galois theory -...
TRANSCRIPT
GALOIS THEORY
Alessio Corti∗
v0.94, 29th April 2020
Contents
1 Elementary topics 2
2 Axiomatics 4
3 Fundamental Theorem 5
4 Philosophical considerations 8
5 Proofs of the Axioms 9
6 Biquadratic extensions 11
7 Normal extensions 17
8 The separable degree 18
9 Separable extensions 19
10 Finite fields 21
11 Frobenius lifts 21
12 Cyclotomic polynomials over Q 24
13 Texts 26
14 Questions 26
15 Answers 37
∗This is a preliminary version. Updates will keep appearing here. The text is a concise version of Galoistheory as presented in my MP11 lecture course at Imperial College London (Easter terms 2018, 2019 & 2020).I thank the students who attended those courses, and my class assistant Giulia Gugiatti for correcting manymistakes.
1
1 Elementary topics
1.1 The category of fields and finite extensions
There is a category of fields where: objects are fields, morphisms are finite extensions.In fact, we almost always work with a variant of this, namely we fix a field k and we
work in the category whose objects are finite extensions of k, and for two objects k ⊂ Kand k ⊂ L, morphisms are understood as morphisms that restrict to the identity on k.
Remark 1. Every morphism of fields is injective.
For two fieldsK, L, we denote by Emb(K,L) the set of field homomorphisms (Embeddings)from K to L, and by
Embk(K,L) = f ∈ Emb(K,L) | ∀a ∈ k, f(a) = a
the set of field homomorphisms f : K → L that are the identity on k, and we call the elementsof this set k-embeddings of K in L.
1.2 Degree and the tower law
An important observation The earlier you understand this, the better: The field struc-ture is complicated and strangely baroque: there are two operations and they satisfy thisweird distributive property. Vector spaces, groups and their representations are simplerstructures and hence they are much easier to work with. There are fewer ways to go down arabbit-hole and hence it is easier to keep on the right path.
If K ⊂ L is a field extension, then L is a K-vector space.
Definition 2. The degree of a finite extension K ⊂ L is the quantity:
[L : K] = dimK L
(the dimension of L as a vector space over K).
Remark 3. (a) If [L : K] = 1 then K = L;
(b) For all k ⊂ K, every element of Embk(K,K) is surjective,1 hence an isomorphism. ThusG = Embk(K,K) is a group, and the k-vector space K is a k-linear representationof G.
Theorem 4 (Tower law). For a tower k ⊂ K ⊂ L
[L : k] = [L : K][K : k]
1This follows immediately from the rank-nullity formula.
2
1.3 Elementary topics
Elsewhere in the text I use the following freely and refer to them by bold numerals like so[ix].2 The topics are: minimal polynomials [i-v]; splitting fields [vi-ix]; separablepolynomials [x-xiii]
(i) For K ⊂ L, all a ∈ L have a minimal polynomial fa ∈ K[X]. By definition, fa isthe (unique) monic generator of the kernel ideal of the evaluation-at-a homomorphismϕa : k[X] → L. By definition, the field K(a) ⊂ L is the image of ϕa; the evaluationhomomorphism ϕa induces (by passing to the quotient) a field isomorphism
[ϕa] : K[X]/(fa)∼=−→ k(a)
Because [ϕa] is (by construction) injective, and because L is a field, hence as a ring itis an integral domain, it follows that k[X]/(fa) is also an integral domain, and hence(fa) is a prime ideal, and hence fa is irreducible;
(ii) Conversely, given an irreducible f ∈ K[X] there is a field extension K ⊂ K(a) =K[X]/(f) such that a = [x] is a root of f . If f is also monic, then f is the minimalpolynomial of a;
(iii) For all K ⊂ L, there is a canonical bijection from the set of roots of fa in L toEmbK(K(a), L) that maps a root b to the unique K-embedding ϕ : K(a) → L suchthat ϕ(a) = b;
(iv) In [i] and [ii] [K(a) : K] = deg fa;
(v) If k ⊂ K ⊂ L and a ∈ L then the minimal polynomial of a over K divides the minimalpolynomial of a over k;
(vi) For all f ∈ K[X] (not necessarily irreducible) there is a field extension K ⊂ L suchthat:
(a) The polynomial f splits completely in L[X];
(b) L = K(a1, . . . an) where the ai are the roots of f in L.
An extension K ⊂ L satisfying properties (a) and (b) is called a splitting field of f ; Anytwo splitting fields of f ∈ K[X] are K-isomorphic; the isomorphism is not canonical,but it sends a root of f to a root of f .
(vii) More generally, let f ∈ K[x] be a polynomial, K ⊂ L a splitting field of f , and K ⊂ Eany field extension. If a1, . . . , am ∈ E are roots of f , then there is a K-embeddingK(a1, . . . , am) ⊂ L;
(viii) If K ⊂ L is a splitting field of a polynomial f ∈ K[X] then the group G = EmbK(L,L)is a subgroup of the symmetric group on the roots of f . If, in addition, f ∈ K[X] isirreducible, then this group acts transitively on the roots;
2In the course, I introduce and explain all of these points in detail when they are needed.
3
(ix) Let K ⊂ L be an extension of fields; note that this induces an inclusion of polynomialrings K[X] ⊂ L[X]. The following are equivalent for two polynomials f, g ∈ K[X]:
(a) The polynomials f , g are coprime as elements of K[X];
(b) The polynomials f , g are coprime as elements of L[X];
(c) For all finite extensions K ⊂ E, the polynomials f , g have no common root in E;
(d) If K ⊂ E is a field extension in which both f and g split completely, then f and ghave no common root in E;
(x) By definition a polynomial f ∈ K[X] is separable if it has n = deg f distinct roots (ina — and hence by [vii] in all — splitting field K ⊂ L). The Jacobian criterion: apolynomial f is separable if and only if f and Df (the derivative of f) are coprime;
(xi) An irreducible polynomial f ∈ K[X] is not separable if and only if (a) K has character-istic p > 0 and (b) there is a polynomial (necessarily irreducible) h ∈ K[X] such thatf(X) = h(Xp);
(xii) For K ⊂ L, an element a ∈ L is separable over K if the minimal polynomial f ∈ K[X]of a over K is a separable polynomial;
(xiii) If k ⊂ K ⊂ L and a ∈ L is separable over k, then by [v] it is also separable over K.
2 Axiomatics
The following correspond roughly to Grothendieck’s axioms for a Galois category. The onlynontrivial ones are Axiom 1, Axiom 4 and Axiom 5. The proof is postponed till Section 5.
Axiom 1 Fix a field k. The category of algebraic field extensions k ⊂ K finite over k hasan initial object (the field k) and for all pairs of objects k ⊂ K and k ⊂ L, Embk(K,L)is finite.3
Axiom 2 Every morphism is injective. Also: every k-morphism from K to K is an isomor-phism, that is, Embk(K,K) is a group. For all fields Ω the (right) action of Embk(K,K)on Embk(K,Ω) is free.4
Axiom 3 Fibered products exist. (These are just “intersections” in a bigger field.) Also,“framed” pushouts exist. (If x1 : K → L1, x2 : K → L2 are contained in a biggerfield Ω, then it makes sense to take the “product”—which in fact categorically is apushout—L1L2 ⊂ Ω: this is the subfield of Ω generated by L1 and L2 applying all fieldoperations; alternatively it is the smallest subfield of Ω containing both L1, L2.)
3In fact, it is shown in Lemma 14 that |Embk(K,L)| ≤ [K : k].4f : K → Ω is injective if for all pairs of morphisms g1, g2 : L→ K, f g1 = f g2 implies g1 = g2. Now
f ∈ Embk(K,Ω) is fixed by some g ∈ Embk(K,K) if and only if f g = f idK and then, because f isinjective, g = idK .
4
Axiom 4 For all pairs of finite extensions K ⊂ L, K ⊂M there is a field extension K ⊂ Ω,and a commutative diagram:5
L
K
>>
Ω
M
>>
Axiom 5 For every field L and finite subgroup G ≤ AutL, it makes sense to take the fixedfield K = LG, and the natural inclusion G → EmbK(L,L) is an isomorphism.6
3 Fundamental Theorem
The aim of this Section is to state and prove the Fundamental Theorem of Galois Theoryfrom the axioms.
Definition 5. k ⊂ K is normal if: For all Ω any two k-embeddings x1, x2 : K → Ω differ bya k-automorphism of K. More formally, for all x ∈ Embk(K,Ω), the naturally induced
x? : Embk(K,K)→ Embk(K,Ω) defined as: x? : σ 7→ x σ
is bijective.7
In simpler words, k ⊂ K is normal if and only if for all pairs of k-embeddings x1, x2 : K →Ω, we have
x1(K) ⊂ x2(K)
Informally: no matter where I go I can not “displace” K away from itself.
Lemma 6. Splitting fields [vi] are normal.
Proof. Let k ⊂ K be a splitting field of the polynomial f ∈ k[X], K ⊂ Ω and x : K → Ω ak-embedding. We aim to prove that x(K) ⊂ K. If a ∈ K is a root of f , then x(a) ∈ Ω is alsoa root of f , and hence x(a) ∈ K. But K is generated by roots of f , hence x(K) ⊂ K.
5This field has no universal property and it is not unique. Indeed the intersection and product of L andM in Ω are not determined (even up to noncanonical isomorphism) by L and M . This axiom substitutes forthe non-existent fibered co-product. Indeed the tensor product of rings K ⊗k L is not a field.
6This is half the fundamental Theorem. This axiom corresponds exactly to Grothendieck’s axiom (G5)for a Galois category stating “F [...] commutes with taking the quotient by a finite group action.” Recallthat F is the fibre functor: for fields F (K) = Embk(K,Ω) where Ω is an algebraic closure of k. Suppose thatG is a finite group acting on K, then by naturality G acts on F (K) by composing on the left. (And in factthe axioms imply that this action is free.) Axiom (G5) then states F (KG) = F (K)/G. Counting elements
we obtain [KG : k] = [K:k]|G| and using the tower law then [K : KG] = |G|. Even Grothendieck sneaks half the
fundamental Theorem into an axiom!7The function (x, σ) 7→ x σ is in fact a right group action X × G → G, where X = Embk(K,Ω) and
G = Embk(K,K). Axiom 2 states that this action is free. An extension k ⊂ K is normal if and only if, forall Ω, X = Embk(K,Ω) is a G-torsor.
5
The following statement is immediate from the definition:
Lemma 7. Suppose given k ⊂ K ⊂ L: If k ⊂ L is normal, then K ⊂ L is normal.
(There are counterexamples to all other obvious wishful statements.)
Definition 8. k ⊂ K is separable if: For every tower of subfields
k ⊂ K1 ( K2 ⊂ K
there exist: a field Ω, and at least two distinct K1-embeddings x, y : K2 → Ω.8
The slogan is: k ⊂ K is separable if embeddings separate subfields.
Lemma 9. If k ⊂ K is a splitting field of a separable polynomial [x], then k ⊂ K is separable.
Sketch of proof. This is Corollary 37, but: that is not for some time, the result is not verydifficult, and it is needed to make sense of the statement of the Fundamental Theorem, so Igive a sketch now, with details to be discussed later.
In Definition 31 we define the separable degree [K : K]s of an extension k ⊂ K to be thenumber of elements of the set Embk(K,Ω) of k-embeddings of K into a field Ω such thatk ⊂ Ω is a normal extension that also contains K. In Section 8 it is shown that the separabledegree is well-defined and that it satisfies the tower law. Note that by [iv] [K : k]s ≤ [K : k].
It is more-or-less obvious (and proved as Step 1 in the proof of Theorem 36) that if[K : k]s = [K : k] then k ⊂ K is separable.
Now if a is separable over k, then by [iii] [k(a) : k]s = [k(a) : k].
The following is immediate from the definition:
Lemma 10. For k ⊂ K ⊂ L, if k ⊂ L is separable, then K ⊂ L and k ⊂ K are separable.
(The converse is also true, see Theorem 36(II).)
Theorem 11 (Fundamental Theorem of Galois Theory). If k ⊂ K is normal and separablethen the correspondence between subfields and subgroups holds.9
Proof. Write G = Embk(K,K) the Galois group of the extension. One defines functions:
For H ≤ G : H 7→ H? =a ∈ K | ∀ g ∈ H, g(a) = a
and
For k ⊂ L ⊂ K : L 7→ L† = g ∈ G | ∀ a ∈ L, g(a) = aand the task is naturally split into two halves:
8This definition substitutes for Grothendieck’s axiom (G6) for a Galois category (Stating that if F (u) isan isomorphism then u is an isomorphism). The category of separable field extensions is a Galois categoryin the sense of Grothendieck but the category of all extensions (separable and inseparable) is not.
9In fact more is true, namely there is an equivalence of categories between the category of fieldextensions (intermediate between k and K) and the category of subgroups of the Galois group. This fact,essential though it is, is seldom pursued to the bitter end: for subgroups H1, H2 of G, what is the correctdefinition of Mor(H1, H2) that makes this equivalence work? An important consequence of the equivalenceof categories is: given k ⊂ L ⊂ K, then k ⊂ L is normal if and only if the corresponding subgroup H ≤ G(of elements that fix L) is a normal subgroup, AND, in that case, the Galois group of k ⊂ L is the quotientgroup G/H.
6
First Half For all H, (H?)† = H. This is just Axiom 5;
Second Half For all L, (L†)? = L.
We need to prove the second half. Since L ⊂ K is normal and separable, it is enough toshow that G? = k.
Let k ⊂ F = G?; I show that for all K ⊂ Ω, the set Embk(F,Ω) consists of the obviousinclusion F ⊂ K ⊂ Ω and hence — because we are assuming that the extension is separable— it must be that k = F .
Indeed consider the tower k ⊂ F ⊂ K ⊂ Ω. For clarity denote by ιF,Ω : F → Ω the inclu-sion in this tower and similarly all other inclusions in the tower. Consider y ∈ Embk(F,Ω).By Lemma 12(B) below, there is x ∈ Embk(K,Ω) such that y = x|F . Because k ⊂ K isnormal, there exists σ ∈ Embk(K,K) such that y = ιK,Ω σ. By construction of F σ|F isthe identity, and therefore x = y|F = ιF,Ω.
Lemma 12. Suppose that k ⊂ K is normal. Then for all given towers:
k ⊂ F ⊂ K ⊂ Ω
we have:
(A) the natural Embk(F,K) → Embk(F,Ω) (compose with the given inclusion K ⊂ Ω) issurjective;
(B) the natural Embk(K,Ω)→ Embk(F,Ω) (restrict to the given subfield F ⊂ K) is surjec-tive.
Remark 13. (i) (A) states that every k-embedding x : F → Ω in fact lands F in K: x(F ) ⊂K. When F = K, this is just the definition of k ⊂ K normal.
(ii) (B) states that every k-embedding x : F → Ω extends to a k-embedding x : K → Ω(which, in fact, lands K in itself, but this fact is here besides the point):
Kx
F
x //
OO
Ω
k
OO >>
The case K = Ω is especially significant: If k ⊂ K is normal then every embeddingx : F → K extends to an automorphism σ : K → K: this fact was crucial in the proofof the fundamental Theorem.
(iii) It is manifestly the case that, assuming as we are that k ⊂ K normal, (B)⇒(A): Indeed,given x : F → Ω, extend it to x : K → Ω, and then observe that by normality x(K) ⊂ Kand hence a fortiori x(F ) ⊂ K, that is, F landed in the given inclusion of K in Ω.
7
Proof. By Remark 13(iii) we only need to prove (B). Let x : F → Ω: we want to show that xis the restriction of some x : K → Ω. First, use Axiom 4 to construct a commutative diagram:
Ky2 // Ω
F
OO
x// Ω
z
OO
Note that we have TWO k-embedding K → Ω: one is y2 : K → Ω in the diagram above; theother is obtained composing z with the given inclusion K ⊂ Ω:
y1 : K ⊂ Ωz→ Ω
It follows from the normality of k ⊂ K that y2(K) ⊂ y1(K) ⊂ z(Ω). In other words, y2
landed K in Ω, that is, it gave the sought-for extension y2 = x.10
4 Philosophical considerations
In teaching the course these were my aims:
(i) Work with finite extensions only; avoid constructing an algebraic closure. (Even ifhaving one helps a great deal.) There should be no need of discussing infinite algebraicextensions if one is only interested in finite ones.
(ii) Discuss characteristic p and the phenomenon of inseparability. Develop the theoryof Frobenius lifts and use these to give a transparent proof of the irreducibility ofcyclotomic polynomials over Q (a shockingly deep Theorem of Dedekind).
(iii) Avoid copying Emil Artin like everybody else does. Aim to follow Grothendieck in spirit:uncompromisingly express all important definitions and statements in pure categoricalterms; have a clean set of axioms.
(iv) Most books spend a lot of time developing many things; the fundamental Theoremcomes at the very end when one has seen so much detail that one can not see whatreally makes it work. I wanted to do the opposite: prove the fundamental Theorem assoon as possible and develop the theory later.
There are three levels of abstraction:
(i) At the beginning one is interested in polynomials and their roots and in proving thatthere is no general formula in radicals for the roots of polynomials of degree ≥ 5;
(ii) One then discovers that it is helpful to keep roots in the fields that they inhabit, andthat one is interested only in those permutations of roots that are automorphisms ofthese fields;
(iii) The third level of abstraction is arrows in a category, and the discipline of phrasingeverything in terms of these. My innovation in this course is to access this level andtry to convince you of the benefits.
10In fact, it is even true that y2(K) ⊂ K, which implies (A).
8
5 Proofs of the Axioms
5.1 Proofs of Axioms 1 and 4
Lemma 14 (Axiom 1). Embk(K,L) has at most [K : k] elements.
Proof. We know that [iii, iv] that in the case of a primitive extension k ⊂ k(a) Embk(k(a), L)has at most deg fa = [k(a) : k] elements, hence the statement is true in this case. We willreduce the general case to the primitive case by tower law and induction on the degree of theextension.11 Pick a ∈ K \ k (if there is no such a then we have nothing to do). Consider thetower k ⊂ k(a) ⊂ K. We have an obvious restriction map:
ρ : Embk(K,L)→ Embk (k(a), L)
andfor x ∈ Embk (k(a), L), ρ−1(x) = Embk(a)(K,L)
hence|Embk (K,L) | =
∑x∈Embk(k(a),L)
|ρ−1(x)| ≤ [k(a) : k][K : k(a)] = [K : k]
by the primitive case, induction, and the tower law.
Proof of Axiom 4. We want to create a field Ω completing to a commutative diagram:
L
K
>>
Ω
M
>>
The proof is by induction on [L : K]. The base of the induction: If [L : K] = 1, then K = Land we take Ω = M .
If [L : K] > 0, pick a ∈ L \ K and let f(X) ∈ K[X] be the minimal polynomial [i] ofa over K. It is an elementary fact [ii] that there is a (possibly trivial) extension E = M(b)in which f(X) has a root b ∈ E; and we know [iii] that there is a (unique) K-embedding
11This is my preferred scheme of proof in this lecture course: to show that something holds for an arbitraryextension k ⊂ K, we show that it hods for a primitive extension, and then reduce to the primitive extensioncase by tower law and induction on degree. This scheme is not super-elementary but it is never very difficultto implement. I must tell you that every finite field extension is primitive: this is a very nontrivial fact knownas the primitive element theorem; its proof is elementary but difficult.
9
x : K(a)→ E such that x(a) = b:
L
K(a)
<<
x
""K
<<
""
E
M
<<
Since by the tower law [L : K(a)] < [L : K], we may assume by induction that there exists afinite field extension K(a) ⊂ Ω:
L
K(a)
<<
x
""
Ω
E
??
and by composing the relevant arrows K ⊂ Ω proves the statement.
5.2 Proof of Axiom 5
Proposition 15 (Axiom 5). Let L be a field, G ⊂ AutL a finite subgroup, and write K = LG
the subfield fixed by G. Then
(i) [L : K] = |G|;
(ii) The manifest inclusion G ⊂ EmbK(L,L) is an isomorphism.
Proof. We already know from Lemma 14 that
|EmbK(L,L)| ≤ [L : K]
and obviously |G| ≤ |EmbK(L,L)| so (i) and (ii) follow from: [L : K] ≤ |G|, which is provedin the next lemma.
Lemma 16. Let L be a field, G ⊂ AutL a finite subgroup, and write K = LG the subfieldfixed by G. Then [L : K] ≤ |G|.
Proof. Suppose that G = σ1, . . . , σn, we want to show that all (n+1)-tuples a1, . . . , an+1 ∈L are linearly dependent over K.
Indeed, to start with, the n+ 1 vectors in Ln:
a1 =
σ1(a1)...
σn(a1)
, . . . , an+1 =
σ1(an+1)...
σn(an+1)
10
are linearly dependent over L. Let k ≤ n + 1 be the smallest number of summands in anontrivial linear dependence between the ai; by rearranging the indices we may assume thatsuch a dependence holds between a1, . . . , ak:
x1a1 + · · ·+ xkak = 0 (1)
Because this is a nontrivial linear dependence with the smallest number of summands, allxi 6= 0, and by rescaling we may also assume that x1 = 1.
We can re-word the linear dependence by stating that x1, . . . , xk is a solution of the linearsystem of equations:
∀ j ∈ [n],k∑i=1
xiσj(ai) = 0 (2)
Applying σ ∈ G to this reshuffles the j and from this we conclude that σ(x1), . . . , σ(xn)is another solution of the system of Equations 2. So it is the old solution (otherwise bysubtracting it from the old solution—since x1 = 1 and so also σ(x1) = 1—we would obtain anontrivial linear dependence with a smaller number of summands), and then all xi ∈ K = LG.
But then the equation corresponding to σj = e states:
k∑i=1
xiai = 0
and this is the sought-for linear dependence over K.
Remark 17. There is a perspective that makes it clear why we expect a solution with allxi ∈ K. If V is a K-vector space we can extend scalars to L: VL = L⊗KV , and make a vectorspace VL over L. When K ⊂ L is the extension R ⊂ C you are familiar with this constructionunder the name of complexification of a real vector space. Given an L-vector space VL, wesay that VL descends to K when there is a K-vector space V such that VL = L ⊗K V asabove. So how can we tell if VL descends to K? A descent datum is a G-action on VL suchthat: For all g ∈ G, for all λ, µ ∈ L, for all v, w ∈ VL: g(λv + µw) = g(λ)g(v) + g(µ)g(w).It is a general fact of linear algebra (which you can show by adapting the ideas of the proofof Lemma 16) that the category of K-vector spaces is equivalent to the category of L-vectorspaces equipped with descent datum. Now the set of solutions of Equation 2 is a (nontrivial)L-vector subspace of Ln+1 with descent datum inherited from the standard descent datumof Ln+1. This vector subspace descends to a nontrivial K-vector subspace V ⊂ Kn+1 and anonzero element of V is the same as a solution x1, . . . , xn+1 ∈ Kn+1 of Equation 2.
6 Biquadratic extensions
6.1 The key statement
This is a piece de resistance that every beginner in Galois theory needs to master completely.In this Section we fix a field K of characteristic 6= 2. For a, b ∈ K× we study the field
extension
L = K(
√a±√b)
11
Below we always assume that b is not a square in K — otherwise, L is not very interesting.(The case a = 0, on the other hand, is perfectly interesting.)
Remark 18. (i) L is the splitting field of the polynomial:
f(X) = x4 − 2aX2 + c ∈ K[X] where c = a2 − b
(ii) If charK 6= 2, then f(X) is separable. Indeed, under these assumptions, Df = 4X3 −4aX = 4X(X2 − a) manifestly has no roots in common with f(X).
(iii) It is not super-obvious, but important, that L is also the splitting field of the companionpolynomial :
g(Y ) = Y 4 − 4aY 2 + 4b ∈ K[Y ]
You can either prove this now, or leave it as a mystery to unveil later. In case youwonder, there are certain annoying factors of 2 in this story and I don’t think you canget rid of them.
Theorem 19. Let K be a field of characteristic 6= 2; let a, b ∈ K× with b not a square in K.Consider the normal and separable extension
K ⊂ L = K(
√a±√b) and set c = a2 − b
Denote by G the Galois group of the extension. Then
(i) If bc, c are not squares in K, then [L : K] = 8 and G = D8. The discussion in Sec-tion 6.3 below identifies all the intermediate fields, displays the Galois correspondence,and details the action of G on the roots of f(X) and g(Y ).
(ii) If bc is a square in K (and then c is not a square in K, for if both bc and c are squares,then b is also square, which it isn’t), then [L : K] = 4 and G = C4.
(iii) If c is a square in K (and then bc is not a square in K), then either
(iiia) Neither 2(a +√c) nor 2(a −
√c) are squares in K. In this case [L : K] = 4 and
G = C2 × C2; or
(iiib) One of 2(a +√c), 2(a −
√c) is a square in K (but not both). In this case L =
K(√b), and G = C2.
It will be clear that f(X) is irreducible in cases (i), (ii), (iiia), and it splits into twoquadratic polynomials in case (iiib). See Lemma 21 for a discussion of this point.
6.2 Initial considerations and set-up
I summarise all of the key algebra. Invest the time to familiarise yourself with it now.In the discussion below we make the following choices:
(1) Choose β ∈ L such that β2 = b (there are two choices, make one);
12
(2) Next, choose α, α′ ∈ L such that α2 = a+β and α′ 2 = a−β. It is clear that L = K(α, α′).The roots of f(X) are ±α,±α′ ∈ L.
The following quantities will be used throughout:
(3) γ = αα′. Note that γ2 = (a+ β)(a− β) = a2 − b = c;
(4) δ = α + α′ and δ′ = α− α′. Note that
δ2 = α2 + α′ 2 + 2γ = 2(a+ γ)
andδ′ 2 = α2 + α′ 2 − 2γ = 2(a− γ)
Exercise 20. 1. Write formulas for β, α, α′ in terms of γ, δ, δ′;
2. Convince yourself that K ⊂ L is the splitting field of the companion polynomial g(Y ).
We haven’t yet understood if or when the polynomial f(X) ∈ K[X] is irreducible.
Lemma 21. Let K be a field of characteristic 6= 2. Consider f(X) = X4−2aX2 + c ∈ K[X]where c = a2− b and b not a square in K. Then f(X) is reducible if and only if c is a squarein K and either 2(a+
√c) or 2(a−
√c) is a square in K (but not both).
Proof. We havef(X) = (X − α)(X + α)(X − α′)(X + α′)
where none of the roots is in K (otherwise, for example, b is a square in K). If f(x) isreducible, then either (X − α)(X − α′) ∈ K[X] or (X − α)(X + α′) ∈ K[X].
Case 1 (X − α)(X − α′) ∈ K[X]:
(X − α)(X − α′) = X2 − δX + γ and f(X) = (X2 − δX + γ)(X2 + δX + γ)
since γ2 = c, we get that c is a square in K and also δ2 = 2(a+ γ) is a square in K.Case 2 (X − α)(X + α′) ∈ K[X]:
(X − α)(X + α′) = X2 − δ′X − γ andf(X) = (X2 − δ′X − γ)(X2 + δ′X − γ)
and in this case c is a square in K and also δ′ 2 = 2(a− γ) is a square in K.Note that 2(a± γ) are not both squares in K, for otherwise the product
(a+ γ)(a− γ) = a2 − c = b
is also a square in K.
It really can happen that f is reducible:
Example 22. Consider f(X) = X4 − 6X + 1 ∈ Q[X], so a = 3, b = 8 (not a square in Q)
and c = a2 − b = 1. Now f(X) has innocent-looking roots ±√
3± 2√
2 but
f(X) = (X2 − 2X − 1)(X2 + 2X − 1)
So in fact the splitting field is L = Q(√
2) and the four roots are
1+√
2 =
√3 + 2
√2, −1+
√2 =
√3− 2
√2, 1−
√2 = −
√3− 2
√2, −1−
√2 = −
√3 + 2
√2
13
Remark 23. Recall that the companion polynomial of f(X) = X4−2aX2+c is the polynomial
g(Y ) = Y 4 − 4aY 2 + 4b ∈ K[Y ]
with roots ±√
2(a±√c). The previous discussion shows that f splits in K[X] if and only if
the companion polynomial has a root in K.
We will need the following:
Lemma 24. Let F be a field and A,B ∈ F . If A is a square in K(√B) then either A or
AB is a square in K.
6.3 The generic case: bc, c not squares in K
Step 1 We show: [L : K] = 8 in this case.
L
K(α) K(α′)
K(β)
K
Write K1 = K(β); by assumption [K1 : K] = 2. I claim that a+ β is not a square in K1. Ifit were, there would be x, y ∈ K such that
a+ β = (x+ yβ)2 = (x2 + by2) + 2xyβ, and then a− β = (x− yβ)2
would also be a square in K1, and then
c = a2 − b = (a+ β)(a− β) = (x+ yβ)2(x− yβ)2 = (x2 − y2b)2
would be a square in K, which it isn’t. A similar argument shows that a− β is not a squarein K1. We conclude that [K1(α) : K1] = [K1(α′) : K1] = 2.
To finish Step 1 we just need to argue that K1(α) 6= K1(α′), that is, for example, a − βis not a square in K1(α). Apply Lemma 24 with E = K1, A = a − β, B = a + β. Ifa − β were a square in K1(α), then either a − β is a square in K1, which it is not, or(a − β)(a + β) = a2 − b = c is a square in K1. We need to exclude this last possibility. Weapply again Lemma 24, this time with E = K, A = c, B = b. If c were a square in K1, theneither c is a square in K, or bc is a square in K, but we are assuming that neither is.
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Step 2 Action of the Galois group on roots. We will show: the Galois group acts as thegroup of symmetries of the square:
α′
−α α
−α′
and this identifies the group with D8.
Below I work with the following generators of D8: τ is the reflection in the horizontalline, and σ the counterclockwise π/2-rotation.
Indeed let g ∈ G be any element. Clearly g(β) = ±β. If g(β) = β, then g(α) = ±α andg(α′) = ±α′. On the other hand if g(β) = −β, then g(α) = ±α′ and g(α′) = ±α. There isa total of 8 possibilities and they all are in D8. Hence G acts on the roots as a subgroup ofD8. On the other hand [viii] G is a subgroup of the permutation group S4 on the roots off , and by Proposition 15 it has order 8 = [L : K], hence G = D8.
Step 3 Picture of the Galois correspondence. This is the lattice of fields lying betweenK and L and below the corresponding lattice of subgroups of D8:
L
K(α) K(α′) K(β, γ) K(δ) K(δ′)
K(β) K(βγ) K(γ)
K
〈e〉
〈τ〉 〈σ2τ〉 〈σ2〉 〈στ〉 〈σ3τ〉
〈σ2, τ〉 〈σ〉 〈σ2, στ〉
D8
To establish the whole picture is straightforward. For example it is easy to see that σ2τ(α′) =α′ (and σ2τ(α) = −α) thus the corresponding fixed field is K(α′); similarly, στ(α) = α′,στ(α′) = α, hence the fixed field is K(δ); etcetera.
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6.4 bc square in K
The situation with fields is as in the following diagram:
L = K(α) = K(α′) = K(δ) = K(δ′)
K(β) = K(γ)
K = K(βγ)
where all arrows are degree-2 extensions. Indeed, first of all, (βγ)2 = bc so βγ ∈ K and henceK(β) = K(γ). I claim that L = K(α): clearly β ∈ K(α) so also γ ∈ K(α) and then so
α′ = γ/α ∈ K(α)
This shows that L = K(α). Similar arguments show that L = K(α′) = K(δ) = K(δ′).We already know that in this case f(X) is irreducible, so then [L : K] = [K(α) : K] =deg f(X) = 4.
I claim that the Galois group G is a cyclic group of order 4 acting on the set of rootspictured above as rotations. Indeed consider an element g ∈ G. Then g(β) = ±β and westudy the two possibilities in detail:
(1) Suppose that g(β) = β. Now βγ ∈ K so we must also have g(γ) = γ. If g(α) = α thenalso g(α′) = α′ (recall that γ = αα′), i.e., g is the identity. Similarly, if g(α) = −α, thenalso g(α′) = −α′: g is a π-rotation.
(2) Now suppose that g(β) = −β and hence also g(γ) = −γ. If g(α) = α′ then
α′ =γ
αso g(α′) =
g(γ)
g(α)=−γα′
= −α
hence g is a rotation in this case. A similar argument shows that if g(α) = −α′, then inthat case also g is a rotation.
6.5 c square in K
We are saying here that γ ∈ K. It is best to work with the companion polynomial:
Y 4 − 4aY 2 + 4b = (Y 2 − 2a− 2γ)(Y 2 − 2a+ 2γ)
In Case (iiia) the two quadratic factors of the companion polynomial are both irreducibleover K. The situation with fields is as in the following diagram:
L
K(β) K(δ) K(δ′)
K
16
where all arrows are degree-2 extensions. Indeed, for example, 2(a − γ) is not a square inK(δ): if it were then — by Lemma 24 — either it is already a square in K, and we areassuming that it isn’t, or the product (a+ γ)(a− γ) = a2 − c = b is a square in K, which itisn’t.
6.6 Examples of biquadratic extensions
Example 25. If K is the splitting field of X4 − 2 over Q, then G = D8.
Example 26. If K is the splitting field of X4−4X2 + 2 over Q, then K = Q(√
2 +√
2) andG = C4.
Example 27. If K is the splitting field of Φ12(X) = X4−X2 +1 (the cyclotomic polynomial)over Q, then G = C2 × C2.
Example 28. If K is the splitting field of the polynomial X4 − 10X2 + 1 over Q, thenG = C2 × C2. In fact K = Q(
√2,√
3), which – inter alia – explains the identity:√5 + 2
√6 =√
2 +√
3
7 Normal extensions
We can not go very far without some practical understanding of normal and separable ex-tensions. Here I just sketch the bare bones.
Theorem 29 (Characterisation of normal extensions). For a finite extension of fields k ⊂ KTFAE:
(I) for all f ∈ k[X] irreducible, either f has no root in K or f splits completely in K[X];
(II) there exists f ∈ k[X] (not necessarily irreducible) such that K is the splitting field [vi]of f ;
(III) k ⊂ K is normal.
Proof. I show (I)⇒(II)⇒(III)⇒(I), in that order.
(I)⇒(II) If K = k(a1, . . . , an), let fi ∈ k[X] be the minimal polynomial [i] of ai. It isclear that k ⊂ K is the splitting field [vi] of f =
∏fi.
(II)⇒(III) Suppose that K = k(a1, . . . , an) is the splitting field [vi] of f ∈ k[X] where infact
f =∏
(X − ai) ∈ K[X]
(here f ∈ k[X] is not necessarily irreducible and the ai are not necessarily pairwise distinct).Suppose that K ⊂ Ω and let x : K → Ω be a k-embedding. It is an elementary fact [vii] thatfor all i x(ai) is a root of f , that is, x(K) ⊂ K.
(III)⇒(I) Now assume that k ⊂ K is normal. Let f ∈ k[X] be an irreducible polynomialwith a root a ∈ K: we will show that f splits completely in K. To this end, let K ⊂ Ω be
17
the splitting field [vi] of f—seen as an element of K[X]—and let b ∈ Ω be a root of f : wewant to show that b ∈ K. With F = k(a) consider the tower:
k ⊂ F ⊂ K ⊂ Ω
It is an elementary fact [iii] that there is a unique k-embedding x : F → Ω such that x(a) = b.By Lemma 12(A), x(F ) ⊂ K, that is, b ∈ K as was to be shown.
Lemma 30. For all k ⊂ K there is K ⊂ L such that k ⊂ L is normal.
Proof. The easiest way to prove this is to use the part of Theorem 29 stating that splittingfields [vi] are normal: there are elements a1, . . . , an ∈ K such that K = k(a1, . . . , an), letfi ∈ k[X] be the minimal polynomial [i] of ai, take K ⊂ L the splitting field [vi] of f =
∏fi,
then k ⊂ L is also the splitting field [vi] of f over k and hence k ⊂ L is normal.
8 The separable degree
Definition 31. For k ⊂ K, the separable degree [K : k]s is the number of elements of the(finite) set Embk(K,Ω) where k ⊂ Ω is a normal field extension that also contains K.12
Remark 32. (i) By Lemma 14 [K : k]s ≤ [K : k];
(ii) By Lemma 30 the separable degree is defined, but we don’t yet know that it is welldefined, that is, at the moment [K : k]s a priori depends on Ω.
Lemma 33. [K : k]s does not depend on Ω.
Proof. Suppose that K ⊂ Ω1 and K ⊂ Ω2 are two field extensions and k ⊂ Ω1 and k ⊂ Ω2
are normal. Let Ω be an over-field of Ω1, Ω2 whose existence is guaranteed by Axiom 4; byLemma 12(A) we have
Embk(K,Ω1) = Embk(K,Ω) = Embk(K,Ω2)
Remark 34. We can rephrase the definition of separable extensions as follows: k ⊂ K isseparable if and only if for all towers of subfields: k ⊂ K1 ⊂ K2 ⊂ K, if [K2 : K1]s = 1, thenK2 = K1.
Theorem 35 (Tower law for the separable degree). For a tower k ⊂ K ⊂ L
[L : k]s = [L : K]s[K : k]s
12As a geometer, the degree of a covering is the number of geometric points in the fibre of a geometricpoint! The dimension of a vector space is a much more mysterious invariant.
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Proof. Consider a tower k ⊂ K ⊂ L and use Lemma 30 to make an extension L ⊂ Ω suchthat k ⊂ Ω is normal. The key point is:
Claim the natural restriction:
ρ : Embk(L,Ω)→ Embk(K,Ω)
is surjective, that is every k-embedding σ : K → Ω can be extended to a k-embeddingσ : L→ Ω.
Indeed by Lemma 12(B) σ extends to all of Ω hence a fortiori to L.
Now for all y ∈ Embk(K,Ω), ρ−1(y) is the set of K-embeddings x : L → Ω, and K ⊂ Ωis normal, hence ρ−1(y) consists of [L : K]s elements. The formula follows from countingelements of Embk(L,Ω):
[L : k]s = |Embk(L,Ω)| =∑
y∈Embk(K,Ω)
|ρ−1(y)| = |Embk(K,Ω)|[L : K]s = [K : k]s[L : K]s
9 Separable extensions
Recall [x] that a polynomial is separable if it has distinct roots, and an element is separableover k [xii] if its minimal polynomial over k is separable.
Theorem 36 (Characterisation of separable extensions). (I) k ⊂ K is separable if andonly if [K : k]s = [K : k];
(II) For all tower k ⊂ K ⊂L, if k ⊂ K and K ⊂ L are separable, then k ⊂ L is separable;
(III) k ⊂ K is separable if and only if every a ∈ K is separable over k.
Corollary 37. (i) If a is separable over k, then k ⊂ k(a) is separable.
(ii) If k ⊂ K is a splitting field of a separable polynomial, then k ⊂ K is separable.
Proof. To prove statement (i) just observe that, by [iii], [k(a) : k]s = [k(a) : k]; the statementthen follows from part (I) of the Theorem.
Let us prove statement (ii). We can realise k ⊂ K as a tower of extensions:
k ⊂ k(a1) ⊂ k(a1, a2) ⊂ · · · ⊂ k(a1, a2, . . . , an) = K
where all ai are roots of f . By [v] for all i the minimal polynomial of ai over k(a1, . . . , ai−1)is a factor of f , and hence it is a separable polynomial. By [iii] for all i
[k(a1, . . . , ai) : k(a1, . . . , ai−1)]s = [k(a1, . . . , ai) : k(a1, . . . , ai−1)]
and then [K : k]s = [K : k] by repeated application of two tower laws.
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Proof of Theorem 36. The proof is in SIX steps:
Step 1 We show: If [K : k]s = [K : k], then k ⊂ K is separable.
Indeed applying two tower laws to the tower
k ⊂ K1 ⊂ K2 ⊂ K
(and remembering that [:]s ≤ [:]) we get: if [K2 : K1]s = 1 then [K2 : K1] = 1 and thenK1 = K2.
Step 2 We show: If K ⊂ K(a) is separable, then the element a is separable over K.
Indeed, let f ∈ K[X] be the minimal polynomial [i] of a over K and suppose for acontradiction that f is not a separable polynomial. It is then an elementary fact [xi] thatthere exists h ∈ K[X] irreducible such that f(X) = h(Xp). Now b = ap ∈ K(a) is a root ofh: h(b) = h(ap) = f(a) = 0. We can form the tower
K ⊂ K1 = K(b) ⊂ K2 = K(a)
By the tower law
p deg h = deg f = [K2 : K] = [K2 : K1][K1 : K] = [K2 : K1] deg h
hence [K2 : K1] = p. It follows that Xp − b ∈ K1[X] is the minimal polynomial [i] of a overK1: indeed a is a root of this polynomial and the extension has degree p. BUT
Xp − b = (X − a)p
hence by [iii] [K2 : K1]s = 1 and this fact contradicts the separability of k ⊂ k(a) becausemanifestly K1 6= K2 (for instance because [K2 : K1] = p).
Step 3 We show: If K ⊂ K(a) is separable, then [K(a) : K]s = [K(a) : K].
Indeed, we know from Step 2 that a is separable over K. In other words, the minimalpolynomial [i] f of a has distinct roots and hence [iii, v]
[K(a) : K] = deg f = |roots of f| = [K(a) : K]s
Step 4 We show: If k ⊂ K is separable, then [K : k]s = [K : k]. Together with Step 1,concludes the proof of part (I) of the Theorem.
Pick a ∈ K \ k and consider the tower
k ⊂ k(a) ⊂ K
we know (trivially from the definition of separable extension of fields) that k ⊂ k(a) andk(a) ⊂ K are both separable. We have just shown that [k(a) : k]s = [k(a) : k]; on the otherhand definitely [K : k(a)] < [K : k] hence we may assume inductively that [K : k(a)]s = [K :k(a)], hence by two tower laws [K : k]s = [K : k].
Step 5 We show: (II).
(II) follows easily from (I) and two tower laws.
Step 6 We show: (III).
20
This is easy to put together given all the above. (Don’t read my proof, just do it in yourhead.)
Indeed, if k ⊂ K is separable, let a ∈ K. By (II) k ⊂ k(a) is separable, and hence byStep 2 a is separable over k.
Conversely, let k ⊂ K be an extension and suppose that every a ∈ K is separable over k.Pick a ∈ K \ k; because a is separable over k we have that [k(a) : k]s = [k(a) : k] and henceby Step 1 k ⊂ k(a) is separable. By (II), to show that k ⊂ K is separable, it suffices to showthat k(a) ⊂ K is separable. By assumption every element b ∈ K is separable over k, andhence a fortiori it is also separable over k(a). Since [K : k(a)] < [K : k], we may assume byinduction on degree that k(a) ⊂ K is separable.
10 Finite fields
If F is a finite field, then for some prime p > 0 charF = p and Fp ⊂ F . Since F is finite, theextension Fp ⊂ F is finite and hence if m = dimFp F = [F : Fp], then |F | = q = pm.
Theorem 38. Fix a prime p > 0. For all integer m > 0 there exists a field Fq with q = pm
elements, unique up to isomorphism. The field Fq is the splitting field of the (separable)polynomial Xq −X ∈ Fp[X]. The Galois group of the extension Fp ⊂ Fq is a cyclic group oforder m generated by the Frobenius automorphism:
Frp : a 7→ ap
Proof. Suppose such a field F exists. The multiplicative group F× has q− 1 elements, hencethey all satisfy the equation Xq = X and hence Fp ⊂ F is the splitting field of the polynomialXq −X and in particular this shows uniqueness up to isomorphism [vii].
On the other hand let F be a splitting field of the polynomial Xq −X. By the Jacobiancriterion [x] this polynomial is separable and hence it has q distinct roots in F . Now comesthe key observation. Write
µq−1(F ) = z ∈ F | zq−1 = 1
Because (a+b)q = aq+bq in F , and because µq−1(F ) is a group under multiplication, it followsthat the set of roots of Xq −X is a field, and hence by property (b) of the characterisationof splitting fields [vi] it must be all of F , and this shows that F has q elements.
Finally it is clear that the Frobenius automorphism Frp has order m and hence it is all ofthe Galois group.
11 Frobenius lifts
Definition 39. A monoid is a commutative semigroup with identity. (For example, N is amonoid.)
Let P be a monoid and K be a field. A function χ : P → K is a (multiplicative) characterif: χ(0) = 1 and for all p1, p2 ∈ P , χ(p1 + p2) = χ(p1)χ(p2).13
13For us, it will always be the case that χ(P ) ⊂ K×, but I am not requiring this in the definition.
21
Theorem 40 (Linear independence of characters, aka Dedekind independence Theorem).Let K be a field, P a monoid. Any set
χ1, . . . , χn : P → K
of pairwise distinct nontrivial characters is a linearly independent subset of the K-vectorspace of (set-theoretic) functions f : P → K.
Proof. Work by induction on n. Assume for a contradiction a linear relation:∑λiχi = 0
necessarily all λi 6= 0. Find p ∈ P such that χ1(p) 6= χ2(p) and then write a new relation:∑λiχi(p)χi = 0
By induction, the new relation relation must be a multiple of the old relation. The tworelations are
(λ1, λ2, . . . , λn) and (λ1χ1(p), λ2χ2(p), . . . , λnχn(p))
Note that the new relation is not the zero relation, because if it were, then we would have forall i λi(p) = 0, contradicting λ1(p) 6= λ2(p). Hence for all i χi(p) = χ1(p), which contradictsχ2(p) 6= χ1(p).
To make a pedagogical point, we show as a Corollary that if L is a field, G a finite groupof automorphisms of L, and K ⊂ LG, then |G| ≤ [L : K]. Recall that we proved this factearlier in the course by a different method. The proof follows from considering two differentinterpretations of the elements σ1, . . . , σn of G:
(I) In the first interpretation, an element σ ∈ G is a character
σ : P = L× → L
therefore σ1, . . . , σn are linearly independent in the L-vector space Fun(L×, L) of set-theoretic functions f : L× → L. Note that it is the target L that makes any space offunctions Fun(S, L) an L-vector space (S any set);
(II) On the other hand, each σi : L→ L is a K-linear function. Identify the source L withKm (for example by choosing a basis) then an element σ ∈ G is a K-linear map
σ : Km → L
and hence, by extending scalars at the source, σ extends to a L-linear map σ : Lm → L,in other words an element of the vector space
(Lm)∨ = HomL(Lm, L)
Now the σi are linearly independent as elements of (Lm)∨, because a linear dependencebetween the σi implies by restricting scalars a linear dependence between the σi. Itfollows that n ≤ m.
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Theorem 41. Let f(x) ∈ Z[x] be degree n monic; Q ⊂ K a splitting field; p a prime suchthat the reduction fp of f mod p has n distinct roots; Fp ⊂ F a splitting field of fp. Letλ1, . . . , λn ∈ K and let
R = Z[λ1, . . . λn]
We have:
(i) There is a ring homomorphism ψ : R→ F ;
(ii) Any such ψ gives a bijection from the set Z of roots of f in K to the set Zp of roots offp in F ;
(iii) A function ψ′ : R→ F is a ring homomorphism if and only if there exists σ ∈ Gal(K/Q)such that ψ′ = ψσ. In particular there exists σ ∈ Gal(K/Q) such that Frp ψ = ψσ, whereFrp ∈ Gal(F/Fp) is the Frobenius automorphism.
Definition 42. The element σ ∈ Gal(K/Q) such that Frp ψ = ψσ in the above Theorem (iii)is called a Frobenius lift.
Proof. Step 1 R is a finitely generated free Z-module.
Let λ1, . . . , λn ∈ K be the roots of f . It is more-or-less obvious that R is generated as aZ-module by the set of λk11 . . . λknn where all 0 ≤ ki ≤ n− 1. (Use the equation.)
Step 2 Let u1, . . . , ud be a basis of R as a Z-module. This is also a basis of K as aQ-vector space and hence d = [K : Q].
Indeed, it is clear that the u1, . . . , ud are Q-linearly independent (clear denominators).Next QR ⊂ K is a subring of K containing Q, hence it is a field.14 Because it contains allthe roots of f , QR = K and this implies that the u1, . . . , ud generate K as a Q-vector space.
Step 3 (i) holds.
Let R ⊃ m ⊃ (p) be a maximal ideal; then R/m ⊃ Fp is a field generated by the roots offp; hence it is isomorphic to F .
Step 4 (ii) holds.
This is basically obvious. Applying ψ to the identity f(X) =∏
(X − λi) we get fp(X) =∏(X − ψ(λi)), hence the ψ(λi) are the roots of fp. No two are the same, since we are
assuming that fp is a separable polynomial.
Step 5 (iii) holds.
It is clear that G acts on R as a group of ring automorphisms. It follows that for all σ ∈ G,ψσ : R → F is a ring homomorphism. Now fix ψ : R → F and consider ψ1 = ψσ1, . . . , ψd =
14In general if E ⊂ L is an algebraic extension and E ⊂ R ⊂ L is a ring, then R is a field. Indeed if a ∈ R,then a is the root of a polynomial
f(X) = a0XN + a1X
N−1 + · · ·+ 1
with coefficients in E and hence in R. Thus
1/a = −a0aN−1 − a1aN−2 − · · · − aN−1 ∈ R
23
ψσd : R → F : by (ii) and because G ⊂ Sn (the group of all permutations of the roots of f)these are all distinct (because they are distinct on the set of roots).
Suppose now that ψd+1 : R→ F is one other homomorphism. Fix a basis u1, . . . , ud of Ras a Z-module (we showed in Step 2 that |G| = [K : Q] = rkZR!). By linear algebra we cansolve for λi ∈ F :
∀j,d+1∑i=1
λiψi(uj) = 0
and this in fact implies∑λiψi = 0, contradicting Dedekind independence (with monoid
P = R \ 0 and field F ).
Corollary 43. Let f(x) ∈ Z[x] be degree n monic; Q ⊂ K a splitting field; p a prime suchthat the reduction fp of f mod p has n distinct roots and factors as a product of irreduciblefactors of degree n1, . . . , nk. Then the Galois group G of Q ⊂ K contains a permutation ofthe roots of f whose cycle decomposition is (n1)(n2) · · · (nk).
Proof. Any Frobenius lift will do.
12 Cyclotomic polynomials over QDefinition 44. For n > 0 integer, the n-th cyclotomic polynomial is the polynomial:
Φn(X) =∏
k∈(Z/nZ)×
(X − e
2πk in
)Lemma 45. For all n > 0 integer,
Φn(X) ∈ Z[X]
Proof. Denote by µn ⊂ C× the group of roots of unity and by Q ⊂ Q(µn) the splitting fieldof the polynomial Xn − 1 ∈ Q[X]; a priori Φn(X) ∈ Q(µn)[X].
It is easy to argue that Φn(X) ∈ Q[X]. Indeed, the Galois group G of the extension is asubgroup of Autµn = (Z/nZ)× and hence the polynomial Φn is Galois-invariant and hencein Q[X].
It is a little bit more subtle to show that Φn(X) has integer coefficients. Fix n > 0. It isclear from the definition that
Xn − 1 =∏d|n
Φd(X)
If we assume by induction that for all d|n, d 6= n, Φd(X) ∈ Z[X] then it follows from theGauss Lemma that Φn(X) ∈ Z[X].
I conclude the course with the proof of the following very deep theorem of Dedekind. I amshocked and surprised by how deep this theorem is. There really seems to be no elementaryproof of the irreducibility of Φn(X) ∈ Z[X] (if n is prime, or the power of a prime, there isan elementary proof of irreducibility using the Eisenstein criterion).
24
Theorem 46. Fix an integer n > 0, denote by µn the group of n-th roots of unity, byQ ⊂ Q(µn) the splitting field of the polynomial xn − 1, and by Gn the Galois group of theextension Q ⊂ Q(µn).15 The following equivalent facts hold:
(1) The cyclotomic polynomial Φn(X) is irreducible;
(2) [Q(µn) : Q] = ϕ(n) where
ϕ(n) = |(Z/nZ)×| = |k | 0 ≤ k < n and hcf(k, n) = 1|
is Euler’s function;
(3) The canonical injective group homomorphism ρ : Gn → Autµn = (Z/nZ)× is an isomor-phism.
Proof. It is easy to see that the three statements are equivalent. I prove statement (3).To show that the inclusion
ρ : Gn → (Z/nZ)×
is surjective, we need to construct enough elements of Gn that they generate all of (Z/nZ)×,and this we do by Frobenius lifts. The idea is clear enough and the implementation, as weshall see momentarily, straightforward.
Below for a ring A we write:
µn(A) = z ∈ A | zn = 1
Note that this assignment is functorial: if ψ : A→ B is a ring homomorphism, then it inducesa group homomorphism ψ : µn(A)→ µn(B).
If p is a prime not dividing n, then xn − 1 ∈ Fp[X] is separable, and hence Theorem 41applies. We work with the notation of Theorem 41; in particular,
R = Z[ζn] where ζn = e2π in
Fp ⊂ F is the splitting field of xn − 1 ∈ Fp[X], and ψ : R → F the ring homomorphismwhose existence is proved in Theorem 41. According to that theorem ψ gives a set-theoreticbijection from the set of roots of xn− 1 in Q(µn) to the set of roots of xn− 1 in Fp[X]; thesesets are the groups µn = µn(R) (recall that R is the Z-subalgebra of K generated by theroots of xn − 1 in K) and µn(F ) and by what we said above about functoriality of µn
ψ : µn(R)→ µn(F )
is a group homomorphism. But we just said that it is also a set-theoretic bijection, thereforeit is a group isomorphism. Because µn = µn(R) is a cyclic group of order n (generated, for
instance, by ζn = e2π in ), this implies that µn(F ) is also cyclic of order n.16 For both groups,
the group of automorphisms is canonically (Z/nZ)×.17
15I should say “a” splitting field but I have in mind the model Q(µn) = Q(ζn) ⊂ C where ζn = e2π in .
16We knew this already — every finite group of the multiplicative group of a field is cyclic — but this is anew proof.
17I don’t want to make too much of a big deal, but the point is this. Let Cn be a cyclic group of order n. Ihave not chosen a generator. An element k ∈ (Z/nZ)× acts on Cn as g 7→ gk. This gives (Z/nZ)× = AutCn.
25
Consider the Frobenius automorphism Frp : F → F . Frp acts on µn(F ) as the groupautomorphism
Frp : z 7→ zp
By Theorem 41 there exists σ ∈ Gn such that Frp ψ = ψσ, and σ acts on µn = µn(R) also asz 7→ zp.
To summarise we have shown: for every prime p not dividing n, there is an element ofGn that acts on µn as z 7→ zp.
All we need is to show that (Z/nZ)× is generated as a group by the classes of primes p suchthat hcf(p, n) = 1, and this is completely obvious: take any k ∈ (Z/nZ)× and decomposeit into primes: by the argument with Frobenius lifts all those primes are in Gn ⊂ (Z/nZ)×,hence also k ∈ Gn, hence Gn = (Z/nZ)×.
13 Texts
All undergraduate texts on Galois Theory go back to Emil Artin’s treatment [Art44]. Becauseof this fact, almost any book will do, in that it is probably not much better than a moreor less good copy of Artin. In my day I studied [Her75] and I still like it very much. Ialso recommend the notes by my friend and long-time collaborator Miles Reid [Rei], whichyou can find at https://homepages.warwick.ac.uk/~masda/MA3D5/. I am a big fan ofthe Expository papers by Keith Conrad, see https://kconrad.math.uconn.edu/blurbs/,where you will find several articles on Galois theory.
A big step forward was taken by Grothendieck [GR, Expose V] with his theory of theetale fundamental group (the axioms of a Galois category are listed at the beginning of§ 4). Perhaps surprisingly, his treatment did not yet, as far as I know, “trickle down” toundergraduate texts on the subject.
My treatment in this course is close in spirit to Grothendieck’s: I take an uncompromis-ingly “categorical” point of view where I express key definitions and statements in terms offield inclusions; never in terms of elements.
14 Questions
14.1 Worksheet 1
Q 1. This question is designed to get you up to speed with arithmetic in Q[x].
(a) Find the quotient and remainder when x5 + x+ 1 is divided by x2 + 1.
(b) Find the remainder when x2019 + 32x53 + 8 is divided by x− 1.
(c) Find polynomials s(x) and t(x) such that
(2x3 + 2x2 + 3x+ 2)s(x) + (x2 + 1)t(x) = 1.
[Bonus question: how did I know for sure that these polynomials were coprime?]
(d) Find a hcf for x4 + 4 and x3 − 2x+ 4. Express it as a(x)(x4 + 4) + b(x)(x3 − 2x+ 4).
(e) Find polynomials λ(x) and µ(x) such that (1 + x)λ(x) + (x3 − 2)µ(x) = 1.
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Q 2. Write ξ = 3√
2. This question is about understanding the field operations in Q(ξ)explicitly.
(a) Find rational numbers a, b and c such that a+ bξ + cξ2 = 1/(1 + ξ).[Hint : use Part (e) of the previous question.]
(b) Fix rational numbers A, B. Find rational numbers a, b and c (depending on A, B)such that a+ bξ + cξ2 = 1/(A+Bξ + ξ2).
Q 3. Prove that if f, g ∈ K[x] and at least one is non-zero, and if s, t are both hcf’s of f andg, then s = λt for some λ ∈ K×.
Q 4. (a) We know that whether or not a polynomial is irreducible depends on which fieldit’s considered as being over: for example x2 + 1 is irreducible in Q[x] but not in C[x]. Butshow that the notion of divisibility does not depend on such issues. More precisely show thatif K ⊆ L are fields, if f, g ∈ K[x], and if f | g in L[x] then f | g in K[x].
(b) Is it true that if f, g ∈ Z[x] and f | g in Q[x] then f | g in Z[x]? [Hint : No.] Is it trueunder the extra assumption that f is monic? [Hint : Yes.]
Q 5. (a) Prove that if n ∈ Z and√n 6∈ Z then
√n 6∈ Q.
(b) Prove that√
3 6∈ Q(√
2). What is the minimum polynomial of√
3 over Q(√
2)?
(c) Use the Tower Law to prove that [Q(√
2,√
3) : Q] = 4. Write down a basis for theQ-vector space Q(
√2,√
3).
Q 6. (a) Prove that Q(√
2,√
3) = Q(√
2 +√
3).[Hint : the smallest subfield of the complex numbers containing Q and
√2 +√
3 mustcontain loads of other things too: write some of them down.]
(b) Deduce that x4 − 10x2 + 1 is irreducible over Q.
Q 7. Is√
10 ∈ Q(√
6,√
15)?
Q 8. Prove that: If K ⊆ L ⊆ E are fields, and one of [L : K] or [E : L] is infinite, then[E : K] is infinite.
Remark: for those unfamiliar with infinite-dimensional vector spaces, a vector space Vover a field is infinite-dimensional iff it has no finite spanning set, iff for every n ≥ 1 thereexist n elements v1, v2, . . . , vn which are linearly independent.
Q 9. (a) Prove that if K ⊆ L is a finite extension of fields and V/L is a finite-dimensionalvector space then dimK(V ) = [L : K] dimL(V ).
(b) Prove that if K ⊆ L ⊆ E and [E : K] = [L : K] is finite, then L = E.
Q 10. Let K be a field with char(K) 6= 3 and such that f(x) = x3 − 3x + 1 ∈ K[x] isirreducible. Let L = K(α) where α is a root of f(x). Show that f splits completely over L.
[Hint : Factor f over L[x] as (x − α)g(x). Now solve for g(x) = 0 in L observing that12− 3α2 = (−4 + α + 2α2)2.]
27
Q 11 (†). Let K be a field of characteristic 0 containing an element ω ∈ K with
ω2 + ω + 1 = 0 .
(For example you can take K = Q(ω) where ω = exp 2πi3
.) In this question we carve atrick-free path to the formula for the solutions of the equation
y3 + 3px+ 2q = 0 (†)
(where p, q ∈ K) that only involves taking radicals (i.e., n√
of something).We assume that K ⊂ L is the splitting field of the polynomial of Equation (†) and we
denote by α1, α2, α3 ∈ L the three roots. (You can already prove that such a field extensionexists but I don’t care that you do this here.)
We know that the Galois group G permutes the three roots.
(a) Write the action of the cyclic permutation σ = (123) on the elements
u = α1 + ωα2 + ω2α3, v = α1 + ω2α2 + ωα3 .
and conclude that σ(u) = ωu and σ(v) = ω2v.18
(b) Find a formula expressing the three roots α1, α2, α3 in terms of u and v.[Hint : α1 + α2 + α3 = 0.]
(c) Consider the transposition τ = (23): show that τ(u) = v and τ(v) = u, and henceargue that u3 + v3 and u3v3 are fixed by all of S3 — and hence by all of G, irrespectiveof what G is. In other words, it follows from the Galois Correspondence that u3 + v3 andu3v3 ∈ K: show that this is indeed the case by finding explicit formulas for these quantities.Thus write down an explicit quadratic polynomial in K[X] of which u3, v3 are the two roots.Solve the quadratic equation, and combine with (b) to derive the cubic formula.
14.2 Worksheet 2
Q 12. In this question, if α ∈ R>0 and n ∈ Z≥1 then by α1/n or n√α I mean the unique
positive real number β with βn = α. (This removes ambiguities about a general complexnumber having n complex roots in this question).
(i) Set γ = (1 +√
3)1/3. Prove that γ is algebraic19. What is its degree over Q? What isits degree over Q(
√3)?
(ii) Set δ = (10 + 6√
3)1/3. Prove that δ is algebraic. What is its degree over Q? What isits degree over Q(
√2)?
Q 13. In this question we’ll find an explicit complex number z such that z 6∈ Q(z) (by z Imean the complex conjugate of z.)
(a) Set ω = e2πi/3, so ω3 = 1, and say α = 21/3 ∈ R the real cube root of 2. Set z = ωα.What is [Q(z) : Q]? [Hint: minimal polynomial ].
(b) What is [Q(ω) : Q]?
18Whether or not there is an element of G that acts as σ on the three roots is not relevant at this point.Such an element may or may not exist.
19By definition a complex number z ∈ C is algebraic if it is the root of a polynomial with coefficients in Q.
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(c) Let’s assume temporarily that z ∈ Q(z). Show that this implies ω ∈ Q(z). Why doesthis contradict the tower law? Deduce z 6∈ Q(z).
(d) Let’s write z = x+ iy. Prove that none of x, i or y are in Q(z).
Q 14. The purpose of this question is to make the proof of the Gauss’ Lemma more digestible.
(a) Prove that if R is a commutative ring with 1, and x ∈ R then 0x = 0.
(b) Prove that if K is a field and a, b ∈ K are both non-zero, then ab 6= 0.
(c) If K is a field and f =∑d
i=0 aixi ∈ K[x] is a non-zero polynomial, then (by choosing
d sensibly) we may assume ad 6= 0; we call adxd the leading term of f , and d the degree of f ,
and we write d = deg(f). Prove that if f, g ∈ K[x] are non-zero, then fg is also non-zero,and deg(fg) = deg(f) + deg(g).
(d) Prove that if f, g, h ∈ K[x] and h 6= 0 and fh = gh, then f = g (the cancellationproperty for polynomial rings).
[Some of you will know that f, g 6= 0 =⇒ fg 6= 0 is the assertion that K[x] is an integraldomain. Note that we used in class the fact that Fp[X] is an integral domain in the proofof the Gauss Lemma.]
Q 15. Factor the following polynomials in Q[x] into irreducible ones, giving proofs that yourfactors really are irreducible.
(i) x3 − 8;
(ii) x1000 − 6;
(iii) x4 + 4;
(iv) 2x3 + 5x2 + 5x+ 3;
(v) x5 + 6x2 − 9x+ 12;
(vi) x73 − 1;
(vii) x73 + 1;
(viii) x12 − 1.
Q 16. Prove that if α = 21/10 then Q(α) has a basis 1, α, α2, . . . , α9.
Q 17. Let A denote the set of complex numbers that are algebraic (over Q).
(i) Prove that A is a field.
(ii) Prove that [A : Q] = ∞. [Hint: you can use Eisenstein to construct irreduciblepolynomials of large degree.]
(iii) Prove that A is a countable set.
(iv) Prove that [C : A] =∞.
Q 18. The Eisenstein criterion is not a brilliant way to decide if a polynomial f(x) ∈ Z[x] isirreducible. A much better strategy is to choose a prime p and show that the reduction of fmodulo p is irreducible in Fp[x].
Make a list of all irreducible polynomials of degree ≤ 3 in F2[x], F3[x], and F5[x]. (Thisis most definitely NOT a stupid thing to do.)
29
Q 19 (†). (a) Suppose that a, b ∈ K are such that a is a square in K(√b).20 Prove that
either a or ab is a square in K.
(b) Let a, b ∈ K and suppose that b is NOT a square in K; let L = K(β) with β2 = b.Prove that: If one of a + β, a − β is a square in L, then so is the other, and deduce thatc = a2 − b is a square in K.
(c) Let a, b ∈ K and set c = a2 − b; suppose that none of b, c, or bc is a square in K. IfL is a splitting field of the polynomial:
(x2 − a)2 − b ∈ K[x],
prove that [L : K] = 8.[Hint: use Part (a) and Part (b) repeatedly.]
Q 20 (†). Suppose that char(K) 6= 2, and let K ⊂ L be a field extension of degree 4. Provethat the following two conditions are equivalent:
(i) There exists a (nontrivial) intermediate field K ⊂ E ⊂ L;
(ii) L = K(α) for some α ∈ L having minimal polynomial over K of the form:
f = x4 + ax2 + b ∈ K[x] .
Q 21. Say F is the splitting field of x3− 11 over Q. Figure out the Galois group Gal(F/Q).List all the subfields of F , all the subgroups of the Galois group, and draw a picture of theGalois correspondence.
Q 22. Say E = Q and let F be the splitting field of x4 − p, where p is a prime number.What is [F : E]? What is Gal(F/E)? [Hint: you can just go ahead and do the question, butyou may find Question 19 helpful.]
14.3 Worksheet 3
Q 23. Prove that Q ⊂ Q(√
2,√−3, 3√
5) is a normal extension. What is its degree?
Q 24. Establish (with proofs) whether the following extensions of Q are normal or not:
(i) Q(√
6);
(ii) Q(√
2,√
3);
(iii) Q(71/3);
(iv) Q(71/3, e2πi/3);
(v) Q(√
1 +√
7);
20In general if K is a field we say that a ∈ K is a square iff the polynomial X2 − a ∈ K[X] splits into twolinear factors, or, equivalently, there exists α ∈ K such that α2 = a.
30
(vi) Q(√
2 +√
2).
Q 25. (a) Prove that if E ⊂ F and [F : E] = 2 then the extension is normal.(b) Prove that every index 2 subgroup of a group G is normal.
Q 26. Say E = Q and let F be the splitting field of xp− 1, where p is an odd prime number.
(i) What is [F : E]? What is Gal(F/E)?
(ii) Prove that there is a unique subfield K of F with [K : Q] = 2 [Hint: Part (i), plus thefact that (Z/pZ)× is cyclic]. Show that all such extensions are of the form K = Q(
√n) where
n ∈ Z and |n| is squarefree.21 Figure out n when p = 3. Figure out n when p = 5 [Hint: whatis cos(2πi/5)? ]. What do you think the answer is in general? (This is a number-theoreticquestion rather than a field-theoretic one so don’t get frustrated if you see a good-lookingstatement but you can’t prove it: there are tricks but they’re tough to spot even for me.)
Q 27. (i) Say a, b > 1 are distinct squarefree integers. Prove x2 − a is irreducible, so Q(√a)
has degree 2 over Q. Now prove that√b 6∈ Q(
√a).
(ii) Let F be the splitting field of (x2 − a)(x2 − b) over Q. What is Gal(F/Q)? Use thefundamental theorem of Galois theory to find all the fields K with Q ⊆ K ⊆ F . Which onesare normal over Q?
(iii) Prove that F = Q(√a +√b). [Hint: figure out which subgroup of the Galois group
this field corresponds to.]
(iv) Let p, q and r be distinct primes. Prove√r 6∈ Q(
√p,√q). [Hint: use one of the
previous parts.]
(v) Conclude that if F = Q(√p,√q,√r) then [F : Q] = 8. What is Gal(F/Q)?
(vi) Use the fundamental theorem of Galois theory to write down all the intermediatesubfields between Q and F . If you can’t then just write down the subfields E of F with[E : Q] = 2.
(vii) Show that (notation as in the previous part) F = Q(√p+√q +√r).
(viii) Prove that if p1, p2,. . . ,pn are distinct primes, then Q(√p1,√p2, . . . ,
√pn) has degree
2n over Q, and equals Q(√p1 +
√p2 + · · ·+√pn).
Q 28. Say r =11√
51/3 +√
81/5 + 6 + 91/7. Find a sequence of fields Q = F0 ⊆ F1 ⊆ F2 ⊆· · · ⊆ Fn with r ∈ Fn and such that for all i we have Fi = Fi−1(αi) with αnii ∈ Fi−1 for somepositive integer ni.
Q 29. Fix a normal and separable extension of fields K ⊂ L and let G be the Galois group.Recall the standard notation of the Galois correspondence: for K ⊂ F ⊂ L, F † ⊂ G is thegroup that fixes F ; for H ≤ G, H? is the fixed field of H.
(a) Let K ⊂ F be an intermediate field. Let X = EmbK(F,L).22 Observe that composi-tion of functions gives a natural (left) action of G on X. Show that this action is transitive,that is, for all x, y ∈ X there is g ∈ G with gx = y. Why does this generalise the statement
21A natural number is squarefree if it is the product of distinct primes.22Feel free to assume that X is nonempty.
31
about the transitive action of G on the roots of a polynomial? For x in X denote by G thestabiliser of x:
Gx = x ∈ G | gx = xProve that Gx = x†, i.e., the group that fixes F where F is viewed as an intermediate fieldvia the K-inclusion x : F → L.
(b) Let K ⊂ F ⊂ L be an intermediate field and H = F † the corresponding subgroup,i.e., H ≤ G is the subgroup that fixes F and F = H? is the fixed field of H. Show thatK ⊂ F is normal if and only if H ≤ G is a normal subgroup. Show that in this case K ⊂ Fis separable (obvious) and EmbK(F, F ) = H\G.
(c) More generally show that for all K ⊂ F ⊂ G and H = F †:
EmbK(F, F ) = H\NG(H)
where NG(H) = g ∈ G | gHg−1 = H is the normaliser of H in G.23 (By constructionH ≤ N(H) is a normal subgroup and we are allowed to form the quotient group H\N(H).)
(d) Here and below, for H1, H2 ≤ G, write
N(H1, H2) = g ∈ G | gH1g−1 ⊃ H2
(I don’t know what this thing is called in Algebra.) Show that the assignment
g, h1, g 7→ gh1
defines a right action N(H1, H2)×H1 → N(H1, H2). Here and below, denote by
Mor(H2, H1) = N(H1, H2)/H1
the quotient set.
Now suppose given two intermediate fields, K ⊂ F1 ⊂ L and K ⊂ F2 ⊂ L. As usual forclarity denote by x1 : F1 → L and x2 : F2 → L the two inclusions, and let H1 = x†1, H2 = x†2.Prove that
EmbK(F1, F2) = Mor(H1, H2)
(e) In this Part, G is a group and H1, H2, etc. are subgroups of G. Show that thefunction:24
T : N(H1, H2)→ Fun(H2, H1)
where N(H1, H2) is as in Part (d) and T : g 7→ Tg, the function such that
Tg(h) = g−1hg,
in fact lands in the set Hom(H2, H1) of group homomorphisms from H2 to H1.Note that the set N(H1, H2) is in general not a group, but that there is a natural com-
position law:N(H1, H2)×N(H2, H3)→ N(H1, H3)
23Those of you who are taking Algebraic Topology should compare this statement with a similar statementin the theory of covering spaces.
24For X1, X2 sets, I denote by Fun(X1, X2) the set of functions from X1 to X2.
32
Recall that the centraliser of H ≤ G is the subgroup
C(H) = g ∈ G | for all h ∈ H, hg = gh
show that g, z 7→ gz defines a left action C(H2) × N(H1, H2) → N(H1, H2) of C(H1) onN(H1, H2), and that for all g1, g2 ∈ N(H1, H2), Tg1 = Tg2 if and only if there exists z ∈ C(H2)such that g2 = zg1.
(f †) As in Part (b), for subgroups H1, H2 of G write:
Mor(H1, H2) = N(H2, H1)/H2
the quotient set. Show that there is a natural composition Mor(H1, H2) ×Mor(H2, H3) →Mor(H1, H3) that makes the set of subgroups of G into a category.
(g †) Show that the Galois correspondence is a contravariant equivalence of categories,from the category whose objects are intermediate fields K ⊂ F ⊂ L, and where the set ofmorphisms from F1 to F2 is EmbK(F1, F2), to the category of subgroups defined in Part (e).In other words we have identifications
EmbK(F1, F2) = Mor(H2, H1)
compatible with composition.
Q 30. Let K ⊂ F be a field extension and K ⊂ L a normal field extension. Assume giventwo K-embeddings x1 ∈ EmbK(F,L), x2 ∈ EmbK(F,L):
L L
F
x1
``
x2
??
K
OO
(so I am saying that xi|K is the inclusion of K in L given at the beginning).(a) Show that there is a K-embedding y : L→ L such that y x1 = x2.(b) In the same situation as above, let now F ⊂ E be a field extension. For i = 1, 2 denote
by Embxi(E,L) the set of field homomorphisms x : E → L such that x|F = xi. Use part(a) to produce a bijective correspondence from Embx1(E,L) to Embx2(E,L). (In particular,this shows that one set is empty if and only if the other is empty.)
Q 31. Let K ⊂ F ⊂ L be a tower of field extensions. As we stated in class, it is immediatefrom the definition that: If K ⊂ L is normal, then F ⊂ L is also normal.
(a) If K = Q, F = Q(21/3) and L = Q(21/3, ω) with ω = e2πi/3, then show that K ⊂ L isnormal, but K ⊂ F is not normal.
(b) If K = Q, F = Q(√
2) and E = Q(21/4), show that K ⊂ F and F ⊂ L are normal,but K ⊂ L is not normal.
(c) Say H ⊆ K ⊆ G are groups. Prove that if H is normal in G then H is normal inK. Give an example of groups with H is normal in G but K not normal in G. Now give anexample with H normal in K, K normal in G, but H not normal in G. Now wonder whetherthis is all a coincidence or not.
33
14.4 Worksheet 4
Q 32. Say E ⊆ F is an extension of fields with [F : E] finite, and M,N are both subfieldsof F containing E. Assume that M/E and N/E are both normal.
(i) Prove that (M ∩N)/E is normal.(ii) Prove that MN (this notation means “the smallest subfield of F containing both M
and N”) is normal over E as well.
Q 33 (†). (a) Let K ⊂ L be a finite field extension. A finite extension L ⊂ Ω is a normalclosure of K ⊂ L if
1. K ⊂ Ω is normal, and
2. Ω is generated (as a ring, or field) by ∪σσ(L), the union over all K-embeddings σ : L→Ω.
Prove that a normal closure always exists, and that any two normal closures are isomorphicover L.
(b) Let K ⊂ L ⊂ Ω be finite field extensions. Assume:
1. Ω is generated (as a ring, or field) by ∪σσ(L), the union over all K-embeddings σ : L→Ω.
2. The set EmbK(L,Ω) has [L : K]s embeddings.
Then K ⊂ Ω is normal, and hence it is a normal closure of K ⊂ L.
Q 34. Here I ask you again to go through the example of an inseparable extension given inclass.
Let k be any field of characteristic p (for example Fp = Z/pZ), and let L = k(T ), thefield of fractions of the polynomial ring k[T ]. This means that a typical element of L is ofthe form f(T )/g(T ) with f and g polynomials, and g 6= 0. You can convince yourself thatthis is a field by checking that the sum, product etc of such things is of the same form.
Set K = k(T p), the subfield of L consisting of ratios f(T p)/g(T p).
(a) Convince yourself that K really is a subfield of L;
(b) Check that L = K(T ), the smallest subfield of L containing K and T ;
(c) Check that T is algebraic over K and hence [L : K] is finite;
(d) Check that q(x) = xp − T p is an irreducible element of K[x];25
(e) Deduce that q(x) is the min poly of T over K, and is also an inseparable polynomial inK[x];
(f) Deduce that L/K is not a separable extension.
25Hint: suppose it was reducible, and factor it in K[x]. The same factorization would work in L[x]. ButL[x] is a unique factorization domain. Spot that p(x) = (x − T )p in L[x]. By looking at constant terms,convince yourself that this gives a contradiction.
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Q 35. Say E ⊆ F , and L and M are intermediate fields (i.e. E ⊆ L,M ⊆ F ). Let N := LMdenote the smallest subfield of F containing L and M .
(i) If L = E(α1, . . . , αn) then prove N = M(α1, . . . , αn).
(ii) Now assume L/E and M/E are finite and normal. Prove N/E is finite and normal.(hint: splitting field). Next assume L/E and M/E are finite, normal and separable. Provethat N/E is finite, normal and separable.
(iii) Prove that restriction of functions gives a natural injective group homomorphismfrom Gal(N/E) to Gal(L/E)×Gal(M/E). Is it always surjective?
Q 36. (a) Prove that the polynomials
f(x) = x3 + x+ 1, g(x) = x3 + x2 + 1 ∈ F2[x]
are irreducible. Consider the fields K = F2(α), L = F2(β) where α, β are roots of f , g. Ifσ : K → L is a field isomorphism, what are the possible values of σ(α) ∈ L written in thebasis 1, β, β2 of L as a F2-vector space? Explain why K and L are isomorphic. How manyfield isomorphisms σ : K → L are there?
(b) Let L be the same as in Part (a). Consider the polynomial
h(x) = x4 + x+ 1 ∈ F2[x] .
Prove that h is irreducible in F2[x], or else exhibit a factorisation. Let L ⊂ E be the splittingfield of h — seen as a polynomial in L[x]. Is the extension F2 ⊂ E normal? Is it separable?What is the degree [E : F2]? Prove that h ∈ L[x] is irreducible, or else exhibit a factorisation.
Q 37. Show that if G is a transitive subgroup of Sn containing a (n − 1)-cycle and atransposition, then G = Sn.
Q 38. Consider the polynomial:
f(x) = x6 − 12x4 + 15x3 − 6x2 + 15x+ 12
(a) By considering how f(x) factorises in Fp[x] for small primes p, either prove thatf(x) ∈ Q[x] is irreducible, or exhibit a factorisation.
(b) Let Q ⊂ K be the splitting field of the polynomial in (a). Determine the Galois groupof the extension Q ⊂ K.
Q 39. Consider the polynomial
f(x) = x4 + x2 + x+ 1 ∈ Q[x]
(a) By considering how f(x) factorises in Fp[x] for small primes p, either prove thatf(x) ∈ Q[x] is irreducible, or exhibit a factorisation.
(b) Let Q ⊂ K be the splitting field of the polynomial in (a). Determine the Galois groupof the extension Q ⊂ K.
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Q 40. Consider the polynomial
f(x) = x4 + 3x+ 1 ∈ Q[x]
(a) Show that f(x) is irreducible in F2[x] and compute its prime factorisation in F5[x].
(b) Show that: if G is a transitive subgroup of S4 that contains a 4-cycle and a 3-cycle,then G = S4.
(c) Determine the structure of the Galois group of the splitting field of f over Q.
Q 41. (a) Show that for all prime p and all integer n > 0 there exists an irreducible monicpolynomial of degree n in Fp[x].
(b) Let g(x) ∈ F2[x] be an irreducible monic polynomial of degree n; h(x) ∈ F3[x] anirreducible monic polynomial of degree (n − 1); p > n − 2 a prime and k(x) ∈ Fp[x] anirreducible monic quadratic polynomial. Show that there is a monic polynomial f(x) ∈ Z[x]such that f(x) ≡ g(x) mod 2, f(x) ≡ xh(x) mod 3, and f(x) ≡ x(x+ 1) · · · (x+n− 3)k(x)mod p.
[Hint. Chinese remainder theorem.]
(c) If f is the polynomial in (b), show that the Galois group of the splitting field over Qof f is Sn.
Q 42. In this question ζ = e2π i
6 .
(a) Factorise the polynomial x6 − 1 ∈ Q[x]. Hence or otherwise determine the degree[Q(ζ) : Q].
(b) Show that the polynomial f(x) = x6 + 3 ∈ Q[x] is irreducible. Let Q ⊂ K be thesplitting field of f(x). What is the degree [K : Q]?. Determine the Galois group G of theextension Q ⊂ K and describe, perhaps by drawing some picture(s), the action of G on theset of roots of f(x).
[Hint. Consider first the field Q(α) where f(α) = 0 and study the intersection Q(α) ∩Q(ζ).]
(c) Let Q ⊂ L be the splitting field of the polynomial g(x) = x6−3 ∈ Q[x]. Compute thedegree [L : Q], determine the Galois group G of the extension Q ⊂ L and describe, perhapsby drawing some picture(s), the action of G on the set of roots of g(x).
Q 43. For all integers 3 ≤ n ≤ 16, draw pictures illustrating the lattice of subgroups of theGalois group of the cyclotomic extension Q ⊂ Q(µn). Draw the corresponding picture ofsubfields Q ⊂ F ⊂ Q(µn). For each of these subfields, find “natural” generators.
If you feel brave, then do the case n = 17. (The Galois group (Z/17Z)× = C16 is notin and of itself very complicated. The field Q(µ17) is a tower of quadratic extensions but ittakes some elbow grease to determine at each stage what you are taking the square root of;in particular this leads to a formula for cos 2π
17involving just iterated square roots of rational
numbers. Gauss did this calculation in his teens and it led him to a construction of theregular 17-gon with ruler and compass. You don’t yourself need to get to the bitter end ofthe calculation: do the first couple of steps and then look up the last steps on Google.)
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15 Answers
15.1 Worksheet 1
A 1. (a) Doing long division we see x5 + x + 1 = (x3 − x)(x2 + 1) + 2x + 1 so the quotientis x3 − x and the remainder is 2x+ 1.
(b) If x2019 +32x53 +8 = q(x)(x−1)+r(x) then either r(x) = 0 or deg(r) < deg(x−1) = 1so in either case r is a constant. Evaluating the equation at x = 1 shows us that r(x) =1 + 32 + 8 = 41.
(c)
2x3 + 2x2 + 3x+ 2 = (2x+ 2)(x2 + 1) + x
x2 + 1 = (x)(x) + 1
x = x× 1 + 0
so the last non-zero remainder is 1. Now working backwards,
1 = (x2 + 1)− (x)(x)
= (x2 + 1)− x[2x3 + 2x2 + 3x+ 2− (2x+ 2)(x2 + 1)]
= (2x2 + 2x+ 1)(x2 + 1)− x(2x3 + 2x2 + 3x+ 2)
so, if I got it right, one possibility is s(x) = −x and t(x) = 2x2 + 2x+ 1. If you got anothersolution it doesn’t mean you are wrong, because there is more than one answer to this sortof question just as in the case of usual integers—for example, you can add x2 + 1 to s andsubtract 2x3 + 2x2 + 3x + 2 from t and get a new solution that still works (another bonusquestion: what’s the most general solution? Can you prove it?).
Bonus part: I knew they were coprime in Q[x] because they have no roots in common inthe bigger ring C[x] – it’s easy to check this because the roots of x2 + 1 are ±i and neitherof these is a root of 2x3 + 2x2 + 3x+ 2, as you can see by substituting in. Can you see whythis is enough?
(d) Euclid again:
x4 + 4 = x(x3 − 2x+ 4) + 2x2 − 4x+ 4
x3 − 2x+ 4 = (x/2 + 1)(2x2 − 4x+ 4) + 0
and after that mercifully short procedure we see that the last non-zero remainder is 2x2 −4x + 4. Now hcf’s don’t really care about constants, so x2 − 2x + 2 is another hcf which iskind of nicer (in my opinion), but let’s work with what we have and go backwards:
2x2 − 4x+ 4 = (x4 + 4)− x(x3 − 2x+ 4)
oh and that’s it isn’t it – there are serious advantages to Euclid only taking 2 steps! Soa(x) = 1 and b(x) = −x. Actually I see now that the “nicer” hcf wasn’t perhaps so nicebecause then we would have had fractions in a and b.
(e) Just one long division gives:
−1
3(x3 − 2) +
x2 − x+ 1
3(x+ 1) = 1
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A 2. (a) The previous question, part (e), suggests that we take a = 1/3, b = −1/3, c = 1/3.
(b) The matrix of multiplication by A+Bξ + ξ2 in the basis 1, ξ, ξ2 is:
T =
A 2 2BB A 21 B A
We find a, b, c by solving the system:
T
abc
=
100
The Cramer rule gives, setting
D = detT = A3 + 2B3 − 6AB + 4,
the expressions:
a =A2 − 2B
D, b =
−AB + 2
D, c =
B2 − AD
(What does D = 0 mean?)
A 3. The hcf has the property that all other common divisors divide it. So by definitions | t and t | s, so looking at top degree terms we deduce that the degrees of s and t must beequal, and s = tr for a polynomial r of degree 0, that is, a non-zero constant.
A 4. (a) Divide g by f in K[x] and get a quotient and a remainder, and then pretendeveything is in L[x] and use uniqueness of quotient and remainder to do this part immediately.
(b) First part no, e.g. 2x + 2 | x + 1 in Q[x]. Second part yes, and again prove it byfiguring out q(x) such that g(x) = f(x)q(x) by long division and noting that you only everhave to divide by 1 when figuring out the coefficients of g.
A 5. (a) If√n = p/q in lowest terms (with p, q ∈ Z and q 6= 0) then we deduce that nq2 = p2.
In particular q2 divides p2 – but q2 and p2 are coprime, so q2 = 1, so p/q ∈ Z.
(b) We know Q(√
2) = a+b√
2 : a, b ∈ Q. We now prove√
3 6∈ Q(√
2) by contradiction.If√
3 = a + b√
2 with a, b ∈ Q then squaring both sides and tidying up, we deduce2ab√
2 ∈ Q. But√
2 6∈ Q by part (a), so 2ab = 0, so either a = 0 or b = 0. If b = 0 then√3 ∈ Q, contradicting part (a). If a = 0 then
√3 = b
√2 and multiplying both sides by
√2
we deduce√
6 ∈ Q, also contradicting part (a). Either way we’re there, so√
3 6∈ Q(√
2).The min poly of
√3 over Q(
√2) must then be x2−3. Why? It’s monic, and has coefficients
in the right field, so the only issue is whether it’s irreducible. And it is, because if it factoredthen it would have to factor into two linear factors, and one of them would be (up to aconstant) x −
√3, but we’ve just shown that this polynomial does not have coefficients in
Q(√
2).
(c) [Q(√
2,√
3) : Q] = [Q(√
2,√
3)/Q(√
2)][Q(√
2) : Q] = 2 × 2 = 4. We know bothextensions on the right have degree 2; for one it’s clear and for the other it comes frompart (b) and a result proved in class ([K(λ) : K] is the degree of the minimal polynomial ofλ over K).
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A 6. (a) If α =√
2 +√
3 then α2 = 5 + 2√
6 and hence√
6 = (α2 − 5)/2 ∈ Q(α). Henceβ :=
√6α =
√12 +
√18 = 2
√3 + 3
√2 ∈ Q(α). So
√2 = β − 2α ∈ Q(α) and now√
3 = α−√
2 ∈ Q(α).We deduce that Q(α) contains
√2 and
√3, so it contains Q(
√2,√
3). The converseinclusion is obvious, so the two fields are equal.
(b) p(x) = x4 − 10x2 + 1 can be checked to be a polynomial in Q[x] such that p(α) = 0.Hence it is a multiple of the minimal polynomial of α. But part (a) and [Q(α) : Q] =deg(minimal poly. of a) imply that the degree of the min poly of α is 4, so p(x) must be aconstant multiple of this min poly, so it must be the min poly, so it must be irreducible.
A 7. Answer is yes! It’s 13
√6√
15.
A 8. If [L : K] is infinite then E has an infinite-dimensional K-subspace and hence must beinfinite-dimensional over K (for any n ≥ 1 we can choose n K-linearly independent elementsof L and these give n K-linearly independent elements of E).
If [E : L] is infinite, then [E : K] must also be infinite, because for any n we can choosen elements of E that are L-linearly independent, and these are easily checked to also beK-linearly independent.
A 9. (a) This is a variant of the Tower Law argument: Let e1, . . . , en be a basis for L/Kand f1, . . . , fm be a basis for V/L. Then ei ∈ L and fj ∈ V , and V is an L-vector space, sogij = eifj makes sense. Of course the claim is that the gij form a basis for V considered as aK-vector space, and the same proof as in the tower law works: the gij span because if v ∈ Vthen write v as an L-linear combination of the fj and then write each coefficient as a K-linearcombination of the ei, and multiply out. For linear independence, if a linear combination∑
i,j µijgij = 0 then write this as∑
j(∑
i µijei)fj =∑
j λjfj and by linear independence ofthe fj over L we know the λj must be zero, and this means the µij are all zero by linearindependence of the ei over K.
(b) So?
A 10. Long division (for example) of f(X) by X − α yields:
f(X) = (X − α)(X2 + αX + (−3 + α2)
)∈ L[X]
The quadratic formula for g(X) needs the square root of
∆ = b2 − 4ac = α2 − 4(−3 + α2) = 12− 3α2
which is explicitly shown to be a square in the hint.
[Note: if char(K) = 3, then f(x) = x3 + 1 = (x+ 1)(x2 − x+ 1) is not irreducible.]
A 11. (a) We haveu 7→ α2 + ωα3 + ω2α1 = ωu
and, similarly, v 7→ ω2v.
(b) Using that ω + ω2 = −1 and α1 + α2 + α3 = 0, we get, for example:
u+ v
3=α1 + α1 − α2 − α3
3= α1
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and, similarly, α2 = ω2u+ωv3
, α3 = ωu+ω2v3
.
(c) It is pretty obvious that τ(u) = v and τ(v) = u. The rest of this question requiresconsiderable work and we may return to this point later in the lectures, when we study theGalois group of splitting fields of cubic polynomials in general.
We must use the following facts from M1F:
α1 + α2 + α3 = 0, (α2α3 + α1α3 + α1α2) = 3p, α1α2α3 = −2q
We will also need to use the (elementary) algebraic identity:
(z1 + z2 + z3)(z2z3 + z1z3 + z1z2) = (z21z2 + z2
1z3 + z1z22 + z2
2z3 + z1z23 + z2z
23) + 3z1z2z3
We compute by brute force uv and u3 + v3: from these quantities it is easy to construct thesought-for quadratic equation. A direct calculation (using ω + ω2 = −1!) shows that:
uv = α21 + α2
2 + α23 − α1α2 − α1α3 − α2α3 = (α1 + α2 + α3)2 − 3(α1α2 + α1α3 + α2α3) = −9p
and:
u3 + v3 = 2(α31 + α3
2 + α33)− 3(α2
1α2 + α21α3 + α1α
22 + α2
2α3 + α1α23 + α2α
23) + 12α1α2α3 =
= 2(α1 + α2 + α3)2 − 9(α21α2 + α2
1α3 + α1α22 + α2
2α3 + α1α23 + α2α
23) =
= 27α1α2α3 = −27× 2q
Now write down the quadratic equation and deduce the cubic formula!
15.2 Worksheet 2
A 12. (i) γ clearly satisfies (γ3− 1)2 = 3, so it’s a root of the polynomial (x3− 1)2− 3 whichis x6 − 2x3 − 2. By the Eisenstein criterion this polynomial is irreducible, so it must be themin poly of γ, and the degree of γ over Q is 6.
Note that√
3 = γ3−1 ∈ Q(γ) so if F = Q(γ) and K = Q(√
3) we must have Q ⊆ K ⊆ Fand the tower law gives 2[F : K] = [K : Q][F : K] = [F : Q] = 6, and we deduce [F : K] = 3.Because F contains
√3 it must contain K and it’s not hard to deduce that F = K(γ). By
the tower law again, the degree of γ over K must then be 3.Note that if one could show that x3 − (1 +
√3) were irreducible in K[x] then this would
be another way to do the question, but I did not explain any techniques for tackling this.
(ii) Even more evil trick question. Turns out δ = 1 +√
3 (cube it out to check) so thedegree is 2 over Q and also over Q(
√2), the latter because we saw in some previous question
that δ 6∈ Q(√
2) (it would imply√
3 ∈ Q(√
2)).
A 13. (a) Well z3 = ω3α3 = 1 × 2 = 2 so z is a root of x3 − 2 = 0, which is irreducibleover Q because it has no root in Q, so x3 − 2 is the min poly of z, and by what we did inclass this means [Q(z) : Q] = 3. Although we don’t need it, we can note that in fact Q(z) isisomorphic to, but not equal to, Q(α), as an abstract field.
(b) We know ω3 = 1 but ω 6= 1 so ω is a root of (x3 − 1)/(x − 1) = x2 + x + 1. Thispolynomial is irreducible as it has no rational (because no real) roots, so [Q(ω) : Q] = 2.
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Note also while we’re here that solving the quadratic gives ω = −1+i√
32
(plus sign becausethe imaginary part of ω is positive; the other root is ω2).
(c) We have α ∈ R. Furthermore ω is another cube root of 1 so it must be ω2. Hencez = ωα = ω2α = ωz. In particular if z ∈ Q(z) then ω = z/z ∈ Q(z). This meansQ(ω) ⊆ Q(z), and by the first two parts and the tower law we deduce [Q(z) : Q(ω)] = 3
2,
which is nonsense because the dimension of a (finite-dimensional) vector space is a wholenumber.
(d) If x ∈ Q(z) then z = −z + 2x ∈ Q(z), contradiction. So x is not in. If i ∈ Q(z) thenQ(i) ⊆ Q(z) and this contradicts the tower law like in part(c). Finally because the imaginarypart of ω is
√3/2 we see y = α
√3/2, so if y ∈ Q(ω) then y3 = 3α3/8
√3 = 3/4
√3 ∈ Q(z),
implying√
3 ∈ Q(z) which again contradicts the tower law.
A 14. (a) We know 0 is the additive identity in R so 0+0 = 0. Hence 0x = (0+0)x = 0x+0xand subtracting 0x (which we can do, because (R,+) is a group so 0x has an additive inverse)we deduce 0 = 0x.
(b) If a 6= 0 and b 6= 0 then there exist multiplicative inverses a−1 and b−1, and nowabb−1a−1 = 1 × 1 = 1. However if ab = 0 then we deduce 0(b−1a−1) = 1 which contradictspart (a) (as 0 6= 1 in a field).
(c) Look at top degree terms.
(d) fh = gh implies (f − g)h = 0, and if h 6= 0 we must have f − g = 0 by (c).
A 15. (i) Spot root x = 2; so x3 − 8 = (x − 2)(x2 + 2x + 4) and roots of the quadratic arenon-real and hence non-rational, so the quadratic must be irreducible (as any factors wouldbe linear).
(ii) Irreducible by Eisenstein (p = 2 or p = 3).
(iii) The polynomial x2−2x+2 is a factor; dividing out we see x4 +4 = (x2−2x+2)(x2 +2x + 2). Easy check now that both quadratics have non-real and hence non-rational roots,so must be irreducible.
(iv) Either this is irreducible over Q, or there is a root in Q (because any factorizationmust involve a linear term). So let’s substitute in x = p/q in lowest terms (i.e. gcd(p, q) = 1)and see what happens. Clearing denominators we get
2p3 + 5p2q + 5pq2 + 3q3 = 0.
Now p divides the first three terms of the left hand side, so must divide the fourth which is3q3. But p and q are coprime! So p must divide 3. A similar argument shows that q mustdivide 2. So p = ±1 or ±3 and q = ±1 or ±2. Clearly no positive rational is a root (as allthe coefficients are positive) so we are left with the possibilities x = −1,−1/2,−3,−3/2 andwe just try all of them. Miraculously x = −3/2 does work! Pulling off the correspondinglinear factor gives
2x3 + 5x2 + 5x+ 3 = (2x+ 3)(x2 + x+ 1)
and the quadratic term has no real roots and hence no rational ones, so this is the factorizationinto irreducibles.
(v) This one is irreducible by Eisenstein with p = 3.
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(vi) There’s an obvious factor of x − 1 and the other factor x72 + x71 + · · · + x + 1 isirreducible. To see this first substitute y = x− 1, then apply Eisenstein with p = 73 prime.
(vii) This polynomial is obtainable from the polynomial in part (vi): start with the part(vi) polynomial, change x to −x and then change the sign of the polynomial. These sortsof things do not affect things like irreducibility and factorization, so the factorization will be(x+ 1)(x72 − x71 + . . .− x+ 1) and the degree 72 polynomial will be irreducible.
(viii) Spot roots x = 1 and x = −1. Over the complexes we have more roots too, like ±iand so on – how do these control factorization over the rationals? Well (x− i) and (x+ i) arefactors over the complexes, so their product x2 + 1 is a factor over the complexes and hencealso over the rationals. Similarly the two complex cube roots of 1 are complex conjugatesand are the two roots of x2 + x + 1, and the two 6th roots of 1 that we haven’t mentionedyet (e
2π i6 and its complex conjugate) are roots of x2 − x + 1. So we’ve just spotted factors
whose degrees add up to 8. Let’s see what we have so far then: the factors we have spottedare
(x+ 1)(x− 1)(x2 + 1)(x2 + x+ 1)(x2 − x+ 1)
= (x2 − 1)(x2 + 1)(x2 + x+ 1)(x2 − x+ 1)
= (x4 − 1)(x4 + x2 + 1)
and so what is left is
(x12 − 1)/(x4 − 1)(x4 + x2 + 1)
= (x8 + x4 + 1)/(x4 + x2 + 1)
= x4 − x2 + 1
The hardest part of this question is figuring out whether that last polynomial x4 − x2 + 1factors.
A 16. The min poly of α must be x10 − 2 because this is irreducible over Q (by Eisenstein)and has α as a root. In particular there is no non-zero polynomial of degree at most 9 withrational coefficients and α as a root, so 1, α, α2, . . . , α9 are linearly independent elementsin a vector space of dimension 10, and hence are a basis.
A 17. (i) To check that a subset of a field is a subfield all we need to do is to check 0 and 1are in, and that the subset is closed under addition, subtraction, multiplication, and division-by-things-that-aren’t-zero. These things follow from the tower law: if α, β are algebraic then[Q(α) : Q] <∞ and [Q(β) : Q] <∞, but then
[Q(α, β) : Q] = [Q(α)(β) : Q(α)][Q(α) : Q] ≤ [Q(β) : Q][Q(α) : Q] <∞
But then for all λ ∈ Q(α, β) [Q(λ) : Q] <∞, i.e., λ is algebraic.
(ii) Say for a contradiction that [A : Q] = n <∞. Let p(x) = xn+1 − 2 and let α ∈ C bea root. Then α is algebraic and its min poly must be p(x) as p(x) is monic and irreducible.So n = [A : Q] = [A : Q(α)][Q(α) : Q] = (n+ 1)[A : Q(α)] ≥ n+ 1 > n, a contradiction.
(iii) For each n there are only countably many elements of Q[x] with degree at mostn, and a countable union of countable sets is countable, so there are only countably many
42
polynomials. Each algebraic number is a root of a non-zero polynomial in Q[x] and such apolynomial has only finitely many roots, and a countable union of finite sets is countable, soA is countable.
(iv) If [C : A] were finite then C would be isomorphic to An for some n ∈ Z≥1 and henceC would be countable, a contradiction.
A 18. This is tedious and I am not going to do it; but I will make a few comments.
A polynomial of degree ≤ 3 in Fp[x] is irreducible if and only if it has no roots in Fp, andthis can be checked by evaluating at all elements 0, . . . , p− 1 ∈ Fp.
For example with p = 2 the only irreducible quadratic polynomial is:
x2 + x+ 1
and the irreducible cubic polynomials are:
x3 + x+ 1, x3 + x2 + 1
For p = 3 there are 3 irreducible monic quadratic polynomials; they are:
x2 + 1, x2 + x− 1, x2 − x− 1
On the other hand, if you understand some of what we (by now) said about finite fields,there are (27− 3)/3 = 8 irreducible monic cubic polynomial, and it should not be too time-consuming to write them all down.
For p = 5, again if you understand the last part of the question in Test 2, there are(25−5)/2 = 10 irreducible monic quadratic (not too bad to list them all) and (125−5)/3 = 40irreducible monic cubic polynomials in F5[x] (OK so to do this by hand would be a bitridiculous. You can write your own computer program if you wish, or find a table on the’net).
A 19. (a) The statement is obvious if b is a square in K so let us assume that it is not.Suppose that there are x, y ∈ K such that
a = (x+ y√b)2 = (x2 + by2) + 2xy
√b
Since 1,√b are linearly independent over K, we must have that either
(i) y = 0, in which case a = x2 is a square in K, or
(ii) x = 0, in which case a = y2b and then ab = (yb)2 is a square in K.
(b) Suppose say that a + β is a square in L. This means that there are x, y ∈ K suchthat
a+ β = (x+ yβ)2 = (x2 + y2b) + 2xyβ
but then a− β = (x− yβ)2 is also a square in L, and
c = a2 − b = (a+ β)(a− β) = [(x+ yβ)(x− yβ)]2 = [x2 − y2b]2
43
is a square in L.
(c) The roots are ±√a±√b; so choose β, α, α′ ∈ L such that β2 = b, α2 = a + β,
α′ 2 = a− β. We work with the diagram
L
K(α)
::
K(α′)
dd
K(β)
cc ::
K
OO
First, [K(β) : K] = 2 since we are assuming that b is not a square in K.Write K1 = K(β). I claim that [K(α) : K1] = 2. Indeed, by Part (b), if a + β were a
square in K1, then also a− β would be a square in K1 and then c = (a+ β)(a− β) = a2 − bis a square in K, contradicting one of our assumptions.
Similarly, also [K(α′) : K1] = 2.The conclusion of Part (c) follows from the tower law and the new claim: K1(α) 6=
K1(α′). Indeed suppose for a contradiction that α′ ∈ K1(α): this is saying that a − β is asquare in K1(
√a+ β). From Part (a) with u = a− β and v = a+ β in K1, we conclude that
either:
(i) a− β is a square in K1, contradicting the claim proved that [K1(α′) : K1] = 2, or:
(ii) c = (a− β)(a+ β) = a2 − b is a square in K1.
Since the first alternative led to a contradiction, it must be that c is a square in K1. Weapply Part (a) again with u = c, v = b in K. We have c a square in K(
√b), that is, either
c or cb is a square in K, contradicting our assumptions. This final contradiction shows thatK1(α) 6= K1(α′) and finishes Part (c).
A 20. It is easy to see that (ii) implies (i) and here I focus on proving that (i) implies (ii).The key thing to understand is this: Claim If char(K) 6= 2 then every extension K ⊂ L
of degree [L : K] = 2 is of the form L = K(α) for some α ∈ L such that α2 ∈ K. I am goingto leave out the proof of the Claim (hint: quadratic formula) and I will use it to answer thequestion.
So assume (i), then by the tower law [L : E] = 2 and [E : K] = 2 and by the ClaimL = E(α) for some α ∈ L with α2 ∈ E. Also E = K(β) where β2 ∈ K. Hence we can writeα2 = u+ vβ with u, v ∈ K, so
(α2 − u)2 = v2β2 ∈ Khence α is a root of the polynomial
f(X) = (X2 − u)2 − v2β2 = X4 − 2uX2 + (u2 − v2β2) ∈ K[X]
which is of the required form. If f(X) ∈ K[X] is irreducible then we are done.
44
So what if f(X) is not irreducible? This is really awkward ! In that case by the tower law[K(α) : K] = 2 and the minimal polynomial of α over K is a quadratic polynomial
X2 + cX + d ∈ K[X]
and necessarily c = 0, otherwise α = −α2−dc∈ E, a contradiction. Hence in fact α2 ∈ K and
we have extensions:
L = K(α, β)
E = K(β)
77
F = K(α)
gg
K
66hh
where β2 = b ∈ K and α2 = a ∈ K BUT also, clearly, α 6∈ K(β) and β 6∈ K(α).
Remark there is a third field, G = K(αβ), distinct from E, F , and also of degree[G : K] = 2. Note also that (αβ)2 = ab ∈ K. (I leave all this to you to sort out.)
I now want to work with the element α+ β ∈ L: I claim that it has degree 4 over K, andthen L = K(α + β) and, since
(α + β)2 = a+ b+ 2αβ ∈ G, (3)
the argument above shows that the minimal polynomial of α + β has the required form.Suppose for a contradiction that α + β satisfies a quadratic polynomial
X2 + AX +B ∈ K[X]
If A = 0 then we have that (α + β)2 = −B ∈ K, and this implies (by Equation 3) that
αβ ∈ K, a contradiction. If A 6= 0 then α + β = −(α+β)2−BA
∈ G (Equation 3 again) and thepolynomial
g(X) = (X − α)(X − β) = X2 − (α + β)X + αβ
is in G[X]. This polynomial is irreducible, otherwise its roots α, β already belong to G, soL = G and we get a contradiction in too many ways (for instance [L : K] = [G : K] = 2).But then g(X) equals X2 − a, the minimal polynomial of α over K[X], and this then leadsto a contradiction in too many ways (for instance it implies that α = −β).
A 21. This is not difficult at all. Go back to your notes of the discussion of X3 − 2 at thebeginning of the course and make the appropriate minor changes.
A 22. Let’s start by adjoining one root of x4 − p, say, α, the positive real 4th root of p. Weget a field K = Q(α). By Eisenstein, x4 − p is irreducible over Q, so [K : Q] = 4. Is K asplitting field? No, because it’s a subfield of the reals, and x4 − p has some non-real roots(namely ±iα). However K does contain two roots of x4 − p, namely ±α, so x4 − p mustfactor as (x+α)(x−α)q(x), with q(x) ∈ K[x] of degree 2 and irreducible (as no roots in K).
45
If β = iα is a root of q(x) and F = K(β) then [F : K] = 2 so [F : Q] = 8 by the tower law.We can alternatively write F = K(i) as β = iα, so F = Q(i, α).
F is a splitting field over Q so it’s finite, normal and separable (separability isn’t anissue as we’re in characteristic 0). So we know Gal(F/Q) has size 8. We also know that ifτ : F → F is an isomorphism then τ(α) had better be a 4th root of τ(p) = p, so it’s ±αor ±iα; there are at most 4 choices for τ(α). Similarly τ(i) = ±i so there are at most 2choices for τ(i). This gives at most 8 choices for τ ; however we know that Gal(F/Q) hassize 8, so all eight choices must work. It is not hard now to convince yourself that Gal(F/Q)is isomorphic to D8 (think of a square with corners labelled α, iα,−α,−iα).
15.3 Worksheet 3
A 23. Write K = Q(√
2,√−3, 3√
5). Observe that
X3 − 5 = (x− 3√
5)(x− ω 3√
5)(x− ω2 3√
5)
where
ω =−1 +
√−3
2
is a primitive cube root of unity. It follows from this that K is the splitting field of thepolynomial
f(X) = (X2 − 2)(X3 − 5) ∈ Q[X]
indeed the polynomial splits completely in K and K is generated by the roots (if 3√
5 andω 3√
5 are both in F , then clearly ω is also in F ). Hence Q ⊂ L is a normal extension.
Now let us count degrees. First, let us state that√
2 6∈ Q, hence [Q(√
2) : Q] = 2. Nextconsider the field L = Q(
√−3,√
2). It is clear that, say,√−3 6∈ Q(
√2)—for example,
√−3
is purely imaginary while Q(√
2) ⊂ R. If you don’t like this, suppose for a contradiction that√−3 ∈ Q(
√2), that is there exist rational numbers x, y ∈ Q such that
−3 = (x+ y√
2)2 = x2 + 2y2 + 2xy√
2
since this is an identity in a 2-dimensional vector space over Q with basis 1,√
2 we must haveeither x = 0 or y = 0. If y = 0, then x2 = −3, x ∈ Q leads easily to a contradiction. If x = 0then −3 = 2y2. Writing y = p/q with p, q coprime integers, we have
−3q2 = 2p2
and we easily get a contradiction working 2− or 3−adically.26 By a simple application of thetower law then [L : Q] = 4.
26I am deliberately avoiding reaching a contradiction by means of the order structure of the rationals: theleft hand side is negative, the right hand side is positive. This would be reproducing the argument in termsof imaginary numbers that we wanted to avoid.
46
Finally let us consider our field K = L( 3√
5) and the diagram of field extensions:
K
L
@@
Q( 3√
5)
cc
Q
<<^^
I claim that x3 − 5 is irreducible in L[X] and hence [K : L] = 3 and then [K : Q] = [K :L][L : Q] = 3 × 4 = 12. Indeed if x3 − 5 were not irreducible in L[X] then it would havea root α ∈ L; and then from Q ⊂ Q(α) ⊂ L we would conclude from the tower law that[Q(α) : Q] = 3 divides [L : Q] = 4, a contradiction. Hence [K : Q] = 12.
A 24. (i) Q(√
6) is the splitting field of the polynomial x2 − 6 and is hence normal over Q.
(ii) Q(√
2,√
3) is the splitting field of (x2 − 2)(x2 − 3) and is hence normal.
(iii) Q(71/3) contains one, but not all, roots of the irreducible polynomial x3− 7 (becausethe other roots are not even real), so it is not normal over Q.
(iv) Q(71/3, e2πi/3) is the splitting field of x3 − 7 and is hence normal.
(v) Q(√
1 +√
7) is not normal over Q. Here’s why. If α =√
1 +√
7 then α2 − 1 =√
7,so (α2 − 1)2 = 7 and α is hence a root of x4 − 2x2 − 6 = 0. We can spot the four complex
roots of this polynomial; they are ±√
1±√
7 (just substitute in to see that all of these areroots). Two of these numbers are real and two pure imaginary; in particular not all of them
are in Q(√
1 +√
7), which is a subfield of the reals. However, x4− 2x2− 6 = 0 is irreducibleover Q (one can use the Eisenstein criterion, which I haven’t done yet, or argue in an adhocmanner, or, at this point, use the theory developed in class on biquadratic extensions), so
this polynomial has some but not all roots in Q(√
1 +√
7) which is hence not normal overQ.
(vi) Q(√
2 +√
2) is normal over Q, despite the formal similarity with part (v). If α =√2 +√
2 then (as in the previous question) we see (α2 − 2)2 = 2 and hence α is a root of
x4− 4x2 + 2 = 0. This polynomial is irreducible by Eisenstein, but in this case Q(√
2 +√
2)
is actually its splitting field. For two of its roots are ±α and the other two are ±√
2−√
2
and if β =√
2−√
2 then we see αβ =√
2 = α2 − 2, and hence β = (α2 − 2)/α ∈ Q(α)! Sothe extension is a splitting field and hence normal.
A 25. (a) If [F : E] = 2 let α ∈ F \ E, then consider the tower of field extensions E ⊂E(α) ⊂ F . As a simple consequence of the tower law we get that F = E(α). The minimalpolynomial of α over E has degree 2:
f(X) = X2 + aX + b ∈ E[X]
and X −α divides f(X) in F [X] hence f(X) splits completely in F , hence F is the splittingfield of f(X) hence E ⊂ F is a normal extension.
47
(b) Suppose that H ≤ G has index two. This means that there are two elements (cosets)in the quotient set X = H\G and also in the quotient set Y = G/H. Let g ∈ G be anyelement: if g ∈ H then clearly g−1Hg = H, so let us assume that g 6∈ H. It must be the casethat Hg = G \H AND gH = G \H; therefore Hg = gH.27
A 26. (i) We know xp−1 = (x−1)(1+x+x2 + · · ·+xp−1), and f(x) := 1+x+x2 + · · ·+xp−1
is irreducible over Q (by Eisenstein after a coordinate change). Hence if ζ = e2π i/p then f(x)must be the min poly of ζ. Note that the roots of p(x) are just the roots of xp− 1 other thanx = 1, so they’re ζj for 1 ≤ j ≤ p− 1. Moreover if F = Q(ζ) then [F : Q] = deg(f) = p− 1,and K contains ζj for all j, so xp − 1 splits completely in K. Hence K is the splitting fieldof xp − 1 and it has degree p− 1.
Now F/Q is finite, normal and separable, so the fundamental theorem applies, so we knowGal(F/Q) will have size p − 1. If τ ∈ Gal(F/Q) then, because F = Q(ζ), τ is determinedby τ(ζ), which is a root of τ(f) = f , so is ζj for some 1 ≤ j ≤ p − 1. It’s perhaps notimmediately clear that, given j, some field automorphism τ of F sending ζ to ζj will exist– but it has to exist because we know there are p− 1 field automorphisms. So the elementsof the Galois group can be called τj for 1 ≤ j ≤ p − 1. The remaining question is whatthis group is. We can figure out the group law thus: τi τj – where does this send ζ? Wellτj(ζ) = ζj, and τi(ζ) = ζ i so τi(ζ
j) = ζ ij as τi is a field homomorphism. Note finally that ζ ij
only depends on ij mod p, as ζp = 1. So if we identify Gal(F/Q) with 1, 2, . . . , p− 1 thenthe group law is just “multiplication mod p” , and we see Gal(F/Q) ∼= (Z/pZ)×.
If you will, check that our isomorphism (which seemed to depend on a choice of ζ, our pthroot of unity) is in fact independent of that choice, so Gal(F/Q) is canonically isomorphic to(Z/pZ)×. The notation in mathematics for a canonical isomorphism is “=”, so we can writeGal(F/Q) = (Z/pZ)× in this situation. This concludes part (i). I want to add to this that,in fact, (Z/pZ)× ∼= Cp−1 is always a cyclic group. (This is a non-completely trivial fact. Ingeneral, every finite subgroup of the multiplicative group of a field is cyclic. I don’t normallylike to prove this result — sometimes I give it as a worksheet question — but I encourageyou to look it up.)
As for part (ii), I discuss some ideas without giving a complete proof.When p = 3, K = F and hence K = Q(
√−3).
When p = 5, I claim thatK = Q(√
5). Indeed from part (i)G = (Z/5Z)× = 1, 2,−2,−1.It is clear that H = 1,−1 ⊂ G has index 2 and that K = H? in the notation of the Galois
correspondence. Writing as in part (i) ζ = e2π i5 , it is clear that
α = ζ +1
ζ∈ H?
and it is reasonable to guess K = Q(α). It is easy to finish from here:
α2 + α− 1 = 1 + ζ + ζ2 + ζ3 + ζ4 = 0
hence α = −1+√
52
and from this we conclude that K = Q(√
5).
27You are supposed to “see” that the two parts of the question correspond under the Galois correspondence.
48
For p general, writing as above ζ = e2πip , and denoting by H ≤ (Z/pZ)× the unique
subgroup of index 2, we want to evaluate something like∑h∈H
h(ζ)
because this thing being the average over all of H is manifestly H-invariant. The nextobservation is that H is the image of the “squaring homomorphism”
(Z/pZ)× 3 k 7→ k2 ∈ (Z/pZ)×
so we are led to evaluating:
α =
p−12∑
k=0
e2πik2
p
You can find this thing in number theory books under the name of “quadratic Gauss sum”and the upshot is
K =
Q(√−p) if p ≡ 3 mod 4
Q(√p) if p ≡ 1 mod 4
(The exact evaluation of the Gauss sum is a bit tricky, but you may be able to evaluate itup to sign, and this is enough to determine K. This, however, is a number theory question,not a Galois theory question.)
A 27. (i) a > 1 so a has a prime divisor p; now use Eisenstein. Or use uniqueness offactorization to prove
√a 6∈ Q.
Next, if√b ∈ Q(
√a) then write
√b = x + y
√a; square, and use the fact that
√a is
irrational to deduce that 2xy = 0. Hence either y = 0 (contradiction, as√b 6∈ Q) or x = 0
(contradiction, as we can write ab = cd2 with c squarefree, and a 6= b so c 6= 1, and again√c 6∈ Q).
(ii) F = Q(√a,√b) and the preceding part, plus the tower law, shows that [F : Q] = 4.
Now F is a splitting field in characteristic zero, so it’s finite, normal and separable. Bythe fundamental theorem, Gal(F/Q) must be a finite group of order 4, so it’s either C4 orC2×C2. There are lots of ways of seeing that it is actually C2×C2. Here are two that springto mind: firstly, C4 only has one subgroup of order 2, whereas F has at least two subfieldsof degree 2 over Q, namely Q(
√a) and Q(
√b), so by the correspondence in the fundamental
theorem, C4 is ruled out. And another way – if we set K = Q(√a) then F/K is normal and
separable and [F : K] = 2, so Gal(F/K) is cyclic of order 2 by the fundamental theorem,and the Galois group permutes the roots of x2− b. We deduce that there must be an elementof Gal(F/K), and thus a field automorphism ga of F , that sends +
√b to −
√b and fixes
√a
(as it fixes K). Similarly there’s an automorphism gb of F that sends +√a to −
√a and fixes√
b. This gives us two elements of order 2 in Gal(F/Q), which must then be C2 × C2. Ofcourse their product, gagb, sends
√a to −
√a and
√b to −
√b, so it fixes
√ab and is the third
non-trivial element of Gal(F/Q).The subgroups of C2 × C2 are: the subgroup of order 1 (corresponding to F ), the group
itself, of order 4 (corresponding to Q) (both of these because the Galois correspondence is
49
order-reversing, so i.e. sends the biggest things to the smallest things and vice-versa), andthen there are three subgroups of order 2, corresponding to Q(
√a), Q(
√b) and Q(
√ab). One
way to see this for sure is, for example, that ga fixes√a, so the subfield corresponding to
〈ga〉 definitely contains√a, but has degree 2 over Q by the tower law and so must be Q(
√a).
Arguing like this will show everything rigorously.Finally, all of the subfields are normal over Q, because all subgroups of Gal(F/Q) are
normal (as it’s abelian).
(iii) Every element of Gal(F/Q) sends√a+√b to something else! (for example ga sends
it to√a −√b). So the subgroup of Gal(F/Q) corresponding to Q(
√a +√b) must be the
identity, which corresponds to F , and so F = Q(√a+√b).
(iv) If√r ∈ Q(
√p,√q) then Q(
√r) must be one of the quadratic subfields of Q(
√p,√q),
and hence it must be either Q(√p), Q(
√q) or Q(
√pq) by part (ii). But by part (i)
√r is not
in any of these fields! Done.
(v) [F : Q(√p,√q)] must be 2 (as it isn’t 1) and now use the tower law. The Galois
group – we know firstly that any element of the Galois group will be determined by what itdoes to
√p,√q and
√r, and of course
√n must be sent to ±
√n for any n ∈ Q, so there are
at most eight possibilities for Gal(F/Q), corresponding to the 8 = 23 choices we have for thesigns. However we know the size of Gal(F/Q) is eight, so all eight possibilities must occurand the group must be C2 × C2 × C2.
Let me stress here, for want of a better place, that you cannot just say “clearly√p,√q
and√r are “independent” so we can move them around as we please” – one really has to
come up with some sort of an argument to prove that there really is a field automorphism ofF sending, for example,
√p to −√p, √q to +
√q and
√r to −
√r. You can build it explicitly
from explicit elements you can write down in the Galois group using degree 4 subfields, or youcan get it via the counting argument I just explained, but you can’t just say “it’s obvious”because Galois theory is offering you precisely the framework to make the arguments rigorousand I don’t think it is obvious without this framework.
(vi) Think of the Galois group as a 3-dimensional vector space over the field with twoelements. There are seven 1-dimensional subspaces (each cyclic of order 2 and generatedby the seven non-trivial elements), and there are also seven 2-dimensional subspaces, byarguing for example on the dual vector space – or by arguing that any subgroup of order 4of C2 × C2 × C2 is the kernel of a group homomorphism to C2 and such a homomorphism isdetermined by where the three generators go; there are eight choices, one of which gives thetrivial homomorphism and the other seven of which give order 4 subgroups.
Hence other than F and Q there are 14 fields; seven have degree 2 and seven have degree 4.The degree 2 ones are Q(
√paqbrc) as a, b, c each run through 0 and 1, but not all zero. The
degree 4 ones are Q(√paqbrc,
√pdqerf ) as (a, b, c), (d, e, f) run through bases of the seven
2-dimensional subspaces of the Galois group considered as a vector space of dimension 3 overthe field with 2 elements.
(vii) We know all seven non-trivial elements of the Galois group, and none of them fix√p +√q +√r (because if you think of it as a real number, they all send it to something
strictly smaller), so the subgroup corresponding to Q(√p +√q +√r) is trivial and we’re
home.
(viii) Induction and the argument in (v) gives the degree; considering possibilities of signs
50
gives that the Galois group is what you think it is, acting how you think it acts, and the lastpart again follows by observing that Q(
√p1 +
√p2 + · · · + √pn) corresponds to the trivial
subgroup.
A 28.
Q ⊆ Q(81/5
)⊆ Q
(81/5,
√81/5 + 6
)⊆ Q
(81/5,
√81/5 + 6, 51/3
)⊆ Q
(81/5,
√81/5 + 6, 51/3,
11
√51/3 +
√81/5 + 6
)⊆ Q
(81/5,
√81/5 + 6, 51/3,
11
√51/3 +
√81/5 + 6, 91/7
)A 29. This question is not for you to literally do it but to show you what is possible. Infact the statements are very tedious to prove and not super-useful, which is why they aretypically omitted from lecture courses.28
(a) To sow that the action is transitive is to show that given two K-embeddings x, y : F →L, there exists a K-embedding g : L→ L — that is to say, an element g of the Galois group— such that y = gx. This is Question 8, part (a) below. The fact just proven generalizes the
statement: Let K be a field, f(x) ∈ K[x] an irreducible polynomial, and K ⊂ L the splittingfield of f(x). Then G = EmbK(L,L) acts transitively on the roots of f(x). To derivethis from the abstract statement about fields, just note that roots of f(x) are in one-to-onecorrespondence with K-embeddings
x : F → L
where F = K[x]/f(x).
The last statement is a tautology: we have an injection i : F → L by means of which weconsider F as a subset of L, i.e. the elements of F are elements of L. For g ∈ G to say thatgi = i is exactly to say that g|F is the identity on F , in other words g ∈ F †.
(b) Suppose thatK ⊂ F is normal. Let g ∈ G, h ∈ H and consider g−1hg. Since g : L→ Lis a K-embedding, it follows that g|F : F → L is also a K-embedding, and then, becauseK ⊂ F is normal, we have g(F ) ⊂ F (this is exactly our definition of normal extension offields). So for all a ∈ F , g(a) ∈ F and hence h(g(a)) = g(a) and hence g−1hg (a) = a, thatis g−1hg ∈ F † = H or, in other words, H is a normal subgroup of G.
Suppose now that H ≤ G is a normal subgroup. For clarity let me name x : F → L thegiven embedding. For all g ∈ G, we want to show that gx(F ) ⊂ x(F ).
In general, if a group G acts on a set X, then for all g ∈ G and x ∈ X, Ggx = gGxg−1.
What we want now follows from part (a):
x† = H = Gx = gGxg−1 = Ggx = (gx)†
28I had to revise this question significantly, see v2 of the Worksheet.
51
therefore F and gF are the same field, because they have the same dagger H (Galois corre-spondence).
Finally, ifK ⊂ F is normal, restriction gives a group homomorphism ρ : G→ EmbK(F, F ).The kernel is clearly H; and ρ is surjective by part (a).
(c) This is a small step from (b). For any K ⊂ F ⊂ L for clarity denote by x : F → Lthe given inclusion. Claim: For all g ∈ G, g(F ) ⊂ F if and only if g ∈ NG(H).
Indeed, suppose that g(F ) ⊂ F . Then in fact g(F ) = F (an injective linear map betweenfinite dimensional vector spaces of the same dimension is an isomorphism) and hence (gx)† =x†, which implies as above
gHg−1 = gGxg−1 = Ggx = (gx)† = x† = Gx = H
and hence g ∈ NG(H). Conversely and similarly, if gHg−1 = H, then (gx)† = x†, hencegx(F ) and x(F ) are the same subfields of L, that is g(F ) = F . This shows the claim.
From the claim it follows that restriction is a group homomorphism ρ : NG(H)→ EmbK(F, F );the kernel is obviously H and the image is everything: if u ∈ EmbK(F, F ) then by part (a)there is g ∈ G such that gx = xu, in other words g|F = u: by what we said earlier g ∈ NG(H)and by what we just said ρ(g) = u.
(d) This part is a minor variation on part (c). Here we start from two K-embeddingsx1 : F1 → L, x2 : F2 → L and set H1 = x†1, H2 = x†2. Claim: For all g ∈ G, gx1(F1) ⊂ x2(F2)if and only if g ∈ N(H1, H2). Indeed, by the Galois correspondence, gx1(F ) ⊂ x2(F ) if andonly if (gx1)† ⊃ x†2 if and only if gH1g
−1 ⊃ H2.From the claim we construct a restriction map
ρ : N(H1, H2)→ EmbK(x1, x2)
which is surjective by part (a). Suppose that g, g′ ∈ N(H1, H2). This just means thatgx1 = g′x1 or in other words g−1g′ ∈ H1.
(e) This is really pretty easy. I show the last bit: suppose that g1, g2 ∈ N(H1, H2) andthat Tg1 = Tg2 . This means that for all h ∈ H2, g−1
1 hg1 = g−12 hg2 or, equivalently
g2g−11 h = hg2g
−11 , that is z = g2g
−11 ∈ C(H2)
(f) This is not hard but it is boring. The composition we are talking about is inheritedfrom the composition of part (e). You need to check that the composition of part (e) iscompatible with various equivalence relations.
(g) This is all an elaborate way to rephrase part (d).
A 30. (a) By assumption K ⊂ L (there is only one such inclusion so I don’t need to callit anything) is normal, hence it is the splitting field of a polynomial f(x) ∈ K[x]; so nowxi : F → L is also a splitting field of f(x), and the first half of part (a) (existence of y) followsfrom uniqueness of splitting fields over F .
(The fact that y is a field automorphism follows from a familiar argument: it is injectivebecause every field homomorphism is, and it is surjective by the rank-nullity theorem, becauseit is an injective K-linear endomorphism of a finite dimensional K-vector space.)
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(b) Define a set-theoretic function
y? : Embx1(E,L)→ Embx2(E,L)
as follows:y?(x) = y x
Indeed, suppose that a ∈ F , then
y?(x)(a) = y(x(a)
)= y(x1(a)
)= x2(a)
therefore, as claimed, y?(x) ∈ Embx2(E,L).Finally y? is a bijective correspondence because it has an inverse given by (y−1)?.
A 31. (a) First note that if α = 21/3 then L is the splitting field of x3− 2 over Q; indeed thesplitting field is by definition Q(α, ωα, ω2α) (as these are the roots), and this field must beQ(α, ω) because each of the generators of one field can be easily checked to be in the other.
We immediately deduce that K ⊂ L and F ⊂ L are normal because (both are the splittingfield of X3− 2, seen as a polynomial in either K[X] or F [X]. (We can also deduce normalityof F ⊂ L from normality of K ⊂ L). However K ⊂ F is not normal, because x3 − 2 isirreducible over K and has one, but not all, roots in F .
(b) Let’s first compute some degrees. We know the min poly of√
2 over Q has degree 2,so [F : K] = 2. Also the min poly of 21/4 over Q must be x4 − 2 (because this poly is irredby Eisenstein), and hence [L : K] = 4. By the tower law we deduce [L : F ] = 2 (and hencethat x2 −
√2 must be the min poly of 21/4 over F , but we don’t need this). We could use
Question 3 to deduce that K ⊂ F and F ⊂ L are normal, but we could also see it directly:F is the splitting field of x2 − 2 over K and L is the splitting field of x2 −
√2 over F , so
they’re both normal. However x4− 2 is irreducible over K and has one root in L (in fact tworoots in L) but not all its roots (as two are not real, whereas L ⊆ R so K ⊂ L is not normal.
(c) If H is normal in G then g−1Hg = H for all g ∈ H, so trivially g−1Hg = H for allg ∈ K, so H is normal in K.
Examples: H = 1 ⊆ K = 〈(1 2〉 ⊆ S3 for the first, and H = 〈σ〉 ⊆ K = 〈σ, ρ2〉 ⊆ G =D8 for the second, with D8 = 〈ρ, σ〉 the dihedral group generated by a rotation ρ of order 4and a reflection σ of order 2.
15.4 Worksheet 4
A 32. (i) If p(x) is irreducible over E and has a root in M ∩N , then it has a root in M anda root in N , and hence splits completely in both M and N . So all the roots of p(x) in F arein both M and N , and hence in M ∩N . Hence M ∩N is normal.
(ii) If M is the splitting field of f(x) and N is the splitting field of g(x) then MN is thesplitting field of f(x)g(x) (not hard to check). Hence MN is normal.
A 33. This is an unreasonably difficult question, but let’s at least see what I can do.29
29Don’t worry too much if you don’t follow the proof. I forbid you to spend more than five hours tryingto understand this.
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(a) Pick λ1, . . . , λn ∈ L such that L = K(λ1, . . . , λn). For all i = 1, . . . , n, let fi(x) ∈ K[x]be the minimal polynomial of λi over K, let
f(x) =n∏i=1
fi(x) ∈ K[x]
and let L ⊂ Ω be the splitting field of f(x) now thought of as a polynomial in L[x]. ComposingK ⊂ L and L ⊂ Ω we have K ⊂ Ω and It is pretty clear that this is also the splitting fieldof f(x) over K. (Why?) I claim that L ⊂ Ω is a normal closure of K ⊂ L.
To check this I need to verify the two defining properties of a normal closure. The first isjust saying that K ⊂ Ω is normal, and it is because it is a splitting field.
It remains to verify that Ω is generated over K by the set:
Λ = σ(λi) | i = 1, . . . , n, σ ∈ EmbK(L,Ω)
(Please convince yourselves that this is so.) On the other hand we know that Ω is generatedover K by the set of the roots of f(x):
Z = µ ∈ Ω | f(µ) = 0
so we will be done if we show that Λ = Z. Now we know that if λ is a root of f andσ ∈ EmbK(L,Ω) then σ(λ) is also a root of f , that is, Λ ⊂ Z. Let us now show that Z ⊂ Λ.Let µ ∈ Ω be a root of f . Then for some i = 1, . . . , n µ is a root of fi. We know that thereis an embedding
K(λi)ϕ // Ω
K
OO <<
such that ϕ(λi) = µ; and by Lemma 16 (B) this ϕ extends to ϕ : Ω→ Ω:
Ωϕ
""K(λi)
OO
ϕ // Ω
K
OO <<
Now consider σ = ϕ|L ∈ EmbK(L,Ω); by construction σ(λi) = µ. This shows that µ ∈ Λand that Z ⊂ Λ and finishes Part(a).
(b) We need to show that K ⊂ Ω is normal.
Claim For all λ ∈ L let f(x) ∈ K(x) be its minimal polynomial: then f(x) ∈ Ω[x] splitscompletely.
The claim and property 1. imply: Choose λ1, . . . , λn such that L = K(λ1, . . . , λn) andcall fi ∈ K[x] the minimal polynomial of λi, then Ω is the splitting field of f =
∏ni=1 fi
30,and hence K ⊂ Ω is normal.
30this requires some argument that I am not spelling out: please convince yourself that this is true
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Let us prove the claim. We don’t yet know that K ⊂ Ω is normal, but we can alwaysmake a bigger field Ω ⊂ Ω such that K ⊂ Ω is a normal extension. Now here comes the keypoint. Composing with the inclusion Ω→ Ω we obtain a natural inclusion
EmbK(L,Ω)→ EmbK(L, Ω)
and this inclusion is a bijection because by 2. the two sets have the same number [L : K]s of
elements. In other words (and this is the key point): every K-embedding σ : L → Ω in factlands in Ω. Now we are ready to prove the claim. We know that f(x) splits completely in
Ω. Thus, it is enough to show that if µ ∈ Ω is a root off , then in fact µ ∈ Ω. Arguing asin Part (a) we can construct an embedding σ : L → Ω such that σ(λ) = µ. But as we saidσ(L) ⊂ Ω so in fact µ = σ(λ) ∈ Ω and we are done.
A 34. (a) It’s the field of fractions of k[T p]. Or, check explicitly that if S = T p then this isjust the field of fractions of k[S]. Or check that it’s a subset containing 0 and 1 and closedunder +−×/.
(b) In fact any subfield of L containing k and T must contain f(T ) for any polynomialf ∈ k[T ] and hence it must contain f(T )/g(T ) if g is a non-zero polynomial. Hence L = k(T )in the sense that it’s the smallest subfield of L containing k and T , so L = k(T ) ⊆ K(T ) ⊆ Land all inclusions are equalities.
(c) T is a root of the polynomial xp − T p ∈ K[x].
(d) If q(x) = xp − T p factored in K[x] into two factors f and g of degrees a and b, witha + b = p and 0 < a, b < p, then by rescaling we can assume both factors are monic. Nowconsider the factorization q(x) = (x − T )p in L[x]. This is the factorization of q(x) intoprimes in L[x], and there’s only one prime involved, namely x− T . Because q = fg in L[x],we must have f(x) = (x− T )a and g(x) = (x− T )b – anything else would contradict uniquefactorization. But this means the constant term of f(x) is ±T a and because 0 < a < p weknow a isn’t a multiple of p and hence ±T a 6∈ K and so f(x) 6∈ K[x], a contradiction.
(e) q(x) is irreducible in K[x] and T is a root, so it’s the min poly. It’s not separablebecause it is irreducible over K but has repeated roots in L (namely T , p times).
(f) T ∈ L is not separable over K because its min poly isn’t. Hence L/K is not separable,because L contains an element which is not separable over K.
A 35. (i) If L = E(α1, . . . , αn) then for E ⊆ K ⊆ F we have that K contains L iff Kcontains all the αi. So if E ⊆ K ⊆ F then E contains N iff E contains M and the αi iff Econtains M and L; hence N is the smallest subfield of F containing M and L.
(ii) If L is the splitting field of p(x) ∈ E[x] and M is the splitting field of q(x) ∈ E[x](these polynomials exist by normality) then I claim N is the splitting field of p(x)q(x); indeedif the αi are the roots of p and βj are the roots of q then by the first part N is the fieldgenerated by the αi and the βj. Now N is finite and normal; moreover each of the αi andthe βj are separable over E (as each is contained in either L or M) and hence each time weadjoin one we get a separable extension; finally a separable extension of a separable extensionis separable (by comparing degrees and separable degrees).
(iii) If g ∈ Gal(N/E) then g(L) = L by 6.7 and hence the restriction of g to L is inGal(L/E). Similar for M/E. So we get a map Gal(N/E) → Gal(L/E) × Gal(M/E). This
55
is easily checked to be a group homomorphism. It’s injective because anything in the kernelfixes L and M pointwise, so fixes LM pointwise; but LM = N .
It’s not always surjective though – for example if L = M then it hardly ever is. Moregenerally if L∩M 6= E then there will be problems. However if L∩M = E then the map isa bijection.
A 36. (a) The two polynomials have degree 3 and have no roots in F2 (just plug x = 0, 1)hence they are irreducible.
If σ : K → L then σ(α) is a root of f(x) in L; and f(x) has three roots in L:
β + 1; β2 + 1; β2 + β
indeed, for example, we can check directly that:
(β + 1)3 = β3 + β2 + β + 1 = (β2 + 1) + β2 + β + 1 = β = (β + 1) + 1
that is, β + 1 is a root of f(x). The other roots of f are Fr2(β + 1) = β2 + 1 andFr2(β2 + 1) = β4 + 1 = β(β2 + 1) + 1 = β2 + 1 + β + 1 = β2 + β. (But one can also checkdirectly.)
A basic result about fields states that a morphism from K to L is the same as a root off(x) in L and there are 3 of these. As f and g are irreducible we know that K and Lhave degree 3 over F2 and we have shown in class that any two finite fields of the samedegree over the base field are isomorphic. Since both fields have degree 3 over the basefield F2, all morphisms from K to L are isomorphisms hence there are 3 of these. (Thisgives another reason why K and L are isomorphic.)
(b) h(x) ∈ F2[x] is irreducible because: it has no roots (plug x = 0 and x = 1) in F2 AND itis not divisible by x2 + x + 1, the only irreducible degree two polynomial in F2[x] — ascan be checked by performing long division in F2[x].
Let L ⊂ E be the splitting field of h(x) as a polynomial in L[x]. The extension F2 ⊂ Eis normal and separable because ALL finite extensions of finite fields are. Clearly Econtains the splitting field F2 ⊂ F of h(x) ∈ F2[x]:
E
L
3
F
2
F4
We know that h(x) ∈ F2[x] is irreducible; hence if γ ∈ F is a root of h, then [F2(γ) :F2] = 4. We know that every finite extension of a finite field is normal and separable,therefore F2 ⊂ F is normal and hence (by a known characterisation of normal extensions)h(x) splits completely in F2(γ)[x] — because it is irreducible over F2 and has a root inF2(γ) — hence actually F = F2(γ) and then, as indicated in the diagram, [F : F2] =[F2(γ) : F2] = 4.
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The tower law implies that 3|[E : F2] and 4|[E : F2] hence 12|[E : F2]. But clearly alsoE = L(γ) and then [E : L] is the degree of the minimal polynomial of γ over L, which isa factor of h, hence [E : F2] = [E : L][L : F2] ≤ 12. So in fact [E : F2] = 12; [E : L] = 4,h ∈ L[x] is the minimal polynomial of γ and it is therefore irreducible.
A 37. (This is a pure algebra question.) The (n − 1)-cycle c must fix an element of [n]31
which we may well assume to be 1, and then after re-labelling the elements of [n] we mayassume that c = (23 . . . n). Let t be the transposition; then:
Either t involves 1, and then by further relabelling elements we may assume c = (23 · · ·n),t = (12), and it is easy to conclude from here;
Or t = (ab) where 1 < a < b: this is what we assume from now on.
Because G is transitive, it must contain an element σ such that σ(a) = 1, but then σtσ−1 =(1σ(b)) and we are back in the previous case.
A 38. We look at the polynomial modulo small primes:32 Modulo p = 2 we get:
f(x) = x6 − 12x4 + 15x3 − 6x2 + 15x+ 12 ≡ x(x5 + x2 + 1) mod 2
where the second polynomial r(x) = x5 + x2 + 1 is irreducible because if it weren’t it wouldsplit an irreducible degree two polynomial, but the only such polynomial is x2 + x+ 1 whichdoes not divide into r(x) (direct inspection). By the theorem in the footnote, the Galoisgroup G contains a 5-cycle.
Eisenstein at p = 3 shows that f(x) is irreducible in Q[x] and in turn this implies that Gis transitive.
Next:f(x) ≡ (x+ 1)(x+ 2)(x+ 3)(x+ 4)(x2 + 3) mod 5
thus by the theorem in the footnote G contains a transposition.
By Question 6 G = S6.
A 39. (a) Let us first consider the polynomial in F2[x]. Clearly x = 1 is a root of f(x) anda small calculation shows
x4 + x2 + x+ 1 = (x+ 1)(x3 + x2 + 1) in F2[x]
and then x3 +x+1 ∈ F2[x] is irreducible because it has no roots in F2 (just plug in x = 0and x = 1).
31Notation: [n] = 1, 2, . . . , n is the set with n elements.32In this and the next questions we use the following result which was proved in class: THEOREM
Let f(X) ∈ Z[X] be monic of degree n, p ≥ 1 a prime such that the reduction mod p f(X) ∈ Fp[X] hasdistinct roots and factors as a product of irreducible factors of degree n1, . . . , nk. Then the Galois groupG of the splitting field Q ⊂ L of f contains a permutation of the roots of f whose cycle decomposition is(n1)(n2) · · · (nk).
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Next, we work in F3[x]. A quick inspection shows that f(x) has no roots in F3: just plugx = 0, 1,−1. To show that the polynomial f(x) ∈ F3[x] is irreducible, we show that it isnot divisible by any of the three irreducible degree 2 polynomial in F3[x]: these are:
x2 + 1, x2 + x− 1, x2 − x− 1
Performing three long divisions in F3[x] we see:
x4 + x2 + x+ 1 = (x2 + 1)(x2) + x+ 1
x4 + x2 + x+ 1 = (x2 + x− 1)(x2 − x) + 1
x4 + x2 + x+ 1 = (x2 − x− 1)(x2 + x)− x+ 1
these calculations show that f is irreducible in F3[x].
(b) A result proved in class implies that the Galois group G of the splitting field Q ⊂ Kcontains a 3-cycle and a 4-cycle. If a subgroup G of S4 contains a 3-cycle and a 4-cyclethen G = S4. (See Question 9 below.) Therefore, G = S4.
A 40. (a) First, working modulo 2,
f(X) ≡ X4 + 3X + 1 ∈ F2[X]
is irreducible. Indeed, by inspection, it does not have a root in F2, and it is not divisible bythe only irreducible degree 2 monic polynomial x2 +x+1 ∈ F2[X]. In fact long division gives
X4 +X + 1 = (X2 +X + 1)(X2 +X) + 1
Next, it is easy to factor f(X) mod 5:33
f(X) ≡ (X − 1)(X3 +X2 +X − 1) ∈ F5[X]
where the degree 3 factor is irreducible because, by inspection, it has no root in F5.
(b) Suppose that G ⊂ S4 contains a 4-cycle and a 3-cycle. Let the 4-cycle be s = (abcd).Note that we can write s = (dabc), etc. Thus, we may assume that the 3-cycle t fixes the lastletter d in the 4-cycle. Now either t = (abc) or t = (acb), but then t2 = (abc). The conclusionis that we may assume s = (1234), t = (123). You take it from here.
(c) By Part (a) and the theorem in the footnote, the Galois group contains a 4-cycle anda 3-cycle hence, by Part (b) it must be all of S4.
A 41. With all the hints and the examples, this should not be too hard. You do it (or elseignore this question).
33Working mod 3 is not going to lead to useful information: it is clear by inspection that f(X) has no rootin F3 and then either f(X) is irreducible (no useful conclusion) or it splits into two quadratic polynomials(again no useful conclusion).
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A 42. (a) x6 − 1 = (x− 1)(x+ 1)(x2 + x+ 1)(x2 − x+ 1) and
ζ =1 + i
√3
2is a root of Φ6(x) = x2 − x+ 1 ∈ Q[x]
Since Q ⊂ Q(ζ) is the splitting field of Φ6(x), we have [Q(ζ) : Q] = 2.
(b) The polynomial x6 +3 ∈ Q[x] is irreducible by the Eisenstein criterion. By an elementaryfact on fields, if α ∈ K is a root, that is α6 = −3, then [Q(α) : Q] = 6. Now β = α3
satisfies:β2 = −3,
in other words, β = ±i√
3 and we have a tower of fields:
Q ⊂ Q(ζ) = Q(i√
3) ⊂ Q(α)
Since K = Q(ζ, α) and ζ ∈ Q(α), we have in fact K = Q(α), so from above [K : Q] = 6.
The Galois group G permutes the roots of f(x) = x6 + 3 so consider σ ∈ G, thenσ(α) = ζkα for a unique k ∈ Z/6Z, and — ideally — I would like you to have checkedthat the assignment σ 7→ k is an isomorphism G ∼= Z/6Z. (It is an injective grouphomomorphism and |G| = 6.)
(c) As before x6 − 3 ∈ Q[x] is irreducible. Let α ∈ R, α6 = 3. As before [Q(α) : Q] = 6 butnow, because Q(α) ⊂ R, ζ 6∈ Q(α) and Φ6(x) remains irreducible in Q(α). We have adiagram of fields
K
Q(α)
2
6
Q(i√
3)
2
Q
The diagram shows that [K : Q] = 12. The situation at this point is similar to x4 − 2 ∈Q[x] — which was discussed at length in class — and you can treat it in a similar fashion:an element σ ∈ G is completely determined once you know: σ(α) (6 possibilities) and σ(ζ)(two possibilities) for a total of 12 possibilities. Because |G| = 12 all these possibilitiesare realised, and it is not hard to see that one gets the dihedral group D12.
A 43. I am sorry, I can’t write this down for you. You do it: it is fun!
References
[Art44] Emil Artin. Galois Theory. Notre Dame Mathematical Lectures, no. 2. Universityof Notre Dame, Notre Dame, Ind., second edition, 1944.
[GR] Alexander Grothendieck and Michele Raynaud. Revetements etales et groupe fon-damental (SGA 1). arXiv:math/0206203.
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[Her75] Israel Nathan Herstein. Topics in Algebra. Xerox College Publishing, Lexington,Mass.-Toronto, Ont., second edition, 1975.
[Rei] Miles Reid. MA3D5 Galois Theory. Available from https://homepages.warwick.
ac.uk/~masda/MA3D5/.
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