TITLE: Is an algebraic bijection from a projective variety to itself necessarily an isomorphism?
QUESTION [8 upvotes]: Let $X$ be a projective variety. Assume there is an algebraic map $f: X \rightarrow X$ that is a bijection. I am thinking of $X$ as a variety, not a scheme, so by a bijection I mean a bijection on closed points. Most likely I am working over the complex numbers, so if you like I mean a bijection on complex points. Can you conclude that $f$ has an algebraic inverse? 
I think this is not immediately obvious, since it is not true that any algebraic bijection between two projective varieties is an isomorphism. For instance, there is an algebraic bijection from ${\Bbb P}^1$ to a cuspidal cubic in ${\Bbb P}^2$ given by $[x,y] \rightarrow [x^3, x^2y, y^3]$. So if this is true one must use the fact that the map is from $X$ to itself.
I am interested in cases where $X$ is both singular and reducible (although is of pure dimension, if that helps), so a complete answer would cover any such case. Alternatively, if it is not true that such a map has an algebraic inverse, I would like an explicit counter example.

REPLY [8 votes]: I am indeed claiming that it works over any field, but with the additional assumption that the morphism f be separable (this is required to make the induced morphism between the normalizations an isomorphism, rather than simply a finite morphism).  Also, the Frobenius morphism is actually proper.
The main emphasis of the argument above though was that it did not require any form of irreducibility on the schemes, which I thought was important.  d