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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 $k$.)

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    $\begingroup$ is infinitely generated = not finitely generated? $\endgroup$
    – Wanderer
    Commented Oct 31, 2011 at 0:47
  • $\begingroup$ Yes, that's what I meant. $\endgroup$
    – James D. Taylor
    Commented Oct 31, 2011 at 1:01
  • $\begingroup$ I've made the body of the question reflect this. $\endgroup$
    – James D. Taylor
    Commented Oct 31, 2011 at 2:10
  • $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Commented Oct 31, 2011 at 2:43
  • $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Commented Oct 31, 2011 at 2:45

2 Answers 2

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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.

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  • $\begingroup$ Well, that was way easier than I expected... $\endgroup$
    – James D. Taylor
    Commented Oct 31, 2011 at 2:15
  • $\begingroup$ I don't understand; $K$ isn't an algebraic extension of $k$. I think you want to take $K = \mathbb{R} \cap \bar{\mathbb{Q}}$. $\endgroup$
    – Qiaochu Yuan
    Commented Oct 31, 2011 at 2:22
  • $\begingroup$ You're right. But that's just as easy. $\endgroup$
    – James D. Taylor
    Commented Oct 31, 2011 at 2:31
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    $\begingroup$ But $x^2+y^2=0$ is reducible over $\bar k$, so this is still not that great an example... $\endgroup$
    – Noam D. Elkies
    Commented Oct 31, 2011 at 3:21
  • $\begingroup$ $x^4+x^2+y^2=0$, or something along those lines, should fix that. $\endgroup$
    – Will Sawin
    Commented Oct 31, 2011 at 5:51
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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.

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  • $\begingroup$ How is this any different from a "large field" in the sense of Florian Pop? $\endgroup$
    – Jason Starr
    Commented Oct 31, 2011 at 23:29
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    $\begingroup$ Dear Jason, it isn't. There are several names in common use. $\endgroup$
    – Alex B.
    Commented Nov 5, 2011 at 14:00

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