Fiberwise isomorphic condition open or closed?

Question

Let $X\to Z$ and $Y\to Z$ be two (flat and projective) families over $Z$, and let $W\subset Z$ be the subscheme on which the fibers of the two families are isomorphic. Then, is $W$ an open or closed subset of $Z$? (or neither)

Answer 1

You can't guarantee either. Any projective flat family $X\to Y$ which has some proper closed non-open subvariety $Z\subset Y$ so that the fibers over $Y\setminus Z$ and $Z$ are not isomorphic will give a counterexample to both statements. Why? Pick $y_1\in Z$ and $y_2\notin Z$, then $Y\times f^{-1}(y_1)\to Y$ and $Y\times f^{-1}(y_2)\to Y$ are both flat and projective but agree with $X\to Y$ on a closed and not open and open not closed subscheme, respectively.

Here is one such family. Let $Y=\operatorname{Spec} k[t]$ and $X=\operatorname{Spec} k[t,u]/(u^2-t)$ with the map being $\operatorname{Spec}$ of the map $k[t]\to k[t,u]/(u^2-t)$ by $t\mapsto t$. This is a finite flat map of rings, since the target is a free module with basis $\{1,u\}$, thus it is a projective and flat map of schemes. But the fibers over nonzero points are not isomorphic to the fiber over $0$, since the former are reduced and the latter is not.