For finite flat surjective maps $X\rightarrow Y\rightarrow Z$ of regular schemes, does etaleness of the composite imply etaleness of each piece? Let $f : X\rightarrow Y$ and $g : Y\rightarrow Z$ be finite flat surjective morphisms of regular schemes.
Suppose $g\circ f : X\rightarrow Z$ is etale. Must $f$ and $g$ both be etale?
I believe the answer is yes. Purity of the branch locus turns this into a problem about extensions of complete discrete valuation rings, but then ramification indices are multiplicative, so any ramification that happens in $f$ or $g$ should be visible in $g\circ f$.
I just want to make sure I'm not making a subtle mistake.
 A: This sounds right to me. 
Here's an alternative argument: recall that a morphism $X\to S$ is etale iff the fiber $X_s$ over every point $s\in S$ is a disjoint union of finite separable extensions of $\kappa(s)$ (cf Stacks). As base change preserves finite, flat, and surjective, we may assume that $Z$ is the spectrum of a field.
As finite morphisms are affine, we have that $X,Y$ are both spectra of module-finite (thus artinian) $k$-algebras $B, A$ respectively. By etaleness, $B$ is a product of finite separable extensions of $k$, and we note that as faithfully flat is exactly flat plus surjective and faithfully flat maps are injective, all three maps $k\to B$, $k\to A$, and $A\to B$ are injective. As $B$ has no nilpotents, we see that $A$ has no nilpotents. By the structure theorem for artinian rings, we see that $A$ is a finite product of field extensions of $k$.
Localizing at a point of $X$, we see that our composition $k\to A\to B$ turns in to a sequence of field extensions. As $E\subset F\subset K$ is finite separable iff each composite step is, we see the result.
