MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A field $K$ is called ample if every smooth curve over $K$ that has a $K$-point has infinitely many $K$-points. Examples include fraction fields of henselian rings. (This is far from trivial, but was proven in the paper "henselian implies large". "Large" is a synonym for "ample".) For example the field of $p$-adics, for any prime $p$, is ample.

I have known for a long time that $\mathbb{Q}^{ab}$ is conjectured to be ample, but I don't know how much evidence there is for this conjecture. Can you direct me to some references supporting this conjecture, or suggest heuristic arguments that would explain why this conjecture is reasonable?

share|cite|improve this question
The solvable variant seems more like a question than a conjecture, as it is just an optimistic guess with no compelling evidence beyond that it sure would be useful if it were true (validity in low-genus cases is not particularly convincing, even if rather nontrivial for genus 1). Where in the literature has someone been bold enough to pose the "abelian" version as a question, let alone as a conjecture? – user29720 Jun 23 '13 at 20:09
kreck, I think you are confusing between "ample" and "PAC" (Pseudo Algebraically Closed). The maximal solvable extension of $\mathbb{Q}$ is thought to perhaps be PAC. But ampleness is a much weaker condition. – Makhalan Duff Jun 23 '13 at 20:14
Aha, right you are. Well, the "ampleness" condition is new to me, and I am amazed to hear that it is conjectured to be true. – user29720 Jun 24 '13 at 0:19

It seem like most of the evidence is rather indirect. The ampleness of $\mathbb{Q}^{ab}$ would settle other open conjectures/questions:

  • By a result of Pop, it would imply the following conjecture of Shafarevich:

Conjecture: The Galois group of $\mathbb{Q}^{ab}$ is a free profinite group.

  • By a result of Fehm and Petersen, the ampleness conjecture would imply an affirmative answer to the following question of Frey and Jarden:

Question: Is the rank of $A(\mathbb{Q}^{ab})$ infinite for every non-trivial abelian variety $A/\mathbb{Q}$?

There are partial results and evidence for both of these conjectures/question that are independent of the ampleness conjecture. For example the answer to the above question is known to be "yes" for elliptic curves, and for some other families of abelian varieties.

You can read more about this in this survey of Fehm and Bary-Soroker on ample fields and in the references therein.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.