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the following statement appeared implicitly in a text I read and maybe you could just give me a hint how to see this resp. give a reference:

If you have two k-varieties $X$ and $Y$ (sufficiently nice) and you have a morphism

$f:\ X \rightarrow Y$

between them, which is surjective and injective, then it is an isomorphism if $k$ is of characteristic zero.

Thank you!

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up vote 33 down vote accepted

This is false.

Consider a characteristic zero field $k$ and the cusp $C\subset \mathbb A^2_k$ with equation $y^2=x^3$ .
Its normalization $n: \mathbb A^1_k \to C: t\mapsto (t^2, t^3)$ is bijective but not an isomorphism.

"Ah, but Georges", you will say, "be attentive! The OP said nice varieties. Yours is ugly!"

In that case Zariski's main theorem will come to your rescue. One version says that a birational morphism $f:Y\to X$ of $k$-varieties (in any characteristic) with finite fibers and $X$ normal is an isomorphism of $Y$ onto an open subset of $X$, hence an isomorphism if $f$ is bijective.

"Aw, come on Georges, admit that you just dug up this birational stuff to make yourself look important!"

Well, the theorem no longer holds without some such hypothesis, even in dimension zero.
Just consider the bijective $\mathbb Q$-morphism $Spec(\mathbb Q(\sqrt 2)) \to Spec(\mathbb Q)$ of (singleton!) smooth schemes, which is not an isomorphism because it is not birational.(Of course you can inflate this to counterexamples in all dimensions)

Some other counter-examples of bijective morphisms which are not isomorphisms, even over $\mathbb C$, are $Spec \mathbb C[\epsilon] \to Spec(\mathbb C)$ and $\mathbb G_m \bigsqcup Spec(\mathbb C)\to \mathbb A^1_\mathbb C$ (the evident morphism from the disjoint sum of a punctured affine line and a point onto the affine line).However the sources of those morphisms are respectively non reduced and reducible.

Edit: Our friend Akhil gives a great argument (see his answer) showing that in ernest's case birationality is automatic. So, to sum up, we have the precise statement answering ernest's question:
Proposition Let $k$ be an an algebraically closed field of characteristic zero and $f:X\to Y$ a morphism between integral $k$-schemes of finite type over $k$. Then if $f$ is bijective and $Y$ normal the morphism $f$ is an isomorphism.

The case of characteristic $p$ As Akhil remarks, the Proposition is false in characteristic $p$. Consider an algebraically closed field $k$ of characteristic $p$ and the Frobenius morphism $f:\mathbb A^1_k \to \mathbb A^1_k:x\mapsto x^p$ with associated ring morphism $\phi:k[T] \to k[T]:P(T)\mapsto P(T^p)$ The morphism $f$ is bijective, but all fibers at closed points are non reduced of degree $p$ over $k$ .Indeed, let me denote for clarity by $A$ the $k$-algebra $\phi:k[T] \to A=k[T]$ above. Then the fibre of $f$ at the closed point $(T-a)$ in $\mathbb A^1_k$ is the affine $k$-scheme with algebra $A\otimes _{k[T]} \frac{k[T]}{(T-a)}=\frac{k[T]}{(T^p-a)}= \frac{k[T]}{(T-\sqrt[p] a)^p}$, so that the fiber is a single point but with non-reduced structure.
This is an example where Grothendieck's introduction of non-reduced schemes helps dissipate a mystery: how can a bijective morphism have degree $p\gt 1$ ?

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Dear Georges: In your last paragraph, ZMT says that $f$ factors as an open immersion followed by a finite morphism (assuming $X, Y$ separated). So if we have $X \to Z \stackrel{p}{\to} Y$ then as $Z \to Y$ is finite (we can assume $Z$ integral), and if $char(k)=0$ we get $Z = Y$ (as $p_*(\mathcal{O}_Z)$ is finite over $\mathcal{O}_Y$ with the same quot. field b/c the generic fiber size is the degree, hence equal to it). Am I missing something here? (Of course we need char. 0 because otherwise the Frobenius morphism.) – Akhil Mathew Aug 21 '11 at 15:35
I.e. I think birationality is automatic because of the fact I quoted about finite separable morphisms of varieties. – Akhil Mathew Aug 21 '11 at 15:36
Dear Akhil, you write "Am I missing something here?". On the contrary: it was I who missed your great remark that you could compute the degreee of the extension of rational function fields of the varieties by looking at the cardinality of the fibers over the closed points, thus proving birationality since that cardinality is $1$ here. May I suggest that you expand your comment to an answer, so that I (and hopefully many others) can give you a very deserved upvote? – Georges Elencwajg Aug 21 '11 at 17:38
Done as you suggested. I'm glad my comment was helpful. – Akhil Mathew Aug 21 '11 at 18:35
@ Georges and Akhil: greatest thanks for your exhaustive and ultimately helpful comments, you answered my question in any respect. Best Greetings! – ernest Aug 21 '11 at 19:43

According to Georges Elencwajg's request, here is an expanded version of my comment:

Given a dominant morphism $f: X \to Y$ of varieties over the algebraically closed field $k$ whose degree is $d$ (that is, $[k(X): k(Y)] = d$) and which is separable (i.e., the same field extension is separable), then the degree of the generic fiber $f^{-1}(y)$ consists of $d$ points.

(I think of this is the analog of the fact in algebraic number theory that, given a finite extension of Dedekind domains with one the integral closure of the other in a finite sep. extension, almost all primes are unramified. Of course, not all primes split into the full number of factors possible -- the degree of the extension of quotient fields --- but that's because the extension on residue fields might be nontrivial. That doesn't happen in the case of varieties over an alg. closed field: the residue fields on closed points are all the same.)

Proof: $f$ is generically finite, so shrinking $X, Y$, we can assume $f$ finite. Shrinking $Y$, we can also assume $Y$ normal (e.g. smooth). Shrinking further, we can assume $k[X]$ is generated by one element over $k[Y]$, so $k[X] = (k[Y])/(P)$ for some polynomial $P$ of degree $d$. Then at all maximal ideals $\mathfrak{M}$ of $k[Y]$ (i.e. closed points of the scheme $Y$ -- or points of the variety $Y$), such that the discriminant of $P$ does not lie in said ideal, the fiber consists of $d$ points: this follows because the fiber is $\mathrm{Spec} k[Y]/(\mathfrak{M}, P)$.

So this implies that a morphism of varieties in characteristic zero which is injective is birational. In particular, the argument given by Georges Elencwajg shows that a bijective morphism of varieties over an alg. closed field of characteristic zero, with the target normal, is an isomorphism.

Slightly fancier argument: an injective morphism of varieties over an algebraically closed field is radicial: consequently, if it is dominant, the map from the generic point of $X$ to the generic point of $Y$ is radicial. But radicial extensions are trivial in characteristic zero.

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They must be extremely nice, as the projection of the cusp $y^2=x^3$ onto the $x$-axis is already a counterexample. You want the varieties normal?

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Let $\pi:X\rightarrow Y$ be a morphism or relative dimension $r$ of smooth varieties over an algebraically closed field. Assume that the relative cotangent sheaf $\Omega_{X/Y}$ is locally free of rank $r$ on $X$. We have an exact sequence $$\pi^{*}\Omega_{Y}\rightarrow\Omega_{X}\rightarrow\Omega_{X/Y}\mapsto 0.$$ Let $k(x)$ be the residue field at a closed point $x$. Tensorizing we get
$$\pi^{*}\Omega_{Y}\otimes k(x)\rightarrow\Omega_{X}\otimes k(x)\rightarrow\Omega_{X/Y}\otimes k(x)\mapsto 0.$$ Since $X$ and $Y$ are smooth and $\Omega_{X/Y}$ is locally free of rank $r$ these three vector spaces are of dimension $dim(Y),dim(X),r$ respectively. So the first map is injective and we have $$0\mapsto\pi^{*}\Omega_{Y}\otimes k(x)\rightarrow\Omega_{X}\otimes k(x)\rightarrow\Omega_{X/Y}\otimes k(x)\mapsto 0.$$ For any closed point $x\in X$ we have $k(x)\cong k$. Therefore we can identify the injective map $\pi^{*}\Omega_{Y}\otimes k\rightarrow\Omega_{X}\otimes k$ with the map between the cotangent spaces $\mathfrak{m}_y/\mathfrak{m}^2_y\rightarrow\mathfrak{m}_x/\mathfrak{m}^2_x$ where $y = \pi(x)$. Dualizing we have that the differential $T_\pi(x):T_xX\rightarrow T_yY$ is surjective. Therefore $\pi$ is a smooth morphism.

In particular, if any fiber of $\pi$ is just one point, then $\Omega_{X/Y}$ is locally free of rank $r = 0$ and $\pi$ is a smooth morphism of relative dimension zero. Finally, since $\pi$ is surjective it is an isomorphism.

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