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 $M$?

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When $R$ and $S$ are $k$-algebras, an $(R,S)$-bimodule is supposed to be $k$-symmetric, that is, the actions of $k$ on the left and on the right induced by restriction from those of $R$ and $S$ coincide. If it isn't, then it is not true that it is a $R\otimes_kS$-module in a sensible way: take $R=S=k=\mathbb Z[X]$ and $M=\mathbb Z$ with $R$- and $S$-module structures which are different in the way $X$ acts. – Mariano Suárez-Alvarez Aug 21 '12 at 19:22
what do you mean supposed to be? should we include $k$-symmetric structure in the hypothesis? – Paul Slevin Aug 22 '12 at 13:21
my book Rotman just claims that every $(R,S)$-bimodule $M$ is a left $R \otimes_k S$-module – Paul Slevin Aug 22 '12 at 14:33

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.

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"we assume both $k$ module structures coincide" is this part of the hypotheses? otherwise why is it true? – Paul Slevin Aug 22 '12 at 12:04
@PaulSlevin , you need to assume this as part of the hypothesis, otherwise your claim is not true. This is what Mariano Suarez-Alvarez was saying above. – Manny Reyes Aug 22 '12 at 14:54
Thanks for clarifying. There must be a mistake in my book. – Paul Slevin Aug 22 '12 at 14:59