    # Fundamental Theorem of Galois Theory

The greatest enemy of knowledge is not ignorance, it is the illusion of knowledge, Stephen Hawking

Theorem. Let K/F be a finite extension. The following statements are equivalent:

1. K/F is Galois.
2. F is the fixed field of Aut(K/F), i.e., F = KGal(K/F)
3. A finite extension K/F is Galois iff the order of the Galois group equals the degree of the extension, i.e., |Gal(K/F)| = [K : F]
4. K/F is a normal, finite, and separable extension.
5. K is the splitting field of a separable polynomial f ∈ F[x] over F.
6. Let Char(F)=0 or F be finite (or more generally F is perfect). Then, a finite extension K/F is Galois if and only if K/F is normal.

Lemma. Let K/F be a finite Galois extension. Suppose an intermediate field L (F ⊆ L ⊆ K). Then, L is Galois over K and Gal(K/L) is a subgroup of G.

Proof.

1. K/F is Galois ⇒ K/F is both normal and separable. Since K is normal and finite over F, it is the splitting field of a polynomial f ∈ F[x] ⇒ [F ⊆ L ⊆ K] K is also the splitting field of f ∈ L[x] ⇒ L/K is normal.
2. To see L/K is separable, suppose α ∈ L, and consider mFα and mLα to be their minimal polynomials over F and L respectively. Then, the minimal polynomial mLα of α over L divides mFα, and [K/F is separable] mFα has no double roots in its splitting field ⇒ [Recall. Lemma If g divides f (g | f) and if f is separable over F ⇒ g is separable.] mLα is separable, α was taken an arbitrary element of L ⇒ L/F is separable ⇒ Galois.
3. Since L is an intermediate field (F ⊆ L ⊆ K). Then, Gal(K/L) is a subgroup of Gal(K/F) = G because ∀σ ∈ Gal(K/L), σ: K → K is an L-automorphism of K, so σ(α) = α ∀α ∈ L, and in particular, σ(α) = α ∀α ∈ F ⇒ σ ∈ Gal(K/F), and it is also a subgroup Gal(K/L) ≤ Gal(K/F).

# Fundamental Theorem of Galois Theorem

Lemma. Suppose an intermediate field L (F ⊆ L ⊆ K) corresponds to the subgroup H = Gal(K/L) under the Galois correspondence. Then, for each σ ∈ G(K/F), we have Gal(K/σL) = σHσ-1.

Proof:

Let σ ∈ G, i.e., σ is an automorphism of K, i.e., σ: K → K, σ fixes F. Consider σL = [abusing notation] σ|L = {σ(α) = [abusing notation] σα | α ∈ L}

σ|L: L → K, σL = σ|L ⊆ K because σ is an automorphism of K.

Let τ ∈ H = Gal(K/L), α ∈ L, σα ∈ σL ⇒ (στσ-1)(σα) = [We are “just” working with composition of mapping] (στ)(σ-1σα) = (στ)(α) = σ(τα) = [τ ∈ H = Gal(K/L), τ fixes L, and α ∈ L] σ(α) = [abusing notation] σα, and therefore (στσ-1)(σα) = σα ⇒ [στσ-1 is an automorphism of K and fixes σL] ∀τ ∈ H: στσ-1 ∈ Gal(K/σL) ⇒ σHσ-1 ⊆ Gal(K/σL).

We are going to demonstrate that they are indeed equal. Consider that L ≋ σL and this is true because every field homomorphism is injective, and it is also surjective because it goes to σL by definition ⇒ L ≋ σL as F-vector spaces ⇒ [L : F] = [σL : F] ⇒ [K : L] = [K : σL]

|H| = |Gal(K/L)| = [K/F Galois ⇒ K/L Galois, K/L is Galois↭|Gal(K/L)|=[K : L]] [K : L] = [K : σL]

The order of the Galois Group is always at most the degree of the extension, so |Gal(K/σL)| ≤ [K : σL] = |H| = [Group Theory, the conjugate subgroup has the same order as the subgroup; Sketch: f: H → xHx-1, f(h) = xhx-1 is bijective] |σHσ-1|

∀τ ∈ H: στσ-1 ∈ Gal(K/σL) ⇒ σHσ-1 ⊆ Gal(K/σL) ⇒ σHσ-1 is a subgroup of Gal(K/σL) and |Gal(K/σL)| ≤ |σHσ-1| ⇒ [In words, the order of the group is at most the order of the subgroup] σHσ-1 = Gal(K/σL)

Fundamental Theorem of Galois Theorem. Let K/F be a finite separable extension, i.e., a Galois extension. Let G = Gal(K/F) be the Galois group. Then, the following statements holds, (1) There is an inclusion-reversing bijective map or correspondence from subgroups of G and intermediate fields of K/F, given by H → KH, its inverse is defined by L → Gal(K/L). H1 ⊇ H2 ⇒ KH1 ⊆ KH2 and L1 ⊆ L2 ⇒ Gal(K/L1) ⊇ Gal(K/L2). Futhermore, it satisfies the following equality |H| = [K : KH] and [G : H] = [KH : F]

This map bijection fails when K/F is not Galois, e.g. K = $\mathbb{Q}(\sqrt{2}), F = \mathbb{Q}$ is not Galois, G = Gal(K/F) = {1}. There is exactly one subgroup of G, but there are two distinct intermediate fields, namely F and K.

(2) Suppose an intermediate field L (F ⊆ L ⊆ K) corresponds to the subgroup H under the Galois correspondence. K/L is always normal (hence Galois). L is Galois over F if and only if H = Gal(K/L) is a normal subgroup of G, H = Gal(K/L) ◁ G In this case, the Galois group of L/F is isomorphic to the quotient group G/H, i.e., Gal(L/F) ≋ G/Gal(K/L)

Proof.

(1.a) The map described in the theorem is well-defined. Let H be a subgroup of G (H ≤ G), then KH is an intermediate field, F ⊆ KH ⊆ K because every element of KH is an F-automorphism of K, a map K → K, but it fixes F pointwise.

Lemma. Let K/F be a finite Galois extension. Suppose an intermediate field L (F ⊆ L ⊆ K). Then, L is Galois over K and Gal(K/L) is a subgroup of G. Therefore, these maps are both well-defined.

(1.b) Next, we want to show that the map is bijective.

H → KH → Gal(K/KH) = [Let K be any field and let G be a finite group of automorphisms of K. Suppose F is the fixed field of G. Then G = Gal(K/F) = Gal(K/KG).] H

L → Gal(K/L) → KGal(K/L) = [Galois extension F ⊆ L ⊆ K, K/F Galois ⇒ K/L is GaloisK/L Galois, (2) KGal(K/L) = L] L

Hence, both maps are inverses of each other ⇒ there exist a bijection.

(1.c) This is an inclusion reversing bijection:

Let’s prove that H1 ⊆ H2 ⇒ KH1 ⊇ KH2. Let α ∈ KH2 ⇒ σ(α) = α ∀σ ∈ H2 ⇒ σ(α) = α ∀σ ∈ H1 ⇒ α ∈ KH1

Now, we want to demonstrate that L1 ⊆ L2 ⇒ Gal(K/L1) ⊇ Gal(K/L2). Let σ ∈ Gal(K/L2) ⇒ σ: K → K, σ|L2 = id ⇒ [L1 ⊆ L2] σ: K → K, σ|L1 = id ⇒ σ ∈ Gal(K/L1)

(1.d) Let H be a subgroup of G, H ≤ G. Then, [K : KH] = [Degree of Fixed Fields, [K: KH] = |H|, H be a finite group of automorphisms] |H|

We know that |G| = |H|[G : H], [K : F] = [K : KH][KH : F], [K : KH] = |H|, and |G| = [because K/F is Galois] [K : F] ⇒ [G : H] = [KH : F]

(2) Let L be an intermediate field of the extension K/F (F ⊆ L ⊆ K), consider H = Gal(K/L) is a subgroup of G, H = Gal(K/L) ≤ G = Gal(K/F)

Recall Lemma. Suppose an intermediate field L (F ⊆ L ⊆ K) corresponds to the subgroup H = Gal(K/L) under the Galois correspondence. Then, for each σ ∈ G(K/F), we have Gal(K/σL) = σHσ-1.

Suppose L is Galois over F ⇒ σL = L ∀σ ∈ G (😄)

(😄) σ:K → K, L/F Galois ⇒ L/F normal ⇒ [Every embedding of L in $\bar K$ induces an automorphism from L to itself] σ|L: L → L

∀σ ∈ G, σL = L ⇒ Gal(K/L) = Gal(K/σL) ⇒ ∀σ ∈ G, H = Gal(K/L) = Gal(K/σL) = σHσ-1 ⇒ ∀σ ∈ G, H = σHσ-1 ⇒ [By definition] H is normal in G ∎

Suppose that H is normal in G ⇒ [By definition] ∀σ ∈ G, H = σHσ-1 ⇒ ∀σ ∈ G, H = Gal(K/L) = Gal(K/σL) = σHσ-1, that is, Gal(K/L) = Gal(K/σL) ⇒ ∀σ ∈ G, KGal(K/σL) = KGal(K/L) ⇒ [K/σL and K/L are Galois] ∀σ ∈ G, σL = L ⇒ [Separability is transitive, K/F is separable ⇒ K/L and L/F are separable] L is not just normal, but Galois over F.

Therefore, restricting σ to L we get a homomorphism of groups as follows, Φ: G = Gal(K/L) → Gal(L/F), σ → σ|L (Notice that it is well defined because σ: K → L, σ|L: L → σL = L).

Ker(Φ) = {σ ∈ G | σ|L = id} = {σ: K → K | σ|L = id} = Gal(K/L) = H ⇒ [H is normal in G] We have an injective map, G/H → Gal(L/F)

[L : F] = [KH : F] = [Part 1. Fundamental Theorem of Galois theorem] |G : H| ≤ [We have an injective map, G/H → Gal(L/F)] |Gal(L/F)| ≤ [The order of the Galois group is at most the degree of the extension] [L : F] ⇒ [L : F] = |Gal(L/F)| ⇒ [Characterization Galois extension] L/F is Galois.

We have an injective map, G/H → Gal(L/F), but |G : H| = |Gal(L/F)| ⇒ [It is also surjective] G/H ≋ Gal(L/F), that is, G/Gal(K/L) ≋ Gal(L/F)∎

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