Tensor products: proving that $I \otimes_R M \cong IM$ Assume it if it´s neccesarly that the ring has an 1 or is commutative ( I´m not sure if it´s needed)
Given a ring $R$ an ideal $I$ of $R$, and a $R$ module $M$ , prove that:
 $
I \otimes _R M \cong IM
$
 where $
IM = \left\{ {x \in M:x = \sum\limits_{finite} {i_k m_k \,\,\,i_k  \in I\,\,m_k  \in M} } \right\}
$
This is what I did. First I defined the obvious function $
\varphi\colon I\times M \to \,IM
$ which is bilinear, so it defines a R-module-homomorphism $$
\varphi ^ \bullet  \colon I \otimes _R M \to IM
$$
and satisfies $
\varphi ^ \bullet  \left( {i \otimes m} \right) = \varphi \left( {i,m} \right) = im
$
I proved that $
\varphi ^ \bullet  
$ is surjective since, given $
\sum\limits_{finite} {i_k m_k }  \in IM
$ clearly $
\varphi ^ \bullet  \left( {\sum\limits_{finite} {i_k  \otimes m_k } } \right) = \sum\limits_{finite} {i_k m_k } 
$
But the injectivity how can I prove it?

"But the injectivity how can I prove it?"
 Dear Susuk, you can't because it is false and it is a good omen for you  that you couldn't prove it!
  Here is a counterexample:     
Let $k$ be any field . Consider   $R=k[X]/(X^2)=k[\epsilon]$ and let $I$ be the ideal $I=(\mathbb \epsilon)=k\cdot  \epsilon \subset R$ .
 Take $M=I$.   We have $I\cdot M=I^2=(0)$ and in order to show that your map $
\varphi ^ \bullet  :I \otimes _R I \to I^2$ is not injective,  it suffices to prove that $I \otimes _R I\neq 0$.
However, since $I$ is killed by $\epsilon$ we have $I \otimes _R I=I \otimes _{R/(\epsilon)} I=I \otimes _k I$ and the latter vector space is one dimensional over the field $k$, hence non-zero.
(Of course if $M$ is flat over $R$, the isomorphism $I \otimes _R M \stackrel {\simeq} {\to}IM $ holds)
