**Background**: When proving that the group of $k$-isogenies $\mathrm{Hom}\_k(A,B)$ between two abelian varieties is finitely generated, one first shows that the Tate map $$\mathbb{Z}\_\ell\otimes\_{\mathbb{Z}} M \to \mathrm{Hom}\_{\mathbb{Z}\_\ell}(T\_\ell A,T\_\ell B)$$ is injective. Since each Tate module is free of finite rank over $\mathbb{Z}\_\ell$, it follows that the localization $M\_\ell$ is $\mathbb{Z}\_\ell$-finite. One then uses a little trick to deduce the $\mathbb{Z}$-finiteness of $M$ itself. (See Silverman I, for example.)


The above proof needs only a single prime $\ell$,
but disregarding issues of the characteristic of the field (which are apparently surmountable) we actually have an injective Tate map at *every* prime. Thus...


**Question**: Can the $\mathbb{Z}$-finiteness of $M$ be deduced directly from the $\mathbb{Z}\_\ell$-finiteness of $M\_\ell$ for all primes $\ell$?


One can consider this a question about general *torsion-free* abelian groups $M$. A non-counterexample to keep in mind is $M=\mathbb{Z}[1/p]$,
for which $M\_\ell$ is $\mathbb{Z}\_\ell$-finite for all $\ell\neq p$.


(A google search shows that there is actually quite a body of literature on torsion-free abelian groups, so perhaps the answer to this question is well-known, but I'm not sure where to look...)



Is a torsion-free abelian group finitely generated, if all of its localizations at primes $p$ are finitely generated over $\mathbb{Z}_p$?
I don't think so. Consider the $\mathbb{Z}$-module $M$ be the additive subgroup of the rationals consisting of rationals with square-free denominator.


