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The question was a bit long for the title, so let me explain what I mean here. Let $k$ be some field (ideally, a number field). Let $X$ be a curve that is defined over $k$, such that it has a $k$-point. Let $K$ be an algebraic extension of $k$ which is not finitely generated. Does this imply that $X$ has infinitely many $K$-points?

Motivation

The canonical example I have of a field which is infinitely generated over its prime field, together with a curve, such that the curve doesn't have infinitely many rational points is: $x^2+y^2=-1$ where the field is $\mathbb{R}$. However in this case, $\mathbb{R}$ is not an algebraic extension of a field $k$ for which this curve does have a rational point. (Let alone is a not-finitely-generated extension of such a $k$.)

edited Oct 31 2011 at 2:10

asked Oct 31 2011 at 0:15

James D. Taylor
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is infinitely generated = not finitely generated? – Wanderer Oct 31 2011 at 0:47

Yes, that's what I meant. – James D. Taylor Oct 31 2011 at 1:01

I've made the body of the question reflect this. – James D. Taylor Oct 31 2011 at 2:10

Have you looked at the universal divisor in the Hirzebruch surface F1, the blowing up of P^2 at 1 point, which has divisor class d times the pullback of the hyperplane, H, minus 1 times the exceptional divisor, E? Because this the total space of a projective bundle over F1, the Picard group is easy, and small: freely generated by H, E and a relative hyperplane class L of the projective bundle. Every Cartier divisor of relative degree 1 over the base is E + a(H-dE) + bL. It should be relatively easy to compute H^0 of each of these. – Jason Starr Oct 31 2011 at 2:43

Of course when d equals 1 or 2, there are Cartier divisors with global sections, e.g., for d=2 the class E + 1(H-2E). But for d > 2, it appears at first blush that there may be no global sections for a > 0. – Jason Starr Oct 31 2011 at 2:45

Will Sawin · Accepted Answer · 2011-10-31 02:13:24Z UTC

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Let $X$ be the plane curve defined by the equation $x^2+y^2=0$. Let $k=\mathbb Q$ and $K=\mathbb R$.

The curve has exactly one point in both.

answered Oct 31 2011 at 2:13

Will Sawin
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Well, that was way easier than I expected... – James D. Taylor Oct 31 2011 at 2:15

I don't understand; $K$ isn't an algebraic extension of $k$. I think you want to take $K = \mathbb{R} \cap \bar{\mathbb{Q}}$. – Qiaochu Yuan Oct 31 2011 at 2:22

You're right. But that's just as easy. – James D. Taylor Oct 31 2011 at 2:31

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But $x^2+y^2=0$ is reducible over $\bar k$, so this is still not that great an example... – Noam D. Elkies Oct 31 2011 at 3:21

$x^4+x^2+y^2=0$, or something along those lines, should fix that. – Will Sawin Oct 31 2011 at 5:51

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Alex Bartel · Answer 2 · 2011-10-31 02:35:27Z UTC

Mazur in his article "Rational points of abelian varieties with values in towers of number fields", Invent. Math. 18 has produced lots of curves over number fields that have a finite number of points over $\mathbb{Z}_p$ extensions.

Also, there is a theorem due to Kato, Ribet, and Rohrlich that if $E/\mathbb{Q}$ is an elliptic curve and $K$ is the maximal abelian extension of $\mathbb{Q}$ unramified outside a fixed finite set of primes, then $E(K)$ is finitely generated. I imagine that if $E$ was instead hyperelliptic, then the conclusion would be that $E(K)$ is finite.

A trickier question is the following: a field $K$ is called ample if every smooth curve over $K$ that has a $K$-rational point has infinitely many of them (this is a stronger condition than what you want, since you only want this to be true for curves defined over some subfield of $K$). It seems that it is quite difficult to show that a concrete infinitely generated algebraic extension of $\mathbb{Q}$ is non-ample. For example it isn't even known whether $\mathbb{Q}^{ab}$ is ample. You might be interested in this survey on open problems concerning ample fields.

Does a curve have infinitely many $K$-rational points under these hypotheses?

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Motivation

2 Answers

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