An $(R,S)$-bimodule is a left $R \otimes_k S^{\text{op}}$-module Let $k$ be a commutative ring, and let $R,S$ be $k$-algebras. To me "$R$ is a $k$-algebra" means that $R$ is a $k$-module such that $a(rs)=(ar)s=r(as)$ for all $a\in k$ and $r,s \in R$. Let $M$ be a $(R,S)$-bimodule. I am trying to show that $M$ is a  left $R \otimes_k S^{\text{op}}$-module.
Define a map $\phi \colon R \times S^{\text{op}} \to M$ given by $\phi(r,s) = (rm)s$. I believe that for me to progress with the proof this map has to have the property that $\phi (ar,s) = \phi (r, as)$ for $a \in k$. However $\phi(ar,s) = [(ar)m]s$ and $\phi(r, as) = (rm)(as)$. However I have no idea how to get that $a$ to pass to the other side of the $m$. How is this achieved? Is there implicitly a $k$-module structure on $M$? 
 A: We assume $R$ and $S$ are unital.
Hence there exists the canonical homomorphism $k \rightarrow R$.
Hence by restricting the actions of $R$ on $M$ to $k$, $M$ can be a $k$-module.
Similarle $M$ can be a $k$-module through $S$.
We assume the both $k$-module structures on $M$ coincide. 
Let $T = End_k(M)$ be the $k$-algebra of $k$-endomorphisms on $M$.
For $r \in R$, we denote by $\lambda(r) \in T$ the endomorphism $x \rightarrow rx$.
For $s \in S$, we denote by $\rho(s) \in T$ the endomorphism $x \rightarrow xs$.
$\lambda\colon R \rightarrow T$ and $\rho\colon S^{\text{op}} \rightarrow T$ are $k$-algebra homomorphisms.
Define a map $\phi \colon R \times S^{\text{op}} \to T$ by $\phi(r, s) = \lambda(r)\rho(s)$.
For $a \in k$,
$a\lambda(r)\rho(s) = \lambda(ar)\rho(s) = \lambda(r)\rho(as)$.
Hence $a\phi (r,s) = \phi (ar, s) = \phi (r, as)$.
Hence there exists a $k$-linear map $f\colon R \otimes_k S^{\text{op}} \rightarrow T$ such that $f(r\otimes s) = \lambda(r)\rho(s)$.
Since $\lambda(r)\rho(s) = \rho(s)\lambda(r)$, $f$ is a $k$-algebra homomorphism.
Hence $f$ makes $M$ into a left $R \otimes_k S^{\text{op}}$-module.
