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Let $X\subseteq \mathbb{C}^n$ be an irreducible variety defined over $\mathbb{Q}$. I would like to show that for all but finitely many prime $p$ the variety $X(\mathbb{F}_p)$, defined over $\mathbb{F}_p$, is geometrically irreducible, i.e., $X(\overline{\mathbb{F}}_p)$ is irreducible. Can someone give me a hint or a reference?

Thanks

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2 Answers 2

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I agree with the OP that there is no need to use fancy and sophisticated language and tools. (Maybe what he calls variety is nowadays better called algebraic set.)

(a) The case that $X$ is a hypersurface is well known, some people call it the Bertini-Noether Theorem. Here one has to show that an absolutely irreducible polynomial $f(\bf x)\in{\mathbb Q}[\bf x]$ remains absolutely irreducible modulo all but finitely many primes. That follows from a straightforward application of Hilbert's Nullstellensatz, see e.g. Chapter VIII, $\S$5, Prop. 7 (page 157) of Lang's Diophantine Geometry or Chapter IX, $\S$5, Prop. 5.3 (page 241) of Lang's Fundamentals of Diophantine Geometry.

(b) The general case can be reduced to the case of a hypersurface because $X$ is birationally equivalent to a hypersurface (this uses the primitive element theorem). Indeed, a generalization of the OP's question is Prop. 10.4.2 in the second and third edition of Field Arithmetic by Fried and Jarden.

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    $\begingroup$ If you look at the proof of Théorème (9.7.7) in EGAIV that I reference in my answer, then you will see that this is exactly the same proof that you give. The proof there reduces to the case of hypersurfaces using birational equivalence, with the case of hypersurfaces being handled by Lemme (9.7.5). $\endgroup$
    – Daniel Loughran
    Commented Oct 28, 2015 at 16:22
  • $\begingroup$ Dear Peter and Dear Daniel: Many thanks for your comments and references. $\endgroup$
    – M.B
    Commented Oct 30, 2015 at 0:06
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Note that there is some small ambiguity here, as to talk about the reduction of $X/\mathbb{Q}$ modulo a prime $p$, one needs to choose a model $X$. i.e., a scheme $\mathcal{X} \to \mathbb{Z}$ whose generic fibre is isomorphic to $X$.

Anyway, if $X/\mathbb{Q}$ is geometrically integral, then the same holds over $\mathbb{F}_p$ for all but finitely many primes $p$. This follows from fact that being geometrically integral is a constructible property (see Section 9 of EGAIV). One applies this to the morphism $\mathcal{X} \to \mathbb{Z}$.

Precise references:

Definition of a constructible property: EGAIV Définition (9.2.1)

Proof that being geometrically integral is a constructible property: EGAIV Théorème (9.7.7).

You can also find a nice treatment of this in Chapter 10 of the book

Ulrich Görtz, Torsten Wedhorn - Algebraic Geometry: Part I: Schemes. With Examples and Exercises By

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    $\begingroup$ Section 9 of EGA IV is rather long, so to avoid the advice being a needle in a haystack it would be better to provide the precise reference for the result in question (constructibility of the locus of geometrically irreducible fibers for a finitely presented morphism of schemes). $\endgroup$
    – nfdc23
    Commented Oct 27, 2015 at 15:07
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    $\begingroup$ I would hope that the OP would have the initiative to find it themselves, but I have edited my question to include some more precise references. Once one knows that the relevant key phrase is "constructible property", it is quite easy to find treatments of this by searching on google. $\endgroup$
    – Daniel Loughran
    Commented Oct 27, 2015 at 17:07
  • $\begingroup$ @Daniel: Many thanks for references. But to be honest I don't think one needs heavy machinery for this. Maybe I am wrong. For me an algebraic variety is a set that is defined by polynomials and these polynomials are defined over rational. So if I take prime p big enough I can talk about the reduction of these polynomials modulo p. Where is the ambiguity? $\endgroup$
    – M.B
    Commented Oct 27, 2015 at 18:28
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    $\begingroup$ The ambiguity is that an algebraic variety is actually an isomorphism class of sets (locally) defined by polynomials, with the isomorphism being some algebraic change of coordinates which has an algebraic inverse. The problem is that the isomorphism might involve denominators. For example x^2+y^2=1 and X^2+Y^2=p^2 define isomorphic varieties (X=px, Y=py) and these might become non-isomorphic if you reduce the equations mod p. $\endgroup$
    – eric
    Commented Oct 27, 2015 at 19:16
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    $\begingroup$ @M.B: An variety over a field $k$ is a scheme $X$ equipped with a finite type separated morphism $X \to \mathrm{Spec}(k)$ (some authors also require that $X$ be integral). So given a variety $X \to \mathrm{Spec}(\mathbb{Q})$, there is no way to canonically obtain a variety over $\mathbb{F}_p$ for all primes $p$. But you may choose equations for $X$ and "clear denominators" to assume that these equations are integral, then one can reduce these equations modulo each prime $p$. This corresponds to choosing a model $\mathcal{X} \to \mathrm{Spec}(\mathbb{Z})$ for $X$ as I explain in my answer. $\endgroup$
    – Daniel Loughran
    Commented Oct 28, 2015 at 8:32

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