Let $f:X \to Y$ be a flat morphism of projective noetherian integral schemes. Is there any known condition on a morphism $Z \to Y$ under which the resulting fiber product $X \times_Y Z$ is still integral?
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1$\begingroup$ Projective is a property of a morphism, not of a scheme, so it doesn't make sense to say that $X$ and $Y$ are projective without specifying the base. $\endgroup$– Kestutis CesnaviciusCommented Apr 24, 2015 at 15:42
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$\begingroup$ A standard use of the term "Projective scheme" is to mean a closed subscheme of a projective space. Here the morphisms are implicit, not necessary for the problem. $\endgroup$– RonCommented Apr 24, 2015 at 15:52
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2$\begingroup$ But saying "projective space" without specifying the base doesn't make sense either. $\endgroup$– Kestutis CesnaviciusCommented Apr 24, 2015 at 18:24
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$\begingroup$ You can take the base to be $\mathrm{Spec} R$ for a complete DVR $R$. $\endgroup$– RonCommented Apr 24, 2015 at 18:45
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If $Z$ is integral, $Z \to Y$ is dominant, and the field of fractions extension $k(Y) \to k(Z)$ is purely transcendental, you're OK.
First, note that the map from the fiber product to $Z$ is still flat, hence torsion free, so if there are zero divisors on any fiber they are on the generic fiber. So we can replace the issue with base change from $k(Y)$ to $k(Z)$. We can embed the ring of functions on $X$ into $k(X)$ and work with $k(X) \otimes_{k(Y)} k(Z)$. But this embeds into the field given by adjoining the generators of $k(Z)$ as transcendentals independent from the elements of $k(X)$, so is an integral domain.
On the other hand if the field of fractions extension contains an algebraic extension you're in trouble, because you can take $X$ to be the normalization of $Y$ in that algebraic extension.
If the map is not dominant then I'm not sure if there are always counterexamples but there are certainly many non-dominant maps which don't preserve integrality.
answered Apr 25, 2015 at 15:43
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$\begingroup$ What if $k(Y)$ is only algebraically closed in $k(Z)$ (without the pure transcendentality condition)? In that case, the normalization of $Y$ in $k(Z)$ is generically just $Y$, right? I'd like to know whether the fiber product is integral in that case too. $\endgroup$ Commented Jan 10, 2017 at 20:26