Lecture. Normal extensions and Galois extensions (cont.) [l12]

May 11, 2026

Proposition 1. $\mathrm{Aut}(K / F) \to F$-embeddings [2207]

Let $K / F$ be an algebraic field extension.

Fix an algebraic closure of $F$ with $F \subseteq K \subseteq \overline{F}$. (By (2) of Proposition 1606, one can simply take $\overline{F} = \overline{K}$). Then we have the following bijective map$$\begin{align*}\mathrm{Aut}(K / F) &\to \left\{ F\text{-embeddings }\varphi : K \to \overline{F} \text{ s.t. } \varphi(K) = K\right\}, \\ \sigma &\mapsto \varphi_{\sigma} \text{ defined as } \varphi_{\sigma}(\alpha) = \sigma(\alpha).\end{align*}$$
If $K / F$ is finite, then$$\lvert \mathrm{Aut}(K / F) \rvert \leq \# \{ F\text{-embeddings from } K \text{ to } \overline{F} \}$$and the equality holds if and only if $K / F$ is normal.

Example 2. $\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p}$ [2208]

For all $a \in \mathbb{F}_{p}^{\times}$, $x^{p} - x + a$ is irreducible in $\mathbb{F}_{p}[x]$.

Proof. sketch [2208A]

Let $\alpha \in \overline{\mathbb{F}_{p}}$ be a root of $x^{p} - x + a$. To show $x^{p} - x + a$ is irreducible, it suffices to show that $m_{\alpha, \mathbb{F}_{p}}(x) = x^{p} - x + a$. Moreover, since $m_{\alpha, \mathbb{F}_{p}}(x) \mid (x^{p} - x + a)$, it suffices to show that $\deg m_{\alpha, \mathbb{F}_{p}}(x) \geq p$. We also know that $[\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}] = \deg m_{\alpha, \mathbb{F}_{p}}(x)$, so we only need to prove that $[\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}] \geq p$. This can be done by showing that $$ p \leq \lvert \mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p}) \rvert \leq \#\{ \mathbb{F}_{p}\text{-embeddings }\varphi : \mathbb{F}_{p}(\alpha) \to \overline{\mathbb{F}_{p}} \} \leq [\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}]. $$ $\operatorname{Fr}$ is a good instance of $\mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p})$. If $\operatorname{id}_{\mathbb{F}_{p}(\alpha)}, \operatorname{Fr}, \dots, \operatorname{Fr}^{p-1}$ are different, then of course $\lvert \mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p}) \rvert \geq p$ for that $\mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p})$ is a group. Let us check it on $\alpha$. First, since $\alpha$ is a root of $x^{p} - x + a$, we have $$ \operatorname{Fr}(\alpha) = \alpha^{p} = \alpha - a. $$ Then $$ \operatorname{Fr}^{2}(\alpha) = \operatorname{Fr}(\alpha - a) = \operatorname{Fr}(\alpha) - \operatorname{Fr}(a) = \alpha - a - a^{p} = \alpha - 2a. $$ Using induction on $n$, we can prove that $\operatorname{Fr}^{n}(\alpha) = \alpha - na$. Therefore, $\operatorname{id}_{\mathbb{F}_{p}(\alpha)}(\alpha), \operatorname{Fr}(\alpha), \dots, \operatorname{Fr}^{p-1}(\alpha)$ are differentiable, so are those automorphisms.

Example 3. $\mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q}$ and $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ [2209]

$\mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q}$ is not normal. $m_{\sqrt[3]{ 2 }, \mathbb{Q}}(x) = x^{3} - 2$ does not split completely in $\mathbb{Q}(\sqrt[3]{ 2 })$ since the other two roots $\sqrt[3]{ 2 } \frac{-1 \pm \sqrt{ -3 }}{2}$ do not belong to $\mathbb{Q}(\sqrt[3]{ 2 })$.
$\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ is normal.
1. Use (3) of Proposition 2205: $m_{\sqrt[3]{ 2 }, \mathbb{Q}}(x) = x^{3} - 2$ and $m_{\sqrt{ -3 }, \mathbb{Q}}(x) = x^{2} + 3$ both split completely over $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$.
2. Use (4) of Proposition 2205: $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ is the splitting field of $x^{3} - 2 \in \mathbb{Q}[x]$.

Definition 4. Galois extension [22010]

An algebraic field extension is called Galois if it is both separable and normal.

Definition 5. Galois group [22011]

If $K / F$ is a Galois extension, we call the group $\mathrm{Aut}(K / F)$ the Galois group of $K / F$ and denote it by $\mathrm{Gal}(K / F)$.

Example 6. $\mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q}$ and $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ [22012]

$\mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q}$ is not Galois because it is not normal as seen in (1) of Example 2209.
In (2) of Exmaple 2209, we know that $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ is normal. It is also separable because $\operatorname{ch}(\mathbb{Q}) = 0$ and by Proposition 1704. Thus $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ is a Galois extension. Moreover, by (4) of Example 2106, we have$$\mathrm{Gal}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) \cong S_{3}.$$

Proposition 7. necessary and sufficient conditions for finite $K / F$ to be Galois [22013]

A finite extension $K / F$ is Galois if and only if $\lvert \mathrm{Aut}(K / F) \rvert = [K : F]$.

Proof. sketch [22013A]

$\implies$: Since $K / F$ is Galois, it is normal and separable. By (2) of Proposition 2207, $\lvert \mathrm{Aut}(K / F) \rvert = \# \{ F\text{-embeddings }\varphi : K \to \overline{F} \}$. By Proposition 1707, $\#\{ F\text{-embeddings }\varphi : K \to \overline{F} \} = [K : F]$. Hence $\lvert \mathrm{Aut}(K / F) \rvert = [K : F]$.

$\impliedby$: By (2) of Proposition 2207, $\lvert \mathrm{Aut}(K / F)\rvert = [K : F] \leq \# \{ F\text{-embeddings }\varphi : K \to \overline{F} \}$. By Proposition 1707, $[K : F] \geq \#\{ F\text{-embeddings }\varphi : K \to \overline{F} \}$. Hence $[K : F] = \#\{ F\text{-embeddings }\varphi : K \to \overline{F} \} \implies$ $K / F$ is separable. At the same time, $\lvert \mathrm{Aut}(K / F) \rvert = \#\{ F\text{-embeddings }\varphi : K \to \overline{F} \} \implies$ $K / F$ is normal. Therefore, $K / F$ is Galois.

Proposition 8. necessary and sufficient conditions for finite $K / F$ to be Galois [22013]

A finite extension $K / F$ is Galois if and only if $\lvert \mathrm{Aut}(K / F) \rvert = [K : F]$.

Proof. sketch [22013A]

Remark 9. the textbook definition for Galois extensions [22014]

The textbook uses $\lvert \mathrm{Aut}(K / F) \rvert = [K : F]$ as the definition of Galois extensions.

Proposition 10. necessary and sufficient conditions for $K / F$ to be Galois [22015]

An algebraic extension $K / F$ is Galois if and only if $K$ is the splitting field for a subset of separable polynomials in $F[x]$.

Proof. sketch [22015A]

Fix $\overline{F} = \overline{K}$. Then $F \subseteq K \subseteq \overline{F}$.

$\implies$: Let $A = \{ m_{\alpha, F}(x) \in F[x] \mid \alpha \in K \}$. First, we show that $K$ is the splitting field of $A \subseteq F[x]$. Let $S = \{ \text{roots of } \overline{F} \text{ of polynomials in } A \}$. Then $F(S)$ is the splitting field for $A$. We need to show $K = F(S)$. Given $\alpha \in K$, by the definition of $A$, $m_{\alpha, F}(x) \in A$ and by the definition of $S$, $\alpha$ is a root of $m_{\alpha, F}(x) \in A$, so $\alpha \in S$. Thus, $K \subseteq S \subseteq F(S)$. Conversely, if $\beta \in S$, then $\beta$ is a root of $m_{\alpha, F}(x)$ for some $\alpha \in K$. Since $K / F$ is normal, $K$ contains all the roots of $m_{\alpha, F}(x)$ in $\overline{F}$, so $K$ contains $\beta$. This shows that $S \subseteq K$, so $F(S) \subseteq K$. We have proved $K = F(S)$, the splitting field for $A \subseteq F[x]$.

Also, because $K / F$ is separable, we know that every $\alpha \in K$ is separable over $F$ and $m_{\alpha, F}(x)$ is separable. Hence $A$ is a subset of separable polynomials in $F[x]$.

$\impliedby$: Suppose $A$ is the subset of separable polynomials in $F[x]$ and $K$ is the splitting field for $A \subseteq F[x]$. We show that $K / F$ is both normal and separable. Proposition 2205 $\implies$ $K / F$ is normal. By the proof of Proposition 2204, $K \cong F(S)$ where $S \subseteq \overline{F}$ is the set of roots of polynomials in $A$. The polynomials in $A$ are separable $\implies$ every element in $S$ is separable over $F$. Then Proposition 17012 $\implies$ $K = F(S)$ is a separable extension of $F$.