Field automorphism - Imported Knowlpedia Content

Let $L$ be a field. A field automorphism of $L$ is a bijective field homomorphism $\sigma:L\to L$ . The set $\mathrm{Aut}(L)$ of all field automorphisms is a group under composition (an instance of automorphism group).

If $L/K$ is a field extension, a $K$ -automorphism of $L$ is a field automorphism $\sigma\in\mathrm{Aut}(L)$ such that $\sigma|_K=\mathrm{id}_K$ . The group of all $K$ -automorphisms is denoted $\mathrm{Aut}_K(L)$ , and when $L/K$ is Galois it is the Galois group $\mathrm{Gal}(L/K)$ .

Remarks

Automorphisms are the “symmetries” that permute conjugates, and they control invariants such as fixed fields, trace, and norm.

Examples

Complex conjugation. The map $\sigma:\mathbb{C}\to\mathbb{C}$ , $\sigma(a+bi)=a-bi$ , is a field automorphism. It is a $\mathbb{R}$ -automorphism of $\mathbb{C}/\mathbb{R}$ .

Quadratic extension. For $L=\mathbb{Q}(\sqrt{d})$ , the map $\sigma(a+b\sqrt{d})=a-b\sqrt{d}$ is a nontrivial $\mathbb{Q}$ -automorphism. Thus $\mathrm{Aut}_\mathbb{Q}(L)\cong C_2$ .

Finite fields and Frobenius. If $L=\mathbb{F}_{p^n}$ is a finite field, then the Frobenius map $x\mapsto x^p$ is an automorphism of $L$ , and its powers generate $\mathrm{Gal}(L/\mathbb{F}_p)$ (see finite-field Galois groups are cyclic).