Let $g$ be a positive integer, and let $G$ be a commutative group with the following constraint on its torsion subgroup: there is an injection $G[\operatorname{tors}] \hookrightarrow (\mathbb{Q}/\mathbb{Z})^{2g}$. Must there be subgroups $G_1,\ldots,G_g$ of $G$ such that

(i) $G = G_1 \times \ldots \times G_g$ (internal direct product), and
(ii) For all $1 \leq i \leq g$, there is an injection $G_i[\operatorname{tors}] \hookrightarrow (\mathbb{Q}/\mathbb{Z})^2$?

Motivation: If this is true, then it reduces the "Inverse Mordell-Weil Problem for Abelian Varieties" to the "Inverse Mordell-Weil Problem for Elliptic Curves".

Thus although the given question certain has an affirmative answer in many special cases -- e.g. it is a triviality if $G$ is finitely generated -- I am not really interested in that. But it would be "lucky for me" if the answer turns out to be affirmative in the general case, so it's worth asking.

Added: This previous question contains some information on when the torsion subgroup of a commutative group is a direct summand.

  • $\begingroup$ I've been told the torsion subgroup of an abelian group need not be a direct summand, but I don't think I actually know any examples. It seems like a reasonable strategy here to construct a counterexample would be to find a group with torsion subgroup, say, $(\mathbb{Q}/\mathbb{Z})^3$ which isn't a direct summand. $\endgroup$ – Qiaochu Yuan Jun 29 '13 at 4:01
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    $\begingroup$ @QiaochuYuan That's impossible. $(\mathbb{Q}/\mathbb{Z})^3$ is divisible and hence injective as $\mathbb{Z}$-module. $\endgroup$ – Kevin Ventullo Jun 29 '13 at 4:02
  • $\begingroup$ I'm a bit confused about the motivation; in the Mordell-Weil situation the group is finitely generated, right? Or is the idea that you want to show any such group can arise as the $K$-points of an ab. var. for some possibly infinite extension $K$? In that case, you may assume $G[tors]$ is finite, basically since $\mathbb{Q}_p/\mathbb{Z}_p$ is divisible. $\endgroup$ – Kevin Ventullo Jun 29 '13 at 4:03
  • $\begingroup$ @Qiaochu: indeed the torsion subgroup need not be a direct summand. I asked an MO question about this a while back. The easiest example is probably $\prod_{\ell} \mathbb{Z}/\ell \mathbb{Z}$. The precise result is a theorem of Baer: for a torsion commutative group $T$, every (commutative!) extension of a torsionfree group by $T$ is split iff $T$ is the direct sum of a divisible $T_1$ and a group $T_2$ of bounded exponent (i.e., $T_2 = T_2[n]$ for some $n$). $\endgroup$ – Pete L. Clark Jun 29 '13 at 4:27
  • $\begingroup$ @Kevin: Here, by "Mordell-Weil group" I mean any group of the form $A(K)$ for an abelian variety over any field $K$. (And it is not enough to assume that the torsion subgroup is finite. In fact that case is significantly easier than the general case, because the torsion subgroup is necessarily a direct factor.) $\endgroup$ – Pete L. Clark Jun 29 '13 at 4:29

The answer is negative.

Define $S=\bigoplus_p C_p$, where $C_p$ is cyclic of order $p$ and $p$ ranges over prime numbers. Fix an integer $n\ge 1$. Define $P=\prod_p C_p$, and consider a maximal $\mathbf{Z}$-free family in $P^n/S^n$, generating a free abelian group (of continuum rank) $L/S^n$. Fix a prime $q$ and let $U/S^n$ be (*) a subgroup of $P^n/S^n$ isomorphic to $\mathbf{Z}_q$ (the group of $q$-adic numbers), with $L\subset U$.

Let's check that for every direct decomposition $U=V\oplus W$, either $V$ or $W$ is finite. Since $T(U)\simeq S^n$, this implies that $U$ answers negatively your question (as soon as $n\ge 3$).

First observe that $\mathbf{Z}_q$ is indecomposable (i.e. has no nontrivial decomposition as direct product, easy exercise (**)). Since $U/T(U)\simeq\mathbf{Z}_q$, it follows that either $V$ or $W$, say $W$, is torsion. So $V/T(V)\simeq\mathbf{Z}_q$.

Write $T(V)=\bigoplus A_p$ and $T(W)=\bigoplus B_p$, so that $A_p\oplus B_p=C_p^n$ for all $p$.

We have $P^n=(\prod A_p)\times (\prod B_p)$. So $P^n/T(V)=(\prod A_p/\bigoplus A_p)\times \prod B_p$. Since there is no nonzero homomorphism $\mathbf{Z}_q\to C_p$ for $p\neq q$ and since the subgroup $V/T(V)$ of $P^n/T(V)$ is isomorphic to $\mathbf{Z}_q$, the projection of $V/T(V)$ into $\prod B_p$ is contained in $B_q$. Thus in the above decomposition, $V\subset (\prod A_p)\times B_q$ and $W\subset (\bigoplus A_p)\times (\bigoplus B_p)$. So $U\subset (\prod A_p)\times (\bigoplus B_p)$. Since $P^n/U$ is torsion (because it is a quotient of $P^n/M$ which is torsion), it follows that $\prod B_p/\bigoplus B_p$ is torsion. This means that $B_p=0$ for large $p$. It follows that $W$ is finite.

Here are the two easy verifications:

(*): pick in $\mathbf{Z}_q$ a maximal $\mathbf{Z}$-free family, generating an abelian free subgroup $Z$; since $P^n/S^n$ is a rational vector space, any isomorphism $Z\to L/S^n$ (it exists since both groups are free abelian of continuum rank) can be extended to an injective homomorphism $\mathbf{Z}_q\to P^n/S^n$.

(**): $\mathbf{Z}_q$ is $p$-divisible for all $p\neq q$ and $\mathbf{Z}_q/q\mathbf{Z}_q$ has order $q$, so for every direct decomposition of $\mathbf{Z}_q$, at least one factor is divisible. But $\mathbf{Z}_q$ does not contain any copy of $\mathbf{Q}$.

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  • $\begingroup$ Thanks very much. (This is very picky, but: I think you're not giving a counterexample but rather a negative answer to a question.) $\endgroup$ – Pete L. Clark Jun 29 '13 at 17:52
  • $\begingroup$ you're right; I'll fix this if I need to edit something else. $\endgroup$ – YCor Jun 29 '13 at 18:55
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    $\begingroup$ Now fixed (after 5 years of intense thinking...) $\endgroup$ – YCor May 22 '18 at 10:58
  • $\begingroup$ With some more efforts maybe one could get a countable example. We can replace the $p$-adics by rational $p$-adics, but then argument does not carry over when we use that $P^n/M$ is torsion... but it maybe possibly still holds that every decomposition has a finite summand. $\endgroup$ – YCor May 22 '18 at 11:01

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