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Let $X \subset \mathbb{C}^n$ be an affine variety which is defined over $\mathbb{Q}$ (i.e. the zero set of some finite collection of polynomials with coefficients in $\mathbb{Q}$). Define $X'$ to be the subset of points of $X$ whose coordinates lie in the algebraic closure $\overline{\mathbb{Q}}$. Question : why is $X'$ Zariski dense in $X$? In there words, if $f(x_1,\ldots,x_n)$ is a polynomial with $\mathbb{C}$-coefficients which vanishes on $X'$, why does $f$ have to vanish on all of $X$?

I came across this question while reading Rapinchuk's article "On strong approximation for algebraic groups", where he uses it while proving the easy direction of the strong approximation theorem. I've found some proofs of it elsewhere on the internet, but they were written in a very abstract language, so I had trouble following them (I'm a topologist and not an algebraic geometer).

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  • $\begingroup$ Let $A$ be a reduced finite generated algebra over an algebraically closed field $k$, $K/k$ an extension field. Let $X={\rm{MaxSpec}}(A)$. The map $A \rightarrow \prod_xk$ defined by $a\mapsto (a(x))_x$ is injective, so it remains injective after scalar extension to $K$. For arbitrary $k$-vector spaces, $V \otimes_k(\prod_i W_i) \rightarrow\prod_i (V\otimes_k W_i)$ is injective, even with infinitely many $W_i$'s (use direct limits in $V$ to reduce to $\dim V < \infty$, and then to $\dim_k V = 1$). Setting $V=K$ and $\{W_i\}=\{k\}_{x\in X}$, $K\otimes_kA\rightarrow \prod_x K$ is injective. QED $\endgroup$
    – user76758
    Commented Jan 31, 2014 at 6:21
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    $\begingroup$ Choose a basis $(e_i)$ for $\mathbb{C}$ over $\bar{\mathbb{Q}}$, and write $f=\sum e_i f_i$ where the $f_i$ are polynomials with coefficients in $\bar{\mathbb{Q}}$. Because $f$ vanishes on the $\bar{\mathbb{Q}}$-points of $X$, so does each $f_i$, and so lies in the (radical of) the ideal defining $X$ over $\bar{\mathbb{Q}}$. Hence vanishes on the $\mathbb{C}$ points. $\endgroup$
    – abz
    Commented Jan 31, 2014 at 6:53
  • $\begingroup$ @abz : Thanks! That's a really elegant way of seeing this. $\endgroup$
    – Brian
    Commented Jan 31, 2014 at 16:03

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