My question might be very naive.

Let $X$ and $Y$ be reduced schemes of finite type over an algebraically closed field $k$ and $f:X\longrightarrow Y$ a morphism of $k$-schemes. Let us assume that $Y$ is normal, connected and that $f$ induces a bijection on closed points. Can we conclude from these assumptions that $f$ is a local isomorphism?

If I assume that $X$ is also connected, may I conclude that $f$ is an isomorphism?

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    $\begingroup$ No. Let $Y= \mathbb{A}^1_k$, and $X$ the disjoint union of $ \mathbb{A}^1_k-\{0\}$ and $\{0\}$ with $f$ the obvious map. $\endgroup$ – Donu Arapura Apr 18 '19 at 22:05
  • $\begingroup$ Ok, I just edited my question. $\endgroup$ – Gaussian Apr 18 '19 at 22:09

If $k$ is algebraically closed of non-zero characteristic $p$, then bijectivity on closed points is not enough (take the morphism $\mathrm{Spec}\,k[x]\rightarrow \mathrm{Spec}\,k[x]$ corresponding to $f(T)\rightarrow f(T^p)$).

If $k$ is algebraically closed of characteristic 0, then bijectivity on closed points is enough (if your target is normal, that is).

If you do not make any assumption on characteristic, then the following result is true. A quasi-finite birational morphism between integral schemes of finite type over an algebraically closed field, with target normal, is an open immersion. If your morphism is also surjective, it is an isomorphism (since surjective open immersions are isomorphisms). In particular, if your morphism is bijective birational, it is an isomorphism.

The reason why characteristics 0 and $p$ apparently behave differently is that only in $\mathrm{char}\,p$ you have this funny thing called purely inseparable field extension.

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  • $\begingroup$ I was thinking of a field of characteristic zero, do this answers my question. Thanks! $\endgroup$ – Gaussian Apr 18 '19 at 22:51

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