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Suppose $X \rightarrow Y$ is a map of projective schemes over a field $k$. Is $\{y \in Y: \pi^{-1}(y) \text{ is irreducible}\}$ a constructible subset of $Y$?

Note: One cannot hope to do "better" than constructible. That is, if we let H be the hilbert scheme of conics in $\mathbb P^2$ and let $\mathscr C$ be the universal curve over $H$, then $\mathscr C \rightarrow H$ has irreducible fibers consisting of smooth conics and double lines, which is constructible but not locally closed.

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  • $\begingroup$ Nitpicking: you need to change "irreducible" to "geometrically irreducible" if you want this to be true. $\endgroup$ Sep 4, 2015 at 23:29
  • $\begingroup$ This, more or less, follows from EGA IV, Prop. 4.5.9. In, "Low Degree Complete Intersections are Rationally Simply Connected", de Jong and I had reason to spell this out further, in Lemma 3.2. You might as well base change so that $f : X\to Y$ is proper and flat, with $X$ and $Y$ irreducible normal, and with geometrically irreducible and normal generic fiber. Moreover, by further base change to $\text{pr}_2:X\times_Y X \to X$, you may assume there is a section, namely the diagonal. Now restrict over the locus in the base where the section lands in the normal locus and apply Lemma 3.2. $\endgroup$ Sep 4, 2015 at 23:38
  • $\begingroup$ I also have a vague recollection that this is somewhere in the first chapter of Jouanolou, "Th'eor`emes de Bertini et Applications". $\endgroup$ Sep 5, 2015 at 0:59
  • $\begingroup$ Dear Jason, can we always base change a flat and proper morphism $f:X\rightarrow Y$, such that $f^\prime:X^\prime\rightarrow Y^\prime$ is proper and flat with $X^\prime,Y^\prime$ irreducible normal? $\endgroup$
    – user42804
    Jan 15, 2016 at 4:14

1 Answer 1

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I double-checked. This is Théorème 4.10, p. 36 of the following.

MR0725671 (86b:13007) Reviewed
Jouanolou, Jean-Pierre
Théorèmes de Bertini et applications. (French)
Progress in Mathematics, 42. Birkhäuser Boston, Inc., Boston, MA, 1983. ii+127 pp.
ISBN: 0-8176-3164-X

As mentioned above, you need to replace "irreducible" by "geometrically irreducible". You can also prove this from Lemma 3.2 as I sketched above. But Lemma 3.2 is really designed for a different purpose.

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    $\begingroup$ The correct EGA reference is IV$_3$, 9.7.7 (for geometric fibers of a finitely presented morphisms between arbitrary schemes -- Jouanolou considers just schemes of finite type over a field, if I remember correctly). In contrast, IV$_2$, 4.5.9 mentioned above is concerned with the (certainly important) equivalence of several possible definitions of "geometrically irreducible" rather than the harder fact of its (local) constructibility as a condition on fibers of a finitely presented morphism. In particular, properness is not relevant. $\endgroup$
    – grghxy
    Sep 5, 2015 at 7:22
  • $\begingroup$ @grghxy: Of course you are correct that the result holds without properness, just assuming finite presentation. You can use Lemma 3.2 from that paper with de Jong to prove construcibility, as I sketched above, and that only uses IV_2, 4.5.9 to know that the property may be checked after appropriate base change. Maybe Lemma 3.2 and the sketch above are repeating steps from IV_3, 9.7.7. Somehow I feel like I have had this conversation several times before (with Osserman, Anders Buch, ...). $\endgroup$ Sep 5, 2015 at 10:15
  • $\begingroup$ Your Lemma 3.2 with dJ (which has some unstated finite-type hypotheses on the morphisms $e$ and $i$ there) is very different from the argument in EGA. The basic principle for those results in EGA is to reduce to the case of hypersurfaces in affine spaces and consider the Nullstellensatz applied to the task of factoring a multivariable polynomial in terms of lower-degree polynomials. I don't think there is any proof in EGA which uses the normality technique that you use (and for sure it never comes up in that part of IV$_3$, unless my memory is completely broken). $\endgroup$
    – grghxy
    Sep 7, 2015 at 0:35
  • $\begingroup$ Thank you grghxy. I will re-read that part of EGA IV. I am due for a refresher anyway. $\endgroup$ Sep 7, 2015 at 11:04

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