How to prove $S \otimes_R M$ is an $S$-module. Assume $R$ a not commutative ring and $S$ a commutative ring, $f:R\to S$ is a ring homomorphism, and $M$ is an $R$-module. My question is 
Prove that $S \otimes_R M$ is an $S$-module, where $S$ -action on $S \otimes_R M$ is defined by $$ s.(s' \otimes_R m) = (ss')\otimes_R m$$
I hope that someone can help. Thanks!
 A: It is sometimes useful to consider a left $S$-module $N$ as an (additive) abelian group together with a ring homomorphism $S\to\operatorname{End}(N)$, where the codomain is the ring of endomorphisms of $N$ (as abelian group).
Thus our purpose is to define, for every $s\in S$, an endomorphism $\hat{s}\colon S\otimes_RM\to S\otimes_RM$ and then to prove that the map $s\mapsto\hat{s}$ is a ring homomorphism.
In order to define $\hat{s}$ we can use the universal property of the tensor product: first define $\tilde{s}\colon S\times M\to S\otimes_RM$ by
$$
\tilde{s}(t,x)=(st)\otimes x
$$
It's easy to verify that $\tilde{s}$ is balanced, so it uniquely defines a group homomorphism $\hat{s}\colon S\otimes_RM\to S\otimes_RM$ such that
$$
\hat{s}(t\otimes x)=(st)\otimes x
$$
Now it remains to prove that $\widehat{s_1+s_2}=\widehat{s_1}+\widehat{s_2}$ and $\widehat{s_1s_2}=\widehat{s_1}\circ\widehat{s_2}$, which are direct verifications:
\begin{align}
\widehat{s_1+s_2}(t\otimes x)
&=((s_1+s_2)t)\otimes x \\
&=(s_1t)\otimes x+(s_2t)\otimes x \\
&=\widehat{s_1}(t\otimes x)+\widehat{s_2}(t\otimes x) \\
&=(\widehat{s_1}+\widehat{s_2})(t\otimes x)
\end{align}
for the sum and
\begin{align}
\widehat{s_1s_2}(t\otimes x)
&=((s_1s_2)t)\otimes x \\
&=(s_1(s_2t))\otimes x \\
&=\widehat{s_1}\bigl((s_2t)\otimes x\bigr) \\
&=\widehat{s_1}\bigl(\widehat{s_2}(t\otimes x)\bigr) \\
&=(\widehat{s_1}\circ\widehat{s_2})(t\otimes x)
\end{align}
for the product/composition.
We just need to check the identities on elements of the form $t\otimes x$ because they generate $S\otimes_RM$.
A: I think something like this works.
First, note that $S\otimes_R M$ is generated by elements of the form $s\otimes m$, so every element of the tensor is of the form $\sum_i s_i \otimes m_i$. 
So our would-be module action of $S$ on $S\otimes_R M$ is:
$$p \bullet \left(\sum_i s_i \otimes m_i\right) = \sum_i p s_i \otimes m_i \ \forall p\in S.$$
By construction, we immediately get
\begin{align*}p\bullet \left( \sum_i s_i \otimes m_i + \sum_j s_j \otimes m_j \right)&=\sum_i ps_i \otimes m_i  + \sum_j ps_j \otimes m_j \\ &= p\bullet \sum_i s_i \otimes m_i + p\bullet \sum_j s_j \otimes m_j.
\end{align*}
Next, we have
\begin{align*}
(p+q) \bullet \left(\sum_i s_i \otimes m_i \right) &= \sum_i (p+q)s_i \otimes m_i \\
&=\sum_i(ps_i + qs_i) \otimes m_i \\
&=\sum_ips_i \otimes m_i + \sum_i qs_i \otimes m_i \\
&=p\bullet \sum_i s_i \otimes m_i + q\bullet \sum_i s_i \otimes m_i.
\end{align*}
\begin{align*}
(pq)\bullet \sum_i s_i \otimes m_i &=\sum_i (pq)s_i \otimes m_i \\
&=p\bullet \sum_i qs_i \otimes m_I \\
&=p\bullet \left( q\bullet \sum_i s_i \otimes m_i \right)
\end{align*}
