Section 14.2. Let K/F be a Galois extension. The main goal of this lecture is to prove the Funda- mental Theorem of Galois Theory, which makes precise the relation between the subgroups H ⊂ Gal(K/F ) and the subfields F ⊂ E ⊂ K. Let us do some preparatory work first. Lemma 1. A finite-dimensional vector space over an infinite field can not be presented as a union of a finite number of its proper subspaces. s Proof. Let V = ∪i=1Vs, where Vs are proper subspaces of the vector space V . For every Vi let us choose a non-zero linear function li on V , which turns into zero on Vi. Now, let us Qs consider a polynomial p = i=1 li. Then for any v ∈ V we must have p(v) = 0 which would imply that p is constantly zero, and we arrive at a contradiction.  Theorem 2. Let K/F be a field extension of degree n, and G ⊂ Aut(K/F ) be a subgroup of Aut(K/F ). Denote by KG the fixed field of G. Then KG = F if and only if |G| = n. Moreover, if KG = F then for any fields P and Q such that F ⊂ P ⊂ Q ⊂ K, any homomorphism ϕ: P → K extends to a homomorphism ψ : Q → K in precisely |Q : P | ways. Proof. By the definition of a fixed field, we have G ⊂ Aut(K/KG). Therefore, G |G| 6 |K : K | 6 |K : F | = n. Then |G| = n implies KG = F . Conversely, let KG = F . We shall first show that G = Aut(K/F ), and then prove the equality | Aut(K/F )| = n. For any element α ∈ K let {α1, . . . , αm} ⊂ K be its G-orbit. Then m Y G f(x) = (x − αi) ∈ K [x] = F [x](∗) i=1 is the minimal polynomial of the element α ∈ K. Note that for any ϕ ∈ Aut(K/F ), the ele- ment ϕ(α) is also a root of f(x). Then, there exists an element gα ∈ G such that ϕ(α) = gα(α). Now, if K is a finite field, we can choose α to be the generator of the cyclic group K×. Then ϕ = gα ∈ G. If K is an infinite field, then for any g ∈ G let us set

Kg = {α ∈ K | ϕ(α) = g(α)} ⊂ K.

Clearly, Kg is a vector subspace (in fact, even a subfield) of K, and it follows from the above discussion that [ K = Kg. g∈G By previous lemma we have K = Kg for some element g ∈ G, which in turn implies G = Aut(K/F ). Now, let us pause for a second and prove the second statement of the theorem. Since any finite field extension can be obtained as a chain of simple extensions, it is enough to prove the statement in the case when Q = P (α) is a simple field extension. Let p(x) be the minimal polynomial for α over P . Then p(x) divides f(x) in the ring P [x], where f(x) is the minimal polynomial of α over F . Now, let pϕ(x) ∈ K[x] be the polynomial obtained from p(x) by applying the homomorphism ϕ to its coefficients. Then pϕ(x) divides f(x) in the ring K[x] = ϕ(P )[x] and therefore decomposes into linear factors in K[x]. The latter implies that the homomorphism ϕ: P → K extends to a homomorphism ψ : Q → K in precisely |Q : P | ways. Applying the second statement of the theorem to the case P = F and Q = K, we obtain | Aut(K/F )| = n.  1 2

Remark 3. Note that in the course of proving the previous theorem we showed that if K is a field, G ⊂ Aut(K) is a finite subgroup, and KG is the fixed subfield of G, then (1) K is algebraic over KG; (2) G = Aut(K/KG). Corollary 4. If K is a field, G ⊂ Aut(K) is a finite subgroup, and F = KG is the fixed subfield of G, then K is a finite extension of F . Proof. Assume that |K : F | = ∞. Since K is algebraic over F , there exists an infinite chain

F = F0 ⊂ F1 ⊂ F2 ⊂ ... of simple extensions Fi+1/Fi, where Fi ⊂ K for all i. Note that for any N ∈ Z+ there exists a number i such that |Fi : F | > N. Since G = Aut(K/F ) and | Aut(K/F )| > | Aut(Fi/F )| for any i, we have | Aut(G)| = ∞ which contradicts our assumption.  Corollary 5. If K is a field, G ⊂ Aut(K) is a finite subgroup, and KG is the fixed subfield of G, then K/KG is a Galois extension and G is its . Proof. Since K/KG is a finite extension, we can apply Theorem 2 and conclude that G G G G |G| = |K : K |. Since |G| 6 | Aut(K/K ))| 6 |K : K |, we have G = Aut(K/K ) G G and | Aut(K/K )| = |K/K | in the statement of the Corollary follows. 

Corollary 6. Let K be a field and G1,G2 ⊂ Aut(K) be a pair of distinct subgroups. Then the fixed fields KG1 and KG2 are also distinct. G G G G Proof. If K 1 = K 2 , then K 1 is fixed by G2 and G2 ⊂ G1 = Aut(K/K 1 ). Similarly, we get G1 ⊂ G2. Then we have G1 = G2 which contradicts our assumption.  Last time we proved that the splitting field K of a f(x) ∈ F [x] is a Galois extension of F . Let us now prove the converse, which will characterize Galois extensions. Theorem 7. The extension K/F is Galois if and only if K is the splitting field of some separable polynomial f(x) ∈ F [x]. Furthermore, if K/F is Galois then every irreducible polynomial p(x) ∈ F [x] which has a root in K is separable and has all its roots in K. Proof. As we just mentioned, we only need to prove the “only if” statement. Let K/F be a Galois extension, G = Gal(K/F ), and α ∈ K be a root of an irreducible polynomial p(x) ∈ F [x]. Then we have p(x) = f(x) ∈ KG[x] = F [x], where the polynomial f(x) is defined by (∗), which implies the second statement of the Theorem. Now, let β1, . . . , βn be a basis of K over F . For each i = 1, . . . , n, let pi(x) ∈ F [x] be the minimal polynomial of βi. Define g(x) ∈ F [x] to be a polynomial obtained by removing any multiple factors from the product p1(x) . . . pn(x). Then g(x) is separable and K is its splitting field.  Recall that an extension K/F is normal if every irreducible polynomial p(x) ∈ F [x] which has a root in K splits into linear factors over K. Then previous results yield the following equivalent descriptions of Galois extensions. Corollary 8. Let K/F be a field extension with | Aut(K/F )| < ∞. Then the following 4 statements are equivalent: (1) K/F is a Galois extension, that is | Aut(K/F )| = |K : F |; (2) K is a splitting field of a separable polynomial in F [x]; (3) F is the fixed field of Aut(K/F ); 3

(4) K is a finite, normal, and of F .

Now we are ready to prove the Fundamental Theorem of Galois Theory. Let K/F be a Galois extension, and G = Gal(K/F ). To any subfield F ⊂ E ⊂ K we associate a subgroup

GE = {g ∈ G | g(α) = α ∀α ∈ E} ⊂ G of elements which act trivially on E. On the other hand, to any subgroup H ⊂ G we associate its fixed subfield

KH = {α ∈ K | h(α) = α ∀h ∈ H} ⊂ K.

Theorem 9 (Fundamental Theorem of Galois Theory). Let K/F be a Galois extension, H and G = Gal(K/F ). Then mappings E 7→ GE and H 7→ K are mutually inverse, and therefore establish a bijction between the subfields of K which contain F and the subgroups of G. Moreover, under this bijection subfields E ⊂ K, which are Galois over F correspond to normal subgroups of G.

Proof. First, let us note that by Theorem 2 we have

H |K : K | = |H| and |GE| = |K : E|.

G G Now, on one hand we have E ⊂ K E , on the other |K : K E | = |GE| = |K : E| which GE H implies E = K . Similarly, H ⊂ GKH , and |GKH | = |K : K | = |H| implies H = GKH . This proves the first statement. Now, by Theorem 2 every ϕ ∈ Aut(E/F ) extends to an automorphism ψ ∈ Aut(K/F ) in exactly |K : E| = |GE| ways, namely one can compose ϕ with any element σ ∈ GE to obtain σ ◦ ϕ ∈ G. In general, | Aut(E/F )| 6 |E : F | and the condi- tion that E/F is Galois is equivalent to the statement that the subgroup which preserves (but not necessarily fixes) E induces exactly |E : F | distinct on E. Now, |E : F | = |K : F |/|K : E| = |G : GE|, and therefore if E/F is Galois the subgroup which preserves E would have |G : GE| · |GE| = |G| elements. Thus we conclude that the whole Galois group G = Gal(K/F ) has to preserve E in order for E/F to be Galois. −1 Condition E = KGE implies that for any g ∈ G we have g(E) = KgGE g . Then the subfield E is preserved by all elements g ∈ G if and only if the subgroup GE ⊂ G is normal, which completes the proof.  √ √ Example. (1) Last time we showed that the extension Q( 2, 3)/Q is Galois. In this case, the Galois group is isomorphic to the Klein 4-group K4. Recall that K4 is an abelian group with 2 generators of order√ √ 2, that is K4 '{1, σ, τ, στ}. Moreover, let us fix the isomorphsm K4 ' Gal(Q( 2, 3)/Q) by setting √ √ √ √ √ √ √ √ σ( 2) = − 2, σ( 3) = 3, and τ( 3) = − 3, τ( 2) = 2.

Since the extension in question√ is√ Galois, there is a bijection between the subgroups of K4 and subfields of E ⊂ Q( 2, 3) such that Q ⊂ E. Moreover, the fact that K4 is abelian implies that every subgroup of K4 is normal, and therefore every extension 4

E/Q is Galois. The diagrams of subgroups of K4 and the subfields E are shown below √ √ Q( 2, 3) {1}

√ √ √ Q( 2) Q( 6) Q( 3) hτi hστi hσi

Q K4 √ 3 (2) We also showed last time that the extension K/Q is a Galois, where K = Q( 2, ζ3), and that Gal(K/Q) ' S3. If we choose the generators σ, τ of Gal(K/Q) to be √ √ √ √ 3 3 3 3 2 σ( 2) = ζ3 2, σ(ζ3) = ζ3, and τ( 2) = 2, τ(ζ3) = ζ3 , hσi we have the subfield K = Q(ζ3). Note that hσi ⊂ S3 is the only normal subgroup of S3. Therefore the only subfield E which fits into the chain Q ⊂ E ⊂ K and is Galois over Q is E = Q(ζ3), and the Galois group of E/Q is Gal(E/Q) ' Z/2Z. The correspondence between subfields of E ⊂ K and subgroups H ⊂ S3 is shown below. √ 3 Q( 2, ζ3) √ √ √ 3 3 2 3 Q( 2) Q(ζ3 2) Q(ζ3 2)

Q(ζ3)

Q

{1}

hτi hτσi τσ2

hσi

S3 √ √ √ √ (3) Let K √= Q√( 2 + 3). Clearly,√ K √⊂ Q( 2, 3) and since the only subgroup of Gal(Q( 2, 3)/Q) which fixes 2 + 3 is the trivial one, we conclude that √ √ √ √ Q( 2 + 3) = Q( 2, 3). √ √ Moreover, one can easily find the minimal polynomial√ √f(x) ∈ Q[x] for 2 +√ 3.√ In- deed, the√ roots√ of f(x) are√ elements√ of the Gal(Q( 2, 3)/Q)-orbit through 2+ 3, that is 2 ± 3 and − 2 ± 3. Therefore, √ √ √ √ √ √ √ √ f(x) = (x − 2 − 3)(x − 2 + 3)(x + 2 − 3)(x + 2 + 3) = x4 − 10x2 + 1. 5

8 (4) Let K be the splitting field of x − 2 over Q. Then the extension K/Q is Galois and

√8 √8 √ √8 K = Q( 2, ζ8) = Q( 2, 2, i) = Q( 2, i). √ 8 8 2 Minimal polynomials√ for √2 and i over Q are respectively x − 2 and x + 1. Since 8 8 the extension Q( 2, i)/Q( 2) is quadratic, we see that

√8 √8 |K : Q| = |K : Q( 2)| · |Q( 2) : Q| = 16,

and therefore the Galous group √G = Gal(K/Q) has order 16. Any element of G is determined by the images of 8 2 and i, which has to be roots of corresponding minimal polynomials x2 − 8 and x2 + 1. Therefore, we are left with 16 possibilities, all of which could be realized via elements σ, τ ∈ G defined by √ √ √ √ σ( 8 2) = ζ 8 2, σ(i) = i, and τ( 8 2) = 8 2, τ(i) = −i, √ 2 where we set ζ = ζ8. Therefore, G = hσ, τi. Now, ζ = (1 + i) 2 hence √ √ σ( 8 2)4 2 σ(ζ) = (1 + i) = ζ5 = −ζ and τ(ζ) = (1 − i) = ζ7, 2 2 and we can explicitly compute the action on K of every element in G. It is then easy to check that στ = τσ3 and that relations G = σ, τ | σ8 = τ 2 = 1, στ = τσ3 . Determining the lattice of subgroups of G is a straightforward, although rather boring, problem with the following answer: G

σ2, τ hσi σ2, τσ3

σ4, τσ6 σ4, τ σ2 τσ3 hτσi

τσ2 τσ6 τσ4 hτi σ4

1

H An observation that for a subgroup H ⊂ G one has |K : Q| = |G : H| becomes handy in determining the corresponding lattice of subfields. Indeed, to find KH we only need to find a subfield of K which is fixed by H, and which is an extension of Q of degree |G : H|. For example, the subfield Q(i) is fixed by hσi, and |Q(i): Q| = |G : hσi | = 2, hσi from which we conclude that Q(i) = K . However, for some subgroups we have to work harder to determine the corresponding subfield. For example, let us consider H = τσ3 ⊂ G, a subgroup of order 4. To find a subfield fixed by H, let us first note that σ4 ⊂ H is a subgroup of order 2, with the two cosets of H/ σ4 are 1 6 √ √ √ and τσ3. Now, σ4( 8 2) = − 8 2 and σ4(i) = i. Therefore, 4 2 is fixed by σ. We can now consider an element √ √ α = (1 + τσ3)( 4 2) = (1 + i) 4 2 ∈ K and notice that α is preserved by σ4 and by τσ3. Indeed, √ √ σ4(α) = σ4(1 + τσ3)( 4 2) = (1 + τσ3)σ4( 4 2) = α and √ √ √ τσ3(α) = τσ3(1 + τσ3)( 4 2) = (τσ3 + (τσ3)2)( 4 2) = (τσ3 + σ4)( 4 2) = α. Therefore, Q(α) is fixed by H and it remains to notice that |Q(α): Q| = 4. Proceeding in a similar fashion, we arrive at the following lattice of subfields of K, which extend Q: Q

√ √ Q( 2) Q(i) Q(− 2)

√ √ √ √ √ 4 4 4 4 Q(i 2) Q( 2) Q(i, 2) Q((1 + i) 2) Q((1 − i) 2)

√ √ √ √ √ 8 3 8 2 8 8 4 Q(ζ 2) Q(ζ 2) Q(ζ 2) Q( 2) Q(i, 2)

√ 8 Q(i, 2)