Base change and étale morphism

Question

Let $f:X\to Y$ be a finite surjective morphism between projective normal varieties, where $Y$ is smooth. Let $g:Z\to Y$ is a finite surjective morphism from a projective normal variety $Z$. Consider the base change $Z\times_YX\to Z$. If $(Z\times_YX)_{\rm red}\to Z$ is étale outside $g^{-1}(T)\subset Z$ for a proper subvariety $T\subset Y$, can we conclude that $f$ is étale outside $T$?

Answer 1

This cannot be true exactly as stated. For example, if $X=Z=Y$ and $f$ and $g$ are both the Frobenius morphism then the reduced part of $X \times_Y Z$ maps isomorphically to $Z$ but $X \to Y$ is nowhere étale.

The statement is true if $(Z \times_Y X)_{\rm red}$ is replaced by $Z \times_Y X$. We can simply the question by replacing $Y$ with $Y-T$. We lose the projectivity, but it is not necessary. A morphism is étale if and only if it is flat and unramified. Unramifiedness depends only on the fibers of $f$, so it can be verified after any surjective base change. We consider the flatness. Since $f$ is finite, it is flat if and only if $f_\ast \mathcal O_X$ is locally free. Since $Y$ is a variety, we can check this by checking that the rank of $f_\ast \mathcal O_X$ is locally constant. Equivalently, we want to check that the subset $U$ where $f_\ast \mathcal O_X$ has rank $\geq n$, which is automatically closed, is also open. Since $g$ is finite and surjective, $U$ is open if and only if $g^{-1} U$ is open. But $f$ is affine, so the formation of $f_\ast \mathcal O_X$ commutes with base change. Therefore $g^{-1} U$ is the locus in $Z$ where $f_\ast \mathcal O_{Z \times_Y X}$ has rank $\geq n$. If $Z \times_Y X \to Z$ was assumed to be flat then this locus will be open in $Z$ and we may conclude that $U$ is open in $Y$.