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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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up vote 4 down vote accepted

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.

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In the affine case case if a polynomial vanishes on $\overline{ \mathbb{Q} }$-points then it must be identically zero by the Nullstellensatz, so it vanishes on $\mathbb{C}$-points. – Qiaochu Yuan Aug 3 '12 at 17:13
@Qiaochu:So how about if you want the polynomial to be non-zero? Is this more interesting? – Peter Aug 3 '12 at 17:24
@Peter: I am not sure I understand the question. – Qiaochu Yuan Aug 3 '12 at 17:29
@Qiaochu: I meant that if I edited my question and added a hypothesis that the variety was defined by a non zero polynomial then might it be true, or at least more interesting? – Peter Aug 3 '12 at 17:43
@Peter: I think you are misinterpreting my claim. "Polynomial" here means "polynomial function on a given affine variety over $\mathbb{Q}$" and "identically zero" means "identically zero on the affine variety." – Qiaochu Yuan Aug 3 '12 at 18:03

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.

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"but $J$ is not a product of one-dimensional curves". As you say, it might be, so better phrasing would be something like "not necessarily a product of one-dimensional curves". (Or, more ambitiously, one might try to make the point that this is generically not the case...) – Pete L. Clark Aug 3 '12 at 16:29
Further, $X(\overline{\mathbb{Q}})$ is certainly not Zariski dense in $J_{\mathbb{C}}$: rather the Zariski closure is $X(\mathbb{C})$. – Pete L. Clark Aug 3 '12 at 16:30
@firstcomment. You're right. The phrasing is very bad. @ secondcomment. My example is wrong for very silly reasons. The embedding of $X$ in $J$ is a closed immersion. I don't know why I thought $X$ would be dense in $J$...I was probably confused by the following statement: The group generated by $X$ in $J$ is $J$. (Is this statement true???) That might have been the reason for my confusion. – Harry Aug 3 '12 at 16:55
Yes, the statement in your last comment is true. In fact it was my guess that that was what you had in mind, and I almost mentioned that in my comment. – Pete L. Clark Aug 3 '12 at 20:21

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