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Is $\lim_{n \rightarrow \infty} |\{(x,y) \in \mathbb{Q}(\zeta_n)^2 : y^3 = x^3 + x + 1\}| < \infty ?$ where $\zeta_n$ is a primitive $n$-th root of unity.

That is, I am asking whether the number of $\mathbb{Q}^{\text{ab}}$-rational points of $y^3 - x^3 - x - 1$ is finite.

If the answer is negative, then a slight augmentation of the polynomial to a one that makes the limit finite or a bound on the rate of growth will be very much appreciated.

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    $\begingroup$ Isn't this question very close to mathoverflow.net/questions/187163/… ? It's best not to ask multiple related questions simultaneously. $\endgroup$ Nov 14, 2014 at 19:15
  • $\begingroup$ @FelipeVoloch you are totally right. I just wanted to have a shorter and more explicit/specific version of the question. $\endgroup$
    – Pablo
    Nov 14, 2014 at 19:17
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    $\begingroup$ Do you know any example of a geometrically irreducible curve over $\mathbb{Q}$ that can be proved to contain only finitely many $\mathbb{Q}^{\mathrm{ab}}$-rational points? How about one that is smooth of genus $> 1$ but contains infinitely many such points? $\endgroup$ Nov 14, 2014 at 19:18
  • $\begingroup$ @VesselinDimitrov : For your second question I think that there are abelian varieties (which are not elliptic so the genus is not $1$) which are known to acquire infinite rank when we pass to $\mathbb{Q}^{\text{ab}}$. $\endgroup$
    – Pablo
    Nov 14, 2014 at 19:24
  • $\begingroup$ But curves? By the way, under certain circumstances (modularity of the Jacobian), a deep theorem of Kato about modular forms implies the finiteness of the set of points in $\mathbb{Q}_S^{\mathrm{ab}}$, the compositum of all abelian extensions with ramification limited to a given finite set $S$ of primes (adjoining roots of unity of order divisible only by primes in $S$). Perhaps one can expect this statement in general too (i.e. without the a priori modularity assumption). $\endgroup$ Nov 14, 2014 at 19:30

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To answer your concrete question, first, your curve (or rather, its projective closure) is isomorphic to the elliptic curve 8649b1, $y^2 + y = x^3 - 8$. Already over $\mathbb Q$, it has rank two and therefore infinitely many rational points.

More generally, for any elliptic curve $E$ over $\mathbb Q$, you should be able to find a quadratic twist $E^{(d)}$ with positive rank; then $E({\mathbb Q}(\sqrt{d}))$ is infinite, and since ${\mathbb Q}(\sqrt{d})$ is contained in a cyclotomic field, your limit is infinite.

For curves of higher genus, the question might be more interesting.

If you have a hyperelliptic curve $C \colon y^2 = f(x)$, then you will get infinitely many points with rational $x$-coordinate and $y$-coordinate in a quadratic number field. Since the compositum of all quadratic fields is contained in ${\mathbb Q}^{\text{ab}}$, this shows that hyperelliptic curves also have infinitely many points over this field.

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  • $\begingroup$ Ah, yes of course - we will have infinitely many quadratic points this way. On the other hand, we know after Siegel that for a given finite set $S$ of primes the equation $f(x) = S-\mathrm{unit}$ has finitely many rational solutions, hence finitely many points in $\mathbb{Q}_S^{\mathrm{ab}}$ arise this way. I wonder if the statement "$|C(\mathbb{Q}_S^{\mathrm{ab}})| < \infty$" has been voiced or considered anywhere (for $C$ of genus $> 1$). $\endgroup$ Nov 14, 2014 at 19:52
  • $\begingroup$ Can you give a candidate for such a curve (a curve of genus > 1 which we could suspect to be nonempty and finite over $\mathbb{Q}^{\text{ab}}$ )? I mistakenly thought this curve to be of genus 2... $\endgroup$
    – Pablo
    Nov 14, 2014 at 20:07
  • $\begingroup$ @VesselinDimitrov : Do you have any reason to believe the number of solutions will not grow? It is natural to ask now whether there is some explicit form of Falting's theorem with which we can give a bound for each cyclotomic field and then let this go to infinity. Is there a relatively simple/explicit version of Falting's theorem for cyclotomic fields? $\endgroup$
    – Pablo
    Nov 14, 2014 at 20:19
  • $\begingroup$ @Pablo: (about your first question) - How about any smooth and geometrically irreducible equation $f(x) = g(y)$ with $\deg{f}, \deg{g} > 3$ such that the Galois groups of the equations $f(x) = T$ and $g(y) = T$ over $\mathbb{C}(T)$ are not abelian? $\endgroup$ Nov 14, 2014 at 20:22
  • $\begingroup$ @VesselinDimitrov this is a good idea but maybe insufficient because it is possible to have such an example with infinitely many values of $y$ (and $x$) giving an abelian group after substitution... $\endgroup$
    – Pablo
    Nov 14, 2014 at 20:25