Timeline for Geometrically irreducible variety
Current License: CC BY-SA 3.0
13 events
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| Oct 29, 2015 at 8:26 | comment | added | Ariyan Javanpeykar | @MartinBright If $X$ is embedded in $\mathbb A^n_{\mathbb Q}$ there is no ambiguity, as you say. But the OP starts by saying that $X $ is embedded in $\mathbb A^{n}_{\mathbb C}$ and has a model over $\mathbb Q$. The ambiguity now lies (essentially only) in the choice of model for $X$ over $\mathbb Q$. | |
| Oct 28, 2015 at 16:32 | comment | added | pro | Came here just to write "Gortz-Wedhorn rules!"; done. | |
| Oct 28, 2015 at 16:21 | comment | added | Martin Bright | I don't think there's any ambiguity here. The variety is embedded in $\mathbb{A}^n_{\mathbb{Q}}$, so it comes with a natural model obtained by taking the closure in $\mathbb{A}^n_{\mathbb{Z}}$. If explicit defining equations are given, then for almost all p this corresponds to just reducing the equations mod p. In general one might need to saturate at the ideal $(p)$. | |
| Oct 28, 2015 at 8:38 | comment | added | Daniel Loughran | @M.B: Quite possibly there is an "elementary" proof of this result in your special case. But the problem you are asking about is non-trivial, as you want to know how irreducibility behaves in families. Being irreducible is not a nice condition in general, unlike say properness, as, for example, it is not preserved under base-change. Using the fact that this is a constructible property seems like the most natural way to approach this to me. | |
| Oct 28, 2015 at 8:34 | comment | added | Daniel Loughran | Given any two choices of model, the reductions modulo all but finitely many primes are isomorphic. This is why I said that the ambiguity was "small". | |
| Oct 28, 2015 at 8:32 | comment | added | Daniel Loughran | @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. | |
| Oct 27, 2015 at 19:16 | comment | added | eric | 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. | |
| Oct 27, 2015 at 18:28 | comment | added | M.B | @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? | |
| Oct 27, 2015 at 17:07 | comment | added | Daniel Loughran | 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. | |
| Oct 27, 2015 at 17:06 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
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| Oct 27, 2015 at 15:07 | comment | added | nfdc23 | 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). | |
| Oct 27, 2015 at 9:00 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
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| Oct 27, 2015 at 8:52 | history | answered | Daniel Loughran | CC BY-SA 3.0 |