Characterization of an Abelian surface

I have a smooth projective surface $$X$$, and two flat family of elliptic curves on it: $$E_{1,t}$$ and $$E_{2,t}$$, (I don't know what either $$t$$ runs through!) such that

(1), for any i={1，2}, the closed points of $$X$$ are the disjoint union of closed points of all $$E_{i,t}$$.

(2), the intersection number of $$E_{1,t}$$ and $$E_{2,t}$$ is always 1.

Can we conclude that $$X$$ is an abelian surface?

I think that the answer is yes, at least if you are working over an algebraically closed field of char $$0$$. Let me try to give an argument, I hope that there are no mistakes.

You have two flat families $$E_1 \to C_1$$ and $$E_2 \to C_2$$ of elliptic curves over some unknown bases $$C_1, C_2$$.

The first assumption says that the map $$E_1 \to X$$ induces a bijection on closed points. If the field is of characteristic 0, one can use Zariski's main theorem to conclude that $$E_1 \to X$$ is an isomorphism of varieties (see e.g. Bijection implies isomorphism for algebraic varieties). The same holds for $$E_2 \to X$$. Using these identifications $$X \cong E_1$$ and $$X \cong E_2$$, we get two flat projections $$X \to C_1$$ and $$X \to C_2$$, where the fibers are elliptic curves. In particular $$C_1$$ and $$C_2$$ must be smooth (by flat descent of smoothness).

I am interpreting the assumption that the intersection numbers are always 1 to mean that the fibers of the induced morphism $$X \to C_1 \times C_2$$ are singleton points, i.e. any two fibers of the families $$E_1$$ and $$E_2$$ respectively intersect in a single point (is this the precise condition you have in mind?). Since $$C_1 \times C_2$$ is smooth, this implies again that $$X \to C_1 \times C_2$$ is an isomorphism. But now by choosing a closed point $$x_1 \in C_1$$, we see that $$C_2 = x_1 \times C_2$$ is isomorphic to the fiber of $$E_1 \to C_1$$ at $$x_1$$. This means that $$C_1$$ is an elliptic curve. The same holds for $$C_2$$. Therefore $$X \cong C_1 \times C_2$$ is a product of elliptic curves.

• I think we just solved the 'four-distance' problem using this result. Although I don't know whether or not it is solved yet. Can we talk by email? Please contact me at ucahans@ucl.ac.uk Nov 30 '21 at 2:02
• @Yuan It sounds like your application might involve considering varieties defined over $\mathbb{Q}$? The solution afh gives here depends quite heavily on the algebraically closed assumption - there are a lot of subtleties that arise if you're trying to work over number fields (even the problem statement, as it's currently given, isn't well-posed). Dec 1 '21 at 1:49
• @JonathanLove Yes indeed. I‘ve made quite a few mistakes. It's more complicated than I thought. Dec 1 '21 at 7:52
• @JonathanLove My current status is: I have an algebraic surface in $P^1*P^1*P^1*P^1$, and I have families of elliptic curves(I actually have $P^1$ families of them), but for each family, these curves are not disjoint-they actually all intersect in one point, at one of the infinity point. But indeed, any two family intersect at only one point. Dec 1 '21 at 8:02
• @JonathanLove For non-closeness of Q: if the variety base change to $\bar{Q}$ is an abelian variety, then either: it doesn’t have a rational point; or it’s an abelian variety over Q, too. Dec 1 '21 at 8:17