Ring Bimodule

Let $R,S$ be a pair of Rings. An R-S-bimodule is an Abelian Group $M$ such that:

• $M$ is a Left R-Module
• $M$ is a Right S-Module
• For all $r:R,x:M,s:S$, $(r\cdot x)\cdot s=r\cdot(x\cdot s)$

Note that this is the ring-theoretic analog of a Profunctor, or, more accurately, a Two-Sided Fibration.

Examples

• Every Ring $R$ is an $(R,R)$-bimodules, with addition given by $R$, and left and right actions given by multiplication in $R$; the compatibility condition here just reduces to Associativity of multiplication.
• Every Module over a Commutative Ring is a bimodule. This means that many theorems about modules over Commutative Rings are best viewed as theorems about bimodules when we pass to non-commutative settings.
• Every Left R-Module $M$ is a $(R,\mathbb{Z})$-bimodule, as every abelian group is canonically a right $\mathbb{Z}$-module, and the compatibility condition reduces to distributivity of the left action of $R$ on $M$. A dual result holds for Right R-Modules.
• A bimodule over the Ring of Integers is an Abelian Group.

Via Ternary Actions

We can equivalently present an R-S bimodule $M$ as an Abelian Group with a trilinear Ternary Operation $(-)\cdot(-)\cdot(-):R\times M\times S\to M$ such that

• $1\cdot x\cdot 1=x$
• $r_{1}\cdot(r_{2}\cdot x\cdot s_{1})\cdot s_{2}=(r_{1}r_{2})\cdot x\cdot(s_{1}s _{2})$
• $(r_{1}+r_{2})\cdot x\cdot s=r_{1}\cdot x\cdot s+r_{2}\cdot x\cdot s$
• $r\cdot(x_{1}+x_{2})\cdot s=r\cdot x_{1}\cdot s+r\cdot x_{2}\cdot s$
• $r\cdot x\cdot(s_{1}+s_{2})=r\cdot x\cdot s_{1}+r\cdot x\cdot s_{2}$

This alternative presentation encodes the left and right actions simultaneously, which avoids the interchange law. This is also closer to the Profunctorial point of view.