# 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 = $\mathbb{Q}(\sqrt[3]{2}), F = \mathbb{Q}$ 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 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)∎

# Bibliography

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