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Let $K$ be a finite field of characteristic $p$ and let $L$ be a finite extension of $K$. Then $L$ has an absolute Frobenius morphism which is given by the $p$th power map. Moreover, we have a map of $K$ schemes $\mathrm{spec} (L) \to \mathrm{spec} (L)^p$ called the relative Frobenius morphism. Now $L^{p}$ is isomorphic to $L$ and so we should be able to think of the relative Frobenius as an automorphism of $L$, in which case I want to know which automorphism it is.

My question is: can one naturally identify $L$ and $L^{p}$ and if so, what is the relative Frobenius morphism?

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If $K=\mathbb F_{p^n}$, the relative Frobenius on $L$ is given by the ring map $x \mapsto x^{p^n}$, while the absolute Frobenius (which is the relative Frobenius for the case $K=\mathbb F_p$) is given by the ring map $x \mapsto x^p$.

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  • $\begingroup$ How does one show that that relative Frobenius agrees with the scheme-theoretic definition? $\endgroup$ – Alexander Apr 21 '16 at 17:41

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