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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?

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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.

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  • $\begingroup$ 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 [email protected] $\endgroup$
    – Yuan Yang
    Commented Nov 30, 2021 at 2:02
  • $\begingroup$ @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). $\endgroup$
    – Jonathan Love
    Commented Dec 1, 2021 at 1:49
  • $\begingroup$ @JonathanLove Yes indeed. I‘ve made quite a few mistakes. It's more complicated than I thought. $\endgroup$
    – Yuan Yang
    Commented Dec 1, 2021 at 7:52
  • $\begingroup$ @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. $\endgroup$
    – Yuan Yang
    Commented Dec 1, 2021 at 8:02
  • $\begingroup$ @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. $\endgroup$
    – Yuan Yang
    Commented Dec 1, 2021 at 8:17

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