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Let $p$ be a prime number and let $F_p$ be the Frobenius automorphism of $\overline{\mathbb F_p}$.

Given an explicit element $x $ of $\overline{\mathbb Q_p}$, how do I compute $F_p(x)$?

Does it even make sense?

Har deevos
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    The field $\Bbb{Q}_p$ has no non-trivial automorphisms, so the Frobenius automorphism in particular is the identity there. – Jyrki Lahtonen May 25 '14 at 15:34
  • @JyrkiLahtonen The Frobenius is never an automorphism in characteristic 0, so that proof doesn't apply. The map still makes sense in many situations. The question only claims that it is an automorphism of $\mathbb{F}_p$. – Matt May 25 '14 at 15:44
  • @JyrkiLahtonen Thank you for your comment. I wanted to consider the algebraic closures. I edited the question. – Har deevos May 25 '14 at 15:45
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    I'm not sure whether the algebraic closure has a Frobenius automorphism. On the other hand there is a Frobenius automorphism for any unramified finite extension of $\Bbb{Q}_p$. If $K$ is a finite unramified extension of degree $f$, then it is gotten by adjoining a primitive root of unity $\zeta$ of order $q^f-1$. The automorphism determined by $\zeta\mapsto\zeta^p$ is then the Frobenius automorphism. – Jyrki Lahtonen May 25 '14 at 15:54
  • @Matt: There is something called Frobenius automorphism for many a field of characteristic zero, for example number fields and a $p$-adic fields. When defined it induces the usual characteristic $p$ Frobenius in a quotient field of the ring of integers. I guess your point was that one should not try to use the same simple formula that works in characteristic $p$. That is, of course, correct. – Jyrki Lahtonen May 25 '14 at 15:59
  • (cont'd) so Matt is right in that you should not confuse the Frobenius automorphism of $\Bbb{F}_p$ or its closure and automorphisms with the same name in the context of $p$-adics. – Jyrki Lahtonen May 25 '14 at 16:00
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    I think usually the Frobenius is only defined on the maximal unramified extension $\mathbb{Q}p^\text{unr} = \mathbb{Q}_p\left(\bigcup{(n,p)=1} \mu_n\right)$, i.e. adjoin all the $n$th roots of unity with $n$ prime to $p$. – John M May 29 '14 at 02:51

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