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

Question

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

Answer 1

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.