Every automorphism a of a finite field with characteristic p $\mathbb{F}$ we have that $a(r)=r$ $\forall r\in\mathbb{Z}_{p}$

Asked Jun 10 '23 at 16:11

Active Jun 10 '23 at 16:11

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Hi i need to prove the following:

$\mathbb{F}$ is a finite field with characteristic p.Show that for every automorphism a of the field $\mathbb{F}$ we have that $a(r)=r$ $\forall r\in\mathbb{Z}_{p}$

I am a little confused on how to start it,I know there is a simiral prove in galois theory for fields with characteristic 0 that automorphisms have fixed points in $\mathbb{Q}$ but I dont understant how to do it generally.Thanks

asked Jun 10 '23 at 16:11

freida ef

By definition $a(1)=1$, and $a(1+1)=a(1)+a(1)$... Can you take on? – Desperado Jun 10 '23 at 16:16
Any automorphism will fix the prime field elementwise. So $\alpha(x)=x$ for all $x\in \Bbb F_p$. – Dietrich Burde Jun 10 '23 at 16:19
thanks a lot .I think I got it – freida ef Jun 10 '23 at 16:22

Every automorphism a of a finite field with characteristic p $\mathbb{F}$ we have that $a(r)=r$ $\forall r\in\mathbb{Z}_{p}$

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