Does there exist a finite morphism $f: X \to X$ from an integral scheme of finite type over a field to itself such that $f$ induces an identity map of topological spaces on a dense open $U \subset X$?
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1Assuming that $X$ is separated, then the answer should be no by standard theory (i.e. the closed subscheme where $f$ and the identity agree is closed, and contains a dense open, so is everything, but then by integrality that implies it's equal to $X$). – Alex Youcis Jan 28 '21 at 17:11

dear @AlexYoucis, sorry, of course you are right, I guess I am just too tired today to have missed that. I wonder if I should remove the question to stop embarassing myself :) – Dima Sustretov Jan 28 '21 at 17:36

3That's up to you. Asking silly questions never changes, no matter how much you know, so I wouldn't stress over it. – Alex Youcis Jan 28 '21 at 18:30