8
$\begingroup$

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.

Improve this question
$\endgroup$
7
  • $\begingroup$ A bijective regular morphism between irreducible varieties in characteristic zero is biregular. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 1, 2010 at 21:41
  • $\begingroup$ So, are you OK with the smooth case? That is much easier and I can sketch it if you want. $\endgroup$
    – Andrea Ferretti
    Commented Mar 1, 2010 at 23:43
  • $\begingroup$ Basically the smooth case follows from a computation of differential and Zariski main theorem. $\endgroup$
    – Andrea Ferretti
    Commented Mar 1, 2010 at 23:44
  • $\begingroup$ I don't think it's true: $\mathbb{A}^2$ is irreducible, but consider the map from $\mathbb{A}^2$ to itself that sends each vertical line into a "translated" cuspidal cubic... $\endgroup$
    – Qfwfq
    Commented Mar 21, 2010 at 17:51
  • $\begingroup$ Something like: $(x,y)\mapsto(y^2+x,y^3)$ $\endgroup$
    – Qfwfq
    Commented Mar 21, 2010 at 17:55

3 Answers 3

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ To avoid confusion, I suggest writing out the correct version of "Mariano's statement" (I am sure he meant to say that the two varieties are the same). $\endgroup$
    – Bjorn Poonen
    Commented Mar 2, 2010 at 0:15
  • $\begingroup$ Ah. What I actually had in mind was the statement that a bijective regular morphism $X\to Y$ between irreducible varieties in characteristic zero is biregular if $Y$ is smooth, and I had forgotten about the smoothness hypothesis. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 2, 2010 at 12:22
8
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yes, not requiring irreducibility is important. Frank's argument seems to be more or less the same as the one in Kaliman's paper, but perhaps you are right that their varieties are irreducible.. $\endgroup$
    – Peter Tingley
    Commented Mar 2, 2010 at 20:38
  • $\begingroup$ I have looked at the argument in Kaliman's paper and indeed it is very similar to the one given below. The reason for which Kaliman needs irreducible varieties is that he is interested in showing something different. The example on the first page of Kaliman's paper is exactly the main issue: he wants to be able to forget codimension two subschemes and still retain enough information. In particular, he cannot distinguish a scheme that becomes disconnected by removing a codimension two subscheme from its normalization. $\endgroup$
    – damiano
    Commented Mar 3, 2010 at 14:16
7
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ A p-adic field has characteristic 0. $\endgroup$
    – Pete L. Clark
    Commented Mar 3, 2010 at 6:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .