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?



Which base change preserves integrality of schemes
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. 


