Geometrically integral Here is a stupid question about the notion of geometric integrality.
Say I have a smooth, projective variety $X$ over a some field $k$, equipped with a morphism $f: X \to C$ to a smooth, projective curve $C$, such that the generic fibre is geometrically integral.
Assume that there exists a finite (dominant) morphism $\varphi: C \to C$ of degree at least $2$.
Is it true that the generic fibre of the composition $\varphi \circ f$ is not geometrically integral?

Yes, absolutely. The generic fiber $Y$ is an algebraic variety over the function field $K$ of $C$. Consider $K$ as a finite non-trivial extension of a subfield $L$, then $Y \times_L (K \otimes_L \overline{L})$, and $K \otimes_L \overline{L}$ is never integral.
