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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 $K$.

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Pablo
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    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 03 '15 at 13:14
  • @VesselinDimitrov sadly enough, it seems difficult to construct ample fields algebraic over $\mathbb{Q}$. Taking some random $\sigma \in \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ and taking its fixed fied will (almost surely) result in a PAC field, and thus, an ample field. Another example is the field of totally real numbers, and you have analogues of it $(\mathbb{Q}_{\mathrm{tot},S})$ for any set $S$ of primes of $\mathbb{Q}$. Besides these, I am not aware of any example, and I do not think that these fields have the Northcott property. Can you prove that a Northcott field is not PAC? – Pablo Aug 03 '15 at 13:37
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    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: https://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 03 '15 at 13:59
  • @VesselinDimitrov It is interesting whether there exists a non-ample (algebraic) field $L/\mathbb{Q}$ in which infinitely many rational primes ramify. I suspect that this can be done by carefully choosing a sequence of primes ${p_n}_{n=1}^\infty$ and taking $L = \mathbb{Q}(\sqrt{p_n} \ : \ n \in \mathbb{N})$. – Pablo Aug 03 '15 at 14:11
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    @VesselinDimitrov Actually the question of Amoroso, David, and Zannier has been settled. In this note: http://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 03 '15 at 20:48
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    @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 03 '15 at 21:00
  • I agree. I wish it wasn't so hard to say anything new about fields with the Northcott property :) – Bobby Grizzard Aug 04 '15 at 05:20
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    @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 04 '15 at 05:28
  • @BobbyGrizzard: I don't know the answer to this; seems like an interesting problem! – Vesselin Dimitrov Aug 04 '15 at 08:37
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    @Vesselin Dimitrov: I had not known that the question of PAC fields with Northcott had actually been asked by someone, but I did construct such an example in "Three counterexamples concerning the Northcott property of fields", see https://www.ems-ph.org/journals/show_abstract.php?issn=1120-6330&vol=29&iss=2&rank=6 – Arno Fehm Sep 07 '20 at 12:53

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