$M/IM$ as $R$-module and $R/I$-module Let $R$ be a commutative ring with 1 and let $M$ be an $R$-module. We know that if we take an ideal $I$ of $R$ we can define the $R$-module $M/IM$, but this is also an $R/I$-module with the operation
$$(r+I)(m+IM)=rm+IM.$$
I have some doubts about the relation between this two modules. In my mind they are practically the same module, if I try to prove a statement for $M/IM$ as $R$-module the proof works also for $M/IM$ as $R/I$-module and viceversa.
For example, if $M/IM$ is finitely generated as $R/I$-module by  $\{\bar{m_1},…,\bar{m_n}\}$ then $M/IM$ is finitely generated as $R$-module by $\{\bar{m_1},…,\bar{m_n}\}$.
The only difference I understand is that if I take $R$ not a field and $I$ a maximal ideal, then $M/IM$ is a $R/I$-vectorial space but not a $R$-vectorial space. I don’t see other difference.
So, my question is:
It is $M/IM$ as $R/I$-module only an artificial construction to work with $M/IM$ as $R$-module in particular condition? Or am I not catching some details?
 A: You are correct that "information" of $M/IM$ as an $R/I$-module all comes from its structure as an $R$-module, for example the submodules are the same.
However, the $R/I$-module structure on $M/IM$ is quite natural. In some sense $M/IM$
is essentially a module over $R/I$. One illustration for this is the isomorphism $M/IM \simeq A/I \otimes_A M$. One may prefer to think $R/I$ is easier to understand than $R$ itself, and so one may extract information from the $R/I$-module structure.
For instance, let $R$ be a local ring and $M$ a finitely generated module over $R$, it is not so clear why every minimal system of generators of $M$ should have the same cardinality, but by Nakayama we know it is $\dim_k M/\mathfrak{m}M$.
A special case one must consider the $R/I$-module on $M/IM$ is when $M = I$.
The module $I/I^2$ is called the conormal sheaf of $R/I$ in $R$. The study of $I/I^2$ as an $R/I$-module is very important in algebraic geometry.
A: We may view the category of $R/I$-modules as a full subcategory of the category of $R$-modules, by sending a $R/I$-module $M$ to the $R$-module $M$ with, the action of $r\in R$ on $x\in M$ defined by $rx:=\overline rx$. The image consists of those $R$-modules $N$ such that for any $r\in I$ and $x\in N$, $rx=0$.
Thus, in a very precise sense, $R/I$-modules can be thought of as $R$-modules satisfying some additional properties.
