When is $(M) / (M ⊗ I)$ $\simeq$ to $M/(MI)$? Let $N$ be a R-Module isomorphic to $R/ker(\pi)$ with $\pi: R \rightarrow N$ the surjection that associate to each element of N an element of the canonical basis of R.
One has the following exact sequence: $0 \rightarrow ker(\pi) \rightarrow R \rightarrow N \rightarrow 0 $
Let $M$ be a R-Module.
Tensoring by $M$ gives the exact sequences : $M⊗ker(\pi) \rightarrow M⊗R \rightarrow M⊗N \rightarrow 0 $
$1⊗u: M⊗R \rightarrow M⊗N$
Then one has: $M⊗N$ $\simeq$ $(M⊗R)/ker(1⊗u)$ $\simeq$ $(M⊗R)/Im(1⊗i)$ not necessary $\simeq$ $(M⊗R)/(M⊗ker(\pi))$ as $1⊗i$ is not necessary injective.
Otherwise it would mean that: (making the same reasoning for $N$ $\simeq$ to $R/I$ with $I$ an ideal of R.
M/(M⊗I) $\simeq$ $(M ⊗ R) / (M ⊗ I)$ ?$\simeq$? $M ⊗ (R/I)$ $\simeq$ $(M/MI)$
This last isomorphism if it stands seems really weird to me: M/(M⊗I) $\simeq$ $(M/MI)$
As I know that if M is flat, free, or projective then $1⊗i$ is injective and so one obtains the isomorphism.
Can someone give me concrete examples and counterexamples or tell me if I made a mistake in the reasoning please ? I need to be more familiar with this, but it seems really strange to me.
 A: 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 $\operatorname{Tor}_R^1(M,R/I)$.
