Making a module a vector space Let $G$ be an abelian group whose elements are all of prime order $p$.  Naturally, $G$ is a  $\Bbb Z$-module.  Now take a maximal ideal $(p)$ and divide the scalar ring with it.  Then one can consider $G$ as a vector space whose scalars are from the field $\Bbb Z/(p)$.
Does this kind of construction of a vector space have a name or is it based on some general principle?
 A: You can always start with a ring $R$ and an $R$-module $M$. Considering the localization $M_{\mathfrak p}$ of the module $M$ at the prime ideal $\mathfrak p$ (which can be realized as $R_{\mathfrak p} \otimes_R M$), you can assume the ring local. Then you can consider the tensor product $R_{\mathfrak p}/\mathfrak p R_{\mathfrak p} \otimes_{R_{\mathfrak p}} M_{\mathfrak p}$, which is naturally a $\kappa(\mathfrak p)$-module (where $\kappa(\mathfrak p) = R_{\mathfrak p} / \mathfrak p R_{\mathfrak p}$). 
For instance, if I start with a finite abelian group $G$, $\mathbb Z/p\mathbb Z$ is already a field, so no need to localize (this is because $\mathbb Z$ is a PID, so all non-zero prime ideals are maximal ; if you had done this construction with $(0)$, you would have no need to quotient, just to localize). If you write 
$$
G = \bigoplus_{i=1}^n \mathbb Z/p_i^{n_i} \mathbb Z, \quad n_i \ge 1
$$
then $\mathbb Z/p\mathbb Z \otimes_{\mathbb Z} G$ is a $\mathbb Z/p \mathbb Z$-vector space with dimension equal to the number of $p_i$'s equal to $p$. 
If you apply this construction to the $R$-module $R$, you get the residue fields $\kappa(\mathfrak p)$. Examples where this construction is relevant are a bit more complicated to explain and I don't know your background so I can't think of anything relevant at the moment.
Hope that helps, 
