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Let $k$ be a field of characteristic $p>0$ and $\phi:\mathbb{A}^1\to\mathbb{A}^1$ be the Frobenius morphism $\phi(x)=x^p$.

According to Hartshorne exercise I.3.2, $\phi$ is not an isomorphism of varieties. Why is this?

I think $\phi$ is the identity map (by Fermat's little theorem), so $\phi$ is its own inverse morphism. Thus $\phi$ is an isomorphism of varieties. Where is my error?

rpf
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    I think the reason is that the inverse of $\phi$, that map $x$ to $\sqrt[p]{x}$, is not a morphism of varieties. Also, $\phi$ is only the identity on $\mathbb{F}_p$. I guess $k$ is an algebraically closed field of characteristic $p$. – Huy Dang Oct 07 '17 at 19:04
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    A morphism $\phi:X\to Y$ of affine varieties induces a morphism $\phi^:\Gamma(Y)\to\Gamma(X)$ of rings. $\phi$ is an isomorphism if and only if $\phi^$ is an isomorphism of rings. In the above situation, it is not. – Mohan Oct 07 '17 at 19:08
  • @Mohan That seems like it would make for a good answer. – Tobias Kildetoft Oct 07 '17 at 19:14
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    $\phi$ is the identity map on the "$\mathbb{F}_p$ rational points," $0, 1, ... , p-1$. But there are the only elements of $k$ which satisfy $x^p = x$. – D_S Oct 07 '17 at 23:19
  • @Mohan Can you please explain why it's not an isomorphism of rings? – BKFH Nov 26 '23 at 22:18
  • @BKFH The map $k[x]\to k[x]$ , $x\mapsto x^p$ is not an isomorphism. – Mohan Nov 27 '23 at 02:36

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