## 9.21 Galois theory

Here is the definition.

Definition 9.21.1. A field extension $E/F$ is called Galois if it is algebraic, separable, and normal.

It turns out that a finite extension is Galois if and only if it has the “correct” number of automorphisms.

Lemma 9.21.2. Let $E/F$ be a finite extension of fields. Then $E$ is Galois over $F$ if and only if $|\text{Aut}(E/F)| = [E : F]$.

Proof. Assume $|\text{Aut}(E/F)| = [E : F]$. By Lemma 9.15.9 this implies that $E/F$ is separable and normal, hence Galois. Conversely, if $E/F$ is separable then $[E : F] = [E : F]_ s$ and if $E/F$ is in addition normal, then Lemma 9.15.9 implies that $|\text{Aut}(E/F)| = [E : F]$. $\square$

Motivated by the lemma above we introduce the Galois group as follows.

Definition 9.21.3. If $E/F$ is a Galois extension, then the group $\text{Aut}(E/F)$ is called the Galois group and it is denoted $\text{Gal}(E/F)$.

If $L/K$ is an infinite Galois extension, then one should think of the Galois group as a topological group. We will return to this in Section 9.22.

Lemma 9.21.4. Let $K/E/F$ be a tower of algebraic field extensions. If $K$ is Galois over $F$, then $K$ is Galois over $E$.

Lemma 9.21.5. Let $L/K$ be a finite separable extension of fields. Let $M$ be the normal closure of $L$ over $K$ (Definition 9.16.4). Then $M/K$ is Galois.

Proof. The subextension $M/M_{sep}/K$ of Lemma 9.14.6 is normal by Lemma 9.15.4. Since $L/K$ is separable we have $L \subset M_{sep}$. By minimality $M = M_{sep}$ and the proof is done. $\square$

Let $G$ be a group acting on a field $K$ (by field automorphisms). We will often use the notation

$K^ G = \{ x \in K \mid \sigma (x) = x \ \forall \sigma \in G\}$

and we will call this the fixed field for the action of $G$ on $K$.

Lemma 9.21.6. Let $K$ be a field. Let $G$ be a finite group acting faithfully on $K$. Then the extension $K/K^ G$ is Galois, we have $[K : K^ G] = |G|$, and the Galois group of the extension is $G$.

Proof. Given $\alpha \in K$ consider the orbit $G \cdot \alpha \subset K$ of $\alpha$ under the group action. Consider the polynomial

$P = \prod \nolimits _{\beta \in G \cdot \alpha } (x - \beta ) \in K[x]$

The key to the whole lemma is that this polynomial is invariant under the action of $G$ and hence has coefficients in $K^ G$. Namely, for $\tau \in G$ we have

$P^\tau = \prod \nolimits _{\beta \in G \cdot \alpha } (x - \tau (\beta )) = \prod \nolimits _{\beta \in G \cdot \alpha } (x - \beta ) = P$

because the map $\beta \mapsto \tau (\beta )$ is a permutation of the orbit $G \cdot \alpha$. Thus $P \in K^ G[x]$. Since also $P(\alpha ) = 0$ as $\alpha$ is an element of its orbit we conclude that the extension $K/K^ G$ is algebraic. Moreover, the minimal polynomial $Q$ of $\alpha$ over $K^ G$ divides the polynomial $P$ just constructed. Hence $Q$ is separable (by Lemma 9.12.4 for example) and we conclude that $K/K^ G$ is separable. Thus $K/K^ G$ is Galois. To finish the proof it suffices to show that $[K : K^ G] = |G|$ since then $G$ will be the Galois group by Lemma 9.21.2.

Pick finitely many elements $\alpha _ i \in K$, $i = 1, \ldots , n$ such that $\sigma (\alpha _ i) = \alpha _ i$ for $i = 1, \ldots , n$ implies $\sigma$ is the neutral element of $G$. Set

$L = K^ G(\{ \sigma (\alpha _ i); 1 \leq i \leq n, \sigma \in G\} ) \subset K$

and observe that the action of $G$ on $K$ induces an action of $G$ on $L$. We will show that $L$ has degree $|G|$ over $K^ G$. This will finish the proof, since if $L \subset K$ is proper, then we can add an element $\alpha \in K$, $\alpha \not\in L$ to our list of elements $\alpha _1, \ldots , \alpha _ n$ without increasing $L$ which is absurd. This reduces us to the case that $K/K^ G$ is finite which is treated in the next paragraph.

Assume $K/K^ G$ is finite. By Lemma 9.19.1 we can find $\alpha \in K$ such that $K = K^ G(\alpha )$. By the construction in the first paragraph of this proof we see that $\alpha$ has degree at most $|G|$ over $K$. However, the degree cannot be less than $|G|$ as $G$ acts faithfully on $K^ G(\alpha ) = L$ by construction and the inequality of Lemma 9.15.9. $\square$

Theorem 9.21.7 (Fundamental theorem of Galois theory). Let $L/K$ be a finite Galois extension with Galois group $G$. Then we have $K = L^ G$ and the map

$\{ \text{subgroups of }G\} \longrightarrow \{ \text{subextensions }L/M/K\} ,\quad H \longmapsto L^ H$

is a bijection whose inverse maps $M$ to $\text{Gal}(L/M)$. The normal subgroups $H$ of $G$ correspond exactly to those subextensions $M$ with $M/K$ Galois.

Proof. By Lemma 9.21.4 given a subextension $L/M/K$ the extension $L/M$ is Galois. Of course $L/M$ is also finite (Lemma 9.7.3). Thus $|\text{Gal}(L/M)| = [L : M]$ by Lemma 9.21.2. Conversely, if $H \subset G$ is a finite subgroup, then $[L : L^ H] = |H|$ by Lemma 9.21.6. It follows formally from these two observations that we obtain a bijective correspondence as in the theorem.

If $H \subset G$ is normal, then $L^ H$ is fixed by the action of $G$ and we obtain a canonical map $G/H \to \text{Aut}(L^ H/K)$. This map has to be injective as $\text{Gal}(L/L^ H) = H$. Hence $|G/H| = [L^ H : K]$ and $L^ H$ is Galois by Lemma 9.21.2.

Conversely, assume that $K \subset M \subset L$ with $M/K$ Galois. By Lemma 9.15.7 we see that every element $\tau \in \text{Gal}(L/K)$ induces an element $\tau |_ M \in \text{Gal}(M/K)$. This induces a homomorphism of Galois groups $\text{Gal}(L/K) \to \text{Gal}(M/K)$ whose kernel is $H$. Thus $H$ is a normal subgroup. $\square$

Lemma 9.21.8. Let $L/M/K$ be a tower of fields. Assume $L/K$ and $M/K$ are finite Galois. Then we obtain a short exact sequence

$1 \to \text{Gal}(L/M) \to \text{Gal}(L/K) \to \text{Gal}(M/K) \to 1$

of finite groups.

Proof. Namely, by Lemma 9.15.7 we see that every element $\tau \in \text{Gal}(L/K)$ induces an element $\tau |_ M \in \text{Gal}(M/K)$ which gives us the homomorphism on the right. The map on the left identifies the left group with the kernel of the right arrow. The sequence is exact because the sizes of the groups work out correctly by multiplicativity of degrees in towers of finite extensions (Lemma 9.7.7). One can also use Lemma 9.15.7 directly to see that the map on the right is surjective. $\square$

Comment #4405 by TS on

In Lemma 09I3, "acting faithfully on K..." seems to be "acting freely on K...".

Comment #4507 by on

@TS: no. The Galois group of an extension never acts freely.

Comment #6984 by YNChen on

In general for an algebraic field extension $K/F$, does the Galois closure exist? Or if $K/F$ is finite and not necessarly separable, is the Galois closure of it finite?

Comment #6985 by on

Finite inseparable field extensions have a normal closure but it isn't Galois. The simplest example of this is the following fact: any degree $p$ purely inseparable extension $L/K$ is normal but not Galois (good exercise). If you have any (possibly infinite) algebraic field extension, then you can define and construct a normal closure, but again this need not be Galois, and in fact it will be Galois if and only if you start with a separable algebraic extension.

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