Automorphism of algebraic closure $\overline{{\bf F}}_p$.

Question

Problem : I want to give an concrete example of automorphism of $\overline{{\bf F}}_p$ which fixes ${\bf F}_p$, where $$\overline{{\bf F}}_p =\bigcup_{n\geq 1} {\bf F}_{p^n} $$ and ${\bf F}_{p^n} $ is finite extension over a finite field ${\bf F}_p$, i.e., splitting of $$ x^{p^n} -x =0$$

Automorphism of ${\bf F}_{p^n}$ : Let $\alpha$ be some solution of the equation. Then ${\bf F}_{p^n}$ is a vector space over a basis $$ \alpha,\ \alpha^p,\ \alpha^{p^2},\ \cdots , \ \alpha^{p^{n-1}} $$

Hence any element has the form $$ \sum_{i=0}^{n-1} a_i \alpha^{p^i} ,\ a_i\in {\bf F}_i$$

Here $Frobenius$ map, which is a generator of ${\bf Z}_n={\rm Aut}\ ({\bf F}_{p^n}/{\bf F}_p) $, is $$ \sum_{i=0}^{n-1} a_i \alpha^{p^i} \mapsto\sum_{i=0}^{n-1} a_i \alpha^{p^{i+1}} $$

Note that here $a_i \alpha^{p^i},\ a_i\neq 0$ is not fixed.

Proof : Recall the fact that $$ {\bf F}_{p^n}\subseteq {\bf F}_{p^{nt}} $$

Consider a equation $x^{p^{nt}} -x=0$. Let $\beta$ be a solution which plays a role of $\alpha$. Then $$ \alpha =\beta +\beta^{p^n}+\cdots + \beta^{p(t-1)n } $$

We have a claim : If $\sigma\in {\rm Aut}\ ({\bf F}_{p^{nt}}/{\bf F}_p)$ and $\sigma(\alpha)=\alpha^p$, then $$\sigma(\beta)=\beta^p$$

If this is true, $q$ is a prime, and $f\in {\rm Aut}\ (\overline{{\bf F}}_p/{\bf F}_p)$ s.t. $$ f(x)=x^p,\ x\in {\bf F}_{p^{n}} $$ then by the above argument we have $$ y\mapsto y^p,\ y\in {\bf F}_{p^{qn}}$$ Hence $f$ is $$ y\mapsto y^p,\ \forall y\in \overline{{\bf F}}_p$$

Am I right ? If so, how can we prove a claim ?

Answer 1

The claim is not necessarily true. If we assume $\sigma(\beta) = \beta^{p^i}$, then we get the condition $\alpha^p = \alpha^{p^i}$ which implies $i \equiv 1 \pmod{n}$. This congruence is not enough to imply $\beta^{p^i} = \beta^p$, since $\beta$ is in a larger field (of size $p^{nt}$) than $\alpha$ (size $p^n$).

Having said that, it certainly could be true that $\sigma(\beta) = \beta^p$. If you are looking for a concrete automorphism of the algebraic closure, then the map $y \mapsto y^p$ certainly is one. But $\sigma(\beta) = \beta^p$ does not have to be true.

In general, for $\mathbb{F}_{p^n} \subset \mathbb{F}_{p^m}$, an automorphism $\sigma$ of $\mathbb{F}_{p^n}$ with $\sigma(x) = x^{p^i}$ extends to an automorphism of $\mathbb{F}_{p^m}$ with $\sigma(x) = x^{p^j}$ iff $i \equiv j \pmod{n}$. So an automorphism of the algebraic closure is described by choosing a compatible system of values $x_n \in \mathbb{Z}/n\mathbb{Z}$ for every positive integer $n$, where the compatibility condition is $x_n \equiv x_m \pmod{n}$ whenever $n|m$. Such compatible systems form a ring called $\hat{\mathbb{Z}}$. The automorphism then sends $x \in \mathbb{F}_{p^n}$ to $x^{p^{x_n}}$.

Answer 2

Your candidate choice of automorphism is correct. More generally:

Proposition: If $k$ is any field of characteristic $p$, then the Frobenius map $\sigma : k \to k$, $\sigma(x) = x^p$, is an injective ring homomorphism.

Proof: Additivity follows from the binomial formula which takes the form $(x+y)^p = x^p + y^p$ in characteristic $p$, multiplicativity $(xy)^p = x^py^p$ holds since $k$ is commutative, and $1^p = 1$. Thus $\sigma$ is a ring homomorphism. Since $k$ is a field, any ring homomorphism on $k$ is automatically injective.

It remains to show that $\sigma : \overline{\mathbb{F}_p} \to \overline{\mathbb{F}_p}$ is surjective. Take $x \in \overline{\mathbb{F}_p}$, so $x \in \mathbb{F}_{p^n}$ for some $n$. Now $\sigma|_{\mathbb{F}_{p^n}} : \mathbb{F}_{p^n} \to \mathbb{F}_{p^n}$ is an injection and $\mathbb{F}_{p^n}$ is finite, so $\sigma|_{\mathbb{F}_{p^n}}$ is surjective. Thus $x \in \text{im}(\sigma|_{\mathbb{F}_{p^n}}) \subseteq \text{im}(\sigma)$, hence $\sigma : \overline{\mathbb{F}_p} \to \overline{\mathbb{F}_p}$ is surjective.