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Let $X$ be a variety (integral scheme of finite type) over $\overline{\mathbb Q}$. We may endow the sets $X(\overline{\mathbb Q})$ and $X(\mathbb C)$ of $\overline{\mathbb Q}$- resp. $\mathbb C$-valued points of $X$ with the topologies induced by the topology on $\overline{\mathbb Q} \subset \mathbb C$. We have $X(\overline{\mathbb Q}) \subset X(\mathbb C)$.

Q: Is $X(\overline{\mathbb Q})$ dense in $X(\mathbb C)$?

For a smooth $X$ I can show this. For $\dim X = 0$ this is also true.

Ansatz: The question is local on $X$, so we can assume that $X = \operatorname{Spec} A$, $A = \overline{\mathbb Q}[X_1, \ldots, X_n]/(f_1, \ldots, f_r)$ for polynomials $f_i$ in variables $X_1, \ldots, X_n$ over $\overline{\mathbb Q}$. So the question is, whether $$ \{ (x_1, \ldots, x_n) \in \overline{\mathbb Q}^n~|~f_i(x_1, \ldots, x_n) = 0~\text{for all}~i \} $$ is dense in $$ \{ (x_1, \ldots, x_n) \in \mathbb C^n~|~f_i(x_1, \ldots, x_n) = 0~\text{for all}~i \}. $$

boxdot
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  • I take it that by that bar over the Q you mean to denote the algebraic numbers. So, you are not actually asking about rational-valued points, but algebraic-valued points. If I have that right, maybe you would want to edit the title of the question. – Gerry Myerson Sep 25 '13 at 13:13
  • @GerryMyerson: Done, thank you. – boxdot Sep 25 '13 at 14:25
  • How do you prove the result in the smooth case? – Georges Elencwajg Sep 25 '13 at 17:02
  • @finite: want a direct proof without reducing to the smooth case ? – Cantlog Sep 25 '13 at 17:04
  • @GeorgesElencwajg: By considering étale neighborhoods from $\mathbb A^n_{\overline{\mathbb Q}}$ which are local isomorphisms on $\mathbb C$-valued points by implicit function theorem. They give charts to open sets in $\mathbb C^n$. Further, I approximate a point by points in this charts with coordinates in $\overline{\mathbb Q}$. This is independent of the chosen chart, since the change to a different chart is given by a $\overline{\mathbb Q}$-morphism. – boxdot Sep 25 '13 at 17:52
  • Thanks for answering, @finite. – Georges Elencwajg Sep 25 '13 at 17:54
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    @Cantlog: although you are not asking me, I certainly would love to see a direct proof ! – Georges Elencwajg Sep 25 '13 at 17:56
  • Dear @GeorgesElencwajg Georges, I thought I had a simple proof, but it is essentially the same as that of "finite": consider a Noether normalization map $X\to \mathbb A^n_{\overline{\mathbb Q}}$, remove the nonwhere Zariski (hence complex) dense ramification locus and we are reduced to the 'etale situation. – Cantlog Sep 25 '13 at 19:18
  • @GeorgesElencwajg The answer of Pete Clark to this question http://math.stackexchange.com/questions/178461/complex-variety-with-zariski-dense-set-of-algebraic-points might be useful here. It seems that he has a direct proof (which I do not understand) of density of algebraic points, which does not involve generic regularity. – Moishe Kohan Oct 16 '13 at 14:58
  • Also, Martin Brandenburg seems to know an easy proof of generic regularity (see his comment here http://math.stackexchange.com/questions/435510/generic-regularity-of-affine-varieties) which I do not understand at all. – Moishe Kohan Oct 16 '13 at 15:00

2 Answers2

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The smooth locus of $X(\mathbb C)$ is Zarisk dense, hence dense for the complex topology, in $X(\mathbb C)$. So it is enough to prove the density for smooth varieties, and you already known how to do it.

Community
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Cantlog
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It seems to me we can prove this without the reduction to the smooth case using Noether normalization. Assuming WLOG that $X$ is affine, by Noether normalization it admits a surjective finite morphism $f : X \to \mathbb{A}^d$ to an affine space over $\overline{\mathbb{Q}}$. This induces a compatible pair of maps on $\overline{\mathbb{Q}}$-points and on $\mathbb{C}$-points, and now we can reduce to the case of affine space: since $\overline{\mathbb{Q}}^d$ is dense in $\mathbb{C}^d$, if $x \in X(\mathbb{C})$ is a $\mathbb{C}$-point we can write down a sequence of $\overline{\mathbb{Q}}$-points converging to $f(x) \in \mathbb{C}^d$, and the preimages of these points in $X$, which remain in $X(\overline{\mathbb{Q}})$ since $X$ is defined over $\overline{\mathbb{Q}}$, contain $x$ in their closure.

Qiaochu Yuan
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  • Hey Qiaochu. I'm not sure I follow the last argument. It almost sounds like you're saying that if $f\colon X\to Y$ is a map of topological spaces and $S\subseteq Y$ is dense, then $f^{-1}(S)$ is dense. Of course, this is false as the inclusion $\mathbb{Q}\to \mathbb{R}$ and $S=\mathbb{R}-\mathbb{Q}$ shows. That said, I guess this is true if $f$ is an open map. So, perhaps your argument can be saved with some machinery. Namely, I think one can assume WLOG that $X$ is irreducible. – Alex Youcis Jul 08 '24 at 08:20
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    Then, one can apply the fact that your map $f$ induces an open map on analytifications: https://mathoverflow.net/questions/21676/in-the-analytic-category-finite-morphisms-are-open-maps – Alex Youcis Jul 08 '24 at 08:20
  • @Alex: well, I'm certainly not claiming that in general, at the very least I need $f$ to be surjective. I didn't think about exactly what hypotheses I needed for the general statement but surjective + finite seemed safe to me, I'll have to think about what exactly is needed here. – Qiaochu Yuan Jul 08 '24 at 08:27
  • Okay, yes, it looks like we need $f$ open, thanks for the correction. – Qiaochu Yuan Jul 08 '24 at 08:29
  • OK. Cool. I like this alternative proof. Thanks for writing it. – Alex Youcis Jul 08 '24 at 09:10