📂Abstract Algebra

Automorphisms of a Field

Definition¹

Let $E$ be an extension field of $F$.

1. For a field $E$, an isomorphism $\sigma : E \to E$ is called an automorphism, and the set of automorphisms of $E$ is denoted by $\text{Auto} (E)$.
2. For $\sigma \in \text{Auto} (E)$, if $\sigma ( a ) = a$, we say that $\sigma$ leaves $a$ fixed.
3. Let $S \subset \text{Auto} (E)$. If every $\sigma \in S$ leaves every $a \in F$ fixed, we say that $S$ leaves the subfield $F$ fixed.
4. If $\left\{ \sigma \right\} \subset \text{Auto} (E)$ leaves $F$ fixed, we say that $\sigma$ leaves $F$ fixed.
5. If $\left\{ \sigma \right\} \subset \text{Auto} (E)$ leaves the field $E_{ \left\{ \sigma \right\} }$ fixed, we say that $\sigma$ leaves the field $E_{\sigma}$ fixed.
6. The set of all automorphisms of $E$ leaving $F$ fixed is denoted by $G ( E / F )$ and is called the group of $E$ over $F$.

Theorem

• [1]: $\left< \text{Auto} ( E ) , \circ \right>$ is a group.
• [2]: $G ( E / F) \le \text{Auto} ( E )$

Example

The wording is quite difficult and convoluted, so it is better to understand the concept through an example.

$$ F = \mathbb{Q} ( \sqrt{2} ) \le \mathbb{Q} ( \sqrt{2} , \sqrt{3} ) = E $$ If we set this, then $(x^2 - 3) \in F [ x ]$ is irreducible over $F$, so $\sqrt{3} , \sqrt{-3}$ are conjugates over $E$. By the conjugation isomorphism theorem, $\psi_{ \sqrt{3} , - \sqrt{3} } : E \to E$ is an isomorphism, and therefore we can see that $ \psi_{ \sqrt{3} , - \sqrt{3} } \in \text{Auto} (E) $.

Let us actually take the function $\psi_{ \sqrt{3} , - \sqrt{3} }$. For $a,b,c,d \in \mathbb{Q}$, $$ \psi_{ \sqrt{3} , - \sqrt{3} } ( a+ b \sqrt{2} + c \sqrt{3} + d \sqrt{6} ) = a+ b \sqrt{2} - c \sqrt{3} - c \sqrt{2} \sqrt{3} $$ but for every $( x + y \sqrt{2} ) \in \mathbb{Q} ( \sqrt{2} )$,

$$ \psi_{ \sqrt{3} , - \sqrt{3} } ( x + y \sqrt{2} ) = x + y \sqrt{2} $$ so we can say that $\psi_{ \sqrt{3} , - \sqrt{3} }$ leaves $\mathbb{Q} ( \sqrt{2} )$ fixed. Simply put, $\mathbb{Q} ( \sqrt{2} )$ can be seen as a subfield unaffected by $\psi_{ \sqrt{3} , - \sqrt{3} }$. It is in this sense that expressions like ‘fixed’ and ’leaves’ are used.

Meanwhile, for the identity map $I$ and the composition operation of functions $\circ$, $$ \left( \psi_{ \alpha , - \alpha } \circ \psi_{ \alpha , - \alpha } \right) = I $$ so $$ \left< \left\{ I, \psi_{ \sqrt{2} , - \sqrt{2} }, \psi_{ \sqrt{3} , - \sqrt{3} } , ( \psi_{ \sqrt{2} , - \sqrt{2} } \circ \psi_{ \sqrt{3} , - \sqrt{3} } ) \right\} , \circ \right> $$ not only forms a group, but in particular is isomorphic to the Klein four-group.

Proof

[1]

• (i): Composition of functions $\circ$ satisfies the associative law, and composing functions in $\text{Auto} ( E )$ yields an automorphism of $E$.
• (ii): The identity map $I : E \to E$ satisfies $I (a) = a$ for every $a \in E$, so it is an automorphism and $I \in \text{Auto} ( E )$.
• (iii): Since $\text{Auto} ( E )$ is the set of automorphisms, for any $\sigma$ there exists its inverse map $\sigma^{-1} \in \text{Auto} ( E )$.

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[2]

• (i): Composition of functions $\circ$ satisfies the associative law, and for $\sigma , \tau \in G ( E / F )$ and $a \in F$, $$ (\sigma \tau) = \sigma ( \tau ( a ) ) = \sigma (a) = a $$ so $(\sigma \tau) \in G ( E / F )$.
• (ii): The identity map $I$ is the identity element of $ G ( E / F )$.
• (iii): If $\sigma ( a ) = a$ then $a = \sigma^{-1} (a)$, so $\sigma^{-1} \in G ( E / F )$.

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