Rings, Modules, Fields

14 Galois extensions

Definition 14.1. An extension is Galois if it is normal and separable.

Corollary 14.2. The composita and intersections of Galois extensions is Galois. If \(M/L/K\) are extensions, and \(M/K\) is Galois, \(M/L\) is Galois.

Proof. The corresponding properties hold for normal and separable extensions. □

For a Galois extension we write \(\Gal (L/K)\) for \(\Aut (L/K)\), and call it the Galois group. The normal closure for a separable extension is called the Galois closure.

For a subgroup \(H \subset \Aut (L/K)\), we let \(L^H\), denote the fixed field, or the subfield fixed by \(H\).

Proposition 14.3. TFAE for a finite extension (the first three are equivalent in general):
- 1. \(L/K\) is Galois.
- 2. \(L\) is the splitting field of some separable polynomials.
- 3. \(L^{\Gal (L/K)} = K\).
- 4. \(|\Gal (L/K)| = [L:K]\).

Proof. The first two are clearly equivalent. The third is implied by the first since the Galois group acts transitively on the conjugates and every element not in \(K\) has more than one conjugate iff the extension is separable. Conversely, if \(L^{\Aut (L/K)} = K\), for any \(\alpha \in L\) where \(\alpha _i\) are its conjugates in \(L\), the polynomial \(\Pi _i(t-\alpha _i)\) is \(\Aut (L/K)\)-invariant, so has coefficients in \(K\), so \(\alpha \) is separable and all its conjugates are in \(L\). For the fourth, pick a primitive element, and note that the Galois group acts simply and transitively on its conjugates. For the converse, we can assume \(L\) is separable, since the automorphism group of an extension is a subgroup of the automorphism group of the separable part. The automorphism group acts on the conjugates of a primitive element simply, so if there are \([L:K]\) automorphisms, the extension is normal. □

If \(M/L/K\) are normal extensions, there is a map \(\Aut (M/K) \to \Aut (L/K)\) given by restriction, which is surjective by the extension lemma. We get the exact sequence:

(-tikz- diagram)

We can endow \(\Aut (M/K)\) with the profinite topology, where we view it as the limit of all the finite normal intermediate extensions. If the extension is finite this is the discrete topology. The Galois group of the algebraic closure with this topology is called the absolute Galois group of the field.

We can slightly strengthen the statement of \((3)\) in the previous proposition. This is the key to Galois Theory.

Lemma 14.4. If \(L/K\) is a finite Galois extension, and \(H\) is a subgroup of \(\Gal (L/K)\), then \([\Gal (L/K):H] = [L^H:K]\).

Proof. Choose a primitive element \(\alpha \) of \(L/L^H\), and consider \(\Pi _i(t-\alpha _i)\) as before where the product is over the \(H\)-conjugates of \(\alpha \). This polynomial has coefficients in \(L^H\), so is the minimal polynomial of \(\alpha \). \(H\) acts simply so we get the equality. □

Theorem 14.5 (Fundamental Theorem of Galois Theory). Let \(M/K\) be a Galois extension. Then, there is a correspondence between closed subgroups of \(\Gal (M/K)\) and intermediate extensions given by taking the fixed field, and taking the Galois group over the intermediate extension. The degree of the extension matches the index of the subgroup. This is a bijection, and restricts to a bijective correspondence between normal intermediate extensions and normal closed subgroups.

Proof. First assume our extension is finite.

For a subset \(H \subset \Gal (M/K)\) we can ask which elements of \(\Gal (M/K)\) fix \(M^H\). Certainly the subgroup generated by \(H\) does, and the previous results give \([\Gal (M/K):H] = [M^H:K] = [M:K]/[M:M^H] = [\Gal (M/K):\Gal (M^H/K)]\), so \(H = \Gal (M^H/K)\). Conversely, given an intermediate extension, by the extension lemma it is the fixed field of the automorphisms that fix it.

In this correspondence, our subgroup \(H\) is normal iff for an automorphism \(\tau \) that fixes \(M^H\), and \(\sigma \) that fixes \(K\), \(\tau \sigma M^H = \sigma M^H\) iff automorphisms fixing \(M^H/H\) fix every conjugate of \(M^H/H\) iff \(M^H\) contains its conjugates iff \(M^H\) is normal.

The extension to infinite groups is purely formal: it is easy to see that the fixed field of any intermediate extension is closed, since if \(\tau _i \to \tau \) and each \(\tau _i\) fixes \(\alpha \), then \(\tau \) does too. Now if there is an element not in a closed subgroup \(H\) fixing its fixed field, there is some open set containing \(H\) missing it, which arises from picking finitely many subsets of the Galois group of finite sub extensions, which amounts to picking some subset of the Galois group of their compositum. Then the element doesn’t fix something in this finite subextension by finite Galois theory. The rest of the argument proceeds as before. □

Corollary 14.6. Given an intermediate extension \(L\) of a Galois extension \(M/K\) corresponding to subgroup \(H\), \(\Aut (L/K) = N_{\Gal (M/K)}(H)/H\).

Proof. Automorphisms of \(M\) that fix \(L\) as a set are the same as \(\sigma \in \Gal (M/K)\) such that \(\tau \sigma L = \sigma L\) for each automorphism \(\tau \) fixing \(L\). This is exactly the normalizer of \(H\). □