When is $(M) / (M ⊗ I)$ $\simeq$ to $M/(MI)$?

Solution 1:

You need to be careful with what you're writing here: if $I$ is an ideal of $R$ (say a left ideal) and if $M$ is a right $R$-module, then tensoring the inclusion $I\to R$ with $M$ will generically not give an injective map (as you noted)

$$\iota : M \otimes_R I \longrightarrow M \otimes_R R = M$$

and so $M / M\otimes_R I$ has no meaning, other than $M/ \iota(M\otimes_R I) = \operatorname{Coker}(\iota)$.

However, this map has image $IM$ in $M$ ---more or less by definition of what this map does--- so it is always true that the cokernel of $\iota$ is $M/MI = M\otimes_R R/I$.

Note that in your example $N=R/I$, and that you are tensoring the short exact sequence

$$0 \to I \to R \to R/I \to 0$$

with $M$. What you get is either a right exact sequence (because $-\otimes_R M$ is right exact)

$$ M\otimes_R I \to M \to M/IM \to 0$$

or, more generally, a longer (but not long, because $R$ is free) exact sequence

$$0\to \operatorname{Tor}_R^1(M,R/I) \to M\otimes_R I \to M \to M/IM \to 0$$

so in this case the kernel of the map $\iota$ is precisely $\operatorname{Tor}_R^1(M,R/I)$.