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I'm studying a bit of field theory, and had a quick question about the Frobenius endomorphism, $\phi : K \rightarrow K$. For any prime $p$, is the Frobenius endomorphism on $F_p$ (the finite field with $p$ elements) the identity automorphism?

My reasoning is from working in $\mathbb{Z}/p\mathbb{Z}$, as $F_p \cong \mathbb{Z}/p\mathbb{Z}$. By Fermat's Little Theorem, $\forall x\in \mathbb{Z}$, $x^p\equiv x \mod p$, so $ \phi(x) = x^p \mod p = x \mod p $ and therefore $\phi$ fixes all elements of $\mathbb{Z}/p\mathbb{Z}$ and is the identity on $F_p$.

Is this correct? I'm asking because I haven't been able to find anything to support this claim anywhere-- I'm not sure if I've simply created a false statement and made an error in my justification for it. Thanks!

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    $\begingroup$ The Frobenius automorphism is the identity on $\Bbb{F}_p$ due to Fermat's Little Theorem as you noted, $x^p\equiv x\bmod p$. When you start looking at field extensions of $\Bbb{F}_p$, you will use the fact that the Frobenius fixes $\Bbb{F}_p$. $\endgroup$ – sharding4 Jun 5 '17 at 2:13
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    $\begingroup$ You'll show that $\mathbb{F}_{p^n}$ is the splitting field of the polynomial $x^{p^n}-x \in \mathbb{F}_p[x]$. Thus, if $K$ is a field of characteristic $p$, then the field morphism $\varphi(a) = a^{p^n}$ fixes only the elements of $\mathbb{F}_{p^n}$ $\endgroup$ – reuns Jun 5 '17 at 3:13
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    $\begingroup$ Slightly more generally: any automorphism of any field will fix the prime field (=the minimal subfield) elementwise. The prime field is either $\Bbb{F}_p$ or $\Bbb{Q}$ depending on the characteristic. $\endgroup$ – Jyrki Lahtonen Jun 5 '17 at 10:12

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