11

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 ?

Thanks.

Community
  • 1
bytrz
  • 1,242
  • 1
  • 14
  • 26

2 Answers2

18

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.

Jyrki Lahtonen
  • 140,891
  • Why $[L:K∩L]=l$? – user302934 Oct 17 '19 at 17:20
  • @Comol If $G$ is a finite group of automorphisms of a field $K$, and $K^G$ is the fixed field, then always $[K:K^G]=|G|$. – Jyrki Lahtonen Oct 17 '19 at 17:27
  • Not sure if I'm missing something simple but why is $\sigma$ restricted to $L$ an automorphism? That is, I don't see why $\sigma(x) \in L$. – Aryaman Maithani Jun 04 '20 at 08:52
  • 1
    @AryamanMaithani $L$ is Galois over $\Bbb{F}_p$, so any $\Bbb{F}_p$-automorphism of any field containing $L$ will map $L$ to itself. Alternatively, $L$ is the set of zeros of $x^{p^r}-x$, and $\sigma$ has to map a zero of that polynomial to another zero. – Jyrki Lahtonen Jun 04 '20 at 08:56
  • 1
    Compare with extensions of rationals: Any automorphism of any field containing $\Bbb{Q}(\sqrt2)$ has to map $\Bbb{Q}(\sqrt2)$ to itself, because it is Galois over $\Bbb{Q}$. On the other hand, an automorphism of a field containing $\Bbb{Q}(\root3\of2)$ may map it to one of its conjugate fields $\Bbb{Q}(\omega^k\root3\of2)$. – Jyrki Lahtonen Jun 04 '20 at 08:58
  • Yep, got it. Thanks! – Aryaman Maithani Jun 04 '20 at 09:27
  • @JyrkiLahtonen "By replacing σ with its appropriate power, we can then assume that its order is a prime number ℓ". Why is replacing $\sigma $ with an appropriate power allowed and why do you want order to be a prime number? –  Oct 17 '21 at 12:42
  • @Avenger. An exercise for you: If a group has an element of a finite order, then it also has an element such that the order is a prime number. I wanted a prime because the rest of the argument needs it. – Jyrki Lahtonen Oct 17 '21 at 13:10
6

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.

Community
  • 1
Dietrich Burde
  • 140,055
  • 1
    Why do you need Artin-Schreier to conclude that $\widehat{\mathbf Z}$ is torsion-free?... – Bruno Joyal Nov 29 '13 at 22:04
  • I do not need Artin-Schreier for that, but it is a result which is relevant anyway for such questions (see the other example). – Dietrich Burde Nov 30 '13 at 08:46
  • 1
    Even if the only element of finite order in the absolute Galois group $G$ of $\Bbb Q$ is, up to conjugation, complex conjugation, I think that the torsion subgroup of $G$ is infinite (and not cyclic of order $2$ as claimed), since there are many embeddings $\overline{\Bbb Q} \to \Bbb C$. – Watson Jun 14 '17 at 21:16
  • 1
    Just noticed your comment, @Watson. There is a torsion subset, but it is not a subgroup. In a nonabelian group, product of two torsion elements need not be torsion. – Lubin Jan 08 '18 at 02:04
  • @Lubin : thank you for this correction. I was probably confused by the sentence "Here it is cyclic of order $2$." in the answer. – Watson Jan 08 '18 at 09:45