Generalizing the notion of a module, we can consider an abelian group endowed with both a left- and right-action. Such a structure is called a bimodule.

Definition of bimodule

Let $R$ and $S$ be rings (with unity). An $(R,S)$-bimodule consists of an abelian group $M$ with both the structure of a left $R$-module and right $S$-module, such that $r(ms)=(rm)s$ for every $r\xe2\x88\x88R$, $s\xe2\x88\x88S$, and $m\xe2\x88\x88M$.

This might seem like a lot to expect out of an abelian group, but it turns out that there are at least as many bimodules as there are modules, as the following examples illustrate.

Examples of bimodules

Abelian groups are $(\mathbf{Z},\mathbf{Z})$-bimodules

Every abelian group $A$ has a unique $(\mathbf{Z},\mathbf{Z})$-module structure, since there is a unique left $\mathbf{Z}$-action on $A$ as well as a unique right $\mathbf{Z}$-module structure. As an illustration, given an element $a\xe2\x88\x88A$, the action of the integer $2$ on the left is (and must be!) $2\xe2\x8b\x85a=a+a$, while the action of the integer $\xe2\x88\x923$ on the right is (and must be!) $a\xe2\x8b\x85(\xe2\x88\x923)=\xe2\x88\x92a\xe2\x88\x92a\xe2\x88\x92a.$ These two actions satisfy the required "associativity" condition:

We will freely switch between referring to $A$ as an abelian group and as a $(\mathbf{Z},\mathbf{Z})$-bimodule, according to whatever is most appropriate in the current context.

From modules to bimodules

Left $R$-modules are $(R,\mathbf{Z})$-bimodules

Similar to the previous example, suppose $M$ is any left $R$-module. Then there is a unique right $\mathbf{Z}$-action on $M$ (coming directly from the abelian group structure on $M$), and this gives $M$ the structure of an $(R,\mathbf{Z})$-bimodule. Analogously, every right $R$-module $M$ also has a unique $(\mathbf{Z},R)$-bimodule structure.

As with abelian groups, we will freely move between referring to $M$ as a left $R$-module (resp., right $R$-module) and as a $(R,\mathbf{Z})$-bimodule (resp., $(\mathbf{Z},R)$-bimodule).

From $R$-modules to $(R,R)$-bimodules

Suppose $R$ is a commutative ring and $M$ is a left $R$-module. We can define a right $R$-action on $M$ by $m\xe2\x8b\x85r=rm$. Note that the "associativity" condition of the ring action requires^{[1]} the commutativity of $R$ in order for this to truly define a right $R$-action on $M$, since in general we only have

The commutativity in $R$ ensures the last term is equal to $({r}_{1}{r}_{2})m=m\xe2\x8b\x85({r}_{1}{r}_{2})$, as required. Of course, we can do the same for any right $R$-module, $M$.

We call this $(R,R)$-bimodule structure on $M$ the standard $(R,R)$-bimodule structure on the $R$-module $M$.

Note

We should really be defining a functor from the category of $R$-modules to the category of $(R,R)$-bimodules.

From bimodules to modules

For any $(R,S)$-bimodule $M$ we can always forget either the left $R$- or right $S$-action, resulting in a right $S$- or left $R$-module, respectively. Although technically we should define a forgetful functor from one category to another, in practice most mathematicians usually just wave their hands and simply say "considered only as a left $R$-module" and what not.

Note that this gives us a way to talk about morphisms between bimodules that do not have any actual bimodule morphisms between them; e.g., a morphism $M\xe2\x86\x92N$ where $M$ is an $(R,S)$-bimodule and $N$ is an $(R,T)$-bimodule. Of course, if we were being carefully we would note that we're actually talking about a morphism ${U}_{1}(M)\xe2\x86\x92{U}_{2}(N)$ in the category of left $R$-modules, where ${U}_{1}$ and ${U}_{2}$ are the appropriate forgetful functors.

Rings and bimodules

From a ring to a bimodule

Suppose $R$ is a ring (with unity). Then $R$ acts on itself on the left (and the right) by multiplication. This gives $R$ the structure of an $(R,R)$-bimodule. As with all bimodules, we can forget part of this action and retain an $(R,\mathbf{Z})$-bimodule structure (i.e., a left $R$-module structure), a $(\mathbf{Z},R)$-bimodule structure (i.e., a right $R$-module structure), or a $(\mathbf{Z},\mathbf{Z})$-bimodule structure (i.e., an abelian group structure).

As usual, all of this forgetting is really just using the appropriate forgetful functors.

Ring homomorphisms create bimodule structures

Suppose $f:R\xe2\x86\x92S$ is a ring morphism (between rings with unity, and so we also assume $f({1}_{R})={1}_{S}$). We can then endow $S$ with an $(S,R)$-bimodule structure, using the left action of $S$ on itself by left multiplication, and the right action of $R$ on $S$ by $s\xe2\x8b\x85r=sf(r)$.

Quotient rings have a natural bimodule structure

Suppose $R$ is a ring and $I\xe2\x8a\x86R$ is a two-sided ideal. By considering the canonical projection morphism $\mathrm{\u012a\x80}:R\xe2\x86\x92R/I$, the previous example gives the quotient ring $R/I$ the structure of an $(R/I,R)$-bimodule.