# Rational points on algebraic curves over $\mathbb Q^\text{ab}$

Motivation:

Let $$\mathbb{Q}_{\infty,p}$$ be the field obtained by adjoining to $$\mathbb{Q}$$ all $$p$$-power roots of unity for a prime $$p$$. The union of these fields for all primes is the maximal cyclotomic extension $$\mathbb{Q}^\text{cycl}$$ of $$\mathbb{Q}$$. By Kronecker–Weber, $$\mathbb{Q}^\text{cycl}$$ is also the maximal abelian extension $$\mathbb{Q}^\text{ab}$$ of $$\mathbb{Q}$$.

A well known conjecture due to Mazur (with known examples) asserts, for an elliptic curve $$E$$ with certain conditions, that $$E(\mathbb{Q}_{\infty,p})$$ is finitely generated. This is the group of rational points of $$E$$ over $$\mathbb{Q}_{\infty,p}$$ (not a number field!).

A theorem due to Ribet asserts the finiteness of the torsion subgroup $$E(\mathbb{Q}^\text{ab})$$ for certain elliptic curves.

Questions:

(a) Can one expect to find elliptic curves (or abelian varieties) $$A$$ with $$A(\mathbb{Q}^\text{ab})$$ finitely generated?

(c) Can one expect to find curves $$C$$ of genus $$g >1$$ with $$C(\mathbb{Q}^\text{ab})$$ finite?

• Re question (c), Pete Clark has shown there are curves of every genus $g\geq 4$ with no points over ${\mathbb Q}^{ab}$: math.uga.edu/~pete/plclarkarxiv8v2.pdf. Fingers crossed and he'll appear here soon...
– dke
Apr 17, 2011 at 3:14
• Well, here I am, but you've already said most of what I would have. Nevertheless I left an answer, the main point being that there is something to say in the genus one case as well... Apr 17, 2011 at 7:58
• To answer (c), $\mathbb{Q}^{ab}$ is a large field (in the sense of Florian Pop) so any variety that has a $\mathbb{Q}^{ab}$-rational point, has infinitely many. So the answer to (c) is that either, as people pointed out, there are no points -- or there are nec. infinitely many. Apr 17, 2011 at 16:44
• @Duff: Nice, did not know this. Thanks! Could you please point me to a reference? Also, this seems to have a consequence for (a). Namely, that the group $A(Q^{ab})$ has positive rank for any $A$. Maybe this is known already?
– SGP
Apr 17, 2011 at 17:36
• $\mathbb{Q}^{ab}$ is only conjectured to be large. Otherwise Shafarevich's conjecture would be known (see math.upenn.edu/~harbater/patch35.pdf , pages 55-56.) Apr 17, 2011 at 18:20

Actually Ken Ribet proved that if $K$ is a number field and $K(\mu_{\infty})$ is its infinite cyclotomic extension generated by all roots of unity then for every abelian variety $A$ over $K$ the torsion subgroup of $A(K(\mu_{\infty}))$ is finite: http://math.berkeley.edu/~ribet/Articles/kl.pdf .

On the other hand, Alosha Parshin conjectured that if $K_{p}$ is the extension of $K$ generated by all $p$-power roots of unity (for a given prime $p$) then the set $C(K_{p})$ is finite for every $K$-curve $C$ of genus $>1$: http://arxiv.org/abs/0912.4325 (see also http://arxiv.org/abs/1001.3424 ).

• Thanks a lot for these very interesting papers! And thanks for pointing out the correct version of Ribet's result (I had the mistaken impression that it was only for elliptic curves).
– SGP
Apr 17, 2011 at 12:37
• I think I am misremembering things and this is right and what I wrote in my answer (which I am going to delete) is wrong. Apr 17, 2011 at 16:08

As dke mentioned, I have a paper in which I construct various kinds of varieties $$X_{/\mathbb{Q}}$$ without abelian points (i.e., with $$X(\mathbb{Q}^{\operatorname{ab}}) = \varnothing$$). Here is a brief summary:

If $$X$$ admits a $$2:1$$ map to a variety $$Y$$ with infinitely many $$\mathbb{Q}$$-rational points, then $$X$$ itself has infinitely many quadratic points -- i.e., points defined over the union of all quadratic extensions of $$\mathbb{Q}$$. This certainly lives inside $$\mathbb{Q}^{\operatorname{ab}}$$, so gives infinitely many abelian points.

Now I call a curve $$X$$ hyperelliptic if it admits a $$2:1$$ map down to $$\mathbb{P}^1$$. (I say "I call" because I am not making any genus restrictions and requiring the map to be defined over $$\mathbb{Q}$$. Standard terminology is taking a little while to catch up to me here...) Now any curve of genus $$0$$ or $$2$$ is hyperelliptic, as is any curve of genus one with a rational point. So they all have infinitely many abelian points.

If $$E$$ is an elliptic curve over $$\mathbb{Q}$$, then what I'm saying is that if it is given as $$y^2 = x^3 + Ax + b$$, then take $$x$$ to be any rational number and extract the square root: that will give you an abelian point. One can see that only finitely many of these quadratic points are torsion points, so we are certainly getting positive rank this way. Do these quadratic points already give infinite rank? I'm not sure (but I feel like I am forgetting something here). [Added: I think I was forgetting what is in Dror Speiser's nice comment below!] Note that here I am -- anemically -- addressing your question a).

A genus one curve without rational points need not be hyperelliptic, and in my paper I construct lots of genus one curves over $$\mathbb{Q}$$ without elliptic points. This is the key part, actually, because using this I construct curves of every genus $$g \geq 4$$ over $$\mathbb{Q}$$ without abelian points.

This leaves genus $$3$$, which I was frustratingly unable to deal with in the paper and still can't. (In an appendix, I show that there are genus $$3$$ curves over some field without points over the maximal abelian extension of that field, unlike in the hyperelliptic cases. So my guess is that this should be possible over $$\mathbb{Q}$$ as well.)

I didn't think at all about the problem of infinitely versus finitely many abelian points, probably because it cannot be attacked using the methods of my paper. But of course it is interesting too.

• "Do these quadratic points already give infinite rank?" - yes. Say we have $n$ linearly independent points over the union of $\mathbb{Q}(\sqrt{d_i})$. Take $p$ a large prime that doesn't divide any one of the $d_i$, and that $x^3+Ax+b=0\pmod{p}$ has a solution $x_0$ (Chebotarev). Then $(x_0,\sqrt{f(x_0)})$ is a new point, and these $n+1$ points are again linearly independent: if a relation exists, just add to it its galois conjugate relation ($\sqrt{f(x_0)}\mapsto -\sqrt{f(x_0)}$), getting a relation on the $n$ points. The answer to a) (for jacobians of hyperelliptic curves) is "no". Apr 17, 2011 at 9:14
• @Speiser: thanks for the explanation!
– SGP
Apr 17, 2011 at 12:07

I admit that I haven't read it carefully, but in this paper E. Kobayashi conjectures that $$E(\mathbb Q^{\rm ab})$$ has infinite rank for all elliptic curves $$E$$ defined over $$\mathbb Q^{\rm ab}$$. In particular, assuming the "weak" Birch and Swinnerton-Dyer conjecture for $$E$$ and certain properties of twisted Hasse-Weil $$L$$-functions of $$E$$, she shows that if $$E$$ is defined over a number field $$K$$ of odd degree then $$E(K\cdot\mathbb Q^{\rm ab})$$ has infinite rank (Theorem 2 in Kobayashi's article).

This result for elliptic curves seems to suggest that the answer to your question (a) might be "no".

As for a simple abelian variety $$A$$ over a number field $$K$$, it is perhaps worth pointing out that Zarhin proved here that the torsion subgroup of $$A(K^{\rm ab})$$ is finite if and only if $$A$$ is not of CM-type over $$K$$ (when $$K=\mathbb Q$$ this reduces to an earlier theorem of Ribet).

• Thanks! I did not know any of these results!
– SGP
Apr 17, 2011 at 12:00
• As for a simple abelian variety $A$ over a number field $K$, the torsion subgroup of $A(K^{ab})$ is finite if and only if $A$ is not of CM-type over" $K$. (in other words, $\bar{K}$ on the last line of your comment should be replaced by $K$.) Apr 17, 2011 at 14:54
• @Yuri Zarhin: Actually, there is no $\bar{K}$ in the last line of my comment: it's just the underline to the word "here" in the second to last line. Or perhaps I'm misunderstanding what you mean, sorry... Apr 17, 2011 at 20:10
• OOPS! Sorry. I've mistook the underline as bar over K. Apr 17, 2011 at 20:27