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 $l$?

(Edited to reflect that $\mathbb Z_{l}$ denotes the $l$-adic integers instead of $\mathbb Z/l\mathbb Z$.)
No.  Let $G$ be the additive group of the ring $\mathbb Z[1/p]$, where $p \neq l$ is a prime.  Then you can check that $\mathbb Z[1/p] \otimes_{\mathbb Z} \mathbb Z_l = \mathbb Z_l$, essentially because $1/p \in \mathbb Z_l$.  This is free of rank one over $\mathbb Z_l$, even though $\mathbb Z[1/p]$ is not finitely-generated as an abelian group.
