📂Abstract Algebra

Automorphisms of a Body

Definition ¹

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

1. An isomorphism $\sigma : E \to E$ of field $E$ to itself is called an automorphism, and the set of all automorphisms of $E$ is denoted as $\text{Auto} (E)$.
2. If $\sigma \in \text{Auto} (E)$ then $\sigma ( a ) = a$, it is said that $\sigma$ fixes $a$.
3. Let $S \subset \text{Auto} (E)$. If for all $a \in F$, all $\sigma \in S$ fixes $a$, it is said that $S$ leaves a fixed subfield $F$.
4. If $\left\{ \sigma \right\} \subset \text{Auto} (E)$ leaves a fixed $F$, it is said that $\sigma$ fixes $F$.
5. If $\left\{ \sigma \right\} \subset \text{Auto} (E)$ leaves a fixed field $E_{ \left\{ \sigma \right\} }$, it is said that $\sigma$ fixes field $E_{\sigma}$.
6. The set of all automorphisms of $E$ that leaves $F$ is denoted as $G ( E / F )$ and 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 concept is very difficult and complex, so it is better understood through examples.

If $$ F = \mathbb{Q} ( \sqrt{2} ) \le \mathbb{Q} ( \sqrt{2} , \sqrt{3} ) = E $$, then since $(x^2 - 3) \in F [ x ]$ is an irreducible element over $F$, $\sqrt{3} , \sqrt{-3}$ is prime over $E$. By the conjugate homomorphism theorem, $\psi_{ \sqrt{3} , - \sqrt{3} } : E \to E$ is an isomorphism, and therefore we know that $ \psi_{ \sqrt{3} , - \sqrt{3} } \in \text{Auto} (E) $.

Actually, take the function $\psi_{ \sqrt{3} , - \sqrt{3} }$. For $a,b,c,d \in \mathbb{Q}$, $$ \psi_{ \sqrt{3} , - \sqrt{3} } ( a+ b \sqrt{2} + c \sqrt{3} + d \sqrt{6} ) = a+ b \sqrt{2} - c \sqrt{3} - c \sqrt{2} \sqrt{3} $$ since for all $( x + y \sqrt{2} ) \in \mathbb{Q} ( \sqrt{2} )$,

$$ \psi_{ \sqrt{3} , - \sqrt{3} } ( x + y \sqrt{2} ) = x + y \sqrt{2} $$ hence, it can be said that $\psi_{ \sqrt{3} , - \sqrt{3} }$ leaves a fixed $\mathbb{Q} ( \sqrt{2} )$. Simply put, $\mathbb{Q} ( \sqrt{2} )$ can be considered as a subfield not affected by $\psi_{ \sqrt{3} , - \sqrt{3} }$. This is the sense in which terms like ‘fixed’ or ’leaves’ are used.

Furthermore, regarding the identity mapping $I$ and the operation of function composition $\circ$, $$ \left( \psi_{ \alpha , - \alpha } \circ \psi_{ \alpha , - \alpha } \right) = I $$ hence, $$ \left< \left\{ I, \psi_{ \sqrt{2} , - \sqrt{2} }, \psi_{ \sqrt{3} , - \sqrt{3} } , ( \psi_{ \sqrt{2} , - \sqrt{2} } \circ \psi_{ \sqrt{3} , - \sqrt{3} } ) \right\} , \circ \right> $$ forms not only a group but is specifically isomorphic to the Klein four-group.

Proof

[1]

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

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[2]

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

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