Since F is normal over K, one has σ(F

finite dimensional Galois extension

In this section, we are going to prove the fundamental theorem for finite dimensional Galois extension.

Let F/K be an field extension, we define the Galois group of F over K, denoted Gal_F/K or G_F/K or Aut_K(F ), as

Gal_F/K := {σ|σ ∈ AutF, σ|_K = 1_K}.

It’s clear that for σ ∈ Gal_F/K and u ∈ F algebraic over K with mini- mal polynomial p(x), then σ(u) satisfies the same minimal polynomial.

On the other hand, if F/K is normal, let u, v be two elements having the same minimal polynomial p(x), then we claim that there is an σ ∈ GalF/K such that σ(u) = v. To see this, we fix an algebraic closure K containing F . There is an K-isomorphism σ0 : K(u) → K(v) which extends to an embedding σ : F → K . Since F is normal over K, one has σ(F ) ⊂ F . And hence σ ∈ AutF .

Example 0.1. Consider the field F := Q(√³

2, ω) which is a splitting field of x³−2 over Q. Thus it’s normal over Q. One can check that the Galois group Gal_F/Q is generated by σ, τ that σ(√³

2) = √³

2ω, σ(ω) = ω, and τ (√³

2) = √³

2, τ (ω) = ω². It’s easy to check that Gal_F/Q∼= S3. Example 0.2. Consider the field F := Q(√³

2) over Q. Then it’s easy to check that Gal_F/Q = {1_F}.

There is a natural correspondence between subgroups of Galois groups and intermediate fields. To be precise, fix an extension F/K. Let H < G := GalF/K be a subgroup. One can define

H⁰ := {u ∈ F |σ(u) = u, ∀σ ∈ H}.

It’s clear that this is a field. On the other hand, given and intermediate field L such that K ⊂ L ⊂ F , then one can define

L⁰ := {σ ∈ Gal_F/K|σ(u) = u, ∀u ∈ L} = {σ ∈ Gal_F/K|σ_|L= 1_L}.

It’s easy to check the following properties:

Proposition 0.3. Let F/K be an extension with Galois group G. Let L be an intermediate field, i.e. K ⊂ L ⊂ F , and H < G is a subgroup.

(1) F⁰ = {1_F}, K⁰ = G, and {1_F}⁰ = F . (2) For any L, one has L ⊂ L⁰⁰, L⁰ = L⁰⁰⁰. (3) For any H, one has H < H⁰⁰, H⁰ = H⁰⁰⁰.

(4) For any intermediate fields L ⊂ M, one has M⁰ < L⁰. (5) For any subgroups J < H, one has H⁰ ⊂ J⁰.

1

Proof. Most of the proof follows directly from the definition. We only sketch the proof for L⁰ = L⁰⁰⁰.

By L ⊂ L⁰⁰ and (4), one has

(L⁰⁰)⁰ < L⁰. On the other hand, by (5), one has

L⁰ < (L⁰)⁰⁰.

We are done. ¤

Proposition 0.4. There is a one-to-one correspondence between {L|K ⊂ L ⊂ F, L⁰⁰ = L} ↔ {H|H < G, H⁰⁰ = H}.

Proof. The correspondence is given by L 7→ L⁰ (or H 7→ H⁰).

To show the injective, one sees that if L⁰₁ = L⁰₂, then L₁ = L⁰⁰₁ = L⁰⁰₂ = L₂.

For any H with H⁰⁰ = H, we take L = H⁰, then H = L⁰. It suffices to check that L⁰⁰ = L. This follows from the fact that H⁰⁰⁰ = H⁰. ¤ In the proposition, one might expect that G⁰ = K. However, this is not always the case (see e.g. Example 2). For extension with this property, we call it Galois. It turns out that this naive definition is a very delicate one which leads to some nice properties.

Definition 0.5. An extension F/K is said to be Galois if (Gal_F/K)⁰ = K.

Theorem 0.6 (Fundamental theorem of finite dimensional Galois ex- tension). Let F/K be a finite dimensional Galois extension with Galois group G, then

(1) There is an one-to-one correspondence between {L|K ⊂ L ⊂ F } ↔ {H|H < G}.

(2) The corresponding degree are equal. That is, if K ⊂ L ⊂ M ⊂ F , then [M : L] = [L⁰ : M⁰]. And if J < H < G, then [H : J] = [J⁰ : H⁰].

(3) An intermediate field E is Galois over K if and only if E⁰C G.

And in this case, Gal_E/K ∼= G/E⁰. Proof. Step 1. [M : L] ≥ [L⁰ : M⁰].

We prove the case that M = L(u) for some u ∈ M and by induction on [M : L], we are done. Suppose now that M = L(u) and let p(x) be the minimal polynomial of u over L. Let S be the set of roots of p(x) in F . Then one has a map

Φ : L⁰ → S, σ 7→ σ(u).

One can check that Φ induces an injective map L⁰/M⁰ → S. Hence one has

[L⁰ : M⁰] = |L⁰/M⁰| ≤ |S| ≤ deg(p(x)) = [M : L].

Step 2. [H : J] ≥ [J⁰ : H⁰].

Let n = [H : J]. Suppose on the contrary that there are n + 1 elements u1, ..., un+1 ∈ J⁰ linearly independent over H⁰.

We consider the equation P_n+1

i=1 u_ix_i = 0 in F Consider now a set of representative of H/J, denoted {e = σ₁, ..., σ_n}. By applying σ_i to the above equation. Then one has a system of linear equations in F .

(∗)









σ1(u1)x1+ σ1(u2)x2+ ... + σ1(un+1)xn+1 = 0 σ2(u1)x1+ σ2(u2)x2+ ... + σ2(un+1)xn+1 = 0 ...

σ_n(u₁)x₁+ σ_n(u₂)x₂ + ... + σ_n(u_n+1)x_n+1 = 0

Pick a solution in F with smallest number of non-zero a_i’s, may assume it’s (a₁, ..., a_s, 0..., 0) and a₁ = 1.

If there is an τ ∈ H such that τ (a₂) 6= a₂, then by applying τ to the system (∗), one get the same system of equations with a solution (τ (a₁), τ (a₂), ..., τ (a_s), 0, ..., 0 . Hence

(a1, ..., as, 0..., 0) − (τ (a1), τ (a2), ..., τ (as), 0, ..., 0) = (0, a2− τ (a2), ..., 0) is a non-zero solution of smaller length. This is the required contradic- tion.

To find τ . We look at u₁a₁ + ... + u_sa_s = 0. Since {u₁, ..., u_s} is independent over H⁰, not all a₁ is in H⁰. We may assume that a₂ 6∈ H⁰. Hence there is a τ ∈ H such that τ (a₂) 6= a₂. We are done.

Step 3. We show that every intermediate field L, L⁰⁰ = L. And every subgroup H < G, H⁰⁰ = H.

By Step 1, one has

[L⁰⁰ : K] = [L⁰⁰ : K⁰⁰] ≤ [K⁰ : L⁰] ≤ [L : K],

however, one has L ⊂ L⁰⁰. Thus one has L = L⁰⁰. Similarly, one can prove that H⁰⁰= H by considering [H⁰⁰: {1_F}].

Step 4. [M : L] = [L⁰ : M⁰] and [H : J] = [J⁰ : H⁰].

This follows from [M : L] = [M : K]/[L : K] = [K⁰ : M⁰]/[K⁰ : L⁰] = [L⁰ : M⁰]. And the other one is similar.

Step 5. F/K is normal and separable.

Given u ∈ F , with minimal polynomial p(x) over K. As in the proof of Step 1. One has [K(u)⁰ : K⁰] ≤ |S| ≤ deg(p(x)) = [K(u) : K]. By Step 4, they are equalities. In particular, every root of p(x) is in F and there is no multiple roots. Thus F is normal and separable over K.

Step 6. If N C G, then N⁰ is stable. That is, for all σ ∈ G, σ(N⁰) ⊂ N⁰ (indeed = N⁰).

Since N C G, for all σ ∈ G and for all τ ∈ N, one has σ⁻¹τ σ ∈ N.

Thus, σ⁻¹τ σ(N⁰) = N⁰. It follows that τ σ(N⁰) = σ(N⁰), for all τ ∈ N.

Hence σ(N⁰) is fixed by all N and thus σ(N⁰) ⊂ N⁰.

Step 7. If E is a stable intermediate subfield. Then the restriction map Gal_F/K → Gal_E/K is well-defined and surjective.

Since E is stable, then σ_|E ∈ Gal_E/K for any σ ∈ Gal_F/K. Moreover, let τ ∈ Gal_E/K, by the extension theorem, there is an extension τ : F → K. Since F is normal over K, τ is in fact an automorphism of F .

Step 8. If an intermediate field E is stable, then E/K is Galois.

To see this, it suffices to show that for any u ∈ E −K, there is an σ ∈ Gal_E/K such that σ(u) 6= u. Fix any F 3 v 6= u with the same minimal polynomial as u. There is an K-isomorphism σ₀ : K(u) → K(v) such that σ(u) = v. σ can be extended to an embedding σ : F → K, which gives an automorphism of F . The restriction σ = σ_|E gives an automorphism of E that σ(u) 6= u.

Step 9. If E/K is Galois, then E is stable.

One first notices that E/K is normal. For every σ ∈ Gal_F/K, σ gives an embedding σ|E : E → K. Since E/K is normal, σ|E is an automorphism of E. And hence E is stable under the Galois group Gal_F/K action.

Step 10. If E is stable, then E⁰ is normal.

This can be checked directly. For all σ ∈ G and τ ∈ E⁰ and for all u ∈ E,

σ⁻¹τ σ(u) = σ⁻¹τ (σ(u)) = σ⁻¹σ(u) = u,

since σ(u) ∈ E. Therefore, σ⁻¹τ σ ∈ E⁰. ¤ Remark 0.7. Some of the result we proved still true in a more general setting. We list some here:

(1) If F/K is an extension, and an intermediate field E is stable, then E⁰C GalF/K.

(2) Let F/K be an extension. If N C Gal_F/K, then H⁰ is stable.

(3) If F/K is Galois, and E is a stable intermediate field, then E is Galois over K. (finite-dimensional assumption is unnecessary here)

(4) An intermediate field E is algebraic and Galois over K, then E is stable.

We conclude this section with the following theorem concerning the relation between Galois extension, normal extension and splitting fields.

Theorem 0.8. Let F/K be an extension, then the following are equiv- alent

(1) F is algebraic and Galois over K.

(2) F is separable over K and F is a splitting field over K of a set S of polynomials.

(3) F is a splitting field of separable polynomials in K[X].

(4) F/K is normal and separable.

Proof. Fix u ∈ F with minimal polynomail p(x) over K. Let {u = u₁, ..., u_r} be distinct roots of p(x) in F . For any σ, then σ permutes {u = u₁, ..., u_r}. Thus f (x) := Q_r

i=1(x − u_i) is invariant under σ.

Hence f (x) ∈ K[x]. It follows that f (x) = p(x). This proved that (1) ⇒ (2), (3), (4).

One notices that (2) ⇔ (4). Thus it remains to show that (2) ⇒ (3), and (3) ⇒ (1).

For (2) ⇒ (3), let f (x) ∈ S and let g(x) be an monic irreducible component of f (x). Since f (x) splits in F , it’s clear that g(x) is an minimal polynomial of some element in F . Moreover, since F/K is separable, g(x) is separable. One sees that F is in fact a splitting field of such g(x)’s.

For (3) ⇒ (1), we first note that F/K is algebraic since F is a split- ting field. We shall prove that (4) ⇒ (1). The implication (3) → (4) follows from a general fact about separable extension that an algebraic extension F/K is separable if F is generated by separable elements.

To this end, pick any u ∈ F − K, with minimal polynomial p(x) of degree ≥ 2 and separable. Hence there is a different root, say v, of p(x) in F . It’s natural to consider the K-isomorphism σ : K(u) → K(v).

Which can be extended to ¯σ : F → K. Since F is normal, ¯σ is an automorphism of F , hence in Gal_F/K sending u to v 6= u. So F/K is Galois.

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