This is a cross-post! For the original post on SE (9 upvotes, no answer) see:


Let $X$ be a complex algebraic set, i.e. the (not necessarily irreducible) vanishing set of some polynomials in $\mathbb{C}[X_1,\ldots,X_n]$. If $X$ contains a Zariski dense subset of points with coordinates in the algebraric numbers $\bar{\mathbb{Q}}$, is it true that $X$ is defined over $\bar{\mathbb{Q}}$, i.e. that $X$ can be defined as the vanishing set of some polynomials in $\bar{\mathbb{Q}}[X_1,\ldots,X_n]$?

It seems to be the case that the converse of this stament holds, as discussed here. I read that the implication in my question follows from Lagrange interpolation, but it is not clear to me how Lagrange interpolation works in the present multivariable case and how it can be used to answer my question.

For context: I try to understand a proof of Mann's theorem in this survey on applications of o-minimality (from mathematical logic) to diophantine geometry. We want to prove the following:

Theorem. Let $Y\subseteq\mathbb{C}^n$ be an algebraic set and consider $\mathbb{G}=(\mathbb{C}^\times)^n$, the $n$th power of the multiplicative group of units in $\mathbb{C}$. Then the set $Y\cap\mathbb{G}^\mathrm{tor}$ of points on $Y$ such that each coordinate is a root of unity is a finite union of cosets of subgroups of $\mathbb{G}$.

At some point on page 15, we want to use bounds from Galois theory and need to reduce to the case that $Y$ is defined over a number field. In the paragraph in paranthesis on p. 15, the following argument is sketched: From basic properties of the Zariski topology using that $\mathbb{G}$ is open, we can see that $Y\cap\mathbb{G}^\mathrm{tor}=\overline{Y\cap\mathbb{G}^\mathrm{tor}}\cap\mathbb{G}^\mathrm{tor}$, and may thus assume that $Y=\overline{Y\cap\mathbb{G}^\mathrm{tor}}$. Then $Y$ contains a dense set of points with algebraic coordinates, since the elements of $\mathbb{G}^\mathrm{tor}$ have algebraic coordinates. Now the author Scanlon claims that it follows from Lagrange interpolation that $Y$ is defined over $\bar{\mathbb{Q}}$. The fact that $Y$ is defined over a number field then just follows from Noetherianess of $\bar{\mathbb{Q}}[X_1,\ldots,X_n]$.

A similar reduction also seems to work in the case of torsion points on abelian varieties defined over $\mathbb{C}$. I would also be grateful for details or references regarding this case.

Thanks for your help!

  • $\begingroup$ would you be interested in a model-theoretic argument? although it's probably an overpowered approach, if I'm not mistaken this can be proved quickly using saturation of $(\mathbb{C},\cdot,+)$ $\endgroup$ Aug 12 at 19:41
  • 3
    $\begingroup$ Yes, see Proposition 16.8 in jmilne.org/math/CourseNotes/AG16.pdf $\endgroup$
    – anon
    Aug 12 at 19:42
  • 1
    $\begingroup$ @AtticusStonestrom I would definitely be interested in a model-theoretic argument! Could you elaborate on your idea? $\endgroup$
    – Luvath
    Aug 12 at 19:52
  • $\begingroup$ @Luvath okay, I've written it up – hope it helps! :) $\endgroup$ Aug 12 at 21:28
  • $\begingroup$ I imagine there is probably a much more elementary argument if Tom is mentioning Lagrange interpolation, but hopefully this is an interesting approach nonetheless $\endgroup$ Aug 12 at 21:29

1 Answer 1


More is true. Let $K/F$ be a field extension. Let $X$ be the vanishing set of some polynomials in $K[X_1,\dots, X_n]$. If $X$ contains a Zariski dense set of points with coordinates in $F$, then $X$ is the vanishing set of some polynomials in $F$.

Proof: Choose a basis $\alpha_i$ for $K/F$. Every polynomial in $K[X_1,\dots, X_n]$ vanishing on $X$ can be written as a (finite) combination $\sum_i \alpha_i f_i$ with $f_i \in F[X_1,\dots, X_n]$, by decomposing every coefficient in this basis.

For $x_1,\dots, x_n\in F$, if $ \sum_i \alpha_i f_i(x_1,\dots, x_n)=0$ then $f_i (x_1,\dots, x_n)=0$ for each $i$, by the definition of basis. So the $f_i$ are polynomials vanishing on a Zariski dense set of points in $X$ and thus vanish on $X$.

Thus the vanishing locus of the $f_i$, for all the polynomials $\sum_i \alpha_i f_i$ in a set of polynomials defining $X$, is $X$, so $X$ is defined over $F$.

  • $\begingroup$ Thank you very much for this elementary argument! $\endgroup$
    – Luvath
    Aug 13 at 9:35
  • $\begingroup$ this is a wonderful argument, much better than mine! (+1) $\endgroup$ Aug 15 at 18:17

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