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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 $k$.)

Community
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James D. Taylor
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    is infinitely generated = not finitely generated? – Wanderer Oct 31 '11 at 00:47
  • Yes, that's what I meant. – James D. Taylor Oct 31 '11 at 01:01
  • I've made the body of the question reflect this. – James D. Taylor Oct 31 '11 at 02:10
  • 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 '11 at 02:43
  • 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 '11 at 02:45

2 Answers2

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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.

Alex B.
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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.

Will Sawin
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