nLab bimodule

Contents

Context

Algebra

higher algebra

universal algebra

Contents

Idea

A bimodule is a module in two compatible ways over two rings.

Definition

With a left action and a right action

Given two rings $R$ and $S$, a $R$-$S$-bimodule is an abelian group $B$ with a bilinear left $R$-action $\alpha_R:R \times B \to B$ and a bilinear right $S$-action $\alpha_S:B \times S \to B$ such that for all $r \in R$, $b \in B$, and $s \in S$, $\alpha_R(r, \alpha_S(b, s)) = \alpha_S(\alpha_R(r, b), s)$.

With a biaction

Equivalently, given two rings $R$ and $S$, a $R$-$S$-bimodule is an abelian group $B$ with a trilinear $R$-$S$-biaction, a function $(-)(-)(-):R \times B \times S \to B$ such that

• for all $b \in B$, $1_R b 1_S = b$

• for all $b \in B$, $r_1 \in R$, $r_2 \in R$, $s_1 \in S$, $s_2 \in S$, $r_1 (r_2 b s_1) s_2 = (r_1 \cdot_R r_2) b (s_1 \cdot_S s_2)$

• for all $r_1 \in R$, $r_2 \in R$, $b \in B$, $s \in S$, $(r_1 + r_2) b s = r_1 b s + r_2 b s$

• for all $r \in R$, $b_1 \in B$, $b_2 \in B$, $s \in S$, $r (b_1 + b_2) s = r b_1 s + r b_2 s$

• for all $r \in R$, $b \in B$, $s_1 \in S$, $s_2 \in S$, $r b (s_1 + s_2) = r b s_1 + r b s_2$

representing simultaneous left multiplication by scalars $r \in R$ and right multiplication by scalars $s \in S$.

Properties

Biactions, left actions, and right actions

Let $R$ and $S$ be rings, and let $B$ be a $R$-$S$-bimodule.

Given a left $R$-action $\alpha_R$ and a right $S$-action $\alpha_S$ of a $R$-$S$-bimodule, the biaction $(-)(-)(-):R \times B \times S \to B$ is defined as

$r b s \coloneqq \alpha_R(r, \alpha_S(b, s)) = \alpha_S(\alpha_R(r, b), s)$

The biaction is trilinear because the left $R$-action and right $S$-action are bilinear.

On the other hand, given an $R$-$S$-biaction $\alpha$ of a $R$-$S$-bimodule, the left $R$-action is defined from the $R$-$S$-biaction as

$\alpha_R(r, b) \coloneqq r b 1_S$

for all $r \in R$ and $b \in B$. It is a left action because

$\alpha_R(1_R, b) = 1_R b 1_S = m$
$\alpha_R(r_1, \alpha_L(r_2, b)) = r_1 (r_2 b 1_S) 1_S = (r_1 \cdot_R r_2) b (1_S \cdot_S 1_S) = (r_1 \cdot_R r_2) b 1_S = \alpha_R(r_1 \cdot_R r_2, b)$

The right $S$-action is defined from the $R$-$S$-biaction as

$\alpha_S(b, s) \coloneqq 1_R b s$

for all $s \in S$ and $b \in B$. It is a right action because

$\alpha_S(b, 1_S) = 1_R b 1_S = m$
$\alpha_S(\alpha_S(b, s_1), s_2) = 1_R (1_R, b, s_1) s_2 = (1_R \cdot_R 1_R) b (s_1 \cdot_S s_2) = 1_S b (s_1 \cdot_S s_2) = \alpha_S(b, s_1 \cdot_S s_2)$

The left $R$-action and right $S$-action satisfy the following identity:

• for all $b \in B$, $r \in R$ and $s \in S$, $\alpha_R(r, \alpha_S(b, s)) = \alpha_S(\alpha_R(r, b), s)$.

This is because when expanded out, the identity becomes:

$\alpha(r, \alpha(1_R, b, s), 1_S) = \alpha(1_R, \alpha(r, b, 1_S), s)$
$(r \cdot_R 1_R) b (s \cdot_S 1_S) = (1_R \cdot_R r) b (1_S \cdot_S s)$
$r b s = r b s$

The left $R$-action and right $S$-action are bilinear because the original biaction is trilinear.

Linear maps

Let $R$ and $S$ be rings. A $R$-$S$-linear map or $R$-$S$-bimodule homomorphism between two $R$-$S$-bimodules $A$ and $B$ is an abelian group homomorphism $f:A \to B$ such that for all $a \in A$, $r \in R$, and $s \in S$,

$f(r a s) = r f(a) s$

A $R$-$S$-linear map $f:A \to B$ is monic or an $R$-$S$-bimodule monomorphism if for every other $R$-$S$-bimodule $C$ and $R$-$S$-linear maps $h:C \to A$ and $k:C \to A$, $f \circ h = f \circ k$ implies that $h = k$.

A sub-$R$-$S$-bimodule of a $R$-$S$-bimodule $B$ is a $R$-$S$-bimodule $A$ with a monic linear map $i:A \hookrightarrow B$.

A $R$-$S$-linear map $f:A \to B$ is invertible or an $R$-$S$-bimodule isomorphism if there exists a $R$-$S$-linear map $g:B \to A$ such that $g \circ f = id_A$ and $f \circ g = id_B$, where $id_A$ and $id_B$ are the identity linear maps on $A$ and $B$ respectively.

Tensor product of bimodules

Given rings $R$ and $S$, the tensor product of $R$-$S$-bimodules $A$ and $B$ is the quotient of the tensor product of abelian groups $A\otimes B$ underlying them by the $R$-$S$-biaction; that is,

$A\otimes_{R,S} B = A\otimes B / (a,r b s) \sim (r a s,b)$

Two-sided ideals of a ring

Every ring $R$ is a $R$-$R$-bimodule, with the biaction $(-)(-)(-):R \times R \times R \to R$ defined by the ternary product $a b c \coloneqq a \cdot b \cdot c$ for elements $a \in R$, $b \in R$, $c \in R$.

Given a ring $R$, a two-sided ideal of $R$ is a sub-$R$-$R$-bimodule of $R$.

Rings over a ring

Let $R$ be a ring. An $R$-ring $S$ is a $R$-$R$-bimodule with a bilinear function $(-)\cdot(-):S \times S \to S$ and an element $1 \in S$ such that $(S, \cdot, 1)$ forms a monoid.

Categories of bimodules

The 1-category of bimodules and intertwiners

Definition

Write $BMod$ for the category whose

• objects are triples $(R,S,B)$ where $R$ and $S$ are rings and where $B$ is an $R$-$S$-bimodule;

• morphisms are triples $(f,g, \phi)$ consisting of two ring homomorphisms $f \colon R \to R'$ and $g \colon S \to S'$ and an intertwiner of $R$-$S'$-bimodules $\phi \colon B \cdot g \to f \cdot B'$. This we may depict as a

$\array{ R &\stackrel{B}{\to}& S \\ {}^{\mathllap{f}}\downarrow &\Downarrow_{\phi}& \downarrow^{\mathrlap{g}} \\ R' &\stackrel{B'}{\to}& S' } \,.$
Remark

As this notation suggests, $BMod$ is naturally the vertical category of a pseudo double category whose horizontal composition is given by tensor product of bimodules.

The 2-category of rings, bimodules, and intertwiners

Consider bimodules over rings.

Proposition

There is a 2-category whose

The composition of 1-morphisms is given by the tensor product of modules over the middle algebra.

Proposition

There is a 2-functor from the above 2-category of rings and bimodules to Cat which

• sends an ring $R$ to its category of modules $Mod_R$;

• sends a $R$-$S$-bimodule $B$ to the tensor product functor

$(-)\otimes_{R} B \;\colon\; Mod_{R} \to Mod_{S}$
• sends an intertwiner to the evident natural transformation of the above functors.

Proposition

This construction has as its image precisely the colimit-preserving functors between categories of modules.

This is the Eilenberg-Watts theorem.

Remark

In the context of higher category theory/higher algebra one may interpret this as says that the 2-category of those 2-modules over the given ring which are equivalent to a category of modules is that of rings, bimodules and intertwiners. See also at 2-ring.

Remark

The 2-category of rings and bimodules is an archtypical example for a 2-category with proarrow equipment, hence for a pseudo double category with niche-fillers. Or in the language of internal (infinity,1)-category-theory: it naturally induces the structure of a simplicial object in the (2,1)-category $Cat$

$\left( \cdots \stackrel{\to}{\stackrel{\to}{\to}} X_1 \stackrel{\overset{\partial_1}{\to}}{\underset{\partial_0}{\to}} X_0 \right) \in Cat^{\Delta^{op}}$

which satisfies the Segal conditions. Here

$X_0 = Ring$

is the category of rings and homomorphisms between them, while

$X_1 = BMod$

is the category of def. , whose objects are pairs consisting of two rings $A$ and $B$ and an $A$-$B$ bimodule $N$ between them, and whose morphisms are pairs consisting of two ring homomorphisms $f \colon A \to A'$ and $g \colon B \to B'$ and an intertwiner $N \cdot (g) \to (f) \cdot N'$.

The $(\infty,2)$-category of $\infty$-algebras and $\infty$-bimodules

The above has a generalization to (infinity,1)-bimodules. See there for more.

References

The 2-category of bimodules in its incarnation as a 2-category with proarrow equipment appears as example 2.3 in

Bimodules in homotopy theory/higher algebra are discussed in section 4.3 of

For more on that see at (∞,1)-bimodule.

Last revised on December 14, 2022 at 22:35:14. See the history of this page for a list of all contributions to it.