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:

- K/F is Galois.
- F is the fixed field of Aut(K/F), i.e., F = K
^{Gal(K/F)} - 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]
- K/F is a normal, finite, and separable extension.
- K is the splitting field of a separable polynomial f ∈ F[x] over F.
- 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.

- 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**. - To see L/K is separable, suppose α ∈ L, and consider m
_{F}α and m_{L}α to be their minimal polynomials over F and L respectively. Then, the minimal polynomial m_{L}α of α over L divides m_{F}α, and [K/F is separable] m_{F}α 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.] m^{L}α is separable, α was taken an arbitrary element of L ⇒**L/F is separable ⇒ Galois**. - 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)**.

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^{-1}σα) = (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 → K^{H}, its inverse is defined by L → Gal(K/L). H_{1} ⊇ H_{2} ⇒ K^{H1} ⊆ K^{H2} and L_{1} ⊆ L_{2} ⇒ Gal(K/L_{1}) ⊇ Gal(K/L_{2}). Futhermore, it satisfies the following equality |H| = [K : K^{H}] and [G : H] = [K^{H} : F]

This map bijection fails when K/F is not Galois, e.g. K = $\mathbb{Q}(\sqrt[3]{2}), F = \mathbb{Q}$ 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 K^{H} is an intermediate field, F ⊆ K^{H} ⊆ K because every element of K^{H} 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 → K^{H} → Gal(K/K^{H}) = [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/K^{G}).] H

L → Gal(K/L) → K^{Gal(K/L)} = [Galois extension F ⊆ L ⊆ K, K/F Galois ⇒ K/L is Galois ↭_{K/L Galois, (2)} K^{Gal(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 H_{1} ⊆ H_{2} ⇒ K^{H1} ⊇ K^{H2}. Let α ∈ K^{H2} ⇒ σ(α) = α ∀σ ∈ H_{2} ⇒ σ(α) = α ∀σ ∈ H_{1} ⇒ α ∈ K^{H1}

Now, we want to demonstrate that L_{1} ⊆ L_{2} ⇒ Gal(K/L_{1}) ⊇ Gal(K/L_{2}). Let σ ∈ Gal(K/L_{2}) ⇒ σ: K → K, σ|_{L2} = id ⇒ [L_{1} ⊆ L_{2}] σ: K → K, σ|_{L1} = id ⇒ σ ∈ Gal(K/L_{1})

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

We know that *|G|* = **|H|**[G : H], *[K : F]* = **[K : K ^{H}]**[K

(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, K^{Gal(K/σL)} = K^{Gal(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] = [K^{H} : 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)∎

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Fields and Galois Theory. Howie, John. M. Springer Undergraduate Mathematics Series.
- Fields and Galois Theory. Morandi. P., Springer.
- Fields and Galois Theory. By Evan Dummit, 2020.