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Lecture-14_2.pdf

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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.

Proof. Let V = ∪si=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 consider a polynomial p =

∏s 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|6 |K : KG|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

f(x) =

m∏ i=1

(x−αi) ∈ KG[x] = F[x] (∗)

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 = ⋃ g∈G

Kg.

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. �

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 ⊂ . . .

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 Galois group.

Proof. Since K/KG is a finite extension, we can apply Theorem 2 and conclude that |G| = |K : KG|. Since |G| 6 |Aut(K/KG))| 6 |K : KG|, we have G = Aut(K/KG) and |Aut(K/KG)| = |K/KG| 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.

Proof. If KG1 = KG2 , then KG1 is fixed by G2 and G2 ⊂ G1 = Aut(K/KG1 ). 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 separable polynomial 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);

(4) K is a finite, normal, and separable extension 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, and G = Gal(K/F). Then mappings E 7→ GE and H 7→ KH 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

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

Now, on one hand we have E ⊂ KGE , on the other |K : KGE| = |GE| = |K : E| which implies E = KGE . Similarly, H ⊂ GKH , and |GKH| = |K : KH| = |H| implies H = GKH . This proves the first statement.

Now, by Theorem 2 every automorphism 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 automorphisms 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.

Condition E = KGE implies that for any g ∈ G we have g(E) = KgGEg −1

. 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) = √

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

E/Q is Galois. The diagrams of subgroups of K4 and the subfields E are shown below

Q( √

2, √

Q( √

2) Q( √

6) Q( √

{1}

〈τ〉〈στ〉〈σ〉

(2) We also showed last time that the extension K/Q is a Galois, where K = Q( 3 √

2,ζ3), and that Gal(K/Q) ' S3. If we choose the generators σ,τ of Gal(K/Q) to be

σ( 3 √

2) = ζ3 3 √

2, σ(ζ3) = ζ3, and τ( 3 √

2) = 3 √

2, τ(ζ3) = ζ 2 3,

we have the subfield K〈σ〉 = Q(ζ3). Note that 〈σ〉 ⊂ 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.

Q( 3 √

2,ζ3)

Q( 3 √

2) Q(ζ3 3 √

2) Q(ζ23 3 √

Q(ζ3)

{1}

〈τ〉〈τσ〉 ⟨ τσ2

⟩ 〈σ〉

(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 + √

= x4 − 10x2 + 1.

(4) Let K be the splitting field of x8 − 2 over Q. Then the extension K/Q is Galois and

K = Q( 8 √

2,ζ8) = Q( 8 √

2, √

2, i) = Q( 8 √

2, i).

Minimal polynomials for 8 √

2 and i over Q are respectively x8 − 2 and x2 + 1. Since the extension Q( 8

√ 2, i)/Q( 8

√ 2) is quadratic, we see that

|K : Q| = |K : Q( 8 √

2)| · |Q( 8 √

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,

where we set ζ = ζ8. Therefore, G = 〈σ,τ〉. Now, ζ = (1 + i) √ 2 2

hence

σ(ζ) = (1 + i) σ(

8 √

2)4

2 = ζ5 = −ζ and τ(ζ) = (1 − i)

√ 2

2 = ζ7,

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 = τ2 = 1, στ = τσ3

⟩ .

Determining the lattice of subgroups of G is a straightforward, although rather boring, problem with the following answer:

⟨ σ2,τ

⟩ 〈σ〉

⟨ σ2,τσ3

⟩ ⟨ σ4,τσ6

⟩ ⟨ σ4,τ

⟩ ⟨ σ2 ⟩ ⟨

τσ3 ⟩

〈τσ〉

⟨ τσ2

⟩ ⟨ τσ6

⟩ ⟨ τσ4

⟩ 〈τ〉

⟨ σ4 ⟩

An observation that for a subgroup H ⊂ G one has |KH : 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 〈σ〉, and |Q(i) : Q| = |G : 〈σ〉 | = 2, from which we conclude that Q(i) = K〈σ〉. 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

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( √

2) Q(i) Q(− √

Q(i 4 √

2) Q( 4 √

2) Q(i, √

2) Q((1 + i) 4 √

2) Q((1 − i) 4 √

Q(ζ 8 √

2) Q(ζ3 8 √

2) Q(ζ2 8 √

2) Q( 8 √

2) Q(i, 4 √

Q(i, 8 √