Fermat's last theorem over larger fields 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.
 A: 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 .
A: 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.


*
*$X$ 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 $X(({\mathbb Q}^{\text{ab}})^{\text{ab}})$.

*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.
A: 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...)
A: 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.
