Is an algebraic bijection from a projective variety to itself necessarily an isomorphism? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:21:10Z http://mathoverflow.net/feeds/question/16786 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16786/is-an-algebraic-bijection-from-a-projective-variety-to-itself-necessarily-an-isom Is an algebraic bijection from a projective variety to itself necessarily an isomorphism? Peter Tingley 2010-03-01T21:35:28Z 2010-03-02T18:56:20Z <p>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? </p> <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/16786/is-an-algebraic-bijection-from-a-projective-variety-to-itself-necessarily-an-isom/16803#16803 Answer by Pavel Etingof for Is an algebraic bijection from a projective variety to itself necessarily an isomorphism? Pavel Etingof 2010-03-01T23:27:51Z 2010-03-02T02:29:35Z <p>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. </p> http://mathoverflow.net/questions/16786/is-an-algebraic-bijection-from-a-projective-variety-to-itself-necessarily-an-isom/16809#16809 Answer by Frank for Is an algebraic bijection from a projective variety to itself necessarily an isomorphism? Frank 2010-03-02T00:03:30Z 2010-03-02T00:03:30Z <p>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</p> <p>$\textbf{Theorem:}$ 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 $X'=f'^{-1}(U)$. </p> <p>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.</p> http://mathoverflow.net/questions/16786/is-an-algebraic-bijection-from-a-projective-variety-to-itself-necessarily-an-isom/16896#16896 Answer by damiano for Is an algebraic bijection from a projective variety to itself necessarily an isomorphism? damiano 2010-03-02T18:56:20Z 2010-03-02T18:56:20Z <p>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.</p> <p>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</p>