Lecture. Groups of automorphisms of fields [l10]

Apr 29, 2026

Groups of automorphisms of fields

Definition 1. automorphism [2101]

Let $F$ be a field.

An automorphism of $F$ is an isomorphism from $F$ to itself. We denote by $\mathrm{Aut}(F)$ the set of all automorphisms of $F$.
An automorphism $\sigma$ of $F$ is said to fix $\alpha \in F$ if $\sigma(\alpha) = \alpha$, and is said to fix a subset $S \subseteq F$ if $\sigma(\alpha) = \alpha$ for all $\alpha \in S$.

Example 2. some examples [2102]

For every field $F$, $\operatorname{id}_{F}$ is an automorphism of $F$.
If $\sigma$ is an automorphism of $F$, then $\sigma(1) = 1$, and $\sigma(1 + 1 + \cdots + 1) = 1 + 1 + \cdots + 1$. Hence, $\sigma$ fixes the prime field of $F$.
Let $F$ be a field of characteristic $p$ and $\mathrm{Fr} : F \to F$ be the Frobenius endomorphism (sending $a \in F$ to $a^{p}$). $\mathrm{Fr}$ is a field homomorphism, so it is injective. It is surjective if and only if every element in $F$ is a $p$-th power, i.e. if and only if $F$ is perfect by Proposition 1902. Thus, $\mathrm{Fr}$ is an automorphism if and only if the field is perfect.

Definition 3. $\mathrm{Aut}(K/F)$ [2103]

Let $K / F$ be a field extension. $\mathrm{Aut}(K / F)$ is defined to be the set of all automorphisms of $K$ which fix $F$.

Proposition 4. groups of automorphisms of fields [2104]

Let $K / F$ be a field extension.

$\mathrm{Aut}(K / F)$ is a group under composition and $\mathrm{Aut}(K / F)$ is a subgroup of $\mathrm{Aut}(K)$.
If $L$ is a field and $F \subseteq K \subseteq L$, then$$\mathrm{Aut}(L / K) \leq \mathrm{Aut}(L / F).$$

Proposition 5. fixed field [2105]

Let $K / F$ be a field extension.

Given $H \leq \mathrm{Aut}(K / F)$, define$$K^{H} = \{ \alpha \in K \mid \sigma(\alpha) = \alpha \text{ for all } \sigma \in H \}.$$Then $K^{H}$ is a subfield of $K$ and $H \leq \mathrm{Aut}(K / K^{H})$. $K^{H}$ is called the fixed field of $H$.
For $H_{1} \leq H_{2} \leq \mathrm{Aut}(K / F), K^{H_{1}} \supseteq K^{H_{2}}$.

Example 6. some examples [2106]

Example 6.1. [2106A]

$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt{ 2 })$. One can check that $$ \tau : \mathbb{Q}(\sqrt{ 2 }) \to \mathbb{Q}(\sqrt{ 2 }), \qquad \tau(a + b\sqrt{ 2 }) = a - b\sqrt{ 2 }, \, a, b \in \mathbb{Q} $$ is an automorphism in $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$. If $\sigma \in \mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$, then $\sigma(\sqrt{ 2 })^{2} = \sigma(\sqrt{ 2 }^{2}) = \sigma(2) = 2$. Hence, $\sigma(\sqrt{ 2 }) = \pm \sqrt{ 2 }$. If $\sigma(\sqrt{ 2 }) = \sqrt{ 2 }$, then $\sigma = \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}$, then $\sigma = \tau$. Thus, $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right) = \{ \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}, \tau \} \cong \mathbb{Z} / 2\mathbb{Z}$ ($\tau \circ \tau = \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}$).

Example 6.2. [2106B]

Let $K / F$ be an algebraic field extension. If $\alpha \in K$ with $m_{\alpha, F}(x)$ and $\sigma \in \mathrm{Aut}(K / F)$, then $\varphi(\alpha)$ is a root of $m_{\alpha, F}(x)$. Therefore, we have the map $$ \mathrm{Aut}\left( F(\alpha) / F \right) \to \{ \text{root of }m_{\alpha, F}(x) \text{ in }F(\alpha) \}, \qquad \sigma \mapsto \sigma(\alpha). $$ The map is injective because an automorphism $\sigma \in \mathrm{Aut}\left( F(\alpha) / F \right)$ is completely determined by $\sigma(\alpha)$. Next, we show that it is also surjective. If $\beta \in F(\alpha)$ is a root of $m_{\alpha, F}(x)$, then $F(\beta) = F(\alpha)$ ($[F(\alpha) : F(\beta)] = \frac{[F(\alpha) : F]}{F(\beta) : F} = \frac{\deg m_{\alpha, F}(x)}{\deg m_{\beta, F}(x)} = 1$), and $$ \sigma: F(\alpha) \to F(\beta) = F(\alpha), \qquad a(\alpha) \mapsto a(\beta) \, \forall a(x) \in F[x] $$ defines an automorphism of $F(\alpha)$ fixing $F$ by Theorem 1408. Thus, $\beta = \sigma(\alpha)$ belongs to the image of the map. This shows that it is surjective.

Example 2106 (1) can be viewed as a special case of this. The minimal polynomial of $\sqrt{ 2 }$ over $\mathbb{Q}$ is $x^{2} - 2$, and the set $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$ is in bijection with the set of roots of $x^{2} - 2$ in $\mathbb{Q}(\sqrt{ 2 })$ which is $\{ \pm \sqrt{ 2 } \}$.

Example 6.3. [2106C]

$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt[3]{ 2 })$. The minimal polynomial of $\sqrt[3]{ 2 }$ over $\mathbb{Q}$ is $x^{3} - 2$. $x^{3} - 2$ has three roots $$ \theta_{1} = \sqrt[3]{ 2 }, \quad \theta_{2} = \sqrt[3]{ 2 }e^{ 2 \pi i / 3 } = \sqrt[3]{ 2 } \frac{-1 + \sqrt{ 3 }}{2}, \quad \theta_{3} = \sqrt[3]{ 2 } e^{ 4 \pi i / 3 } = -\frac{\sqrt[3]{ 2 }(1 + \sqrt{ 3 })}{2} $$ in $\overline{\mathbb{Q}}$, but only $\sqrt[3]{ 2 }$ belongs to $\mathbb{Q}(\sqrt[3]{ 2 })$. $x^{3} - 2$ has only one root in $\mathbb{Q}(\sqrt[3]{ 2 })$. By Example 2106 (2), $$ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q} \right) = \{ 1 \}, \qquad \text{the trivial group}. $$

Example 6.4. [2106D]

$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$, which is the splitting field of $x^{3} - 2$ over $\mathbb{Q}$ with $[\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}] = 6$ (part 2 of Example 1505). We show that $\mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) \cong S_{3}$.

Proof. sketch [2106DA]

Denote the roots of $x^{3} - 2$ by $\theta_{1}, \theta_{2}, \theta_{3}$. Define the map as $$ \begin{equation} \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) \to S_{3}, \qquad \sigma \mapsto \tau_{\sigma} \end{equation} $$ such that for $\sigma \in \mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right)$, the permutation $\tau_{\sigma} \in S_{3}$ is defined as $\tau_{\sigma}(i) = j$ if and only if $\sigma(\theta_{i}) = \theta_{j}$.

First we need to check that the map $(1)$ is indeed a group homomorphism. Then simply noticing that $\theta_{1}, \theta_{2}, \theta_{3}$ generate $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ over $\mathbb{Q}$ since $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ is the splitting field of $x^{3} - 2$ can show that $(1)$ is injective. To show the surjectivity, it remains to prove that every $\tau \in S_{3}$ corresponds to a valid $\sigma \in \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right)$ such that $\tau = \sigma_{\tau}$.

Example 6.5. [2106E]

Keep the notation in Example 2106 (4). We have $$ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) = \left< \sigma_{1}, \sigma_{2} \right> \cong S_{3}, $$ with $$ \begin{align*} \sigma_1 : && \sqrt[3]{2} &\longmapsto \sqrt[3]{2}\frac{-1 + \sqrt{-3}}{2}, & \sigma_2 : && \sqrt[3]{2} &\longmapsto \sqrt[3]{2}, \\ && \frac{-1 + \sqrt{-3}}{2} &\longmapsto \frac{-1 + \sqrt{-3}}{2}, & && \frac{-1 + \sqrt{-3}}{2} &\longmapsto \frac{-1 - \sqrt{-3}}{2}. \end{align*} $$ One can check that $$ \begin{align*} \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}(\sqrt[3]{ 2 }) \right) &= \left< \sigma_{2} \right> \quad (\cong \mathbb{Z} / 2\mathbb{Z}), \\ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 + \sqrt{ -3 }}{2} \right) \right) & = \left< \sigma_{2} \sigma_{1} \right> \quad (\cong \mathbb{Z} / 2\mathbb{Z}), \\ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 - \sqrt{ -3 }}{2} \right) \right) & = \left< \sigma_{2} \sigma_{1}^{2} \right> \quad (\cong \mathbb{Z} / 2\mathbb{Z}), \\ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}(\sqrt{ -3 }) \right) & = \left< \sigma_{1} \right> \quad (\cong \mathbb{Z} / 3\mathbb{Z}), \end{align*} $$ and $$ \begin{align*} \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{2} \right> } &= \mathbb{Q}(\sqrt[3]{ 2 }), \\ \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{2} \sigma_{1} \right> } &= \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 + \sqrt{ -3 }}{2} \right), \\ \mathbb{Q}(\sqrt[3]{ 2 }. \sqrt{ -3 })^{\left< \sigma_{2}\sigma_{1}^{2} \right> } &= \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 - \sqrt{ -3 }}{2} \right), \\ \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{1} \right> } &= \mathbb{Q}(\sqrt{ -3 }), \\ \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{1}, \sigma_{2} \right> } &= \mathbb{Q}. \end{align*} $$