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I want to use the following statement concerning varieties, but I do not know why it is true.

Claim. Let $V \subset \mathbb{C}^n$ be a variety defined over $\mathbb{Q}$. Then the set $V \cap \bar{\mathbb{Q}}$ is dense in $V$ with respect to the Hausdorff topology (not the Zariski topology).

Here, $\bar{\mathbb{Q}}$ denotes the algebraic closure of the rationals $\mathbb{Q}$.

It was pointed out to me that one can show this using the so-called Tarski-Seidenberg Principle, in particular using Proposition 5.3.5 of Real Algebraic Geometry by Bochnak, Coste, Roy.

Let $R$ be a real closed field, $A\subset R^m$ and $B\subset R^n$ semialgebraic sets, and $f:A\to B$ a semialgebraic map with graph $G\subset A\times B$. Let $K$ be a real closed extension of $R$, and denote the extension of a semialgebraic set $S$ defined over $R$ to $K$ as $S_K$.

Proposition 5.3.5

i) The semialgebraic set $A$ is open (resp. closed) in $R^m$ iff $A_K$ is open (resp. closed) in $K^m$. More generally, $clos(A_K)=(clos(A))_K$.

ii) The semialgebraic mapping $f$ is continuous iff $f_k$ is continuous.

I do not see how my statement follows. Does anyone have experience with applying this principle to this kind of situation? Or can you think of another approach leading to a proof of the claim?

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Levi
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    I've edited the question to include the proposition you reference. In the future, please include information like this in the question statement - it will help other users write good answers to your question (and not everyone has this book immediately on hand!). – KReiser Nov 28 '18 at 00:04
  • You can pick a transcendental basis to put the function field in the form $K(V) \cong K(u_1,\ldots,u_m)[z_1,\ldots,z_l]/J$ ? – reuns Nov 28 '18 at 03:29
  • @reuns Is $\overline{\mathbb{Q}}$ even real closed field?. Or one should apply the theorem to different fields? – random123 Nov 28 '18 at 05:50
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    $\overline{\Bbb Q}$ is not real closed (it has $i$, for instance). Techniques from semialgebraic geometry will need a slight adjustment here. The right way around this is to decompose in to real and imaginary parts, then show that the $\Bbb R_{alg}$ points are dense in the $\Bbb R$ points of $V$ regarded as a semialgebraic set in $\Bbb R^{2n}$. This is a place where you may use an extension of real closed fields: $\Bbb R_{alg} \subset \Bbb R$. Unfortunately I don't have time to write a full solution right now. – KReiser Nov 28 '18 at 06:10
  • @KReiser It seems that then one will have to use that $X(\mathbb{R})$ is dense in $X(\mathbb{C})$. Or is that obvious? – random123 Nov 28 '18 at 07:27
  • One cannot use this as it is not true: consider $V(x^2+1)$ or $V(x^2+y^2)$, for instance. One potential solution would be to use a cell decomposition and reduce proving the claim to showing that $\Bbb R_{alg}$ is dense in $\Bbb R$ as a subset of $(0,1)^d$, but this seems perhaps a little inelegant. – KReiser Nov 28 '18 at 07:34
  • @KReiser Right! The step I am not following is how to conclude from here that $X(\mathbb{R})$ is dense in $X(\mathbb{C})$. I suppose that is the final aim. – random123 Nov 28 '18 at 09:02
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    @random123 what? I don't see how that's relevant here. What I'm saying is that one writes the coordinates $z_j$ of $\Bbb C^n$ as $z_j=x_j+iy_j$ and considers $V$ as a semialgebraic set in $\Bbb R^{2n}$ with coordinates $x_j$ and $y_j$. Now the $\overline{\Bbb Q}$ points of the original $V$ are points in the semialgebraic set in $\Bbb R^{2n}$ which have all their coordinates in $\Bbb R_{alg}$ while the $\Bbb C$ points of the original $V$ are all points in the semialgebraic set. – KReiser Nov 28 '18 at 09:18
  • @KReiser Thanks! That clear things up. – random123 Nov 28 '18 at 09:18

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A proof originally developed in the comments:

Let $\Bbb R_{alg}$ denote the set of real algebraic numbers. Let $z_j$ be the coordinates on $\Bbb C^n$. Write $z_j=x_j+iy_j$ which identifies $\Bbb C^n$ with $\Bbb R^{2n}$, and identifies the $\overline{\Bbb Q}$ points of $V$ with the $\Bbb R_{alg}$ points of the semialgebraic set given by all points of $V$ under the identification of $\Bbb C^n$ with $\Bbb R^{2n}$.

From here, if one can show that the $\Bbb R_{alg}$ points of a semialgebraic set $S\subset \Bbb R_{alg}^n$ are dense in the $\Bbb R$ points, then everything is fine. By a cell decomposition, we may obtain a finite list of semialgebraic homeomorphisms of $(0,1)^d$ with semialgebraic subsets of $S$ with each map defined over $\Bbb R_{alg}$. Then the problem reduces to showing that the $\Bbb R_{alg}$ points of $(0,1)^d$ are dense in the $\Bbb R$ points of $(0,1)^d$, which is clear as $\Bbb Q\subset \Bbb R_{alg}$ and the $\Bbb Q$ points are clearly dense in $(0,1)^d$.

This feels a little inelegant, but it works.

Alternatively, picking up from the point where we need to prove that the $\Bbb R_{alg}$ points are dense in the $\Bbb R$ points: Suppose there was a point in $S(\Bbb R)$ which was not in the closure of $S(\Bbb R_{alg})$. Then this point cannot be in any of the semialgebraic components of $S(\Bbb R_{alg})$, so by careful selecting of components of $S$, one would obtain a set which is semialgebraically connected over $\Bbb R_{alg}$ but semialgebraically disconnected over $\Bbb R$, which violates Proposition 5.3.6, a direct corollary of 5.3.5. (I should also mention the authors of the book prove this claim essentially by cell decomposition, so it's the same argument as the first part of this post, just with a different ending.)

KReiser
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    "and and the ℚ-points are clearly dense in (0,1)^d." Why? – Asal Beag Dubh Nov 28 '18 at 15:40
  • @AsalBeagDubh because $\Bbb Q$ is dense in $\Bbb R$? This is a standard fact from real analysis. – KReiser Nov 28 '18 at 18:58
  • An affine elliptic curve can have no rational points (but a continuum of real points) – reuns Nov 29 '18 at 04:51
  • @reuns and...? The rational points in $(0,1)^d$ are not necessarily sent to rational points by the homeomorphisms from the cell decomposition: as stated the maps are defined over $\Bbb R_{alg}$ (and even if one defines them over $\Bbb Q$, they aren't even necessarily polynomial mappings, so the result you quote is not necessarily a contradiction). But it is true that the image of a dense set under a homeomorphism is again a dense set. – KReiser Nov 29 '18 at 05:17
  • What is it ? To me the density of algebraic points is obtained from a $\overline{\mathbb{Q}}$ birational map obtained from $\overline{\mathbb{Q}}(V) \cong \overline{\mathbb{Q}}(u_1,\ldots,u_m)[z_1,\ldots,z_l]/J= \overline{\mathbb{Q}}(S)$ and the points in $S(\overline{\mathbb{Q}})$ are easy to find and clearly dense in $S(\mathbb{C})$. – reuns Nov 29 '18 at 05:22
  • Then you should write this as a separate answer. The OP asked for a perspective from semialgebraic geometry, which is what I have provided here. (I really do not understand the point of your comments - what are you getting at here?) – KReiser Nov 29 '18 at 05:29
  • Explain us how you know the $\mathbb{R}_{alg}$-points are dense in $(0,1)^d$ which is the core of OP's question. You are not answering to the OP but to every MSE user – reuns Nov 29 '18 at 05:37
  • $\Bbb Q\subset \Bbb R$ is dense. $\Bbb Q \cap (0,1) \subset (0,1)$ is dense. $(\Bbb Q\cap (0,1))^d \subset (0,1)^d$ is dense by the definition of the product topology. Any set containing a dense set is also dense. These are elementary proofs that anyone who has completed a course in real analysis and/or topology should be familiar with - in particular, anyone browsing the algebraic-geometry tag. Alternatively, you could read the second proof! – KReiser Nov 29 '18 at 05:47
  • ??? We are talking of solutions of algebraic equations. So how do you reduce it to something about $(0,1)^d$ or $\mathbb{R}^d$. For example with an elliptic curve how would you do it. – reuns Nov 29 '18 at 07:04
  • Let us continue this discussion in chat. – KReiser Nov 29 '18 at 08:23
  • @KReiser: sorry, I misread what you were saying. – Asal Beag Dubh Nov 29 '18 at 09:40
  • @KReiser Does this result also hold for an algebraic variety defined over an algebraic number field? – Rajath Radhakrishnan Dec 07 '20 at 16:19
  • @RajathRadhakrishnan I don't see why not - the portion of the argument which depends on the field of definition is the one where we turn our variety in to a semialgebraic set, and the method of separating real and imaginary parts will work just as well over a number field. Do you have a specific concern? – KReiser Dec 07 '20 at 21:15
  • @KReiser I didn't have a specific concern. Just wanted to double check with you. On a related note, (and as someone new to algebraic geometry) I have a reference request for a proof of the statement: "Algebraic points of a complex affine algebraic variety defined over $\overline{\mathbb{Q}}$ are dense in the Euclidean and hence the Zariski topology". Can you suggest a reference for a proof for this statement if this is standard material? Thank you. – Rajath Radhakrishnan Dec 07 '20 at 23:02
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    @RajathRadhakrishnan Same proof as above applies, perhaps with a reduction to the affine case. Alternatively, here and here give solutions of a different flavor. I would not count this as standard material (it's not in Hartshorne, Vakil, Harris & Griffiths, etc), but it's not difficult. If you're asking in order to solve a specific problem, it might be worth writing a post about it so you can explain what's going in more detail than the 600 characters for comments. – KReiser Dec 07 '20 at 23:22
  • @KReiser Thank you for the comment and links. That really helps! – Rajath Radhakrishnan Dec 07 '20 at 23:38