Given a group homomorphism $R \otimes_{\Bbb{Z}} M \rightarrow M$, I need to show that this makes $M$ into a left $R$-module. 
Given a group homomorphism $R \otimes_{\Bbb{Z}} M \rightarrow M$, I need to show that this makes $M$ into a left $R$-module. 

We can see that
$\bullet$ $r(m+m') = r \otimes (m+m') = r \otimes m + r \otimes m' = rm + rm'$
$\bullet$ $(r+r')m = (r+r') \otimes m = r \otimes m + r' \otimes m = rm + r'm$
$\bullet$ $(rr')m = rr' \otimes m = r \otimes r'm = r(r'm)$
$\bullet$ $1m = 1 \otimes m = m$
I have a question about the third one, where I used $rr' \otimes m = r \otimes r'm$. Is this valid? Because we are tensoring over $\Bbb{Z}$ and not over $R$, right? 
Thanks in advance 
 A: I don't think this is true, take for instance $R=\Bbb Z[X]$, then you can define a morphism $\Bbb Z[X]\otimes_{\Bbb Z}M$ by sending $P\otimes m$ to $$\left(\sum_{i\geq 0} a_ic_i\right)m$$ where $(c_n)_{n\in\Bbb N}$ is some sequence of integers, and $P=\sum_{i\geq 0}a_iX^i$. This, in general, does not define a $\Bbb Z[X]$-module structure on $M$.

Actually, there is no reason to assume a general map of groups should map $1\otimes m$ to $m$ either. For instance, this fails when $R=\Bbb Z/2\Bbb Z$ and $M=\Bbb Z/3\Bbb Z$, since $\Bbb Z/2\Bbb Z\otimes_{\Bbb Z}\Bbb Z/3\Bbb Z=0$.
A: One could try with $rm=\varphi(r\otimes m)$, where $\varphi\colon R\otimes_{\mathbb{Z}}M\to M$ is the given group homomorphism.
Then $(r+s)m=\varphi((r+s)\otimes m))$, $r(m+n)=\varphi(r\otimes(m+n))$ don't pose problems.
The problem is with
$$
r(sm)=\varphi(r\otimes(sm))=\varphi(r\otimes(\varphi(s\otimes m)))
$$
while
$$
(rs)m=\varphi((rs)\otimes m))
$$
So, without some more hypotheses on $\varphi$, I'd say this is not true.
