Timeline for Is the compositum of all quadratic extensions of the rationals an ample field?
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| Sep 7, 2020 at 12:53 | comment | added | Arno Fehm | @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 ems-ph.org/journals/… | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Aug 4, 2015 at 8:37 | comment | added | Vesselin Dimitrov | @BobbyGrizzard: I don't know the answer to this; seems like an interesting problem! | |
| Aug 4, 2015 at 5:28 | comment | added | Bobby Grizzard | @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? | |
| Aug 4, 2015 at 5:20 | comment | added | Bobby Grizzard | I agree. I wish it wasn't so hard to say anything new about fields with the Northcott property :) | |
| Aug 3, 2015 at 21:00 | comment | added | Vesselin Dimitrov | @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). | |
| Aug 3, 2015 at 20:48 | comment | added | Bobby Grizzard | @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). | |
| Aug 3, 2015 at 14:11 | comment | added | Pablo | @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 $L = \mathbb{Q}(\sqrt{p_n} \ : \ n \in \mathbb{N})$. | |
| Aug 3, 2015 at 13:59 | comment | added | Vesselin Dimitrov | 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 $T > 0$.) | |
| Aug 3, 2015 at 13:37 | comment | added | Pablo | @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? | |
| Aug 3, 2015 at 13:14 | comment | added | Vesselin Dimitrov | 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? | |
| Aug 3, 2015 at 11:28 | history | edited | Pablo | CC BY-SA 3.0 |
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| Aug 3, 2015 at 9:53 | history | asked | Pablo | CC BY-SA 3.0 |