Part III - Algebra

Lectured by Chris Brookes

Comments or corrections should be sent to wz302@cam.ac.uk.

Last updated: April 2019

Contents

0 Introduction 2Link between AG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1 Examples, localizations, tensor products 2Localization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Local properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Tensor product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Ideal structure 8Nilradical and Jacobson radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Minimal and associated primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Dimension 12Integral dependence/extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4 Height 17

5 Artin rings 18Artin-Wedderburn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

6 Homological algebra 22Hochschild (co)homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

7 Graded rings and filtrations 28Filtration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Appendices 33

A Cayley-Hamilton and the determinant trick 33

B More on homological algebra 33

C Hilbert functions and Hilbert polynomials 34

D Rees algebra and algebraic geometry 36

1

0 Introduction

Example 0.1. 1 We look at the ring k[x1, . . . , xn]. The symmetric group Σn acts on k[x1, . . . , xn] as ringautomorphisms. Let S be the set of all symmetric polynomials, i.e. those that are fixed by all g ∈ Σn.Then we have some elementary symmetric polynomials:

f1 = x1 + x2 + · · ·+ xnf2 =

∑xixj

...

S is then generated by f1, . . . , fn, and S ∼= k[Y1, . . . , Yn] under the map fi 7→ Yi.

We are going to see four main results in this course, namely:

1. Hilbert’s basis theorem.

2. Nullstellensatz.

3. Polynomial nature of a function; Hilbert function (and the degree is the start of dimension theory)

4. Syzygy theorem (resolutions).

Basis theorem is about Noetherian rings.

Definition 0.2. A commutative ring R is Noetherian iff all ideals are finitely generated (abbr. fg.).

Theorem 0.3. If R is a commutative Noetherian ring, then R[x] is also commutative Noetherian.

Corollary 0.4. k[x1, . . . , xn] also commutative Noetherian.

In a noetherian ring, we have some ideal structures: no prime decomposition, but primary decompo-sition.

Link between AG

FTA: f ∈ C[x] is determined by a scalar and the multiplicities of its zeros. Define Z(I) := {(a1, . . . ,an) ∈Cn : f(a1, . . . ,an) = 0 ∀f ∈ I}. This is an affine algebraic set and is closed in Zariski topology. We can, ofcourse, replace I by the ideal generated by I.

For a set S ⊂ Cn, define I(S) = {f ∈ C[x1, . . . , xn] : f(a) = 0 ∀a ∈ S}. This is a radical ideal, i.e.fn ∈ I⇒ f ∈ I.

Then one version of Nullstellensatz is that radical ideals of C[x1, . . . , xn]↔ algebraic subsets of Cn.We will also talk about dimension of commutative rings:

1. Length of chains of prime ideals.

2. Growth rate (degree of Hilbert’s functions).

3. Transcendantal degree of the fraction field of an ID.

For polynomial rings and their quotients, these will be equivalent definitions.

1 Examples, localizations, tensor products

Throughout this chapter, R are commutative rings with identity.

Lemma 1.1. LetM be a left R-module. The following are equivalent:

1I might have messed up with the numbering of theorems and stuff in the first few chapters

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1. every submodule is finitely generated.

2. there does not exist an infinite strictly ascending chain of submodules.

3. every non-empty subset of submodules has a maximal element.

Proof is left as an exercise.

Definition 1.2. R-moduleM satisfying these conditions is called noetherian.

Definition 1.3. A commutative ring R is noetherian if it is a noetherian (left) R-module.

Lemma 1.4. Let N be a submodule ofM. ThenM noetherian iff N andM/N are noetherian

Exercises.

Remark.

1. Therefore images of noetherian modules are noetherian.

2. Ring images of noetherian rings are noetherian.

Lemma 1.5. Let R be a noetherian ring. Then any finitely generated R-module is a noetherian module.

Exercise.

Example 1.6.

1. Fields are noetherian.

2. PID are noetherian.

3. The ring {q ∈ Q |q = m/nwhere p - n for a fixed p}.. This is an example of localization of Z. Ingeneral, localization of noetherian rings are noetherian.

4. k[x1, . . . , xn], Z[x1, . . . , xn] are noetherian, results of basis theorem.

5. k[x1, . . . ] countably many variables is not noetherian. We can produce an ascending chain

(x1) ⊂ (x1, x2) ⊂ · · ·

6. Finitely generated commutative rings (as Z-module) are noetherian. Take a generating set{a1, . . . ,an}, then there is a surjection Z[x1, . . . , xn] → R, mapping xi 7→ ai. So R is a ringimage of a noetherian ring.

Remark. We will meet two classes of non-commutative rings that are both left and right noetherian:

1. The enveloping algebra of a finite dimensional Lie algebra.

2. Iwasawa algebras of compact p-adic groups.

Theorem 1.7. (Basis theorem) Let R be a noetherian ring. Then R[x] is noetherian.

Proof. (sketch) we show every ideal I of R[x] is finitely generated. Define

I(n) = {elements of Iwith degree ≤ n}.

These are non-empty since 0 ∈ I(n). Define

R(n) = {leading coefficients of xn appearing in elements in I(n)}.

Then R(n) are ideals, and R(0) ⊂ R(1) ⊂ · · · (since if f ∈ I(n), then Xf ∈ I(n+ 1)). So R(n) = R(N) forn ≥ N. Now each R(n) is finitely generated by an,1, . . . ,an,mn , and there are polynomials fn,m(x) =an,mx

n + · · · ∈ I. Claim: the set of all fi,j(x) is finite and generates I. Exercise.

Remark. In computations, it’s good to deal with generating set without redundancies. This is known asGrobner bases in computer algorithms.

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Theorem 1.8. If R is noetherian, then R[[x]] is noetherian.

Proof. Either similar to the proof of basis theorem, using trailing coefficients (the lowest appearing),which is left as an exercise, or use the following two results.

Theorem 1.9. R is noetherian iff all prime ideals of R are finitely generated.

Lemma 1.10. Let P be a prime ideal of R[[x]], and θ : R[[x]] → R be the map taking constant term. Then P isfinitely generated iff θ(P) is finitely generated ideal of R.

Proof. (to theorem 1.9) Suppose R is not noetherian. The set of non-finitely generated ideals is non-empty.Partially order them by inclusion, and every chain is bounded by the union of all (which is still nonfinitely generated). Zorn’s lemma implies there is a maximal element P. We claim P is prime.

By contradiction, suppose a,b 6∈ P, and ab ∈ P. Then P+ (a) is larger than P, so finitely generatedby pi + ria for 1 ≤ i ≤ n. Also J = (P : a) = {r ∈ R | ra ∈ P} contains P and b. Then:

P = Rp1 + · · ·+ Rpn + Ja.

(Take t ∈ P, t ∈ P+ (a), write t =∑ui(pi + ria), but then

∑uiri ∈ J)

Proof. (to lemma 1.10) Suppose θ(P) is finitely generated by α1, . . . ,αn. If x ∈ P, then P = (X,α1, . . . ,αn).Otherwise, take fi ∈ P with constant term αi. Take g ∈ P. Then the constant term of g is of the form∑aiαi. So g−

∑aifi = x · g1. Now P is a prime ideal, x 6∈ P, so g1 ∈ P. We can repeat this process:

g1 =∑bifi + x · g2, and etc. This gives hi = ai + bix + cix

2 + · · · , and g =∑hifi. Therefore

P = (f1, . . . , fn).

Localization

Definition 1.11. S is multiplicatively closed of R if S is closed under multiplication and 1 ∈ S.

Define a relation on R× S: (r1, s1) ∼ (r2, s2) iff ∃x ∈ S with x(r1s2 − r2s1) = 0. Write S−1R for theequivalence classes. This is a ring, and there is a ring homomorphism θ : R→ S−1R given by r 7→ r/1.

This new ring has a universal property.

Lemma 1.12. If φ : R → T ring homomorphism such that φ(s) are invertible, then φ extends uniquely toα : S−1R→ T .

Proof. uniqueness: α(r/s) = φ(r) ·φ(s)−1. Existence: α is well defined.

Example 1.13.

1. Field of fractions of an integral domain R. S = R \ {0}.

2. S−1R is the zero ring iff 0 ∈ S.

3. I any ideal, then S = 1+ I is closed.

4. Rf where S = fn |n ≥ 0.

5. If P is a prime ideal, S = R \ P, write RP for S−1R in this case. The process R → RP is calledlocalization.

The elements r/s with r ∈ P form an ideal of RP, called PP. This is the unique maximal ideal. Ifr/s 6∈ PP, then it is invertible.

Definition 1.14. R is a local ring if it has a unique maximal ideal.

Example 1.15.

1. R = Z, P = (p) for a prime p. Then ZP = {m/n : p - n}. And the unique maximal ideal isPP = {m/n : p | m,p - n}.

2. R = k[x1, . . . , xn]. P = (x1 − a1, . . . , xn − an). Then RP is the subring of k(x1, . . . , xn) consistingof those rational functions that are defined at (a1, . . . ,an). The unique maximal ideal consists ofrational functions that are non-zero at (a1, . . . ,an).

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Similarly, for an R-moduleM, we may define an equivalence relation onM× S for a multiplicativelyclosed subset S of R by (m1, s1) ∼ (m2, s2) iff ∃s ∈ S such that s(m1s2 −m2s1) = 0. Denote by S−1Mthe set of all equivalence classes. WriteMP when S = R \ P.

If θ : M1 → M is an R-module map, define S−1θ : S−1M1 → S−1M by sending m/s 7→ θ(m)/s.This will be an S−1R-module map. S−1(φ ◦ θ) = S−1φ ◦ S−1θ.

Definition 1.16. A sequence of R-modulesM1θ−→M

φ−→M2 is exact atM iff Im θ = kerφ. A short exact

sequence (abbr. ses) is of the form 0→M1θ−→M

φ−→M2 → 0with exactness atM1,M,M2, that is, θ is

injective, φ is surjective, and Im θ = kerφ. ThusM2 ∼=M/M1.

Lemma 1.17. S−1· is an exact functor.

Proof. S−1φ ◦ S−1θ = S−1(φ ◦ θ) = S−1(0) = 0. So Im⊂ ker. Conversely if S−1φ(m/s) = 0 = 0/1,then ∃t ∈ S with tφ(m) = 0. So φ(tm) =, and therefore tm ∈ Im θ. Write tm = θ(m1). ThenS−1θ(m1/ts) = tm/ts = m/s.

Corollary 1.18. If N ⊂M, then S−1(M/N) ∼= S−1M/S−1N.

Proof. Apply lemma to 0→ N→M→M/N→ 0.

Remark.

1. If N ⊂M, then S−1N can be regarded as a submodule of S−1M.

2. If I is an R-ideal, then S−1I is a S−1R-ideal.

Proposition 1.19. Every ideal J of S−1R is of the form S−1I for some R ideal I, where I = {r ∈ R : r/1 ∈ J}.Furthermore, there is a 1-to-1 correspondence between prime ideals of S−1R and prime ideals of R which avoids S.

Remark. Think of taking image in S−1R of an ideal as an operation f, and the process of recovering Ifrom J an operation g. The first part is saying fg = idideals of S−1R.

Example 1.20.

1. When P is a prime ideal, set S = R \ P. Then prime ideals of S−1R correspond to prime ideals of Rthat are contained in P. If R = k[x1, . . . , xn], and P = (x1 − a1, . . . , xn − an), then prime ideals ofS−1R correspond to prime ideals consisting of polynomials that vanish at (a1, . . . ,an).

2. R = Z/6,P = 2Z/6 a prime ideal, S = {1, 3, 5} ⊂ R multiplicatively closed. PP = 0. Take theses 0 → P → R → R/P ∼= Z/2 → 0. Localize at S, we get a ses 0 → 0 → RP → (Z/2)P → 0.Thus RP = (Z/2)P = Z/2. Note that bijection between prime ideals does not extend to all ideals.Localize at 0 ideal we would still get 0 = PP.

Proof. (i) Let J be an ideal of S−1R. I = {r : r/1 ∈ J}. If r/s ∈ J, then r/1 ∈ J, and therefore r ∈ I. SoJ ⊂ S−1I. Conversely if r ∈ I, then r/1 ∈ J, and r/s ∈ J for all s. So S−1I ⊂ J

(ii) Let Q be a prime ideal of S−1R. Set P = {r : r/1 ∈ Q}. If xy ∈ P, then xy/1 = x/1 · y/1 ∈ Q.So x/1 or y/1 is in Q. Thus x or y is in P. Also if s ∈ S ∩ P, then s/1 · 1/s = 1 ∈ Q. Therefore P is aprime ideal avoiding S. Conversely, if P is such a prime ideal, Q = S−1P, and if p1/s1 · p2/s2 ∈ Q, thenp1p2/s1s2 ∈ Q. So sp1p2 ∈ P for some s ∈ S. But s 6∈ P, so p1p2 ∈ P. P is prime, so p1/s1 ∈ Q orp2/s2 ∈ Q.

Lemma 1.21. If R is noetherian, then S−1R is noetherian.

Lemma 1.22. Let P be a prime ideal of R, S multiplicatively closed disjoint from P, and S−1P is a prime ideal ofS−1R. Then (S−1P)S−1P

∼= R/P. In particular, if P ⊂ Q, and S = R \Q, then (RQ)PQ = R/P.

Local properties

Definition 1.23. A property P of a ring R of R-moduleM is local if: R (orM) has property P iff RP (orMP) has P for all prime ideal P of R.

Lemma 1.24. (Example) The following are equivalent:

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1. M = 0.

2. MP = 0 for all prime ideals P.

3. MP = 0 for all maximal ideals P.

Proof. (3 =⇒ 1): Pick 0 6= x ∈ M. Then Ann(x) ⊂ P for some maximal ideal P. There is a surjectionφ : M1 = R/Ann (x) → R/P. Localize the ses to get 0 → (kerφ)P → M1,P → RP/PP → 0. But thensince RP/PP 6= 0,M1,P 6= 0. ThenM1,P ⊂MP 6= 0. Contradiction.

Question 7 on sheet 1 has a few more local properties. For an R-module map φ : M → N, phi isinjective iff φP injective. Same for being surjective.1

Tensor product

Definition 1.25. L,M,N be R-modules. φ :M×N→ L is bilinear if

φ(r1m1 + r2m2,n) = r1φ(m1,n) + r2φ(m2,n)φ(m, r1nn + r2n2) = r1φ(m,n1) + r2φ(m,n2).

The idea of tensor product is to replace the discussion of multilinear maps by one about linear maps.If φ :M×N→ T is R-bilinear, and θ : T → L is linear, then θ ◦φ is R-bilinearM×N→ L. So we get

φ∗ : {R-module maps T → L}→ {R-bilinear mapsM×N→ L}.

φ is universal if φ∗ is a bijection.

Lemma 1.26. Given M,N, ∃ an R-module T and a universal map φ : M ×N → T , i.e. given two such

φ1 :M×N→ T1, φ2 :M×N→ T2, there exists a unique isomorphism T1β−→ T2 with φ2 = β ◦φ1.

Proof. Let F be the free R-module generated by em,n indexed by pair (m,n) ∈M×N. Let X ⊂ F be thesubmodule generated by elements of the form:

er1m1+r2m2,n − r1em1,n − r2em2,n,em,r1n1+r2n2 − r1em1,n − r2em2,n.

Then set T = F/X, and we writem⊗ n the image of em,n under the quotient map. Set φ :M×N→ Tby sending (m,n)→ m⊗ n. Note that T is generated bym⊗ n as an R-module, and φ is bilinear. Anymap α :M×N→ L extends to an R-linear map α : F→ L. If α is bilinear, then α vanishes on X. So wehave an induced map α ′ : T = F/X → L. α ′ is uniquely determined (to make the diagram commute),and the universality follows.

Definition 1.27. T above is the tensor product M⊗R N of M and N over R. Usually we omit R if it’sclear from context.

Warning: It is not necessarily the case that an element of M⊗N is of the form m⊗ n. The generalform is

∑ni=1mi ⊗ ni (finite sum since the free module F is a direct sum).

Remark. One can define tensor product over non-commutative rings whereM is right R-module, and Nis left R-module. In this case,M⊗N is only an abelian group, not necessarily an R-module. One takes thefree Z-module on em,n and X is the Z-submodule generated by em1+m2,n− em1,n− em2,n, em,n1+n2 −em1,n − em2,n, emr,n − em,rn. If M is an R− S bimodule (left R and right S module), N is an S− Tbimodule, thenM⊗N is an R− T bimodule (for rings R,S, T ).

Lemma 1.28. There are unique isomorphisms (R-module maps):

1. M⊗N→ N⊗M,m⊗ n 7→ n⊗m.

2. (M⊗N)⊗ L→M⊗ (N⊗ L), (m⊗ n)⊗ l 7→ m⊗ (n⊗ l).1As a funny side fact, being noetherian is not a local property: consider the ring kN.

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3. (M⊕N)⊗ L→M⊗ L⊕N⊗ L.

4. R⊗RM→M, r⊗m 7→ rm.

If φ : R→ T ring homomorphism, and N is a T -module, then N can be regarded as an Rmodule viar ∗m = φ(r)m, called restriction of scalars. Similarly given an R-moduleM, we can define a T -modulestructure on T ⊗RM via t ∗ (t0 ⊗m) = (tt0)⊗m, called extension of scalars.

Example 1.29. Localization may be regarded as the extension of scalars via R → S−1R. Given R-

module M, multiplicatively closed S, there exists a unique isomorphism S−1R⊗RM∼=−→ S−1M, since

S−1R×M→ S−1M sending (r/s,m) to rm/s is bilinear, and then the universality gives the R-modulemap.

Definition 1.30. Given θ :M1 → N1, φ :M2 → N2 R-module maps. Define:

θ⊗φ :M1 ⊗M2 → N1 ⊗N2m1 ⊗m2 7→ θ(m1)⊗φ(m2)

NoteM1 ×M2 → N1 ⊗N2, (m1,m2) 7→ θ(m1)⊗φ(m2) is bilinear, so θ⊗φ is well defined.

Given a ses 0→M1θ−→M

φ−→M2 → 0, consider

0→ N⊗M1id⊗θ−−−→ N⊗M id⊗φ

−−−−→ N⊗M2 → 0.

This is NOT always exact.1

Example 1.31. R = Z. 0 → Z×2−−→ Z

mod2−−−−→ Z/2 → 0, and let N = Z/2. Then the complex after

tensoring is

0→ Z/2 0−→ Z/2→ Z/2→ 0.

So we don’t have injectivity on the left. It is, however, always right exact.

Definition 1.32. An R-module N is flat if N⊗− retains exactness for all short exact sequences.

Example 1.33.

1. S−1R is flat R-module2 (we proved S−1− is an exact functor, and use the fact that S−1R⊗RM =S−1M)

2. R is flat (R⊗R − = id).

3. Any free R-module is flat.

Definition 1.34. Tensor product of algebras. Given φ : R → T ring homomorphism, we say T is anR-algebra. Given φi : R→ Ti, i = 1, 2, we can take the tensor product T1 ⊗R T2 as R-modules, and define

(t1 ⊗ t2) ∗ (t ′1 ⊗ t ′2) = t1t ′1 ⊗ t2t ′2.

Then T1 ⊗R T2 is an R-algebra via φ : R→ T1 ⊗R T2 by r 7→ φ1(r)⊗ 1 = 1⊗φ2(r) = r ∗ (1⊗ 1). (checkthings are well defined, and 1⊗ 1 is the multiplicative identity)

Example 1.35.

1. k field. k[x1]⊗k k[x2] = k[x1, x2].

2. Q[x]/(x2 + 1)⊗Q C = C[x]/(x2 + 1).

3. k[x1]/(f(x1))⊗k k[x2]/(g(x2)) = k[x1, x2]/(f(x1),g(x2)).1It is always exact except at N⊗M1.2A stronger statement: if R→ A is a ring map, M an A-module. Then M is a flat R-module iff Mm is flat RP module (with

P = R∩m) for all maximal ideals m of A.

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2 Ideal structure

Throughout R is commutative ring with identity.

Nilradical and Jacobson radical

Lemma 2.1. The set N(R) of nilpotents of R form an ideal. R/N(R) has no nilpotents.

Definition 2.2. N(R) is the nilradical of R.

Lemma 2.3. N(R) =⋂P prime ideals.

Proof. If xn = 0 ∈ P, then x ∈ P. If xn 6= 0 for all n, let S = {ideals that doesn’t contain xn for any n}.Then (0) ∈ S, and so S 6= ∅. Union gives an upper bound, so by Zorn’s lemma, there exists a maximalelement J1. But J1 must be prime: if y, z 6∈ J1 but yz ∈ J1, then xm ∈ J1 + Ry, xn ∈ J1 + Rz, andxm+n ∈ J1 + Ryz = J1. This is a contradiction.

Definition 2.4. For an ideal I, the radical√I = {r ∈ R | rn ∈ I for some n}. Note

√I/I = N(R/I).

Definition 2.5. The Jacobson radical of R is the intersection of all maximal ideals. N(R) ⊂ J(R).

e.g. R = Z(p) the localization at prime ideal (p). The unique maximal ideal P = {m/n : p | m,p - n}.N(R) = 0.

Lemma 2.6. (Nakayama) LetM be a finitely generated R-module. Then J(R)M =M iffM = 0.

Proof. SupposeM 6= 0. Then the set of proper submodules N 6=M has a maximal member N1 (union ofchains each of which does not contain a given finite generating set also doesn’t contain the generatingset, thus the upper bound is still proper)1. By maximality M/N1 is simple. Take 0 6= m ∈ M/N1. Itgenerates M/N1 (otherwise Rm is a proper submodule, contradicting the simplicity), and we have asurjection θ : R→M/N1 by r 7→ rm. ThusM/N1 ∼= R/ ker(θ). So ker θ is a maximal submodule (ideal)of R. Write P = ker θ. Then PM ⊂ N1. Hence J(R)M ⊂ PM ⊂ N1 (M.

Remark.

1. This is not the normal proof you would find in AM. This is adaptable to non-commutative rings.

2. M = 0 iffMP =M for all prime ideals P (M finitely generated R-module).

Theorem 2.7. (Weak Nullstellensatz) Let k be a field, T a finitely generated k-algebra. Q a maximal ideal. ThenT/Q is a finite field extension of k. In particular, if k is algebraically closed, and T is the polynomial algebra, thenmaximal ideals Q is of the form (x1 − a1, . . . , xn − an).

We need two results to prove this:

Lemma 2.8. (Artin-Tate) Let R ⊂ S ⊂ T be rings. Suppose R is noetherian, and T is finitely generated by Rand x1, . . . , xm as a ring. Suppose further that T is a finitely generated S-module. Then S is generated by R andfinitely many elements as an R-algebra.2

Proof. Proof is very easy actually. Let x1, . . . , xm generate T as an R-algebra, y1, . . . ,yn generate T as anS-module. Write xi =

∑j bijyj, and yiyj =

∑k bijkyk, with b’s in S.

Consider the R-algebra S0 generated by those b’s. Then T is finite (fg as module) over S0: any elementin T is a polynomial in xi with coefficients in R, thus a polynomial in yi with coefficients in S0, and thusa linear combination of yi’s with coefficients in S0. Basis theorem says S0 is noetherian, and thus T isnoetherian S0-module. But S is a S0-submodule of T . By noetherian definition, S is finitely generatedS0-module. But S0 is finitely generated R-algebra, so S surely is a finitely generated R-algebra.

Proposition 2.9. (Zariski’s lemma) k field, R finitely generated k-algebra. If R is a field, then R/k is a finitealgebraic extension (that is, finitely generated k-vector space)

1For a finitely generated module, every submodule is contained in a maximal submodule. However this is not generally true:every projective module has a maximal submodule, but not necessarily a submodule is contained in a maximal submodule

2An R algebra T is finite if it is finitely generated as an R-module; is of finite type if it is finitely generated as an R-algebra. So ifR noetherian, T over R is of finite type, T over S is finite, then S over R is of finite type.

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Proof. Suppose R is generated by x1, . . . , xn, and suppose R/k is not finite algebraic. Reorder to assumex1, . . . , xm are algebraically independent and xm+1, . . . , xn are algebraic over k(x1, . . . , xm) =: F. Byassumption, m ≥ 1. Hence R is finite extension over F, and R is a finitely generated F-module (vectorspace). Apply Artin-Tate to k ⊂ F ⊂ R to get that F is a finitely generated k-algebra, say by q1, . . . ,qk ∈ F.Write qi = fi/gi with fi,gi ∈ k[x1, . . . , xm], there is a polynomial h coprime to each gi (by all meanstake

∏gi + 1). Then 1/h cannot be in the ring generated by k and those qi’s. This is a contradiction. So

m = 0.Proof. (to weak Nullstellensatz) LetQ be a maximal ideal of a finitely generated k-algebra T . Set R = T/Q.By Zariski’s lemma, T/Q over k is a finite algebraic extension. If k is algebraically closed, then k = T/Q.Set π : T → kwith kerπ = Q. Then we claim kerπ = (x1 − π(x1), . . . , xn − π(xn)).

Too see this, notice π fixes elements of k, so RHS is in the kernel. Conversely, T/(x1 − π(x1), . . . , xn −π(xn)) is a 1-dimensional k-vector space. So kernel must also contained in RHS.

Thus Q is of the form required.

Now go back to what we saw in the intro: there is a bijection between

{radical ideals in C[x1, . . . , xn]}↔ {affine algebraic sets of Cn}.

This can now be rephrased as: letQ(a1,...,an) = (x1 −a1, . . . , xn −an). Then there is a bijection between

{radical ideals}←→ {algebraic subsets}I 7→ {(a1, . . . ,an) | I ⊂ Q(a1,...,an)}⋂

(a1,...,an)∈SQ(a1,...,an) ← [ S.

Theorem 2.10. (Nullstellensatz) Let R be a finitely generated k-algebra. ThenN(R) = J(R). Thus if I is a radicalideal of k[x1, . . . , xn] and R = k[x1, . . . , xn]/I, then the intersection of maximal ideals of R is 0. Hence I is theintersection of all maximal ideals containing I.

For this we need the following lemma.

Lemma 2.11. Let k be a field, R a k-algebra which is also an integral domain. If R is finite dimensional vectorspace over k, then R is a field.

Proof. Take 0 6= r ∈ R. Multiplication by r is k-linear. R being integral domain implies this map isinjective. Finite dimensional implies any injective map is also surjective. Thus r has an inverse.Proof. (Nullstellensatz) Let P be a prime ideal of R, and take s ∈ R \ P. Let S = {1, s, s2, . . . }. Then we canform S−1R, and θ : R → S−1R. Since R is finitely generated k-algebra, and S−1R is generated by θ(R)and 1/s (as k-algebra), S−1R is also finitely generated. Take a maximal ideal Q of S−1R containing S−1Pand apply weak Nullstellensatz: S−1R/Q is a finite extension of k.

By correspondence of prime ideals, we know Q corresponds to some P1 = θ−1(Q) prime idealcontaining P disjoint from S. Thus θ induces embedding R/P1 ↪→ S−1R/Q. But S−1R/Q is a finitedimensional k-vector space, so R/P1 must also be finite dimensional.

Now apply the previous lemma, P1 is actually maximal. So we have found, for any s 6∈ P, a maximalideal containing P but not containing s. This shows

⋂{maximal ideals containing P} ⊂ P. But the

converse direction is trivial. So P = {maximal ideals containing P}.Now N(R) is the intersection of all prime ideals. So we are done.

Minimal and associated primes

Lemma 2.12. If R is noetherian, then any ideal I contains a power of its radical√I. In particular, N(R) is

nilpotent (i.e. N(R)m = 0 for somem) (take I = (0)).

Proof. Suppose x1, . . . , xm ∈√I generate

√I as an ideal. Then xmii ∈ I. Set m to be sufficiently large,

and then√Im

is generated by xr11 · · · xrmm with

∑ri = m. When m is sufficiently large, at least one

ri ≥ mi. So√Im ⊂ I.

Lemma 2.13. If R is noetherian, a radical ideal is the intersection of finitely many primes.

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Proof. Suppose for a contradiction, and take a maximal element I from the set of radicals that are not theintersection of finitely many primes. Claim: I is prime.

If not, there exist1 J1, J2 6⊂ I with J1J2 ⊂ I. Taking our new Ji to be Ji + I if necessary, we can assumewe have I ( J1, J2 with J1J2 ⊂ I. Maximality implies that√

J1 = Q1 ∩ · · · ∩Qs√J2 = Q ′1 ∩ · · · ∩Q ′t,

as prime intersections. Then consider J =√J1 ∩√J2. By definition, J ⊂

√J1, so Jm1 ⊂ J1, and Jm2 ⊂ J2.

Then Jm1+m2 ⊂ J1J2 ⊂ I. But I is radical, so J ⊂ I. However, I ⊂ Ji ⊂√Ji. So I ⊂ J. Thus I = J.

Now suppose I is an ideal of a noetherian ring R, and√I = P1 ∩ · · · ∩ Pm. We may assume Pi

are distinct, and no Pi contains any other Pj. Note that if P is a prime containing I, we would have∏Pi ⊂

⋂Pi =

√I ⊂ P. So some Pi ⊂ P.

Definition 2.14. The minimal primes over an ideal I of a noetherian ring R are those primes, i.e. I ⊂ P isminimal if for any other prime I ⊂ Q ⊂ P, we have Q = P.

Lemma 2.15. I is an ideal of a noetherian ring R. Then√I is the intersection of the minimal primes over I, and I

contains a finite product of them, perhaps with repetitions.

Proof. Each minimal prime over I contains√I, and so over I is the same thing as over

√I. The latter

assertion follows trivially from lemma 2.12.

Definition 2.16. Let M be a finitely generated R-module with R noetherian. Then an associatedprime P ⊂ R of M is one such that P = Ann(m) = {r : rm = 0} for some m ∈ M. Ass(M) ={all associated primes ofM}.2

e.g. Ass(R/P) = {P} if P is prime, since the annihilator of a ∈ R/P can either be whole of R, which isnot a prime; or P.

Definition 2.17. A submodule N ofM is P-primary if Ass(M/N) = {P}. An ideal I of R is P-primary ifAss(R/I) = {P}.

Remark.

1. If P is evident, just say primary.

2. There’s an equivalent definition of P-primary, cf sheet 2.

Lemma 2.18. If Ann(M) := {r : rm = 0 ∀m ∈M} =⋂m∈M Ann(m) is a prime P. Then P ∈ Ass(M).

Proof. Let m1, . . . ,mn generate M, and Ii = Ann(m).∏Ii annihilates all mi, and thus

∏Ii ⊂

Ann(M) = P. Definition of prime ideal gives Ij ⊂ P. However, P = Ann(M) ⊂ Ann(mj) = Ij.Thus P = Ij.

We can always be sure that Ass(M) 6= ∅.

Lemma 2.19. Let Q be maximal among {Ann(x) : 0 6= x ∈ M}. Then Q is prime, and therefore by definitionQ ∈ Ass(M).

Proof. Say Q = Ann(m), r1r2 ∈ Q, and r2 6∈ Q. We need to show r1 ∈ Q. Now r1r2m = 0, sor1 ∈ Ann(r2m). r2 6∈ Q implies that r2m 6= 0. But Q ⊂ Ann(r2m). So by maximality of Q, we musthave Q = Ann(r2m) 3 r1.

Thus if R is noetherian, and M 6= 0, the above set of ideals must have a maximal element, andAss(M) 6= ∅.

Our next task is to show: 1. Ass(M) is in fact finite, and 2. the minimal ideals over I ⊂ Ass(R/I).Before this, let’s see one trivial example.

1You should think of ideal inclusion being ‘divisible by’. So I ⊂ J should be thought of as J | I. This will make things a loteasier. In particular, P is prime ideal iff IJ ⊂ P implies I ⊂ P or J ⊂ P.

2Old day joke is that this is called the assassin – due to Bourbaki.

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Example 2.20. R = k[x,y],Q = (x,y) ⊃ P = (x). Let I = PQ = (x2, xy). We could show that1

Ass(R/I) = {P,Q}.

Now the only minimal prime over I is P (since√I = P). Note that I is not primary, as there are two

primes in Ass(R/I); but we can write

I = (x2, xy,y2)∩ (x).

Then (x2, xy,y2) = (x,y)2 is Q-primary2, and (x) is P-primary. This is an example of primary decompo-sition.

Let’s make this idea more precise.

Definition 2.21. LetM be a finitely generated R-module with R noetherian. N ⊂M a submodule. Thenthere exist N1, . . . ,Ns submodules ofM containing N such that Ni is Pi-primary, Pi distinct, and

N = N1 ∩ · · ·Ns.

ThusM/N ↪→ ⊕M/Ni.

Note that primary decompositions are not unique. However, there are two uniqueness theorem (forfinitely generated module over noetherian ring).

Theorem 2.22. The Pi occurring in a primary decomposition are unique. They are precisely Ass(M/N).

Theorem 2.23. If Pj are minimal among all occurring Pi’s, then the corresponding Nj are unique. If Pj are notminimal, called embedded, then Nj can vary.

Proofs are omitted. It’s just a repetitive use of localizations. You can find proofs in any commutativealgebra book.

In particular, in the previous example, I = (x2, xy,y2) ∩ (x) is a primary decomposition, with Pminimal, Q embedded. Thus the ideal (x) in the decomposition is unique, and Ass(R/I) = {P,Q}.

Aim: 1. Ass(M) is finite, and 2. the minimal ideals over I ⊂ Ass(R/I). We show both of these withoutinvoking the above uniqueness theorems (otherwise they are obvious...)

Lemma 2.24. For a non-zero finitely generated R-moduleM with R noetherian, there exists a chain

0 =M0 (M1 (M2 · · · (Ms =M,

withMi/Mi−1 ∼= R/Pi for some prime ideal Pi of R. Those Pi need not be distinct.

Proof. Since M 6= 0, there3 exists 0 6= m1 ∈M with Ann(m1) = P1. Set M1 = Rm1. Then R/P1 ∼=M1.Repeat this process. SinceM is noetherian, this process stops.

Lemma 2.25. If N ⊂M, then Ass(M) ⊂ Ass(N)∪Ass(M/N).

Proof. From above we know M contains a submodule isomorphic to R/P. There are two cases: eitherM1 ∩N = 0; then the image of M1 in M/N is isomorphic to R/P, and hence P ∈ Ass(M/N). OrM1 ∩N 6= 0; then any 0 6= x ∈M1 ∩N has Ann(x) = P, giving P ∈ Ass(N).

Lemma 2.26. Ass(M) is finite for any finitely generated module over a noetherian ring.

Proof. Use two lemmas above. Ass(M) ⊂ {P1, . . . ,Ps}.

Lemma 2.27. {minimal primes over I} ⊂ Ass(R/I).

Proof. By lemma 2.15, there exists a product of minimal primes over I contained in I. Write

Ps11 · · · P

snn ⊂ I.

LetM = Ps22 · · · Psnn + I/I. Consider J = Ann(M).

1I am not sure how to see this immediately. It is obvious, however, from primary decomposition.2powers of maximal ideal are primary; not true for prime ideals.3This is an equivalent definition for associate primes: P ∈ Ass(M) iff P prime andM contains a submodule isomorphic to R/P.

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Certainly Ps11 ⊂ J. Also JPs22 · · · Psnn ⊂ I ⊂ P1, which gives J ⊂ P1, since P1 certainly doesn’t divide

any other Pi. In particular, J 6= R, and thereforeM 6= 0. Thus forMwe have a chain

0 (M1 (M2 · · · (Mt =M

with Mj/Mj−1 = R/Qj for some primes Qj. But Ps11 annihilates M, and hence every Mj/Mj−1 =R/Qj, and so Ps11 ⊂ Qj. Since Qj are prime, we have P1 ⊂ Qj for every j. Not all P1 ( Qj, since∏Qj ⊂ Ann(M) = J ⊂ P1 (because any m ∈ Mj \Mj−1 for the largest j can be killed by

∏jk=1Qk).

Take the smallest j such that Qj = P1, and then∏i<jQi 6⊂ P1. Finally, take any x ∈Mj \Mj−1. We are

going to show P1 ∈ Ass(M).If j = 1, then Ann(x) = Q1 = P1 and so P1 ∈ Ass(R/I). If j ≥ 1, pick r ∈

∏i<jQi \ P1, and note

that r(sx) = 0 for any s ∈ P1 = Qj. So s(rx) = 0, and P1 ⊂ Ann(rx). However, rx 6∈ Mj−1 sincer 6∈ P1 and Ann(rx+Mj−1) = P1 (Mj/Mj−1 = R/P1). So Ann(rx) ⊂ P1. Thus Ann(rx) = P1 andP1 ∈ Ass(R/I).

3 Dimension

Throughout this section, R is commutative ring with identity.

Definition 3.1. The prime spectrum of a ring is the set of all prime ideals, denoted by Spec(R).

Definition 3.2. The length of a chain of prime ideals P0 ( P1 ( · · · ( Pn with each Pi prime is n.

Definition 3.3. The Krull dimension of R, written as dimR, is the supremum of the lengths of all chainsof prime ideals. If it doesn’t exist, dimR =∞.

Definition 3.4. The height of a prime ideal P, written as ht(P), is the supremum of the lengths of allchains of prime ideals that end in P.

Note: correspondence of prime ideals of R and RP implies dimRP = dimR.

Example 3.5.

1. A noetherian ring Rwith dimension 0 is artin.

2. dim Z = 1. dimR[x] = 1 for any field k.

3. dimk[x1, . . . , xn] = n. It is easy to see dim ≥ n, since 0 ⊂ (x1) ⊂ (x1, x2) ⊂ · · · ⊂ (x1, x2, . . . , xn)is a chain of prime ideals. Equality requires more work.

Lemma 3.6. ht 1 primes in k[x1, . . . , xn] is of the form (f) for f irreducible (or equivalently prime).

Integral dependence/extensions

Definition 3.7. Let R ⊂ T be rings. We say x ∈ T is integral if x satisfies a monic polynomial withcoefficients in R. T over R is integral if all element are integral.

Lemma 3.8. The following are equivalent:

1. x ∈ T is integral over R.

2. R[x] is a finitely generated R-module (by construction it is always a fg R-algebra).

3. There exists a subring T1 which is also a finitely generated R-module with R[x] ⊂ T1 ⊂ T .

Proof. 1 =⇒ 2 =⇒ 3 is clear. Consider the map f : T1 → T1,m 7→ xm. Now since T1 is afinitely generated R-module, and by picking an ideal I = R, we can use the determinant trick fromAppendix A to deduce that f satisfies a monic polynomial with coefficients from Ii = R. However,fn(1) = 1 · x · x · · · · x = xn. Taking value 1 gives a monic polynomial satisfied by x.

Lemma 3.9. If x1, . . . , xn are integral over R, then R[x1, . . . , xn] is integral over R.

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Lemma 3.10. The set of integral elements of T over R form a subring T1 of T .

Proof. Follows directly from lemma above.

Definition 3.11. The above T1 is called the integral closure of R in T . If T1 = R, then R is called integrallyclosed. If T1 = T , then T is integral over R. If R is an integral domain, we say R is integrally closed if it isintegrally closed in its fraction field.

E.g. Z is integrally closed in Q. k[x1, . . . , xn] integrally closed in k(x1, . . . , xn). Any ring of integersOK is the integral closure of Z in a number field K.

Remark.

1. Being integrally closed is a local property.

2. Noether’s normalization lemma: for a finitely generated k-algebra T where k is a field, T contains ak-subalgebra R that is isomorphic to a polynomial algebra, over which it is integral. Furthermore,if T is an integral domain, its integral closure T1 in Frac T is a finitely generated T -module.

The geometric property analogous to being integrally closed is being normal. The fibers of theserestriction maps are finite and these maps are onto:

Spec T1 −→ Spec T −→ SpecRQ 7→ Q∩ R

3. The integral closure of an integral domain in its fraction field has another definition: the intersectionof all valuation rings containing R.

We need to understand better those restriction maps.

Lemma 3.12. If R ⊂ T1 ⊂ T , T1 integral over R, and T integral over T1, then T integral over R.

Lemma 3.13. Let R ⊂ T , with T integral over R.

1. If J is an ideal of T , then R/(J∩ R) = (R+ J)/J ⊂ T/J is integral.

2. If S is multiplicatively closed subset of R, then S−1T is integral over S−1R

Proof. (1) Pass the monic polynomial to T/J. (2) Any x/s would satisfy the monic polynomial of xmultiplied by sn, which is again monic.

Lemma 3.14. R ⊂ T integral with T ,R both integral domains. Then T is a field iff R is a field.

Proof. If R is a field, pick any 0 6= t ∈ T . Then tn+ · · ·+ r0 = 0. Now −r−10 (tn−1+ · · · ) ∈ T is an inversefor t.

If T is a field, and 0 6= s ∈ R, s has an inverse in T , and s−1 would satisfy s−n + · · ·+ r0 = 0. Nows−1 = s−n · sn−1 = −(rn−1 + rn−2s+ · · · ) ∈ R.

Corollary 3.15. Let R ⊂ T be integral, andQ a prime ideal in T , P = Q∩ R. ThenQ is maximal iff P is maximal.

Theorem 3.16. (Incomparability) Let R ⊂ T be integral, Q ⊂ Q1 prime ideals of T , and suppose Q ∩ R =Q1 ∩ R = P, then Q = Q1. Hence a strict chain of prime ideals in T will induce a strict chain of primes in R, andthus dimR ≥ dim T .

Proof. We have the localization RP ⊂ TP is integral (TP is a slight abuse of notation, as there is noguarantee that P will remain a (prime) ideal of T ; nevertheless S = R \ P is multiplicatively closed subsetof both R and T ). Take the unique maximal ideal PP in the local ring RP. There are also S−1Q,S−1Q1 inTP. By construction,

S−1Q∩ S−1R = S−1P = S−1Q1 ∩ S−1R,

since S−1P ⊂ S−1Q ∩ S−1R (trivially) and S−1Q ∩ S−1R is a proper ideal containing the maximalideal PP. By corollary above, S−1Q and S−1Q1 must both be maximal ideals in TP. Since Q ⊂ Q1,S−1Q = S−1Q1. But the 1-1 correspondence between primes in localization further impliesQ = Q1.

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Theorem 3.17. (Lying over) Let R ⊂ T be integral. Take P prime ideal of R. Then there is a prime ideal Q of Twith Q∩ R = P. In other words, the restriction map

Spec T res−→ SpecR

is onto.

Proof. TP is integral over RP. Take a maximal ideal (not unique) of TP. It is of the form S−1Q for someprime Q of T . Then S−1Q ∩ S−1R is maximal by corollary 3.15. So S−1Q ∩ S−1R = PP. This wouldimply Q∩ R = P.

Theorem 3.18. (Going up) Let R ⊂ T be integral, P1 ⊂ · · · ⊂ Pn a chain in SpecR, and Q1 ⊂ · · · ⊂ Qma chain in Spec T . If m < n and Qi ∩ R = Pi for all 1 ≤ i ≤ m, then the chain Qi can go up: there existQm+1 ⊂ · · · ⊂ Qn extending the chain with Qi ∩ R = Pi for all i ≤ n.

Theorem 3.19. (Going down) Let R ⊂ T be integral, and R, T are integral domains with R integrally closed(in FracR). Let P1 ⊃ · · · ⊃ Pn a chain in SpecR, and Q1 ⊃ · · · ⊃ Qm a chain in Spec T . If m < n andQi ∩ R = Pi for all 1 ≤ i ≤ n, then the chain Qi can go down: there exist Qm+1 ⊃ · · · ⊃ Qn extending thechain with Qi ∩ R = Pi for all i ≤ n.

Assuming these theorems, we would have something marvellous.

Corollary 3.20. If R ⊂ T is integral, then dimR = dim T .

Proof. Take a prime chain in T . Incomparability implies dimR ≥ dim T . But conversely take a primechain in R, there is a Q0 lying over P0. Now going up gives dim T ≥ dimR.

Corollary 3.21. Let R ⊂ T be integral, R, T integral domains and R integrally closed. If Q is a prime ideal in T ,then ht(Q∩ R) = ht(Q).

Proof. Take a prime chain Q0 ( · · · ( Qn = Q in T . Incomparability says Q0 ∩ R ( · · · ( Qn ∩ R =Q ∩ R. So ht(Q ∩ R) ≥ ht(Q). Conversely if P0 ( · · · ( Pn = Q ∩ R in R, then start with the largestprime Q, going down implies a chain of primes contained in Q of length n.

Now we prove both going-up and going-down. Going-up is easy to prove, whereas going-downrequires a bit more work.Proof. (to going up theorem) By induction, we reduce the case tom = 1,n = 2. Write R for R/P1, T forT/Q1, where Q1 ∩ R = P1. Then R ↪→ T and this is an integral extension by lemma 3.13. By lying over,there exists a Q2 of T st Q2 ∩ R = P2. Lift this back to give Q2 of T , and we have Q2 ∩ R = P2.

Definition 3.22. If I is an ideal of R, and R ⊂ T , x ∈ T is integral over I if x satisfies some monicpolynomial f(x) ∈ I[x].Lemma 3.23. Let R ⊂ T be integral, and I is an ideal of R. Then the integral closure of I in T is the radical ideal√TI, where TI is an ideal of T . Consequently, this integral closure is closed under + and · . In particular, if R = T ,

we get the integral closure of I in R is√I.

Proof. If x is integral over I, write xn + a1xn−1 + · · ·+ an = 0. So xn ∈ TI, and thus x ∈

√TI.

Conversely, if xn =∑tiri with ri ∈ I, we know each ti is integral over R, and thusM := R[t1, . . . , tn]

is a finitely generated R-module. Now multiplication by xn is a map such that xnM ⊂ IM (since(∑tiri) · f =

∑ri(ti · f) is in IM). By the determinant trick mentioned in appendix, xn satisfies a monic

polynomial with coefficients in I.

Lemma 3.24. Let R ⊂ T be integral domains and R integrally closed. If x ∈ T is integral over an R-ideal I, then xis algebraic over FracR =: K, with minimal polynomial xn + rn−1x

n−1 + · · ·+ r0 = 0 for ri ∈√I.

Proof. x is certainly algebraic with coefficients of its minimal polynomial in K. Claim: coefficients ri ∈ Rand are integral over I.

Take a field extension K ⊂ L containing all conjugate x1, . . . , xn, e.g. splitting field of the minimalpolynomial of x. Then there exist K-automorphisms of L sending x 7→ xi. So if x is integral over Iwithmonic equation xm + am−1x

m−1 + · · ·+ a0 = 0, then so are xi’s. Thus each conjugate is integral overI, and in particular lies in the integral closure of R in L. However, the coefficients ri are obtained byproducts and sums of those xi’s. So each coefficient ri ∈ K are integral over I. In particular, ri are integralover R. But we assumed R is integrally closed (in K), so in fact ri ∈ R.

Now use the previous lemma with T = R, and we obtain that ri ∈√I.

14

Now we are ready to prove going down.

Proof. Again by induction, we’ve reduced the case to P1 ⊃ P2, and Q1 with Q1 ∩ R = P1. We need toshow there is a prime Q2 ⊂ Q1, and Q2 ∩ R = P2.

Let S2 = R \ P2, S1 = T \Q1. Set S = S1S2 = {rt : r ∈ S1, t ∈ S2}. This is multiplicatively closed, andcontains both S1,S2. We prove TP2 ∩ S = ∅.

Assuming this, we know TP2 is an ideal of T , and so S−1(TP2) is a proper ideal of S−1T . It is thuscontained in some maximal ideal of the formm = S−1Q2 for some prime Q2 ⊂ T with Q2 ∩ S = ∅ andTP2 ⊂ Q2. Hence P2 ⊂ TP2 ∩ R ⊂ Q2 ∩ R. Since Q2 ∩ S = ∅, and S2 = R \ P2 ⊂ S, Q2 ∩ R ⊂ P2, andthus Q2 ∩ R = P2. Similarly S1 = T \Q1 ⊂ S, and so Q2 ⊂ Q1. We would then be done.

We prove the claim by contradiction. Take 0 6= x ∈ TP2 ∩ S. Apply lemma 3.23 with I = P2, we havex ∈ TP2 ⊂

√TP2. So x is in the integral closure of P2 in T . Lemma 3.24 says x is algebraic over the

fraction field K of R, with minimal polynomial xn + · · ·+ r0 with coefficients in√P2 = P2. But x ∈ S, so

we write x = rtwith r ∈ S2, t ∈ S1. Then t = x/rwould have a minimal polynomial

Tn +rn−1rTn−1 + · · ·+ r0

rn= 0

with coefficients in R by lemma 3.24 with I = R. Write r ′i = ri/rn−i ∈ R. The equation r ′i · rn−i = ri ∈ P2

and the fact that r 6∈ P2 imply that r ′i ∈ P2. By definition, t is therefore integral over P2. So lemma 3.23again says t ∈

√TP2. This is a contradiction since t ∈ S1 = T \Q1, but TP2 (and hence

√TP2 because

Q1 is prime) lies in Q1 (as P2 ⊂ P1 ⊂ Q1).

Now we move to the key result.

Theorem 3.25. (Noether’s normalization lemma) Let T be a finitely generated k-algebra. Then T is integral (in factfinite) over some subalgebra R which is isomorphic to a polynomial algebra1 (R = k[X1, . . . ,Xr]→ k[x1, . . . , xr]is an isomorphism, say x1, . . . , xr are algebraically independent over k).

Proof. Prove by induction on the number of generators a1, . . . ,an for T . Note if all generators arealgebraic over k, then taking the set {a1, . . . ,am11 ,a2, . . . ,am22 , . . . ,an, . . . ,amnn } as a basis, T will be afinite dimensional k-vector space. Then taking R = kwill suffice. If they are all algebraically independent,then taking R = T will suffice. Otherwise, we assume a1, . . . ,ar are algebraically independent. Nowtake 0 6= f ∈ k[X1, . . . ,Xr,Xn] with f(a1, . . . ,ar,an) = 0. Write f as a sum of terms λ~lX

l11 X

l22 · · ·X

lrr X

lnn

where~l = (l1, . . . , lr, ln).

Key claim: there exist m1, . . . ,mr ≥ 0 such that th map φ :~l 7→ m1l1 + · · ·+mrlr + ln is injectivefor those~lwith λ~l 6= 0.

Assuming this, now we put g(X1, . . . ,Xr,Xn) = f(X1 +Xm1n , . . . ,Xr +Xmrn ,Xn). This will be a sum

of termsλ~l(X1 +X

m1n )l1(X2 +X

m2n )l2 · · · (Xr +Xmrn )lrXlnn .

View this as a function in Xn. By construction of φ, each term will have different degree. Since weassumedmi ≥ 0, the leading term will have coefficient one of the λ~l ∈ k.

If we put bi = ai − amin , and consider

h(Xn) = g(b1, · · · ,br,Xn).

The leading coefficient will be in k, and all coefficients are in k[b1, · · · ,br], with

h(an) = g(b1, . . . ,br,an)

= f(a1 − am1n + am1n , . . . ,ar − amrn + amrn ,an)

= f(a1, . . . ,ar,an) = 0.

Dividing by the leading coefficient, this shows an is integral over k[b1, . . . ,br]. Now consider the subringT0 = k[b1, . . . ,br,ar+1, . . . ,an−1] ⊂ T which is a subalgebra with fewer generators. And ai = bi + a

min

is integral over T0 for 1 ≤ i ≤ r. So T is integral over T0. Now by induction, T0 would be integralover a polynomial subalgebra, but being integral is obviously transitive. This would then complete theinduction.

1The dimension of this polynomial algebra rwill be the Krull dimension of T .

15

Proof. (Claim): There are finitely many possibilities for differences d = l− l ′ with λl 6= 0 6= λl ′ (onlyfinite such l’s). Note d is of the form (d1, . . . ,dr,dn), and we consider the finitely many non-zero(d1, . . . ,dr) ∈ Zr. Vectors in Qr orthogonal to one of these lie in finitely many (r− 1)-dimension rationalsubspaces, and thus their union cannot be the whole space. Thus we can pick (q1, . . . ,qr) with eachqi > 0 such that

∑qidi 6= 0 for any of those finitely many non-zero (d1, . . . ,dr). Clearing out the

denominator, we have ∑midi 6= 0

for all of those finitely many non-zero d. Thus for thesemi’s, if φ(l) = φ(l ′), then∑midi = 0, and thus

(d1, . . . ,dr) = 0.

As in linear algebra, we can also study maximal algebraically independent set.

Definition 3.26. Let x1, . . . , xr be algebraically independent over k. If the canonical map k[X1, . . . ,Xr]→k[x1, . . . , xr] is an isomorphism as k-algebra maps, then these are called transcendence basis of a field Lcontaining k. There is also an exchange lemma which tells that every maximal algebraically independentset would have the same cardinality. Call this number the transcendence degree of L over k, written astrdegk(L).

As an analogy in terminology,

linear indepedent set←→ algebraically independent setspan of a set←→ algebraic closure of a set

For more references, see Stewart’s Galois Theory page 131-133.

Theorem 3.27. Let T be a finitely generated k-algebra that is a domain with field of fraction L. Then dim T =trdegk(L).

Proof. Apply noether’s normalization. There is a subalgebra R of T which is isomorphic to k[x1, . . . , xr],and T is integral over R. By corollary of going up, dim T = dimR. Thus any finitely generated k-algebrahas dimension equal to that of this polynomial subalgebra. But the transcendence degree is exactly thenumber of variables r, so we need to prove dimk[x1, . . . , xr] = r. (Note we saw earlier the dimension isat least r.)

We prove by induction. If r = 0, it is clear. Otherwise, let P0 ⊂ · · · ⊂ Ps be a chain of prime ideals.Assume P0 = (0) since R is an integral domain. Since k[x1, . . . , xr] is a UFD, P1 contains a principal primeideal (f) for some f irreducible element. Assume P1 = (f). Then trdegk(Frac(k[x1, . . . , xr]/(f))) = r− 1(check!). Apply noether’s normalization again to k[x1, . . . , xr]/(f). This ring would have dimension thesame as some subalgebra k[Y1, . . . , Yt]. Thus

dimk[x1, . . . , xr]/(f) = dimk[Y1, . . . , Yt] = trdegkk(Y1, . . . , Yt) = r− 1,

where the second equality comes from induction. So quotient out by (f), we get a chain of prime ideals0 ⊂ P2/(f) ⊂ · · · ⊂ Ps/(f), with length s− 1. But dimension if r− 1. So s− 1 ≤ r− 1, and therefores ≤ r.

Theorem 3.28. Let R be a noetherian integral domain, integrally closed in its fraction field K. Let L be a separableextension over K, and let T1 be the integral closure of R in L. Then T1 is a finitely generated R-module.

Note: separability always holds in characteristic 0.

Corollary 3.29. If R = Z, which is integrally closed (UFD), and L any number field, then the integral closure ofZ in L is a finitely generated abelian group.

Corollary 3.30. Assume char k = 0. Let T be a finitely generated k-algebra which is an integral domain, integralover a polynomial algebra R. Let L be the fraction field of T . Then the integral closure of T in L, denoted by T1, is afinitely generated R-module. The fibres of these restriction maps are finite:

Spec T1 −→ Spec T −→ SpecR

Proof. See example sheet.

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Proof. (to theorem) We will need the trace function. Define

TrL/K(x) = |L : K(x)| · (−1 · next-to-top coefficient of the minimal polynomial of x over K)

for any finite field extension L/K. If L is Galois over K, then

Tr(x) =∑g∈G

g(x)

where G = Gal(L/K). Note this is a sum of conjugates but there may be repetitions.Quote: if L/K is separable, then

L× L→ K

x · y 7→ Tr(xy)

gives a non-degenerated symmetric K-bilinear form (Reid 8.13). Pick a K-vector space basis of L,y1, . . . ,yn. BY multiplying an appropriate element of K, assume each yi ∈ T1. Since Tr(xy) defines anondegenerate symmetric bilinear form, we can find x1, . . . , xn such that Tr(xiyy) = δij. Take x ∈ T1,then x =

∑λxi with λi ∈ K. So Tr(xyj) =

∑λTr(xiyj) =

∑λiδij = λj. But xyj ∈ T1. By lemma 3.24

with I = R, the minimal polynomial of xyj would have coefficients in R. Thus λj ∈ R, and x ∈∑Rxi

which implies T1 ⊂ Rxi. So T1 lies in a finitely generated R-module with R noetherian. Therefore T1 is afinitely generated R-module.

4 Height

Our main aim is to prove:

Theorem 4.1. (Krull’s Hauptidealsatz) Let R be a noetherian ring, and a is a non-unit of R. Then the minimalprimes P over (a) satisfy ht(P) ≤ 1.

Remark. In UFD, the height one primes are precisely the principal primes (f) for f irreducible.

Corollary 4.2. Let R be a noetherian ring, I a proper ideal generated by n elements. Then ht(P) ≤ n for anyminimal prime over I.

Prove this by induction, and the first step is precisely theorem 4.1.

Corollary 4.3.

1. Every prime ideal of a noetherian ring has finite height.

2. Every noetherian local ring R has finite dimension ≤ the minimal number of generators of the uniquemaximal ideal P = the R/P-vector space dimension of P/P2.

Definition 4.4. A regular local ring is one where dimR = dimR/P P/P2 where P is the unique maximalideal. In geometry, this corresponds to a localization at a non-singular point P. P/P2 is the cotangentspace at the point P.

Remark. Actually we have dimR ≤minimal number of generators of any ideal I with√I = P (in a local

ring) from 4.2, and in fact there exists such an Iwith dimR = minimal number of generators of Iwith√I = P.

Proof. (4.3). 1. Any prime ideal is finitely generated, and it is the unique minimal ideal over itself.2. For a local ring R, dimR = ht(P). By 1, ht(P) ≤number of generators of P. The last equality follows

from Nakayama:Claim: P is generated by x1, . . . , xs iff P/P2 is generated by x1, . . . , xs as R/P-vector space.⇒ is obvious. Conversely, suppose x1, . . . , xs generate P/P2 for some x1, . . . , xs ∈ P. Consider the

ideal I = (x1, . . . , xs) ⊂ P. Clearly I+P2 = P (since P/P2 gen. by xi+P2), so P(P/I) = (I+P2)/I = P/I(easy check). Thus by nayakama, P/I = 0.

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Proof. (4.1 Hauptidealsatz) First we show we can assume R is local. Take a non-unit a of a noetherianring, and let P be a minimal ideal over (a). We localize at P. RP is a local ring with unique maximalideal PP. Write S = R \ P, and therefore S−1P is minimal over S−1(a). So we may assume that R is anoetherian local ring.

Suppose ht(P) > 1, and there is a chain Q∗ ( Q ( P. Consider R/(a). Since R is local, P/(a) is theunique maximal ideal, which is moreover also a minimal prime, i.e. it is the only prime of R/(a). ThusP/(a), being the nilradical of a noetherian ring, is nilpotent. Write Pn ⊂ (a).

We now show R/(a) satisfies descending chain condition on ideals. Consider the descending chain:

R ⊃ P ⊃ P2 ⊂ · · ·

Each factor Pi/Pi+1 is a finite dimensional R/P-vector space, and therefore satisfy the descending chaincondition on subspaces. This implies, by an easy induction, that R/(a) satisfies DCC on ideals1.

Now let’s consider Im = {r ∈ R : r/1 ∈ S−1Qm}, here S = R \Q. Note S−1Qm = (S−1Q)m. Then:

Q = I1 ⊃ I2 ⊃ · · · (*)

From this we get a chain I1 + (a)/(a) ⊃ I2 + (a)/(a) ⊃ · · · of ideals in R/(a). Thus Im + (a) =Im+1 + (a) = · · · . We now show (*) terminates.

Let r ∈ Im, then we can write r = t+ xa for t ∈ Im+1, x ∈ R. So xa = r− t ∈ Im. But a 6∈ Q (since Pis the minimal prime containing a). This implies x ∈ Im (sinceQ is prime), and hence Im = Im+1 +aIm.But a ∈ P, so quotient out by Im+1 we have Im/Im+1 = PIm+1. Now nakayama gives Im = Im+1.

From here, (S−1Q)m = S−1Qm = S−1Im = S−1Im+1 = S−1Qm+1 = (S−1Q)m+1, where thesecond equality comes from the correspondence of ideals in localizations. This is an equality of ideals inS−1R. Since S−1Q is the unique maximal ideal of this local ring S−1R, nakayama gives (S−1Q)m = 0.

Correspondence of prime ideals says S−1Q ′ ( S−1Q; however, 0 = (S−1Q)m ⊂ S−1Q ′. Bydefinition of prime ideals, S−1Q ⊂ S−1Q ′. This is a contradiction, and thus concludes the proof.

Proof. (4.2) By induction. n = 1 is clear.As in the proof above, we assume R is local, with unique maximal ideal P. Pick any primeQmaximal

among thoseQ ( P. Thus P is the only prime in R strictly containingQ. Claim: ht(Q) ≤ n− 1. Note it isenough to do this for all such Q to deduce that ht(P) ≤ n.

By assumption, I = (a1, . . . ,an. Assume an 6∈ Q. Since P is the only prime strictly containing Q(and an), P/(Q+ (an)) is the only prime ideal of R/(Q+ (an)), and is therefore nilpotent. So there is anm such that ami ⊂ Q+ (an) for all 1 ≤ m < n. Write ami = xi + rian. Any prime containing all the xi’sand an will also contain a1, . . . ,an. Note (x1, . . . , xn−1) ⊂ Q.

Claim: Q is a minimal prime over (x1, . . . , xn−1). Indeed, write R = R/(x1, . . . , xn−1), and bars forimages in R; the unique maximal ideal P is a minimal prime over (an). By 4.1, P would have height atmost 1. Hence height of Q is at most 0 as Q ( P. So Q is minimal over (x1, . . . , xn−1).

5 Artin rings

In this chapter, R is not assumed to be commutative.

Definition 5.1. R is right Artinian if it satisfies descending chain condition on its right ideals. Similarlyfor modules.

Example.

1. Any k-algebra which is a finite dimensional k-vector space, e.g. Mn(k) of all n× nmatrices.

2. The group algebra kG := {∑λggwith g ∈ G} for a finite group G. It is both left and right Artin.

1This argument can be generalized to the following: if in a ring R (we’re considering R/(a)) here) there exist maximal idealswhose product is 0 (here Pn ⊂ (a) thus is 0), then R noetherian iff R artin. Or even more trivially, R/(a) is a noetherian ring withonly one prime (maximal) ideal, thus it is artin.

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3. R = {

(q r0 s

): q ∈ Q, r, s ∈ R}. It is only right Artin.

Warning: for a right R-module M and m ∈ M, the subset AnnR(m) = {r ∈ R : mr = 0} is onlya right ideal, not necessarily a two-sided one. However, AnnR(M) = {r ∈ R : mr = 0 ∀m ∈ M} =⋂m∈MAnnR(m) is two-sided.

Definition 5.2. An R-moduleM is simple (or irreducible) if its only submodules are 0 andM.

Remark. It’s clear that a submodule N ⊂M is maximal iffM/N is simple.

Definition 5.3. The Jacobson radical Jac(R) is the intersection of all maximal right ideals.

Remark. We could define using all left ideals, and we would get the same thing.

Definition 5.4. R is semi-simple if Jac(R) = 0. (Note that R/Jac(R) is always semi-simple.

Lemma 5.5. The following are equivalent for a right ideal I:

1. I ⊂ Jac(R).

2. IfM is finitely generated R-module, N ≤M with N+MI =M, thenM = N.

3. {1+ x : x ∈ I} = G is a subgroup of the unit group of R.

Proof. (1) =⇒ (2): same as the proof in Nakayama’s lemma.(2) =⇒ (3): Let x ∈ I and y = 1+ x. Then 1 = y− x ∈ yR+ I and therefore yR+ I = R. Set M = R

and N = yR, apply (2), we get R = yR. So we could write 1 = yz for some z ∈ R. This shows y is a unit.It remains to show its inverse z is also in G, i.e. we could write z = 1+ x ′ for some x ′ ∈ I. For this,

note that yz = z+ xz and thus z = yz− xz = 1− xz. I is a right ideal and therefore xz ∈ I.(3) =⇒ (1): Suppose I1 is a right maximal ideal. If x ∈ I and x 6∈ I1, then by maximality I1 + xR = R.

So we could write 1 = y+ xz for some y ∈ I1 and z ∈ R. But then y = 1− xz, by (3), is a unit and I1 = R.And we have a contradiction.

Remark. From this, we see Jac(R) is the largest ideal such that 1+ I is a subgroup of the units. This isindependent of left/right ideals.

Definition 5.6. An R-module is completely reducible if it is a sum (or equivalently direct1) of simplemodules. If so, by Zorn’s lemma, any submodule has a complement. Furthermore, if it is also Artin, thenit’s a direct sum of finitely many simples.

Example. R =Mn(D) for a division ring D, e.g. D = H quaternions. Then a right ideal generated by amatrix A is

AR = {B : columns of B lie is the (right) span of columns of A}.

A maximal right ideal could be like 0 · · · 0∗ · · · ∗

. . .∗ · · · ∗

i.e. A1j = 0. A minimal right ideal could be like

∗ · · · ∗0 · · · 0

. . .0 · · · 0

i.e. Aij = 0 for all i ≥ 2. ThenMn(D) is semisimple (obvious), and is completely reducible (direct sumof some minimals). The only 2-sided ideals of R are 0 and R.

1Consider the maximal sub-sum which is direct by Zorn’s lemma and obtain a contradiction. Converse is a little bit more work.First show there exists a simple submodule, and then consider the sum of all of those, and show it is our original module.

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Definition 5.7. A simple ring R is one whose only (2-sided) ideals are 0 and R.

The above example showsMn(D) is a simple ring.

Lemma 5.8. If R is semisimple, right Artin, then every right R-moduleM is completely reducible. Furthermore,ifM is an Artin module, then it is a direct sum of finitely many simple modules.

Proof. Since R is semisimple and Artin, the chain R ⊃ m1 ⊃ m1 ∩m2 ∩ · · · terminates at 0 after a finitenumber of steps. Assume no redundancies, writem1 ∩ · · ·mn = 0 and Ui =

⋂j 6=imj 6= 0. Then Ui are

submodules ofM, and we have a composition of maps

Ui ↪→M→M/mi.

The kernel of this composition is 0, so Ui injects into M/mi. But M/mi is simple, and thereforeUi =M/mi is simple. We can write R = ⊕ni=1Ui, and the R-module R is completely reducible.

Now M is a quotient of a free R-module. Since sum and quotient of completely reducibles arecompletely reducible, this showsM is completely reducible.

Examples.

1. kG is semisimple and completely reducible if chark = 0 and G finite. This is known as Maschke’stheorem.

2. Let G be the cyclic group of order p, and consider FpG = Fp[x]/(xp − 1). Consider the ideal(x− 1) mod (xp − 1). This is the kernel of the map

FpG→ Fp∑λgg 7→∑ λg

and is the Jacobson radical of FpG. This is because FpG is a local ring with a unique maximal idealwhose p-th power is 0: (x− 1)p = xp − 1. Thus FpG is not semisimple.

Definition 5.9. The socle, soc(M), of a non-zero Artin R-moduleM is the sum of all minimal submodulesofM.

Remark. Minimal submodules exist because of Artin property, and are necessarily simple.

Lemma 5.10. soc(M) = {m ∈M : mJac(R) = 0}.

Proof. Each minimal submoduleM ′ ofM is simple, and thereforeM ′ ∼= R/AnnR(m ′) for any 0 6= m ′ ∈M ′, because the map R→M ′ sending r 7→ m ′r is necessarily surjective. Therefore Ann(m ′) is a maximalright ideal, and J ⊂ Ann(m ′) for anym ′. Hence J annihilates any minimal submodule, and therefore thesum soc(M).

Conversely, ifmJ = 0, thenmRmaybe viewed as an R/J-module. But R/J is (always) semisimple andArtin, and therefore mR is completely reducible by Lemma 5.8. Thus mR is a sum of simple modules,and by definitionmR ⊂ soc(M).

Lemma 5.11. Let R be right Artin. Then

1. Jac(R) is nilpotent.

2. R is right Noetherian.

To prove this lemma, we need to use the socle series.

Definition 5.12. The socle series is defind inductively. soc1(M) = soc(M), and

soci(M)/soci−1(M) = soc(M/soci−1(M)).

Remark. 0 ( soc1(M) ( soc2(M) ( · · · is a chain of strict inclusion unless soci(M) = M. Also,soci(M) = {m : mJi = 0}.

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Proof. Consider R ⊃ J ⊃ J2 ⊃ · · · . So Jn = Jn+1. By remark above, socn(R) = socn+1(R), and we’veobserved that this can happen iff socn(R) = R. In particular, 1, in socn(R), is annihilated by Jn, and thusJn must be 0. We also observe that each factor in the socle series (as R/Jmodule) is a finite direct sumof simple R-modules, and is therefore Noetherian. Since R = socn(R) for large enough n, R must beNoetherian.

Artin-Wedderburn

Next we want to prove the Artin-Wedderburn theorem, and for that, we want to learn about endomor-phisms.

Lemma 5.13. (Schur’s lemma) If S is a simple right R-module, then EndR(S) is a division ring. Furthermore, ifS1, S2 are simple R-modules, either they are isomorphic, or HomR(S1,S2) = 0.

Proof. Letφ : S1 → S2. Supposeφ 6= 0. Then 0 6= Im(φ) ⊂ S2 andφmust be onto. Also ker(φ) = 0, sinceif ker = S1 then φ = 0. So φ is an isomorphism. If S1 = S2, we see that any non-zero endomorphism isactually an automorphism. So it has inverses. Thus EndR(S) is a division ring.

Lemma 5.14. If we regard R as a right module (usually denoted by RR), then EndR(RR) ∼= R.

Proof. The map is given by

EndR(RR) ∼= R

φ 7→ φ(1).

φ(1) determines the endomorphism φ, and given r ∈ R, multiply on the left by r gives an endomorphismof RR.

Theorem 5.15. (Artin-Wedderburn) Let R be a semisimple, right Artin ring. Then R = ⊕ri=1Ri, whereRi =Mni(Di) for some division ringDi. Ri are uniquely determined, and R has exactly r isomorphism classes ofsimple modules Si with EndR(Si) = Di and dimDi(Si) = ni.

Furthermore, if R is a k-algebra which is a finite dimensional vector space, then Di is a finite dimensionalk-vector space. If k is algebraically closed, then Di = k. In particular, CG for a finite group G can be written asCG ∼= ⊕ri=1Mni(C) where CG has r isomorphism classes of simple modules.

Remark. If k is a finite field, there’s another Wedderburn theorem says finite division rings are commuta-tive, i.e. fields (1905).

Corollary 5.16. G finite group, then Z(CG) is an r-dimensional C-vector space, where r is the number ofisomorphism classes of simple modules, which equals the number of conjugacy classes.

Proof. Conjugacy classes sum∑g ′∈conj. class ∈ Z(CG) and observe that they are linearly independent

and span Z(CG). So dimZ(CG) is the number of conjugacy classes. Now apply Artin-Wedderburn. CGis a sum of matrix algebrasMni(C), and Z(Mni(C)) consists of scalar matrices, and is one dimensional.So dimZ(CG) is also the number of matrix algebras appearing, which is the number of isomorphismclasses of simple modules.

Proof. (to 5.15) R is semisimple, right Artin and therefore completely reducible as a right module. WriteRR as a finite direct sum of simple modules, and take isomorphism class representatives S1, . . . ,Sr, i.e.

RR = (S1,1 ⊕ S1,2 ⊕ · · · ⊕ S1,n1)⊕ (S2,1 ⊕ S2,2 ⊕ · · · ⊕ S2,n2)⊕ · · · ⊕ (Sr,1 ⊕ Sr,2 ⊕ · · · ⊕ Sr,nr)

where each Si,j ∼= Si. Consider EndR(RR) = R by Lemma 5.14, and EndR(Si,1 ⊕ Si,2 ⊕ · · · ⊕ Si,ni) =Mni(Di) where Di = EndR(Si) is a division ring by Schur, Lemma 5.13, and φ ∈ EndR(Si,1 ⊕ Si,2 ⊕· · · ⊕ Si,ni) =Mni(Di) is represented by a matrix (φml) where φml ∈ Hom(Si,l,Si,m). So we get

R =

Mn1(D1) 0

Mn2(D2). . .

0 Mnr(Dr)

21

where the 0 block is because Hom(Si,l,Sj,m) = 0 as Si 6∼= Sj.What about dimSi? We saw the minimal right ideals ofMn(D) consist of matrices whose columns

are scalar multiples (on the right) of a fixed column d, and so are n-dimensional over D (because ofnumber of basis element)1.

Lastly, to show Ri are uniquely determined, again write R as the direct sum of simples

R = RR = (S1,1 ⊕ S1,2 ⊕ · · · ⊕ S1,n1)⊕ (S2,1 ⊕ S2,2 ⊕ · · · ⊕ S2,n2)⊕ · · · ⊕ (Sr,1 ⊕ Sr,2 ⊕ · · · ⊕ Sr,nr).

Take a simple submodule S, and consider the projections

πil : R→ Sil, π|S : S→ Sil.

Schur’s lemma implies that this restriction is either 0 or an isomorphism. If S is isomorphic to an Sil, itcannot be isomorphic to any other Sjl. However, one of the projections must be non-zero, and S ∼= Sifor exactly one Si, and Ri is the sum of all simple submodules isomorphic to Si. So Ri are uniquelydetermined.

Example 5.17. Consider kS3 where chark = 3. We know in char 0 case, there are 3 isomorphism classesof simple modules:

1. U1 trivial, one dimensional, and group acts trivially.

2. U2 also one dimensional, group elements acts by multiplication by sgn of the permutation.

3. U3 two dimensional. Write g = (12) and h = (123), Then g acts by(−1 0−1 1

)and h acts by(

−1 1−1 0

).

Now we work in characteristic 3, and denote by ∗. U1 and U2 are one dimensional simple modules.

U3 is not simple.(21

)is a common eigenvalue of g and h, and therefore there is a one dimensional

submodule.Consider

α : kS3 → kC2

g 7→ generator of C2h 7→ 1,

where kC2 is the direct sum of U1 and U2. Then kerα = (h− 1), and this kernel is nilpotent: (h− 1)3 =h3 − 1 = 0. Furthermore, Jac(kS3) = kerα, and (h− 1)2 = (h2 + h+ 1) is a direct sum of simples(∼= kC2). The socle series will be

0 ⊂ (h2 + h+ 1) ⊂ (h− 1) ⊂ kS3.

Only two one-dimensional simples appear in the factors.

6 Homological algebra

In this chapter, R need not be commutative.We’ve already seen that N⊗− need not preserve short exact sequences. If it does, then N is flat.

Now consider HomR(N,−) and HomR(−,N). These, likewise, do not necessarily preserve short exactsequences; e.g. abelian groups 0 → 2Z/4 → Z/4 → Z/2 → 0, and apply Hom(Z/2,−): the mapHom(N, Z/4) → Hom(N, Z/2) is the zero map, but hom(N, Z/2) 6= 0, and the map is therefore notsurjective.

1Alternatively, note thatMni(Di) ∼= End(Sni

1 ) ∼= Snii , and thus n2i = ni dimSi.

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Definition 6.1. A module P is projective if Hom(P,−) preserves all short exact sequences. In otherwords, given ses 0→M1 →M→M2 → 0, every map P →M2 lifts to a map P →M.

P

M M2 0

Definition 6.2. A module E is injective if, given ses 0 → M1 → M → M2 → 0, every map M1 → Eextends to a mapM→ E

0 M1 M

E

In other words, Hom(−,E) preserves exactness.

Example.

1. Free modules are projective.

2. Fraction fields of integral domains R are injective R-modules.

Lemma 6.3. For an R-module P, the following are equivalent:

1. P is projective.

2. Hom(P,−) is an exact functor.

3. If α :M→ P is surjective, then there exists β : P →M such that αβ = id.

4. P is a direct summand in every module of which it is a quotient.

5. P is a direct summand of a free module.

Proof. (1) =⇒ (2) obvious. (2) =⇒ (3): take 0 → kerα → Mα−→ P → 0, and apply Hom(P,−): then

Hom(P,M)→ Hom(P,P) is onto, and thus some map β : P →M composes with α to get identity.(3) =⇒ (4): Every 0→M1 →M→ P → 0 splits. SoM =M1 ⊕ P.(4) =⇒ (5): Given a generating set {xλ} for P, there exists a surjective map F→ P, where F is the free

module on {eλ}, and eλ 7→ xλ. Apply (4) to get the desired result.(5) =⇒ (1): Write F = P ⊕Q. Since F is projective, and Hom behaves well under direct sums, we

deduce that the direct summand P is projective.

Remark.

1. ⊗ behaves well under direct sums and so direct summands of free modules are flat, and soprojective modules are flat.

2. If R is a PID, every finitely generated projective module is free (using the structure theorem forfinitely generated modules over PID). 1

3. There is a similar lemma characterizing injective modules. See example sheet (or appendix).

Definition 6.4. A projective presentation ofM is a ses of the form

0→ K→ P →M→ 0

with P projective. If P is actually free, then we have a free representation.1In fact, projective is the same as free over PID. See appendix.

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Definition 6.5. Given a projective presentation ofM, apply N⊗− to get an exact sequence:

N⊗ K→ N⊗ P → N⊗M→ 0.

Define TorR(N,M) = ker(N⊗ K→ N⊗ P). In particular, if N is flat, then Tor(N,M) = 0 for allM. If weapply Hom(−,N), we get an exact sequence

0→ Hom(M,N)→ Hom(P,N)→ Hom(K,N)

and define ExtR(M,N) = coker(Hom(P,N) → Hom(K,N)). In particular, if N is injective, thenExt(M,N) = 0 for allM.

Remark.

1. TorR(M,N) = TorR(N,M), ExtR(M,N) = ExtR(N,M). Naturally suppress R if we can.

2. If P is projective, and 0→ K→ P1 → P → 0 is a projective presentation, then P is a direct summandof P1. So applying Hom(−,N) for any N is going to give an exact sequence. So Ext(P,N) = 0 forall P projective modules.

3. These Ext and Tor are independent of the choice of presentation.

4. Similarly we could take a projective presentation for N and apply −⊗M. This gives the sameresult.

5. We can take a ses 0 → N → E → L → 0 with E injective. Apply Hom(M,−) to get 0 →Hom(M,N)→ Hom(M,E)→ Hom(M,L). Then coker is Ext(M,N).

6. To give a ses in 5, we need to show there are enough injectives (see example sheet).

7. Ext comes from extension. We can define Ext by defining an equivalence relation on all extensions ofM by N, 0→ N→ X→M→ 0. The zero of Ext(M,N) represents the split 0→ N⊕M→M→ 0.So if Ext=0, then the only extensions are split ones, i.e. must be a direct sum.

8. Tor comes from torsion. If R = Z, then Tor(Q/Z,A) is the torsion subgroup of the abelian groupA.

9. Tor and Ext are abelian groups. If R commutative, then they are R-modules. The point isM⊗RNis R-module if R commutative, but not necessarily otherwise.

Examples.

1. If R = Z, and a free presentation for the R-module Z/2:

0→ Z×2−−→ Z

mod 2−−−−→ Z/2→ 0.

Apply Z/2⊗− (use the fact that Z/n⊗Z A = A/n):

Z/2 ×2−−→ Z/2→ Z/2→ 0.

So Tor(Z/2, Z/2) = Z/2. Similarly apply HomZ(−,N), then

Ext(Z/2,N) = coker(HomZ(Z,N)×2−−→ HomZ(Z,N))

= coker(N ×2−−→ N)

= N/2.

If we apply with N = Z/2, then Ext(Z/2, Z/2) = Z/2. So two classes of extensions:

split: 0→ Z/2→ Z/2⊗Z/2→ Z/2→ 0

non-split: 0→ Z/2→ Z/4→ Z/2→ 0.

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2. R = k[X], still PID. We have a free presentation of the trivial module k (X acts like 0, i.e. f(x) ∗ a =f(0) · k):

0→ (X)→ k[X]→ k→ 0,

where (X), as a k[X]-module, is isomorphic to k[X] (check the k[X]-module map k[X]→ (X) wheref 7→ X · f is an isomorphism). So the presentation is:

0→ k[X]×X−−→ k[X]→ k→ 0.

3. R = k[X, Y], and consider the trivial module k.

0→ (X, Y)→ k[X, Y]→ k→ 0.

By same reasoning, (X, Y) is isomorphic to k. So we take a free presentation for k:

0 k[X, Y] k[X, Y]⊕ k[X, Y] (X, Y) 0

f (Yf,−Xf)

(g,h) Xg+ Yh

Combine these two together, we have:

0→ k[X, Y]→ k[X, Y]⊕ k[X, Y]→ k[X, Y]→ k→ 0.

Here k = (X, Y) is the first syzygy module.

Definition 6.6. A projective resolution ofM is an exact sequence of the form

· · ·→ Pn → Pn−1 → · · ·→ P1 → P0 →M→ 0

where each Pi is projective. If they are actually free, this is a free resolution.

Remark.

1. The one in above example is a free resolution for kwhere R = k[X, Y].

2. For R noetherian, and a finitely generated R-module M, we can take a finite generating set forM, and produce a free presentation 0 → K1 → F → M → 0, where F is the free module on thegenerating set, thus of finite rank. K1 is isomorphic to a submodule of F, which is noetherian sinceit is of finite rank. So K1 is again finitely generated. Repeat this process to give free presentation ofK1: 0→ K2 → F1 → K1 → 0. So we have a free resolution forM:

· · ·→ F3 → F2 → F1 → F0 →M→ 0.

3. Many authors prefer to write P2 → P1 → P0 → 0 and the last map is not exact. Then ker(P0 →0)/Im(P1 → P0) =M.

Definition 6.7. The Koszul complex gives a free resolution of the trivial module k for the polynomialalgebra in n variables X1, . . . ,Xn. (We’ve already seen when n = 1, 2)

Define Fi to be the firee module on basis {ej1,...,ji }, where j1, . . . , ji is a subset of size i in {1, 2, . . . ,n}.Define the map

Fi → Fi−1

ej1,...,ji 7→ i∑l=1

(−1)l−1Xlej1,...,jl−1,jl+1,...,ji

This gives a free resolution for k.

Now we want to define higher Tor and Ext groups. Firstly make iterative definitions.

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Definition 6.8. For an R-module M, take a projective presentation 0 → K1 → P0 → M → 0. DefineTor0(N,M) = N⊗M, Tor1(N,M) = Tor(N,M). For i ≥ 2, define Tori(N,M) = Tori−1(N,K1).

Definition 6.9. Similarly, define Ext0(M,N) = Hom(M,N), Ext1(M,N) = Ext(M,N), and Exti(M,N) =

Exti−1(K,N).

Remark. These are independent of the choice of projective presentations. The alternate approach is byconsidering a projective resolution forM:

· · ·→ P2 → P1 → P0 →M→ 0

and then applying N⊗− to get

· · ·N⊗ P2 → N⊗ P1 → N⊗ P0 → 0.

This is a chain complex. The homology groups of the chain complex at N⊗ Pi is ker(N⊗ Pi → N⊗Pi−1)/Im(N⊗ Pi+1 → N⊗ Pi), which is defined to be Tori(N,M). Similarly if we apply Hom(−,N),

0→ Hom(P0,N)→ Hom(P1,N)→ . . .

Then Exti(M,N) =cohomology group at Hom(Pi,N).

Lemma 6.10. The following are equivalent:

1. Extn+1(M,N) = 0 for all N.

2. M has a projective resolution of length n (i.e. 0→ Pn → Pn−1 → · · ·→ P0 →M→ 0).

Proof. We’ve observed that Ext1(M,N) = 0 for all N iffM is projective.

Consider Ext2(M,N) = 0 for all N. This happens iff Ext1(K1,N) = 0 for all N. So K1 is projective,and 0→ K1 → F0 →M→ 0 is a resolution. Apply dimension shifting to get the result inductively.

Definition 6.11. The projective dimension of M is n if Extn+1(M,N) = 0 for all N, and there exists Nwith Extn(M,N) 6= 0. The global dimension of R, denoted by gl dim(R) is the supremum of all projectivedimensions over allM. (We can talk about left and right if R is non-commutative)

Remark. There is an analogous concept for injective dimension.

Examples.

1. For a field k, the global dimension is 0. All k-modules are free.

2. (Assume R is artin) The condition that global dimension being 0 is equivalent to any submodule ofR is a direct summand. Thus it happens iff R is semisimple, e.g. CG for a finite group G.

3. gl dim of any PID which isn’t a field is 1 (syzygy projectives).

4. We’ve going to prove Hilbert’s syzygy theorem which gives projective resolutions of length ≤ nfor any finitely generated module over k[x1, . . . , xn] (and so, gl dim(R) = n).

Hochschild (co)homology

This is the homological theory for bimodules. A k-algebra R here are not assumed to be commutative.An R− R bimodule is one where R acts both on the left and on the right, and the two actions commute.One can reformulate in terms of a right module for R⊗k Rop, where Rop is defined on the same set as R,but we define x · y in Rop to be yx ∈ R. A left R-module can be made into a right R-module.

So an R− R bimodule can be turned into a right R⊗ Rop module1 by:

r ∗m ∗ s = m ∗ (s⊗ r).1Since module is a linear structure, the most natural way to put two rings together will be their tensor instead of the usual

direct sum.

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And similarly, it is a left R⊗ Rop module by:

r ∗m ∗ s = (r⊗ s) ∗m.

If R is commutative, then Rop = R. If R = kG, then Rop ∼= R by∑λgg 7→∑ λgg

−1.Examples.

1. R itself is an R− R bimodule.

2. R⊗k R is a bimodule. Similarly, R⊗k · · · ⊗k R is a bimodule.

3. There is a bimodule presentation of R:

0→ kerµ→ R⊗ R µ, mult−−−−−→ R→ 0,

where µ(r⊗ s) = rs. R⊗k R is a free bimodule (as an R⊗Rop module). It is free of rank 1, generatedby 1⊗ 1.

Definition 6.12. Given an RR-bimodule M, the Hochschild homology is defined to be HHn(R,M) =

TorR−Rn (R,M), and the Hochschild cohomology is HHn(R,M) = ExtnR−R(R,M).

In particular, HH0(R,M) = HomR−R(R,M) = {m ∈ M : rm = mr ∀r ∈ R} (the map is determinedby where 1 is sent to). And HH0(R,R) = {s ∈ R : sr = rs ∀r ∈ R} = Z(R) the center.

Also HH0(R,M) = R⊗R−RM = R⊗R⊗Rop M (here R is a right module, and M is a left module)= M/ < rm−mr : m ∈ M, r ∈ R >. And in particular, HH0(R,R) = R/[R,R] where [r, s] = rs− srthe Lie bracket on R, cf. question 4 on example sheet (where we need the extra condition that R issemisimple).

Definition 6.13. The Hochschild chain complex gives a free resolution for the bimodule R:

· · ·→ R⊗ R⊗ R d0−−→ R⊗ R µ−→ R→ 0,

where

dn−1(r0 ⊗ · · · ⊗ rn+1) =n∑i=0

(−1)ir0 ⊗ · · · ⊗ riri+1 ⊗ · · · ⊗ rn+1.

Definition 6.14. The (Hochschild cohomology) dimension DimR of R is

sup {n : HHn(R,M) 6= 0 for some bimoduleM}.

Definition 6.15. The k-algebras R of DimR = 0 are precisely those where the bimodule R is projective.This is when R is a direct summand of R⊗ R, i.e. there exists β : R→ R⊗ R such that µ ◦ β = idR (µ isdefined above to be the multiplication map). These algebras are called k-separable. This generalizes thenotion of a separable field extension, but note that k-separable algebras must be finite dimensional ask-vector spaces. (See example sheet)

Examples.

1. Mn(k) is k-separable. Given the map β : R → R ⊗ R, the image of 1 is called a separatingidempotent. ForMn(k), a separating idempotent is obtained as follows:

Take Eij to be the elementary matrix with 1 in the ij-entry, and 0 otherwise. Fix j, and consider∑i Eij ⊗ Eji. This is a separating idempotent. Note that the image of this under µ is

∑Eii = 1 the

identity matrix.

2. CG for a finite group G. CG⊗CGop ∼= CG⊗CG ∼= C(G×G), which is semisimple, completelyirreducible. So all submodules are direct summands. In particular, CG is a direct summand ofC(G×G). Thus Dim(CG) = 0.

Now consider higher degrees. Note that HomR−R(R⊗ R,M) ∼= Homk(k,M). A map is determinedby the image of 1⊗ 1 (since it is free of rank 1 on this generator). HomR−R(R⊗ R⊗ · · · ⊗ R,M) ∼=Homk(R⊗ · · · ⊗ R,M), where on the left hand side there are n+ 2 copies of R, and on the right handside there are n copies.

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Definition 6.16. The Hochschild cochain complex is:

M ∼= Homk(k,M)δ0−→ Homk(R,M)

δ1−→ Homk(R⊗ R,M)→ · · ·where

(δ0f)(r) = rf(1) − f(1)r,(δ1f)(r1 ⊗ r2) = r1f(r2) − f(r1r2) + f(r1)r2,

(δ2f)(r1 ⊗ r2 ⊗ r3) = r1f(r2 ⊗ r3) − f(r1r2 ⊗ r3) + f(r1 ⊗ r2r3) − f(r1 ⊗ r2)r3

and etc. For HH1(R,M) we want ker δ1/Im δ0.

Definition 6.17. ker δ1 = {f ∈ Homk(R,M) : f(r1r2) = r1f(r2) + f(r1)r)2} is the derivations from R toM, written as Der(R,M). Im δ0 = {f ∈ Homk(R,M) of the form r 7→ rm−mr for somem}, called theinner derivations, written as Innder(R,M). So HH1(R,M) = Der(R,M)/Innder(R,M).

If M = R, we get HH1(R,R) = Der(R)/Innder(R). If R is commutative, then Innder(R) = 0, andtherefore HH1 = Der(R). In general, Der(R) form a Lie algebra. If D1,D2 are derivations R→ R, thenD1D2 −D2D1 ∈ Endk(R) is also a derivation (easy check).

Example. If R = k[x], then Der(R) = {p(x)d

dx: p(x) ∈ k[x]} (here we assume char 0, otherwise

differentiation doesn’t behave very well).If we think about the case where M = R, a R− R bimodule, if R is commutative, we have that the

derivations are examples of differential operators:For commutative k-algebra R, define differential operators iteratively D(R) =

⋃Di(R).D0(R) = {D ∈ Endk(R) : [r,D] = 0 ∀r ∈ R}D1(R) = {D ∈ Endk(R) : [r,D]]inD′(R) ∀r ∈ R}

and so on. Side: D-modules are modules over differential operators.Example continued. D(k[x]) = k[x,d/dx] ⊂ Endk(k[x]). If R = k[x1, . . . , xn] and char 0, thenD(k[x1, . . . , xn]) = k[x1,∂/∂x1, . . . , xn,∂/∂xn] (induction).

Definition 6.18. Given an R−R bimoduleM, we can form a semidirect product RnM, addition on pairsas usual, and multiplication is (r1,m1) · (r2,m2) = (r1r2, r1m2 +m1r2). Or think of this as R+Mε,where ε2 = 0, and ε commutes with everything. Mε ideal with (Mε)2 = 0.

Lemma 6.19. Der(R,M) ∼= {algebra complement toM in RnM}.

Proof. A complement toM is an embedded copy of R in RnM.

R ↪→ RnMr 7→ (r,D(r))

This function D : R→M is a derivation.Conversely, D : R→M gives such an embedding.

Inner derivations in this context?

Corollary 6.20. Der(R,M) = {automorphisms of RnM = R +Mε}, where on RnM, r 7→ (r,D(r) andm 7→ m correspond to on R +Mε, r 7→ r +D(r)ε,mε 7→ mε. Innder(R,M) = {automorphisms of RnM of the form obtained by conjugation by 1+mε}.

7 Graded rings and filtrations

Definition 7.1. A (Z-)graded ring R is of the form R = ⊕i∈ZRi, where Ri is called the i-th homogeneouscomponent, such that RiRj ⊂ Ri+j. Then R0 is a subring, and each Ri is a R0 submodule. A graded idealI ⊂ R is of the form ⊕iIi for each Ii ⊂ Ri.

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Note: if such a graded ideal has a finite generating set, then it must have a finite generating set ofhomogeneous elements.

A graded R-module V is of the form ⊕Vj such that RiVj ⊂ Vi+j. Note each Vi is an abelian group; atmost an R0-module, as R0Vi ⊂ Vi.

Similarly, we can have gradings with respect to different indexed sets, e.g. positively graded rings⊕i≥0Ri.

Now suppose1 R is graded, with R = ⊕i≥0Ri, R commutative and noetherian, and is generated (asR0-algebra) by R0 and x1, . . . , xm of degree k1, . . . ,km ≥ 1. Thus R0 as a quotient is noetherian. Let λ bean additive integer-valued function on finitely generated R0-modules, i.e. if we have a ses of finitelygenerated R0-modules 0→ U1 → U→ U2 → 0, then λ(U) = λ(U1) + λ(U2).

Example 7.2.

1. If R0 = k is a field, then R0 modules are k-vector spaces. We could take λ to be the k-vector spacedimensions.

2. R0 is local, Artin, with unique maximal ideal P. Then with any fg R0 module U, we could builda chain U ⊃ U1 ⊃ U2 · · · ⊃ Um = 0 with each Ui/Ui+1 = R/P. Set λ(U) to be the number offactors in this chain, called the composition (series) length of U.

Given an additive function λ, the Poincare series of a fg. graded module V = ⊕i≥0Vi for R = ⊕i≥0Riis the power series (in variable t):

P(V , t) =∑i≥0

λ(Vi)ti ∈ Z[[t]].

Theorem 7.3. (Hilbert-Serre) In this situation, P(V , t) is a rational function f(t)/(∏mj=1(1 − t

kj)), wheref(t) ∈ Z[t] is a polynomial, and kj are the degrees of the generators xj.

Corollary 7.4. If each kj = 1, then for large enough i, λ(Vi) = φ(i) for some rational polynomial φ(t) ∈ Q[t] ofdegree d− 1, where d is the order of pole at t = 1. Moreover,

∑ij=0 λ(Vj) = χ(i) for some χ(t) ∈ Q[t] of degree

d.

Definition 7.5. φ(t) is the Hilbert polynomial, and χ(t) is the Samuel polynomial. The degree of Samuelpolynomial gives us yet another dimension d(V).

Example. Take R to be the polynomial ring in m variables, R0 = k is a field, V = R, and λ be the k-vsdimension. Then λ(Vi) is the number of monomials of degree i, which is

φ(i) =

(i+m− 1

m− 1

)=

1

(m− 1)!(i+m− 1) · · · (i+ 1).

The degree of Hilbert polynomial is thereforem− 1, and so d(V) = m.Proof. (Hilbert-Serre) By induction on the number of generators.

If m = 0, then R = R0, and V fg R0-module. So Vj = 0 for large enough j, and then P(V , t) isautomatically a polynomial.

If m > 0, multiplication by xm of degree km defines a map · xm : Vi → Vi+km . Let Ki = ker(· xm),Li+km = coker(· xm), and K = ⊕Ki, L = ⊕Li. Then K is a graded submodule of V = ⊕Vi, and so a fgR-module. L = V/xmV and is also a fg R-module. Observe that xm acts trivially on both. So K,L can beregarded as R0[x1, . . . , xm−1]-module. Apply λ to the following exact sequence and use induction:

0→ Ki → Vi → Vi+km → Li+km → 0

λ(Ki) − λ(Vi) + λ(Vi+km) − λ(Li+km) = 0. Multiply by ti+km , and sum over all i ≥ 0, we have:

tkmP(K, t) − tkmP(V , t) + P(V , t) − P(L, t) = g(t)

where g(t) ∈ Z[t] is a polynomial arising to compensate the first km terms in the Poincare polynomialsof Vi+km and Li+km . Induction on P(K, t) and P(L, t), and we get the result.

1See theorem 2 in appendix on Hilbert functions for more detail.

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Proof. (Corollary) P(V , t) = f(t)/(1− t)d for some d, and f(1) 6= 0.Since (1 − t)−d =

∑i≥0

(d+i−1d−1

)ti, if f(t) = a0 + a1t + · · · + asts, then λ(Vi) = a0

(d+i−1d−1

)+

a1(d+i−2d−1

)+ · · · + as

(d+i−s+1d−1

), where

(rd−1

)= 0 if r < d − 1. Now rearrange to give a rational

polynomial φ(i) valid for i large enough: d+ i− s+ 1 ≥ d− 1.

φ(i) =f(1)

(d− 1)!td−1 + . . .

So degree of φ(i) is d− 1. Now using the identity(mn

)=(m−1n

)+(m−1n−1

), we have

i∑j=0

(d+ j− 1

d− 1

)=

(d+ i

d

).

Soi∑j=0

λ(Vj) = a0

(d+ i

d

)+ a1

(d+ i− 1

d

)+ · · ·+ as

(d+ i− s

d

)for i ≥ s. This is a rational polynomial χ(i) ∈ Q[t] of degree d for i large.

Filtration

Definition 7.6. A (Z)-filtered ring is one containing a filtration by additive subgroups

· · · ⊂ R−1 ⊂ R0 ⊂ R1 ⊂ · · ·

such that RiRj ⊂ Ri+j, and 1 ∈ R0.

Usually we impose two more conditions:

1.⋃Ri is a subring, and we assume

⋃Ri = R (exhaustive).

2. R0 is a subring, and⋂Ri is an ideal of R0. We assume

⋂Ri = 0 (separated).

Example

1. The I-adic filtration. Ri = R for i ≥ 0. R−j = Ij for j > 0. If R is a local ring, we are interested in theP-adic filtration where P is the unique maximal ideal.

2. If R is a k-algebra generated by generators x1, . . . , xm, set R−j = 0 for j > 0, R0 = span {1},R1 = span {1, x1, . . . , xm}, R2 = span {polynomial expressions of degree ≤ 2}, and etc.

One can have good filtrations for non-commutative rings.

1. Differential operator ring, e.g. k[x1,∂/∂x1, . . . , xn,∂/∂xn].

2. Enveloping algebras of Lie algebras of finite dimension.

3. Iwasawa algebras. These have a maximal ideal P, and we can consider the P-adic filtration.

Definition 7.7. Given a filtered ring, one can form the associated graded ring

grR = ⊕i∈ZRi/Ri−1

as additive group, with (r+ Ri−1)(s+ Rj−1) = rs+ Ri+j−1.

E.g. for the P-adic filtration, grR = ⊕Pj/Pj+1 the (−j)-th component. This is negatively graded, butby renumbering, one can regard it as positively graded. If k = R/P where P is a maximal ideal, thenPi/Pi+1 are k-vector spaces, and we can use λ = k-vs dimension and apply Hilbert-Serre if the gradedring is commutative.

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Remark. For certain Iwasawa algebras R, one we filter, grR is commutative. Similarly, grR is commutativefor both differential operator rings and enveloping algebras.

Definition 7.8. Let R be a filtered ring with filtration Ri. M be an R-module. Then M is a filteredR-module if · · · ⊂ M−1 ⊂ M0 ⊂ M1 ⊂ · · · as additive groups with RjMi ⊂ Mi+j. We can form theassociated graded module grM = ⊕Mi/Mi−1. This is a graded grR module, with (r+ Ri−1)(m+Mj−1) = rm+Mi+j−1.

Lemma 7.9. If 0→ N→M→M/N→ 0 ses, andM is filtered, then we can filterN by usingN∩Mi,M/Nby using Mi +N/N. Form grN, grM, and grM/N. Then 0 → grN → grM → grM/N → 0 is exact.(Example sheet)

Definition 7.10. The Rees ring of the filtration {Ri} of R is a subring of the Laurent polynomials R[T , T−1]in variable T :

⊕j∈ZRjTj

Since 1 ∈ R0 ⊂ R1, T ∈ Rees(R). Then R = Rees(R)/(T − 1), and grR = Rees(R)/(T).Similarly the we have form a Rees(R)-module called the Rees module, Rees(M) = ⊕MjT j, and

Rees(M)/TRees(M) = grM.

Definition 7.11. For a ring (not necessarily commutative) Rwith maximal ideal P and the residue fieldk = R/P,form a graded ring grR associated to the P-adic filtration. Assume grR is commutative and ispositively graded (renumbered). Take λ to be the k-vector space dimension. Take any finitely generatedR-module and form grM (positively graded). Apply Hilbert-Serre1 to get dimkMPj/MPj+1 = φ(i) fori large enough, and some φ(t) ∈ Q[t]. We also know that

∑ij=0 dimkMPj/MPj+1 = χ(i) for i large

enough and some χ(t) ∈ Q[t].But∑ij=0 dimkMPj/MPj+1 is the composition length2 of M/MPi+1,

χ(t) is of degree d. Finally, set d(M) = d, another dimension. In particular, this gives d(R) (at this stage,d(M) is P-dependent).

Remark.

1. For a noetherian commutative local integral domain, d(R) = dimR.

2. For a finitely generated k-algebra which is an ID, we saw that all maximal ideals have the sameheight, namely the dimension. dimR = dimRP for any maximal ideal P (dimension of anylocalization is the height)= d(R) with respect to the P-adic filtration.

Theorem 7.12. (Hilbert’s Syzygy theorem) Let k be a field, R = k[x1, . . . , xn] considered as a graded ring wrt.total degree. Let M be a finitely generated R-module. Then there is a free resolution of M of length less or equal ton.

Proof. First we recall that the Koszul complex gives a free resolution of a trivial module k of length n.We are going to view Torn(k,M) in two different ways: either apply · ⊗M to the Koszul complex (freeresolution), or apply k⊗ · to a free resolution ofM. We know that the homology groups arising fromtwo approaches are the same. Now say a free resolution is given by

→ Fn → Fn−1 → · · ·→ F1 → F0 →M→ 0

with each F finite rank (noetherian). Assume it is also minimal, i.e. with lowest ranks, rk (F0) is theminimal number of generators (minimal free resolutions). Now the crucial point is that if we consider0 → K → F0 → M → 0, where K is the first syzygy module, then apply k⊗ ·, we get a (non-exact)sequence 0 → k⊗ K → k⊗ F0 → k⊗M → 0. Minimality would imply that any basis element mustbe mapped to an element with no constant term in any coefficient, i.e. if e ∈ Fk+1 s mapped to(p1, . . . ,pl) ∈ Fk, then none of pi is constant. This would further imply that the map k⊗ K→ k⊗ F0 isthe zero map. Hence the sequence

→ k⊗ Fn → k⊗ Fn−1 → · · ·→ k⊗ F0 → k⊗M

are all 0 except the last one, and therefore homology groups are k⊗ Fi, which is a k-vector space withdimension the rank of Fi. However, with · ⊗M to the Koszul complex, we learn that Tori(k,M) = 0∀i > n. So rk Fi = 0 =⇒ Fi = 0 ∀i > n.

1At least for finitely generated k-algebras2well they are vector spaces, so dim corresponds to the number of extra factors that can be put into the series

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–End of lecture notes–

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Appendices

A Cayley-Hamilton and the determinant trick

This section is largely copied from Reid’s Undergraduate Commutative Algebra, section 2.6 and 2.7.

Theorem A.1. Let R be a commutative ring, N = (aij) is an n× n matrix with entries in R. Write pN(x) =det(x · In −N). This is a polynomial of degree n in x with coefficient in R and leading coefficient 1. ThenpN(N) = 0 as N×N matrix.

This should remind you of elements being integral. Indeed, when a process involves integralextension, the following trick is quite often surprisingly useful and powerful.

Theorem A.2. Let M be a finitely generated R-module by n elements, and φ : M → M a homomorphism.Suppose that I is an ideal of R such that φ(M) ⊂ IM, then φ satisfies a relation of the form

φn + r1φn−1 + · · ·+ rn−1φ+ r0 = 0,

where ri ∈ Ii. Here we interpret both sides as mapsM→M.

Proof. SayM is generated bym1, . . . ,mn. Since φ(mi) ∈ IM, we can write

φ(mi) =∑

aijmj.

This can be rewritten as ∑i

(δijφ− aij)mj = 0.

Then consider the matrix ∆ij = δijφ− aij. Multiply by the adjoint matrix, we have (det∆)mk = 0 foreach k. Thus det∆ = 0. Expand out to get the required relation.

This has many applications, such as the standard proof in AM about Nakayama’s lemma, and the3 =⇒ 1 direction in checking element being integral.

B More on homological algebra

This section talks about more homological algebra results in a rather dense way without too much proof.See, e.g. Weibel, for more detail.

First is about projective/injective modules in general.

Lemma B.1. A direct sum of modules is projective iff each summand is projective. A direct product is injective iffeach factor is injective.

Theorem B.2. (Criterion of projectivity) A module is projective iff it is a direct summand of a free module.

Theorem B.3. (Baer’s criterion on injectivity) A moduleM is injective iff for any map of left R-modules f : I→M,where I is a left ideal of R, there is an elementm ∈M such that f has the form f(x) = xm. Equivalently, any mapof left R-modules I→M extends to a map of left R-modules R→M.

This result is on example sheet. In particular, consider the pairs (A, f ′) that extends f, and use Zorn’slemma to come up with a maximal one.

There are special properties if R is a principal ideal domain.

Theorem B.4. Any projective module is torsion-free, any injective module is divisible. Over PID, projective andfree are the same thing; injective and divisible are the same thing; flat and torsion-free are the same thing.

To do homological algebra (where we need to take an injective resolution), we need to show:

Theorem B.5. There are enough injectives in the category of R-modules. In other words, every left R-module isisomorphic to a submodule of an injective module.

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The name Tor and Ext have explicit meanings. The origin comes from dealing with torsion Z-modules.

Lemma B.6. For any abelian group B, we have:

1. TorZ0 (Z/n,B) = Z/n⊗ B = B/nB.

2. TorZ1 (Z/n,B) = B[n].

3. TorZ≥2(Z/n,B) = 0.

Thus the name Tor for torsion.

For example, if 0→ A→ B→ C→ 0 is a ses, then applying the derived functors of Z/n⊗Z − weget the exact sequence

0→ A[n]→ B[n]→ C[n]→ A/n→ B/n→ C/n→ 0

although this can be slickly proved by snake’s lemma...In fact:

Theorem B.7. For any abelian groups A,B, we have TorZ>1(A,B) = 0, and Ext>1Z (A,B) = 0.

Ext for extension is much less obvious.

Definition B.8. An extension of C by A is a ses 0→ A→ B→ C→ 0. Two extensions are equivalent ifthere is a commutative diagram

0 A B C 0

0 A B ′ C 0

It is clear that all maps in the above diagram are isomorphisms. It follows that this is indeed anequivalence relation.

Definition B.9. Given two extensions of R-modules 0→ Aα−→ B

β−→ C→ 0 and 0→ A

α ′−→ B ′

β ′−→ C→ 0,

the Bare sum is the extension 0→ A→ X→ C→ 0, where X is the homology group of the complex

A(α,−α ′)−−−−−−→ B⊕ B ′ β−β

′−−−−→ C.

The map A→ X is obtained from (α, 0), and the map X→ C is obtained from β.

Lemma B.10. The Bear sum turns the set of equivalence classes of extensions into an abelian group; the zeroelement is the class of split extensions.

Definition B.11. The class of an extension 0 → A → B → C → 0 is the image of idC ∈ HomR(C,C)under the connecting homomorphism

∂ : HomR(C,C)→ Ext1R(C,A)

appearing in the long exact sequence of derived functors. Thus equivalent extensions have the sameclass.

Theorem B.12. The class defines a bijection between the equivalence classes of extensions of C by A andExt1R(C,A). It is a group isomorphism under the Baer sum of extensions.

C Hilbert functions and Hilbert polynomials

We start by noticing a corollary to the basis theorem.

Corollary C.1. Any homomorphic image of a noetherian ring is noetherian. And therefore, any finitely generatedalgebra over a noetherian ring is noetherian.

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So in particular, any finitely generated algebra over a field is noetherian. The converse is obviouslyfalse. However, it is true for graded rings.

Theorem C.2. For a graded ring R = R0 ⊕ R1 ⊕ · · · , the following are equivalent:

1. R is noetherian.

2. R0 is noetherian, and the irrelevant ideal R+ is finitely generated.

3. R0 is noetherian, and R is finitely generated R0-algebra.

Proof. (1 =⇒ 2): R0 = R/R+ is noetherian, and any ideal is finitely generated.(2 =⇒ 3): suppose S generate R+ as an ideal, we claim S generate R as an R0-algebra. To do this, let T

be the R0-algebra generated by S, and we show inductively Rn ⊂ T . Obviously it is true for n = 0. Thenfor any f ∈ Rn ⊂ R+, write f =

∑aisi for si ∈ Swith degree di, and ai ∈ Rn−di ⊂ T . So f ∈ T .

(3 =⇒ 1): follows from corollary above.

We know a subring of a noetherian ring needn’t be noetherian, it is if the subring if a summand.

Proposition C.3. Let R ⊂ S be rings, and assume R is a summand of S as an R-module, i.e. there exists a retracthomomorphism φ : S→ R of R-modules fixing R. Then if S is noetherian, R is noetherian.

Proof. If I is an R-ideal, and SI the S-ideal generated by S, it can be easily shown that φ(SI) = I. Thus achain of ideals in R gives rises to a chain of ideals in S.

We have a similar results for graded ring.

Proposition C.4. Let k be a field, and S = k[s1, . . . , xr] a polynomial ring graded by degree. Let R be a k-subalgebra of S. If R is a summand of S, in the sense that there is a retract map of R-modules φ : S→ R preservingdegrees and fixing R, then R is a finitely generated k-algebra.

Proof. Let I be the ideal generated by all homogeneous elements of R of strictly positive degree. SinceS is noetherian, SI has a finite set of generators, chosen to be homogeneous f1, . . . , fs. We show theseelements generate R as a k-algebra.

Let T be the k-subalgebra generated by these elements. Again show by induction on the degree ofany f ∈ R that f ∈ T . If degree is 0, then f ∈ k ⊂ T . Otherwise, f ∈ R has strictly positive degree, so thatf ∈ I. Write f =

∑gifi, where each gi is homogeneous with degree

deggi = deg f− deg fi < deg f.

Applying φ, and that f, fi ∈ R so that they are fixed, we have f =∑φ(gi)fi. Since φ(gi) has lower

degree than f, φ(gi) ∈ T . Thus f ∈ T .

The original motivation for Hilbert function arises in this context.

Definition C.5. Let M be a finitely generated graded module over k[x1, . . . ,dr], with grading by degree.In particular, R0 = k. In the context of original lecture notes, take λ to be the k-vector space dimensions.Then the Hilbert function is the numerical function

HM(s) := dimkMs = λ(Ms).

Note that these values must be finite, since ifMs is not finite dimensional, then the submodule ⊕∞s Mi

would not be finitely generated, contradicting the basis theorem.

Theorem C.6. IfM is a finitely generated graded module over k[x1, . . . , xr], then for large s, HM(s) agrees witha polynomial of degree ≤ r− 1. This polynomial, denoted PM(s), is called the Hilbert polynomial ofM.

From here, we shall assume all rings are noetherian. Write R for a local ring with maximal ideal m.All modules are finitely generated R-modules unless otherwise stated.

A slight generalization of the definition of Hilbert function is that, for a finitely generated R-module,define

HM(s) := dimR/mmnM/mn+1M.

If we turnM into a graded module, it is easy to see that two Hilbert functions agree.We state without proof here a theorem.

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Theorem C.7. If R is a local ring, then dimR = 1+ degPR.

In particular, dimR = 0 iff m is nilpotent.

D Rees algebra and algebraic geometry1

In this section, all algebras are over C.Typically when we try to give a graded algebra, we assign each generator {xi} a degree, and some

relations that are homogeneous. On the other hand, filtered algebras are easier to come up with: we cansimply define the degree of an element f to be the smallest degree of any polynomial that represents f.

Recall that a filtered algebra is an algebra Rwith a sequence of vector spaces Fn such that Fn ⊂ Fn+1,FnFm ⊂ Fn+m, and

⋂Fn = R. We now concentrate on positive filtered (index set are N) algebra with

F0 = C. These two conditions together are usually called being connected. You should think of anelement in Fn \ Fn−1 to be of degree n.

We constructed two graded algebras out of a filtered algebra. The first one is the associated gradedalgebra, following the degree definition above:

grR = ⊕nFn/Fn−1.

Each summand is the collection of degree n terms. There is a set map R→ grR that sends an element rof degree n to its image in Fn/Fn−1. Note this is not even additive. By five lemma, for a map of filteredring to be an isomorphism, it’s necessary for the induced map on associated graded algebra to be anisomorphism.

The second algebra is the Rees algebra:

Rees(R) := ⊕nFntn ⊂ R[t]

(remember that we are considering a positively filtered algebra). The very first observation is that theRees algebra contains the information in R and grR.

Lemma D.1. R = Rees(R)/(t− 1) and grR = Rees(R)/t.

Proof. The map Rees(R)→ R given by ∑ant

n 7→∑an

is surjective. Any element x = (t− 1)∑ant

n = (0− a0) + (a0 − a1)t+ (a1 − a2)t2 + . . . is mapped to

0; and if∑an = 0, we can write (0− b0) = a0, and (bn − bn+1) = an. Then

∑ant

n = (t− 1)∑bnt

n.Thus by isomorphism theorem we get the first equality.

Similarly for second one, where we define Rees(R)→ gr (R) by sending antn to the image of an inFn/Fn−1, and extend this map linearly. The rest follows easily.

But actually we have more. Rees(R) is naturally a C[t]-algebra essentially by inclusion (since C =F0 ⊂ Fn for any n). Such a ring map gives a scheme map of Spec(Rees(R)) to the affine line.

Recall that the fibre is defined to be Spec(Rees(R))×Spec(C[t]) Speck(p). For a point p = (t−a) wherea ∈ C, the fibre over this (simple affine case) is

Spec(C[t]/(t− a))⊗C[t] Rees(R)) = Spec(Rees(R)/(t− a)).

So by the lemma above, the fibre over t = 0 is Spec(gr (R)), and over any other point is Spec(R).What about Proj(Rees(R))? Notice first that t = 0 is a homogeneous relation (t = 1 · t ∈ F1t1). So it

cuts out a closed subscheme, which is Proj(Rees(R)/t) = Proj(grR).Its complement is given by t 6= 0, which is a distinguished open set. By definition of proj, this is

isomorphic to the Spec of the ring

{f

tr: f is homogeneous of degree r}.

1This section is largely a copy from https://cornellmath.wordpress.com/2007/08/28/filtrations-in-algebraic-geometry/.

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But if you carefully think about it, any homogeneous term of degree r is precisely of the form artr, and

thus this gives a subring of R. In fact it is R. So this open subscheme is isomorphic to Spec(R).The upshot is that Proj(Rees(R)) is the union of two pieces, Proj(gr (R)) and Spec(R). Taking Proj of

the Rees algebra is the closure of Spec(R), with the extra piece added being Proj(gr (R)).Relating back to filtration. Giving a connected filtration is then taking a compactification of Spec(R).

Once you took the compactification, what do we look at? The whole compact thing, which is Reesalgebra, and the extra piece added, the associated graded algebra.

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