# Can an abelian variety/Q have no non-trivial points over Q_sol?

Let $A/\mathbb{Q}$ be an abelian variety. Must there be a finite solvable extension $K/\mathbb{Q}$ such that $A(K)$ is nontrivial?

This follows from the conjecture that the maximal (pro-)solvable extension $\mathbb{Q}^{\mathrm{sol}}$ of $\mathbb{Q}$ (the field generated over $\mathbb{Q}$ by radicals) is PAC (Pseudo Algebraically Closed - each geometrically irreducible variety has infinitely many points). It even follows from the much weaker conjecture that $\mathbb{Q}^{\mathrm{sol}}$ is large/ample (every smooth variety with a point has infinitely many points).

It is not difficult to see that the answer is positive if $A$ is an elliptic curve, and I guess that for some other families of abelian varieties this can be not too complicated.

Therefore, I will be very happy to see something nonconjectural, which could work for general abelian varieties.

• Rather than elliptic curves (where you could just adjoin the points of order two), much more interesting is the case of smooth genus one curves over $\mathbb{Q}$. For these, as you probably know, there is the Ciperiani-Wiles theorem proving the existence of a $\mathbb{Q}^{\mathrm{sol}}$ point under some (fairly mild) technical conditions. (There should be a $\mathbb{Q}_p$-point for all $p$, and the Jacobian, an elliptic curve over $\mathbb{Q}$, should be semistable.) – Vesselin Dimitrov Mar 1 '15 at 1:45
• See Theorem 0.0.2 in their paper "Solvable points on genus one curves" in the Duke Math. Journal, 2008. While the statement is very likely to be true without the condition of existence of points in all $\mathbb{Q}_p$, the latter is quite essential to their proof. – Vesselin Dimitrov Mar 1 '15 at 7:51
• By definition, an abelian variety over a field K has a rational point over K, so in your question, you presumably mean a homogenous space for your abelian variety. – Laurent Berger Mar 1 '15 at 8:47
• @LaurentBerger I explicitly wrote "$A(K)$ is nontrivial" and not "$A(K)$ is nonempty" in the body of the question in grey. That is, I ask whether the group $A(K)$ may be trivial for every finite solvable extension $K/\mathbb{Q}$. – Pablo Mar 1 '15 at 8:50
• I believe this is not known, though it is even conjectured that $A(F)\ne\{0\}$ over the maximal abelian extension of $F$ for any number field $F$ (Fehm, Petersen, On the rank of abelian varieties over ample fields, IJNT 6 (2010), no. 3, 579–586). And for "generic" abelian varieties the existence of such a $K$ would follow from the parity conjecture. But, unconditionally, I am not sure you can prove this except for "easy" families, e.g. Jacobians of hyperelliptic curves. – Tim Dokchitser Mar 1 '15 at 8:59