Checking that a torsion-free abelian group has finite rank Suppose $G$ is a torsion-free abelian group and that $G \otimes_\mathbb{Z}\mathbb{Z}_l$ is free of finite rank as a $\mathbb{Z}_l$-module, where $\mathbb{Z}_l$ denotes the $l$-adic integers and $l $ is a fixed prime number. Can we conclude that $G$ is free of finite rank? Additionally, what if the statement is true for all primes $l$?

The result is still false if you require that $G\otimes_{\mathbb Z}\mathbb Z_l$ is free of finite rank for every prime $l$.
Let $G$ be the subgroup of $\mathbb Q$ generated by $1/p$ for each prime $p$. Equivalently, it consists of precisely the rational numbers with squarefree denominator.
G is not finitely generated: indeed in any finitely generated subgroup of $G$ the elements are of bounded denominator. Thus it cannot be free, since it is rank one.
I claim $G\otimes_{\mathbb Z}\mathbb Z_l=\mathbb Z_l$ for any prime $l$. One has a short exact sequence $$0\rightarrow \mathbb Z\rightarrow G\rightarrow \bigoplus_{p\; \mathrm{prime}}\mathbb Z/p\mathbb Z\rightarrow0,$$ which after applying $-\otimes_\mathbb Z \mathbb Z_l$ yields $$0\rightarrow \mathbb Z_l\rightarrow G\otimes_{\mathbb Z}\mathbb Z_l\rightarrow \mathbb Z/l\mathbb Z\rightarrow0$$
again exact. But now $G\otimes_{\mathbb Z}\mathbb Z_l$ is a finitely generated $\mathbb Z_l$-module which is torsion-free as an abelian group. We deduce from the classification that $G\otimes_{\mathbb Z}\mathbb Z_l=\mathbb Z_l$ as desired.
