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I'm working on a problem that could use this as a lemma, and therefore I'm wondering if this is true. let $F$ be a finite field of characteristic $p$. Let $\phi: F \to F$ be given by

$$ \phi = x \mapsto x^p$$

The main question is, is $\phi$ always surjective? I'm not super sure how to go about this. I've looked at the multiplication tables for $GF(2), GF(4), GF(8), $ and $GF(9)$, and this seems to be true.

I'm not super sure how to show this. I know that the set of units in $F$, $F^\times$ forms a cyclic multiplicative group of order $p^r - 1$, and is therefore isomorphic to $\mathbb Z/(p^r - 1)\mathbb Z$. If $\psi$ is an isomorphism $F^\times \to \mathbb Z/(p^r - 1)\mathbb Z$, then showing $\phi$ is surjective should be equivalent to showing the map

$$ n \mapsto \phi(\psi(\phi^{-1}(n))) = n \mapsto p \cdot n$$

Is surjective, where $n \in \mathbb Z / (p^r - 1) \mathbb Z$ and $n = p^r - 1$. But I don't know where to go from here.

Any help would be appreciated. Thanks!

Update

This is known as a Frobenius endomorphism, and it is an automorphism from $F \to F$ when $F$ is finite.

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  • $\begingroup$ I assume $GF(10)$ was a typo. $\endgroup$ – Andreas Blass Apr 18 '17 at 23:22
  • $\begingroup$ Sorry, I meant $GF(9)$. $\endgroup$ – Enrico Borba Apr 18 '17 at 23:22
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Any field homomorphism is injective. An injective map from a finite set to itself is always surjective. (By the pigeon-hole principle)

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  • $\begingroup$ But the map $x \mapsto x^2$ is not injective. Specifically in the case of $\mathbb F_3$ since $1^2 = 1$ and $2^2 = 1$. $\endgroup$ – Enrico Borba Apr 18 '17 at 20:23
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    $\begingroup$ Yes, but it is injective if $\operatorname{char} F = 2$, because then it is a field homomorphism. $\endgroup$ – MatheinBoulomenos Apr 18 '17 at 20:25
  • $\begingroup$ So, why is $x \mapsto x^p$ for fields of characteristic $p$ a field homomorphism? $\endgroup$ – Enrico Borba Apr 18 '17 at 20:25
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    $\begingroup$ @EnricoBorba Just use the binomial expansion and notice that most of the binomial coefficients are divisible by $p$. For details, see here or here. $\endgroup$ – André 3000 Apr 18 '17 at 20:27
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    $\begingroup$ @EnricoBorba $(a+b)^p=a^p+b^p$ in $F_{p^n}$, and $(ab)^p=a^pb^p$ for any field. So it is a homomorphism. $\endgroup$ – Thomas Andrews Apr 18 '17 at 20:30
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Note that $x^{p^n}-x$ splits in this field (in fact this is how we characterize finite fields of a given order since their unit sets are groups and we have Lagrange's theorem), therefore all elements of the field are the roots of that polynomial. But then $x^{p^n}$ is the identity map, so your original map is bijective, hence surjective. That it is a field homomorphism follows immediately from the binomial theorem, so it is an automorphism.

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