Is every complex affine/projective variety isomorphic/birational to one defined by an ideal $I\subset \mathbb{Z}[x_1,\dots,x_n]$?

Question

For a while now I've been thinking about this question, but I have no idea how to go about it:

Is every complex affine/projective variety isomorphic/birational to one defined by an ideal $I\subset \mathbb{Z}[x_1,\dots,x_n]$?

So I'm interested in all $4$ combinations of affine/projective and isomorphic/birational. So to make sure the question is clear: given an ideal $J\subset \mathbb{C}[x_1,\dots,x_n]$, is there an ideal $I\subset \mathbb{Z}[x_1,\dots,x_m]$ (we don't require $n=m$) such that $Z(J)\cong Z(I)$ as varieties?

My only idea here was that if this is true then there are only countably many isomorphism classes of varieties. I'm not sure if this can lead to a contradiction, since e.g. the Euler characteristic is only able to distinguish countably many, and I don't know of an invariant that can do better (that's not to say there is none, just that I don't know much).

edit: the same question but with $\mathbb{Z}[x_1,\dots,x_m]$ replaced by $\mathbb{Z}[i][x_1,\dots,x_m]$ would also be interesting.

Answer 1

The answer is at least a partial no. Consider complex elliptic curves- no two complex elliptic curves are isomorphic without having the same $j$-invariant, which is a complex number and may take uncountably many values. Therefore there are at least uncountably many isomorphism classes of complex projective varieties. My guess is that there might be an easy way to extend this or something like it to the affine or birational questions you ask, but I do not know it off the top of my head.

I am curious about your question once one removes the possibility for such cardinality trickery- perhaps there's something to be said for varieties defined over $\overline{\mathbb{Q}}$. Then the countable/uncountable distinction doesn't appear.