Galois group of algebraic closure of a finite field

Is there any element of finite order of the Galois group of algebraic closure of a finite field, and if there is how can I construct it ?


Solutions Collecting From Web of "Galois group of algebraic closure of a finite field"

Undoubtedly you are only interested in non-trivial such elements. They do not exist.

Assume that $\sigma$ is such an automorphism of $\overline{\Bbb{F}_p}$. By replacing $\sigma$ with its appropriate power, we can then assume that its order is a prime number $\ell$. Let $K$ be the fixed field of $\sigma$.

Let $x\in\overline{\Bbb{F}_p}$ be such an element that $\sigma(x)\neq x$. Then
$L:=\Bbb{F}_p[x]=\Bbb{F}_{p^r}$ for some integer $r>1$. The restriction of $\sigma$
to $L$ is then a non-trivial automorphism, so the order of $\sigma\vert_L$ is necessarily also $\ell$, and $[L:K\cap L]=\ell$.

Let $m=r\ell, E=\Bbb{F}_{p^m}$. The restriction $\sigma\vert_E$ is non-trivial, because $x\in E$. Therefore also $[E:E\cap K]=\ell$. But $L$ is the only subfield of $E$ such that $[E:L]=\ell$. Therefore the fixed field of $\sigma\vert_E$ should be $L$ contradicting the fact that $\sigma\vert_L$ is non-trivial.

The same conclusion follows from the observation that $\hat{\Bbb{Z}}$ has no non-trivial torsion elements.

No, there are not. Let $\overline{\mathbb{F}_p}$ be an algebraic closure of $\mathbb{F}_p$. Then $\overline{\mathbb{F}_p}/\mathbb{F}_p$ is an infinite Galois extension (infinite, algebraic, normal, separable) with pro-finite Galois group
Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)=\widehat{\mathbb{Z}}=\varprojlim \mathbb{Z}/n\mathbb{Z}.
This is the pro-finite completion of $\mathbb{Z}$. By Artin-Schreier, the torsion subgroup
must be trivial or cyclic of order $2$. It is trivial, so there are no
no non-trivial elements of finite order.
One could ask the same question for the absolute Galois group over $\mathbb{Q}$. Here it is cyclic of order $2$.
The only element of finite order in the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ is, up to conjugation, complex conjugation.

For details on the group $\widehat{\mathbb{Z}}$ see Elements in $\hat{\mathbb{Z}}$, the profinite completion of the integers.