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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.

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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 11 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 $B$.

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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

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