Galois Extensions

Definition.
A field extension $E/F$ is called a Galois extension if it is both normal and separable. In this case, we denote the group $\operatorname{Aut}(E/F)$ by $\operatorname{Gal}(E/F)$ and call it the Galois group of the extension.

Theorem (Characterization of Finite Galois Extensions).
Let $E/F$ be a finite field extension. The following are equivalent:

$E/F$ is a Galois extension.
$E$ is the splitting field of a separable polynomial $p(x) \in F[x]$.
$|\operatorname{Aut}(E/F)| = [E:F]$.
$F = E^{\operatorname{Aut}(E/F)} := \{ e \in E : \sigma(e) = e \text{ for all } \sigma \in \operatorname{Aut}(E/F) \}$.

Proof.
$1 \Rightarrow 2$: Since $E/F$ is normal and finite, it is the splitting field of some polynomial $p(x) \in F[x]$ (see Normal Extension). As $E/F$ is also separable, each irreducible factor of $p(x)$ is a scalar multiple of some minimal polynomials of elements in $E$, and hence separable. Therefore, $p(x)$ is separable.

$2 \Rightarrow 3$ and $3\Rightarrow 1$: This will be proved in the following theore,

We will prove the equivalence with (4) in separate theorems later.

Theorem (Number of Automorphisms Characterizes Galois Extensions).
Let $E/F$ be a finite field extension, and let $\sigma : F \hookrightarrow E$ be an injective field homomorphism. Define:

$\operatorname{Iso}_\sigma(E, E) = \{ \hat\sigma : E \to E \mid \hat\sigma|_F = \sigma \}$
$\operatorname{Embed}_\sigma(F(\alpha), E) = \{ \bar\sigma : F(\alpha) \hookrightarrow E \mid \bar\sigma|_F = \sigma \}$

Then,

\[|\operatorname{Iso}_\sigma(E, E)| \leq [E : F]\]

with equality if and only if $E/F$ is a Galois extension. Taking $\sigma = \operatorname{id}_F$ proves the equivalence of (1) and (3) in the previous theorem.

Proof.
We proceed by strong induction on $[E : F]$. The base case $[E : F] = 1$ is trivial. Suppose $[E : F] > 1$, and choose $\alpha \in E \setminus F$.

Every extension $\hat\sigma \in \operatorname{Iso}_\sigma(E, E)$ restricts to $\bar\sigma = \hat\sigma|_{F(\alpha)} \in \operatorname{Embed}_\sigma(F(\alpha), E)$. By inductive hypothesis,
\[|\operatorname{Iso}_\sigma(E, E)| = \sum_{\bar\sigma \in \operatorname{Embed}_\sigma(F(\alpha), E)} |\operatorname{Iso}_{\bar\sigma}(E, E)| = |\operatorname{Embed}_\sigma(F(\alpha), E)| \cdot [E : F(\alpha)].\]
Each $\bar\sigma$ is determined by $\bar\sigma(\alpha)$, which must be a root of $\sigma(m_{\alpha, F}(x))$ in $E$. Therefore,
\[|\operatorname{Embed}_\sigma(F(\alpha), E)| \leq \deg m_{\alpha, F}(x) = [F(\alpha) : F],\]
and hence,
\[|\operatorname{Iso}_\sigma(E, E)| \leq [F(\alpha) : F] \cdot [E : F(\alpha)] = [E : F].\]
Equality holds if and only if $\sigma(m_{\alpha, F}(x))$ has all of its roots in $E$, and they are distinct. This is true for all $\alpha\in E$. Thus, we know that $E/F$ is normal and separable. If $E/F$ is the splitting field of a separable polynomial, the above equality will also hold by the same argument.

Lemma (Artin’s Lemma).
Let $E$ be a field, and let $G \subset \operatorname{Aut}(E)$ be a finite subgroup. Define the fixed field of $G$ by

\[E^G := \{ e \in E : \sigma(e) = e \text{ for all } \sigma \in G \}.\]

Then $E^G$ is a subfield of $E$, and

\[[E : E^G] \leq |G|.\]

Proof.
It is straightforward to verify that $E^G$ is a subfield of $E$.

Suppose $|G| = n$ with $G = \{ \sigma_1, \ldots, \sigma_n \}$, where $\sigma_1 = \operatorname{id}$. We will show that any $n+1$ elements $e_1, \ldots, e_{n+1} \in E$ are linearly dependent over $E^G$.

Let $V \subset E^{n+1}$ be the kernel of the $n \times (n+1)$ matrix:
\[\begin{bmatrix} \sigma_1(e_1) & \cdots & \sigma_1(e_{n+1}) \\ \vdots & \ddots & \vdots \\ \sigma_n(e_1) & \cdots & \sigma_n(e_{n+1}) \end{bmatrix}.\]
Since this matrix has more columns than rows, the kernel $V$ has dimension at least 1.

Let $v = (c_1,\ldots, c_{n+1}) \in V$ be a nonzero vector of minimal weight (number of nonzero entries), and scale $v$ so one coordinate is 1. Then, for any $\sigma \in G$, we have $\sigma(v) \in V$, so $\sigma(v) - v \in V$. We will show that $v\in (E^G)^{n+1}$. Then, the first row of the matrix multiplication by $v$ yields a non-trivial linear relation among $e_1,\ldots, e_{n+1}$ over $E^G$.

If $v \notin (E^G)^{n+1}$, then $\sigma(v) \ne v$ for some $\sigma\in G$, and $\sigma(v) - v$ is a nonzero vector of smaller weight—contradicting minimality. Thus, $v \in (E^G)^{n+1}$, and $e_1, \ldots, e_{n+1}$ are linearly dependent over $E^G$.

Theorem (Galois Extension from Automorphism Group).
Let $E$ be a field, and $G \subset \operatorname{Aut}(E)$ be a finite group of automorphisms. Then, if $F = E^G$ is the fixed field of $G$, the extension $E/F$ is Galois and

\[\operatorname{Gal}(E/F) = G.\]

Proof.
By Artin’s Lemma, $[E : F] \leq |G|$, and $G \subset \operatorname{Aut}(E/F)$. Thus,
\[|G| \leq |\operatorname{Aut}(E/F)| \leq [E : F] \leq |G|,\]
so all inequalities are equalities. Hence, $|\operatorname{Aut}(E/F)| = [E:F]$, which implies $E/F$ is Galois. Moreover, since $|G| = |\operatorname{Aut}(E/F)|$ and $G \subset \operatorname{Aut}(E/F)$, we conclude that $\operatorname{Gal}(E/F) = G$.

Corollary (Fix Field Characterizes Galois Extension).
Let $E/F$ be a finite extension. Then $E/F$ is Galois if and only if

\[F = E^{\operatorname{Aut}(E/F)}.\]

Proof.
If $E/F$ is Galois, then $\operatorname{Aut}(E/F)$ has size $[E : F]$. Since $F\subset E^{\operatorname{Aut}(E/F)}$, we have
\[|\operatorname{Aut}(E/F)| = [E: F]\geq [E:E^{\text{Aut}(E/F)}] = |\operatorname{Aut}(E/F)|,\]
and hence must be $F = E^{\text{Aut}(E/F)}$.

The converse direction is proved in the above theorem with $G = \operatorname{Aut}(E/F)$.