Complex variety with Zariski dense set of algebraic points

Let $V$ be an irreducible algebraic variety in $\mathbb{C}^n$ containing a Zariski dense set of points such that every coordinate is algebraic. Then is $V$ a product of one dimensional components?

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No. If you take any (say projective, irreducible ) variety $X$ defined over $\overline{\mathbb{Q}}$, then its $\overline{\mathbb{Q}}$-points — i.e., points in which every coordinate in a suitable projective embedding is $\overline{\mathbb{Q}}$-rational — are Zariski dense in its $\mathbb{C}$-points. You can see this e.g. by noting that the dimension of the closure in each case is the transcendence degree of the function field, and the transcendence degree of a field extension is unchanged by base extension. There are many other ways as well…

So it comes down to showing that there are varieties over $\overline{\mathbb{Q}}$ which are not products of one-dimensional varieties. The easiest such example seems to be the projective plane $\mathbb{P}^2$: the fact that $H^2(\mathbb{P}^2(\mathbb{C}),\mathbb{C}) = 1$ means, by the Kunneth formula, that it cannot be a product of curves.

No, but the argument below doesn’t work as Pete L. Clark points out.

Let $X$ be a curve of genus at least two. Let $J$ be its Jacobian and let $X\to J$ be the Jacobian embedding. Then $X$ is Zariski dense in $J$, but $J$ is not a product of one-dimensional curves.

In fact, if $J$ were a product of curves then these would all have to be elliptic curves (because of the group structure on $J$). It is well-known that there are Jacobians which aren’t isomorphic (even isogenous) to a product of elliptic curves.

I just saw that you want the coordinates to be algebraic:

You can do the above starting from a curve $X$ defined over $\overline{\mathbf{Q}}\subset \mathbf{C}$. The points $X(\overline{\mathbf{Q}})$ are dense in $J_{\mathbf{C}}$, I believe.