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Let $ X, Y $ be separated finite type schemes over an algebraically closed field $ k $. Assume that $ Y $ is reduced. Let $ \phi : X \rightarrow Y $ be a morphism of schemes. Suppose that $ \phi $ gives a bijection on $ k $-points and an injection on $ S$-points for all $k$-schemes $ S$. Prove or disprove that $ \phi $ is an isomorphism (or add some extra hypotheses to ensure that $ \phi $ is an isomorphism).

When $ k = \mathbb{C} $, $ X $ is normal, an $ Y $ is normal and irreducible, then I have a proof which uses the following crazy fact: If $ X $ and $ Y $ are irreducible varieties over $\mathbb{C} $ and $ Y $ is normal, then a morphism $ \phi $ inducing a bijection on $ \mathbb{C} $-points is an isomorphism. My original question follows from this fact via a small tweaking of the usual Yoneda argument.

If anyone can give me a proof or reference for this last fact, I would be grateful too. I read it in the appendix of Kumar's book on Kac-Moody groups.

Edit: In light of some counterexamples, let me assume that X is irreducible and Y is normal and irreducible.

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    $\begingroup$ Why can't $X$ be the disjoint union of the affine line and a point, and $Y$ be the projective line (and the map be the obvious thing) for a counterexample? $\endgroup$
    – Kevin Buzzard
    Sep 6, 2010 at 14:14
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    $\begingroup$ <TeX-pedant>It is usually much better to write `\$X\$, \$Y\$' than '\$X, Y\$', because of the resulting spacing and, in some cases, the fact that the math comma may be different from the text comma.</TeX-pedant> $\endgroup$
    – Mariano Suárez-Álvarez
    Sep 6, 2010 at 14:24
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    $\begingroup$ See mathoverflow.net/questions/12767 for a related question. $\endgroup$
    – VA.
    Sep 6, 2010 at 14:37
  • $\begingroup$ And also mathoverflow.net/questions/16786 $\endgroup$
    – Frank
    Sep 6, 2010 at 14:40

2 Answers 2

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Trivial counterexample when $X$ is not connected: let $F$ be closed an nonempty in $Y$, $U:=Y\setminus F$ (assumed nonempty), and $X$ the disjoint sum of $U$ and $F$.

A bit less trivial with $X$ and $Y$ irreducible: $Y$= an irreducible curve with a node $y$, $X'$:=its normalization, $X$= $X'$ with one of the two points over $y$ removed.

The property holds, indeed, if $X$ is irreducible and reduced and $Y$ normal, assuming only that $X(\mathbb{C})\to Y(\mathbb{C})$ is bijective. In fact, in this case $f$ must be quasifinite and birational, hence an isomorphism if $Y$ is normal.

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[This is a minor comment on Laurent Moret-Bailly's answer; I'm too new to leave comments]

That the map $f$ in Laurent Moret-Bailly's answer is birational follows from Proposition 7.16 in [J. Harris, Algebraic Geometry, A First Course, GTM 133, 1992]; that it is an isomorphism from Zariski's Main Theorem.

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