Here’s a funny question. Let f in C[x] be a squarefree polynomial of degree at least 6. Let S be the set of complex numbers t such that the Jacobian of the hyperelliptic curve

is not simple. Is S always finite? Even more, is there a bound on |S| which doesn’t depend on f, or depends only on the degree of f?

This question comes from the introduction to “Non-simple abelian varieties in a family: geometric and analytic approaches” , a new paper by me, Christian Elsholtz, Chris Hall, and Emmanuel Kowalski. In its original form this was a four-author, six-page paper — fortunately we’ve now added enough material to make the ratio a bit more respectable!

The paper isn’t about complex algebraic geometry at all — it explains how to get bounds on S when f has *rational *coefficients and t ranges over rational numbers, which is quite a different story. The point of the paper is partly to prove some theorems and partly to make a metamathematical point — that problems of this kind can be approached via either arithmetic geometry or analytic number theory, and that the two approaches have complementary strengths and weaknesses. Bounds from arithmetic geometry are stronger but less uniform; bounds from analytic number theory are weaker but have better uniformity.

Here’s my favorite example of this phenomenon. Let X be a smooth plane curve over Q of degree d at least 4. Then by Faltings’ Theorem we know that X has only finitely many rational points.

On the other hand, a beautiful theorem of Heath-Brown tells us that the number of rational points on X with coordinates of height at most B is at most C B^(2/d), for some constant C depending only on d. At first, this seems to give much less than Faltings. After all, as B gets larger and larger, the upper bound given by Heath-Brown gets arbitrarily large — whereas we know by Faltings that there are only finitely many points on the whole curve, no matter how large we allow the coordinates to be.

But note that the constant in Heath-Brown’s result *doesn’t depend on the curve X.* It is what’s called a *uniform* bound. Faltings’ theorem, by contrast, gives an upper bound on the number of points which depends very badly on the choice of X. Depending on what you’re trying to accomplish, you might be willing to sacrifice uniformity to get finiteness — or the reverse. But it’s best to have both options at hand.

Is it possible to have uniformity and finiteness simultaneously? Conjecturally, yes. Caporaso, Harris, and Mazur showed that, conditional on Lang’s conjecture, there is a constant B(g) such that every genus-g curve X/Q has at most B(g) rational points. The Caporaso-Harris-Mazur paper came out when I was in graduate school, and the idea of such a uniform bound was considered so wacky that CHM was thought of as evidence *against* Lang’s conjecture. Joe Harris used to wander around the department, buttonholing graduate students and encouraging us to cook up examples of genus-g curves with arbitrarily many points, thus disproving Lang. We all tried, and we all failed — as did many more experienced people. And nowadays, the idea that there might be a uniform bound for the number of rational points on a genus-g curve is considered fairly reputable, even among people who have their doubts about Lang’s conjecture. As far as I know, the world record for the number of rational points on a genus-2 curve is 588, due to Kulesz. Can you beat it?