Let $L/\mathbb{Q}$ be the field generated over $\mathbb{Q}$ by all of the (projective) coordinates of all of the torsion points of all abelian varieties defined over $\mathbb{Q}$. Is $L$ algebraically closed?

$\begingroup$ I have deleted my answer, since I realized that the argument does not work (subgroups may have different composition factors than the group they are contained in). Still, I have the feeling that the answer is no. $\endgroup$ – Michael Stoll Dec 2 '14 at 10:50

1$\begingroup$ My feeling was apparently wrong  nice and short answer! $\endgroup$ – Michael Stoll Dec 2 '14 at 13:42

$\begingroup$ @MichaelStoll Still, why have you considered GSp and not GL? $\endgroup$ – Pablo Dec 2 '14 at 13:45

3$\begingroup$ Because of the Weil pairing: there is a Galoisequivariant alternating pairing $A[n] \times A[n] \to \mu_n$, so the image of Galois in $\operatorname{GL}(2d,{\mathbb Z}/n{\mathbb Z})$ is contained in $\operatorname{GSp}$. $\endgroup$ – Michael Stoll Dec 2 '14 at 13:55
If $\lambda\in\overline{\mathbb{Q}}$, the elliptic curve $$ E_\lambda\colon y^2=x(x1)(x\lambda) $$ has $(\lambda,0)$ as $2$torsion point and is defined over (a subfield of) $L=\mathbb{Q}(\lambda)$. Its Weil restriction $A_\lambda:=\operatorname{Res}_{L/\mathbb{Q}}(E_\lambda)$ is an abelian variety defined over $\mathbb{Q}$ and shares the same points of $E_\lambda$, including their torsion structure, so $\lambda\in \mathbb{Q}(A_\lambda[2])$.
EDIT As Urlich observed, what I wrote was wrong/useless: indeed, the defining property of Weil restricition is that the $\mathbb{Q}$points of $\operatorname{Res}_{L/\mathbb{Q}}$ coincide with the $L$ones of $E_\lambda$, so the point corresponding to $(\lambda,0)$ is defined over $\mathbb{Q}$. Instead of $\operatorname{Res}_{L/\mathbb{Q}}$, (assume $L/\mathbb{Q}$ Galois, which makes no harm, and) consider $A_\lambda=\prod_{\sigma\in G(L/\mathbb{Q})}E^\sigma$, which is an abelian variety over $\mathbb{Q}$ and has the required property: the point $$ \big((\lambda,0),O,O,\dots,O\big) $$ where $O$ is the unit has the required property (namely, it generates $L$ and is torsion): thanks to Felipe for the solution.

4$\begingroup$ The point $(\lambda, 0)$ is an $L$rational point of $E_{\lambda}$ so corresponds to a $\mathbb{Q}$rational point of $A_{\lambda}$. $\endgroup$ – ulrich Dec 2 '14 at 13:06

2$\begingroup$ The dimension of the resulting abelian variety $A_\lambda$ is unbounded when one ranges over $\lambda$, right? $\endgroup$ – Pablo Dec 2 '14 at 13:08

6$\begingroup$ Ulrich is correct, but Filippo is almost correct. Instead of restriction of scalars use the product of the conjugates. $\endgroup$ – Felipe Voloch Dec 2 '14 at 14:41

2$\begingroup$ @Felipe: I am lost here. I always thought the product of the conjugates was the Weil restriction. $\endgroup$ – Laurent MoretBailly Dec 3 '14 at 14:01

2$\begingroup$ @Felipe: But I cannot think of two natural Galois actions making this product "defined over $\mathbb{Q}$", unless of course the original $E_\lambda$ is. $\endgroup$ – Laurent MoretBailly Dec 3 '14 at 17:20
Another proof that $L = \,\overline{\bf \!Q\!}\,$: Clearly $L$ is contained in $\,\overline{\bf \!Q\!}\,$, so we need only show $L$ contains every algebraic number $x \notin \bf Q$. Let $P(X)$ be the minimal polynomial of $x$. If $\deg P$ is odd, then the class of $((x,0))  (\infty)$ is a $2$torsion point on the Jacobian of the elliptic or hyperelliptic curve $y^2 = P(x)$. If $\deg P$ is even, then the class of $((x,0))  (\infty)$ is a $2$torsion point on the Jacobian of the elliptic or hyperelliptic curve $y^2 = x P(x)$. QED

$\begingroup$ Excellent. Now, to answer the further question, can you modify this so that the associated Jacobians are geometrically simple? Most of the ones you construct will be, but not all, I think. Maybe use $y^2=P(x)Q(x)$ for a generic $Q(x)$ so that $\deg(PQ)$ is odd? $\endgroup$ – Joe Silverman Dec 3 '14 at 23:23

$\begingroup$ Thanks. Yes, multiplying $P(x)$ by some auxiliary polynomial with degree of the appropriate parity must work, though proving it might be beyond the scope of a reasonable MO answer, unless some helpful result is already known (like a strong enough upper bound on the proportion of Jacobians of hyperlliptic genus$g$ curves of height $\leq H$ that are not geometrically simple). $\endgroup$ – Noam D. Elkies Dec 4 '14 at 1:04

$\begingroup$ The following paper might help: MR1361754 Masser, D. W.(CHBASL); Wüstholz, G.(CHETHZ) Factorization estimates for abelian varieties. Inst. Hautes Études Sci. Publ. Math. No. 81 (1995), 5–24. They give effective estimates for the degree of the isogeny and fields of definition for the factorization of $A$ into simple factors. So one might be able to use that to prove that almost all of the Jacobians of $y^2=PQ$, counted by height, are simple. It wouldn't follow directly, but might be the right tool to apply. (Just a thought.) $\endgroup$ – Joe Silverman Dec 4 '14 at 3:19

$\begingroup$ @NoamD.Elkies this is a very nice and explicit construction. Following Silverman, and Dimitrov I would like to know if you can think of some natural "big" family of abelian varieties (with unbounded ranks) such that adding all their torsion points to the rationals will not result in an algebraically closed field? That is, something like the simple varieties... $\endgroup$ – Pablo Dec 4 '14 at 7:09