9
$\begingroup$

Let $f:X\rightarrow Y$ be a flat morphism of normal projective varieties with fibers of positive dimension (in particular all the fibers are connected and of the same dimension).

Let $g:X\rightarrow Z$ be another morphism with connected fibers of positive dimension. Assume that $g$ contracts the general fiber of $f$ to a point of $Z$. Is it true that then $g$ must contract any fiber of $f$ to a point of $Z$?

In other words, is the map

$$Y\rightarrow \mathbb{Z},\: y\mapsto dim(g(f^{-1}(y)))$$

continuous?

$\endgroup$
6
$\begingroup$

Another proof: The assumptions imply that $X\times_YX$ is irreducible and that the two composite maps $X\times_YX\rightrightarrows X\xrightarrow{g}Z$ coincide over the generic point of $Y$. By density they must be equal (I have to assume $Z$ separated here), so $g$ is constant along the fibers of $f$ (and, by descent, $g$ factors through $Y$).

$\endgroup$
3
$\begingroup$

For each point in $Y$, take a curve $C \subseteq Y$ that passes through that point and the generic point. Let $X'= X \times_{Y} C$. Then the generic fiber of $f': X' \to C$ is still irreducible, and because $f$ is flat no irreducible component is contained in a special fiber, so $X'$ is irreducible.

Consider the image of $X'$ in $C \times Z$. It is closed and irreducible, and is zero-dimensional over the generic point of $C$. Hence it is an irreducible curve with a nonconstant map to $C$. Thus every fiber of the map to $C$ is zero-dimensional.

Hence every fiber of the map from the image of $X$ in $Y \times Z$ to $Y$ is zero-dimensional, as desired.

$\endgroup$
  • $\begingroup$ What do you mean exactly by no irreducible component is contained in a special fiber? What's the map between the image of $X^{'}$ in $C\times Z$ and $C$? $\endgroup$ – user58018 Apr 12 '15 at 21:01
  • $\begingroup$ 1. I mean that every irreducible component of X' maps surjectively to C - a general property of flat maps. Because the generic fiber is irreducible, there is only one irreducible component. 2. The first projection to $C$. $\endgroup$ – Will Sawin Apr 12 '15 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.