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 $X$ and $Y$ be affine varieties over $\mathbb C$, and consider a morphism $f:X\to Y$ and the induced homomorhism

$$ \varphi=f^*:B=\mathbb C[Y]\to A=\mathbb C[X]. $$

It is very easy to see that if $\varphi$ is surjective then $f$ is injective. The converse is not true, as the inclusion $\mathbb C\setminus\{0\}\to\mathbb C$ shows.

My question is:

Assume $f$ is a finite injective morphism. Is it true that $\varphi$ must be surjective?

Here finite means that $A$ is integral over $B$ or, equivalently, that $A$ is a fintely generated $B$-module. Further, we are allowed to assume that $X$ and $Y$ are irreducible, and that $A$ and $B$ are free polynomial rings, if that helps.

One idea was to localize at an arbitrary maximal ideal $\mathfrak m$ of $B$, but I do not know if this works.

share|cite|improve this question
Just a quick comment that in characteristic $p > 0$ there isn't any hope, consider the Frobenius morphism. As is already pointed out in auniket's nice answer, things are better behaved over $\mathbb{C}$ – Karl Schwede Feb 1 '14 at 14:46
up vote 10 down vote accepted

No - consider the normalization of a cuspidal rational curve.

I don't understand your condition that $A$ and $B$ are free polynomial rings: do you mean that $X$ and $Y$ are isomorphic to $\mathbb{C}^n$ for some $n$? In that case the answer is positive. More precisely, the answer is positive if $B$ is integrally closed. Indeed, the injectivity and finiteness of $f$ implies that the quotient fields of $A$ and $B$ are the same, so that $A$ is contained in the integral closure of $B$.

share|cite|improve this answer
Thanks a lot for your comments. So your argument shows that a finite bijective morphism $f:X\to Y$ between affine varieties where $Y$ is normal must be an isomorphism. I think now I understand better my problem and maybe I will reformulate the question in anedit. – Claudio Gorodski Jan 31 '14 at 16:42
Just wanted to add that in your first statement above, "affine" is not necessary. Essentially the same argument shows that "a finite bijective morphism f:X→Y between varieties where Y is normal must be an isomorphism." – auniket Jan 31 '14 at 21:40

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.