3
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Cantlog Jun 11 '14 at 19:22
  • $\begingroup$ @Cantlog: I believe your comment should be an answer. $\endgroup$ – RghtHndSd Jun 12 '14 at 15:21
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.