Feeds:
Posts

## Morita equivalence and the bicategory of bimodules

In the previous post we learned that it is possible to recover the center $Z(R)$ of a ring $R$ from its category $R\text{-Mod}$ of left modules (as an $\text{Ab}$-enriched category). For commutative rings, this justifies the idea that it is sensible to study a ring by studying its modules (since the modules know everything about the ring).

For noncommutative rings, the situation is more interesting. Two rings $R, S$ are said to be Morita equivalent if the categories $R\text{-Mod}, S\text{-Mod}$ are equivalent as $\text{Ab}$-enriched categories. As it turns out, there exist examples of rings which are non-isomorphic but which are Morita equivalent, so Morita equivalence is a strictly coarser equivalence relation on rings than isomorphism. However, many important properties of a ring are invariant under Morita equivalence, and studying Morita equivalence offers an interesting perspective on rings on general.

Moreover, Morita equivalence can be thought of in the context of a fascinating larger structure, the bicategory of bimodules, which we briefly describe.

Fiber functors

Suppose that, in addition to the $\text{Ab}$-enriched category $R\text{-Mod}$, we remember the forgetful functor $R\text{-Mod} \to \text{Ab}$. This functor is representable by $R$, so by the Yoneda lemma we conclude that its monoid of natural endomorphisms is $\text{End}_R(R) \cong R^{op}$; in other words, it is indeed possible to recover $R$ if we know in addition what every left $R$-module looks like as an abelian group.

For the rest of this discussion, we won’t assume we’re given a forgetful functor.

Some preliminary categorical remarks

When we defined a 2-category in the previous post, we implicitly required that the functor $\circ : \text{Hom}(A, B) \times \text{Hom}(B, C) \to \text{Hom}(A, C)$ be associative on the nose. However, many natural examples of categories that one would want to be 2-categories don’t satisfy this requirement. For example, the monoidal category $(\text{Vect}, \otimes)$ ought to be a 2-category with one object, except that the tensor product isn’t strictly associative: we don’t literally have an equality

$\displaystyle \displaystyle (U \otimes V) \otimes W = U \otimes (V \otimes W)$

but rather isomorphisms called associators:

$\displaystyle \alpha_{U, V, W} : (U \otimes V) \otimes W \cong U \otimes (V \otimes W)$.

Moreover, the associators $\alpha_{U, V, W}$ satisfy some a priori complicated compatibility relations, since it ought to be the case that any two ways of moving between two different parenthesizations of a tensor product ought to be given by the same isomorphism. We also need to include some compatibility maps to describe the identity object. As it turns out, two compatibility relations called the pentagon identity and triangle identity generate all the other compatibility relations we want, giving us a weak monoidal category. The adjective “weak” is used whenever we wish to explicitly distinguish between the weak and “strict” version, although

1. the term “monoidal category” without qualification seems to more commonly refer to the weak version, and
2. monoidal categories have a natural notion of equivalence (monoidal equivalence) with respect to which every monoidal category turns out to be monoidally equivalent to a strict monoidal category (I believe this result is due to Mac Lane).

Generalizing to 2-categories, we obtain the notion of a bicategory, which is widely regarded as the “correct” next step after ordinary categories. As above, however, there is a natural notion of equivalence (biequivalence) between bicategories, and every bicategory is biequivalent to a 2-category.

The upshot of all of this is that later in this post we will define a bicategory and pretend that it is a 2-category, and nothing too horrible will happen to us.

Some elementary observations about Morita equivalence

Given a ring $R$, what can we say about recovering $R$ from $R\text{-Mod}$? As a first observation, if we could isolate the module $R \in R\text{-Mod}$, then we know that $\text{End}_R(R) \cong R^{op}$, so we would be done. The problem is that there’s no obvious way to isolate $R$! Abstractly, $R$ has the property that the functor

$\displaystyle \text{Hom}(R, -) : R\text{-Mod} \to \text{Ab}$

is faithful; that is, $R$ is a generator. However, this is also true of the modules $R^n, n \in \mathbb{N}$, among others.

If $R\text{-Mod}$ happens to be isomorphic to $S\text{-Mod}$ for some other ring $S$, it follows that there should exist a generator $M \in R\text{-Mod}$ (the image of $S$ under an isomorphism $S\text{-Mod} \to R\text{-Mod}$) such that $\text{End}_R(M) \cong S^{op}$.

$M$ is in particular an abelian group with a left action of $R$ and a right action of $S$ such that these actions are compatible (that is, the left action is a right $S$-module homomorphism and vice versa); such a thing is called an $(R, S)$bimodule. One usually writes $_RM_S$ to emphasize the left and right actions on $M$.

Moreover, we have $\text{End}_R(M) \cong S^{op}$ and, dually, $\text{End}_S(M) \cong R$, so $M$ is in addition a faithfully balanced bimodule. (If we only have that the natural maps $S^{op} \to \text{End}_R(M)$ and $R \to \text{End}_S(M)$ are surjective, we call $M$ balanced).

Hom and tensor

To find some nontrivial examples of Morita equivalences, we should find a large supply of functors between module categories. A simple such collection of functors is as follows: if $f : R \to S$ is a ring homomorphism, then precomposition by $f$ defines a functor $S\text{-Mod} \to R\text{-Mod}$ called restriction of scalars. In order for this to be an equivalence of categories, $S$ (as an $(R, S)$-bimodule) needs to be faithfully balanced, so we need the natural map $R \to \text{End}_S(S) \cong S$ to be an isomorphism. However, this map is just $f$, so if this map is an isomorphism then $R, S$ are already isomorphic, and we get no nontrivial examples of Morita equivalence this way.

A more promising option is to think about Hom functors. Let $_RM_S$ be an $(R, S)$-bimodule; any such bimodule defines a Hom functor

$\displaystyle \text{Hom}_R(_RM_S, -) : R\text{-Mod} \to S\text{-Mod}$

where $S$ acts by precomposition. If this functor is supposed to be an equivalence, then it must in particular have a left adjoint (which will be its inverse if it exists). Let $U : S\text{-Mod} \to R\text{-Mod}$ be such a left adjoint, if it exists; then we need an isomorphism

$\displaystyle \text{Hom}_R(U(_SN), _RP) \cong \text{Hom}_S(_SN, \text{Hom}_R(_RM_S, _RP))$.

The RHS describes the set of maps $f : M \times N \to P$ which are $R$-linear in the first variable, linear in the second variable, and which identify the two actions of $S$ in the sense that $f(ms, n) = f(m, sn)$. This is reminiscent of the universal property of the tensor product, and indeed we can define a functor $U(-) = M \otimes_S - : S\text{-Mod} \to R\text{-Mod}$, the tensor product functor over $S$, with the desired property. Explicitly, $M \otimes_S N$ is the abelian group spanned by formal symbols $m \otimes n, m \in M, n \in N$ subject to the relations

1. $(m_1 + m_2) \otimes n = m_1 \otimes n + m_2 \otimes n$,
2. $m \otimes (n_1 + n_2) = m \otimes n_1 + m \otimes n_2$,
3. $(ms) \otimes n = m \otimes (sn)$.

Note that these axioms only refer to the structure of $M$ as a right $S$-module and $N$ as a left $S$-module, but it’s not hard to see that the tensor product is a covariant functor in both $M, N$, so the left $R$-modules tructure on $M$ extends to the tensor product. Our desired tensor-Hom adjunction then takes the form (getting rid of some subscripts to improve readability)

$\displaystyle \text{Hom}_R(M \otimes_S N, P) \cong \text{Hom}_S(N, \text{Hom}_R(M, P))$.

It follows in particular that $\text{Hom}_R(_RM_S, -) : R\text{-Mod} \to S\text{-Mod}$ preserves limits and $_RM_S \otimes_S - : S\text{-Mod} \to R\text{-Mod}$ preserves colimits.

Note that the tensor functor sends $S$ to $_RM_S$, hence $_RM_S$ must be faithfully balanced and a generator in order for $_RM_S \otimes_S -$ to be an equivalence. Note also that if $f : R \to S$ is a ring homomorphism, then $\text{Hom}(_SS_R, -) : S\text{-Mod} \to R\text{-Mod}$ is restriction of scalars.

Example. Let $R$ be a commutative ring, $G$ a group, and $R[G]$ the group ring of $G$ over $R$. If $H$ is a subgroup of $G$, then restriction of scalars gives a functor $R[G]\text{-Mod} \to R[H]\text{-Mod}$, and by the above it has a left adjoint

$\displaystyle R[G] \otimes_{R[H]} - : R[H]\text{-Mod} \to R[G]\text{-Mod}$

which abstractly describes induction of representations from $H$ to $G$.

The bicategory of bimodules

If $_RM_S$ is an $(R, S)$-bimodule and $_SN_T$ is an $(S, T)$-bimodule, then the tensor product $M \otimes_S N$ acquires by functoriality the structure of an $(R, T)$-bimodule; moreover, there are natural associativity isomorphisms which can be proven, like in the case of the tensor product of vector spaces, by verifying a corresponding universal property for trilinear maps.

This suggests the following definition. The bicategory of bimodules $\text{Bim}$ is the bicategory whose objects are rings and where $\text{Hom}_{\text{Bim}}(R, S)$ is the category of $(R, S)$-bimodules. Composition $\text{Hom}(R, S) \times \text{Hom}(S, T) \to \text{Hom}(R, T)$ is given by the tensor product over $S$, and the identity $\text{id}_R \in \text{Hom}(R, R)$ is $R$ as an $(R, R)$-bimodule.

This bicategory turns out to be a natural setting to think about Morita equivalence. First, we need the following definition: two objects $R, S$ in a bicategory are equivalent if there exist morphisms $f : R \to S, g : S \to R$ such that $fg \cong \text{id}_S$ (in the category $\text{Hom}(S, S)$) and $gf \cong \text{id}_R$ (in the category $\text{Hom}(R, R)$).

Example. In the 2-category $\text{Cat}$ of categories, functors, and natural transformations, the above reduces to the usual definition of equivalence of categories.

Example. In the 2-category of topological spaces, continuous functions, and homotopy classes of homotopies between continuous functions, the above reduces to the usual definition of homotopy equivalence.

If $R, S$ are equivalent objects in a bicategory (written $R \cong S$), we get natural equivalences $\text{Hom}(R, T) \cong \text{Hom}(S, T)$ for every object $T$. In particular, if $R, S$ are rings in $\text{Bim}$ and $T = \mathbb{Z}$, then an equivalence between $R$ and $S$ defines an equivalence of categories $R\text{-Mod} \to S\text{-Mod}$; in other words, $R, S$ are Morita equivalent! (The converse also turns out to be true, which is a great reason to study the bicategory of bimodules if we’re interested in Morita equivalence, but we won’t prove or use this fact yet.)

Thus to exhibit nontrivial examples of Morita equivalence, it suffices to find a pair of rings $R, S$ and a pair of bimodules $_RM_S, _SN_R$ such that the functors

$(_RM_S) \otimes_S (_SN_R \otimes -) : R\text{-Mod} \to R\text{-Mod}$

and

$(_SN_R) \otimes_R (_RM_S \otimes -) : S\text{-Mod} \to S\text{-Mod}$

are naturally isomorphic to identity functors. Of course, we didn’t need to set up the bicategory of bimodules to observe this, but nevertheless the bicategory of bimodules is the most natural setting for it. Now, thanks to the associativity of the tensor product, the tensor-Hom adjunction, and the (enriched) Yoneda lemma, it is necessary and sufficient that we have isomorphisms

$(_RM_S) \otimes_S (_SN_R) \cong _RR_R, (_SN_R) \otimes_R (_RM_S) \cong _SS_S$.

Matrix rings

We are now in a position to provide our first nontrivial examples of Morita equivalence.

Proposition: Let $R$ be an arbitrary ring, let $S = \mathcal{M}_k(R)$, let $M = R^k$ (row vectors) regarded as an $(R, \mathcal{M}_k(R))$-bimodule, and let $N = R^k$ (column vectors, again) regarded as an $(\mathcal{M}_k(R), R)$-bimodule. Then the above isomorphisms hold. Thus $R \cong \mathcal{M}_k(R)$.

Proof. This is actually a fairly straightforward exercise in matrix multiplication. We first fix some notation. Let $e_1, ... e_k$ be the standard basis of $R^k$ and let $e_{ij} \in \mathcal{M}_k(R)$ be those matrices with only one nonzero entry such that $e_{ij} e_j = e_i$ and such that $e_i e_{ij} = e_j$.

By bilinearity, $M \otimes_{\mathcal{M}_k(R)} N$ is generated as an $(R, R)$-bimodule by elements of the form $e_i \otimes e_j$. Since we can write this as $e_j e_{ji} \otimes e_j = e_j \otimes e_{ji} e_j$, we conclude that these elements are zero unless $i = j$. We can also write $e_i \otimes e_i = e_j e_{ji} \otimes e_i = e_j \otimes e_{ji} e_i = e_j \otimes e_j$, so all of these elements are equal and the entire module is generated by a single element $e_1 \otimes e_1$. Finally, note that defining $\langle e_i, e_j \rangle = \delta_{ij}$ (matrix multiplication of row and column vectors) defines a bilinear map $R^k \times R^k \to R$ which correctly respects all of the relevant actions, and we obtain the first desired isomorphism.

Similarly, $N \otimes_R M$ is generated as an $(\mathcal{M}_n(R), \mathcal{M}_n(R))$-bimodule by elements of the form $e_i \otimes e_j$. Since we can write this as $e_{i1} e_1 \otimes e_1 e_{1j} = e_{i1} (e_1 \otimes e_1) e_{1j}$, the entire module is generated by a single element $e_1 \otimes e_1$. Finally, note that defining $\langle e_i, e_j \rangle = e_{ij}$ (matrix multiplication of column and row vectors) defines a bilinear map $R^k \times R^k \to \mathcal{M}_k(R)$ which correctly respects all of the relevant actions, and we obtain the second desired isomorphism.

Corollary: Any semisimple ring is Morita equivalent to a finite direct product of division rings.

Proof. This follows from the above proposition and Artin-Wedderburn.

### 3 Responses

1. Great post, thanks. Do you have a plan for future posts on rings, or at least some topics you wish to treat?

Also, Morita equivalence is coarser than isomorphism.

• Well, if I get really adventurous, then at some point down the line I’ll talk about the derived bicategory of bimodules…

2. […] (not necessarily commutative) ring does not uniquely determine , since for example we always have Morita equivalences of the […]