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Fermat's last theorem implies that the number of solutions of $x^5 + y^5 = 1$ over $\mathbb{Q}$ is finite.

Is the number of solutions of $x^5 + y^5 = 1$ over $\mathbb{Q}^{\text{ab}}$ finite?

Here $\mathbb{Q}^{\text{ab}}$ is the maximal abelian extension of $\mathbb{Q}$. By Kronecker-Weber, this is the field obtained from $\mathbb{Q}$ by adjoining all roots of unity.

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    $\begingroup$ Just curious, what is the reason for asking about the $n=5$ case specifically? I thought about the $n=4$ case and I couldn't answer that either -- but perhaps you can? $\endgroup$ – RP_ Nov 25 '14 at 16:22
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    $\begingroup$ @René For $n = 4$ there are infinitely many solutions because you have infinitely many Pythagorean triples, which you can lift to your curve using the fact that a square root of an integer is in $\mathbb{Q}^{\text{ab}}$. $\endgroup$ – Pablo Nov 25 '14 at 16:30
  • $\begingroup$ @René I do not know what happens for any $n \geq 5$. $\endgroup$ – Pablo Nov 25 '14 at 16:33
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    $\begingroup$ Trivial comment: A similar argument gives infinitely many points over $k^{\mathrm{ab}}$, where $k$ is the fifth cyclotomic field. $\endgroup$ – Daniel Loughran Nov 25 '14 at 17:11
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    $\begingroup$ How many solutions are known? Besides the trivial I found only one in Q[sqrt(5)]. $\endgroup$ – joro Nov 28 '14 at 14:44
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This is not quite an answer, but not quite a comment either.

We can at least show that $X(({\mathbb Q}^{\text{ab}})^{\text{ab}})$ is infinite (where $X$ is the quintic Fermat curve). There are in fact (at least) two ways of showing this.

  1. $X$ has an automorphism $\tau$ of order 3 defined over $\mathbb Q$ (rotate the projective coordinates). The quotient $X/\langle \tau \rangle$ is a hyperelliptic curve $Y$. As I pointed out in a related thread, $Y({\mathbb Q}^{\text{ab}})$ is infinite for every hyperelliptic curve $Y$. Now the preimages on $X$ of any ${\mathbb Q}^{\text{ab}}$-point on $Y$ are defined over a cyclic degree-3 extension of ${\mathbb Q}^{\text{ab}}$, so are in $X(({\mathbb Q}^{\text{ab}})^{\text{ab}})$.
  2. Fix any $y \in {\mathbb Q}$, then the points $(x:y:1) \in X$ have $x$ in ${\mathbb Q}(\mu_5, \sqrt[5]{1-y^5}) \subset ({\mathbb Q}^{\text{ab}})^{\text{ab}}$.

The second construction obviously generalizes to arbitrary Fermat curves. In my originial post, I claimed that the first one does, too. But this seems to be wrong: the quotient of the degree $n$ Fermat curve $x^n + y^n + z^n = 0$ by the obvious $S_3$-action has positive genus as soon as $n \ge 6$, so the quotient by the cyclic group is no longer obviously hyperelliptic. (It is true, however, that this curve maps to the hyperelliptic curve $y^2 = x^n + \frac{1}{4}$, but this is via the quotient w.r.t. the action of $\mu_n$ such that $\zeta$ sends $(x:y:z)$ to $(\zeta x : \zeta^{-1} y : z)$. This would lead to a weaker conclusion, replacing the double by the triple abelian closure of $\mathbb Q$. I had mixed up the two quotients.)

Note that these constructions use the fact that $X$ has (many) nontrivial automorphisms. So perhaps another interesting question is the following.

Let $C$ be a nice (smooth, projective, geometrically irreducible) curve over $\mathbb Q$. Can we at least show that $C({\mathbb Q}^{\text{sol}})$ is infinite?

Here ${\mathbb Q}^{\text{sol}}$ denotes the union of all finite Galois extensions of $\mathbb Q$ with solvable Galois group.

(Originally the question was formulated for (sufficiently generic) curves of genus 3. However, as René pointed out, we find lots of quartic points by intersecting the plane quartic $C$ with rational lines, and all quartic fields are contained in ${\mathbb Q}^{\text{sol}}$.)

EDIT: The question whether solvable points exist has been studied, see for example a recent preprint by Trevor Wooley and the references given there, in particular this paper by Ambrus Pál.

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    $\begingroup$ I think the answer is yes, since a non-hyperelliptic genus 3 curve is a plane quartic, and every quartic polynomial (in one variable) has solvable Galois group. Nice answer/comment, by the way! $\endgroup$ – RP_ Nov 25 '14 at 17:37
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    $\begingroup$ @René: True. So we should take a curve whose gonality is at least 10, to avoid degree 4 maps to hyperelliptic curves. I'll edit the question in the answer and remove the "genus 3" assumption. $\endgroup$ – Michael Stoll Nov 25 '14 at 17:56
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    $\begingroup$ @MichaelStoll Can we at least show that $C(\mathbb{Q}^{\text{sol}}) \neq \emptyset ?$ $\endgroup$ – Pablo Nov 25 '14 at 18:01
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    $\begingroup$ This is a good question, with implications towards the Langlands program, as shown by work of Taylor, inter al. In the positive direction, a theorem of Çiperani and Wiles proves that a curve of genus 1 with semistable Jacobian possesses a solvable point, Duke Math. J (2008). See ma.utexas.edu/users/mirela/solvable.pdf $\endgroup$ – ACL Nov 25 '14 at 22:11
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    $\begingroup$ For plane quintics and also for generic genus 4 curves, the two questions $C({\mathbb Q}^{\text{sol}}) \neq \emptyset$? and $\#C({\mathbb Q}^{\text{sol}}) = \infty$? are equivalent. Given a ${\mathbb Q}^{\text{sol}}$-point, we can in both cases find a ${\mathbb Q}^{\text{sol}}$-defined degree 4 map to ${\mathbb P}^1$; then René's argument applies again. $\endgroup$ – Michael Stoll Nov 26 '14 at 9:18
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There might well be an elementary construction of infinitely many points (which I cannot think of right now), but in any case, I think that there are experts out there who expect there to be infinitely many points over $\mathbb{Q}^{\rm ab}$. Namely, that field is conjectured to be large (see Pop's survey article), which concretely means that any irreducible curve that has one smooth point over that field is expected to have infinitely many.

As far as I can tell, there isn't all that much evidence in favour of the largeness conjecture, nor against it (in particular Pop is careful to phrase it as a question), so take this heuristic for whatever it's worth. Since you are in Tel-Aviv, you should ask Moshe Jarden what he believes.

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  • $\begingroup$ Dear Alex, I am familiar with the literature on large fields and with some of the people working on these things. The point is that some of these people think that this conjecture is false, even though were it true, the consequences would be far reaching - Shafarevich conjecture on the freeness of $\text{Gal}(\mathbb{Q}^{\text{ab}})$. The point is that there are almost no varieties for which the conjecture holds true (even for abelian varieties this is unknown). This is the reason I am asking this here for specific varieties to see if, even in very special cases, the conjecture shoul hold. $\endgroup$ – Pablo Nov 25 '14 at 14:49
  • $\begingroup$ @Pablo I guess you are then already aware of what I wrote in my "answer", and that it doesn't work in fact (or does it?). $\endgroup$ – Ariyan Javanpeykar Nov 25 '14 at 14:50
  • $\begingroup$ @AriyanJavanpeykar I was not aware of this kind of argument, but I do not see how this can work. It is not clear from your answer which properties of $\mathbb{Q}^{\text{ab}}$ or of the curve you use. I think that there are examples of smooth projective curves with a finite set of points over an infinite extension of $\mathbb{Q}$. Please explain to which polynomials you apply HIT to clarify. $\endgroup$ – Pablo Nov 25 '14 at 14:58
  • $\begingroup$ @Pablo I added some more details to my comment above. $\endgroup$ – Ariyan Javanpeykar Nov 25 '14 at 17:06
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Concerning solutions of Fermat's equation $x^{\ell}+y^{\ell}+z^{\ell}=0$ in $\ell^n$th cyclotomic fields (for regular primes $\ell$) see a Kolyvagin's 2001 Izvestiya paper http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=im&paperid=337&option_lang=eng .

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Edit: This is now just a comment.

You can construct many points of degree $5$ using Hilbert's irreducibility theorem.

Let $X = \{x^5+y^5 =z^5\}$ be your curve over $\mathbb Q$ considered in $\mathbb P^2_{\mathbb Q}$.

To see that $X(\mathbb Q^{ab})$ is infinite, it suffices to show that $X$ has infinitely many "Galois" points of degree $5$. (Here a point is Galois if its residue field is Galois over $\mathbb Q$.)

Note that the projection $X\to \mathbb P^1_{\mathbb Q}$ is of degree $5$. In particular, by Hilbert's irreducibility theorem, the curve $X$ has infinitely many points of degree $5$. As Rene points out, you won't get that these points are Galois though.

N.B. Faltings's theorem implies that the points of degree five on $X$ must lie in infinitely many distinct number fields.

This works in more generality as follows:

If $f:X\to \mathbb P^1_K$ is a finite morphism with $X$ a smooth projective geometrically connected curve over a number field $K$, then $f^{-1}(K)$ contains infinitely many points of degree $\deg f$. (You don't need $f$ to be Galois, but if $f$ is Galois then you will probably get infinitely many Galois points. They don't have to be abelian though...)

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    $\begingroup$ I'm really curious about how you want to apply HIT here! I would think that, on the contrary, no preimage of a rational point will give a Galois point, since you're extracting a fifth root, so you're missing a primitive fifth root of unity in your residue field! $\endgroup$ – RP_ Nov 25 '14 at 15:55
  • $\begingroup$ (OK, except the trivial rational solutions, that is.) $\endgroup$ – RP_ Nov 25 '14 at 16:04
  • $\begingroup$ Ari: I think the problem with your argument is that the cover $X \to \mathbb{P}^1$ not a Galois cover (i.e. the extension of function fields is not Galois). This is due to the reason that René gives, namely $\mathbb{Q}$ does not contain the fifth roots of unity. $\endgroup$ – Daniel Loughran Nov 25 '14 at 16:51
  • $\begingroup$ @René Yes, I was worried about that. HIT just gives you points of degree five in this case, but probably no Galois points. If you pass to the Galois closure of $f$ you will generate many Galois points, but these might not be abelian... $\endgroup$ – Ariyan Javanpeykar Nov 25 '14 at 17:04

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