# Is the compositum of all quadratic extensions of the rationals an ample field?

Let $K$ be the compositum of all quadratic extensions of $\mathbb{Q}$, that is $$K = \mathbb{Q}(\sqrt{d} \ : \ d \in \mathbb{Q}).$$

Is there a (geometrically irreducible) smooth variety $V/\mathbb{Q}$ such that $V(K)$ is nonempty and finite?

This is equivalent to asking whether $K$ is an ample/large/anti-Mordellic/fertile/Pop field.

I do not know the answer even in the (very interesting) case when $V$ is an abelian variety. If $V$ is an elliptic curve then $V(K)$ is infinite by a 1974 result of Jarden and Frey.

For the "much larger" field $K \leq \mathbb{Q}^{\mathrm{ab}}$ (the maximal abelian extension of $\mathbb{Q}$) the analogous question is an open problem of Pop, so I expect the answer to be negative (even for abelian varieties).

There exist finiteness theorems in the spirit of Faltings that give a negative answer if instead of $K$ we take some $L \leq \mathbb{Q}^{\mathrm{ab}}$ which is unramified outside some finite set of rational primes. Therefore, I am also interested in an answer for some other (abelian) extensions of $\mathbb{Q}$ in place of $K$ in which infinitely many rational primes ramify.

The case of curves is also interesting. For a specific curve see Fermat's last theorem over larger fields

What if we take $V$ to be a Fermat curve? Is $V(K)$ nonempty? finite?

Conjectural answers are more than welcome (see also Can an abelian variety/Q have no non-trivial points over Q_sol?).

It is possible that recent results on modularity of elliptic curves over totally real fields can help treat the case $M = \mathbb{Q}(\sqrt{d} \ : \ d \in \mathbb{N})$ in place of $K$.

• The question itself may be difficult to answer (as with many questions on this topic), but this field has the Northoctt finitness property (for every $T$ there are only finitely many elements $\alpha \in K$ with $h(\alpha) < T$), so in a certain (diophantine) sense $K$ is a rather small field. Are you aware of any ample field having the Northcott property? – Vesselin Dimitrov Aug 3 '15 at 13:14
• No, these fields do not have the Northcott property. And no, I cannot prove that PAC fields are not Northcott, but I expect this to be true, and section 6 of this paper by Amoroso, David, and Zannier proposes that a much stronger statement could be true: hal.archives-ouvertes.fr/hal-00649954/document . (Property (B) considered there is the much weaker one stating $\{\alpha \in K \mid h(\alpha) < T\}$ finite for some $T > 0$.) – Vesselin Dimitrov Aug 3 '15 at 13:59
• @VesselinDimitrov Actually the question of Amoroso, David, and Zannier has been settled. In this note: arxiv.org/abs/1408.6411 Lukas Pottmeyer constructs an example of a PAC field with the Bogomolov property (see the last section). – Bobby Grizzard Aug 3 '15 at 20:48
• @BobbyGrizzard: Thanks for the reference! Anyway, the example is not Northcott either, and it seems to me a reasonable question to raise whether all fields with the Nortchott property are not ample/large/fertile (and a fortiori, not PAC). – Vesselin Dimitrov Aug 3 '15 at 21:00
• @VesselinDimitrov: as far as you know, is it possible that there exists a variety $V$ such that $V(F)$ is finite for any field $F$ with the Northcott property? – Bobby Grizzard Aug 4 '15 at 5:28