Let $X,Y,Z$ be reduced algebraic varieties, and let $Y$ and $Z$ be normal. Let $f:X \to Y$ and $g:X \to Z$ two surjective projective morphisms of algebraic varieties such that the geometric fibers of $f$ and $g$ coincide. Is there an isomorphism $h:Y\to Z$ such that $g=h \circ f$?
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$\begingroup$ I don't understand "$Y$ and $Z$ coincide set-theoretically". $\endgroup$– Laurent Moret-BaillyCommented Aug 29, 2011 at 11:01
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$\begingroup$ I mean that there's a one-to-one map between the set of closed points of $Y$ and the set of closed points of $Z$. $\endgroup$– IMeasyCommented Aug 29, 2011 at 11:05
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$\begingroup$ I don't understand, $f$ and $g$ have different targets, so how can they be the same morphism unless $Y=Z$? Do you mean that $Y$ and $Z$ have the same underlying topological space? In that case the answer is still no: Let $Y$ be a non-reduced scheme, $X=Z=Y_{red}$, $f:Y_{red}\to Y$ the inclustion and $g$ the identity. $\endgroup$– J.C. OttemCommented Aug 29, 2011 at 11:35
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1$\begingroup$ I still don't understand. Any positive-dimensional complex algebraic variety $Y$ has the same cardinality as, say, $[0,1]$. Fixing a bijection, we obtain by transport of structure a complex variety $Y'$ isomorphic to $Y$, with underlying set $[0,1]$. Now, starting with $f$ and $g$ but without your assumption on $Y$ and $Z$, we can apply the above construction and get, by composition, $f':X\to Y'$ and $g':X\to Z'$ where $Y'$ and $Z'$ have the same underlying set. In other words, this assumption on $Y$ and $Z$ is not a restriction. $\endgroup$– Laurent Moret-BaillyCommented Aug 29, 2011 at 12:40
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1$\begingroup$ I think you want the maps to be surjective. Also, I suggest you rephrase the question to ask whether there exists an isomorphism $h: Y \to Z$ such that $g = h \circ f$. $\endgroup$– S. Carnahan ♦Commented Aug 29, 2011 at 15:23
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1 Answer
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In positive characteristic, you get a counterexample by taking $X=Y=Z=$ the affine line (say), $f$ the identity and $g$ the Frobenius map.
Assume now that the ground field is algebraically closed of characteristic zero. Consider the map $(f,g):X\to Y\times Z$. Its image $\Gamma$ is a closed subvariety of $Y\times Z$. The assumption on the fibers exactly means that both projections from $\Gamma$ to $Y$ and $Z$ are bijective (on closed points). Since they are proper they must be (in char. zero) finite birational, hence (by normality) isomorphisms. So, $\Gamma$ is the graph of the isomorphism we are looking for.
answered Aug 29, 2011 at 18:36
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2$\begingroup$ Wonderful argument! The confusing part for me is why proper bijective implies finite birational in characteristic zero. I see that proper and quasi-finite implies finite. Because of characteristic zero, we have that the map is smooth on an open upstairs (right?), so it is smooth on an open set of the base (finite bijection implies homeomorphism?). Now there is an open set in the base where the map is finite, smooth (so etale), with connected fibers, so is an isomorphism. Is this the right approach? $\endgroup$ Commented Aug 30, 2011 at 6:34
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$\begingroup$ @Anton: yes, that's also how I see it. $\endgroup$ Commented Aug 30, 2011 at 7:27