn).

Corollary 15. Let R be a ring, and let E be a semisimple R-module, and suppose E is finitely generatedas an R′-module. Then λ : R→ R′′(E) is surjective. That is, for all f ∈ R′′(E), there exists an a ∈ R suchthat f = λa.

Proof. Since E is finitely generated as an R′-module, there exists a finite generating set {x1, ..., xn}. Letf ∈ R′′(E). By the Jacobson Density Theorem, there exists some a ∈ R such that f(xi)λa(xi) for all i.

If x ∈ E, then x =

n∑i=1

sixi for some si ∈ R′. Then f(x) = f(∑

sixi) =∑

sif(xi) =∑

siλa(xi) =

λa(∑

sixi) = λa(x). Thus f = λa.

Corollary 16. Let R be a semisimple ring, and let E 6= 0 be a free R-module (that is, E has an R-basis{eα}α∈I). Then λ : R→ R′′(E) is an isomorphism.

Proof. Let e be any basis element of E. Then R′e ∼= E since one can define an R-linear map by sendinga basis to any set of elements in E. Therefore E is a finitely generated R′ module. Thus E =

⊕R is

semisimple, since R is. Therefore, by the previous corollary, λ : R→ R′′(E) is surjective.Suppose λ(a) = 0. Then in particular ae = 0, so a = 0 since e is a basis element. Thus λ is injective,

hence bijective. Thus λ is an isomorphism.

Corollary 17. Let D be a division ring and let E 6= 0 be a D-vector space. Then λ : D → D′′(E) =EndD′(E) is an isomorphism.

Proof. Since D is a division ring, it is semisimple. Since E is a D-vector space, then it is a free D-module.Then by the previous theorem, λ is an isomorphism.

Corollary 18. Suppose Mn1(D1) ∼= Mn2

(D2) where D1 and D2 are division rings and n1, n2 are positiveintegers. Then D1

∼= D2 and n1 = n2.

53

Proof. Recall that for any ring, Mn(R) ∼= EndR0(Ro)n. Since Do1 and Do

2 are also division rings, by renamingDo

1 to D1 and Do2 to D2, we will simply show that if EndD1 D

n11∼= EndD2 D

n22 , then D1

∼= D2 and n1 = n2.Let R = EndD1

Dn11 . Recall that R is simple and semisimple. Therefore, R has a unique simple left

module (up to isomorphism). This simple left module is isomorphic to the rows of this matrix, which are ofcourse Dn1

1 . Thus Dn11 is a simple. R-module.

Similarly, Dn22 is a simple R-module. Therefore Dn1

1∼= Dn2

2 as left R-modules. Thus EndR(Dn11 ) ∼=

EndR(Dn22 ) as rings. But by Corollary 17, EndR(Dn1

1 ) ∼= D1 and EndR(Dn22 ) ∼= D2. Thus D1

∼= D2. ButDn1

1∼= Dn2

2∼= Dn2

1 , so by taking dimensions, we can see that n1 = n2.

Theorem 50 (Artin-Wedderburn, Part 2). Let R be a semisimple ring. Then R ∼= Mn1(D1) × ... ×

Mnk(Dk) for division rings D1, ..., Dk and positive integers n1, ..., nk. Furthermore, the D1, ..., Dk andn1, ..., nk that let you write R in this way are unique, up to reordering.

Proof. We have previously shown that there exist division rings and integers satisfying this. It then sufficesto show uniqueness.

By a homework problem [edit: see immediately below], the number k is unique, and furthermore therings Mni(Di) are unique up to reordering. But then by the previous corollary, since we know Mni(Di), weknow what ni and Di are. Thus this decomposition is unique.

Homework Problem 7. If A1× ...×Al ∼= B1× ...×Bk where each Ai and Bj are simple rings, then k = land after rearrangement Ai ∼= Bi for all i.

Definition 47. Let R be a commutative ring, and let S be a ring. We say S is an R-algebra if there existsa ring homomorphism φ :→ S such that φ(R) ⊂ Z(S) = {s ∈ S|st = ts for all t ∈ S}.

Remark 47. Let R be a ring, let E be an R-module, and let r ∈ Z(R). Then λr ∈ R′(E), so there exists aring homomorphism λ : Z(R)→ R′(E) by r 7→ λr.

Remark 48. If E is finitely generated over Z(R) (that is, if E = Z(R)u1 + ...+Z(R)un for some u1, ..., un ∈E), then E = R′u1 + ...+R′un. Thus E is finitely generated as a R′-module.

Remark 49. Suppose k is a field, and R is a finite dimensional k-algebra (that is, dimk(R) <∞).Let E be a finitely generated semisimple R-module. Then λ : R→ R′′(E) is surjective.

Proof. Since R is a finite dimensional k-algebra, then R = kv1 + ...+ kvs ⊂ Z(R)v1 + ...+Z(R)vs. However,Z(R)v1 + ...+ Z(R)vs ⊂ R, so Z(R)v1 + ...+ Z(R)vs = R.

Then by substitution, E = Ru1 + ...+Rut =∑i,j

Z(R)viuj , so E is finitely generated as a Z(R) module.

Thus E is finitely generated as an R′ module. Therefore, by Corollary 15, λ is onto.

17 k-Algebras


Let’s do examples of k-algebras!

Example 50. Let k be a field. Then the following are k-algebras: k[x1, ..., xn], k[x1, ..., xn]/I, k(x1, ..., xn),A1(k) (the first Weyl algebra), Mn(k), and k[G] (the group ring). Of these, the finite dimensional ones arek[x1, ..., xn]/I (for certain I), Mn(k), and k[G] (if and only if G is finite).

Lemma 21. Let D be a division ring, and let k be an algebraically closed field such that D is a k-algebra.If dimk(D) <∞, then D = k.

Proof. Let α ∈ D. Then k[α] ⊂ D. Since dimk(D) < ∞, then there exists a linear dependence amongα, α2, α3, .... That is, there exists kn, ..., k0, not all equal to 0, such that 0 = knα

n + ...+ k0. That is, α is aroot of f(x) = knx

n + ...k0. Thus α ∈ k, so k = D.

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Corollary 19. Suppose k is an algebraically closed field and R is a finite dimension k-algebra. Then R issemisimple if and only if R ∼= Mn1(k)× ...×Mnl(k).

Proof. Suppose R ∼= Mn1(k)×...×Mnl(k). Then each of these matrix rings is semisimple, so R is semisimple,

as desired.Suppose instead that R is semisimple. By the Artin-Wedderburn theorem, R ∼= Mn1

(D1)× ...×Mnl(Dl)where D1, ..., Dl are division rings.

Recall that Doi = EndR(Ei), where Ei is a semisimple R-module. Note that EndR(Ei) ⊂ Endk(Ei).

Then since Ei is simple, Ei = Rxi for any xi ∈ Ei \ {0}. Thus dimk(R) < ∞, so dimk Ei < ∞. ThusE ∼= kn. Therefore Endk(Ei) = Mn(k).

But EndR(Ei) ⊂ Endk(Ei), so Doi is finite dimensional as a k-vector space. But Do

i is also a divisionring, so by the lemma, Do

i∼= k. Thus Di ∼= ko = k, so R ∼= Mn1(k)× ...×Mnl(k), as desired.

Theorem 51 (Burnside). Let k be an algebraically closed field, and let V be a finite-dimensional k-vectorspace. Let R be a subalgebra of Endk(V ). Then if V is a simple R-module, then

1. k = EndR(V )

2. R = Endk(V )

Proof. (Proof of 1) Certainly, since k ⊂ R, then EndR(V ) ⊂ Endk(V ). Since V is a finite-dimensionalk-vector space, then Endk(V ) ∼= Mn(k), so EndR(V ) is a finite dimesnional k-algebra. Also, EndR(V ) is adivision ring, as V is a simple R-module. Therefore by the lemma, k ∼= EndR(V ).

(Proof of 2) By the last remark, from last class, the ring homomorphism R → R′′(V ) is surjective. ButR′′(V ) = EndR′(V ), and R′ = EndR(V ) ∼= k, so R′′(V ) = Endk(V ). But R ⊂ Endk(V ), so λ is injective aswell. Therefore λ is an isomorphism, so R = Endk(V ).

Definition 48. Let R be a ring. An element x ∈ R is nilpotent if xn = 0 for some n ∈ N.Let I be a left ideal. Recall In is the left ideal generated by {a1...an|ai ∈ I}. Then we say an ideal I is

nilpotent if In = 0 for some n.A left ideal I is called nil if every element of I is nilpotent.

Remark 50. Every nilpotent left ideal is nil.It need not follow that a nil ideal is nilpotent, even if the ring is commutative. An example of this is

given in the next exercise.

Exercise 16. Let R = k[x1, x2, ...]/(x11, x

22, x

33...), and let I = (x1, x2, ...). Show that I is nil, but not

nilpotent.

Exercise 17. If I is a finitely generated ideal, then I is nilpotent if and only if I is nil.If R is a commutative ring, then the sum of nilpotent elements is nilpotent.However, if R is not commutative, the sum of two nilpotent elements may not be nilpotent. For instance,

let A =

(0 01 0

)and let B =

(0 10 0

). Then A2 = 0 and B2 = 0, but A + B =

(0 11 0

)and

(A+B)2 = I, so A+B is not nilpotent.

History time: Wedderburn’s actual result was classifying all finite dimensional R-algebras such that theonly left ideal which is nil is zero. The criterion of the only nil left ideal being zero is equivalent to afinite-dimensional k-algebra being semisimple.

Artin later gave a more general condition in terms of chain conditions. Later still, Jacobson generalizedthis to a statement about the intersection of all maximal left ideals. This was later named after him.

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18 Jacobson Radical

Definition 49. Let R be a ring. The Jacobson radical is defined to be the intersection of all maximal leftideals. We denote this by J(R).

Lemma 22. Let R be a ring and let y ∈ R. Then the following are equivalent:

1. y ∈ J(R)

2. 1− xy is left-invertible for all x ∈ R

3. yM = 0 for every simple left R-module M .

Proof. (1 ⇒ 2) Suppose y ∈ J(R). Suppose for the sake of contradiction that 1 − xy is not left invertiblefor some x ∈ R. Then R(1 − xy) is a proper ideal of R, so there exists a left-maximal ideal m such thatR(1− xy) ⊂ m. But y ∈ m, so 1− xy + xy = 1 ∈ m. This is a contradiction, so 1− xy is left invertible.

(2⇒ 3) Suppose 1−xy is left invertible for all x ∈ R. Suppose for the sake of contradiction that yM 6= 0for some simple module M . Then RyM = M , and in fact Ryu = M for some u ∈M . Therefore u = ryu forsome r ∈ R, so (1− ry)u = 0. As 1− ry is left invertible, then u = 0. This is a contradiction, so yM = 0.

(3 ⇒ 1) Suppose yM = 0 for all simple left R-modules M . Let m be a maximal left ideal. Then R/mis a simple left R-module. Then y · R/m = 0, so in particular y · 1 = y = 0. Thus y ∈ m, so y is in everymaximal left ideals. Thus y ∈

⋂m maximalm = J(R).


Recall from last class the definition of the Jacobson radical: the intersection of all maximal left ideals. Wedenote it by J(R). Recall also this characterization from last class:

Lemma 23. Let R be a ring and let y ∈ R. Then the following are equivalent:

1. y ∈ J(R)

2. 1− xy is left-invertible for all x ∈ R

3. yM = 0 for every simple left R-module M .

Definition 50. If R is a ring and M is an R-module, then the annihilator of M is AnnR(M) = {r ∈ R|rm =0 for all m ∈M}.

Remark 51. Let R be a ring and M is an R-module. Also let r ∈ R, i ∈ AnnR(M) and m ∈ M . Then(ir)m = i(rm) = 0 since i ∈ AnnR(M), and (ri)m = r(im) = r0 = 0 for the same reason. Thus AnnR(M)is a two-sided ideal.

Corollary 20. The Jacobson radical is always a two-sided ideal.

Proof. By the lemma, J(R) = {y ∈ R|yM = 0 for all simple modules M}. That is, J(R) =⋂M simple AnnR(M).

Thus J(R) is the intersection of two-sided ideals, so it is a two-sided ideal.

Proposition 30. Let R be a ring. Then y ∈ J(R) if and only if 1− xyz is a unit for all x, z ∈ R.

Proof. Suppose 1− xyz is a unit for all x, z ∈ R. Therefore 1− xy is left invertible for all x ∈ R, so by thecharacterization Lemma, y ∈ J(R).

If instead y ∈ J(R), then since J(R) is a two-sided ideal, then yz ∈ J(R) for all z ∈ R. Then by thecharacterization Lemma, 1− xyz is left invertible for all x ∈ R.

let u be the left inverse of 1 − xyz. That is, u − uxyz = 1. Therefore u = 1 + uxyz. But J(R) is atwo-sided ideal, so xyz ∈ J(R). Therefore, 1 + uxyz = 1− (−u)xyz is left invertible by the characterizationlemma. Thus u = 1 + uxyz has a left inverse v. That is, vu = 1. But 1− xyz = vu(1− xyz) = v so 1− xyzhas u as both its left inverse and its right inverse. That is, 1− xyz is invertible.

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Corollary 21. If R is a ring, then J(R) =⋂

m maximalright ideal

m =⋂

m maximalright ideal

AnnR(M).

The proof of this is left as an exercise.

Definition 51. A ring R is called semiprimitive if J(R) = 0.

Example 51. Any simple ring has J(R) = 0, so a simple ring is semiprimitive.

Proposition 31. Let R be a semisimple ring. Then J(R) = 0.

Proof. Since R is a semisimple ring, we can write R = I1 ⊕ ...⊕ Il, where each Ij is a simple left module.Then we write 1 = e1 + ... + el, where each ei ∈ Ii. Let y ∈ J(R). Then yIj = 0 for all j by the

characterization lemma. Thus y = y · 1 = ye1 + ...+ yei = 0. Thus J(R) = 0.

Example 52. If F is a field and A1(F ) is the first Weyl algebra, then J(A1(F )) = 0 since A1(F ) is simple.Since the maximal ideals in Z are generated by primes, and no number except zero is divisible by infinitely

many primes, then J(Z) = 0.By a similar argument, if K is a field, then J(K[x1, ..., xn]) = 0.

Theorem 52. Let R be a ring. Then R is semisimple if and only if R is left Artinian and semiprimitive.

Proof. We have already shown that if R is semisimple, then R is left Artinian. Today in the previousproposition, we showed that if R is semisimple, then J(R) = 0. That is, R is semiprimitive, as desired.

Suppose instead that R is semiprimitive and left Artinian. That is, J(R) = 0. We will first show that Ris semisimple.

To this end, suppose I and L are left ideals of R such that I ⊂ L. Suppose also that I is simple. Since Iis simple it is nonzero, and since J(R) = 0, then I 6⊂ J(R). Thus there exists a maximal left ideal m, withI 6⊂ m. But then I +m = R since m is maximal, and I ∩m = 0 since I is simple and m is maximal. ThusR = I ⊕m. From this, one can verify that L = I ⊕ (L ∩m).

Note that if R is left Artinian, then every nonzero left ideal contains a simple left ideal (just extend thechain as far as possible). Then if L1 is any ideal in R, we find a simple left ideal I1 ⊂ L1. Then there existsan L2 such that L1 = I1 ⊕ L2. If L2 = 0, then we are done. If L2 6= 0, then there exists a simple left idealI2 ⊂ L2.

We then repeat this process until some Ln = 0. If no Li were ever zero, then I1 ⊂ I1 ⊕ I2 ⊂ ... is aninfinite extending chain, which contradicts the assumption that R is left Artinian. Thus there exists someLn = 0. That is, L1 = I1 ⊕ ... ⊕ In. Then we are mostly done. [fix this up; we didn’t actually show itssemisimple.]

Proposition 32. Let I be a nil left ideal. Then I ⊂ J(R).

Proof. Let y ∈ I, and let x ∈ R. Then xy ∈ I, so (xy)n = 0 for some n. Thus 1 − xy is a unit since(1− xy)(1 + xy + (xy)2 + ...+ (xy)n−1) = 1− (xy)n = 1. Therefore y ∈ J(R).

Theorem 53. Suppose R is left Artinian. Then J(R) is nilpotent.

Proof. Let J = J(R). Consider the descending chain J ⊃ J2 ⊃ J3.... As R is left Artinian, there exists ksuch that Jk = Jk+1. Let I = Jk, and note that I = I2.

Suppose for the sake of contradiction that I 6= 0. Then let Λ = {L left ideal|IL 6= 0}. Note that I ∈ Λ,so Λ 6= ∅. Since R is left Artinian, we can choose L ∈ Λ minimal. Since L ∈ Λ, then there exists some y ∈ Lsuch that Iy 6= 0. Certainly, Iy ⊂ L since L is a left ideal. Also, I(Iy) = I2y = Iy 6= 0. Thus Iy = L.

Since y ∈ L, then y = iy for some i ∈ I. Therefore (1− i)y = 0. but 1− i is a unit since i ∈ I = J(R)k.Thus y = 0, but this is a contradiction of the fact that Iy 6= 0. Thus I = 0, but I = J(R)k.

Corollary 22. If R is left Artinian (for instance if R is a finite-dimensional k-algebra), then J(R) is thelargest nil left ideal. Hence J(R) = 0 if and only if 0 is the only nil left ideal.

57

Proposition 33. Let M be a semisimple R-module. Then the following are equivalent

1. M is Artinian.

2. M is Noetherian.

3. M is finitely generated.

4. λR(M) <∞.

Proof. First note that since R is semisimple, then M is semisimple. Write M =⊕i∈Λ

Mi, where each Mi is

simple.Note that λR(Mi) = 1 for each i since Mi is simple, so λR(M) = |Λ|.If λR(M) = |Λ| <∞, then we have already shown that the first three propositions hold.If λ(R) = |Λ| =∞, then note that M1 (M1 ⊕M2 (M1 ⊕M2 ⊕M3 ( ... is a strictly ascending chain of

infinite length so M is not Noetherian. Therefore it is not finitely generated. Similarly, M )M2⊕M3⊕ ... )M3 ⊕ ... ) ... therefore M is not Artinian either.

Note also that if R is semisimple, then every R-module is semisimple, so the previous theorem holds.


Tom just looked up this definition, so it probably isn’t important.

Definition 52. A ring R is left primitive if there exists a simple left R-module M such that AnnR(M) = 0.

Remark 52. Recall that J(R) is the intersection of AnnR(M) over all simple left R-modules M , so beingprimitive implies that J(R) = 0. That is, a ring being primitive implies it is semiprimitive.

Also, R being simple (as a ring) implies that R is primitive.Furthermore, if D is a division ring and V is a D-vector space, then one can show that EndD(V ) is

primitive. Also, one can show that EndD(V ) is simple if and only if dimD(V ) < ∞. Thus if V is infinite-dimensional, then EndD(V ) is primitive but not simple.

We now forever leave behind primitive rings, and focus again on the Jacobson radical. Here we shall useit to show that Artinian rings are Noetherian.

Theorem 54. Let R be a left Artinian ring, and let M be an Artinian left R-module. Then λR(M) <∞.In particular, R is left Noetherian.

Proof. Let J = J(R). Since M is Artinian, then J iM is Artinian for any i ≥ 0 (with the convention thatJ0 = R). Therefore J iM/J i+1M is Artinian.

But note that J ·(J iM/J i+1M) = 0. Therefore J iM/J i+1M is an R/J-module by the action of r ·u = rufor r ∈ R/J and u ∈ J iM/J i+1M .

One can also check (and I should do this) that J(R/J) = 0.Since J(R/J) = 0, thenR/J is semisimple. Therefore, by the last Proposition from last class, λR/J(J iM/J i+1M) <

∞ for all i. But one can check that λR(anything) = λR/J(anything), so λR(J iM/J i−1M) <∞ for all i.We will now show that λR(M/J iM) <∞ for all i. We will use induction for this. The base case is that

M/JM = J0M/J1M has finite length, and this follows from the last statement.In the inductive case, we can make the following exact sequence of modules 0 → J i−1M/J iM →

M/J iM → M/J i−1M → 0. Since length is additive on exact sequences, J i−1M/J iM is finite lengthby previous statement, and M/J i−1M is finite length by the inductive hypothesis, then M/J iM is finitelength. Thus by induction M/J iM is finite length for all i ≥ 0.

However, by a Theorem from last class, Jn = 0 for some n, so M/JnM = M/0 ∼= M is finite length.

Now we will prove Nakayama’s Lemma, which is one of the most important claims built off of the Jacobsonradical.

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Theorem 55 (Nakayama’s Lemma). Let R be a ring, and let M be a finitely generated left R-module.Suppose also that M = J(R)M . Then M = 0.

Proof. Let J = J(R). Suppose for the sake of contradiction that M 6= 0. Let n be the least number ofgenerators for M . Since M 6= 0, then n > 0.

Let M = Rx1 + ... + Rxn for some xi ∈ M . Then JM = Jx1 + ... + Jxn. Since xn ∈ JM = M , thenxn = j1x1 + ...jnxn where each ji ∈ J . Then (1− jn)xn = j1x1 + ...+ xn−1xn−1. However, jn ∈ J , so 1− jnis a unit in R. Thus xn = (1− jn)−1xj1x1 + ...+ (1− jn)−1jn−1xn−1 ∈ Rx1 + ...+Rxn−1.

Therefore M = Rx1 + ...+Rxn−1, which contradicts our choice of n. Thus M = 0.

Remark 53. Nakayama’s lemma fails for non-finitely-generated rings. For instance, if k is a field, andR = k[[x]], then J(R) = m = (x). Then Q(R) = k((x)) is an R-module, and furthermore JQ(R) = Q(R).However, Q(R) 6= 0, so Q(R) is not finitely generated.

This is an example of how Nakayama’s Lemma can let you show something is not finitely generated.

Remark 54. If R is a ring and M is an R-module, then we use the following notation. Define µR(M) =inf{n ≥ 0|M = Rx1 + ...+Rxn for some xi ∈M}. That is, µR(M) is the minimal number of generators ofM .

Lemma 24. Let M be a finitely generatede R-module, let N ⊂ M be a submodule, and let J = J(R).Suppose M = N + JM . Then M = N .

Proof. Note that M = N + JM if and only if M/N = (JM +N)/N = J ·M/N . As M is finitely generated,then so is M/N . Therefore by Nakayama’s Lemma, M/N = 0, so M = N .

Remark 55. The term “Nakayama’s lemma” is used loosely by mathematicians. The lemma has manyimmediate corollaries, and mathematicians may refer to any of them as “Nakayama’s lemma”.

Proposition 34. Let M be a finitely generated R-module, and let x1, ..., xn ∈ M . Let J = J(R). Thenx1, ..., xn generate M if and only if x1, ..., xn generate M/JM .

Proof. Certainly, if x1, ..., xn generate M , then their representatives in M/JM generate M/JM .Conversely, suppose x1, ..., xn generate M/JM . Let N = Rx1 + ... + Rxn. Then (N + JM)/JM =

Rx1 + ...Rxn = M/JM . Therefore N + JM = M . By the previous lemma, N = M , as desired.

Corollary 23. LetM be a finitely generated R-module, and let J = J(R). Then µR(M) = µR/J(M/JM).

Proof. By the previous proposition, generating sets for M as an R-module correspond to generating sets forR/JM as an R/J-module. Thus their lengths are equal.


Recall from last class the following named theorem:

Theorem 56. (Nakayama’s Lemma) Let R be a ring, and let M be a finitely generated left R-module.Suppose also that M = J(R)M . Then M = 0.

Yet another corollary of Nakayama’s Lemma is the following.

Corollary 24. Let R be a commutative ring with a unique maximal ideal m (that is, R is a local ring). LetM be a finitely generated R-module. Then µR(M) = dimR/mM/mM .

Proof. Since m is the unique maximal ideal, then J(R) = m. Therefore µR(M) = µR/m(M/mM) =dimR/m(M/mM).

Definition 53. Let # denote the short exact sequence 0→ Af→ B

g→ C → 0 and let ## denote the short

exact sequence 0→ Lh→M

l→ N → 0. Bothof these are short exact sequences of (left) R-modules.We say # and ## are isomorphic if there exists a commutative diagram [oh boy] (consisting of # above

##, with arrows pointed down from A to L, etc, where each of those arrows is an isomorphism.)

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Definition 54. A short exact sequence of the form 0 → Aρ→ A ⊕ B π→ B → 0 where ρ : a 7→ (a, 0) and

π : (a, b) 7→ b is canonically split .We say a short exact sequence is split if it is isomorphic to a canonically split short exact sequence.

Theorem 57 (The Splitting Theorem). Let # denote the short exact sequence 0→ Af→ B

g→ C → 0.Then the following are equivalent

1. # splits.

2. There exists an isomorphism of commutative diagrams between # and 0 → Al→ A ⊕ C π→ C → 0

where the vertical arrows are 1A : A → A, h : B → A ⊕ C, 1C : C → C, and such that h is anisomorphism.

3. There exists i : B → A such that if = 1A.

4. There exists j : C → B such that gj = 1C .

5. There exists i : B → A and j : C → B such that 1B = fi+ jg.

6. f(A) is a direct summand of B.

Proof. (1⇒ 5) Suppose # splits. That is, there exists an isomorphism between # and 0→ Ll→ L⊕M π→

M → 0 (where the maps are labelled α, β, γ respectively. Then define τ : L ⊕M → L by τ(l,m) = l andδ : M → L⊕M by δ(m) = (0,m). Then 1L⊕M = lτ + δπ. Finally, let i = α−1τβ and let j = β−1δγ. Then

fi+ jg = fα−1τβ + β−1δγg

= β−1lτβ + β−1δπβ

= β−1(lτ + δπ)β

= β−1β

= 1B

as desired.(5⇒ 4) [lost].

(4⇒ 3) Let K = ker g = im f . Then f : A→ K is an isomorphism. Let ρ : K → A where ρ = f−1

. Letb ∈ B. Then g(b− jg(b)) = g(b)− g(b) = 0. Therefore (1B − jg)(b) = b− jb(b) ∈ K so 1− jg : B → K. Leti = ρ(1− jg) : B → A. Then if = ρ(1− jg)f = ρf = 1A.

(3⇒ 2) Define h : B → A⊕ C by h : b 7→ (i(b), g(b)). Then one can verify that the appropriate diagramcommutes. It then suffices to show that h is an isomorphism.

We will now show that h is injective. First, note that if h(b) = 0 then g(b) = 0, so b ∈ im(f). Letb = f(a) for some a ∈ A. Then 0 = i(b) = i(f(a)) = a, so a = 0 and therefore b = f(a) = 0. Thus h isinjective.

We will now show that h is surjective. Let (a, c) ∈ A⊕ C. Since g is onto, then there exists some b ∈ Bsuch that g(b) = c. Let b′ = f(a) + b − f(i(b)). Then i(b′) = i(f(a)) + i(b) − ifi(b) = a + i(b) − i(b) = a.Also, g(b′) = g(b), so h(b′) = (a, c). Thus h is surjective.

Therefore h is bijective, so it is an isomorphism. Thus there exists the desired isomorphism of commutativediagrams.

(2⇒ 1) This is immediate, since such an isomorphism of commutative diagrams is a splitting of #.We have now shown that (1)− (5) are equivalent. It then suffices to show that (6) is equivalent.(6⇒ 3) Suppose B = f(A)⊕D for some submodule D of B. Define i : B → A by f(a) + d 7→ a. This is

well-defined since f is injective and the sum is direct. Also, i(f(a)) = a, so if = 1A.(2 ⇒ 6) Suppose h is an isomorphism from B to A ⊕ C. Then h−1 : A ⊕ C → B is an isomorphism.

Then B = h−1(A)⊕ h−1(C), and h−1(A) = h−1ρ(A) = f(A). Thus B is a direct summand of f(A).

Corollary 25. Suppose 0→ A→ B → C → 0 is split. Then B ∼= A⊕ C.

Proof. This is a direct result of (1⇒ 2).

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19 Projective Modules

Definition 55. An R-module P is called projective if, given any exact sequence Mf→ N → 0 and map

g : P → N , there exists an h : P →M such that fh = g. (This can be represented with a diagram.)

Proposition 35. Any free R-module F is projective.

Proof. Since F is free, there exists an R-basis {eα}. Suppose Mf→ N → 0. And suppose g : F → N . Then

since f is onto, for each α, there exists mα ∈ M such that f(mα) = g(eα). Then define h : F → M byeα 7→ mα. One can then verify that fh = g.


We’re back!Recall the following definition from last time:

Definition 56. An R-module P is called projective if, given any exact sequence Mf→ N → 0 and map

g : P → N , there exists an h : P →M such that fh = g. (This can be represented with a diagram.)

Proposition 36. Let R be a ring, and let P be an R-module. Then the following are equivalent:

1. P is projective.

2. Every short exact sequence of the form 0→ L→M → P → 0 splits.

3. P is a direct summand of a free module.

Proof. (1 ⇒ 2) Take a short exact sequence 0 → L → Mf→ P → 0. Since P is projective and embeds into

itself by 1P , then by the definition of projective, there exists a function j : P →M such that fj = 1P . Thenby one of the criterion for a splitting sequence, the short exact sequence splits.

(2 ⇒ 3) Suppose every short exact sequence with P in it splits. There exists a free module F and aprojection φ : F → P which is free. Let K = ker φ. Then 0→ K → F → P → 0 is a short exact sequence.By (2), this short exact sequence splits. Therefore F ∼= P ⊕K. That is, P is a direct summand of a freemodule.

(3 ⇒ 1) Suppose P is a direct summand of a free module. That is, there exists a free module F and amodule Q such that P ⊕Q = F . Define π : F → P by (p, q) 7→ p and rho : P → F by p 7→ (p, 0). Note thatπρ = 1P .

Consider a diagram Mf→ N → 0, with g : P → N . Then there exists a function h : F → M such that

fh = gπ [why?]. Let h = hρ : P →M . Then fh = fhρ = gπρ = g, so P is projective.

Theorem 58. Let R be a ring. Then R is semisimple if and only if every R-module is projective.

Proof. Suppose R is semisimple. Let M be an R-module. As always, there exists a free module F and a

surjective map φ : F → M . Let K = kerφ. This gives a short exact sequence 0 → K → Fφ→ M → 0.

Since R is semisimple, F is semisimple, so K is a direct summand of F . Therefore the sequnce splits, soF ∼= M ⊕K. By the previous proposition, this is equivalent to M being projective, as desired.

Suppose instead that every R-module is projective. Let I be a left ideal of R. Then consider the shortexact sequence 0→ I → R → R/I → 0. Since R/I is an R-module, it is projective, so this sequence splits.That is R ∼= I ⊕R/I, so R is semisimple.

Theorem 59. Let R be a ring. Then R is a division ring if and only if every R-module is free.

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Proof. Suppose R is a division ring. Then R-modules are R-vector spaces. By elementary linear algebra,every R-vector space is free.

Suppose every R-module is free. Then since free modules are projective, by the previous theorem R issemisimple. That is, R = I1 ⊕ ... ⊕ In where each Ij is a simple left ideal. Then λR(Ij) = 1 for all j, sinceIj is a simple left ideal. Thus λR(R) = n. Since Ij is simple, it is free, so Ij ∼= Rl for some n > 0. Thus1 = λ(Ij) = λ(Rl) = ln, so n = 1. That is, R = I1, and I1 is a simple left ideal, so the only left ideals of Rare 0 and R. Thus R is a division ring.

Example 53. Let D be a division ring and let n > 1. Let R = Mn(D), and let P = Dn. Recall that P is asimple R-module. (In fact, R ∼= P ⊕ ...⊕ P .) Then P is projective but not free.

We can see that P is projective either because it is a direct summand of R, which is a free R-module, orbecause R is semisimple, so all its modules are projective.

However, P is not free since λR(P ) = 1 < n = λ(R).

Example 54. Let R = Z[√−5]. Then I = (2, 1 +

√−5) is projective, but not free.

Example 55. Let R = R[x, y, z]/(x2 + y2 + z2 − 1). Define φ : R → R by (r1, r2, r3) 7→ r1x + r2y + r3z.Note that φ is surjective since φ(rx, ry, rz) = r(x2 + y2 + z2) = r · 1 = r.

Let P = ker φ. We have a short exact sequence 0 → P → R3 φ→ R → 0. As an R-module, R is free, soR is projective as an R-module as well. Therefore this short exact sequence splits, so R3 ∼= P ⊕R. Thus Pis a direct summand of a free module, so P is projective as an R-module. However, P 6∼= R2, so P is not free(this last assertion is nonobvious and the proof uses differential geometry).

Example 56. (Quillen-Suslin, 1976) Let k be a field and let R = k[x1, ..., xn] be a polynomial ring over k.Then every projective R-module is free.

20 Group Rings are Semisimple

Recall what a group ring is: if R is a ring and G is a group, then R[G] is the group ring of formal sums ofelements of G.

20.1 Day 39 - December 2

We will now continue with the proof of Maschke’s Theorem.

Theorem 60 (Maschke’s Theorem). Let k be a field and let G be a finite group. Then R = k[G] issemisimple if and only if char k - |G|.

Proof. Suppose char k - |G|. Let I be a left ideal of R. We need to show that 0→ I → R→ R/I → 0 splits.Since I is an R-module, it is also a k-module. That is, I is a k-vector space, so R is a direct summand

as k-modules.Let π : R→ I be a k-linear function such that π(i) = i for all i ∈ I.Let g ∈ G. We wish to show gπg−1 maps from R into I. Let a ∈ R. Then π(g−1a) ∈ I, so gπ(g−1a) ∈ I

since I is a left ideal. Therefore gπg−1 : R→ I. Also, once can verify that gπg−1 is k-linear.

Lastly, let φ = 1|G|

∑g∈G

gπg−1 : R→ I. Here we used the fact that char k - |G|. Since this is the average

of k-linear maps, this is k-linear. In order to show that φ is R-linear, it suffices to show that it respectsmultiplication by h ∈ G.

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Fix some h ∈ G and a ∈ R. Then

φ(ha) =1

|G|∑g∈G

gπg−1(ha)

=1

|G|∑hg∈G

(hg)π(hg)−1(ha)

= h

1

|G|·∑hg∈G

gπg(a)

= hφ(a)

Also, for all i ∈ I, φ(i) = 1|G|

∑g∈G

gπg−1(i) =1

|G|∑g∈G

i = i, so φ fixes I. Thus since 0→ I → R→ R/I →

0 splits, then R ∼= I ⊕R/I. Since I was arbitrary, then R is semisimple.Conversely, suppose char k

∣∣|G|. Then char k 6= 0, so let p = char k.

Let x =∑g∈G

g. Then x ∈ k[G]. Also note that for all h ∈ G, hx =∑g∈G

hg = x = xh, so x ∈ Z(k[G]).

But also x · x =

∑g∈G

g

x =∑g∈G

gx =∑g∈G

x = |G|x = 0 (the final inequality follows because p∣∣|G|.

Thus x2 = 0, so Rx is a nilpotent left ideal. Therefore Rx ⊂ J(R), but Rx 6= 0, so R is not semisimple sinceit has a nontrivial Jacobson radical.

We can extend the statement of Maschke’s Theorem to infinite groups as well.

Proposition 37. Suppose k is a field and G is an infinite group. Then k[G] is not semisimple.

Proof. Let R = K[G] and define φ : R→ K by∑g∈G

agg 7→∑g∈G

ag. One can verify that φ is a homomorphism

of rings [and I should do this].Let L = ker φ. Since φ is a homomorphism or rings, then L is an ideal of R. Also, L 6= 0 since 1− g ∈ L

for all g ∈ G.Suppose for the sake of contradiction that R is semisimple. Then there exists a left ideal J such that

R = L⊕ J . Since L 6= R, J 6= 0.

Let f be a nonzero element of J , and write f =∑g∈G

agg. Let h ∈ G. Then 1 − h ∈ L, so (1 − h)f ∈ L

and (1− h)f ∈ J . Since L ∩ J = 0, then (1− h)f = 0, so f = hf .

That is, f =∑g∈G

agg = hf =∑g∈G

ag(hg). By comparing coefficients, ag = ah−1g for all g ∈ G. However,

our choice of h was arbitrary as well, so by the correct choice of h, we can show that ag = ag′ for all g, g′ ∈ G.

Since∑g∈G

agg must be a finite sum, then ag = 0 for all g. Thus f = 0, which contradicts our choice of f .

Therefore R is not semisimple.

Let’s recap.

Remark 56. If R is both simple and semisimple as a ring, then R = I ⊕ I ⊕ ...⊕ I for any simple left idealI. We will write R = nI to denote this, when there are n copies of I. Then D = EndR(I) is a division ring,and R ∼= EndD(I).

Furthermore, if R is an arbitrary semsimple ring, then there are finitely many distinct simple left ideals

up to isomorphism. Let us denote them I1, ..., It. For each j, let B(Ij) =∑J∼=Ij

J . Then B(Ij) is a ring,

63

and is both simple and semisimple as a ring. Then R ∼= B(I1) × ... × B(It) ∼= n1I1 ⊕ ... ⊕ ntIt. Also,EndR(Ij) = EndB(Ij)(Ij) = Dj for some division ring Dj . Therefore B(Ij) = EndDj (Ij).

We are interesting in figuring out what the nis are.

Theorem 61. Let k be an algebraically closed field, and let R be a finite dimensional semisimple k-algebra.Write R ∼= n1I1 ⊕ ...⊕ ntIt where the Ijs are simple left ideals of R and distinct including isomorphism.

Then

1. ni = dimk(Ii) for all i.

2. dimk(R) =

t∑i=1

n2i .

3. k = EndR(Ij) for all j = 1, ..., t.

Proof. (Proof of 3) Let Dj = EndR(Ij). Since EndR(Ij) ⊂ Endk(Ij), then dimk(Dj) ≤ dimk(Endk(Ij)) = n2j .

Also, k ⊂ Z(Endk(Ij)), so k ⊂ Z(Dj). Thus D is a finite dimensional extension of k, but since k is analgebraically closed field, Dj = k.

(Proof of 1) Let mj = dimk(Ij) ≤ dimk(R) < ∞. We wish to show that mj = nj . Let B(Ij) = njIj .Since Dj = k, we have that njIj = B(Ij) = Endk(Ij). By comparison of the vector space dimensions,njmj = n2

j , so by cancellation, mj = nj .

(Proof of 2) then dimk(R) =

t∑i=1

ni dimk(Ii) =

t∑i=1

n2i .


Recall this theorem from last class:

Theorem 62. Let k be an algebraically closed field, and let R be a finite dimensional semisimple k-algebra.Write R ∼= n1I1 ⊕ ...⊕ ntIt where the Ijs are simple left ideals of R and distinct including isomorphism.

Then

1. ni = dimk(Ii) for all i.

2. dimk(R) =

t∑i=1

n2i .

3. k = EndR(Ij) for all j = 1, ..., t.

Proposition 38. Let G be a finite group, and k be a field. Let C1, ..., Cr be the distinct conjugacy classes

of G. For each i = 1, ..., r, let zi =∑g∈Ci

g ∈ k[G]. Then {z1, ..., zr} is a k-basis for Z(k[G]).

Proof. Let h ∈ G and let Ci be a conjugacy class. Then hCih−1 = Ci since conjugation by h is a injective

and sends elements of Ci to elements of Ci. Therefore

hzih−1 = h(

∑g∈Ci

gh−1

=∑g∈Ci

hgh−1

=∑g∈Ci

g

= zi

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Therefore hzi = zih for all h ∈ G. Thus zi ∈ Z(k[G]) for all i, so {z1, ..., zr} ⊂ Z(k[G]).Also, since Ci ∩ Cj = ∅ for all i 6= j, then {z1, ..., zr} is linearly independent.

Now suppose w ∈ Z(k[G]), and write w =∑g∈G

agg. Then for all h ∈ G, hw = wh, so hwh−1 = w. Then

∑g∈G

agg = w

= hwh−1

=∑g∈G

aghgh−1

=∑g∈G

ahgh−1g

Then by comparing coefficients, ag = ahgh−1 for all h ∈ G. That is, for all g1, g2 which are conjugates ofeach other, ag1 = ag2 . Thus w ∈ span({z1, ..., zr}, so {z1, ..., zr} is a basis for Z(k[G]).

Proposition 39. Let G be a finite group, let k = k be a field such that char K - |G|, and let R = k[G].Let I1, ..., It be the distinct (up to isomorphism) simple left ideals of R. Then t is the number of conjugacyclasses of G.

Proof. By the Theorem from last class, we can write R = nI1 ⊕ ...ntIt, where ni = dimk(Ii).

Then if we write B(Ii) =∑J∼=Ii

J , then R = B(I1)× ...×B(It) ∼= Mn1(k)× ...×Mnt(k).

Therefore Z(R) = Z(Mn1(k))× ...× Z(Mnt(k)). One can verify that Z(Mni(k)) = k1ni (the center of a

matrix ring are the diagonals matrices with the same thing on the diagonal), so Z(R) = k1n1× ... × k1nt .

Therefore a k-basis for Z(R) is {1n1, ...1nt} (where 1ni) represents the natural embedding of 1ni into R).

Thus dimk(Z(R)) = t.But by the previous theorem, another basis for Z(R) is {z1, ..., zr}, so dimk(Z(R)) = t = r, where r is

the number of conjugacy classes.

Remark 57. (Standard Hypothesis) The next several theorems have a “standard hypothesis” that G is afinite group, k is an algebraically closed field, such that char K - |G|, and R = k[G]. Recall that by aprevious theorem, R is then semisimple. Recall also that we can write R = n1I1 ⊕ ...⊕ ntIt.

Then we can summarize the previous results into the following theorem:

Theorem 63. Suppose the standard hypothesis holds. Then

1. |G| =t∑i=1

n2i .

2. t is the number of conjugacy classes of G.

3. t is the number of simple left ideals of k[G].

4. If I1, .., It are the simple left ideals, then ni = dimk Ii and this equals the number of times Ii appearsin a decomposition of k[G] into simple left ideals.

Corollary 26. Suppose the standard hypothesis holds. Then the following are equivalent

1. G is abelian

2. t = |G|

3. ni = 1 for all i

65

4. Every simple left ideal in R has dimension 1 over k.

Example 57. LetG = S3 under the standard hypothesis. Then the conjugacy classes ofG are {e}, {(12), (23), (13)}, {(123), (132)},so t = 3. Also, |G| = 6 = n2

1 + n22 = n2

3. Therefore n1 = n2 = 1 and n3 = 2. Then we can writek[S3] = I1 ⊕ I2 ⊕ 2I2, where I1, I2, I3 are simple left ideals in R, and dimk(I1) = dimk(I2) = 1, butdimk(I3) = 2.

21 Representaions of Finite Groups

Now let’s talk about linear actions.

Definition 57. Let G be a group and let X be a set. An action of G on X is a map G×X → X which wedenote by (g, u) 7→ gu such that (g1g2)u = g1(g2u) and 1u = u for all u ∈ X and g1, g2 ∈ G.

If V is a vector space, a linear action of G on V is an action of G on V such that g(u1 +u2) = gu1 + gu2

and g(ku) = k(gu) for all g ∈ G, u ∈ V , and k ∈ K.

Definition 58. Suppose G acts linearly on V . For each g ∈ G, let φg : V → V by u 7→ gu. Note that(φg)

−1 = φg−1 . Then φg ∈ Endk(V )∗ = GLk(V ) (this is the “general linear group”).Define ρ : G→ GLk(V ) by g 7→ φg. Then ρ is a group homomorphism. We call ρ a k-linear representation

of G.

Remark 58. Continuing the last definition, if ρ : G → GLk(V ) is a linear representation, we can define alinear action of G on V by gu = ρ(g)(u).

Therefore k-linear representations of G are in correspondence with k-linear actions of G on V .

Remark 59. Given a linear representation of ρ : G→ GLk(V ), we can define a k[G]-module Mρ as follows:The underlying set Mρ = V as a k-vector space. For g ∈ G and u ∈ Mrho, we give the action of g on u

by gu = ρ(g)(u). We then linearly extend this to give an action of k[G] on Mρ.This makes Mρ into a k[G]-module.

Remark 60. Conversely, given a k[G]-module, we can make it into a group action. Let M be a k[G]-module. Let VM be M as the underlying k-vector space. For g ∈ G, let φg : VM → Vm by u 7→ gu. Thenφg ∈ GLk(V ). Define ρM : G→ GLk(V ) by ρ(g) = φg.

Then one can verify that ρM is a k-linear representation of G.

Definition 59. Let ρ : G → GLk(V ) be a linear representation of G. we say ρ is irreducible if and only ifMrho is a simple k[G]-module.

Additionally, the degree of the representation ρ is dimk(V ) = dimk(Mρ).

Remark 61. It follows immediately from the definitions that if ρ has degree 1, then ρ is irreducible.Additionally, under the standard hypothesis, G is abelian if and only if every irreducible k-linear repre-

sentation of G has degree 1.


Recall from last class that there are actions of groups on sets, and linear actions of groups on vector spaces.Also, linear actions of a G on a vector space are in natural correspondence with the module actions of k[G]on that vector space. If ρ is our group action, then we use Mρ to denote the group action. Alternatively, ifM is our module action, then we use ρM to denote the corresponding group action.

Definition 60. Let ρ, φ be two k-linear representations of a group G. Let ρ : G → GLk(V ) and φ : G →GLk(W ). An isomorphism of representations between ρ and φ is a k-vector space isomorphism f : V →Wsuch that for all g ∈ G and u ∈ V , f(ρ(g)(u)) = φ(f(u)).

66

This is equivalent to f inducing a k[G]-module isomorphism between Mρ and Mφ. In fact, we willuse the fact that ρ and φ are isomorphic as representations if and only if Mρ and Mφ are isomorphic ask[G]-modules.

Remark 62. Recall that if ρ is a representation, we say that ρ is irreducible if and only if Mρ is simple asa k[G] module. We also define deg(ρ) = dimk(Mrho).

From this it immediately follows that degree 1 representations are irreducible.

Definition 61. We define the direct sum of two representations as follows: if ρ : G → GLk(V ) andφ : G → GLk(W ) are k-linear representations, then we say ρ ⊕ φ : G → GLk(V ⊕W ) by g 7→ (ρ ⊕ φ)(g) :(v, w) 7→ (ρ(g)(v), φ(g)(w)).

As a matrix, we can represent this as

(ρ(g) 0

0 φ(g)

).

Alternatively, one can “define” ρ⊕ φ by Mρ⊕φ = Mρ ⊕Mφ.

Example 58. (Trivial representation) Let G be a group, and k be a field. Then define ρ : G→ k∗ = GL(k1)by g 7→ 1. Since deg ρ = dimk k = 1, then ρ is irreducible.

Then Mρ = k, and the module action is defined by gα = α for all g ∈ G and α ∈ k.

Example 59. Let G be a finite group, and k is an algebraically closed field such that char k - |G|. Letρ be any representation of G. Then Mρ is simple by a previous theorem [which one?], and since k[G] is

semisimple, then Mρ is isomorphic to a simple left ideal. Let w =∑g∈G

g. Then hw = w for all h ∈ G (since

left multiplication by h simply permutes the elements of G).Let R = k[G], and I = Rw. Since hw = w for all h ∈ G, then Rw = kw. Therefore I is a simple left

ideal, so I ∼= Mρ. Therefore I is the simple left ideal corresponding to the trivial representation.In fact, since dim I = 1, then in the decomposition of k[G] into the direct sum of simple left ideals, there

is only one copy of an ideal isomorphic to I. That is, I is the unique simple left ideal isomorphic to thetrivial representation.

Example 60. (Regular Representation) Let G be a group, and let k be a field. Let V =∑g∈G

kg = k[G] (as

a k-vector space). Define ρ : G → GLk(V ) by g 7→ φg, where φg : V → V is given by u 7→ gu. That is, wehave given ρ by ρ = ρk[G].

Then deg ρ = dimk(k[G]) = |G|. Then, given the standard hypothesis, ρ is irreducible if and only ifG = {1}.

Under the standard hypothesis, we can write k[G] ∼= n1I1 ⊕ ...⊕ ntIt where the Ij are simple left idealswith Ij 6= Il for all j 6= l.

Then we can write ρ = ρk[G]∼= n1ρI1 ⊕ ... ⊕ ntρIt . Here, each ρIj is irreducible since Ij is simple as a

module.Since every simple k[G]-module is isomorphic to one of the Ij , then every irreducible representation of G

is isomorphic to one of the ρIj s. Hence every irreducible representation of G occurs as a direct summand ofthe regular representation.

Example 61. (Cyclic groups) Let G =< α >∼= Cn. Assume the standard hypothesis. To find the numberof irreducible representations, it is the number of conjugacy classes. Since Cn is abelian, then this is simply

n. But at the same time, |G| =t∑i=1

n2i , so each ni = 1.

Then we can explicitly give the representations by ρi : G→ GLk(k) = k∗ by a 7→ λi. Since an = 1, thenλni = ρi(a)n = 1. Therefore, λi must be an n-th root of unity, and since char k - n, there are n distinct rootsof unity. Thus the representations are a 7→ ζin where ζn is a primitive n-th root of unity and 1 ≤ i ≤ n.

67


Let us continue our example from last class. Last time we dealt with it as a linear representation of a group,but we can do it with group rings instead. Let us do so now.

Example 62. (Cyclic groups) Let G =< α >∼= Cn. Assume the standard hypothesis. Then define φ :k[x]→ k[G] by x 7→ a, so f(x) 7→ f(a). Note that φ is surjective.

Note also that xn − 1 ∈ kerφ, so we can mod out by xn − 1. Then consider φ : k[x]/(xn − 1) → k[G].Note that φ is still surjective, and the dimensions on both sides are n. Therefore, since φ is a k-lineartransformation, it has no kernel as a linear transformation. Thus φ is injective as a ring homomorphism.Thus φ is a ring isomorphism.

But by the Chinese Remainder Theorem, k[G] ∼= k[x]/(xn − 1) ∼= k[x]/(x − λ1) × ... × k[x]/(x − λn) ∼=k × ...× k.

Remark 63. For this statement, we used the Chinese Remainder Theorem, which is the following statementfor rings:

Let R be a commutative ring, and let I1, ..., In be ideals in R such that Ii + Ij = 1 for i 6= j. ThenR/(I1...In) = R/(I1∩ ...∩In) ∼= R/I1× ...×R/In. Furthermore, this isomorphism is given by r 7→ (r, ...r).

Remark 64. Let H C G and let ρ be a representation for G/H. That is, ρ : G/H → GLk(V ) is a grouphomomorphism.

Then, if π : G→ G/H is the natural projection, then ρ : G→ GLk(V ) is a group homomorphism. Thatis, it is a representation.

Claim 4. Let H C G and let ρ be a representation for G/H. Let ρ : G → GLk(V ) be the inducedrepresentation on G. Then the ρ-submodules of V are exactly the ρ-submodules of V .

Proof. Let W be a k[G/H]-module of V . Then gW ⊂W for all g ∈ G. But note that the action of g on Wfrom the induced representation is given by gW = gW . Therefore W is a k[G] submodule of V . Similarly,if W is a k[G] submodule of V , then W is a k[G/H] submodule of V .

Corollary 27. Let H C G and let ρ be a representation for G/H. Let ρ : G → GLk(V ) be the inducedrepresentation on G. Then ρ is irreducible if and only if ρ is irreducible.

Proof. Recall that an irreducible representation is one with precisely two submodules, namely the wholespace and the zero space. Since the ρ-submodules and ρ-submodules are the same set, then ρ is irreducibleif and only if ρ is irreducible.

Remark 65. We have a special case if H C G such that [G : H] = 2. Then G/H ∼= C2 =< α >. Thenthere are exactly two one-dimensional representations, ρ1 : G/H → k∗ by a 7→ 1 and ρ2 : G/H → k∗

by a 7→ −1. This gives two irreducible representations of G, the trivial one ρ and also ρ2 : G → k∗ by

g 7→

{1 if g ∈ H−1 if g 6∈ H

.

Example 63. Let G = Sn and H = An. Then [Sn : An] = 2, so the sign sgn∗ : Sn → k∗ by σ 7→ (−1)sgnσ

is the sign representation.

Remark 66. Let’s do another construction of a representation. Let H ≤ G with [G : H] = n. Let C1, ..., Cn

be the two left cosets of H in G. For each i = 1, ..., n, let ui =∑g∈CI

g. For every g ∈ G, gCi = Cj for

some j, so gui = uj . Thus I = ku1 ⊕ ... ⊕ kun is a left ideal of k[G]. Then I is irreducible (if n > 1) sincek(u1 + ...+ un) is a left subideal. (Recall that k(u1 + ...+ un) is the trivial representation.)

Example 64. Let G = S3, and assume the standard hypothesis. Let t be the number of irreduciblerepresentations of G. Then t is the number of conjugacy classes of G, which happens to be 3 in this case.

Since

3∑i=1

n2i = 6, then n1 = 1, n2 = 1, and n3 = 2.

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Let us consider the representations associated with each of these. For n1, this is the trivial representation.For n2, this is the sign representation, discussed earlier.

What about the n3 = 2 representation? This would be a representation in dimension 2. Let H =< (12) >.Then we shall do the previous construction on H ≤ G.

The three cosets of H are C1 = H, C2 = (23)H, and C3 = (13)H. Let ui =∑σ∈C1

σ. Then I =

ku1 ⊕ ku2 ⊕ ku3subsetk[S3]. Then, as before, k(u1 + u2 + u3) is an invariant submodule, so we can writeI = k(u1 + u2 + u3)⊕ J for some invariant submodule J . Note that J is a left ideal of dimension 2.

We wish to show that J is irreducible. Note that if J does reduce, then it would reduce into one-dimensional representations. But we have already found all of the one-dimensional representations, and theyhad multiplicity one, so they can’t appear again! Therefore J is irreducible.

Therefore J ∼= I/k(u1 + u2 + u3) = ku1 ⊕ ku2 where u3 = −u1 − u2.

We can give this explicitly by φ : S3 → GLk(k2) = GL2(k). Then (12) 7→(

1 −10 −1

)and (123) 7→(

−1 1−1 0

)and this defines what φ does for the whole group.


How do you do problems 3 and 4 on Test 2?

Claim 5. (Exam problem 3) Let R be a left Artinian ring. Then R has only finitely many two-sided idealscontaining its Jacobson Radical.

Proof. Recall that for any ringR and any ideal I, there is a natural bijection between {ideals in R containing I}and {ideals in R/I}.

Then we can use this correspondence for the Jacobson radical J(R), so consider the two-sided ideals inR/J(R). Since R is left Artinian, then so is R/J(R). Also, J(R/J(R)) = 0 for some reason. ThereforeR/J(R) is semisimple.

Then we can write R/J(R) ∼= Mn1(D1)× ...×Mnt(Dt) as the product of matrix rings. But each matrix

ring is simple, so it has only two left ideals. In the finite product, the ideals of the product are products ofideals, so there are 2t two-sided ideals.

Claim 6. (Exam problem 4) Let R be a left Artinian ring and let M,N be simple R-modules such thatAnnR(M) ∼= AnnR(N). Then M ∼= N .

Proof. Let J = J(R). Then JM = 0 for all simple modules M . Therefore every simple module is an R/Jmodule by the action r · u = r · u.

One can verify that M ∼= N as R-modules if and only if M ∼= N as R/J-modules.However, R/J is a semisimple ring for the same reasoning above. Therefore the problem reduces to the

case of a semisimple ring. Therefore assume without loss of generality that R is semisimple.It was a theorem that simple modules over semisimple rings are isomorphic to simple left ideals of the

ring. That is, M ∼= I and N ∼= L for some simple left ideals of R.Now consider the decomposition of R into matrix rings given by the Artin-Wedderburn Theorem. If

I 6∼= L, then they would have different annihilators, so M 6∼= N . This is a contradiction, so M ∼= N .

69

Chapter 2

MATH 902

1 Representation Theory

1.1 Day 1 - January 11

What we’ll cover this semester is representation theory, commutative algebra, and maybe a dash of categorytheory.

Definition 62. Let A be an n × n matrix with entries for a ring R (usually, R is commutative). Let the

entries be denoted A = [ai,j ]. Then we say Tr(A) =

n∑i=1

ai,i is the trace of the matrix.

Remark 67. One can easily verify the following facts:

1. Tr(A+B) = Tr(A) + Tr(B).

2. Tr(cA) = cTr(A) for any c ∈ R.

3. If R is a commutative ring, then Tr(AB) = Tr(BA).

4. If R is a commutative ring, then Tr(PAP−1) = Tr(A) for all invertible matrices P .

Using these facts, we can make the following definition.

Definition 63. Let k be a field and let V be a finite dimensional k-vector space. For a linear transformationf : V → V , define the trace of f to be Tr(f) := Tr([f ]β) for any basis β of V . Note that the trace isindependent of our choice of basis by part 4 of the previous remark.

Remark 68. From the previous remark, we get that Tr : Endk(V )→ k is a linear functional.

Definition 64. Let k be a field, let R be a finite-dimensional k-algebra, and let M be a finitely generated(left) R-module. (Note that this implies that M is a finite dimensional k-vector space.)

For each r ∈ R, note that the map rm : M → M by u 7→ ru is k-linear. Define the character χm (of R)associated to M by χm : R→ K by r 7→ Tr(rm).

We say that the degree of χm is dimkM .

Theorem 64. We have the following results:

1. χm(1) = dimk(M) · 1k (and if char k = 0 then this equals degχm).

2. χm : R→ k is a linear functional.

3. If β is a k-basis for R, then χm is determined by χm

∣∣∣β.

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4. χM⊕N = χM + χN .

5. Suppose M ∼= N as R-modules. Then χM = χN .

Proof. (of part 2) χm(r+ s) = Tr((r+ s)m) = Tr(rm + sm) = Tr(rm) + Tr(sm) = χm(r) + χm(s). Similarly,if c ∈ k, then χm(cr) = cχm(r) for all r ∈ R.

Proof. (of part 4) Identify M,N as submodules of M ⊕ N , and let β1 and β2 be bases for M and N

respectively. Then β1∪β2 is a basis for M⊕N . Let r ∈ R. Then [rM⊕N ]β1∪β2=

([rM ]β1 0

0 [rN ]β2

). Thus

Tr(rM⊕N ) = Tr(rM ) + Tr(rN ). But at the same time the lefthand side equals χM+N (r) and the righthandside equals χM (r) + χN (r).

Proof. (of part 5) Let φ : M → N be a k-linear isomorphism. Let β be a basis for M . Then φ(β) is a basis

for N . Let β = {u1, ..., un}. Let r ∈ R. Let [rM ]β = [ai,j ]. Then rM (uj) = ruj =

n∑j=1

ai,juj . If we apply φ

to both sides, we get that rN (φ(ui)) = rφ(ui) = φ(rui) =

n∑j=1

ai,jφ(uj).

Therefore [rN ]φ(β) = [ai,j ]. Hence χM (r) = Tr(rM ) = Tr(rN ) = χN (r) for all r ∈ R.

Example 65. Let R = k[x]/(x2) (note that R is not semisimple). Let M = R. A basis for R is β = {1, x}.

Then [1M ]β =

(1 00 1

), so χM (1) = 2. Also, [xM ]β =

(0 01 0

). Thus χM (x) = 0.

Let N = R/(x) ⊕ R/(x). Then [1N ]β =

(1 00 1

)and [xN ]β =

(0 00 0

). Thus χM (1) = 2 and

χM (x) = 0.Therefore the characters of both of these are equal, but the modules are not isomorphic. After all,

xN = 0, but xM 6= 0.

Proposition 40. Let R be a semisimple finite dimensional k-algebra such that char k = 0. Let M,N befinitely generated modules. Then χM = χN if and only if M ∼= N .

Proof. We have already shown that M ∼= N implies χM = χN .Conversely, suppose χM = χN . Then let R = nI1⊕ ...⊕ntIt = B(I1)× ...×B(It) where the Ij are distinct

simple left ideals. Recall that this is the unique decomposition given by the Artin-Wedderburn theorem.We can write M ∼= m1I1 ⊕ ...⊕mtIt and N ∼= r1I1 ⊕ ...rtIt. Then M ∼= N if and only if mi

∼= ri.Let ei be the identity element of B(Ii). Recall that eiIj = 0 for all j 6= i. Then (ei)M : M → M by

u 7→ eiu. Therefore (ei)M

∣∣∣mjIj

= 0 for j 6= i. However, (ei)m

∣∣∣miIi

= 1miIi . Thus χM (ei) = (dimk(miIi)) ·

1k = mi dimk(Ii) · 1k.Similarly, χN (ei) = ri dimk(Ii)·k. Then we can set these equal to zero, and since 1k 6= 0 and dimk(Ii) 6= 0,

we can cancel by these and get that mi = ri. Since i was arbitrary, then M ∼= N .

Note that this makes us really excited. Now, a finitely generated semisimple ring is uniquely determinedby its character.

Definition 65. Let R be a finite dimensional k-algebra. Then a character for R is a χM for some R-moduleM . A character χ 6= 0 is irreducible if, whenever χ = χ1 + χ2, with both the χis being characters, theneither χ1 = 0 or χ2 = 0.

Corollary 28. Suppose R is a semisimple finite-dimensinal k-algebra with char k = 0. Then a characterχ = χM is irreducible if and only if M is simple.

Proof. Suppose M = N1 ⊕N2. Then χM = χN1+ χN2

. Note that χNi = 0 if and only if Ni = 0. Thereforeany decomposition of the character leads to a decomposition of the module, and vice versa.

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Recall the following theorem:

Theorem 65. Let R be a semisimple, finite-dimensional k-algebra (where k is an algebraically closed fieldof characteristic 0). Then R ∼= n1I1 ⊕ ...⊕ ntIt where Ii 6= Ij for i 6= j. Let χi = χIi for i = 1, ..., t.

Then

1. {χ1, ..., χt} is the complete set of all irreducible characters from R.

2. Every character χ for R can be expressed uniquely in the form m1χ1 + ...mtχt where each mi ≥ 0.

3. ni = χi(1).

4. χR = χ1(1)χ1 + ...+ χt(1)χt.

Proof. (Proof of 1) As proved last class, χ = χM is irreducible if and only if M is simple. But this is thecase if and only if M ∼= Ii for some i, and this is the case if and only if χM = χIi (since isomorphic moduleshave the same character).

(Proof of 2) Let χ = χM where M is a finitely generated R-module. Then M ∼= m1I1⊕ ...⊕mtIt for somemi ≥ 0. (This decomposition is unique by the Jordan-Holder Theorem). Then χM = m1χ1 + ...+mtχt. IfχM = r1χ1 + ...+ rtχt, then M ∼= r1I1 ⊕ ...⊕ rtIt, so mi = ri for all i.

(Proof of 3) Recall that χi(1) is the trace of the map given by left multiplication by 1. But this is theidentity, and the trace of the identity is the dimension of the vector space. That is, χi(1) = dimk(Ii) = ni.

(Proof of 4) By decomposing R into the direct sum of simple modules and using the fact that χM⊕N =χM + χN , we get that χR = n1χ1 + ...ntχt. Then applying part 3, we know that ni = χi(1), so χR =χ1(1)χ1 + ...χt(1)χt, as desired.

Recall the following things about linear representations:

Remark 69. We will usually work under a common set of hypotheses: that k is a field, V is a finitedimensional vector space over k, G is a finite group, and ρ : G → GLk(V ) is a (k-linear) representation forG. The character χρ of ρ is defined by χρ : G→ k by g 7→ Tr(ρ(g)).

Let M be the k[G]-module corresponding to ρ. That is, M has an underlying set equal to V , with thek-vector space structure. We define the action of G on M by ρ, thereby making M a k[G]-module. Thenχρ(g) = χM (g), so we get a one-to-one correspondence between characters of k[G] modules, and charactersof k-linear representations of G.

Remark 70. We will often work under the following hypotheses:Let k be an algebraically closed field. Let G be a finite group such that char k - #G. Then k[G] is a

semisimple ring, so we can decompose it into k[G] = B(I1)× ...×B(It) ∼= n1I1 ⊕ ...⊕ ntIt. Let χi = χIi fori = 1, ..., t.

Let ei ∈ B(Ii) be the identity element. Let c1, ..., ct denote the distinct conjugacy classes of G. Let

mi = |ci|, and let zi =∑g∈ci

g ∈ k[G]. Then Z(k[G]) = ke1 × ...× ket = kz1 ⊕ ...⊕ kzt.

These hypotheses will be referred to by (#).

Theorem 66. Let χ be a character of a representation ρ : G → GLk(V ). Let ∼ denote the equivalencerelation on G given by “is conjugate to”. Then whenever g ∼ h, then χ(g) = χ(h). (Such functions arecalled class functions.)

Proof. If g = xhx−1, then ρ(g) = ρ(xhx−1) = ρ(x)ρ(h)ρ(x−1). So χ(g) = Tr(ρ(g)) = Tr(ρ(h)) = χ(h).

Lemma 25. Suppose (#). Let φ = χk[G]. Then φ(g) =

{|G| if g = 1

0 if g 6= 1

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Proof. Note that φ(g) = χk[G](g) = Tr(gk[G]). We can use any basis, so let’s use the basis given by theelements of g. If g = 1, then gk[G] is the identity, so φ(g) = |G|. If g 6= 1, then gh 6= h for all h ∈ G,so multiplication by g permutes the basis elements without fixed points. Thus the matrix corresponds to apermutation matrix with no 1s on the diagonal, so φ(g) = 0.

Theorem 67. Suppose (#). Then

1. ei = ni|G|

∑g∈G

χi(g−1)g

2. zi = mi

t∑j=1

χj(g)

njej for any g ∈ ci.

Proof. (Proof of 1) Recall that ei ∈ k[G], so ei can be written unique as a linear combination of elements of

g. Thus let ei =∑

agg (where each ag ∈ k. It suffices to show that ag = ni|G|χ(g−1).

Let h ∈ G. Then, let φ = χk[G], we have that φ(eih−1) = φ(

∑g∈G

aggh−1) =

∑g∈G

agφ(gh−1) = ah|G|. But

eih−1Ij = eiIj = 0 when j 6= i. Thus χj(eih

−1) = 0 for j 6= i.But on the other hand,

φ(eih−1) = (n1χ1 + ...+ ntχt)(eih

−1)

= niχi(eih−1)

= niχi(h−1)

Thus ah|G| = niχ(h−1), so ah = ni|G|χ(h−1).


Recall the following set of hypotheses:

Remark 71. We will often work under the following hypotheses:Let k be an algebraically closed field. Let G be a finite group such that char k - #G. Then k[G] is a

semisimple ring, so we can decompose it into k[G] = B(I1)× ...×B(It) ∼= n1I1 ⊕ ...⊕ ntIt. Let χi = χIi fori = 1, ..., t.

Let ei ∈ B(Ii) be the identity element. Let c1, ..., ct denote the distinct conjugacy classes of G. Let

mi = |ci|, and let zi =∑g∈ci

g ∈ k[G]. Then Z(k[G]) = ke1 × ...× ket = kz1 ⊕ ...⊕ kzt.

These hypotheses will be referred to by (#).

Recall also this theorem, which we previously stated and proved the first half of:

Theorem 68. Suppose (#). Then

1. ei = ni|G|

∑g∈G

χi(g−1)g

2. zi = mi

t∑j=1

χj(g)

njej for any g ∈ ci.

We will now prove the second half of this theorem:

73

Proof. (Proof of 2) Note that from the first half of the theorem, since no ei is 0, then ni 6= 0 in k, since

ei = ni|G|

∑g∈G

χi(g−1)g. Thus char k - ni for all i.

Since zi ∈ k[G] = n1I1 ⊕ ...⊕ ntIt, then zi =

t∑j=1

ujej for some uj . Then

χl(zi) = χl(

t∑j=1

ujej)

=

t∑j=1

ujχl(ej)

= ulχl(el)

= ulnl

Note that the penultimate equation is true since χl(ej) = 0 for all j 6= l, and that the last equation is

true since χl(el) = Tr(idIl) = dimk(Il) = nl. Then zi =∑g∈Ci

g, so

χl(zi) = χl(∑g∈Ci

g)

=∑g∈Ci

χl(g)

= miχl(g)

For any g ∈ Ci.Therefore ul = miχl(g)

nl, and this completes the proof.

Corollary 29. Suppose (#). Then

1. For all i, j ∈ {1, ..., t}, we have that∑g∈G

χi(g)χj(g−1) = δi,j |G|.

2. For all g, h ∈ G, we have that

t∑i=1

χi(g)χi(h−1) = δg,h|CG(g)|.

Proof. (Proof of 1) Recall that ei = ni|G|

∑g∈G

χi(g−1)g, so niδi,j = χj(ei) = ni

|G|

∑g∈G

χi(g−1)χk(g). By multi-

plying by |G|ni , we get the desired equation.(Proof of 2) Recall that

zi = mi

t∑j=1

χj(g)

njej

= mi

t∑j=1

χj(g)

nj(nj|G|

∑h∈G

χj(h−1)h)

=mi

|G|∑h∈G

(

t∑j=1

χj(g)χj(h−1))h

74

Also, by definition we have that zi =∑h∈Ci

h, so we shall compare coefficients. By doing so, we get that

if h ∈ Ci (that is, if g ∼ h), then mi|G|

t∑j=1

χj(g)χj(h−1) = 1, so

t∑j=1

χj(g)χj(h−1) =

|G|mi

= |CG(g)|. On the

other hand, if h 6∈ Ci, then

t∑j=1

χj(g)χj(h−1) = 0, as desired.

Remark 72. A special case of interest to us of part (2) in the corollary is that for all g 6= 1,

t∑j=1

χj(g)χj(1) =

0.

Definition 66. A function f : G → k is called a (k-) class function if f(g) = f(h) whenever g ∼ h. LetFk(G) = {f | f is a k-class function on G}.

Then Fk(G) is a k-vector space.

Remark 73. Let fi : G → k be defined by fi(g) = 1 if g ∈ Ci and fi(g) = 0 if g 6∈ Ci. Then {f1, ..., ft} isa basis for Fk(G). Therefore dimk Fk(G) = t (that is, the dimension is the number of conjugacy classes ofG).

Definition 67. Let V be a vector space, and f : V × V → k be a function. We say f is a bilinear form if,whenever you fix one component, f is linear in the other component.

We say a bilinear form f is symmetric if f(x, y) = f(y, x).

Definition 68. Suppose (#). Define the following bilinear form (−,−) : Fk(G) × Fk(G) → K by (φ, ψ) =1|G|

∑g∈G

φ(g)ψ(g−1) for all phi, ψ ∈ Fk(G). One can verify that this is also symmetric.

Proposition 41. Suppose (#). Then (χi, χj) = δi,j , so {χ1, ..., χt} is a basis for Fk(G).

Proof. Suppose c1χ1 + ...+ctχt = 0. Then 0 = (χi, 0) = (χi, c1χ1 + ...+ctχt) =

t∑j=1

ci(χi, χj) = ci. Therefore

χ1, ..., χt is linearly independent. One call also verify that they are spanning, hence a basis.

Lemma 26. Suppose (#) and that k = C. For all i ∈ {1, ..., t} and g ∈ Gi, we have that χi(g−1) = χi(g)

(meaning the complex conjugate).

Proof. Recall that χi(g) = Tr(gIi). Let n = |G|. Since gn = 1, then (gIi)n = 1Ii . Hence, the minimal

polynomial of gIi divides xn − 1, and thus has distinct roots. Hence gIi is diagonalizable as a matrix, so

gIi ∼(λ1...

0

0 .λr

), where each λr is a distinct n-th root of unity, and r = dimC Ii. Let lj be the integers

such that λj = e2πiljn . Then λ−1

j = e−2πiljn = λj .

Then Tr(g−1Ii

) = λ−11 + ...+ λ−1

r = λ1 + ...+ λr = (χi(g)).

Definition 69. Suppose (#) and let k = C. We can define the Hermitian inner product < −,− >:

FC(G)× FC(G)→ C by < φ,ψ >= 1|G|

∑g∈G

φ(g)ψ(g).

Remark 74. The Hermitian inner product has some perks. It is sesquilinear. We have that < φ, φ >= 0 ifand only if φ = 0. Also, < φ,ψ >= < ψ, φ >.

This is an inner product.

Proposition 42. Suppose (#) and that k = C. Then < χi, χj >= δi,j .

75

Proof. Note that < χi, χj >= 1|G|

∑g∈G

χi(g)χj(g) =1

|G|∑g∈G

χi(g)χj(g−1) = (χi, χj) = δi,j .

From this we can conclude that {χ1, ..., χt} are in fact an orthonormal basis for FC(G).


We are now working under the following assumptions:Let G be a finite group, k = C, χ1, ...χt the set of irreducible characters of G. For φ, ψ ∈ FC(G) (the set

of class functions), we define < φ,ψ >= 1|G|

∑g∈G

φ(g)ψ(g). Then we have that < χi, χj >= δi,j .

Let φ : G → GLn(C). Let χ(g) = Tr(φ(g)). If χ is a character of a representation of G, then χ =m1χ1 + ...+mtχt where mi ≥ 0 and mi ∈ Z for all i. Furthermore, mi =< χ,χi >,

Also, < χ,χ >= m21 + ...+m2

t , so < χ,χ >= 1 if and only if χ = χi for some i.We can now move onto character tables. This is a table of characters χi and elements of conjugacy classes

gj , whose entries are χi(gj).1 g1 ... gt−1

χ1 1 1 ... 1χ2 n2

...χt nt

Example 66. Let G = C3 =< a >. Then there are 3 conjugacy classes of elements in G, so t = 3. Thus thereare 3 irreducible representations, all of degree 1. If ω = e

2πi3 , then they are given by φj : C3 → GL1(C) = C∗

by a 7→ ωj . Then χj(a) = Tr(φj(a)) = ωj . Then the character table is1 a a2

χ0 1 1 1χ1 1 ω ω2

χ2 1 ω2 ωNote that columns are always orthogonal because of the orthogonality relation.

Example 67. Let G = V4 = {1, a, b, c}, the Klein 4-group. Recall that if H C G and ρ : G/H → GLn(C)

is a representation, then ρ : G → GLn(C) given by ρ : Gπ→ G/H

ρ→ GLn(C) is also a representation.Furthermore, ρ is irreducible if and only if ρ is irreducible.

In our case, let H =< a >. Then G/H ∼= C2, so let rho : G/H → C∗ by b 7→ −1. Then the inducedrepresentation ρ : G → C∗ is given by 1 7→ 1, a 7→ 1, b 7→ −1 and c 7→ −1. Then let us make our charactertable:

1 a b cχ1 1 1 1 1χ2 1 1 −1 −1χ3 1 −1 1 −1χ4 1 −1 −1 1

Example 68. Let G = S3. Then there are 3 conjugacy classes (corresponding to 1-cycles, 2-cycles, and3-cycles). This gives us the following table:

1 (12) (123)χ1 1 1 1χ2 1 −1 1χ3 2 0 −1

The second row corresponds to modding out by (123) to get C2, and the last row comes from orthogonalitywith the previous ones.

Remark 75. Unfortunately, there exist distinct groups with the same character table. We will talk aboutwhat they are later (after the homework).

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Example 69. Let G = A4. Then H = {(1), (12)(34), (13)(24), (14)(23)} is normal in G. Then there arefour conjugacy classes: (1), H \ {1}, (123)H, and (132)H. We can then build the following character table:

1 (12)(34) (123) (132)χ1 1 1 1 1χ2 1 1 ω ω2

χ3 1 1 ω2 ωχ4 3 −1 0 0

(Where ω is a primitive 3rd root of unity. The first three rows are given by the three representations ofG/H ∼= C3. The last one is given by the columns’ orthogonality relation.


Let’s do the character table for the quaternions!

Example 70. Let G = Q8 (the quaternion group). The conjugacy classes of G are {1}, {−1}, {±i},{±j} and {±k}. Since H = Z(G) = {±1}CG, then we get characters from G/H ∼= V4. This gives us threecharacters, and we get the trivial character for free. Then the final row can be computed by the orthogonalityrelation.

1 −1 i j kχ1 1 1 1 1 1χ2 1 1 1 −1 −1χ3 1 1 −1 1 −1χ4 1 1 −1 −1 1χ5 2 −2 0 0 0

In fact, χ5 is the charcter from the representation of ρ : Q8 → GL2(C) by i→(

0 1−1 0

)and j → 0ii0

and k →(

i 00 i

).

Example 71. LetG = S4. The conjugacy classes are the distinct cycle types (represented by) (1), (12), (12)(34), (123), (1234).We will find characters by looking at a normal subgroup. Let H = {(1), (12)(34), (13)(24), (14)(23)}. ThenH C S4. Furthermore, S4 has no elements of order 6, so S4/H has no element of order 6. ThereforeS4/H ∼= S3. From last class, we know that S3 has three conjugacy classes, giving us three characters forfree.

Also, we know that the sum of the squares of the first column add up to 24, and the first three entriesare 1,1,2, so the last two entries must be 3, 3. This gives us:

1 (12) (123) (12)(34) (1234)χ1 1 1 1 1 1χ2 1 −1 1 1 −1χ3 2 0 −1 2 0χ4 3χ5 3

We can get another representation from the action of S4 on V = Ce1⊕Ce2⊕Ce3⊕Ce4 by σ : e1 → eσ(i).Note that W = C(e1 + e2 + e3 + e4) is fixed under σ, so let’s instead consider the action of G on V/W =Ce1 ⊕Ce2 ⊕Ce3. Then this gives another representation, which happens to be irreducible because its innerproduct with itself is 1. The last row we can get by orthogonality relations. Thus we get this completedcharacter table:

1 (12) (123) (12)(34) (1234)χ1 1 1 1 1 1χ2 1 −1 1 1 −1χ3 2 0 −1 2 0χ4 3 1 0 −1 −1χ5 3 −1 0 −1 1

77

2 Spectrum of a Ring

We now emerge from group theory to commutative algebra!Let R be a commutative ring with unity with 1 6= 0.

Definition 70. An ideal p 6= R is prime if for all a, b ∈ R such that ab ∈ p, then a ∈ p or b ∈ p.

Remark 76. One can show that p is prime if and only if R/p is a domain.

Exercise 18. If p is a prime ideal and p ⊇ I1 ∩ ... ∩ In, then p ⊇ Ii for some i.From this, prove that if p ⊃ I1I2...In, then p ⊃ Ii for some i. In particular, if p ⊃ In, then p ⊃ I.

Exercise 19. Show that maximal ideals are prime.

Definition 71. The prime spectrum of R is Spec R = {p|p is a prime ideal of R}.

Example 72. Let F be a field. Then Spec F = {0}. If R is a ring with Spec R = {0}, then in fact R is afield.

Also, (0) ∈ Spec R if and only if R is a domain.

Example 73. Spec Z = {(0)} ∪ {(p)|p ∈ Z is a prime number}.Let F be a field. Then Spec F [x] = {(0)} ∪ {(f(x))|f(x) ∈ F [x] is irreducible}.


Recall from last class the definition of a prime ideal, and the definition of the spectrum of a ring.

Proposition 43. Let R be a ring, and a1, ..., an ∈ R. Define φ : R[x1, ..., xn] → R by f(x1, ..., xn) 7→f(a1, ..., an). Then φ is a surjective ring homomorphism and kerφ = (x1−a1, ..., xn−an), andR[x1, ..., xn]/(x1−a1, ..., xn − an) ∼= R.

Proof. Certainly, (x1 − a1, ..., xn − an) ⊂ kerφ. To show the other direction, we will use induction on n.For n = 1, f(x) ∈ R[x], so by the division algorithm, f(x) = (x − a)g(x) + r for some r ∈ R. Then

f(a) = 0 if and only if r = 0. Therefore f(a) = 0 if and only if f(x) ∈ (x− a).For the inductive case, suppose the claim holds for n − 1. Suppose f(x1, ..., xn) ∈ R[x1, ..., xn] and

that f(a1, ..., an) = 0. Let g(x1) = f(x1, ..., a2, ..., an) ∈ R[x1]. Then g(a1) = 0. By the n = 1 case,g(x1) = (x1 − a1)`(x1).

Consider h(x2, ..., xn) = f(x1, ..., xn) − g(x1) ∈ S[x2, ..., xn], where S = R[x1]. Then h(a2, ..., an) =(x1, a2, ..., an) − g(x1) = 0. By the n − 1 case applied to S, we get that h(x2, ..., xn) ∈ (x2 − a2, ..., xn −an)S[x2, ..., xn].

Therefore f(x1, ..., xn) − g(x1) = (x2 − a2)u2(x) + ... + (xn − an)u(x). Therefore f(x1, ..., xn) = (x1 −a1)l(x1) + ...+ (xn − an)un(x) ∈ (x1 − a1, ..., xn − an).

Thus kerφ = (x1 − a1, ..., xn − an). The remainder of this theorem follows from the first isomorphismtheorem.

Corollary 30. Let k be a field. For any c1, ..., cn ∈ k, we have that (x1 − c1, ..., xn − cn) = {f(x1, ..., xn) ∈k[x1, ..., xn]|f(c1, ..., cn) = 0}, and that this ideal is maximal in k[x1, ..., xn].

Remark 77. If k is an algebraically closed field, then every maximal ideal of k[x1, ..., xn] has this form (thisis the Nullstellensatz).

Exercise 20. For any c1, ..., cm ∈ k (with m ≤ n), we have that (x1 − c1, ..., xm − cm) is a prime ideal ofk[x1, ..., xn]. (In fact, k[x1, ..., xn](x1 − c1, ..., xm − cm) ∼= k[xm+1, ..., xn].

Example 74. From the previous results, we have that SpecC[x, y] = {(0)} ∪ {(x− c, y − d)|c, d ∈ C} ∪{(f(x, y))|f(x, y) irred}.

Definition 72. For p ∈ SpecR, define the height of p by ht(p) = sup{n|there exists a chain in SpecR with p =pn ) ... ) p0}

78

Example 75. In C[x, y], we have that ht((x, y) ≥ 2 as (x, y) ⊃ (x) ⊃ (0). On the other hand, ht(x) ≥ 1since (x) ⊃ (0).

Definition 73. Let R be a ring. The dimension of R is dimR = sup{ht(p)|p ∈ SpecR}.

Definition 74. Let I be an ideal of R. Define V (I) = {p ∈ SpecR|p ⊃ I}.

Proposition 44. Let I, J, and Iα be ideals in a ring R. Then

1. V (I) ∪ V (J) = V (I ∩ J).

2.⋂α∈Λ V (Iα) = V (

∑α∈Λ

Iα).

3. V ((0)) = SpecR, and V (R) = ∅.

Definition 75. We can define the Zariski topology on SpecR by saying a set F ⊂ SpecR is closed if andonly if F = V (I) for some ideal I. The fact that this satisfies the axioms of a topology is given in theprevious proposition.

Proposition 45. The Zariski Topology is T0 but not T1 (in general).

Proof. Let p 6= q ∈ SpecR = X be two distinct ideals in SpecR. Then one of these ideals does not containthe other, so let p ( q. Let U = X \ V (p). Then p 6∈ U , but q ∈ U . Thus the Zariski topology is T0.

However, if p ⊃ q, every open set containing q also contains p, so X is not T1.

Proposition 46. The Zariski topology is compact (although sometimes this is called “semi-compact” be-cause algebraists use outdated terminology).

Proof. Let X =⋃α∈Λ Uα, where each Uα = X \ V (Iα) for some ideal Iα.

Then the fact that the Uα cover X implies that ∩αV (Iα) = ∅, and by a previous proposition, then

V (∑

Iα) = ∅. Since every proper ideal is contained in a maximal ideal (and maximal ideals are prime),

then it must be that∑

Iα = R.

Thus 1 ∈∑

Iα, so 1 = iα1+ ...iαn for some α1, ..., αn ∈ Λ. Thus

n∑i=1

Iαi = R, so V (

n∑i=1

Iαi) = ∅. Thus

∩V (Iαi) = ∅, so X =⋃ni=1 Uαi .

Definition 76. A nonempty subset S of R is multiplicatively closed if for all a, b ∈ S, ab ∈ S.

Proposition 47. Let S be a multiplicatively closed set of R and assume 0 6∈ S. Then there exists ap ∈ SpecR such that p ∩ S = ∅.

Proof. Let Λ = {I|I is an idealI ∩ S = ∅}. Note that (0) ∈ Λ, so Λ 6= ∅. Note that Λ is partially ordered bysubset.

Also, if C is a chain in Λ, then IC =⋃I∈C I is an ideal. It is also disjoint from S, so IC ∈ Λ. Finally, IC

is an upper bound of C.Thus Zorn’s lemma applies to C. Let p be a maximal element in Λ. We wish to show that p is a prime

ideal.Suppose for the sake of contradiciton it is not. Then [something] and we get a contradiction.


Recall from last class the following theorem:

Proposition 48. Let S be a multiplicatively closed set of R and assume 0 6∈ S. Then there exists ap ∈ SpecR such that p ∩ S = ∅.

We can use this to prove (one of) Krull’s Theorem(s):

79

Theorem 69 (Krull’s Theorem). Let R be a commutative ring. Then Nilrad(R) =⋂

p∈SpecR

p.

Proof. Let x ∈ Nilradical(R). Then xn = 0 ∈ p for any p ∈ SpecR. But since p is prime and x ·x · ... ·x ∈ p,

then x ∈ p. Thus x ∈⋂

p∈SpecR

p.

Conversely, suppose x 6∈ Nilrad(R). Then 0 6∈ S = {xn}n≥1. By the proposition, there exists ap ∈ SpecR such that p ∩ S = ∅, so x 6∈ p.

Proposition 49. Let R be a commutative ring, and let I 6= R be an ideal in R. Then the map f : V (I)→Spec(R/I) by p 7→ p/I is a bijection. (In fact, f is a homeomorphism).

Definition 77. Let I be an ideal in a ring R. Then the radical of I is√I = {r ∈ R| rn ∈ I for some n ∈

N}.

This theorem is also called Krull’s theorem:

Theorem 70. Let I be an ideal in a ring R. Then√I =

⋂p∈V (I)

p.

Proof. Observe that√I/I =

√(0), where (0) is I/I in R/I.

Then by Krull’s theorem about nilradicals and the previous proposition,√I/I =

√(0) = ∩p∈V (I)p/I =

(∩p∈V (I)p)/I. Then by lifting to R, we can see that√I =

⋂p∈V (I)

p

Definition 78. Let p ∈ SpecR. We say p is a minimal prime of R if ht(p) = 0.We denote the set of minimal primes by MinR(R) = {p ∈ SpecR| ht(p) = 0}.Let I be an ideal of a ring R, p ∈ SpecR, and suppose I ⊂ p. Then we say p is minimal over I if

ht(p/I) = 0.We denote the set of minimal primes over I by MinR(R/I) = {p ∈ V (I)|ht(p/I) = 0}.

Exercise 21. Let R be a commutative ring, and let I be an ideal of R with I 6= R. Let p ∈ V (I). Thenp ⊃ q for all q ∈MinR(R/I).

Proof. We apply Zorn’s lemma to Λ = {q ∈ V (I)|q ⊂ p} (with the partial order of reverse inclusion).

Corollary 31. If I is a proper ideal of a ring R, then√I =

⋂p∈MinR(R/I)

p.

Exercise 22. If R is Noetherian, then MinR(R/I) is a finite set.

Remark 78. By combining the previous exercise and previous corollary, we get that in a Noetherian ring,each proper radical ideal is the finite intersection of prime ideals.

Theorem 71. Let R be a commutative ring, and let X = SpecR. Then X is connected if and only if Rhas no nontrivial idempotents. (A trivial idempotent is 0 and 1).

Proof. Recall that X is disconnected if and only if it can be partitioned into two open sets. But if those setsare both open, then they are both closed.

Thus X is disconnected if and only if X = V (I)∪V (J), where V (I) and V (J) are nonempty and disjoint.But if they are disjoint then ∅ = V (I) ∩ V (J) = V (I + J). Since V (I + J) = ∅, then I + J = R. But sinceV (I) and V (J) are nonempty, then I 6= R and J 6= R.

Furthermore, X = V (I) ∪ V (J) = V (IJ). Thus X = V (IJ), so every prime ideal contains IJ . ThusIJ ⊂

√(0).

Thus we have translated the topological property of disconnectedness into finding two ideals I, J suchthat I 6= R, J 6= R, and I + J = R, and IJ ⊂

√(0). We now wish to show that the existence of such ideals

is equivalent to the existence of nontrivial idempotents.

80

Suppose that e is a nontrivial idempotent. Then e2 = e. Furthermore, 1− e is also an idempotent. LetI = (e) and J = (1 − e). Note that I = R if and only if e is a unit. But e2 = e, so if e were a unit,then e = 1. Since e is a nontrivial idempotent, this is not the case, so I 6= R. Similarly, since 1 − e is anontrivial idempotent, then J 6= R. Furthermore, 1 = e + (1 − e, so 1 ∈ I + J . Thus R = I + J . Finally,IJ = (e)(1− e) = (e(1− e)) = (0) ⊂

√(0).

Conversely, suppose we have two ideals satisfying the previously described properties. Then R = I + J ,so 1 = i+ j. Since ij ∈ IJ ⊂

√(0), then there exists an n ∈ N such that (ij)n = 0.

Consider the ideal (in) + (jn). If this ideal were strictly contained in R, there would be a maximal idealm such that (in) + (jm) ⊂ m. Since m is maximal, it is prime, and since in ∈ m, then so is i. Similarly,j ∈ m, so R = (i) + (j) ⊂ m. This is a contradiction of the fact that m is a maximal ideal, so we know that(in) + (jn) = R.

Thus 1 = rin + sjn for some r, s ∈ R. Let e = rin. Then e(1 − e) = (rin)(sjn) = rs(ij)n = 0. Thuse2 = e. However, if e = 1, then 1 ∈ I. If e = 0, then 1 = 1− e ∈ J . Either way, this is a contradiction. Thuse is a nontrivial idempotent.

Example 76. If R is a domain, then it has no nontrivial idempotents, so its spectrum is connected.If R = R1 ×R2, where both R1 and R2 are rings with unit, then R has nontrivial idempotents (such as

(1, 0)), so its spectrum is disconnected.If R is a local ring then R is connected.

Exercise 23. Show that if R is a local ring, then SpecR is connected.

Proposition 50. If R is a ring, and e is an idempotent in R, then φ : R→ (e)× (1−e) by r 7→ (re, r(1−e))is an isomorphism of rings. (Note that (e) is indeed a ring with unit, because e acts as the multiplicativeidentity on it.)

From this we can conclude that a ring R has a disconnected spectrum if and only if it can be factoredinto the product of two rings with unit.


Theorem 72. Let R be a commutative ring. Then R is Artinian if and only if R is Noetherian anddimR = 0.

Proof. Suppose R is Artinian. Then we have already shown that R is Noetherian. We then have to showthat R has dimension 0. Recall that the dimension of a ring is the length of a maximal chain of prime ideals.Therefore a ring is dimension 0 if and only if no prime ideals are contained in any other prime ideals.

To this end, let p be a prime ideal. Then R/p is a domain. Furthermore, since R is Artinian, then R/p isas well. We have previously proven that Artinian domains are fields, so R/p is a field. Thus /p is a maximalideal. Thus no prime ideal contains another one, so dimR = 0.

Conversely, suppose R is Noetherian and that dimR = 0. Therefore SpecR = minR = maxR. SinceR is Noetherian, then by homework problem #5 on problem set 2 [edit: see immediately below], minR isfinite.

Then let SpecR = {m1, ...,mt} (where each mi is maximal and minimal).Let J = ∩ti=1mi. Then J is both the Jacobson radical (the intersection of maximal ideals) and the

Nilradical (the intersection of prime ideals) of R.Note that for i 6= j, we have that mi + mj = R. Therefore by the Chinese Remainder Theorem, we have

that R/J = R/(m1 ∩ ... ∩ mt) ∼= R/m1 × ... × R/mt as rings. Since the righthand side is the finite productof fields, then R/J is semisimple, so it is Artinian.

We now wish to show that R/J i is Artinian for all i, using induction on i.. We have just shown this claimis true for i = 1, completing the base case. Now suppose R/J i−1 is Artinian. Then consider the R/J-moduleJ i−1/J i. Since R is Noetherian, then J i is finitely generated. Hence J i−1/J i is also finitely generated. SinceR/J is semismiple, then J i−1/J i has finite length. Hence J i−1/J i is Artinian.

Furthermore, the sequence 0 → J i−1/J i → R/J i → R/J i−1 → 0. is exact. Then since J i−1/J i isArtinian, and R/J i−1 is Artinian by induction, then so is R/J i (as an R-module). Therefore R/J i isArtinian as a ring.

81

But J is finitely generated, so let J = (a1, ..., an). Since J is also the nilradical, then there exist ni suchthat anii = 0. Let n =

∑ni. Then Jn = 0, so R/J ∼= R. Thus R is Artinian.

Homework Problem 8. Let R be a Noetherian ring, and let I be a proper ideal of R. The the set ofminimal primes over I is finite.

Example 77. Consider R = Z/100Z. Since Z is Noetherian, so is R. Recall that dimZ = 1, but all suchchains contained (0). Since (0) is not a prime ideal in R, then dimZ/(100) = 0, so it is Artinian.

3 Localization

Let’s talk about localization.

Definition 79. Let R be a commutative ring (with 1 6= 0). Let S be a multiplicatively closed set of R. LetR × S = {(r, s)|r ∈ R, s ∈ S}. Define a relation ∼ on R × S by (r1, s1) ∼ (r2, s2) if and only if there existst ∈ S such that t(s2r1 − s1r2) = 0 (or equivalently ts2r1 = ts1r2).

Proposition 51. The relation ∼ defined above is an equivalence relation.

Proof. We will only verify transitivity (the other two properties are easy to check).Suppose (r1, s1) ∼ (r2, s2) and (r2, s2) ∼ (r3, s3). Then there exists a t ∈ S such that ts2r1 = ts1r2, and

there exists a u ∈ S such that us3r2 = us2r3.Let p = uts2. Since u, t, s2 ∈ S, and S is multiplicatively closed, then p ∈ S. Furthermore,

ps3r1 = uts2s3r1

= us3(ts2r1)

= us3(ts1r2)

= ts1(us3r2)

= ts1(us2r3)

= uts2(s1r3)

= ps1r3

Thus (r1, s1) ∼ (r3, s3).

Remark 79. We will use rs to denote the equivalence class of (r, s).

Definition 80. Let RS = { rs |r ∈ R, s ∈ S}. Sometimes we instead denote this by S−1R.We furthermore define two operations on RS by r1

s1+ r2

s2= s2r1+s1r2

s1s2, and r1

s1· r2s2 = r1r2

s1s2. One can verify

that these operations are well-defined.

Theorem 73. The two operations given above make RS a commutative ring with identity.

Remark 80. From now on, we will assume that 1 ∈ S. We will usually assume 0 6∈ S, but it will be possiblefor this to be the case.

Remark 81. If x ∈ R, and we let s = {xn}n≥0, then we will write Rx for RS .If p ∈ SpecR, then we write Rp for RS , where S = R \ p.

Example 78. Let R = Z, and let S = {2n}n≥0. Then Z2 = { a2n |a ∈ Z, n ≥ 0} ⊂ Q.Also, Z(2) = {ab |a, b ∈ Z, b is odd} ⊂ Q.

Example 79. Let R = Z/(6)Z. Let S = {2n}n≥0. Then S = { a2n |a ∈ Z/(6), n ≥ 0}. Then we can do a lotof collapsing, and show that RS ∼= Z/(3).

82

3.1 Day 9 - February 1

Recall that if R is a commutative ring, and S is a multiplicatively closed set, then we use RS to denote thering of fractions.

Recall that we, without loss of generality, let 1 ∈ S.

Remark 82. We have a natural map φ : R→ RS by r 7→ r1 .

Note that this map may have nontrivial kernel, as kerφ = {r ∈ R| there exists t ∈ S such that Tr = 0}.Therefore φ is injective if and only if S contains no zero divisors.

Remark 83. All of these constructions work in the noncommutative case, so long as S ⊂ Z(R).

Proposition 52. Let R be a ring, and let S be a multiplicatively closed set in R. Let f : R→ T be a ringhomomorphism such that f(s) is a unit for all s ∈ S.

Then there exists a unique ring homomorphism ψ : RS → T such that the ψ ◦ φ = f , where φ is thenatural map as before.

Proof. “Define” ψ( rs ) = f(r)f(s)−1. We wish to show that this is well defined.Suppose r1

s1= r2

s2. Then there exists t ∈ s such that ts2r1 = ts1r2. Then f(t)f(s2)f(r1) = f(t)f(s1)f(r2).

By multiplying both sides by f(t)−1f(s1)−1f(s2)−1, then we get that f(r1)f(s1)−1 = f(r2)f(s2)−1. Thusψ is well defined.

One can then verify that ψ is a ring homomorphism.Suppose that another ψ′ meets the required properties. Then for all r ∈ R and s ∈ S,

f(r) = ψ′(r

1)

= ψ′(s

1· rs

)

= ψ′(s

1) · ψ′(r

s)

= f(s)ψ′(r

s)

Since f(s) is a unit in T , then ψ′( rs ) = f(r)f(s)−1 = f(r)f(s) . That is, ψ′ = ψ. This completes the proof of

uniqueness.

Exercise 24. Suppose g : R→ A is a ring homomorphism with the following propoerty: given any f : R→ Tsuch that f(s) is a unit for all s ∈ S, there exists a unique homomorphism ψ : A→ T such that f = ψ ◦ g.Then there exists a unique isomorphism h : RS → A such that g = φ ◦ h.

Remark 84. We use the following notation:If f(R)→ T is a ring homomorphism, then T is naturally an R-module via r · t := f(r)t. If I is an ideal

of R, then IT = {∑iktk|ik ∈ I, tk ∈ T} is an ideal of T . In particular, IT is the ideal of T generated by

f(I).

Proposition 53. Let S be a multiplicatively closed set of R. Let I be an ideal of R, and let φ : R → RSbe the canonical map. Define IS := { is |i ∈ I, s ∈ S}. Then

1. IS = IRS (so IS is an ideal of RS)

2. Every ideal of RS is of the form IS , for some I CR

3. IS = RS if and only if I ∩ S 6= ∅.

Proof. (Proof of (1)) Let is ∈ IS , wher i ∈ IS and s ∈ S. Then i

s = i · 1s . Since i ∈ I and 1

s ∈ RS , thenis ∈ IRS . Thus IS ⊂ IRS .

83

Let∑

ikrksk∈ IRS , where each ik ∈ I, each rk ∈ R, and each sk ∈ S. Without loss of generality, we can

change everything to a common denominator, so si = sj for all i and j. Then we will call all denominators

s. Then∑

ikrksk

=

∑ikrks

∈ IS . Thus IS = IRS .

(Proof of (2)) Let J be an ideal of RS . Let I = φ−1(J) = {a ∈ R|a1 ∈ J}. We wish to show that IS = J .

If is ∈ IS , where i ∈ I and s ∈ S, then i

1 ∈ J , so is = 1

si1 ∈ J . Thus IS ⊂ J . Conversely, let r

s ∈ J . Thenr1 = s

1rs ∈ J . Therefore r ∈ I, so r

s ∈ IS . Thus IS = J .

(Proof of (3)) Suppose IS = RS . Then 11 ∈ IS , so 1

1 = is for some i ∈ I and s ∈ S. Therefore there exists

a t ∈ S such that ti = ts. Then ti ∈ I and ts ∈ S, so ti ∈ I ∩ S. Thus I ∩ S 6= ∅. Conversely, if i ∈ I ∩ S,then 1 = i

i ∈ IS , so IS = RS .

Example 80. Ideals of RS are always of the form IS , but the ideal I which makes this happen might notbe unique. For instance, if R = S = Z, then RS ∼= {0}, and every ideal in R maps down to the same idealin RS .

Corollary 32. If R is Noetherian, so is RS for any multiplicatively closed set S.

Proof. Let J be in ideal of R. Then J = IS = IRS for some ideal I of R. Since R is Noetherian, thenI = (a1, ..., an)R. Therefore J = (a1, ..., an)RS = (a11 , ...,

an1 ). Thus J is finitely generated.

Exercise 25. If R is Artinian, so is RS for all multiplicatively closed sets S.

Exercise 26. If S consists of units, then the map R→ RS by r 7→ r1 .is an isomorphism.

Proposition 54. Let R be a commutative ring, let S be a multiplicatively closed set, and let I be an idealin R. Let S = {s+I|s ∈ S}. Then S is a multiplicatively closed set of R/I. Then the map RS/IS → (R/I)Sby r

s 7→rs is an isomorphism.

Proof. Let π : R → R/I by r 7→ r, and let φ : R/I → (R/I)S . Then, by composition, f = φ ◦ π is a ringhomomorphism, and f(s) = s is a unit for all s ∈ S. x‘ By the universal property that we proved earlier,

there exists a ψ : RS → (R/I)S where ψ( rs ) = f(r)f(s) = r

s . Furthermore, ψ is a surjective ring homomorphism.

Then one can show that kerψ = IS , so by the first isomorphism theorem, we know that RS/IS ∼= (R/I)S .


Recall the following notation: if R is a ring, and S is multiplicatively closed set, then RS is R localized atS, and φ : R→ RS is the natural map given by r 7→ r

1 .

Proposition 55. Let J be an ideal of RS . Then φ−1(J)S = J .

Proof. Let rs ∈ J . Then r

1 ∈ J , so r ∈ φ−1(J), so rs ∈ φ

−1(J)S . Thus J ⊂ φ−1(J)S .Conversely, if r

s ∈ φ−1(J)S , then without loss of generality we can assume r ∈ φ−1(J). Then r

1 ∈ J , sors ∈ J . Thus J = φ−1(J)S .

Remark 85. Let q ∈ SpecRS . Then φ−1(q) ∈ SpecR.

Proposition 56. With the notation as above, there is an inclusion-preserving bijection between {p ∈SpecR|p ∩ S = ∅} ↔ SpecRS given by p 7→ pS and q 7→ φ−1(q).

Proof. If q ∈ SpecRS , then φ−1(q) ∈ SpecR by the second remark. Convervsely, if φ−1(q) ∩ S 6= ∅, lets ∈ φ−1(q) ∩ S. Then s

1 ∈ q, so q = RS , as s1 is a unit. Thus prime ideals (and only prime ideals) in RS are

taken to prime ideals in R which are disjoint from S.Suppose instead that p ∈ SpecR, and p∩S = ∅. Notice that RS/pS ∼= (R/p)S , where S = {s+ p|s ∈ S}.

As P ∩ S = ∅, then 0 6∈ S.Since R/p is a domain, then (R/p)S is a subring of the field of fractions of R/p. Hence (R/p)S is a

domain, so pS is a prime ideal.It follows directly from the definition that these maps are inclusion preserving.

84

Also, we proved in the previous proposition that φ−1(q)S = q. It therefore suffices to show that φ−1(pS) =p. To this end, let a ∈ p. Then a

1 ∈ pS , so a ∈ φ−1(pS). Now let b ∈ φ−1(pS). Then b1 ∈ pS , so b

1 = as

for some a ∈ p, s ∈ S. Therefore there exists t ∈ S such that tsb = ta ∈ p. Since ts ∈ S, then ts 6∈ p, sob ∈ p.

Remark 86. This map above, given by p 7→ pS is a homeomorphism (for the Zariski topology).

Remark 87. We have a special case when S = R \ p, where p is a prime ideal. Recall that Rp denotes RS .Then q ∩ S = ∅ if and only if q ⊂ p. Then SpecRp = {qp = qRp|q ∈ SpecR, q ⊂ p}. Thus Rp has a uniquemaximal ideal, namely pp = pRS .

Furthermore, Rp/pp ∼= (R/p)p = (R/p)(0). This is the field of fractions of R/p.

Definition 81. A quasi-local ring is a commutative ring R with a unique maximal ideal m. To emphasizethis, we write it as (R,m).

The residue field of R is R/m.A local ring is a ring which is both Noetherian and quasi-local.

Example 81. If R is Noetherian, then Rp is a local ring for all p ∈ SpecR.

Example 82. With R = Z and p = (2), then SpecZ(2) = {(0)(2), (2)(2)}. Therefore Z(2) is a local ring.

Definition 82. (Localization of a module) Let M be an R-module, and let S be a multiplicatively closedsubset of R. Let Λ = {(m, s)|m ∈M, s ∈ S}. Define a relation ∼ on Λ by (m1, s1) ∼ (m2, s2) if there existsa t ∈ S such that t(s2m1 − s1m2) = 0.

By the same proofs as before, ∼ is an equivalence relation. Let ms denote the equivalence class of (m, s).

Then let MS (sometimes instead denoted by S−1M be the set {ms |m ∈M, s ∈ S}.We then define m1

s1+ m2

s2= s2m1+s1m2

s1s2, and r1

s1· ms = r1m

s1s. These are well-defined, thereby making MS

and RS-module.

Proposition 57. Let M be an R-module. Then the following are equivalent:

1. M = 0

2. Mp = 0 for all p ∈ SpecR

3. Mm = 0 for all maximal ideals m.

Definition 83. We will often use colon-ideals: if I and J are ideals of R, then (J :R I) = {r ∈ R|rI ⊂ J}.It is a theorem that (J :R I) is an ideal of R containing J .

Proof. (1⇒ 2, 2⇒ 3) Certainly, if M = 0, then Mp = 0 for all ideals p, including the prime ones. This (1)implies (2). Since every maximal ideal is prime, then (2) implies (3).

It then suffices to show that (3) implies (1). To this end, suppose Mm = 0 for all maximal ideals m. Letx ∈ M . Then whenever m is a maximal ideal of R, x

1 ∈ Mm = { 01}, so there exists a tm ∈ R \ m such that

tmx = 0.But AnnR(x) = {r ∈ R|rx = 0} is an ideal of R. Therefore tm ∈ AnnR(x) for all m, so AnnR(x) 6⊂ m

for all m. But the only ideal not contained in any maximal ideal is R itself. Thus AnnR(x) = R, so x = 0.Thus M = 0.

Theorem 74 (Nakayama’s Lemma). Let (R,m) be a quasi-local ring. Then m is the Jacobson radicalof R. Let M be a finite generated R-module. Then the following are all referred to as Nakayama’s Lemma:

1. M = 0 if and only if M = mM .

2. If N ⊂M then M = N if and only if M = N + mM .

85

3. For x1, ..., xn ∈ M , we have that x1, ..., xn generate M if and only if x1, ..., xn spans M/mM as anR/m-vector space.

4. M = Rx1 + ...Rxn if and only if {x1, ..., xn} contains a basis for M/mM . Therefore µR(M) =dimR/mM/mM . (Recall that µR(M) denotes the minimal number of generators for M as an R-module.)

Definition 84. We say {x1, ..., xn} is a minimal generating set for M if M = Rx1 + ... + Rxn and n =µR(M).

Proposition 58. Let (R,m) be a quasi-local ring, and let P be a finitely-generated projective module. ThenP is a free R-module (that is, P ∼= Rn for some n ∈ N).

Proof. Since P is finitely generated, then let n = µR(P ) = dimR/m(P/mP ). Let x1, ..., xn be a minimalgenerating set for P .

Define φ : Rn → P in the natural way. That is, φ : ei 7→ xi. Certainly, this is a surjective ringhomomorphism.

Let K = kerφ. Then we get a short exact sequence

0→ K → Rnφ→ P → 0

Then by a theorem about projective modules, the sequence splits. Thus Rn ∼= P ⊕K. Then Rn/mRn ∼=(P ⊕ K)/m(P ⊕ K). Therefore (R/m)n = P/mP ⊕ K/mK. By comparing dimensions, we have thatdimK/mK = 0, so K = mK, so K = 0 by Nakayama’s Lemma.


When life gives you Lemmas, make Lemma-nade!(All of these lemmas will be stated without proof, but are easy to prove. Also, they will be stated for

modules, but they can also be applied directly to rings.)

Lemma 27. Let S be a multiplicatively closed set of a commutative ring R. Then

1.

(⊕α

Mα

)S

∼=⊕

α(Mα)S (as RS-modules).

2. Suppose S ⊂ T , where T is also a multiplicatively closed set. Then (MS)TS

∼= MT as RT -modules

(where TS = { ts |t ∈ T, s ∈ S}.

3. For a, b ∈ R, Mab∼= (Ma) b

1as Rab-modules.

4. For a ∈ R, Ma∼= Man . (This could be considered a corollary to the previous result.)

5. Let q ⊂ p be prime ideals. Then R \ p ⊂ R \ q, so (Mp)qp∼= Mq.

This lemma is pretty important:

Lemma 28. Let M be an R-module, and let S be a multiplicatively closed set. Then,

1. If AnnRM ∩ S 6= ∅, then MS = 0.

2. If M is finitely generated, then the converse to 1 holds. That is, if MS = 0, then AnnRM ∩ S 6= ∅.

86

Proof. Suppose AnnRM ∩ S 6= ∅. Then there exists some s ∈ AnnRM ∩ S. That is, there exists an s ∈ Ssuch that sM = 0. Then s ·m · 1 = s · 0 · t for all m ∈M and t ∈ S, so m

t = 01 . Thus MS = 0.

Conversely, if M is finitely generated, we can write M = Rx1 + ...+ Rxn. Then xi1 = 0 in MS for all i.

Therefore, for each xi, there exists si ∈ S, such that sixi = 0. Let s = s1...sn. Then s ∈ S and sxi = 0 forall i. Therefore sM = 0.

Let’s look at some not-finitely-generated examples where this fails.

Example 83. Let R = Z, and let M =⊕n≥1

Z/(2n). Then AnnZM = 0, but M2 = 0.

Example 84. Let R = k[x] (for some field k). Then let M = k[x, x−1]/k[x]. Then AnnRM = 0, butMx = 0.

Proposition 59. Let R be a commutative ring, S be a multiplicatively closed subset, and let P be aprojective R-module. Then PS is a projective RS-module.

Proof. Since P is projective, there exists an R-module such thatQ such that P⊕Q ∼=∑α

R. Then PS⊕QS ∼=

⊕αRS . Therefore PS is a direct summand of a free RS-module. In other words, PS is a projective RS-module.

Note that usually PS will not be a projective R-module.

Corollary 33. Let P be a finitely generated projective R-module. Then for all q ∈ SpecR, Pq is a freeRq-module.

Proof. By the previous proposition, Pq is a finitely generated Rq-module. But finitely generated projectivesover quasi-local rings are always free.

Remark 88. If F is a finitely generated free R-module, then F ∼= Rn for a unique number n. We call n therank of F , and write n = rkF .

Combining these two results, for each finitely generated projective P , we have a function fP : SpecR→ N0

by q 7→ rkPq.

Remark 89. If f : M → N is an R-homomorphism and S is a multiplicatively closed set, then f1 : MS → NS

defined by f1 (ms ) = f(m)

s is an RS-homomorphism.

Note also that f1 ◦

g1 = fg

1 . Also note that 1M1 = 1MS

.These properties combined make localization a functor.

Proposition 60. (Localization is exact) If Lf→ M

g→ N is exact, then so is LSf1→ MS

g1→ NS (where R is

a commutative ring, S is a multiplicatively closed subset of R, and L,M, and N are R-modules).

Proof. We know that gf = 0, so g

1 ◦f1 = gf

1 = 0. Thus im g1 ⊂ ker g1 .

Suppose ms ∈ ker g1 . That is, g(m)

s = 0, so there exists a t ∈ S such that tg(m) = 0. Since g is anR-module homomorphism, then g(tm) = 0. Thus tm ∈ ker g = im f . That is, there exists an l ∈ L such that

f(l) = tm. Then f1 ( lst ) = f(l)

st = tmst = m

s . Hence im f1 ⊃ ker g1 .

Thus im f1 = ker g1 , so the sequence is exact.

Remark 90. Recall that we use V (I) to denote the set {p ∈ SpecR|p ⊃ I}. Then every closed set is V (I)for some I, but we don’t have a similar notation for open sets.

We can (partially) fix that: for a ∈ R, let D(a) = SpecR \ V ((a)) = {p ∈ SpecR|a 6∈ p}. Then D(a) isopen.

Theorem 75. Let P be a finitely generated projective R-module. Give N0 the discrete topology. Then themap f : SpecR→ N0 by q 7→ rkPq is continuous.

87

Proof. Since N0 has the discrete topology, then the singletons are a basis for the open sets. Thereforecontinuity is equivalent to the preimage of every singleton being open.

Let n ∈ N0. If f−1(n) = ∅, then this is open, so suppose instead that f−1(n) 6= ∅. Then let q ∈ f−1(n).We’ll show that there exists an a ∈ R such that q ∈ D(a) ⊂ f−1(n).

Since rkPq = n, then Pq∼= Rnq . Then let {u1

s1, ..., unsn } be an Rq-basis for Pq. Then each si 6∈ q for all i.

However, bases are invariant under multiplication by unit scalars, then {u1

1 , ...,un1 } is also a basis.

Define φ : Rn → P by ei → ui. Then φ1 : Rnq → Pq is an isomorphism. Let K = kerφ, and let

C = cokerφ = P/ imφ.Since φ

1 is exact, then Kq = Cq = 0. Since P is finitely generated, then so is C. By the lemma, we knowthat AnnR C ∩R \ q 6= ∅. That is, there exists a b 6∈ q such that bC = 0. Thus Cb = 0.

[Proof to be finished later.]


Recall from last class that we had started the following proof:

Theorem 76. Let P be a finitely generated projective R-module. Give N0 the discrete topology. Then themap f : SpecR→ N0 by q 7→ rkPq is continuous.

Proof. Since N0 has the discrete topology, then the singletons are a basis for the open sets. Thereforecontinuity is equivalent to the preimage of every singleton being open.

Let n ∈ N0. If f−1(n) = ∅, then this is open, so suppose instead that f−1(n) 6= ∅. Then let q ∈ f−1(n).We’ll show that there exists an a ∈ R such that q ∈ D(a) ⊂ f−1(n).

Since rkPq = n, then Pq∼= Rnq . Then let {u1

s1, ..., unsn } be an Rq-basis for Pq. Then each si 6∈ q for all i.

However, bases are invariant under multiplication by unit scalars, then {u1

1 , ...,un1 } is also a basis.

Define φ : Rn → P by ei → ui. Then φ1 : Rnq → Pq is an isomorphism. Let K = kerφ, and let

C = cokerφ = P/ imφ.

That is, we have the exact sequence 0→ K → Rnφ→ P → C → 0.

Since φ1 is exact, then Kq = Cq = 0. Since P is finitely generated, then so is C. By the lemma, we know

that AnnR C ∩R \ q 6= ∅. That is, there exists a b 6∈ q such that bC = 0. Thus Cb = 0.Then by localizing the above exact sequence, we get that 0→ Kb → Rnb → Pb → 0→ 0.[This is where we stopped last time.]From a theorem from last class, since P is projective as an R-module, then Pb is also projective as an

Rb-module.Therefore Kb is finitely generated. Note that (Kb)qb

∼= Kq = 0. Then, since Kb is finitely generated,then by the lemma there exists a c

bn ∈ Rb \ qb such that cbnKb = 0. Thus c

1Kb = 0.Thus (Kb) c1 = 0. But since localization is associative, we know that Kbc = 0. Let a = bc. Since b 6∈ q

and c 6∈ q, and q is a prime ideal, then a 6∈ q. Therefore Ka = 0 and Ca = (Cb) c1 = 0.Suppose p ∈ D(a). Then pa ∈ SpecRa, so Pp

∼= (Pa)pa . Since Ka = Ca = 0, then the exact sequencelocalized at a is 0 → Rna → Pa →. Thus Pa ∼= Rna . Substituting this in, we get that Pp

∼= (Rna )pa∼= Rnp .

Thus rkPp = n.Therefore D(a) ⊂ f−1({n}), so f is continuous.

Definition 85. Let P be a finitely generated projective R-module. Then P is said to be of constant rankif rkPq = rkPp for all p, q ∈ SpecR.

Theorem 77. Let R be a commutative ring. Then every finitely generated projective R-module has constantrank if and only if R has no nontrivial idempotents. (And as we showed before, having no nontrivialidempotents is equivalent to SpecR being connected.)

Proof. Suppose R has no nontrivial idempotents. Then as we showed before, SpecR is connected. Let P bea finitely generated projective R-module.

As before, let f : SpecR → N0 by q 7→ rkPq. Suppose n ∈ im f . Then since N0 has the discretetopology, then {n} is both closed and open. Thus f−1({n}) is both closed and open. But since SpecR isconnected, then either f−1({n}) is either SpecR or the empty set. But since n ∈ im f , it is nonempty. Thusf−1({n}) = SpecR. That is, f is constant, as desired.

88

Conversely, suppose there exist a nontrivial idempotent e ∈ R. Let I = (e) and J = (1 − e). Note thatR = I ⊕ J . Also, I and J are finitely generated (they are generated by a single element), and they areprojectives.

Extend I to a maximal ideal q ∈ SpecR. Then (1− e) 6∈ q. Furthermore, (1− e)I = 0 since (1− e)e = 0.Therefore Iq = 0. Thus f(q) = 0. Now extend J to a maximal ideal p. Then I 6⊂ p, so Ip = Rp, so f(q) > 0.Thus f is not constant.

4 Tensor Products

Now let’s talk about tensor products.

Definition 86. Let R be a ring with unit (not necessarily commutative). Let M be a right R-module, letN be a left R-module, and let A be an abelian group.

A function f : M × N → A is called R-biadditive if for all m,m1,m2 ∈ M , n, n1, n2 ∈ N , and r ∈ R,then we have that

1. f(m,n1 + n2) = f(m,n1) + f(m,n2)

2. f(m1 +m2, n) = f(m1, n) + f(m2, n)

3. f(mr, n) = f(m, rn)

Example 85. One example is f : (R×R)→ R by (r, s) 7→ rs.Another example is f : R2 ×M →M2 by ((r, s),m) 7→ (rm, sm).A third example works for any right ideal I. Let f : R/I ×M →M/IM by (r,m)→ rm.Another example works for any multiplicatively closed set S which is in the center of R. (Note that we

haven’t actually defined how to localize in a non-commutative ring, but one can image that it is doable.)Define f : RS ×M →Ms by ( rs ,m) 7→ rm

s .

We now define tensor products using a universal property.

Definition 87. Let R,M, and N be as above. An abelian group M ⊗R N together with an R-biadditivemap h : M ×N →M ⊗R N is called the tensor product of M and N if it has the following property (whichwe call the universal property): For any abelian group A and R-biadditive map f : M ×N → A, there existsa unique group homomorphism g : M ⊗R N such that f = g ◦ h.

Remark 91. The tensor product (if it exists) is unique, up to isomorphism.We can show this by supposing there exist two “tensor products” T1 with map h1 and T2 with map h2.

Then since h1 and h2 are biadditive maps, there exists φ : T1 → T2 such that h2 ◦ φ = h1 and h1 ◦ ψ = h2.Then h1 ◦ idT1 = h1 ◦ ψ ◦ φ. But we assumed there was a unique group homomorphism which makes thediagram commute, so idT1 = ψ ◦ φ. Thus T1

∼= T2.


Recall that we “defined” the tensor product last class in the following way:

Definition 88. Let R be a commutative ring, let M be a right R-module and let N be a left R-module.An abelian group M ⊗R N together with an R-biadditive map h : M ×N → M ⊗R N is called the tensorproduct of M and N if it has the following property (which we call the universal property): For any abeliangroup A and R-biadditive map f : M × N → A, there exists a unique group homomorphism g : M ⊗R Nsuch that f = g ◦ h.

We showed last time that M ⊗R N is unique (up to isomorphism), if it exists.We will now show existence:

Proposition 61. With R, M , and N , as above, M ⊗R N exists.

89

Proof. Let F be the free abelian groupwith basis of cardinality |M×N |. That is, we write F =⊕

(m,n)∈M×N

Z.

Let [m,n] denote the basis element of F corresponding to (m,n).

Then every element of F can be uniquely written in the form∑finite

ri[mi, ni] where ri ∈ Z.

Let S be the submodule of F generated by all elements of the form

1. [m,n1 + n2]− [m,n1]− [m,n2] for any m ∈M and n1, n2 ∈ N .

2. [m1 +m2, n]− [m1, n]− [m2, n] for any m1,m2 ∈M and n ∈ N .

3. [mr, n]− [m, rn] for any m ∈M , n ∈ n, and r ∈ R.

Then let M ⊗R N = F/S. For m ∈ M and n ∈ N , let m ⊗ n denote [m,n] + S ∈ F/S/ Hence, every

element of M ⊗R N is of the form∑finite

ri(mi ⊗ ni) for ri ∈ Z. Note that there may be more than one way

to write an element now.Note also that

1. m⊗ (n1 + n2) = m⊗ n1 +m⊗ n2 for all m ∈M , n1, n2 ∈ N .

2. (m1 +m2)⊗ n = m1 ⊗ n+m2 ⊗ n for all m1,m2 ∈M and n ∈ N .

3. (mr)⊗ n = m⊗ (rn) for all m ∈M , n ∈ N and r ∈ R.

These follow from the elements generating our submodule.Finally, define h : M × N → M ⊗R N by (m,n) → m ⊗ n. By our previous observations, this map is

R-biadditive.Finally, we must show that M ⊗R N satisfies the universal property. Suppose f : M ⊗ N → A is any

R-biadditive function.Define a group homomorphism g : F → A by [m,n] → f(m,n). Since f is R-biadditive, then each of

the generators for S live in the kernel. Hence S ⊂ ker g. Therefore we get an induced group homomorphismg : F/S → A by m⊗ n 7→ f(m,n) (and this is well-defined).

But then f(m,n) = g(m⊗ n) = g(h((m,n)), so f = g ◦ h.It then suffices to show that this is the unique group homomorphism g for which this is true. To this

end, suppose g1 : M ⊗RN → A is a group homomorphism such that g1 ◦ h = f . Then g1(m⊗ n) = f(m,n),so g1 = g on all generators of M ⊗R N . Thus g1 = g, so g is unique.

Example 86. Consider Z⊗Z Z/(2). Consider the element 1⊗ 1.

Definition 89. Let R,S be rings. An R−S bimodule is an abelian group M such that M is a left R-moduleand a right S-module, and for all r ∈ R, s ∈ S, and m ∈M , we have that (rm)s = r(ms).

Example 87. Let k be a field, and let n and m be positive integers. Then let R = Mm(k), S = Mn(k), andM = Mm×n(k). Then M is an R-S bimodule.

Example 88. Any ring R is an R-R bimodule.

Example 89. Let R be a commutative ring. Let M be a (left) R-module. Then M is an R-R bimodule,where the action on the right is defined to be the same as the action on the left.

Example 90. Suppose R is a commutative ring, M is an R-module, and I is an ideal such that IM = 0.Then M is an R/I-module. That is, M is an R/I-R bimodule (or and R-R/I bimodule).

Example 91. Let R be a commutative ring, with Char R = p (a prime). Let M be a left R-module.We can consider M as an R − R bimodule via the actions r ·m = rm, m · r = rpm. This makes it and

R-R bimodule. This is called the Frobenius bimodule structure.

Proposition 62. Let M be an S-R bimodule and N a left R-module. (so N is an R-Z bimodule). ThenM ⊗R N is a left S-module by the action s · (m⊗ n) = (sm)⊗ n.

90

Proof. We first need to verify that this is well-defined. Fix an s ∈ S. Define a map fs : M×N →M⊗RN bym,n 7→ (sm)⊗ n. One can check that fs is R-biadditive. Then there exists a unique group homomorphismfs : M ⊗R N →M ⊗R N by m× n→ (sm)⊗ n. Hence the multiplication is well-defined!

It is then easy to see that the rest of the module properties hold.


Recall from last time that if M is an S − R bimodule, and N is a left R-module, then M ⊗R N is a leftS-module via the action s(m⊗ n) = (sm)⊗ n.

Similarly, if N is an R− S bimodule, and M is a right R-module, then M ⊗R N is a right S-module viathe action (m⊗ n)s = m⊗ (ns).

Remark 92. We can use this property of tensor products for a “base change”. If φ : R → S is a ringhomomorphismm, then we can give S an S −R bimodule structure by the standard method: S acts on theleft my multiplication, and r ∈ R acts on the right by multiplication by φ(r). Then if M is a left R-module,then S ⊗RM is a left S-module via s′(s⊗m) = (s′s)⊗m.

Example 92. Suppose H ≤ G are are groups, and k is a field. Then k[H] ↪→ k[G] is a ring homomorphism.Given a left k[H]-module M , k[G]⊗k[H] M is a left k[G]-module.

Proposition 63. Let R be a ring. Then

1. If M is a left R-module, then R⊗RM ∼= M as left R-modules (with the map r ⊗m 7→ rm).

2. Let I be a two-sided ideal of R, with M a left R-module. Then R/I⊗RM ∼= M/IM as left R/I-modules(with the map r ⊗m 7→ rm).

3. Let S a multiplicatively closed subset of Z(R), and let M be a left R-module. Then RS ⊗RM ∼= MS

as left R-modules (with the map rs 7→

rms ).

4. If R is commutative, and M and N are R-modules, then M ⊗R N ∼= N ⊗RM as R-modules (with themap m⊗ n→ n⊗m).

Proof. For all of these, we define homomorphisms in both directions, and verify that they are inverses ofeach other.

(Proof of (1)) Define f : R ×M → M by (r,m) 7→ rm. Then f is R-biadditive, so by the universalproperty there is a unique group homomorphism f : R ⊗R M → M by r ⊗ m 7→ rm. Furthermore,f(r′(r ⊗m)) = f((r′r)⊗m) = (rr′)m = r′(rm) = r′f(r ⊗m). Thus f is an R-module homomorphism.

Now define g : M → R⊗RM by m 7→ 1⊗m. Then g(rm) = 1⊗ (rm) = 1 · r ⊗m = r(1⊗m) = rg(m).One can verify that gf and fg are the identity, and this completes the proof.(Proof of part of (2)) Define g : M → R/I ⊗M by m 7→ 1⊗m. Then one can see that g is an R-module

homomorphism.For all i ∈ I and m ∈M , g(im) = 1⊗ im = i×m = 0×m = 0. Therefore IM ⊂ ker g.This is the only hard part of the proof, and the rest is left as an exercise to the reader.The proof of (3) is a homework problem.(Proof of part of (4)) Define f : M×N → N⊗RM by (m,n)→ n⊗m (note that this is heavily dependent

on R being commutative, or this would make no sense). Then we can get that f is an R-biadditive grouphomomorphism. Then there exists a unique R-module homomorphism f : M⊗N → N⊗M by m⊗n→ n⊗m.

Similary, one creates a g : N ×M →M ⊗R N by (n,m)→ m⊗ n, and once gets an g. Then fg and gfare the identities on the appropriate things. Thus M ⊗R N ∼= N ⊗RM .

Proposition 64. Let M be an R-S bimodule, let N be an S − T bimodule, and let P be a left T -module.Then (M ⊗S N)⊗T P ∼= M ⊗S (N ⊗T P ) as left R-modules.

Omitted.

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Remark 93. Here is an application of tensor products. Let R be a commutative ring, and let M and N beR-modules. Let I ⊂ R such that IM = 0 (this implies that M is naturally and R/I-module).

Then

M ⊗N ∼= (M ⊗R/I R/I)⊗R N∼= (M ⊗R/I (R/I ⊗R N)∼= M ⊗R/I N/IN

As R/I-modules. If IN = 0 as well, then M ⊗R N ∼= M ⊗R/I N .

This leads to the following theorem:

Theorem 78. If F and G are free R-modules, with bases {fi}, {gj} respectively, then F ⊗R G is free, anda basis for it is {ei ⊗ fj}.


Remark 94. Let R be a ring, and let f : M → A and g : N → B be homomorphisms of right R-modulesand left R-modules, respectively. Then we can define f × g : M ×N → A⊗R B by (m,n) 7→ f(m)⊗ g(n).

Since f and g are homomorphisms, then f × g is is R-biadditive. Then by the universal property, thereexists a unique group homomorphism f ⊗ g : M ⊗R N → A⊗R B by m⊗ n 7→ f(m)⊗ g(n).

Remark 95. We can expand the previous example to give a module homomorphism under the followingconditions: If M and A are S − R bimodules, and f is an S − R bimodule homomorphism, then f ⊗ g willbe an S-module homomorphism.

Remark 96. We also get a nice result if f and g are isomorphisms. Namely, if f and g are isomorphisms,then f ⊗ g is an isomorphism, since (f ⊗ g)−1 = f−1 ⊗ g−1. [This makes use of the easily-verified fact that(f ⊗ g) ◦ (h⊗ l) = fh⊗ gl.]

Hence if M ∼= A and N ∼= B, then M ⊗R N ∼= A⊗R B.

Remark 97. Let R be a commutative ring, and let M and N be R-modules. If {mα} is a generating setfor M , and {nβ} is a generating set for N , then {mα ⊗ nβ} is a generating set for M ⊗R N .

Proposition 65. LetR be a commutative ring, and let F andG be freeR-modules with bases {fα} and {gβ},respectively. Then F⊗RG is a free R-module with basis {fα⊗gβ}. (So rank(F⊗RG) = rank(F )·rank(G).)

Proof. By the previous remark, we know that {fα ⊗ gβ} generates F ⊗R G. It then suffices to show linearindependence.

Suppose∑

rα,βfα ⊗ gβ = 0. Fix an i ∈ I and j ∈ J . We wish to show ri,j = 0.

Define φ : F → R by fα 7→

{1 if α = i

0 otherwise. Similarly, define ψ : G→ R by gβ 7→

{1 if β = j

0 otherwise.

Then we get an R-module homomorphism F ⊗R G → R ⊗ RR∼=→ R by fα ⊗ gβ 7→ φ(fα) ⊗ φ(gβ) 7→

φ(fα)φ(gβ). In other words, fα ⊗ gβ 7→

{1 if α = i, β = j

0 otherwise.

Then 0 = (φ⊗ψ)(0) = (φ⊗ψ)(∑

rα,βfα⊗gβ) = ri,j . Thus 0 = ri,j , so these are linearly independent.

Corollary 34. Suppose K is a field, and V and W are k-vector spaces. Then dimk V ⊗k W = (dimk V )⊗(dimkW ). In particular, V ⊗k W = 0 if and only if V = 0 or W = 0.

Proposition 66. If R is a commutative ring, and I and J are ideals, then R/I ⊗R R/J ∼= R/(I + J).In particular, if I = m1 and J = m2, a pair of distinct maximal ideals, then R/I 6= 0 and R/J 6= 0, byR/I ⊗R R/J = 0.

Proposition 67. Let A be a right R-module, and {βα}α∈I be a collection of left R-modules. Then A ⊗R(⊕

α βα) ∼=⊕

α(A⊗R βα).

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Proof. For i ∈ I, let ρi : βi →⊕

α βα be the inclusion map, and let Πi : βα → βi be the projection map.Define ψ : A×

⊕α βα → ⊕(A⊗R βα) by (a, v) 7→ (a⊗Πα(v). Then ψ is R-biadditive, so by the universal

property we get an induced map ψ : A⊗⊕Bαrightarrow ⊕ (A⊗R βα) by a⊗ v 7→ (a⊗Πα(v)).

Also, define φα : A×Bα → A⊗R (⊕Bα) by (a, b)→ a⊗ ρα(b).Then by the universal property, this gives a group homomorphism φα : A ⊗R Bα → A ⊗R (⊕Bα) by

a⊗ b→ a⊗ ρα(b). Define φ :⊕

(A⊗R Bα)⊕φα→ A⊗R (⊕Bα) by (uα)→ (φα(uα)).

Then one can check that φ ◦ ψ = 1 and ψ ◦ φ = 1, and this completes the proof.

Exercise 27. How many elements are in the Z-module Z/(100)⊗Z Z3 ⊗Z Z4 ⊗ Z/(150).First, we can recall that Z/(100)⊗Z Z/(150) ∼= Z/(50).Furthermore, Z/(50)⊗Z Z4

∼= Z4/(5)Z4∼= Z4/(25)Z4

∼= (Z/(25))4∼= Z/25.

Finally, by the previous theorem, Z/(25) ⊗Z Z3 ∼= (Z/(25) ⊗Z Z)3 ∼= (Z/(25))3. Therefore this modulehas 253 elements in it.


Great googly-moogly, it’s time for an application! This is apparently used for quantum physics.

Remark 98. Let R be a commutative ring, let A be an m × n matrix, and let B be a an r × s matrix.Define the tensor product (or Kronecker product) A⊗B to be the mr × ns matrix with the form:

a1,1B ... a1,nB

a2,1B . . ....

......

am,1B . . . am,nB

Let F1 and F2 be free modules of rank n and m, respectively. Fix bases β1 and β2 for F1 and F2. Let

β1 = {e1, ..., em} and β2 = {f1, ..., fn}. Define φ : F1 → F2 by ej 7→n∑i=1

ai,jfi. Thus [φ]β2

β1= A.

Let ψ : G1 → G2 be defined similarly with bases β′1 and β′2. Then as we showed in last class, φ ⊗ ψ :F1 ⊗G1 → F2 ⊗G2 is a well-defined function.

Furthermore, β1 ⊗ β′1 is a basis for F1 ⊗G1 and similarly β2 ⊗ β′2 is a basis for F2 ⊗G2. Order the baseslexicographically. Then A⊗B is the matrix for φ⊗ ψ with respect to these matrices.

5 Category Theory and Functors

Now let’s move on to the opposite of an application: categories!

Definition 90. A category C consists of the following:

1. A “class” of objects obj C

2. For any two objects A and B of C, we have a set of morphisms HomC(A,B) (we sometimes writef : A→ B instead of f ∈ HomC(A,B)). These morphisms are sometimes called arrows.

3. For objects A,B,C of C, there is a function ◦ : HomC(B × C) × HomC(A,B) → HomC(A,C). Wewrite (f, g) 7→ f ◦ g = fg, and we call this function composition. Composition satisfies the followingaxioms:

• For all appropriate f, g, h, (fg)h = f(gh).

• For each object A of C, there exists a 1A ∈ HomC(A,A) such that f1A = f and 1Ag = g for allf : A→ B and all g : C → A.

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Example 93. Let C denote the class of all sets, with morphisms being all functions. Then C is a category,and we denote it by 〈〈Sets〉〉.

Example 94. Let C denote the class of all groups, with morphisms being all group homomorphisms. ThenC is a category, and we denote it by 〈〈Groups〉〉.

Example 95. Let C denote the class of all topological spaces, with morphisms being all continuous functions.Then C is a category, and we denote it by 〈〈Top〉〉.

Example 96. Let R be a ring (not necessarily commutative) and let C denote the class of all left R-modules,with morphisms being all homomorphisms of left R-modules. Then C is a category, and we denote it by〈〈R−mod〉〉.

We also use 〈〈mod − R〉〉 to denote the category of right R-modules, and 〈〈S − R mod〉〉 to denote theset of S −R bimodules.

Example 97. Let C denote the class of all short exact sequences, with morphisms being all commuting stuff(chain complexes?). Then C is a category, and we denote it by 〈〈s.e.s.〉〉.

Definition 91. Let C and D be categories. A covariant functor F : C → D is the the following:

1. For each object A of C, there is a unique object, denoted F (A) in D.

2. For every pair of objects A,B of C, there is a function taking HomC(A,B)→ HomD(F (A), F (B)) byf 7→ F (f), such that F (1A) = 1F (A) and F (fg) = F (f)F (g) whenever fg makes sense.

A contravariant functor is the same thing, but we make two modifications. First, F : HomC(A,B) →HomD(F (B), F (A)) (i.e. we switched F (B) and F (A)). Second, F (fg) = F (g)F (f).

Example 98. Define F : 〈〈Groups〉〉 → 〈〈Sets〉〉 by taking the underlying set. Then F is a covariantfunctor, which we call the forgetful functor .

Define F : 〈〈S − R − bimod〉〉 → 〈〈Z − mods〉〉 by taking the underlying abelian group. Then F is acovariant functor, which we also call the forgetful functor.

Example 99. Define F : 〈〈comm rings〉〉 → 〈〈Top〉〉 by F (R) = SpecR. Then F is a contravariantfunctor.

Example 100. Let R be a commutative ring, and S a multiplicatively closed set. Then define F : 〈〈R −mod〉〉 → 〈〈RS − mod〉〉 by F (M) = MS . Then F is a covariant functor which we call the localizationfunctor .

Example 101. Let R be a commutative ring, and let I be an ideal. Then F : 〈〈R−mod〉〉 → 〈〈R/I−mod〉〉by F (M) = M/IM is a covariant functor.

Example 102. Let R be a ring, and let M be a right R-module. Define a functor F = M⊗− : 〈〈R−mod〉〉 →〈〈Z−mod〉〉 by F (N) = M ⊗R N . Then F is a covariant functor.

If N is an S−R bimodule, then F (N) is in the category of S-modules. In particular, if R is commutative,then our M ⊗− maps into the category of R-modules.

In fact, the previous two examples were secretly this one.


Definition 92. A functor of R-modules, F , is called additive if, for all f, g ∈ HomR(M,N), we have thatF (f + g) = F (f) + F (g).

Example 103. Let M be an S−R bimodule. Let F = M⊗R−. Then observe that F (f+g) = 1M⊗(f+g) =1M ⊗ f + 1M ⊗ g = F (f) + F (g). Therefore F is additive.

Definition 93. A functor of R-modules, F , is called multiplicative if, for all r ∈ R and R-modules N , wehave F (µr,N ) = µr,F (N). Where µr,N : N → N is given by n 7→ rn.

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Example 104. Let M be an R-module (where R is commutative). Then let F = M ⊗R −. Then µr,N :

Nr→ N , and 1M ⊗µr,N : M ⊗RN →M ⊗RN by m⊗n 7→ m⊗rn = mr⊗n = r(m⊗n) = µr,M⊗RN (m⊗n).

Thus F is multiplicative.

Definition 94. Let F be an additive covariant functor on module categories. We say F is exact if whenever

Af→ B

g→ C is exact at B, then F (A)F (f)→ F (B)

F (g)→ F (C) is exact at F (B).

Example 105. As we previously showed, the localization functor is exact.

Definition 95. Let F be an additive covariant functor on module categories. Then we say F is left exactif, whenever 0→ A→ B → C is exact, then 0→ F (A)→ F (B)→ F (C) is exact.

Similarly, we say F is right exact if, whenever A→ B → C → 0 is exact, then F (A)→ F (B)→ F (C)→ 0is exact.

Exercise 28. If F is an additive covariant functor on module categories, then F is exact if and only if F isboth left exact and right exact.

Theorem 79. Let M be an S −R bimodule. Then M ⊗R − : 〈〈R−mod〉〉 → 〈〈S −mod〉〉 is right exact.

Proof. Let Af→ B

g→ C → 0 be an exact sequence in 〈〈R −mod〉〉. Apply M ⊗R −, to get: M ⊗R A1⊗f→

M ⊗R B1⊗g→ M ⊗R C → 0.

First we will show that 1 ⊗ g is onto. Let m ⊗ c ∈ M ⊗R C. Since g is onto, then there exists a b ∈ Bsuch that g(b) = c. Then (1⊗ g)(m⊗ c) = m⊗ c. Thus 1⊗ g is onto, so the sequence is exact at M ⊗R C.

We now need to show that im 1⊗ f = ker 1⊗ g.Note that since F is additive, then F (0) = 0 (this is a statement about the zero map). If gf = 0, then

0 = F (gf) = F (g)F (f) (the last equality is due to the fundamental property about composing functors).Thus imF (g) ⊂ kerF (f). [Note that if we stop here, this shows that a functor F takes complexes tocomplexes.] But F (g) = 1M ⊗ g, and F (f) = 1M ⊗ f , so im 1⊗ f ⊂ ker 1⊗ g.

It then suffices to show that ker 1⊗ g ⊂ im 1⊗ f .Now define h : M × C → M ⊗ RB/ im(1 ⊗ f) by (m, c) 7→ m⊗ b, where b is any element such that

g(b) = c. We will first show that h is well defined. Suppose g(b1) = g(b2) = c. Then g(b1 − b2) = 0, sob1 − b2 ∈ ker g = im f . Let a ∈ A be an element such that f(a) = b1 − b2. Then m ⊗ b1 − m ⊗ b2 =m⊗ (b1 − b2) = m⊗ f(a) = (1⊗ f)(m⊗ a) ∈ im(1⊗ f). So m⊗ b1 = m⊗ b2 in M ⊗R B/ im(1⊗ f).

Furthermore, it follows immediately from the definition of h that h is R-biadditive. Therefore there is aninduced S-module homomorphism h : M ⊗R C →M ⊗RB/ im(1⊗ f) by m⊗ c 7→ m⊗ b (where g(b) = c).

Let x =

n∑i=1

mi ⊗ bi ∈M ⊗R B, and suppose x ∈ ker 1⊗ g. Then

0 = h(0)

= h((1⊗ g)(∑

mi ⊗ bi))

= h(∑

mi ⊗ g(bi))

=∑

h(mi ⊗ g(bi))

=∑

mi ⊗ bi

=∑

mi ⊗ bi= x

Therefore, x ∈ im(1⊗ f). That is, ker(1⊗ g) ⊂ im(1⊗ f).Thus ker(1⊗ g) = im(1⊗ f), so the sequence is exact at M ⊗R B, so it is exact.

Example 106. Consider the short exact sequence 0→ Z 2→ Z→ Z/2Z→ 0. Applying Z/2Z⊗Z −, we get

Z/2Z 0→ Z/2Z 1→ Z/2/Z → 0, which is no longer exact.

95

Definition 96. Let R be a commutative ring. Then an R-module M is called flat if M ⊗R − is an exactfunctor.

Example 107. Let R be a commutative ring. Then R is flat as an R-module. Also, RS is flat. Also, anyprojective R-module is flat.


Remark 99. If R is a commutative ring, and A ⊂ B and C ⊂ D are R-modules, then we know thatA× C ⊂ B ×D. We also know that A⊕ C ⊂ B ⊕D. However, it need not follow that A⊗R C ⊂ B ⊗R D.There certainly is a function form the former to the latter, but that function need not be injective, becausetensor products can be tricky.

Definition 97. Let R be a ring, and let M and N be left R-modules. Then we define HomR(M,N) to beHomR(M,N) = {f : M → N |f is a homomorphism of left R-modules}.

Remark 100. We always know that HomR(M,N) is an abelian group. Furthermore, if R is commutative,then HomR(M,N) is an R-module in the natural way:

For r ∈ R, and f : M → N , define rf : M → N by rf(m) = f(rm) = rf(m).More generally, suppose M is an R−S bimodule. Then HomR(M,N) has the structure of a left S-module.

That is, for s ∈ S and f : M → N , define sf : M → N by (sf)(m) = f(ms).Similarly, if N is an R − T bimodule, then HomR(M,N) is a right T -module. That is, for t ∈ T and

f : M → N , define (ft)(m) = f(m)t.

Remark 101. We like Hom because it lets us do a change of base. Let φ : R→ S be a ring homomorphism,and let N be a left R-module. Recall that S has the structure of an R− S bimodule.

Then HomR(S,N) is a left S-module.

Definition 98. Let M be an R−S bimodule. Then define a hom functor F = HomR(M,−) : 〈〈R−mod〉〉 →〈〈S −mod〉〉 by N 7→ HomR(M,N).

Remark 102. Let’s verify that F is indeed a functor. By the previous remarks, F does indeed take objectsfrom 〈〈R−mod〉〉 to 〈〈S −mod〉〉. Let’s check that F behaves well on arrows. If f : N1 → N2, then we havethat F (f) = f∗ : HomR(M,N1)→ HomR(M,N2) by g 7→ fg. Then one can verify that this satisfies all theother requirements of a functor.

Proposition 68. We have the following properties about the HomR(M,−) functor.

1. The HomR(M,−) functor is covariant.

2. The HomR(M,−) functor is additive.

3. If R is commutative, then the HomR(M,−) functor is multiplicative.

(Proof omitted.)

Definition 99. Let N be an R − T bimodule. Define a contravariant functor G = HomR(−, N) : 〈〈R −mod〉〉 → 〈〈mod− T 〉〉 by M 7→ HomR(M,N).

Remark 103. As before, we can verify that this is a contravariant functor.Also, G is additive, and if R is commutative then G is multiplicative.

Proposition 69. Let M be an R− S bimodule. Then HomR(M,−) is left exact.

Proof. Let 0 → Af→ B

g→ C be exact. We need to show that 0 → HomR(M,A)f∗→ HomR(M,B)

g∗→HomR(M,C) is exact.

First we will show that f∗ is injective. Suppose f∗(h) = 0. Then fh = 0. Since f is injective, then h = 0.We will now show that im f∗ ⊂ ker g∗. We can see that g∗f∗ = (gf∗) = (0)∗ = 0.

96

It then suffices to show that ker g∗ ⊂ im f∗. Let h ∈ ker g∗. Then gh = 0. Therefore imh ⊂ ker g = im f .Let m ∈ M . Then h(M) = f(am) for some am ∈ A. Since f is injective, then this am is unique. Define

α : M → A by m 7→ am. Then α ∈ HomR(M,A). Note that f∗(α) = h. In other words, fα = h. Henceh ∈ im f∗.

Definition 100. Let F be a contravariant functor (in 〈〈R − mod〉〉). Then F is sait to be left exact if,whenever A→ B → C → 0 is exact, then 0→ F (C)→ F (B)→ F (A) is exact.

Exercise 29. Let N be an R− T bimodule. Then HomR(−, N) is left exact.

Let’s list some properties of the Hom functors.

Proposition 70. Let M be a left R-module. Then HomR(R,M) ∼= M (as left R-modules).

Proof. Define φ : HomR(R,M) → M by f 7→ f(1). For m ∈ M , we have that f : R → M by r 7→ rm is inHomR(R,M). Therefore φ is surjective.

Also, if φ(f) = 0, then f(1) = 0, so f is the constant 0 map. Thus φ is surjective.Finally, φ(rf) = rf(1) = rφ(f). Thus φ respects multiplication by elements of R. Thus φ is an

isomorphism of left R-modules.

Proposition 71. Let R be a commutative ring. Let I be an ideal of R. Then HomR(R/I,M) ∼= (0 :M I).(Recall that (0 :M I) = {m ∈M |Im = 0}.

Proposition 72. If {Ai}ni=1 is a set of R − S bimodules, then HomR(

n⊕i=1

Ai, B) ∼=n⊕i=1

HomR(Ai, B) (as

left S-modules).

Proposition 73. If {Bi}ni=1 is a set of R − T bimodules, then HomR(A,

n⊕i=1

Bi) ∼=n⊕i=1

HomR(A,Bi) (as

right T -modules).

Example 108. If R is commutative, then HomR(Rn,M) ∼=n⊕i=1

HomR(R,M) ∼= Mn.

If R is commutative, HomR(Rn, Rm) ∼= (Rm)n ∼= Rmn.

Next time: Homs and Tensors and Lions, oh my!


Proposition 74. Let P be a left R-module. Then HomR(P,−) is exact if and only if P is projective.

Proof. Suppose P is projective. We showed last time that HomR(M,−) is always left exact. It then suffices

to show that if Mf→ N → 0 is exact, then HomR(P,M)

f∗→ HomR(P,N)→ 0 is also exact.That is, it suffices to show that f∗ is surjective. Let g ∈ HomR(P,N). That is, g : P → N is an R-module

homomorphism. Since P is projective, then by the definition of projective, there exists an h : P →M suchthat fh = g. Thus g = f∗(h), so f∗ is onto. Thus HomR(P,−) is exact.

Conversely, suppose HomR(P,−) is exact. Choose a free module F and a surjective R-module homo-morphism φ : F → P . Then

0→ Ki→ F

φ→ P → 0

is exact, where K = kerφ. Apply the functor HomR(P−) to this whole thing. Then we get that

0→ HomR(P,K)i∗→ HomR(P, F )

φ∗→ HomR(P, P )→ 0

is also exact. As φ∗ is onto and 1P ∈ HomR(P, P ), there exists a g : P → F such that φg = φ∗(g) = 1P .Therefore F ∼= P ⊕K. Since P is the summand of a free module, it is projective.

Recall that a right R-module F is flat if F ⊗R − is an exact functor.

97

Remark 104. Consider a commutative diagramA −−−−→ B −−−−→ Cy y yD −−−−→ E −−−−→ F

where the vertical arrows are isomorphisms. Then the top row is exact if and

only if the bottom row is exact.The proof of this is left as an exercise for the reader, but is fairly simple diagram chasing.

Proposition 75. Let R be a commutative ring, and S be a multiplicatively closed set. Then RS is a flatR-module (so in particular, R is a flat R-module).

Proof. Let Af→ B

C→ be exact. Then

RS ⊗R A1⊗f−−−−→ RS ⊗R B

1⊗g−−−−→ RS ⊗R Cy y yAS

f1−−−−→ BS

g1−−−−→ CS

is a commutative diagram. Furthermore, we showed that the bottom row is exact. By the homework[edit: see immediately below], the vertical arrows are isomorphisms. Then by the remark, the top row isalso exact. Thus RS is a flat R-module.

Homework Problem 9. Let R be a commutative ring, let S be a multiplicatively closed set in R, and letM be an R-module. Then MS

∼= M ⊗R RS.

Proposition 76. Let {Ai}i∈I be a set of right R-modules. Then ⊕Ai is flat if and only if each Ai is flat.

Proof. Let Lf→M

g→ N be exact. Let (∗)i denote the exact sequence

Ai ⊗R L1i⊗f→ Ai ⊗RM

1i⊗g→ Ai ⊗R N

for each i ∈ I. Let (#) denote the exact sequence

⊕(Ai ⊗R L)⊕(1i⊗f)→ ⊕(Ai ⊗RM)

⊕(1i⊗g)→ ⊕(Ai ⊗R N)

Note that (#) is exact if and only if (∗)i is exact for all i.

(#)⊕ (Ai ⊗R L)⊕(1i⊗f)−−−−−→ ⊕(Ai ⊗RM)

⊕(1i⊗g)−−−−−→ ⊕(Ai ⊗R N)

∼=y ∼=

y ∼=y

(##)⊕ (Ai)⊗R L⊕(1i⊗f)−−−−−→ ⊕(Ai)⊗RM

⊕(1i⊗g)−−−−−→ ⊕(Ai)⊗R NThis diagram commutes because it is so natural it would be weird if it didn’t. Therefore ⊕Ai is flat if

and only if (##) is exact. By the lemma, this is the case if and only if (#) is exact. But this s the case ifand only if (∗)i is exact for all i.

Corollary 35. All free modules F are flat.

Proof. Recall that R is flat. Then F ∼=⊕

iR, so by the previous proposition, F is flat.

Corollary 36. All projective modules P are flat.

Proof. Since P is projective, it is the summand of a free module. That is, F ∼= P ⊕K for some module Kand free module F . But F is flat by the previous corollary. Then by the proposition, P is flat.

98

Lemma 29. (Five Lemma) Consider the following commutative diagram of left R-modules:

A1α1−−−−→ A2

α2−−−−→ A3α3−−−−→ A4

α4−−−−→ A5

γ1

y γ2

y γ3

y γ4

y γ5

yB1

β1−−−−→ B2β2−−−−→ B3

β3−−−−→ B4β4−−−−→ B5

Suppose that each row is exact,and that γ1, γ2, γ4, and γ5 are all isomorphisms. Then γ3 is an isomor-phism.

Proof. (We proceed by diagram chasing.) We will first show that γ3 is injective. Suppose γ3(a3) = 0. Thenγ4α3(a3) = β3(γ3(a3)) = 0. As γ4 is injective, then α3(a3) = 0. Therefore there exists an a2 ∈ A2 suchthat α2(a2) = a3. Then β2(γ2(a2)) = 0, so there exists a b1 ∈ B1 such that β1(b1) = γ2(a2). Since γ1 is anisomorphism, we can choose a1 ∈ A1 such that γ1(a1) = b1. Then γ2(α1(a1)) = γ2(a2). Since γ2 is injective,then a2 = α(a1). Then a3 = α2(a2) = α2(α1(a1)) = 0, since the top row is exact. Therefore γ3 is injective.

One does something similar for surjective.

Proposition 77. We get a special case of the five lemma with a smaller commutative diagram:

Af−−−−→ B

g−−−−→ C −−−−→ 0

φ

y ψ

yL

h−−−−→ Mi−−−−→ N −−−−→ 0

where the top row and bottom row are exact, and φ and ψ are isomorphisms.Then there exists an isomorphism ε : C → N making the diagram commute.

Proof. It suffices to show that a map ε : C → N exists. Then by the 5-lemma on

Af−−−−→ B

g−−−−→ C −−−−→ 0 −−−−→ 0

φ

y ψ

y ε

y y yL

h−−−−→ Mi−−−−→ N −−−−→ 0 −−−−→ 0

this epsilon would be an isomorphism.For c ∈ C, choose b ∈ B such that g(b) = c. Define ε : C → N by ε(c) = iψ(b). It then suffices to show

that ε is well-defined.Suppose b, b′ ∈ B satisfies g(b′) = c. Then g(b− b′) = 0, so b− b′ = f(a) for some a ∈ A. So b = f(a)+ b′.

Then

iψ(b) = iψ(f(a)) + i(ψ(b′)

= ihφ9) + iψ(b′)

= iψ(b′)

since ih = 0. It can easily be shown that ε is R-linear and makes the diagram commute. Therefore, bythe 5-Lemma, ε is an isomorphism.

Definition 101. A left R-module M is called finitely presented if there exists an exact sequence

Rm −−−−→ Rn −−−−→ M −−−−→ 0

99


Let’s talk about the relation between Hom and the tensor product.

Theorem 80 (Hom−⊗ Adjointness/Adjunction). LetR,S, and T be rings. Let A be anR−T bimodule,B be an S−R bimodule, and C be a left S-module. Then the map HomR(A,HomS(B,C))→ HomS(B⊗R

A,C) (which lives in the set of T -module homomorphisms) given by φ 7→

(gφ :

{B ⊗R A→ C

b⊗ a 7→ φ(a)(b)

)is a

well-defined left T -module isomorphism.

Proof. Many of the details of this proof will be omitted.We will first show that this map is well-defined.Fix a φ ∈ HomR(A,HomS(B,C)). Define gφ : B ×A→ C by (b, a) 7→ φ(b)(a).One can check that gφ is R-biadditive. Then by the universal property, there exists a unique gφ which is

a well-defined group homomorphism. Thus this map is well-defined.One can then check that gφ is S-linear.We will now show that the map φ 7→ gφ is T -linear. Define a mapHomS(B⊗RA,C)→ HomR(A,HomS(B,C))

by ψ 7→

(fψ :

{fψ : B → C

b 7→ ψ(b⊗ a)

). By the previous part, this map is well-defined.

One can then check that

• For any a ∈ A, fψ(a) is S-linear.

• fψ is R-linear.

• The map ψ 7→ fψ is T -linear.

We then wish to show that fgφ = φ. Fix an a ∈ A. Then fgφ(a)(b) = gφ(b ⊗ a) = φ(a)(b) for all b ∈ B.Thus fgφ(a) = φ(a) for all a ∈ A.

Thus fgφ = φ. Similarly, gfψ = ψ for all ψ. Thus this map is an isomorphism of T -modules.

Remark 105. This map is a bit stronger than being simply an isomorphism. It is a natural isomorphism,meaning HomR(−, HomS(B,C)) → HomS(B ⊗R −, C) is a functor. It is also a functor if you replace anyother component with a dash.

Remark 106. Let R be a commutative ring, x ∈ R, let M and N be R-modules, and suppose that xM = 0but x is a non-zero-divisor onN . ThenHomR(M,N) ∼= HomR(M⊗R(R/(x)), N) ∼= HomR(M,HomR(R/(x), N)).But HomR(R/(x), N) ∼= (0 :N X) = 0, so HomR(M,N) = 0.

Definition 102. We say an R-module M is finitely presented if there exists an exact sequence of the formRm → Rn →M → 0.

Theorem 81. Let R be a commutative ring and let T be a flat R-algebra (that is, there is a ring homo-morphism φ : R → T ). Suppose M is a finitely presented R-module, and let N be an R-module. Thenψ : T ⊗R HomR(M,N)→ HomT (T ⊗RM,T ) given by t⊗ f 7→ µt ⊗ f is an isomorphism of T -modules.

Proof. We break the problem into several cases.First, suppose M = R. Then HomR(R,N) ∼= N , so T ⊗R HomR(R,N) ∼= T ⊗R (N) ∼= HomT (T, T ⊗R

N) ∼= HomT (T ⊗R R, T ⊗R N). If we follow an element through this chain of equivalences, we get that

t⊗ f 7→ t⊗ f(1) 7→ gt :

{T → T ⊗R N1 7→ t⊗ f(1)

7→ µt ⊗ f . (The last mapping, sending gt to µt ⊗ f , is something

we should check.) Therefore ψR is an isomorphism.Suppose instead that M = F = Rn. Then by using direct sums, we get much the same results. Therefore

ψF is an isomorphism.Finally, consider the general case. Since M is finitely presented, there exists an exact sequence F → G→

M → 0, where F and G are free modules.

100

Now apply T ⊗R − to the entire sequence. Since T is flat, the resulting sequence is still exact. That is,

T ⊗R F → T ⊗R G→ T ⊗RM → 0 (2.1)

is exact. Now we apply HomT (−, T ⊗R N) to the equation. This functor is contravariant, so we get abackwards exact sequence. However, by applying HomT (−, N) and T ⊗R − in the opposite order, we getthe following commutative diagram:

0 −−−−→ 0 −−−−→ T ⊗R HomR(M,N) −−−−→ T ⊗R HomR(G,N) −−−−→ T ⊗R HomR(F,N)

∼=y ∼=

y ∼=y ∼=

y0 −−−−→ 0 −−−−→ HomT (T ⊗RM,T ⊗R N) −−−−→ HomT (T ⊗R G,T ⊗R N) −−−−→ HomT (T ⊗R F, T ⊗R N)

(2.2)Since T is flat and Hom preserves exactness, then both the top and the bottom are exact. Also, the

vertical arrows are isomorphisms, by the previous parts. Furthermore, this is a commutative diagram, sincethese maps are natural. Finally, you can define the middle arrow in the same way as before.

Then, by the 5-Lemma, there is an isomorphism between T⊗RHomR(M,N) andHomT (T⊗RM,T⊗RN).



Let R be a commutative ring. Recall that a finitely generated projective module over a quasi-local ring is infact a free module. What about a converse?

Theorem 82. Let M be a finitely presented R-module. Then the following are equivalent:

1. M is projective.

2. Mp is a free Rp-module for all p ∈ SpecR.

3. Mm is a free Rm-module for all maximal ideals m.

Proof. We have already proven that (1) implies (2). Furthermore, (2) implies (3) follows instantly fromdefinitions, as every maximal ideal is prime.

It then suffices to show that (3) implies (1). To this end, suppose M is a finitely presented R-module,and suppose that Mm is a free Rm-module for all maximal ideals m.

Recall that a module is projective if and only if HomR(M,−) is exact. Let Af→ B → be exact. Then

HomR(M,A)f∗→ HomR(M,B) → C → 0 is exact (where C = coker(f∗)). We wish to show that f∗ is

surjective, and this is equivalent to showing that C = 0.Let’s localize at a maximal ideal m. Then

HomR(M,A)mf∗1−−−−→ HomR(M,B)m −−−−→ Cm −−−−→ 0

∼=y ∼=

y yHomRm

(Mm, Am)( f1 )∗−−−−→ HomRm

(Mm, Bm) −−−−→ 0

Since Mm is free, then it is projective, so the bottom row is exact. The vertical maps are isomorphismssince M is finitely presented. Also, the diagram commutes because the maps are natural. Therefore, by the5-lemma, Cm = 0 for all maximal ideals m. But the only such module is 0, so C = 0. Since C was thecokernel of f∗, then f∗ is surjective. Thus M is projective, by an equivalent condition we proved before.

101

Remark 107. If M is a projective module, it is finitely presented. Therefore, in the previous theorem weneed not check finitely presented if M is projective.

Let’s now make an example to show that finitely presented was a necessary condition in the statementof the previous condition.

Definition 103. Let R be a (not necessarily commutative) ring with unit. We say R is von Neumannregular if, for all a ∈ R, there exists x ∈ R such that a = axa.

Example 109. Any division ring is von Neumann regular.

Example 110. Any product of division rings (infinite or finite) is von Neumann regular.

Exercise 30. Let R be a commutative von Neumann regular ring. Then RS is von Neumann regular forany multiplicatively closed set S.

Proposition 78. Let (R,m) be a commutative, quasi-local, von Neumann regular ring. Then R is a field.

Proof. We know that commutative ring is a field if and only if its only maximal ideal is (0).Fix some a ∈ m. Then there exists x ∈ R such that a = axa = a2x. Then a(1− ax) = 0. If 1− ax ∈ m,

then 1 = (1 − ax) + ax ∈ m, and this is a contradiction. Thus a(1 − ax) 6∈ m. Since R is semi-local, then(1− ax) is a unit. Thus a = 0, so m = (0). Thus R is a field.

Example 111. Let R =

∞∏i=1

C. Then R is von Neumann regular, but not semisimple.

Recall one characterization of a semisimple ring is that every ideal is a direct summand of the ring. SinceR is no semisimple, there exists an ideal I of R such that R/I is not projective.

Let M = R/I, which is finitely generated. But Rm is a field for all maximal ideals m. Therefore Mm isa free Rm-module for all m, and M is not projective.

7 Injective Modules

Definition 104. Let R be a ring and let E be a left R-module. Then E is called injective if, for every

diagram 0→Mg→ N , and f : M → E, there exists an h : N → E such that hg = f .

Theorem 83 (Baer’s Criterion). Let R be a ring, and let E be a left R-module. Then E is injectiveif and only if, for all left ideals I of R, if i : I → R is the inclusion map, and f : I → E is an R-modulehomomorphism, then there exists an h : R→ E such that hi = f .

(That is, every homomorphism f : I → E can be extended onto all of R.)

Proof. Certainly, if E is injective, then the map i has such an h.

Conversely, suppose 0 → Mi→ N is exact, and f : M → E is any R-module homomorphism. Without

loss of generality, we can assume M ⊂ N , and that i is the inclusion map.

We now use Zorn’s Lemma. Let Λ = {(k, hk)|M ⊂ K ⊂ N,hK : K → E such that hK

∣∣∣M

= f}.

Partially order Λ by (K,hK) ≤ (K ′, hK′) if and only if K ⊂ K ′ and hK′∣∣∣K′

= hk.

One can verify that Zorn’s Lemma applies, so there exists a maximal element in Λ. Let (K,hK) be sucha maximal element.

We then wish to show that K = N . Suppose for the sake of contradiction that K ( N . Let x ∈ N \K,and let I = (K :R x) = {r ∈ R|rx ∈ K}.

Define g : I → E by i 7→ hK(ix). One can verify that g is a homomorphism of R-modules. By hypothesis,we can extend g to g. That is, there exists g : R→ E such that g and g agree on I.

Let L = K + Rx. “Define” hL : L → E by k + rx 7→ hK(k) + g(r). We need to show that hL iswell-defined. Suppose k + rx = k′ + r′x. Then (r − r′)x = k′ − k ∈ K. Therefore, r − r′ ∈ I = (K :R X).

102

Therefore,

g(r − r′) = g(r − r′)= hK((r − r′)x)

= hK(k′ − k)

Therefore, hK(k) + g(r) = hK(k′) + g(r′), so hL(k + rx) = hL(k′ + r′x). That is, hL is well-defined.But then (K,hK) < (L, hL). This contradricts the maximality of K, so K = N .

7.1 Day 22 - March 2

Recall the following exercise:

Exercise 31. If R is a ring and E is an R-module, then E is injective if and only if HomR(−, E) is exact.

Recall also Baer’s Criterion:

Theorem 84. Checking the criterion for injective modules is equivalent to checking the same criterion onlyfor ideals of R.

Now we get a new exercise!

Exercise 32. Let {Ei}i∈I be a collection of left R-modules. Then∏i∈I

Ei is injective if and only if each Ei

is injective.From this we get that the direct sum of finitely many injective modules is injective.

Remark 108. The direct sum of an arbitrary family of injectives is injective if and only if R is left Noethe-rian.

Definition 105. We say an R-module M is divisible if, for every u ∈M and non-zero-divisor r ∈ R, thereexists a u′ ∈M such that ru′ = u.

Example 112. If R is a field, every R-vector space is divisible.If R is a domain, then Q, the field of fractions of R, is a divisible R-module. (More generally, if R is a

commutative ring, and S = {non-zero-divisors in R}, then RS is a divisible R-module. We call RS the totalring of fractions of R.)

Sums, products, and quotients of divisible modules are divisible. In particular, Q/Z is a divisible Z-module.

Proposition 79. Let R be a commutative ring, and let E be an injective left R-module. Then E isdivisible.

Proof. Let u ∈ E, and r ∈ R. Suppose r is a non-zero-divisor. Then consider the diagram 0 → Rr→ R.

Define f : R→ E by 1 7→ u.Since E is injective, then there exists an h ∈ H such that hr = f . Then u = f(1) = hr(1) = rh(1). Thus

u′ = h(1) satisfies the definition of divisibility.

Proposition 80. Let R be a PID. Then every divisible module is injective.

Proof. Let E be a divisible module over R. Using Baer’s criterion, it suffices to check the injectivity conditionon ideals of R. Let I be an ideal in R. If I = 0, there is nothing to check, so suppose I 6= 0.

Since R is a PID, then I = (a) for some a ∈ R. Let f : (a) → E be an R-module homomorphism. We

need to find h : R→ E such that h∣∣∣(a)

= f . Let u = f(a).

Note that since a 6= 0 and R is a PID, then a is a non-zero-divisor. Since E is divisible, and a is anon-zero-divisor, then there exists u′ ∈ E such that au′ = u. Define h : R → E by h(x) = xu′. Thenh(ra) = rau′ = ru = rf(a) = f(ra).

103

Thus we have extended f to an h : R→ E. Thus E is injective.

Example 113. Since Z is a domain, then Q, the field of fractions of Z, is divisible as a Z-module.Therefore Q/Z is divisible as a Z-module as well.

Lemma 30. Every Z-module M can be embedded into an injective Z-module. That is, there exists an exactsequence 0→M → E, where E is injective.

Proof. Choose a free module F which surjects onto M . That is, choose a free module F and a surjectiveZ-homomorphism φ : F → M . Let K = kerφ. Then M ∼= F/K. However, K ⊂ F =

⊕Z ⊂

⊕Q = I.

Therefore M ∼= F/K ⊂ I/K, so M embeds into I/K.As Q is divisible, then I is divisible. Thus I/K is divisble. However, Z is a PID, so by the proposition,

I/K is injective.

Proposition 81. Let φ : R → S be a ring homomorphism. Let E be an injective left R-module. ThenHomR(S,E) is an injective left S-module.

Proof. Recall that a left S-module M is injective if and only if HomS(−,M) is exact. However HomS(−,M)is always left exact. It then suffices to show right exactness.

Suppose 0→Mf→ N is exact in 〈〈S −mod〉〉. Then by tensoring with S, we get

0 −−−−→ Mf−−−−→ N

∼=y ∼=

y0 −−−−→ S ⊗S M

1⊗f−−−−→ S ⊗S Nis an exact sequence of R-modules.Then applying HomR(−, E), we get that

HomR(S ⊗S N,E)(1⊗f)∗−−−−→ HomR(S ⊗S M,E) −−−−→ 0y y

HomS(N,HomR(S,E)) −−−−→ HomS(M,HomR(S,E) −−−−→ 0

But the bottom is exact since E is injective. Therefore HomR(S,E) is injective as an S-module.

Theorem 85. Let R be a ring, andM be a left R-module. ThenM embeds into an injective left R-module.

Proof. We know that M is a Z-module, so it imbeds into some E which is an injective Z-module. Letf : M → E be such an embedding. Also, there exists a ring homomorphism from Z→ R.

By the previous proposition, HomZ(R,E) is an injective left R-module. Define φ : M → HomZ(R,E)

by m 7→

{φ(m) : R→ E

r 7→ f(rm).

Then one can verify that φ is an injective R-module homomorphism.


Lemma 31. Suppose E is an injective module, and that A is a direct summand of E. Then A is injective.

Proof. Since A is a direct summand of E, there exists a module B such that E = A⊕B. Let φ : A→ E bethe inclusion taking a 7→ (a, 0) and let π : E → A be the projection (a, b) 7→ a.

Suppose 0 → Mg→ N is an exact sequence and that f : M → A is any map. Then φf is an injection

into E. Since E is an injective, there exists h1 : N → E such that h1g = fφ. Then πh1g = πφf = f (asπφ = 1. Let h = πh1. Then hg = f , so A is injective.

104

Proposition 82. Let R be a ring, and let E be a left R-module. Then E is injective if and only if every

short exact sequence of the form 0→ Ef→M

g→ N → 0 splits.

Proof. Suppose E is an injective. Then 0→ Ef→M is exact, and 1E : E → E, so since E is injective, there

exists an h : M → E such that hf = 1E . In other words, the sequence splits.Conversely, by the last theorem from last class, we know we can imbed E into an injective I. That is,

we can make a short exact sequence of the form 0→ Ef→ I

g→M → 0, where I is injective. This splits, soI ∼= E ⊕M . Since I is injective and E is a summand of I, then by the lemma, E is also an injective.

Definition 106. Let M be an R-module. A projective resolution of M is an exact sequence ... → Pi →Pi−1 → ...→ P0 →M → 0, where each Pi is projective.

We similarly define a free resolution of M to be an exact sequence ...→ Fi → Fi−1 → ...→ F0 →M → 0,where each Fi is free.

The definition of an injective resolution is slightly different. An injective resolution is an exact sequenceof the form 0→M → I0 → I1 → I2 → ..., where each Ij is injective.

Remark 109. Homological algebra is the study of a module and its resolutions. Namely, it attempts tounderstand a module based on its resolutions (for instance, the projective dimension or injective dimension).

8 Integral Extensions

Let’s now change gears and talk about integral extensions. All rings in this case will be commutative.

Definition 107. Let R ⊂ S be rings. An element u ∈ S is integral over R if there exists an equation of theform un + r1u

n−1 + ...+ rn−1u+ rn = 0, with each ri ∈ R.That is, we say u ∈ S is integral if it is the root of some monic polynomial in R[x].

Remark 110. If each Ri is a field, then u ∈ S is integral over R if and only if u is algebraic over R.

Example 114. If R = Z, then√

5 is integral over Z. However, 12 is not integral over Z.

Proposition 83. Let R ⊂ S be rings, and let u ∈ S. The following are equivalent:

1. u is integral over R.

2. R[u] is a finitely generated R-module.

3. There exists a finitely generated R-submodule M of S, such that 1 ∈M and uM ⊂M .

4. There exists a faithful R[u]-module M such that M is finitely generated as an R-module. (Recall thata module is faithful if AnnR[u]M = 0.)

Proof. Suppose (1). Then there exists an equation of the form un + r1un−1 + ... + r0 = 0. Then one can

show that R[u] = R · 1 + Ru + ... + Run−1, so R[u] is finitely generated as an R-module. Thus (1) implies(2).

Suppose (2). Then let M = R[u]. Note that R[u] is a finitely generated R-submodule of S such that1 ∈ R[u] and uR[u] ⊂ R[u]. Thus (2) implies (3).

Suppose (3). Let M be the finitely generated R-submodule M of S such that 1 ∈M and uM ⊂M . Byassumption, M is finitely generated as an R-module, so it suffices to show that M is a faithful R[u]-module.Since uM ⊂ M , then M is an R[u]-module. Furthermore, since 1 ∈ M , if x ∈ AnnR[u](M), then 1 · x = 0,so x = 0. Thus M is faithful. That is, (3) implies (4).

Suppose (4). We use “the determinant trick”. Let m = Rx1 + ...Rxn be the given finitely generated

R-module. since M is an R[u] module, then uM ⊂ M . For j = 1, ..., n, we can write uxj =

n∑i=1

ai,jui for

some ai,j ∈ R. Let A = [ai,j ].

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In matrix form, we have that u 0. . .

0 u

x1

...xn

= A

x1

...xn

Thus (uIn − A)x = 0. Multiplying by adj(uIn − A), we get on the left that det(uIn − A)Inx = 0. Thus

det(uIn−A) ·xi = 0 for all i. Thus det(uIn−A) ∈ AnnR(M), but since M is faithful, then AnnR(M) = 0, sodet(uIn−A) = 0. But det(uIn−A) is a monic polynomial in R[u]! Thus u is the root of a monic polynomialin R[x], so u is integral.


Recall from last class an integral extension:

Definition 108. Let R ⊂ S be rings. An element u ∈ S is integral over R if there exists an equation of theform un + r1u

n−1 + ...+ rn−1u+ rn = 0, with each ri ∈ R.That is, we say u ∈ S is integral if it is the root of some monic polynomial in R[x].

We actually proved that u being integral is equivalent to three other things. Let us not define anotherrelated notion:

Definition 109. Let R ⊂ S be rings. We say S is an integral extension or simply integral if every elements ∈ S is integral over R.

This leads to the following criterion:

Corollary 37. Let R ⊂ S be rings, and let u ∈ S. Then the following are equivalent:

1. u is integral over R.

2. R[u] is a finitely generated R-module.

3. R[u] is integral over R.

Proof. Last time, we showed that (1) implies (2).Suppose (2). Let α ∈ R[u]. Let M = R[u], which by assumption is a finitely generated submodule of S.

Furthermore, 1 ∈M and αM ⊂M . Then by criterion (3) of the conditions from last class, we know that αis integral over R. Since α was arbitrary, then R[u] is integral over R. Thus (2) implies (3).

Suppose (3). Since u ∈ R[u], then u is integral over S.

We can use induction to prove the following corollary:

Corollary 38. Let R ⊂ S be rings. Let u1, ..., un ∈ S. Then the following are equivalent:

1. u1, ..., un are integral over R.

2. R[u1, ..., un] is a finitely generated R-module.

3. R[u1, ..., un] is integral over R

This gives us a neat fact:

Corollary 39. Let R ⊂ S be an integral extension. Then S is finitely generated as an R-algebra if and onlyif S is finitely generated as an R-module.

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Proof. If S is finitely generated as an R-module, then S = Ru1 + ... + Run for some u1, ..., un ∈ S. ThusS = R[u1, ..., un].

Conversely, if S is finitely generated as an R-algebra, then S = R[v1, ..., vn]. But then by the previouscorollary, S is finitely generated as an R-module.

Definition 110. LetR ⊂ S be rings. We say the integral closure ofR over S is T = {u ∈ S|u is integral over R}.If R = T , then we say R is integrally closed .

Proposition 84. Let R ⊂ S be rings, and let T be the integral closure of R over S. Then T is a subring ofS containing R.

Proof. For all r ∈ R, r is the root of x− r ∈ R[x]. Thus r is integral over R, so R ⊂ T .We must now show that T is a subring of S. It suffices to show that T is closed under addition,

subtraction, and multiplication. Let α, β ∈ T . Then α+ β, α− β, αβ ∈ R[α, β]. Since α and β are integral,then by the lemma, R[α, β] is an integral extension of R. Thus α + β, α − β, and αβ are integral. That is,α+ β, α− β, αβ ∈ T . Thus T is closed under addition, subtraction, and multiplication, so T is a subring ofS.

Definition 111. If R is a domain, then we say R is integrally closed (not in the context of any other ring)if R is integrally closed in its field of fractions.

Proposition 85. Let R be a unique factorization domain. Then R is integrally closed.

Proof. Let F be the field of fractions of R. Let α ∈ F be integral over R. Write α = rs where gcd(r, s) = 1

and r, s ∈ R.Then since α is integral over R, there exists a monic polynomial f(x) = xn + c1x

n−1 + ... + cn ∈ R[x]such that f(α) = 0.

Then ( rs )n+c1( rs )n−1 + ...+cn = 0, so rn+c1srn−1 + ...+cns

n = 0. Thus rn = s(−c1rn−1− ...−cnsn−1),so s|rn in R. But gcd(r, s) = 1 = gcd(rn, s).

Thus s is a unit, so α ∈ R. In other words, R is its integral closure over F , so R is integrally closed.

Proposition 86. Let R ⊂ S be rings, and let T be the integral closure of R over S. Let W be a multiplica-tively closed set of R. Then RW ⊂ SW . Furthermore, the integral closure of RW in SW is TW .

Proof. We will first show that TW is the integral closure of RW in SW .Let t

w ∈ TW , so t ∈ T and w ∈W . Then since T is the integral closure of R in S, then there exits ci ∈ Rsuch that tn + c1t

n−1 + ...+ cn = 0.Then, dividing by wn, we get that ( tw )n + c1

w ( tw )n−1 + ...+ cnwn = 0. But each ci

wi ∈ Ri, so tw is integral

over RW .Thus the integral closure of RW over SW contains TW .Conversely, suppose u

w ∈ SW and suppose uw is integral over Rw. Then there exist coefficients ci

v (notethat we can combine to make a common denominator) such that ( uw )n + c1

v ( uw )n−1 + ...+ cn−1

vuw + cn

v = 0.

Multiplying by (wv)n, we get that(uv)n + r1(uv)n−1 + ...+ rn−1(uv) + rn

1=

0

1, where ri = ciw

ivi−1 ∈R. Therefore, there exist w′ ∈W such that w′[(uv)n + r1(uv)n−1 + ...+ rn−1(uv) + rn] = 0.

Therefore (w′n)[(uv)n + r1(uv)n−1 + ...+ rn−1(uv) + rn] = 0 = (uvw′)n + r′1(uvw′)n−1 + ...+ r′n, wherer′i = ri(w

′)i ∈ R.Thus w′uv ∈ T . Let t = w′uv. Then u

1 = tvw′ ∈ TW , so u

w = tvww′ ∈ TW .

Therefore, TW is the integral closure of RW over SW .

Corollary 40. Let R ⊂ S be rings, and let T be the integral closure of R over S. Let W be a multiplicativelyclosed subset of R. Then,

1. If S is integral over R, then SW is integral over RW .

2. If R is integrally closed over S, then RW is integrally closed in SW .

107

3. If R is an integrally closed domain, then so is RW .

Proposition 87. Let R ⊂ S be an integral extension, and let I be an ideal of S. Then I ∩R is an ideal ofR. Furthermore, the map R/(I ∩R)→ S/I by r + I ∩R→ r + I is a well-defined injective ring map.

Finally, S/I is integral over (the canonical image of) R/(I ∩R).

Proof. We will first show that this map is a well-defined injective ring map. First, note that this map iswell-defined since I∩R ⊂ I. Also, it is easy to see that it preserves addition, subtraction, and multiplication.

It then suffices to show that this map is injective. Suppose r + I = 0. Then r ∈ I. Also, r ∈ R, sor ∈ I ∩R. Thus r + I ∩R = I ∩R, so this map is injective.

Now let us show that we have integrality (is that a word?). Let s ∈ S/I. Since S is an integral extensionof R, then sn + r1s

n−1 + ... + rn = 0, then sn + r1sn−1 + ... + rn = 0, so s is integral over R. Thus S/I is

integral over the canonical image of R/(I ∩R).


Let’s continue with integral extensions.

Corollary 41. Let R ⊂ S be an integral extensions. Let p ∈ SpecS. Then p is maximal if and only if p∩Ris maximal.

Proof. If p ∈ SpecS, then S/p is an integral extension of R/(p ∩ R). Then S/p is a field if and only ifR/(p ∩R) is a field.

Lemma 32. LetN1, ..., Nt beR-submodules ofM . LetW be a multiplicatively closed set. Then⋂ti=1(Ni)W =

(∩ti=1Ni)W .

Proof. We use induction. It then suffices to show the case that t = 2.Suppose α ∈ (N1)W ∩ (N2)W . Then α = n1

w1= n2

w2, where n1 ∈ N1, n2 ∈ N2, and w1, w2 ∈W .

Then there exists w3 ∈W such that β = w3w2n1 = w3w1n2. But then β ∈ N1 ∩N2. Thus α = betaw1w2w3

∈(N1 ∩N2)W .

Conversely, suppose γ ∈ (N1 ∩N2)W . Then γ = nw for some n ∈ N1 ∩N2 and w ∈ W . Then γ = n

w is arepresentation of γ as an element of both (N1)W and (N2)W .

Thus (N1 ∩N2)W = (N1)W ∩ (N2)W .

Theorem 86 (Lying Over Theorem). Let R ⊂ S be an integral extensions. Let p ∈ SpecR. Then thereexists q ∈ SpecS such that q ∩R = p. (We say “q lies over p”.)

Proof. We first will consider a special case: suppose (R,m) is a quasi-local ring and suppose that p = m.Then let n be any maximal ideal of S. Then n ∩R is a maximal ideal in R, so n ∩R = m, as desired.

We now move to the general case. Let W = R\p. Then SW is integral over RW . Also, RW is quasi-local,with maximal ideal pW . Then, by the special case, there exists qW ∈ SpecSW such that qW ∩ RW = pW .But by the lemma, qW ∩RW = (q ∩R)W . Thus (q ∩R)W = pW , so q ∩R = p. Also, q is prime since qW isprime.

Theorem 87 (Incomparable). Let R ⊂ S be an integral extension. Let p ∈ SpecR. Suppose q1 6= q2 ∈SpecS lie over p. Then q1 ( q2 and q2 ( q1.

Proof. We again get a special case: suppose (R,m) is quasi-local and that p = m. Then since q1 ∩R = m ismaximal, then by the corollary, q1 is maximal. Similarly, q2 is maximal. Then, since distinct maximal idealsare incomparable, then q1 and q2 are incomparable, as desired.

We now move to the general case. Again, let W = R\p. Then SW is an integral extension over RW . Notethat (q1)W ∩ RW = (q1 ∩ R)W = pW = (q2)W ∩ RW . By the special case, since (RW , pW ) is a quasi-localring, then (q1)W and (q2)W are incomparable. Thus q1 and q2 are incomparable.

108

Theorem 88 (Going Up Theorem). Let R ⊂ S be an integral extension. Suppose p1 ( p2 are primes inSpecR. Let q1 ∈ SpecS lie over p1. Then there exists q2 ) q1 such that q2 ∩R = p2.

Proof. We once again get a special case: suppose (R,m) is quasi-local, and p2 = m. Then from the LyingOver Theorem, there exists a maximal ideal n containing q1. Then n ∩ R = m. Since p1 was not maximal,then q1 isn’t either. Then since q2 is maximal, we get that q2 ( q1.

In the general case, let W = R \ p2. Observer that (q1)W ∩RW = (p1)W . Then RW is a quasi-local ringwith maximal ideal (p2)W . Then by the special case, there exists (q2)W ∈ SpecSW such that q2 ⊃ (q1)Wand (q2)W ∩RW = (p2)W . Therefore qw ) q1 and q2 ∩R = p2.

Theorem 89. Let R ⊂ S be an integral extension. Then dimR = dimS.

Proof. We’ll show that for every n ≥ 0, dimR ≥ n if and only if dimS ≥ n.(⇒) Suppose dimR ≥ n. Then there is a prime chain in SpecR of the form p0 ( p1 ( pn. By the Lying

Over Theorem, there exists q0 ∈ SpecS such that q0 ∩ R = p0. By repeated use of the Going Up theorem,we get that there exist q1 ) q0, q2 ) q1, etc. Thus q0 ( q1 ( ... ( qn is a strictly ascending chain in SpecS.Thus dimS ≥ n.

(⇐) Suppose dimS ≥ n. Then there exists a chain of prime ideals of the form q0 ( q1 ( ... ( qn lyingin S. Then q0 ∩ R ( q1 ∩ R ( ... ( ...qn ∩ R is a chain of prime ideals in SpecR. By the incomparabilitytheorem, the containments are strict. Thus dimR ≥ n.

Thus dimR = dimS.

Example 115. Let F be a subfield of C which is algebraic over Q. Let R be the integral closure of Z in F .(Such an R is called an algebraic number field .) Then dimR = dimZ = 1.

Example 116. Let V = {(z3, z4, z5) ∈ C3|z ∈ C}. Consider I(V ) = {f(x, y, z) ∈ C[x, y, z]|f(p) =0 for all p ∈ V }.

The coordinate ring of V is defined to be C[x, y, z]/I(V ). Note that 1 = x3− yz ∈ I(V ), f2 = z2− x2y ∈I(V ), and f3 = y2 − xz ∈ I(V ). Then it turns out that I(V ) = (f1, f2, f3).

Consider φ : C[x, y, z] → C[t] given by f(x, y, z) → f(t3, t4, t5). Then kerφ = I(V ). ThereforeC[x, y, z]/I(V ) ∼= imφ = C[t3, t4, t5]. Note that C[t] is an integral extension over C[t3, t4, t5]. Then, bythe theorem, 1 = C[t] = dimC[t3, t4, t5] = dimC[x, y, z]/I(V ).

This is a good thing! After all, the dimension should correspond to the intuitive sense of the dimensionof the curve, and V is certainly dimension 1!

8.3 Day 26 - March 11

Last time we proved the Going Up Theorem. Today we will prove the Going Down Theorem, but before wecan do that, we need some lemmas.

Lemma 33. (Division algorithm in polynomial rings) Let R be a ring, and let f(x) ∈ R[x] be a monicpolynomial. Then for all g(x) ∈ R[x], there exist unique q(x), r(x) ∈ R[x] such that g = fq + r anddeg r < deg f .

This is such a classical result, we will not prove it.

Corollary 42. Let R ⊂ S be rings, and let f(x), g(x) ∈ R[x], where f(x) is monic. Then f |g in R[x] if andonly if f |g in S[x].

Proof. Certainly, if f |g in R[x], then f |g in S[x].Conversely, if f |g in S[x], then by the division algorithm, g = fqR + rR in R[x] and g = fqS + rS in S[x].

But both of these lie in R[x], so rR = rS and qR = qS . Since g = fqS + rS and f |g in S[x], then rS = 0.Thus g = fqS = fqR. Since qR ∈ R[x], then f |g in R[x].

Proposition 88. Let R ⊂ S be an extension of domains. Let K = Q(R) ⊂ L = Q(S). Suppose Ris integrally closed in K. Let α ∈ S be integral over R. Then α is algebraic over K. Furthermore, iff(x) = Min(α,K), f(x) ∈ R[x].

109

Proof. Since α ∈ S is integral over R, then there exists a polynomial h(x) ∈ R[x] such that h(α) = 0. Buth ∈ R[x], so h ∈ K[x]. Thus α is algebraic over K.

Let f(x) = Min(α,K). Then, by the definition of the minimal polynomial, f(x)|h(x) in K[x]. Let L bean algebraic closure of L. Then every root of h(x) in L is integral over R (since, by definition, they all arethe roots of a monic polynomial with coefficients in R).

Then, write f(x) = xn + c1xn−1 + ...+ cn ∈ K[x], and also write f(x) = (x− α1)...(x− αn), where each

αi ∈ L. Note that all αi’s are integral over R. But each cj is the product and sum of the αis, so the cjsare integral over R as well. Also, cj ∈ K, so since R is integrally closed in K, then cj ∈ R for all j. Hencef(x) ∈ R[x].

Remark 111. Let R ⊂ S be an extension of domains. Let K = Q(R) ⊂ L = Q(S). Suppose R is integrallyclosed in K. Let α ∈ S be integral over R. By the previous proposition, α is algebraic over K, andf(x) = Min(α,K) ∈ R[x].

Then, if g(x) ∈ R[x] is monic such that g(α) = 0, then f(x)|g(x) in R[x].

Proposition 89. Let R ⊂ S be an integral extension, and let I be an ideal of R. Suppose u ∈ IS. Then uis a root of a polynomial of the form xn + i1x

n−1 + ...+ in where each ij ∈ I. In fact, ij ∈ Ij .

Proof. We again use the “determinant trick”.Since u ∈ IS, then there exist ij ∈ IS and sj ∈ S such that u = i1s1+...+irsr. Let T = R[s1, ..., s−r] ⊂ S.

Note that T is integral over R.Since T is finitely generated as an algebra over R, then T is finitely generated as an R-module. Then

T = Ry1 + ...Ryl for some yi ∈ T .Furthermore, u ∈ IT . Then uyj ∈ ITyj ⊂ IT = Iy1 + ... + Iyl. Thus there exists ai,j ∈ I such that

uyj =

l∑i=1

ai,jyi for all j. Let A = (ai,j), and let y = (y1, ..., yl)T . Then Ay = uy, so (A − uI)(y) = 0.

Therefore det(A − uI) = 0. But det(A − xI) is a monic polynomial of the form xn + i1xn−1 + ... + in. In

fact, each ij ∈ Ij , as desired.

We can now prove the Going Down Theorem:

Theorem 90 (Going Down Theorem). Let R ⊂ S be an integral extension of domains. Suppose that Ris integrally closed in Q(R). Suppose also that p0 ( p1 are primes in SpecR, with q1 ∈ SpecS a prime idealsuch that R ∩ q1 = p1. Then there exists q0 ( q1 such that q0 ∩R = p0.

Proof. Let W = R \ p0, and let T = S \ q1. Then TW = {wt|w ∈ W, t ∈ T} is a multiplicatively closedsubset of S.

We now wish to show that p0S ∩WT = ∅. Suppose for the sake of contradiction that wt ∈ p0S. Then,by the previous proposition, wt is the root of a monic polynomial of the form g(x) = xn + a1x

n−1 + ...+ an,where each aj ∈ p0.

Let h(x) = g(wx). Then h(x) = wnxn + wn−1a1xn−1 + ... + an. Note that h(t) = g(wt) = 0. Let

f(x) = Min(t,K), where K = Q(R). By the first Proposition, since R is integrally closed, then f(x) ∈ R[x].But since h(t) = 0, then by the remark, f(x)|h(x) in R[x]. That is, there exists q(x) ∈ R[x] such thath(x) = f(x)q(x).

Then, in R/p0[x], we have that h(x) = f(x)q(x) = wnxn since the aj ∈ p0.Recall that R/p0 is a domain since p0 is a prime ideal. Recall also that f(x) was monic, so f(x) = xr for

some r. Therefore f(x) = xr + c1xr−1 + ...+ cr, where each ci ∈ p0. Thus 0 = f(t) = tr + c1t

r−1 + ...+ cr.Rearranging, we get that tr = −c1tr−1 − ... − cr ∈ p0S ⊂ p1S ⊂ q1. Since q1 is prime, then t ∈ q1.

However, we defined T = S \ q1, and t ∈ T , so this is a contradiction.[Proof will be completed later.]

110

8.4 Day 27 - March 14

We ended last class halfway through the proof of the Going Down Theorem:

Theorem 91. (Going Down Theorem) Let R ⊂ S be an integral extension of domains. Suppose that R isintegrally closed in Q(R). Suppose also that p0 ( p1 are primes in SpecR, with q1 ∈ SpecS a prime idealsuch that R ∩ q1 = p1. Then there exists q0 ( q1 such that q0 ∩R = p0.

Proof. Let W = R \ p0, and let T = S \ q1. Then TW = {wt|w ∈ W, t ∈ T} is a multiplicatively closedsubset of S.

We showed last time that p0S ∩WT = ∅.Now, let Λ = {I|I ideal of S, I ⊃ p0S, I ∩WT = ∅}. Then p0S ∈ Λ, so Λ is nonempty. Also, any chain

in Λ has an upper bound which is their union.Thus by Zorn’s Lemma, Λ has a maximal element, q0. It must be the case that q0 is prime [why?]. Then

q0 ∈ SpecS.Also, W ⊂ WT , so q0 ∩W ⊂ q0 ∩WT = ∅, so q0 ∩ R ⊂ p0. But q0 ⊃ p0S ⊃ p0, so q0 ∩ R ⊃ p0. Thus

q0 ∩R = p0.Furthermore, since q0 ∩ T = ∅, then q0 ( q1.

Since we can “go down” one step, we can actually “go down” chains of primes:

Corollary 43. Let R ⊂ S be an integral extension of domains, and suppose R is integrally closed. Supposep0 ( p1 ( ... ( pn is a chain of ideals in SpecR. Suppose also that there exists qn ∈ SpecS such thatqn ∩R = pn. Then there exists q0 ( q1 ( ... ( qn−1 ( qn such that qi ∩R = pi for all i.

Corollary 44. Let R ⊂ S be an integral extension of domains. Suppose R is integrally closed, and letq ∈ SpecS. Then ht(q) = ht(q ∩R).

Proof. We will first show that ht(q ∩R) ≥ ht(q) for all integral extensions.Let q0 ( q1 ( ... ( qn = q be a chain of primes in S. Then, contracting to R, we get that q0 ∩ R (

q1 ∩R ( ... ( ...qn ∩R = q∩R (note that we get strict inequalities by the incomparability theorem). Hence,ht(q ∩R) ≥ n.

It then suffices to show that ht(q) ≥ ht(q ∩ R). Let p0 ( p1 ( ... ( pn = p = q ∩ R. By the previouscorollary, there exists q0 ( ... ( qn = q such that qi ∩R = pi for all i. Thus ht(q) ≥ n.

Remark 112. Let R = K[x1, ..., xn], where each xi is an indeterminant, and K is a field. Consider amonomial dxα1

1 ...xαnn , where each αi ≥ 0, and d ∈ K \ {0}. Let a1, ..., an−1 be positive integers.For i = 1, ..., n − 1, let yi = xi − xain , and let yn = xn. Note that K[x1, ..., xn] = k[y1, ..., yn] = R[yn].

Then, dxα11 ...xαnn = d(y1 + yain )α1 ...(yn−1 + y

an−1n )αn−1yαnn .

Note that the highest degree term in yn is dyα1a1+...+αn−1an−1+αnn (and there are other terms as well).

Note that this is “monic” in yn, in the sense that yn has no yi coefficients in front of it.

Let α = α1, ..., αn. Suppose F (x1, ..., xn) =∑finite

dαxα11 ...xαnn , and that F (x1, ..., xn) 6= 0.

We wish to do a change of variables such that this polynomial becomes “monic” in yn . We already canmake each monomial be monic, but this could backfire if the monomials have the same exponents and addup to a leading coefficient of 0.

However, we can make a choice of the ai so that this doesn’t happen. Since F (x1, ..., xn) is a finite sum,then we can choose an integer c which is larger than any αi appearing in a nonzero term in F . Let ai = ci

for i = 1, ..., n− 1.Then, with yi = xi − xcin for i = 1, ..., n − 1, and yn = xn, we get, as before, terms of the form

dαyαnc

0+α1c+...αn−1cn−1

n .But note αnc

0 + α1c+ ...+ αn−1cn−1 = βnc

0 + ...+ βn−1cn−1 with βi < c for all i if and only if αi = βi.

(Because these are essentially base-c expansions for numbers, and these are distinct.)Thus, each nonzero term of F becomes “monic” in yn of distinct degrees in yn. Hence, F becomes

“monic” in yn.

111

9 Affine Rings

Definition 112. Let K be a field. An affine ring (or an affine K-algebra) is a finitely generated K-algebra.That is, an affine ring is a ring of the form K[u1, ..., un], where the uis may be either indeterminant oralgebraic.

Remark 113. If K[u1, ..., un] is an affine ring, then define φ : K[x1, ..., xn]→ K[u1, ..., un] by f(x1, ..., xn)→f(u1, ..., un). Therefore K[u1, ..., un] ∼= K[x1, ..., xn]/I, where I = kerφ.

Thus, an equivalent definition of an affine ring is one which is the homomorphic image of a polynomialring.

Lemma 34. (Noether Normalization Lemma) Let A = K[u1, ..., un] be an affine K-algebra. Then thereexists y1, ..., yr ∈ A which are algebraically independent over K, such that A is integral over the polynomialring R = K[y1, ..., yr].

We will prove the Noether Normalization Lemma next class.

Remark 114. Let A, K, and R be as in the Noether Normalization Lemma. Then, since A is finitelygenerated as a ring over K ⊂ R, then A is finitely generated as a ring over R. Then, since A is also integralover R, then A is finitely generated as an R-module.

Remark 115. Let A = K[u1, ..., un] be an affine K-algebra. Let y1, .., yr be the algebraically independentelements over K guaranteed by the Noether Normalization Lemma. Then dimA = dimR = r.

9.1 Day 28 - March 16

We are about to prove the Noether Normalization Lemma.However, we need an exercise first.

Exercise 33. Let C ⊂ B ⊂ A be rings. Suppose B is integral over C, and A is integral over B. Then A isintegral over C.

Lemma 35. (Noether Normalization Lemma) Let A = K[u1, ..., un] be an affine K-algebra. Then thereexists y1, ..., yr ∈ A which are algebraically independent over K, such that A is integral over the polynomialring R = K[y1, ..., yr].

Proof. Choose a set of y1, .., yr ∈ A such that A is integral over R = K[y1, .., yr] and such that r is the leastpossible. (Note that since u1, ..., un is such a set, then a “least possible” set exists, and r ≤ n).

It then suffices to show that {y1, ..., yr} is algebraically independent over K.Suppose for the sake of contradiction that {y1, ..., yr} is algebraically depended over K. Then there exists

f(T1, ..., Tr) ∈ K[T1, .., Tr] \ {0} such that f(y1, ..., yr) = 0.But by the hideous lemma last class, there exists a change of variables of the formX1 = T1−T a1r , ..., Xr−1 =

Tr−1−T ar−1r , Xr = Tr such that g(X1, ..., Xr) := f(X1 +Xa1

r , ..., Xr) = cXNr + ...+[lower order terms in Xr]

is monic in terms of Xr.Then let z1 = y1 − yra1, ...zr−1 = yr−1 − yar−1

r , zr = yr. Note that g(z1, ..., zr) = f(z1 + za1r , ..., zr) =

f(y1, ..., yr) = 0. But also observe that K[z1, ..., zr] = K[y1, ..., yr]. Since zr is a root of g(z1,...,zr−1,Xr)c , then

zr is integral over K[z1, ..., zr−1].But, by the exercise, integral extensions are transitive, so A is integral over K[z1, ..., zr−1]. Thus our

choice of r was not minimal, and this is a contradiction. Thus {y1, ..., yr} are algebraically independent overK.

Definition 113. If A = K[u1, ..., un] is an affine K-algebra, and y1, ..., yr are a set of algebraically indepen-dent elements guaranteed by the Noether Normalization Lemma, then we say that R = K[y1, ..., yr] is calleda Noether normalization for A.

Remark 116. Suppose A is an affine K-domain, and let K[y1, ..., yr] be a normalization.Then Q(A) is algebraic over k(y1, ..., yr), so {y1, ..., yr} is a transcendence base for Q(A)/K. Therefore

r is the transcendence degree of Q(A)/K.

112

Remark 117. In general, dimA = dimK[y1, ..., yr]. It is a fact that dimK[y1, ..., yr] = r, so by transitivity,dimA = r.

Theorem 92 (Nullstellensatz - Strong Form). Let A be an affine K-algebra which is a field. Then Ais algebraic over K.

Proof. By the Noether Normalization Lemma, there exists a normalization R = k[y1, ..., yr] of A.Then R ⊂ A is an integral extension. We showed that, in an integral extension, the top ring is a field if

and only if the bottom ring is a field. Therefore K[y1, ..., yr] is a field. However, the yi are transcendentalover K, so K[y1, ..., yr is isomorphic to a polynomial ring.

Recall that polynomial rings are fields if and only if they have 0 variables. Thus r = 0, and R = K.Then A is integral over K, so A is algebraic over K.

Corollary 45. Let K be an algebraically closed field, and let R = K[x1, ..., xn] be a polynomial ring. Thenm is a maximal ideal of R if and only if m = (x1 − c1, ..., cn − cn), where c1, ..., cn ∈ K.

Proof. Note that (x1 − c1, ..., xn − cn) is the kernel of the ring surjection K[x1, ..., xn] → K given byf 7→ f(c1, ..., cn). Thus m is always maximal, regardless of the structure of K.

Conversely, suppose m is a maximal ideal of R. Note that there is a ring map K → R→ R/m. Since mis maximal, then m ∩K = (0), so this map is injective.

Consider K ⊂ R/m = K[x1, ..., xn]. Since R/m is a field and finitely generated as a K-algebra, then bythe Strong Form of the Nullstellensatz, R/m is an algebraic extension of K. Since K is algebraically closed,then R/m ∼= K. Therefore the map K → K[x1, ..., xn]/m is an isomorphism. Therefore, for each xi, thereexists ci ∈ K such that xi = xi + m = ci + m. Thus xi − ci ∈ m for all i. Thus m ⊃ (x1 − c1, ..., xn − cn).Since these ideals are both maximal, they are equal. Thus m = (x1 − c1, ..., xn − cn).

9.2 Day 29 - March 18

Theorem 93 (“Trick of Rabinowitsch”). Let R be a commutative ring, and I be an ideal of R, and let

x be an indeterminant over R. Then√I =

⋂m maximal in R[x]

m⊃I

(m ∩R).

Proof. Certainly, I ⊂⋂

m maximal in R[x]m⊃I

(m ∩R). Also, this is a radical ideal, so√I ⊂


m⊃I

(m ∩R).

Let f ∈⋂

m maximal in R[x]m⊃I

(m ∩R), and consider J = (I, fx− 1) ⊂ R[x].

We wish to show that J = R[x]. Suppose for the sake of contradiction that J 6= R. Then let m be amaximal ideal in R[x] such that J ⊂ m. Therefore I ⊂ m, so f ∈ m. But fx − 1 ∈ m, so 1 ∈ m. Thiscontradicts the maximality of m. Thus R = J .

Therefore we can write 1 = i1g1 + ...+ ingn + (fx− 1)gn+1, where gi ∈ R[x] and ij ∈ I for all j. If f is

nilpotent, then f ∈√I, so we’re done. If f is not nilpotent, then Rf 6= 0.

Consider the ring homomorphism φ : R[x]→ Rf given by h(x) 7→ h( 1f ). Now let’s apply φ to the earlier

equation.

Then 1 =i1g′1

fα +i2g′2

fα +...+ing′n

fα in Rf . That is, there exists an fs such that fs+α = i1fsg′1+...+inf

sg′n ∈ I.

Thus f ∈√I.

Thus√I =


m⊃I

(m ∩R).

Proposition 90. Let A be an affine K-algebra. Let x be an indeterminant. Then every maximal ideal ofA[x] contracts to a maximal ideal of A.

113

Proof. Since A is an affine K-algebra, then so is A[x]. Let m be a maximal ideal of A[x].Then A[x]/m is algebraic over K by Hilbert’s Nullstellensatz. Therefore we get the containment chain

K ⊂ A/(m ∩ A) ⊂ A[x]/m. Since the total extension is algebraic, then A/(m ∩ A) is integral over K. Also,A/(m ∩ A) is a domain. Since integral domain extensions over a field are still fields, then A/(m ∩ A) is afield.

Thus m ∩A is a maximal ideal in A.

Theorem 94. Let A be an affine K-algebra, and let I be an ideal of A. Then√I =

⋂m maximal

m⊃I

m.

Proof. By the Trick of Rabinowitsch,√I =


m⊃I

(m ∩R). Then by the previous proposition, if m

is maximal in R[x], then m ∩R is maximal in R. Thus√I =

⋂m maximal

m⊃I

m.

Remark 118. Suppose K = K. Let R = K[x1, ..., xn]. Then every maximal ideal is of the form mp =(x1 − c1, ..., cn − cn) where p = (c1, ..., cn) ∈ Kn. Therefore f ∈ mp if and only if f(p) = 0. Then I ⊂ mp ifand only if f(p) = 0 for all f ∈ I.

10 Algebraic Geometry

10.1 Day 30 - March 28

We’re back from break!

Definition 114. Let K be a field, and let n ≥ 1. Let AnK = Kn = {(a1, ..., an)|ai ∈ K}. Then AnK is calledaffine n-space over K. The elements of AnK are called points.

Definition 115. Let R = K[x1, ..., xn], and let S ⊂ R. The zero set of S is Z(S) = {p ∈ AnK |f(p) =0 for all f ∈ S}. Then Z(S) is called an (affine) algebraic K-variety .

Remark 119. Other people have other terms for varieties. Hartshorne and followers reserve the term“variety” for irreducible varieties. They use the term algebraic set for a not-necessarily-irreducible algebraicvariety.

Example 117. Let K = R. Let f = y − x2 ∈ R[x, y]. Then Z(f) is the graph of the curve y = x2.

Example 118. Let K = R. Let f = x − 12 and g = y2 − y as elements of R[x, y]. Let S = {f, g}. ThenZ(S) = {(1, 0), (1, 1)}.

Remark 120. By generalizing the previous example, one can construct any finite set as an algebraic variety.

Example 119. Let K = R. Let f = z2 − x2 − y2 and let g = z − 1 as elements of R[x, y]. Let S = {f, g}.Then Z(S) is the circle of radius 1 lying in the plane z = 1.

Remark 121. Let K be a field, and let R = K[x1, ..., xn]. Let S ⊂ T ⊂ R. Then observe that Z(T ) ⊂Z(S).

Proposition 91. Let K be the field, and let R = k[x1, ..., xn]. Then,

1. Let {Sα}α∈I , where Sα ⊂ R for all α. Then Z(⋃α Sα) =

⋂α Z(Sα).

2. Let S, T ⊂ R. Then Z(ST ) = Z(S) ∪ Z(T ).

3. Let S ⊂ R. Then Z(S) = Z((S)), where (S) is the ideal generated by S.

114

4. Let I ⊂ R be an ideal. Then Z(I) = Z(√I).

5. AnK = Z((0)) = Z(∅) and ∅ = Z(1).

Proof. It is very easy to prove (1), so we will omit the proof.Let us now prove (2). Certainly, Z(ST ) ⊂ Z(S) ∪ Z(T ). Conversely, suppose p ∈ Z(ST ) by p 6∈ Z(S).

Then there exists an f ∈ S such that f(p) 6= 0.Then, for all g ∈ T , fg ∈ ST , so fg(p) = f(p)g(p) = 0 as p ∈ Z(ST ). Since f(p) 6= 0 and this

multiplication is taking place in k, then g(p) = 0. Since g was arbitrary, then g(p) = 0 for all g ∈ T . Thusp ∈ Z(T ), so Z(ST ) = Z(S) ∪ Z(T ), as desired.

Let us now prove (3). Since S ⊂ (S), then Z((S)) ⊂ Z(S). Let p ∈ Z(S). Then, for all fi ∈ Z(S),fi(p) = 0. Let f ∈ (S). Then f = g1f1 + ...+gtft for some gi ∈ R, so f(p) = g1(p)f1(p)+ ...+gt(p)ft(p) = 0.Thus f(p) = 0 for all f ∈ (S), so p ∈ Z((S)). Thus Z(S) = Z((S)), as desired.

Let us now prove (4). If I ⊂ R is an ideal, then I ⊂√I. Therefore, Z(

√I) ⊂ Z(I).

Let p ∈ Z(I), and let f ∈√I. Then there exists a t such that f t ∈ I, so f t(p) = 0. But this

exponentiation is taking place in K, so f(p) = 0. Thus p ∈ Z(√I), so Z(

√I) = Z(I), as desired.

It is easy to verify (5) on your own.

Corollary 46. The collection of zero sets satisfy the axioms of the closed sets of a topology on AnK .

Definition 116. The Zariski Topology on AnK is the topology whose closed sets are the zero sets of polyno-mials.

Remark 122. The Zariski topology on Rn is strictly coarser than the Euclidean topology.

Corollary 47. Let V = Z(S) be an affine K-variety. Then there exists f1, ..., ft ∈ R = k[x1, ..., xn] suchthat V = Z(f1) ∩ ... ∩ Z(ft). (In fact, one can choose t ≤ n, but this result is hard.)

Proof. Recall that V = Z(S) = Z((S)). Since R = k[x1, .., xn] is Noetherian, then each ideal is finitelygenerated. In particular, (S) = (f1, ..., ft) for some fi ∈ R.

Then, V = Z(S) = Z((S)) = Z((f1, ..., ft)) = Z(f1, ..., ft) = Z(f1) ∩ ... ∩ Z(ft).

Definition 117. Let U ⊂ AnK . Define the vanishing ideal of U to be I(U) = {f ∈ R|f(p) = 0 for all p ∈U}.

Remark 123. Vanishing ideals are, in fact, ideals. Also, 0 ∈ I(U).

Proposition 92. Let U ⊂ AnK . Then, I(U) is a radical ideal.

Proof. Suppose f ∈√I(U). Then fr ∈ I(U) for some r, so fr(p) = 0 for all p ∈ U . This exponentiation is

taking place inside of K, a field, so f(p) = 0. Thus f ∈ I(U). Since f was arbitrary, then√I(U) = I(U).

Thus I(U) is a radical ideal.

Remark 124. We have a correspondence between the set of affine K-varieties, and the set of radical idealsin K[x1, ..., xn].

Proposition 93. For any variety V of AnK , we have that V = Z(I(V )).

[proof omitted]

10.2 Day 31 - March 30

Recall from last class the following results:

Remark 125. We have a correspondence between the set of affine K-varieties, and the set of radical idealsin K[x1, ..., xn].

Proposition 94. For any variety V of AnK , we have that V = Z(I(V )).

115

However, the I(Z(J)) 6=√J for an ideal J . Let’s look at an example.

Example 120. Let k = R, and n = 1. Let J = (x2 + 1) ⊂ R[x]. Then Z(J) = ∅. Therefore I(Z(J)) =I(∅) = R[x]. However,

√(x2 + 1) 6= R[x].

Theorem 95 (Nullstellensatz - geometric version). Let K = K, and let J be an ideal of K[x1, ..., xn].Then I(Z(J)) =

√J .

Proof. If p = (a1, ..., an) ∈ AnK , then let mp = (x1 − a1, ..., xn − an).Recall that f ∈ mp if and only if f(p) = 0.Therefore, for an ideal J , recall that J ⊂ mp if and only if f(p) = 0 for all f ∈ J , which is the case if and

only if p ∈ Z(J).Recall that, as K = K, then every maximal ideal is of the form mp for some p ∈ AnK .

Now we are ready to show that I(Z(J)) =√J . Note that

f ∈ I(Z(J)) ⇐⇒ f(p) = 0 for all p ∈ Z(J)

⇐⇒ f ∈ mp for all p ∈ Z(J)

⇐⇒ f ∈ mp for all mpsupsetZ(J)

⇐⇒ f ∈⋂

p∈Z(J)

mp

However,⋂

p∈Z(J)

mp =⋂

m maximalm⊃J

m =√J .

Thus f ∈ I(Z(J)) if and only if f ∈√J .

Corollary 48. If K = K, then there exists a bijective, inclusion-reversing correspondence between subva-rieties of AnK and radical ideals of K[x1, ..., xn].

And that’s everything we need to know about algebraic geometry!

11 Invariant Theory

Now we get to just talk about a random grabbag of topics! Let’s start with invariant theory.Let K be a field, and let R = K[x1, ..., xn]. Let Sn act on R by σ(f(x1, ..., xn)) = f(xσ(1), ..., xσ(n)).

(Really, what we are doing is defining a σ : R → R for all σ ∈ Sn. Then each σ is an automorphism of Rfixing K, so Sn is (isomorphic to) a subgroup of AutK(R).)

Definition 118. Let RSn := {f ∈ R|σ(f) = f for all σ ∈ Sn}. Then we call RSn the fixed subring of R oran invariant subring of R.

Way back when, we showed thatRSn = K[s1, ..., sn], where the sis are the elementary symmetric functionsin x1, ..., xn.

Remark 126. The “first problem of Invariant Theory” is the following: If R = K[x1, ..., xn], and G is afinite subgroup of AutK(R), then is RG a finitely generated K-algebra?

Hilbert proved the answer was yes in 1890. Later, he showed (in some cases), what the generators wouldbe.

Noether gave a cool proof in the 1920s, so let’s look at that.

Theorem 96. Let R be a finitely generated K-algebra, and let G be a finite subgroup of AutK(R). ThenRG is a finitely generated K-algebra.

116

Proof. Let R = K[u1, ..., un], let r = |G|, and let t be an indeterminant. For each 1 ≤ i ≤ n, consider

fi(t) =∏σ∈G

(t− σ(ui)) = tr + ci,1tr−1 + ...+ ci,r.

Note that fσi (t) = fi(t) for all σ ∈ G. Therefore, σ(ci,j) = ci,j for all i, j, so fi(t) ∈ RG[t].Let S = K[ci,j ]1≤i≤n

1≤j≤r⊂ RG.

Note that fi(ui) = 0 for all i, so each ui is integral over S. Therefore, R = K[u1, ..., un] is integral overS. However, R is a finitely generated S-algebra (in particular, R = S[u1, ..., un]), so R is a finitely generatedS-module.

But since S is a finitely generated K-algebra, then by the Hilbert basis theorem, S is Noetherian.Therefore, R is a Noetherian S-module, so RG is an S-submodule of R. Hence, RG is a finitely generated

S-module. Therefore, RG is a finitely generated S-algebra. Since S is a finitely generated K-algebra, thenby the transitivity of finitely-generatedness, RG is a finitely generated K-algebra.

Corollary 49. With R and G as above, then by the Noether Normalization Lemma, there exists y1, ..., yl ∈RG which are algebraically independent, such that R is a finitely generated K[y1, ..., yl]-module.

Note that this is a bit weaker than what happened when G = Sn, where K[y1, ..., yl] = RG, instead ofthe latter only being finitely generated as a module over the former.

Question 3. Let R be a Noetherian domain, and let K = Q(R). Let R denote the integral closure of R inK. Is R a finitely-generated R-module?

Question 4. Let R be a Noetherian, integrally closed domain. Let L be a finite algebraic field extension ofK = Q(R). Let S be the integral closure of R in L. Is S a finitely generated R-module?

The answer to Question 3 is yes, in the case that R is a finitely generated k-algebra with char k = 0.

12 Extensions Of A Field of Fractions

12.1 Day 32 - April 1

Our final will be on Monday May 2, 10-12. It will be open note, but not open laptop.

Remark 127. Recall the following results about separable extensions: Let E/F be a finite separable ex-tension of fields. Since E/F is finite, it is automatically algebraic.

Let σ1, ..., σr be the distinct field embeddings of E into E fixing F . Recall that TrEF = σ1+...+σr : E → Fis an F -linear functional. Furthermore, recall that since E/F is separable, then TrEF 6= 0.

Theorem 97. Let E/F be a finite separable extension of fields. Let {α1, ..., αr} be an F -basis for E. Thenthere exists a basis {β1, ..., βr} of E such that TrEF (βiαj) = δi,j for all i, j.

Proof. Let E∗ = HomF (E,F ). Then dimF E∗ = dimF (E) · dimF (F ) = dimF E. Define α∗j : E → F by

α∗j : αi 7→

{0 if i 6= j

1 if i = j. That is, α∗j (αi) = δi,j .

(Vocabulary lesson: we say that {α∗1, ..., αr∗} is the dual basis for E∗ corresponding to {α1, ..., αr}.)For each β ∈ E, define fβ : E → F by γ → TrEF (βγ). One can verify that fβ is F -linear, so fβ ∈ E∗.Thus the map φ : E → E∗ by β 7→ fβ is F -linear. We now wish to show that φ is injective. Suppose

φ(β) = 0 for some β ∈ E. Then, for all γ ∈ E, 0 = φ(β)(γ) = fβ(γ) = TrEF (βγ). If β 6= 0, then βE = E, so

it must be that TrEF = 0, which is nonsense. Therefore, β = 0, so φ is injective.Since dimF (E) = dimF (E∗), then φ is also surjective, so φ is an isomorphism.

117

Let βi = φ−1(α∗i ) for all i. Then

TrEF (βiαj) = fβi(αj)

= φ(βi)(αj)

= φ(φ−1(α∗i ))(αj)

= α∗i (αj)

= δi,j

as desired.

Lemma 36. Let R be a Noetherian integrally closed domain. Let K be the field of fractions of R. Let L bea finite separable extension of K. Let S be the integral closure of R in L. Let W = R \ {0}. Then SW = L(and in particular, Q(S) = L).

Proof. Certainly, SW ⊂ L.Conversely, let α ∈ L. Then α is algebraic over K = Q(R), so there exists ci, d ∈ R such that αn +

cn−1

d αn−1 + ...+ c0d = 0. Multiplying by dn, we get that (dα)n+cn−1(dα)n−1 + ...+c1d

n−2(dα)+c0dn−1 = 0.

Thus, dα is integral over R, so dα = f ∈ S. Then α = fd ∈ SW .

Thus L ⊂ SW , so L = SW , as desired.

Lemma 37. Let R be a Noetherian integrally closed domain. Let K be the field of fractions of R. Let Lbe a finite separable extension of K. Let S be the integral closure of R in L. Then TrLK(S) ⊂ R.

Proof. Let u ∈ S. Then u is a root of a monic polynomial f(x) ∈ R[x]. Let σ : L → L be an embeddingfixing K. Since R ⊂ K, then σ fixes the coefficients of f(x). Thus σ(u) is a root of f(x). Therefore σ(u) ∈ Lis integral over R. Therefore TrLK(u) = σ1(u)+ ...+σr(u) is integral over R and is in K. Since R is integrallyclosed in K, then TrLK(u) ∈ R, as desired.

Theorem 98. Let R be a Noetherian integrally closed domain. Let K be the field of fractions of R. Let Lbe a finite separable extension of K. Let S be the integral closure of R in L. Then S is finitely generated asan R-module.

Proof. Let α1, ...αr} be a K-basis for L. By the first lemma, there exist α′i ∈ S and a d ∈ R such that

αi =α′id for i = 1, ..., r. Then {α′1, ..., α′r} is also a K-basis for L¡ so we can assume {α1, ..., αr} ⊂ S.

By the previous theorem, there exists a K-basis for L {β1, ..., βr} such that TrLK(βiαj) = δi,j .We now wish to show that S ⊂ Rβ1 + ...+Rβr.Let u ∈ S. We already know that S ⊂ Kβ1+...+Kβr, so there exists ci ∈ K such that u = c1β1+...+crβr.

Then, uαi = c1β1αi+...+crβrαi, so TrLK(uαi) = ci. Since uαi ∈ S, then by the second lemma, TrLK(uαi) ∈ R.Thus ci ∈ R for all i, so u ∈ Rβ1 + ...+Rβr.

Then, since R is Noetherian, M = Rβ1 + ... + Rβr is a Noetherian R-module. Therefore, S is anR-submodule of M , so S is finitely generated as an R-module.

Theorem 99. Let K be a field with char K = 0, and let R be an affine K-domain. Then the integralclosure of R in its field of fractions is a finitely generated R-module (and is therefore Noetherian).

Proof. By the Noether Normalization Lemma, there exists algebraically independent elements y1, ..., yd ∈ Rsuch that R is a finitely generated module over the subring T = k[y1, ..., yd] (so the extension T ⊂ R isintegral).

Let E be the field of fractions of R, and let S be the integral closure of R in E. Note that S is also theintegral closure of T in E.

Let F be the field of fractions of T . Since R is a finitely generated T -module, then E is a finitely generatedF -module. As char K = 0, then E/F is separable.

Therefore, by the previous theorem, S is a finitely generated T -module. Since T ⊂ R ⊂ S, then S is alsofinitely generated as an R-module.

118

12.2 Day 33 - April 4

Recall the following theorems from last class:

Theorem 100. Let R be a Noetherian integrally closed domain. Let F = Q(R), and let E be a finiteseparable extension of F , and let S be the integral closure of R in E. Then S is finitely generated as anR-module.

Theorem 101. Let K be a field of characteristic 0, and let R be an affine K-domain. Let E = Q(R), andlet S be the integral closure of R in E. Then S is a finitely-generated R-module.

Let’s also talk about some old definitions.

Remark 128. Let F be a field of characteristic p, and let E/F be a field extension. Recall the following:

1. We say α ∈ E is purely inseparable over F if αpn ∈ F for some pn.

2. We say E/F is purely insepable if each α ∈ E is purely inseparable over F .

3. If E/F is finite, then E/F is purely inseparable if and only if Epn ⊂ F for some n.

4. Let F p.i. = {α ∈ E|α is purely inseparable over F}. We called F p.i. the purely inseparable closure ofF in E, andit is a subfield of E containing F .

5. If E/F is normal, then E/F p.i. is separable and F p.i./F is purely inseparable.

Remark 129. Let R be a domain of characteristic p. Let K be an algebraic closure of Q(R). For any c ∈ Rand n ≥ 1, then the polynomial xp

n − c ∈ R[x] ⊂ K[x] has a unique root in K. (In particular, if α is a root,then xp

n − c = (x− α)pn

.)

We will denote this root by c1pn .

Furthermore, for c, d ∈ R, we have that c1pn + d

1pn = (c+ d)

1pn and c

1pn · d

1pn = (c · d)

1pn

Let R1pn = {c

1pn |c ∈ R}. Then R

1pn is a subring of K containing R. Also, R ( R

1p ( R

1p2 ( ... ⊂ K.

However, note that, for all n ≥ 1, R ∼= R1pn by the isomorphism c 7→ c

1pn .

Theorem 102. Let E/F be a field extension, and let char F = p. Suppose {y1, ..., yd} ⊂ E is algebraically

independent over F . Let n ≥ 1. For any algebraic extension L/F , then {y1pn , ..., y

1pn } is algebraically

independent over L.

Proof. Recall that {y1, ..., yd} is algebraically independent over F if and only if Tr . deg. (F (y1, ..., yd)/F ) = d.

It then suffices to show that Tr . deg. (L(y1pn

1 , ..., y1pn

d )/L) = d. However, we can do so by a big complicatedpicture.

From this, we can conclude that the p-th root of an indeterminant will still be an indeterminant.

Lemma 38. Let F be a field of characteristic p > 0. Let x1, ..., xd be indeterminants over F , and let L be

a finite field extension of F . Then L[x1pn , ..., x

1pn ] is a finitely generated F [x1, ..., xd]-module.

[Proof omitted.]

12.3 Day 34 - April 8

(There was no class on Wednesday.)

Hint for a homework problem: Let α ∈⋂

m maximal

Rm. Then let α = ab . We want to show that ((b) :R

a) = R.Recall this theorem:

119

Theorem 103. Let R be a Noetherian integrally closed domain. Let K be the field of fractions of R. LetL be a finite separable extension of K. Let S be the integral closure of R in L. Then S is finitely generatedas an R-module.

Proof.

Theorem 104. Let K be a field, and let A = K[x1, ..., xd] be a polynomial ring over K. Let F = Q(A) =K(x1, ..., xd). Let E be a finite field extension of F , and let B be the integral closure of A in E. Then B isfinitely generated as an A-module.

Proof. We will first show the theorem when E/F is separable or purely inseparable.By the recalled theorem, if E/F is separable, then the theorem is true.

If E/F is purely inseparable, let char K = p, and write E = F (α1, ..., αl) where αpn

i ∈ F for a fixed n

and for all i. Since F is the field of fractions of A = K[x1, ..., xn], then αpn

i = fi(x1,...,xd)gi(x1,...,xd) for some fi, gi ∈

K[x1, ..., xd]. Note that we can in fact commonize the denominators and assume gi(x1, ..., xd) = g(x1, ..., xd)for some g ∈ K[x1, ..., xd].

Then, αi = fi(x1,...,xd)1pn

g(x1,...,xd)1pn

=fi(x

1pn

1 ,...x1pn

d )

g(x1pn

1 ,...x1pn

d )

, where fi, g ∈ K1pn [x

1pn

1 , ...., x1pn

d ].

Let S be the set of all the pnth roots of all the coefficients of the fis and g. Let K ′ = K(S) = K[S]. Then

K ′/K is finite and purely inseparable. Also, fi, g ∈ K ′[x1pn

1 , ..., x1pn

1 ] = T . Let L = Q(T ) = K ′(x1pn

1 , ..., x1pn

1 ).Note that all αi ∈ L.

Then, let C denote the integral closure of A in L. We wish to show that C = T . Certainly, T is integral

over A since (x1pn

i )pn

= xi ∈ A for all i and K ′ps ⊂ K for some s. Thus T ⊂ C. However, T is integrally

closed in Q(T ) = L. Thus T = C.

Recall that C = K ′[x1pn

1 , ..., x1pn

1 ] = T is finitely generated as a module over K ′[x1, .., xd], which in turn isfinitely generated as a module over A (since [K ′ : K] <∞). Then, by transitivity of finitely-generated-ness,C is a finitely-generated A-module. Since B is an A-submodule of C, and A is Noetherian, then B is finitelygenerated as an A-module. This completes the proof in the purely inseparable case.

If E/F is neither separable nor purely inseparable, then we proceed as follows. First, we will show thatwe can assume without loss of generality that E/F is normal.

To this end, let E be the normal closure of E/F . Let B be the integral closure of A in E. SupposeB is a finitely generated A-module. If we can show that B is a Noetherian A-module, then since B is anA-submodule of B, then B is a finitely generated A-module.

Therefore, it suffices to show the case where E/F is finite and normal. Let E′ = F p.i. Then E′/Fis purely inseparable, and E/E′ is separable. Let B′ be the integral closure of A in E′. By the purely-inseparable case, B′ is finitely generated as an A-module. Then B′ is a Noetherian, integrally closed domain(since Q(B′) = E′). Therefore B is the integral closure of B′ in E.

By the separable case, B is a finitely generated B′-module. Since we assumed that B′ is a finitelygenerated A-module, then by transitivity of finitely-generated-ness, B is a finitely generated A-module.

13 Representation Theory Revisited

13.1 Day 35 - April 11

Recall the following things about group-rings.

Remark 130. If G is a finite group, and C is the field of complex numbers, then

C[G] = B(I1)× ...×B(It)∼= n1I1 ⊕ ...⊕ ntIt

, where I1, ..., It represent all the simple left ideals of C[G] up to isomorphism, and B(Ij) is the sum of allthe left ideals of C[G] which are isomorphic to Ij .

120

We then have the following results

1. t is equal to the number of conjugacy classes of G.

2. ni = dimC Ii for 1 ≤ i ≤ t.

3. B(Ii) ∼= Mni(C) for 1 ≤ i ≤ t.

Remark 131. Let R be a commutative ring, and let G be a finite group. Then Z(R[G]) = Rz1 ⊕ ...⊕Rzt,where zi =

∑g∈Ci

, where C1, ..., Ct are the distinct conjugacy classes of G.

In particular, Z(Z[G]) = Zz1 ⊕ ...⊕ Zzt ⊂ Cz1 ⊕ ...⊕ Czt = Z(C[G]).

For the rest of today, let A be the integral closure of Z in C. That is, A denotes the set of algebraicintegers.

We will now build up to the fact that ni

∣∣∣|G| for all i.

Proposition 95. For each i, let ei = ni|G|

∑g∈G

χi(g−1)g. Then zi ∈ Ae1 + ...+Aet.

Proof. Recall that Z(C[G]) ∼= Ce1 × ...× Cet (as an internal direct product).Since zi ∈ Z(C[G]) = Ce1× ...×Cet., then there exist αi ∈ C such that zi = α1e1 + ...+αtet. Note that

zi ∈ Z(Z[G]) = Zz1 + ...Zzt.As Z(Z[G]) is a commutative ring, and is finitely generated as a Z-module, then Z(Z[G]) is integral over

Z. Let f(x) ∈ Z[x] be a monic polynomial such that f(zi) = 0. Then f(α1e1 + ...αtet) = 0.That is,

0 = f(α1e1 + ...αtet)

= f(α1e1) + ...+ f(αtet)

= f(α1)e1 + ...f(αt)et

Note that the second equality is because eiej = 0 for i 6= j, and the third equality is because eiei = ei.But since e1, ...et is linearly independent over C, then each f(αj) = 0 for all j. Thus αj ∈ A for all j, so

zi ∈ Ae1 + ...+Aet.

Lemma 39. Let χ be any character for G and let g ∈ G. Then χ(g) ∈ A for all g ∈ G.

Proof. Recall the definition of χ(g): χ(g) = Tr(ρ(g)) where ρ : G → GLC(V ) is a representation. Sincegn = 1 for some n, then ρ(g)n = 1V . Thus every eigenvalue of ρ(g) is an n-th root of unity.

But each n-th root of unity is an algebraic integer! Also, the trace of a matrix is the sum of its eigenvalues(with some multiplicities), so χ(g) = Tr(ρ(g)) is the sum of some algebraic integers. Since A is closed underaddition (because it is a ring), then χ(g) ∈ A.

Theorem 105. For each i = 1, ..., t, ni

∣∣∣|G|.Proof. Recall that |G|ni ei =

∑g∈G

χi(g−1)g =

t∑j=1

χi(g−1j )zj , where each gj ∈ Cj .

But this is in Az1 + ... + Azt since χi(g−1j ) ∈ A by the Lemma. By the proposition, Az1 + ... + Azt ⊂

Ae1 + ... + Aet. Thus |G|ni ∈ A. However, we can see that |G|ni ∈ Q, so |G|ni ∈ A ∩ Q = Z. In other words,

ni

∣∣∣|G|.Now let’s talk about the representations of products of groups.

121

Definition 119. Let ρ1 : G1 → GLC(V1) and let ρ2 : G2 → GLC(V2) be representations of some groups G1

and G2.Define ρ1 ⊗ ρ2 : G1 ×G2 → GLC(V1 ⊗C V2) by (g1, g2) 7→ ρ1(g1)⊗ ρ2(g2).

Remark 132. One can verify that ρ1 ⊗ ρ2 is a group homomorphism.

Exercise 34. Show that Tr(φ⊗ ψ) = Tr(φ) · Tr(ψ).From this, one gets that χρ1⊗ρ2((g1, g2)) = χρ1(g1) · χρ2(g2).

Corollary 50. The following are equivalent:

1. ρ1 ⊗ ρ2 is irreducible as a representation

2. χρ1⊗ρ2 is irreducible as a character

3. 〈χρ1⊗ρ2 , χρ1⊗ρ2〉 = 1

4. 〈χρ1 , χρ1〉 = 1 and 〈χρ2 , χρ2〉 = 1

5. ρ1 and ρ2 are irreducible as representations.

Proposition 96. If {ρ1, ...ρt} and {φ1, ..., φs} are the irreducible representations for G1 and G respectively,then {ρi ⊗ φj |1 ≤ i ≤ t, 1 ≤ j ≤ s} is the set of irreducible representations for G1 ⊗G2.

Lemma 40. Let φ : G → GLC(V ) be an irreducible representation. Let g ∈ Z(G). Then ρ(g) = λI forsome root of unity λ. In particular, |χ(g)| = χ(1)

Proof. Recall that Z(C[G]) = Ce1 × ... × Cet (as an internal direct product). Then, we can write g =(α1, ..., αt) to represent g = α1e1 + ...+ αtet.

Since V is a simple C[G]-module, and C[G] = B(I1)× ...×B(It), then V ∼= Ij as C[G]-modules, for somej. Without loss of generality, let V ∼= Ij .

Then e1 = (1, 0, ..., 0) acts as the identity on V , so ejV = 0 for all j 6= 1. Let u ∈ V . Then gu =(α1, ..., αt)u = (α1u, 0, ..., 0). Thus ρ(gi) = α1IV . Since |G| is finite, then g has finite order, so αi must be a|G|-th root of unity.

13.2 Day 36 - April 13

Proposition 97. Let R be a Noetherian ring, and let S be a domain containing R. Let u ∈ S and supposethere exists a nonzero r ∈ R such that run ∈ R for all n ≥ 1. Then u is integral over R.

Proof. Recall that u is integral over R if and only if R[u] is finitely generated as an R-module.Consider R · 1

r as an R-submodule of Q(R). Since un ∈ R · 1r for all n, then R[u] ⊂ R · 1

r .Recall that, since R is Noetherian, then any submodule of a finitely generated R-module is also finitely

generated.But R · 1

r is a finitely generated R-module (namely, it is generated by 1r ), so R[u] is a finitely-generated

R-module. Thus u is integral over R.

Theorem 106 (Schur). With the usual notation for a character χ, then ni

∣∣∣[G : Z(G)] for all i.

Proof. For all integers m ≥ 1, let Gm = G×G× ...×G (m times). Let χ be an irreducible character, andlet n = degχ = χ(1). Then there exists a representation ρ : G → GLC(V ) such that Tr(ρ) = χ. We then

wish to show that n∣∣∣[G : Z(G)].

Let ρm = ρ⊗ ρ...⊗ ρ : Gm → GLC(V ⊗C ...⊗C V ) (where both tensor products occur m times).Since ρ is irreducible, then ρm is an irreducible representation for Gm. Also, deg(ρm) = χρm(1) =

Tr(ρm(1)) = Tr(ρ(1))m = nm.

122

Recall from last class that if g ∈ Z(G), then ρ(g) = λgI for some root of unity λ. Define γ : Z(G)→ C∗by g 7→ λg.

Let H = ker γ = {g ∈ Z(G)|ρ(g) = I} = ker ρ ∩ Z(G). Let D = {(g1, ..., gm) ∈ Z(Gm) = Z(G) × ... ×Z(G)|γ(g1, ..., gm) = 1}.

One can check (should check) that D CGm. Moreover, D ⊂ ker ρm since if ρm(g1...gm) = 1 implis thatρm(g1, ..., gm) = ρ(g1)⊗ ...⊗ ρ(gm) = λg1IV ⊗ ...⊗ λgmIV = λg1 ...λgmI = λg1...gmI = I.

But recall that normal subgroups induce representations (and irreducible subgroups induce irreduciblerepresentations). Thus D CGm gives an induced irreducible representation on Gm/D, with ρm : Gm/D →GLC(V ⊗C ...⊗C V ) such that deg ρm = nm. But by the previous theorem, deg ρm

∣∣∣|Gm/D|, so nm∣∣∣|Gm/D|.

We now wish to show that |D| = |Z(G)|m−1 · |H|. In order to do so, let α = (g1, ..., gm) ∈ Z(Gm) =Z(G) × ... × Z(G). Then α ∈ D if and only if g1...gm ∈ H, which is the case if and only if gm ∈(g1...gm−1)−1H. We therefore have a bijective correspondance between D and Z(G)m−1 × H by α =(g1, ..., gm)→ ((g1, ..., gm−1), g1...gm) and (g1, ..., gm−1, (g1...gm)−1h)← ((g1, ..., gm−1, h).

Thus |D| = |Z(G)|m−1 · |H|.Therefore, nm

∣∣∣ |G|m|Z(G)|m−1·|H| . That is, |G|m

|Z(G)|m−1·|H|nm = |Z(G)||H| ·

(|G|

n|Z(G)|m)∈ Z for all m ≥ 1. Then, by

the previous proposition, |G|n|Z(G)| is integral over Z. But it is also a rational number, and the only integral

rational numbers are integers! Thus |G|n|Z(G)| ∈ Z. In other words, n|Z(G)|

∣∣∣|G|, so n∣∣∣ |G||Z(G)| = [G : Z(G)].

Lemma 41. Let G be a finite group and let χ be an irreducible (complex) character of G. Suppose Cis a conjugacy class of G such that gcd(|C|, n) = 1, where n = degχ. Then for all g ∈ C, χ(g) = 0 or|χ(g)| = n.

Proof. Let m = |C|, and let z =∑g∈C

g. Recall that z ∈ Ae1 ⊕ ... ⊕ Zet (were A is the integral closure of Z

in C).

By one of the formulas about characters, mχ(g)n ∈ A for all g ∈ C. Since gcd(m,n) = 1, then there

exists r, s ∈ Z such that 1 = rm + sn. If we multiply this equation by χ(g)n , then we get that χ(g)

n =

r · (mχ(g)n ) + s(χ(g)). Since r, s ∈ Z ⊂ A, and mχ(g)

n , χ(g) ∈ A, then χ(g)n ∈ A.

However, χ(g) = λ1 + ...+ λn for some kth roots of unity λ1, ..., λn.

Let α = χ(g)n = λ1+...+λn

n ∈ A. Then |α| = |λ1+...+λnn | ≤ |λ1|+...+|λn|

n ≤ 1. Furthermore, note that wehave |α| = 1 if and only if λi = λj for all i and j, which is equivalent to |χ(g)| = n.

Let ω be a primitive kth root of unity, and let E = Q(ω). Then let H = Gal(E/Q), and let σ ∈ H. Noteσ(α ∈ A) since σ takes integral elements to integral elements.

Furthermore, for all i, σ(λi = λ′i for some k-th root of unity λ′i. Therefore, σ(α) =λ′1+...+λ′n

n .

Let N = NEQ : E → Q. Then, N(α) =

∏σ∈H

σ(α) ∈ Q ∩ A = Z. Therefore, N(α) =∏σ∈H|σ(α)| ≤ 1. Thus

either N(α) = 0, so α = 0 and χ(g) = 0, or N(α) = 1, so |σ(α)| = 1 for all σ. Namely, this is true σ = id,so |α| = 1. Thus |χ(α)| = n

13.3 Day 37 - April 15

Recall this theorem from last class:

Lemma 42. Let G be a finite group, and let χ be an irreducible C-character of G. Let n = degχ. SupposeC is a conjugacy class of G such that gcd(|C|, n) = 1. Then, for all g ∈ C, χ(g) = 0 or |χ(g)| = n.

Proposition 98. Let G be a finite simple group. Then there does not exist a conjugacy class C such that|C| = pa for some prime p and integer a > 0.

123

Proof. Let χ1, ..., χt be the irreducible (complex) characters of G, and let ρ1, ..., ρt be their associated rep-resentations. Let ni denote the degree of χi. That is, ni = χi(1). Without loss of generality, assume that ρ1

is the trivial representation.Suppose for the sake of contradiction that |C| = pa for some prime p and integer a > 0.First, we will show that if p - ni for some i > 1, then χi(g) = 0 for all g ∈ C.Let Gi = {g ∈ G|ρi(g) = λI for some λ ∈ C}. Note that Gi C G. But by assumption, G is normal, so

either Gi = {1} or Gi = G. Suppose for the sake of contradiction that Gi = G. Since i > 1, then ρi isnot the constant (trivial) homomorphism. Also, since G is simple, then ker ρi = {1}. Then G ∼= ρi(G) =ρi(Gi) = {λI|g ∈ G}. Thus G is abelian, so this contradicts the fact that |C| > 1. Therefore Gi = {1}.

That is, for all g 6= 1, |χi(g)| < ni. As p - ni, then gcd(|C|, ni) = 1. Then, by the lemma, χi(g) = 0or |χi(g)| = ni for all g ∈ C. Since |χi(g)| < ni for all g 6= 1, and 1 6∈ C, then χi(g) = 0 for all g ∈ C, asdesired.

Recall one of the formulae about characters was that, for all g 6= 1,

t∑i=1

χi(1)χi(g) = 0. In particular,

this is true for some g ∈ C. But χi(1) = ni, so

t∑i=1

niχi(g) = 0. Recall that n1 = 1 and χ1(g) = 1, so

1 +

t∑i=2

niχi(g) = 0. By the previous part of the proof, p - ni for all i > 1, so χi(g) = 0 for this g ∈ C. Thus

1 = 0, a contradiction! Thus there is no conjugacy class of size pa. [this is flawed]

Let’s now do one application of character theory. It is Burnside’s Theorem.

Theorem 107. Let G be a group of order paqb, with p and q primes. Then G is solvable.

Proof. Several cases of this are easy. First, if G is abelian, then it is solvable. Also, if a = 0 or b = 0, then Gis, respectively, a p-group or a q-group, and we have already proven such groups are solvable. If a = b = 1,then either it has a unique Sylow p-group or a unique Sylow q-group, which must be normal.

We will now proceed by induction on a + b. If a + b = 2, then a = b = 1 which case we have alreadydiscussed.

Suppose a + b > 2. It then suffices to show that G has a nontrivial normal subgroup. Suppose for thesake of contradiction that G is simple. Let P be a Sylow p-subgroup. Then, by classical results, Z(p) 6= {1},so let x ∈ Z(P ) \ {1}.

Let CG(x) = {y ∈ G|yx = xy} (the centralizer of x in G), and note that P ⊂ CG(x). Let C denotethe conjugacy class of x. Then |C| = [G : CG(x)] = ql for some l ≤ b. But by the previous proposition,l = 0. Therefore, |C| = 1, so C = {x}. In other words, x ∈ Z(G). Thus Z(G) 6= 1, so Z(G) is a nontrivialsubgroup. However, it is a proper subgroup as well, since we already dealt with the case that G is abelian.Thus Z(G) is a proper nontrivial normal subgroup, so by the inductive hypothesis, G is solvable.

And now, a word on Frobenius reciprocity:

Remark 133. Let G be a group, and let H ≤ G. Let χ ∈ char(H). Then, χG induced up to G, so for allφ ∈ char(G), φH ∈ char(H), and 〈χG, φ〉 = 〈χ, φH〉.

This is hard to do normally, but with commutative algebra, this ends up being just Hom−⊗ adjunction!

14 Primary Decompositions

Let’s now talk about primary decompositions.

Definition 120. Let R be a commutative ring. Let N (M be R-modules. For any a ∈ R and any moduleL, let aL : L→ L be the R-module homomorphism given by l 7→ al.

We say that N is a primary submodule of M if, for all a ∈ R, aM/N : M/N →M/N is either injective ornilpotent (i.e. anM/N = 0 for some n).

124

Remark 134. If L ⊂ N (M , then N is primary in M if and only if N/L is primary in M/L. This followsdirectly from the fact that (M/L)/(N/L) ∼= M/N . In particular, N is primary in M if and only if (0) isprimary in M/N .

Proposition 99. Let I ( R be an ideal. Then I is primary (as an R-module) if and only if, wheneverab ∈ I, either a ∈ I or b ∈

√I.

Proof. Suppose I is a primary ideal of R, and suppose ab ∈ I. Then, for all a ∈ R, aR/I is either injectiveor nilpotent.

Either a ∈ I or a 6∈ I. In the former case, we have nothing to show. In the latter case, then a 6= 0 inR/I, but bR/I(a) = ba = 0, so bR/I is not injective. Therefore bR/I is nilpotent, so there exists an n ∈ Nsuch that 0 = bn(1) = b

n. Thus bn ∈ I, or in other words b ∈

√I. Thus either a ∈ I or b ∈

√I.

Conversely, suppose whenever ab ∈ I, then either a ∈ I or b ∈√I. Let a ∈ R and suppose aR/I is not

injective. Then there exists a b 6∈ I such that aR/I(b) = 0 = ab. Thus ba = ab ∈ I. Since b 6∈ I, then a ∈√I.

That is, an ∈ I for some n ∈ N. Thus anR/I = 0, so aR/I is nilpotent.

14.1 Day 38 - April 18

Let’s talk more about primary decompositions.

Proposition 100. Suppose N is a primary submodule of M . Then√

AnnR(M/N) is prime.

Before the proof, let’s define a term.

Definition 121. If p =√

AnnR(M/N), then we say N is p-primary , or p is the prime prime associated toM/N .

Proof. Since N 6= M , then 1 6∈√

AnnR(M/N), so this ideal is not all of R.

Suppose ab ∈√

AnnR(M/N), and suppose a 6∈√

AnnR(M/N). Recall that√

AnnR(M/N) = {r ∈R|rM/N is nilpotent} .

Since N is primary, and a 6∈√

AnnR(M/N), then aM/N is injective. Since ab ∈√

AnnR(M/N), thenthere exists an n ∈ N such that (ab)nM/N = anM/Nb

nM/N = 0 for some n. Since aM/N is injective, then so is

anM/N , so bnM/N = 0. Thus b ∈√

AnnR(M/N).

That is, whenever ab ∈√

AnnR(M/N) and a 6∈√

AnnR(M/N), then b ∈√

AnnR(M/N). Thus thisideal is prime.

Let’s look at a non-example of a primary ideal.

Example 121. Let k be a field, and let R = k[x, y]. Let I = (x2, xy). Then√I = (X). However, xy ∈ I,

by x 6∈ I and y 6∈√I = (X). Thus I is not primary.

Proposition 101. Suppose√

AnnR(M/N) = m is a maximal ideal. Then N is a primary submodule ofM .

Proof. Let a ∈ R, and suppose aM/N is not nilpotent. It then suffices to show that aM/N is injective.

Since a 6∈√

AnnR(M/N) = m, then AnnR(M/N) +Ra = R. Thus 1 = r+ sa for some r ∈ AnnR(M/N)and s ∈ R.

Now suppose aM/N (x) = 0 for some x ∈M/N . Then ax ∈ N . But x = rx+ sax, and both rx ∈ N sincer ∈ AnnR(M/N), and sax ∈ N since ax ∈ N . Thus x ∈ N , so x = 0. That is, aM/N is injectivve. Thus Nis primary.

Remark 135. Let R be a ring, and let I ⊂ R be an ideal. Then AnnR(R/I) = I, and√

AnnR(R/I) =√I.

If I is a primary ideal, then√I is prime. If

√I is maximal, then I is a primary ideal.

Proposition 102. Let R be a PID. Let I be an ideal of R. Then the following are equivalent:

1. I is primary in R.

125

2. I = (0) or I = (fn) for some irreducible element f and n ≥ 1.

3. I = (0) or I = mn for some maximal ideal m and n ≥ 1.

Proof. Let’s first show that (2) implies (1). Certainly, (0) is prime in R (since it is a principal ideal domain),so it is primary. On the other hand, if I = (fn) for some irreducible element f , then

√I =

√(fn) ⊃ (f).

But since f is irreducible, then (f) is maximal, so either√I = (f) or

√I = R. The latter is not the case

because 1 6∈√I, so

√I = (f). Then, by the previous proposition and remark, since

√I is maximal, then I

is primary.Thus (2) implies (1).Now let us show that (1) implies (2). Let I be primary, and suppose I 6= (0). Therefore, by the previous

remark,√I is prime. Recall that the prime ideals in a PID are either 0 or generated by irreducible elements.

Since I 6= 0, then√I 6= 0, so

√I = (f) for some irreducible f ∈ R. Let n be the least number such that

fn ∈ I.We now wish to show that I = (fn). Suppose g ∈ I. Write g = f lh where gcd(f, h) = 1. Then

f lh = g ∈ I, so since I is primary, eiether f l ∈ I or h ∈√I = (f).

But gcd(f, h) = 1, so h 6∈ (f). Thus f l ∈ I. We chose n to be the least element so that fn ∈ I, so sol ≥ n. Thus g = f lh ∈ (fn). Therefore I = (fn).

Thus (1) implies (2).The proof of the equivalence of (2) and (3) is left to the reader.

Corollary 51. If R = Z, then I is primary if and only if I = (0) or I = (pn) for some prime p and n ≥ 1.

Remark 136. People were very interested in generalizing unique prime factorization from Z to other rings.In Z, the statement for ideals goes as follows.

Let R = Z, and let I ( R be a nonzero ideal. Then I = (n) for some n 6= −1, 0, 1.Then, we can write n = pm1

1 ...pmkk , where each pi is prime and mi ≥ 1. Then I = (pm11 ...pmkk ) =

(pm11 ) ∩ ... ∩ (pmkk ) = (p1)m1 ∩ ... ∩ (pk)mk .

However, in order to prove this, you need the fact that Z is height 1, so the chinese remainder theoremapplies nicely. Thus this exact statement of unique prime factorization doesn’t generalize totally.

Example 122. Let k be a field and let R = k[x, y]. Let I = (x2, y). Then√I = (x, y). Since

√I is maximal,

then I is primary. Note that I is not a power of a prime ideal, since it could only be a power of (x, y), andit is not.

Also, powers of prime ideals need not be primary!

Example 123. Let k be a field, and let f = x2 − yz ∈ k[x, y, z]. Let R = k[x, y, z]/(f), and let Letp = (x, z)R = (x, y)/(f). Then p is prime, and p2 = (x2, xz, z2) = (x2, xz, z2, yz)/(f). Thus zy ∈ p2, but

z 6∈ p2 and y 6∈√p2 = p = (x, z). Thus p2 is not prime!

Definition 122. Let M be an R-module, and let N (M be a submodule. Then N is called irreducible inM if, whenever N = L1 ∩ L2 for some submodules L1, L2 ⊂M , then N = L1 or N = L2.

Remark 137. If A ⊂ N ( M , then N is irreducible in M if and only if N/A is irreducible in M/A, whichis the case if and only if (0) is irreducible in M/N .

14.2 Day 39 - April 20

Recall the definition of primary. We will prove some equivalent conditions about it.

Remark 138. Let M be an R-module, and let Q (M be a submodule. Then the following are equivalent:

1. Q is primary. (That is, for all a ∈ R, aM/Q : M/Q→M/Q is either injective as a function, or nilpotentas a function.)

2. For all a ∈ R (Q :M a) = Q (meaning a is a non-zero-divisor on M/Q) or anM ⊂ Q for some n.

126

If M = R is Q = I, an ideal, then we get a further equivalence that

3. For all a ∈ R, a ∈√I (and

√I is a prime ideal), or (I :R: a) = I.

We get a simpler statement if we assume Q = (0).

Remark 139. For any module M , (0) is primary in M if and only if for all a ∈ R, either anM = 0 for somen, or (0 :M a) = 0 (that is, a is a non-zero-divisor on M).

Now let us return to the topic from last class, irreducible submodules of a Noetherian module.

Theorem 108. Let M be a Noetherian R-module, and let Q be an irreducible submodule of M . Then Qis primary.

Proof. By the remark from last class, Q is primary in M if and only if (0) is primary in M/Q. Since M isNoetherian, then M/Q is Noetherian as well. Also, since Q is irreducible in M , then (0) is irreducible inM/Q.

Therefore, we will assume without loss of generality that Q = (0). By the previous remark, it suffices toshow that if a ∈ R and anM 6= 0 for all n, then a is a non-zero-divisor on M .

Suppose for the sake of contradiction that 0 6= (0 :M a). Then consider the ascending chain 0 ( (0 :Ma) ⊂ (0 :M a2) ⊂ (0 :M a3) ⊂ ....

Since M is Noetherian, then this chain eventually stabilizes. That is, (0 :M an) = (0 :M an+j for some nand all j. Suppose for the sake of contradiction that there exists some u ∈ R \ (0 :M an) [other choice of u?].

We then wish to show that Ru ∩ anM = (0).Let x = ru = any for some r ∈ R, y ∈ M . Then ax = rau = an+1y. But rau = 0 since [???]. Thus

an+1y = 0. Thus y ∈ (0 :M an+1) = (0 :M an), so any = 0, so x = 0. Thus Ru ∩ anM = (0).Since (0) is an irreducible submodule, then Ru = 0 or anM = 0. Since Ru 6= 0, then anM = 0.

Definition 123. Let N be a proper submodule of M . A primary decomposition for N (in M) is an equationN = Q1 ∩ ... ∩Qs, where each Qi are primary submodules of M .

Theorem 109. Let M 6= 0 be a Noetherian module. Then every proper submodule of M has a primarydecomposition.

Proof. Let Λ = {N ( M |N does not have a primary decomposition}. Then then suffices to show thatΛ = ∅.

Suppose for the sake of contradiction that Λ 6= ∅. Then since M is Noetherian, we can choose N ∈ Λwhich is maximal. By the contrapositive of the previous theorem, N must be reducible.

That is, there must exist L1, L2 ) N such that N = L1 ∩ L2. But if L1 = M , then N = L1 ∩ L2 = L1.Since this is not the case, then L1 (M . Similarly, L2 (M .

Therefore, L1, L2 ∈ Λ, so they each have a primary decomposition. Then, by intersecting their twoprimary decompositions, we get a primary decomposition for N . This contradicts the fact that N ∈ Λ, soΛ = ∅, as desired.

Lemma 43. Let Q1, ...Qn be p-primary submodules of M . Then Q1 ∩ ... ∩Qn is p-primary as well.

Proof. We must show that Q1 ∩ ... ∩Qn is primary, and that its associated prime is p.For each Qi, since Qi is p-primary, then

√AnnR(M/Qi) = p. But note that AnnR(M/(Q1 ∩ ...∩Qn)) =

AnnR(M/Q1) ∩ ... ∩AnnR(M/Qn).Therefore,

√AnnR(M/(Q1 ∩ ... ∩Qn)) =

√AnnR(M/Q1)∩ ...∩

√AnnR(M/Qn) = p∩ ...∩ p = p. Thus,

if Q1 ∩ ... ∩Qn is primary, it will have the right associated prime.We must now show that Q1 ∩ ... ∩ Qn is primary. Let a ∈ R. As we have just shown, if a ∈ p, then

a ∈√

AnnR(M/(Q1 ∩ ... ∩Qn)), so anM ⊂ Q1∩ ...∩Qn. On the other hand, if a 6∈ p, then (Q1∩ ...∩Qn :Ma) = (Q1 : a) ∩ ... ∩ (Qn :M a) = Q1 ∩ ... ∩Qn (as each Qi is p-primary).

Thus, by the remark above, Q1 ∩ ... ∩ Qn is primary, and since it has the right associated prime, thenQ1 ∩ ... ∩Qn is p-primary.

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Definition 124. Let N ( M . We say an irredundant primary decomposition for N is an equation whereN = Q1 ∩ ... ∩Qs where

1. Each Qi is pi-primary.

2. pi 6= pj for i 6= j.

3. N ( Q1 ∩ ... ∩ Qi ∩ ... ∩Qn for all i.

Proposition 103. If N has a primary decomposition, it has an irredundant primary decomposition.

Proof. If some Qi, Qj violates the second criterion, then they can be combined by the Lemma.If some Qi violates the third criterion, then it can be dropped completely.This eventually terminates, so we are done.

Example 124. Let k be a field, and let R = k[x, y]. Let I = (x2, xy). Note that I = (x) ∩ (x2, y) is anirredundant primary decomposition. However, note that I = (x)∩ (x2, xy, y2) is another irreducible primarydecomposition.

That is, there are two irreducible primary decompostions for I! However, both i.p.d.s have the samecollection of associated primes, namely {(x), (x, y)}. Note also that the number of primary components isthe same in both decompositions, and (x) appears in both. We will formalize the reasons for this later.

Now let’s talk about how localization plays with a primary decomposition.

Proposition 104. Let Q be a p-primary submodule of M , and let S be a multiplicatively closed set of R.Then

1. QS = MS if and only if p ∩ S 6= ∅.

2. If p ∩ S = ∅, then QS is pS-primary.

Proof. Recall that localization commutes with quotienting. Therefore, it suffices to show the statement holdsfor (0) in M/Q. That is, we may assume without loss of generality that Q = 0.

Thus, (0) is primary in M and p =√

AnnR(M).(Proof of 1). If p ∩ S 6= ∅, let t ∈ p ∩ S. Then tnM = 0 for some n. Since t ∈ S, then tn = S. Thus

MS = 0 = QS , as desired.Conversely, suppose QS = MS , and let u ∈M \Q. Then u

1 = qs for some q ∈ Q and s ∈ S. Then, there

exists a t ∈ S such that tsu = tq. Since tq ∈ Q, then u ∈ (Q :M ts) \Q. Therefore, ts ∈ p ∩ S, so p ∩ S 6= ∅.This completes the proof of (1).

We will prove (2) tomorrow.

14.3 Day 40 - April 22

Let’s finish the proof of a proposition from last class.

Proposition 105. Let Q be a p-primary submodule of M , and let S be a multiplicatively closed set of R.Then

1. QS = MS if and only if p ∩ S 6= ∅.

2. If p ∩ S = ∅, then QS is pS-primary.

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Proof. Recall that localization commutes with quotienting. Therefore, it suffices to show the statement holdsfor (0) in M/Q. That is, we may assume without loss of generality that Q = 0.

Thus, (0) is primary in M and p =√

AnnR(M).(Proof of 2). We wish to show that QS is pS-primary. But Q = 0, so QS = 0. Let r

s ∈ RS . Either rs is a

non-zero-divisor on MS or it is not.If r

s is a non-zero divisor on MS , then multiplication by rs is injective on MS .

On the other hand, if rs · ut = 0 for some u

t ∈ MS \ 0, then rust = 0. That is, there exists an s′ ∈ S

such that s′ru = 0. Since ut 6= 0, then u 6= 0, so s′r is a zero-divisor on M . Since M is p-primary, then

all zero divisors are nilpotent. That is, there exists an n such that (s′r)nM = 0. Then ( s′r1 )nMS = 0, so

( rs )nMS = 0. That is, multiplication by rs is nilpotent on MS .

Thus, multiplication by rs is either injective or nilpotent on MS , so (0) is primary in MS .

It then suffices to show that the prime associated to QS = 0 is p.But note that s′r ∈ p [why]? Thus, rs ∈ pS . Therefore,

√AnnRS (MS) ⊂ pS . Thus MS is pS-primary.

Remark 140. Localization distributes across intersections of ideals. That is, (Q1 ∩ ... ∩ Qt)S = (Q1)S ∩... ∩ (Qt)S for any submodules QS , ..., Qt of M . (This is a good exercise to do.)

Suppose N = Q1 ∩ ... ∩Qt is a primary decomposition for N in M , where each Qi is pi-primary. ThenNS = M)S if and only if S ∩ pi 6= 0 for all i, which is true if and only if (Qi)S = (Mi)S for all i.

Proposition 106. Let N = Q1 ∩ ...Qt be an irredundant primary decomposition for N ⊂ M . Let S be amultiplicatively closed set of R such that NS 6= MS . From the previous remark, this implies that S ∩ pi = ∅for at least one i. Reorder the Qi such that S ∩ pi = ∅ for i = 1, ..., r and S ∩ pi 6= 0 for i = r+ 1, ..., t. ThenNS = (Q1)S ∩ ... ∩ (Qr)S is an irredundant primary decomposition.

Proof. By the previous remark, since localization distributes over intersections, then

NS = (Q1 ∩ ... ∩Qt)S= (Q1)S ∩ ... ∩ (Qr)S ∩ (Qr+1)S ∩ ... ∩ (Qt)S

= (Q1)S ∩ ... ∩ (Qr)S ∩MS ∩ ... ∩MS

= (Q1)S ∩ ... ∩ (Qr)S

Note that the third equality is by part 1 of the previous decomposition.Thus, NS = (Q1)S ∩ ... ∩ (Qr)S is a primary decomposition. We must now show that it is irredundant.Because N = Q1∩ ...∩Qt was an irredundant primary decomposition, then they have distinct associated

primes p1, ...pt.. By the previous proposition, the associated prime of (Qi)S is (pi)S for i = 1, ..., r. Sincethe pi are distinct, then the (pi)S are distinct.

It then suffices to show that we cannot do without any of the (Qi)S , for i = 1, ..., r.Suppose for the sake of contradiction that (Qi)S ⊃ (Q1)S∩ ...∩(Qi)S∩ ...∩(Qr)S∩(Qr+1)S∩ ...∩(Qt)S =

(Q1 ∩ ... ∩ Qi ∩Qt)S . We know that Qi 6⊃ Q1 ∩ ... ∩ Qi ∩ .. ∩Qt, since they formed an irredundant primarydecomposition for N . Then, there must be a u ∈ Q1 ∩ ... ∩ Qi ∩ .. ∩Qt \Qi, so u

1 ∈ [something].Hence, w is a zero divisor on M/Qi, so w ∈ pi∩S. This is a contradiction, so the primary decomposition

is irredundant.

Remark 141. In the proposition above, the set of associated primes of NS = Q1 ∩ ... ∩Qt is

{pS |p is an associated prime of NS = (Q1)S ∩ ... ∩ (Qt)S and p ∩ S = ∅}

Theorem 110 (First Uniqueness Theorem). Let R be a Noetherian ring, and let M be a finitely gener-ated R-module. LetN be a proper submodule, and letN = Q1∩...∩Qt be an irredundant primary decomposi-tion. Let pi be the prime associated with Qi. Then {p1, ...pt} = {p ∈ SpecR|p = (N :R x) for some x ∈M}.

Therefore, the set of primes only depends on N , and not on the irredundant primary decomposition. Inpartiuclar, every irreducible primary decomposition for N has the same number of primary components.

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Proof. We can reduce to the case that N = 0.Let 0 = Q1∩ ...∩Qt be an irredundant primary decompostion, where each Qi is pi-primary for i = 1, ..., t.We will first show that p ∈ {p ∈ SpecR|p = (N :R x) for some x ∈M} implies p ∈ {p1, ...pt}.Choose some x ∈ M , and suppose p = (0 :R x) is prime. Then, pRp = (0 :Rp

x1 ). Since pRp 6= Rp, then

x1 6= 0, so Mp 6= 0.

Then (rearranging the order of the irreducible primary decomposition), 0 = (Q1)p ∩ ... ∩ (Qr)p is anirreducible primary decomposition for (0) in Mp, and (Qi)p is piRp-primary.

Since p 6= R, then x 6= 0. Thus x 6∈ Qi for some i. But px = 0 since p = (0 :R x). Thus p consists of zerodivisors on M/Qi. Since Qi is primary, then zero divisors on M/Qi are nilpotent. Thus p ⊂ pi. Therefore,pRp ⊂ piRp. But pRp is a maximal ideal and piRp 6= 0, so pRp = piRp. Thus p = pi.

That is, p ∈ {p1, ..., pt}. Thus {p ∈ SpecR|p = (N :R x) for some x ∈M} ⊂ {p1, ...pt}.Conversely, we wish to show each pi ∈ {p ∈ SpecR|p = (N :R x) for some x ∈ M}. We will show that

p1 ∈ {p ∈ SpecR|p = (N :R x) for some x ∈M}, and an identical argument will apply to the other pi.We have two cases. The first case is that (R,m) is local, and p1 = m. Let Λ = {(0 :R x)|

√(0 :R x) = m}.

We will first show that Λ 6= ∅.Since our decomposition with irredundant, then Q2∩...∩Qt 6⊂ Q1. Thus there exists a u ∈ Q2∩...∩Qt\Q1.Also, since m = p1 is the prime associated to Q1, then m = p1 =

√AnnR(M/Q1). Since R is Noetherian,

then m is finitely generated, so mn ⊂ AnnRM/Q1 for some n. Hence, mnu ⊂ Q1.But since mnu ⊂ Q2 ∩ ... ∩Qt as well, then mnu ⊂ Q1 ∩ ... ∩Qr = 0. Thus mnu = 0, so m ⊂

√(0 :R u).

Since u 6= 0, then√

(0 :R u) 6= R, so m =√

(0 :R u). That is, (0 :R u) ∈ Λ, so Λ 6= 0, as desired.Since R is Noetherian, then there exists a maximal element of Λ. Let (0 :R x) be such an element. We

wish to show that (0 :R x) is prime, and that (0 :R x) = m.Suppose ab ∈ (0 :R x) but that b 6∈ (0 :R x). Then abx = 0, but bx 6= 0. Also, (0 :R x) ⊂ (0 :R bx).

Thus m =√

(0 :R x) ⊂√

(0 :R bx). Since bx 6= 0, then (0 :R bx) ( R, so√

(0 :R bx) ( R. But since m is

maximal, and m ⊂√

(0 :R bx) ( R, then m =√

(0 :R bx). Thus (0 :R bx) ∈ Λ. Since (0 :R x) was maximalin Λ, and (0 :R x) ⊂ (0 :R bx) ∈ Λ, then (0 :R x) = (0 :R bx). But abx = 0, so a ∈ (0 :R bx), so a ∈ (0 :R x).Thus (0 :R x) is prime.

Since (0 :R x) is prime, then it is radical. That is , (0 :R x) =√

(0 :R x) = m. Thus p1 = m ∈ {p ∈SpecR|p = (N :R x) for some x ∈M}. This concludes the specific case.

In the other case, R is not necessarily semi-local. However, by localizing, we get that p1Rp1is an associated

prime of the localized irredundant prime decompositon. Then, by the previous case, p1Rp1= (0 :Rp1

xt ), for

some xt ∈Mp1 .

Since R is Noetherian, then p1 is finitely generated. That is, p1 = (a1, ..., an). Then for each i, ai1 ·xt = 0

1 ,so there exists an si 6∈ p1 such that siaix = 0 for i = 1, ..., n. Let s = s1...sn, and note that s 6∈ p1. Thenai(sx) = 0 for all i = 1, ..., n, so p1 ⊂ (0 :R sx). Then p1 = (0 :R sx) as s 6∈ p1.

Thus p1 ∈ {p ∈ SpecR|p = (N :R x) for some x ∈ M}, so {p1, ...pt} = {p ∈ SpecR|p = (N :Rx) for some x ∈M}, as desired.

15 Associated Primes

15.1 Day 41 - April 25

We showed last class that, in nice cases, any primary decomposition of a specified submodule has the sameset of associated prime ideals. This gives rise to the following definition.

Definition 125. Let R be a Noetherian ring, and N ( M be finitely generated R-modules. Let N =Q1∩ ...∩Qt be an irredunant primary decomposition. For each i, let pi =

√AnnR(M/Qi). Then {p1, ..., pt}

is the set of associated primes of N in M , or of M/N . We denote this set by AssR(M/N).

Remark 142. We proved lass class that AssR(M/N) is well-defined in this context, and that AssR(M/N) ={(N :R x) ∈ SpecR|x ∈M}.

Remark 143. Assume again that Let R be a Noetherian ring, and N (M be finitely generated R-modules.We have the following results:

130

1. Let p ∈ SpecR. Then the following are equivalent:

(a) p ∈ AssR(M) = AssR(M/0).

(b) p = (0 :R x) for some x ∈M .

(c) There exists an injective R-module homomorphism R/p→M (given by 1 7→ x).

(d) HomRp(k(p),Mp) 6= 0.

2. M = 0 if and only if AssRM = ∅.

3. If S is a multiplicatively closed set of R, then AssRSMS = {pS |p ∩ S = ∅ and p ∈ SpecR}.

4. If N ⊂M , then AssRN ⊂ AssRM .

5. Suppose N ⊂M . Then N = 0 if and only if Np = 0 for all p ∈ AssRM .

Let us prove this last item.

Proof. Suppose N ⊂M . Certainly, if N = 0, then Np = 0 for all primes p, including the ones in AssRM .If N 6= 0, then there exists some p ∈ AssRN ⊂ AssRM . But p ∈ AssRN . Therefore, by one of the

remarks, we have that HomRp(k(p), Np) 6= 0. But certainly, if Np = 0, then all homomorphisms into this

would be 0. Since this is not the case, then Np 6= 0.

Proposition 107. Let R be Noetherian, and let M 6= 0 be a finitely generated R-module. Then the set ofzero divisors on M is the union of the associated primes. That is, ZDR(M) := {r ∈ R|ru = 0 for some u ∈M \ {0} =

⋃p∈AssR(M) p.

Proof. Let p ∈ AssRM . Then p = (0 :R x) for some x ∈M . Furthermore, x 6= 0, since p 6= R. Thus px = 0,so p ⊂ ZDR(M). Thus

⋃p∈AssRM ⊂ ZDR(M).

Conversely, suppose r ∈ ZDR(M). Then ru = 0 for some u ∈ M \ {0}. Thus r ∈ (0 :R u). LetΛ = {(0 :R v)|(0 :R u) ⊂ (0 :R v) for some v 6= 0}. Since (0 :R u) ∈ Λ, then Λ is nonzero. Also, R isNoetherian, so one can choose a (0 :R v) which is maximal in Λ.

One can show that (0 :R v) is prime [we made an identical argument previously], so (0 :R v) ∈ AssRM .But since r ∈ (0 :R v), then r ∈

⋃p∈AssR(M) p. Thus ZDR(M) ⊂

⋃p∈AssR(M) p.

Thus ZDR(M) =⋃

p∈AssR(M) p.

Proposition 108. Let R be Noetherian, and let M 6= 0 be a finitely generated R-module. Then the radicalof the annihilator is the intersection of the associated primes. That is,

√AnnRM =

⋂p∈AssRM p.

Proof. Let (0) = Q1 ∩ ... ∩ Qt be an irredundant primary decomposition for (0) in M . Then AnnRM =⋂ti=1 AnnRM/Qi. Thus, by elementary properties of radicals,

√AnnRM =

⋂ti=1

√AnnRM/Qi =

⋂p∈AssRM

p.

Recall that if M is a (finitely generated) R-module, then SuppRM = {p ∈ SpecR|Mp 6= 0}. Recall alsothat if M is finitely generated, then SuppRM = V (AnnRM).

Definition 126. Define the minimal primes of M to be the set of minimal primes in SuppRM .We use the following notation: MinRM = {p ∈ SuppRM |Mq = 0 for all q ⊂ p}.

Remark 144. If M is finitely generated, then MinRM = {p ∈ SpecR|p is minimal over AnnRM}.

Proposition 109. Let R be a Noetherian ring, and let M be a finitely generated R-module. ThenMinRM ⊂ AssRM ⊂ SuppRM .

Proof. Let p ∈ MinRM . Note that p ∈ AssRM if and only if pRp ∈ AssRpMp. But SuppRp

Mp = {pRp}since p ∈MinRM . Thus AssRp

Mp = {pRp}, so p ∈ AssRM .Suppose now that p ∈ AssRM . Then HoMRp

(k(p),Mp) 6= 0, so like we said above, Mp 6= 0. That is,p ∈ SuppRM .

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Lemma 44. Let R be a ring, let I be an ideal, and let M be a finitely generated R-module. If IM = M ,then there exists an s ∈ I such that (1− s)M = 0.

Proof. Suppose p ∈ SpecR, and that I ⊂ p. Then IpMp = Mp. Since I ⊂ p, then Ip ⊂ pRp, so (pRp)Mp =Mp. But pRp is the unique maximal ideal in the semi-local ring Rp, so by Nakayama’s Lemma, we have thatMp = 0. Thus p 6∈ AnnRM .

Thus I + AnnRM = R [why?]. In particular, there exists s ∈ I and t ∈ AnnRM such that s + t = 1.Then t = 1− s ∈ AnnRM .

15.2 Day 42 - April 27

Recall the following Lemma, which we proved last class:

Lemma 45. Let R be a ring, and let I ⊂ R be an ideal. Llet M be a finitely generated R-module, andsuppose IM = M . Then there exists s ∈ I such that (1− s)M = 0.

Last class, we stated Krull’s Intersection Theorem:

Theorem 111 (Krull’s Intersection Theorem). Let R be a Noetherian ring, and let M be a finitelygenerated R-module, and let I be an ideal of I. Then there exists an s ∈ I such that (1−s)(

⋂∞n=1 I

nM) = 0.In particular, if I ⊂ Jac(R), then 1− s is a unit and

⋂∞n=1 I

nM = 0.

Let us now prove it.

Proof. Let N =⋂n≥1 I

nM . Note that N = IM ∩ IN . We then wish to show that there exists an s ∈ Ssuch that (1− s)N = 0.

We have two cases: either IM = M , or IM (M .In the former case, by the Lemma there exists an s ∈ I such that (1− s)M = 0. Therefore

(1− s)N = (1− s)(IM ∩ IN)

= (1− s)IM ∩ (1− s)IN= (1− s)M ∩ (1− s)IN= 0 ∩ (1− s)IN= 0

as desired.If IM 6= M , then IN ⊂ N (M , so IN (M .Therefore, there exists an irredundant primary decompostion IN = Q1 ∩ ... ∩Qt for IN in M .Suppose for the sake of contradiction that IN 6= N . If not, then IN ( N , so N ( Qi for some I. But

IN ⊂ Qi, so I consists of zero divisors on M/Qi. Since Qi is primary, then I ⊂√

AnnR(M/Qi). Since I iffinitely generated, then this implies that In ⊂ AnnR(M/Qi) for some n.

That is, InM ⊂ Qi for some n. But N ⊂ InM ⊂ Qi, so this contradicts the fact that N ( Qi. ThusIN = N .

Since R is Noetherian, and N ⊂ M , a finitely generated R-module, then by the Lemma, there existssome s ∈ I such that (1− s)N = 0.

Corollary 52. If R is Noetherian, and either a domain or a quasi-local ring, and I 6= R, then⋂n≥0 I

n = 0.

Now, let us turn back to primary decompositions. Sometimes, it turns out that the primary ideals in anirredundant primary decomposition are unique. This is summed up in the second uniqueness theorem.

Theorem 112. Let R be a Noetherian ring, and let M be a finitely-generated R-module. Let N (M . LetN = Q1 ∩ ... ∩ Qt be an irredundant primary decomposition, and for i = 1, ..., t, let pi =

√AnnR(M/Qi).

Then, if some pi ∈MinR(M/N), then Qi = φ−1(Npi), where φ : M →Mpi is the natural localization map.Therefore, the pi-primary component for N in M is unique.

132

Proof. Let S = R \ pi. Note that pj ∩ S 6= ∅ for j 6= i. Therefore, (Qj)S = MS for all j 6= i.Therefore,

Npi = (Q1 ∩ ... ∩Qt)pi= (Q1)pi ∩ ... ∩ (Qt)pi

= (Qi)pi

Note also that (Qi)pi is piRpi-primary in Mpi .We now wish to show that Qi = φ−1((Qi)pi).For simplicity, let Q = Qi, and let p = pi. Let u ∈ Q. Then u

1 ∈ Qp, so u ∈ φ−1(Qp). Thus Q ⊂ φ−1(Qp).Conversely, let u ∈ φ−1(Qp). Then u

1 = vs for some v ∈ Q and s 6∈ p. Therefore, there exists some s′ 6∈ p

such that s′u = v ∈ Q. Since s′ 6∈ p =√

AnnR(M/Q), then (Q :M s′) = Q. Since u ∈ (Q :M s′) = Q, thenQ = φ−1(Qp), as desired.

Theorem 113. Let R be Noetherian, and let M 6= 0 be a finitely generated R-module. Then there existsa filtration of submodules of M 0 ⊂ N0 ⊂ ... ⊂ Nt = M such that Ni/Ni−1

∼= Rpi for some pi ∈ SpecR.

Proof. Choose some p1 ∈ AssRM . Then p1 = (0 :R x1) for some x1 ∈ M . Therefore Rx1∼= R/p1 (by the

isomorphism R/p→ Rx1 by r 7→ rx1.Let N1 = Rx1 ⊂M . If N1 = M , then we are done.If N1 6∼= M , then choose p2 ∈ AssR(M/N). Then p2 = (N1 :R x2). Let N2 = N1 + Rx2 = Rx1 + Rx2.

Therefore, N2/N1∼= Rx2

∼= R/p2. We now have a chain 0 = N0 ⊂ N1 ⊂ N2.If N2 6= M , we repeat. Eventually, this must terminate by the ascending chain condition.

Here’s how we normally use this result.

Remark 145. Suppose you have a property that is true for domains, and that it is “preserved by exactsequences” (meaning if 0 → A → B → C → 0 is exact, then the property is true for B whenever it is truefor A and C.)

Then, by the previous theorem, there exists a nice filtration. This gives us a bunch of exact sequences0→ Nk → Nk+1 → Nk+1/Nk → 0. Since Nk+1/Nk is isomorphic to a domain by the previous theorem, ourhypothetical property is true for it. Also, by induction, the property is true for Nk. Thus it is true for Nk+1.

In this way, we can induct and show that this is true for any finitely generated R-module!

Thanks for reading! Good luck in your algebra endeavors!-Robert

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Chapter 3

Appendix

This appendix was originally my ultra-condensed notes for when I was studying for my comprehensiveexam. It consists of raw theorems and results about central concepts (almost all ring/module theory) thatI considered essential to know. You might find it useful too!


Here R is a ring, and P is a projective left R-module.

Definition 127. An left R-module P is called projective if any of the following equivalent conditions aremet:

1. If Mf→ N → 0 is exact and g : P → N , then there exists some h : P →M such that fh = g.

2. Every short exact sequence of the form 0→ L→M → P → 0 splits.

3. P is the direct summand of a free module.

4. HomR(P,−) is exact.

5. Pp is free for all p ∈ SpecR.

6. Pm is free for all m ∈ MaxspecR.

Theorem 114. If R is a polynomial ring, then every f.g. projective module over R is free.

Theorem 115. If R is a semilocal ring, then every f.g. projective module over R is free.

Theorem 116. If R is commutative, S is a multiplicatively closed subset of R, and P is a projectiveR-module, then PS is a projective RS module.

Theorem 117. If P is a finitely generated projective R-module, then the map f : SpecR → N0 given byq 7→ rkPq is continuous (with respect to the discrete topology on N0).

Theorem 118. Projective modules are flat.

Example 125. Free modules are projective. Dn is projective over Mn(D) (where D is a division ring).Summands of free modules are projective. Direct sums of projective modules are projective.

134

2 Injective Modules

Here R is a ring and E is an injective R-module.

Definition 128. A left R-module E is called injective if any of the the following equivalent conditions aremet:

1. If 0→Mg→ N is exact, and f : M → E, then there exists an H : N → E such that hg = f .

2. (Baer’s Criterion) For all left ideals I of R and f : I → E, if i : I → R is the inclusion map, thereexists an h : R→ E such that hi = f .

3. HomR(−, E) is exact.

4. Every short exact sequence of the form 0→ E →M → N → 0 splits.

Theorem 119. Let E =∏i∈I

Ei be a product of left R-modules. Then E is injective if and only if each Ei

is injective.

Theorem 120. E is divisible (i.e. for all non-zero-divisors r ∈ R, and all u ∈ E, there exists u′ ∈ E suchthat u = ru′.

Theorem 121. If R is a PID, then every divisible module is injective.

Example 126. Let R be a PID. Then Q(R) is injective. Q/Z is an injective Z-module. k[x]/x2 is aninjective itself-module. If E is injective, and φ : R→ S is a ring homomorphism (so S is an R-module), thenHomR(S,E) is injective. Every module is contained in an injective hull. A direct summand of an injectivemodule is injective.

3 Semisimple Modules

Definition 129. A (left) R-module M is called semisimple if any of the following conditions are met:

1. Every submodule of M is a direct summand of M .

2. M is the sum of simple submodules.


Theorem 122. Any submodule of a semisimple module is semisimple.

Theorem 123. Any nonzero semisimple module contains a simple submodule.

Theorem 124. If M is semisimple, then the following are equivalent:

1. M is Artinian.

2. M is Noetherian.

3. M is finitely generated.

4. λ(M) <∞.

135

4 Semisimple Rings

Definition 130. A ring R is called semisimple if any of the following conditions are met:

1. R is semisimple as a left- or right- R-module.

2. R is the direct sum of matrix rings over division rings.

3. J(R) = 0 and R is left Artinian.

4. Every R-module is projective.

Theorem 125. A module over a semisimple ring is semisimple.

Theorem 126. If R is semisimple, J(R) = 0.

Theorem 127. Let k be a field of characteristic p (possibly p = 0). Then k[G] is semisimple if and only if|G| <∞ and p - |G|.

5 Localization

Here, R is a commutative ring, and S is a multiplicatively closed subset of R containing 1, and M is anR-module.

Theorem 128. If R is Noetherian (resp. Artinian), then so is RS .

Theorem 129. If I is an ideal in R, then IS = RS if and only if I intersects S.

Theorem 130. SpecRS is in natural bijection with {p ∈ SpecR|p ∩ S = ∅} (by p 7→ pS).

Theorem 131. MS∼= M ⊗R RS .

Theorem 132. RS is a flat R-module (i.e. localizing is exact).

Theorem 133. The following are equivalent:

1. M = 0.

2. Mp = 0 for all p ∈ SpecR.

3. Mm = 0 for all m ∈ MaxspecR.

Theorem 134. If AnnRM intersects S, then MS = 0. If M is finitely generated, then the converse holds(i.e. MS = 0 implies AnnR(M) intersects S).

6 Hom Modules

Definition 131. Let F be an additive covariant functor on module categories. Then F is left exact if,whenever 0→ A→ B → C is exact at B, then 0→ F (A)→ F (B)→ F (C) is exact at F (B).

We say F is right exact if, whenever A→ B → C → 0 is exact at B, then F (A)→ F (B)→ F (C)0→ is

exact at F (B).

Theorem 135. For R a ring and M a module, M ⊗R − is right exact.

Definition 132. We say M is flat if M ⊗R − is exact.

Theorem 136. For any ring R, R and RS are both flat as R-modules.

136

Theorem 137. If R is a ring, then any projective R-module is flat.

Theorem 138. The functors HomR(M,−) and HomR(−, N) are left exact.

Theorem 139. Hom in either coordinate distributes over direct sums.

Theorem 140. For any modules, ⊕Ai is flat if and only if each Ai is flat.

Theorem 141. (Hom−⊗ Adjointness/Adjunction) Let R,S, and T be rings. Let A be an R−T bimodule,B be an S−R bimodule, and C be a left S-module. Then the map HomR(A,HomS(B,C))→ HomS(B⊗R

A,C) (which lives in the set of T -module homomorphisms) given by φ 7→

(gφ :

{B ⊗R A→ C

b⊗ a 7→ φ(a)(b)

)is a

well-defined left T -module isomorphism.

7 Tensors and Flat Modules

Here R is a ring and M is an R-module.

Theorem 142. (Universal Property for Tensor Products) Let h : M ×N →M ⊗R N be the natural map.For any abelian group A and R-biadditive map f : M ×N → A, there exists a unique group homomorphismg : A→M ⊗R N such that f = g ◦ h.

Definition 133. We say M is flat if tensoring by M is exact.

Theorem 143. If I and J are ideals, then R/I ⊗R R/J ∼= R/(I + J).

Theorem 144. Tensors distribute over arbitrary direct sums.

8 Primary Decompositions

Theorem 145. Suppose N is a primary submodule of M . Then√AnnR(M/N) is prime.

Theorem 146. If I ⊂ R is primary, then√I is prime. If

√I is maximal, then I is primary.

Theorem 147. Irreducible submodules of Noetherian modules are primary.

Theorem 148. If M 6= 0, then every proper submodule of M has an (irredundant) primary decomposition.

Theorem 149. Primary submodules play nicely with localization: if Q ⊂ M is p-primary, then QS = MS

if and only if p ∩ S 6= ∅. Also, if p ∩ S = ∅, then QS is pS-[rimary.

Theorem 150 (First Uniqueness Theorem). If R is Noetherian, M is a finitely generated R-module,and N is a proper submodule, then the associated primes of an irredundant primary decomposition of N areindependent of the primary decomposition, and denoted by AnnR(M/N).

Theorem 151 (Second Uniqueness Theorem). If M is a finitely generated module over a Noetherianring, and pi ∈ Min(M/N), then the pi-primary component of N is unique in any irredundant primarydecomposition.

Theorem 152. Let p ∈ SpecR. Then the following are equivalent:

1. p ∈ AssR(M)

2. p =√AnnR(M/Qi) for some primary component Qi in any primary decomposition of 0.

3. p = (0 :R x) for some x ∈M .

4. There exists an injective R-module homomorphism R/p→M given by 1 7→ x.

137

5. HomRp(k(p),Mp) 6= 0.

Theorem 153. We have the following other results:

1. N = 0 if and only if AssR(M/N) = ∅.

2. If S is multiplicatively closed, then AssRS (MS) = {pS |p ∩ S = ∅ and p ∈ SpecR}.

3. If N ⊂M , then AssRN ⊂ AssRM .

4. If N ⊂M , then N = 0 if and only if Np = 0 for all p ∈ AssR(M).

Theorem 154 (Krull’s Intersection Theorem). Let R be a Noetherian ring, and let M be a finitelygenerated R-module, and let I be an ideal of R. Then there exists an s ∈ I such that (1− s)(

⋂InM) = 0.

Corollary 53. If R is Noetherian and either a domain or a quasi-local ring, then for any proper ideal I,⋂In = 0.

9 Integral Extensions

Theorem 155 (Lying Over Theorem). Let S be an integral extension of R. Let p ∈ SpecR. Then thereexists q ∈ SpecS such that p = R ∩ q.

Theorem 156 (Incomparability Theorem). Let S be an integral extension of R. Let q1, q2 lie over p.Then q1 and q2 are incomparable.

Theorem 157 (Going Up Theorem). Let S be an integral extension of R. Then one can “go up” chainsof primes.

Theorem 158 (Going Down Theorem). Let S be an integral extension of R, and assume both R and Sare domains. Suppose also that R is integrally closed in Q(R). Then one can “go down” chains of primes.

10 Techniques For 0

Theorem 159. Let M be an R-module. Then the following are equivalent:

1. M = 0.

2. Mp = 0 for all p ∈ SpecR.

3. Mm = 0 for all m ∈ MaxspecR.

If M is finitely generated, the following are also equivalent:

1. M = 0.

2. (Nakayama’s Lemma) M = JM (where J is the Jacobson radical).

If N ⊂M and R is Noetherian, then the following are equivalent:

1. N = 0

2. AssR(M/N) = ∅

3. Np = 0 for all p ∈ AssR(M).

138

Chapter 4

Index

139

Index

R-algebra, 36, 54R-biadditive, 89R− S bimodule, 90p-primary, 125k-linear representation, 66(Localization of a module), 85

action, 66additive, 94affine K-algebra, 112affine n-space over K, 114affine ring, 112algebraic K-variety, 114algebraic number field, 109algebraic set, 114algebraically dependent over F , 31algebraically independent over F , 31annihilator, 56arrows, 93Artinian, 34associated primes, 130automorphism group, 15

bilinear form, 75

canonically split, 60category, 93character, 70, 71characteristic subgroup, 24class function, 75class functions, 72commutator, 24commutator subgroup, 24composition, 93composition series, 43compositum, 9constant rank, 88contravariant functor, 94covariant functor, 94cyclic (respectively abelian, solvable, nilpotent,

etc.), 22cyclotomic polynomial, 5

degree, 66, 70

derived normal series, 24dimension, 79divisible, 103

equivalent, 42exact, 35, 95exact in degree i+ 1, 35external direct sum, 40

factor modules, 42finite length module, 43finitely presented, 99, 100fixed field, 16fixed subring, 116flat, 96, 97forgetful functor, 94free resolution, 105Frobenius bimodule structure, 90

Galois extension, 15Galois group, 15general equation of degree n, 29

height, 78Hermitian inner product, 75hom functor, 96

ideal, 34, 38injective, 102, 135injective resolution, 105inseparable closure, 13, 15inseparable degree, 12integral, 105, 106integral closure, 107integral extension, 106integrally closed, 107internal direct sum, 40invariant subring, 116irreducible, 66, 71irreducible in M , 126irredundant primary decomposition, 128isomorphic, 59isomorphism of representations, 66

140

Jacobson radical, 56join, 9

left, 34left R-module, 34left exact, 95, 97, 136left primitive, 58left semisimple, 41length, 42, 43linear action, 66local, 85localization functor, 94

minimal generating set, 86minimal over I, 80minimal prime, 80minimal primes, 131morphisms, 93multiple root, 6multiplicative, 94multiplicatively closed, 79

natural isomorphism, 100nil, 55nilpotent, 55Noether normalization, 112Noetherian, 34norm, 19normal closure, 14normal extension, 14

objects, 93opposite ring, 48

perfect, 6points, 114primary decomposition, 127primary submodule of M , 124prime, 78prime associated to M/N, 125prime spectrum, 78primitive n-th roots of unity, 5primitive n-th root of 1, 23projective, 61, 134projective resolution, 105proper refinement, 42purely inseparable, 12

quasi-local, 85

radical, 27, 80rank, 87refinement, 42residue field, 85right, 34right R-module, 34right exact, 95, 136right semisimple, 41ring, 33root tower, 26, 27

semiprimitive, 57semisimple, 39, 135, 136separable, 6separable closure, 9separable degree, 7separably closed, 9series, 42short exact sequence, 35sign representation, 68simple, 38simple root, 6solvable, 24solvable by radicals, 27solvable series, 25split, 60split short exact sequence, 35symmetric, 75

tensor product, 89total ring of fractions, 103trace, 19, 70transcendence base, 32transcendence degree, 33trivial ideals, 38

unital, 34

vanishing ideal, 115von Neumann regular, 102

Zariski Topology, 115Zariski topology, 79zero set of S, 114

141

robert huben with help from matt mills 2015-2016 · 2019. 11. 8. · algebra 901-902 notes robert...

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