Is an algebraic bijection from a projective variety to itself necessarily an isomorphism? - MathOverflow

Is an algebraic bijection from a projective variety to itself necessarily an isomorphism?

2010-03-01T21:35:28Z

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.

Answer by Pavel Etingof for Is an algebraic bijection from a projective variety to itself necessarily an isomorphism?

2010-03-01T23:27:51Z

A reference for a proof that a bijective endomorphism of an algebraic variety over a field of characteristic zero is an automorphism (which is a corrected version of Mariano's statement in the comment): see e.g. S. Kaliman, Proc. Amer. Math. Soc. 133 (2005), 975-977, Lemma 1.

Answer by Frank for Is an algebraic bijection from a projective variety to itself necessarily an isomorphism?

2010-03-02T00:03:30Z

I just asked Damiano Testa this question and he proposed to me the following answer which I write out here. It basically follows from careful application of Grothendieck's take on Zariski's main theorem in EGA III (namely theorem $4.4.1$), which I "paste" here for convenience

$\textbf{Theorem:}$ Let $Y$ is a locally Noetherian prescheme and $f:X\rightarrow Y$ is a proper morphism. Let $X'$ be the set of points $x\in X$ which are isolated in their fibre $f^{-1}(f(x))$. Then $X'$ is an open subset of $X$ and if $f=g\circ f':X\rightarrow Y'\rightarrow Y$ is the Stein factorisation of $f$, the restriction of $f'$ to $X'$ is an isomorphism of $X'$ on a sub-prescheme induced on an open $U$ of $Y'$, namely one has $X'=f'^{-1}(U)$.

In your example of the cusp, the fibre at the singular point is connected but is non-reduced and is so not isolated. So you want to prove that for a self-map, this does not occur. Let $\tilde{X}$ be the normalisation of $X$. Then the ramification locus of the morphism $\tilde{X}\rightarrow X$ induced by $f^n$ is isomorphic to the ramification locus of the morphism induced by the normalisation map $\tilde{X}\rightarrow X$ since the diagram with $\tilde{X}\rightarrow\tilde{X}$ on top and $f^n$ on the bottom (and obvious vertical arrows) commutes, and the top arrow is an isomorphism from Zariski's main theorem. Therefore, if $f$ were ramified the ramification of $f^n$ would increase without bounds which implies that $f$ is unramified. Hence all fibres are isolated and we can apply Grothendieck's theorem.

Answer by damiano for Is an algebraic bijection from a projective variety to itself necessarily an isomorphism?

2010-03-02T18:56:20Z

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