Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
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? – Kevin Buzzard Sep 6 '10 at 14:14
<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> – Mariano Suárez-Alvarez Sep 6 '10 at 14:24
See for a related question. – VA. Sep 6 '10 at 14:37
And also – Frank Sep 6 '10 at 14:40

2 Answers 2

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.

share|cite|improve this answer

[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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.