# Are the algebraic-valued points of a variety dense in complex-valued points?

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 $$\{ (x_1, \ldots, x_n) \in \mathbb C^n~|~f_i(x_1, \ldots, x_n) = 0~\text{for all}~i \}.$$

• 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. Sep 25 '13 at 13:13
• @GerryMyerson: Done, thank you. Sep 25 '13 at 14:25
• How do you prove the result in the smooth case? Sep 25 '13 at 17:02
• @finite: want a direct proof without reducing to the smooth case ? Sep 25 '13 at 17:04
• @Cantlog: although you are not asking me, I certainly would love to see a direct proof ! Sep 25 '13 at 17:56

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.
• Dear @Cantlog, the smooth locus $X^{sm} \subset X$ is Zariski-dense, but why it is also dense in the complex topology? The latter topology is much finer as the Zariski-topology. Sep 26 '13 at 8:41