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

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?)

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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 $Z_l-mod\otimes Q_l$.
Thanks a lot. I guess I was asking a dumb question and I should have figured it out by myself. I guess $\mathbb{Z}_l-mod\otimes \mathbb{Q}_l$ shall be thought as a category whose objects are $\mathbb{Q}_l$ vector spaces and whose morphisms are linear maps which has some multiple can be descent to some $\mathbb{Z}_l$-modules. Correct me if I was wrong. –  temp Mar 5 '12 at 7:27
Hang on: isn't every morphism of $\mathbb{Q}_\ell$-modules a fortiori a morphism of $\mathbb{Z}_\ell$-modules, because $\mathbb{Z}_\ell$ is a subring of $\mathbb{Q}_\ell$? –  David Loeffler Mar 6 '12 at 9:11
David: You're right; in fact I was mistaken in two ways. My original post had $1/2^n$ on the diagonal; it should have been $1/l^n$. I've corrected this. But your point still stands. –  Steven Landsburg Mar 6 '12 at 13:24