$\text{End}(E)$ is the endomorphism ring of an elliptic curve $E$.
Why is it true that
$$\text{rank}_\mathbb{Z}\text{End}(E) = \text{rank}_{\mathbb{Z}_l}\text{End}(E)\otimes\mathbb{Z}_l,$$
in the sense that if one is finite, then the other is finite and they are equal?
This seems to make sense, but I don't know what arguments to use to show that it is true.
The result should follow from the fact that $\text{End}(E)$ is torsion free. I know that because $\text{End}(E)$ is torsion free then if it is finitely generated as a $\mathbb{Z}$-module it is free, i.e. it has a basis, and I probably need to use this?
Thank you for any help.
(From a proof in Silverman, "The Arithmetic of Elliptic Curves")
As $\text{End}(E)$ is a finitely generated Abelian group, this is no more than saying that $$\text{rank}_{\Bbb Z}(\Bbb Z^n)=\text{rank}_{\Bbb Z_l}(\Bbb Z^n\otimes_{\Bbb Z}\Bbb Z_l),$$ i.e., $$\text{rank}_{\Bbb Z}(\Bbb Z^n)=\text{rank}_{\Bbb Z_l}(\Bbb Z_l^n).$$