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

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:X\to Y$ be a surjective morphism of complex varieties with $Y$ affine. Assume that every fiber of this morphism has the property that all the global functions are constant.

What else do I need to assume to conclude that $f$ induces an isomorphism on global functions? Does it enough to assume that $Y$ is smooth in codimension 1?

share|cite|improve this question
Do you assume the fibres are reduced? Without this the condition that $Y$ is smooth in codimension $1$ is not enough. – ulrich Oct 5 '11 at 4:23
Do you have a counterexample in mind? – Roman Fedorov Oct 5 '11 at 4:38
Yes. Let $A=k[x,y]$, $I$ the ideal $(x,y)^2$ and $B$ the subalgebra of $A$ generated by $I$. Let $X=Spec(A)$, $Y = Spec(B)$ and $f$ the morphism induced by the inclusion $ B \subset A$.Then $Y$ is smooth in codimension $1$ and$ f$ is a bijection on points so your condition is satisfied. – ulrich 0 secs ago – ulrich Oct 5 '11 at 5:06
Of course you are right. But I would expect that there is some sufficient condition on $Y$ like being Cohen-Macaulay? – Roman Fedorov Oct 5 '11 at 16:19

Your Answer


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

Browse other questions tagged or ask your own question.