Let $(\mathbb{Z}\_l-\text{mod})\otimes \mathbb{Q}\_l$ be the category whose objects are $\mathbb{Z}\_l$ modules and morphism groups are tensored with $\mathbb{Q}\_l$, my understanding of this is like just kill all the torsion modules and think two morphisms as the same if they differ by a map to a torsion submodule. 


My question is,


Is $(\mathbb{Z}\_l-\text{mod})\otimes \mathbb{Q}\_l\cong \mathbb{Q}\_l-\text{Vect}$ ?


The only thing that makes me worry is whether this is true without any finiteness condition. E.g. I'm convinced that 


$ \text{(Finite rank abelian groups)} \otimes\_\mathbb{Z}\mathbb{Q} \cong \text{Finite-dimensional-} \mathbb{Q}\text{-vector-spaces}$. 


As every $\mathbb{Q}-$ matrix is a $\mathbb{Z}-$matrix times a rational number. But I'm not sure if it is true without the finite-rank condition. 


The question make sense for any domain $R$, I use $\mathbb{Z}\_l$ instead of $R$ just because I thought about this when reading \'etale cohomology. (Maybe dim$R$ = 1 makes a difference?)



Is $(\mathbb{Z}_l-\text{mod})\otimes \mathbb{Q}_l\cong \mathbb{Q}_l-Vect$ ?
Let $V$ be a $Q\_l$ vector space of countable rank. Pick a basis and consider the linear map that is represented by a matrix with zeroes off the main diagonal and $1/l^n$ in the $(n,n)$ place. 


This is clearly not in the image of $Z\_l-mod\otimes Q\_l$.


