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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?

Evariste
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    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 $Y\times_L \bar{L}=Y\times_K (K\otimes_L \bar{L})$, and $K\otimes_L \bar{L}$ is never integral. – Cantlog Jun 11 '14 at 19:22
  • @Cantlog: I believe your comment should be an answer. – RghtHndSd Jun 12 '14 at 15:21

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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.

mathematics2x2life
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