Intereting Posts

Brownian bridge expression for a Brownian motion
Differential equation for Harmonic Motion
In Triangle, $\sin\frac A2\!+\!\sin\frac B2\!+\!\sin\frac C2\!-\!1\!=\!4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$
An exercise about finite intersection property in $T_1$ space
Proving $\sum_{k=0}^n\binom{2n}{2k} = 2^{2n-1}$
How to show that the Laurent series of the Riemann Zeta function has $\gamma$ as its constant term?
The construction of the localization of a category
$\mathbb S_n$ as semidirect product
If $$ is finite, then $ = $ iff $G = HK$ (Hungerford Proposition 4.8, Proof)
How many non-rational complex numbers $x$ have the property that $x^n$ and $(x+1)^n$ are rational?
All tree orders are lattice orders?
The sum of odd powered complex numbers equals zero implies they cancel each other in pairs
Frequency of Math Symbols
What are the generators of $(\mathbb{R},+)$?
Simple Math Problem

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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- Divisor of meromorphic section of point bundle over a Riemann surface
- Determinant of a tensor product of two vector bundles
- Extended ideals and algebraic sets
- Example I.4.9.1 in Hartshorne (blowing-up)

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.

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