Hi,
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!
Best Answer
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 $p\gt 1$ ?