Rank of finitely generated abelian groups Is it true that for every f.g. abelian group $A$ we have 
$\mathrm{rank} \ A=\dim_{\mathbf{R}}\mathrm{Hom}_{\mathbf{Z}}(A,\mathbf{R})$? (Here the vector space structure is defined in a natural way. I made this up myself, and am curious whether this is well-known, such as e.g. that $\mathrm{rank} \ A$ is $\mathrm{dim}_{\mathbf{Q}} A\otimes\mathbf{Q}$)
 A: Yes, it is true. This will work with any field $F$ of characteristic $0$.
First of all, since $\mathbb{R}$ is a torsion-free group (here the 'characteristic $0$' assumption is used), any homomorphism $A \rightarrow \mathbb{R}$ maps the torsion part of $A$ to $0$. So we may WLOG assume that $A$ is torsion-free finitely generated abelian group, i.e. free Abelian group of finite rank, say $n$.
Fix a free basis $X=\{x_1, \dots, x_n\}$ of $A$. Now, every homomorphism $f:A \rightarrow \mathbb{R}$ is uniquely determined by the image of the free basis of $A$, ie by a map $\tilde{f}:X \rightarrow \mathbb{R}$, which can be identified with an $n$-tuple $(\tilde{f}(x_1),\tilde{f}(x_2),\dots, \tilde{f}(x_n)) \in \mathbb{R}^n$. On the other hand, every such $n$-tuple $(a_1, \dots, a_n) \in \mathbb{R}^n$ defines exactly one such homomorphism (induced by the mapping $x_i \mapsto a_i$). The key observation is that these mapping are mutually inverse $\mathbb{R}$-linear isomorphisms between $\mathrm{Hom}_\mathbb{Z}(A,\mathbb{R})$ and $\mathbb{R}^n$. Hence, $\dim_\mathbb{R}\mathrm{Hom}_\mathbb{Z}(A,\mathbb{R})=\mathrm{rank}(A)$.
Remark: Also note that this fact follows from the fact that $\mathrm{rank}(A)=\dim_\mathbb{R}(A \otimes_\mathbb{Z}\mathbb{R})$ (and vice versa) using the tensor-hom adjunction, in particular, using the isomorphism
$$\mathrm{Hom}_\mathbb{R}(A \otimes_\mathbb{Z}\mathbb{R},\mathbb{R})\simeq \mathrm{Hom}_\mathbb{Z} (A, \mathrm{Hom}_\mathbb{R}(\mathbb{R},\mathbb{R})) $$
(and, further, using the fact that a finite-dimensional vector space is isomorphic to its dual).
