Morita Equivalence
Eamon Quinlan

Given a (not necessarily commutative) ring, you can form its category of right modules. Take this category
and replace the names of all the modules with dots. The resulting category is a bunch of dots with a bunchof arrows. The question is: what can you then say about the original ring from this category of dots? Thestudy of this question leads to the notion of Morita equivalence, two rings being Morita equivalent if they
have equivalent categories of right modules. The question then becomes: what properties are preserved
under Morita equivalence? In this talk we will explore this and some other related questions, as well as
explain a theorem of Morita that gives a very nice criterion for when two rings are Morita equivalent.

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Introduction

Notation: rings always have 1, but need not be commutative. Given a ring A, we denote by A − Mod (resp.
Mod −A) the category of left (resp. right) A-modules. Two modules will be considered equal if they areisomorphic.
As in the abstract, suppose we are given a category C which we know is equivalent to the category Mod −A∼via a functor F : Mod −A −→ C. What information about A can we obtain purely based on C? Another way
to think about this question is: suppose we start with our category Mod −A, mod out by isomorphism, anderase all the names of the modules and replace them with a symbol like •. Now all we have left is a bunch of•’s together with a lot of arrows between them; call this category C. What can we recover about our original
category Mod −A, and more generally our original ring A, from this new category?
Exercise 1.1.- (Easy if familiar with some homological algebra, hard otherwise)(i) Show you can find which • corresponded to the zero module.
(ii) Show you can determine which arrows were injective – we can therefore make sense of what subdots
are in our new category of dots.
(iii) Show you can determine which arrows were surjective – we can therefore make sense of what quotient
dots are in our new category of dots.
(iv) Given a (two-ended) chain of subdots of a given dot, we can find the dot corresponding to the union.
(v) Given a (two-ended) chain of subdots of a given dot, we can find the dot corresponding to the intersection.
(vi) Given a dot, we can determine whether its corresponding module was right noetherian.
(vii) Given a dot, we can determine whether its corresponding module was right artinian.
Exercise 1.2.- Show
S that an A-module M is finitely generated if and only if given any chain 0 ⊆ N0 ⊆N1 ⊆ · · · ⊆ M with i∈N Ni = M we have NK = M for some K  0. We can thus determine which dots
corresponded to the finitely generated modules.

1Exercise 1.3.- Show that the following properties of A can be deduced from C.
(i) Whether A is right noetherian. [Recall: A is right noetherian if and only if all finitely generated
A-modules are right noetherian; one can then use 1].
(ii) Whether A is right artinian. [Similar to previous].
(iii) The center of A, up to isomorphism. [Hint: the center of A will be isomorphic to the endomorphismring of the identity functor on Mod −A]. Thus, if we knew A was commutative to start with, we would
be able to recover all of A.
(iv) The characteristic of A.
Exercise 1.4.- The following example shows that not all information can be recovered from this categoryof dots. For this fix a ring R and let Mn (R) be the ring of n by n matrices over R. Show:(i) Given an right R-module M there is a natural right action of Mn (R) on M ⊕n coming from the actionof R.
(ii) This construction actually gives a functor

Mod −R −→ Mod −Mn (R)that is actually an equivalence [Hint: show the functor is fully faithful and essentially surjective toprove it is an equivalence].
(iii) Observe that then after applying our “dot” construction to the rings R and Mn (R) from before get thesame category! For example R could be C and then we would not be able to distinguish whether westarted from the ring C or from the ring Mn (C). As a consequence, whether the ring we started withis commutative cannot be inferred from the category of dots.
(iv) Finally, observe that this example provides a proof that Z(Mn (C)) ∼= C.

2Formalizing the problem

Let’s now try to make the discussion more rigorous. As we have already mentioned or hinted at, whether
two rings R and S do or don’t have the same “category of dots” is in fact testing whether the categories
Mod −R and Mod −S are equivalent. We thus make the following definition.
Definition 2.1.- Two rings R and S are Morita equivalent if the categories Mod −R and Mod −S areequivalent. We denote this by R ∼ S (not standard!). A property P is Morita-invariant if whenever R has
P and R ∼ S we have that S has P .
What we have thus proven is that being right noetherian, right artinian, and having center isomorphic to Z(for a fixed commutative ring Z) are all Morita equivalent properties. We should also remark that in thecase where R and S are commutative, R ∼ S if and only if R ∼= S.
Another important remark is that at this point the relation ∼ should really be called something like “right”Morita equivalence, to reflect the fact that we are testing whether the categories of “right”-modules areequivalent. If you don’t mind spoilers, see Theorem 5.1 to address this concern.

23Bimodules associated to an equivalence

From now on, most of the discussion is extracted from [AF].
We would now like to work towards a criterion due to Morita for when two rings are Morita equivalent. Firstobserve that one of the most fundamental problems of not being able to reconstruct A from Mod −A comesfrom the fact that when we forget the “names” of the modules we can no longer recover where A (as a right∼
A-module) was (c.f. 1.4). In other words, given an equivalence F : Mod −A −→ Mod −B we need not havethat F (A) ∼= B.
However, the right A-module A has some very special properties that do get transferred to the B-module
F (A).
Definition 3.1.- An A-module G is a generator if one of the following equivalent conditions hold:(i) Given two morphisms f 6= g : M → N of right A-modules there exists a morphism h : A → M withthe property that f ◦ h 6= g ◦ h.
(ii) The functor Mod −A → Sets, (M 7→ HomA (G, M )) is faithful.
(iii) Every A-module M is a quotient of a (possibly infinite) direct sum of G’s.
An A-module Q is projective if the following equivalent conditions hold:(i) The A-module Q is a direct summand of a free A-module.
(ii) Given any surjection N → M → 0 of A-modules and a map f : Q → M there is an extension of f toN.
We say a right A-module P is a progenerator if it is a finitely generated projective generator.
Exercise 3.2.-(i) Prove the equivalence of the conditions in the above definition.

(ii) Show that A is a progenerator in Mod −A. (c.f. 1.2 for finitely generated).
(iii) Show that if F : Mod −A → Mod −B is an equivalence then F (A) is a progenerator in Mod −B.
(iv) Observe thatA∼= EndA A ∼= EndB F (A)where all the isomorphisms are isomorphisms of rings.
The above exercise thus proves:
Proposition 3.3.- Suppose A ∼ B. Then there is a progenerator P in Mod −B such that A ∼= EndB P asrings, and a progenerator Q in Mod −A with B ∼= EndA Q.
Finally, observe that because EndB P (resp. EndA Q) act on P (resp. Q) on the left, the isomorphisms above
actually give P (resp. Q) the structures of (A, B)- (resp. (B, A)-) bimodules.
Remark 3.4.- Something that is a bit confusing is why EndA M always acts on M on the left, regardlessof whether M is a left or a right module. The reason is that we always write functions as f (x) as opposedto (x)f . If M is a left-module it becomes a ring-module under (EndA M )op , the opposite ring, which would
be the endomorphism ring if we wrote functions as (x)f . One also observes this symmetry-breaking by thefact that given a ring A we have A ∼= End(AA ) and Aop ∼= End( A A).

34Equivalence from the bimodules

We have associated to an equivalence Mod −A ∼= Mod −B and (A, B)-bimodule P with the property that it isa progenerator as a right B-module, and with A ∼= EndB P . As it turns out from the following proposition,P will also be a progenerator as an A-module. This allows us to start building the symmetry that will
culminate in Morita’s theorem.
Proposition 4.1.- Suppose P is an (A, B)-bimodule, progenerator as a right B-module. Then P is aprogenerator as a left A-module.
Proof. Pick some k so that B ⊕k ∼= P ⊕ P 0 as B-modules. Then:
AP⊕k ∼= HomB (B ⊕k , A PB )∼= HomB (PB ⊕ P 0 , A PB )B∼= A ⊕ A0 .
This shows that P is a generator.
Also observe that for some m we have P ⊕m ∼= B ⊕ B 0 as B-modules. ThenA⊕m ∼= HomB (P, A P )⊕m∼= HomB (P ⊕m , A P )∼= HomB (B ⊕ B 0 , A P )∼= P ⊕ P 0.
which shows that P is projective and finitely generated as a left A-module.
Once we have this result, we can go the other way. That is, we can prove that starting with one of these
special bimodules we can get an equivalence.
Proposition 4.2.- Suppose that P is an (A, B)-bimodule, progenerator on both sides with A ∼= EndB P .
Then Mod A ∼= Mod B via M 7→ M ⊗A P and N 7→ HomB (P, M ).
This proposition is the key to the whole story. Before we can prove it we need a key lemma.
Lemma.- (KEY) Suppose P is a right B-module, progenerator as such, with A ∼= EndB P – in particular,
P is an (A, B)-bimodule. Then B ∼= HomA (P, P ).
Observe that there is certainly a map B → HomA (P, P ) given by right multiplication. We prove this is anisomorphism again as a hinted exercise.
Exercise 4.3.-(i) Recall that for some n  0 we have P ⊕n ∼= B ⊕ B 0 as right B-modules.

(ii) Observe that HomA (P, P ) consists of those Z-linear maps P → P that commute with every B-linearmap.
(iii) Show that if φ ∈ HomA (P, P ) then φ⊕n (B) ⊆ B, and that φ⊕n restricts to a left B-linear map B → B.
Conclude that the map B → HomA (P, P ) is injective.
(iv) Show that given any x ∈ P ⊕n there exists an S-linear map fx : P ⊕n → P ⊕n and sx ∈ S such thatfx (sx ) = x.
(v) Show that if φ⊕n restricts to multiplication by s ∈ S on the right in S then φ⊕n is multiplication by son the right on the whole of P ⊕n . Conclude that the map B → HomA (P, P ) is surjective.
We are now ready to prove the proposition.
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Proof of Proposition 4.2. . Set F : Mod −A → Mod −B by F (M ) = M ⊗A P and G : Mod −B → Mod −Aby G(N ) = HomB (P, N ). Using 20.10 and 20.11 from the book one then has
HomB (P, M ⊗A P ) ∼= M ⊗ HomB (P, P )∼ M ⊗A A =∼M=andHomB (P, M ) ⊗A P ∼= HomA (HomA (P, P ), N )∼= HomA (B, N ) ∼= N.
Therefore, F and G are inverse equivalences.
Observe we have use the fact that P is a progenerator as an A-module as well, to get the first isomorphismin the second chain of isomorphisms.

5Morita’s TheoremWe have thus shown that equivalences Mod −A ∼= Mod −B are in some kind of correspondence with special(A, B)-bimodules P . The facts that make these bimodules special is that A ∼= EndB P and that P is aprogenerator as a B-module. But in the previous section we have proven that whenever we have such abimodule we also get B ∼= EndA P and that P is a progenerator as an A-module.
Proposition 4.2 tells us that we also have A − Mod ∼= B − Mod. We thus arrive at the theorem.
Theorem 5.1.- [Morita] Let A and B be rings. The following are equivalent:(i) Mod −A ∼= Mod −B.
(ii) There exists a (B, A)-bimodule P , with A ∼= EndB P , which is a progenerator on both sides.
(iii) B − Mod ∼= A − Mod.
Example 5.2.- I would like to give one last example to illustrate why this is such a surprising result.
Consider the ring
Z QA=⊆ M2 (Q).
0 Q
Then A is right noetherian but not left noetherian [H]. Now suppose that I give you the “category of dots”
that we gave in the introduction associated to the right modules for A. At the beginning we showed how
once you get this category you can tell that the ring is right noetherian. What Morita’s theorem tells youis that this category has enough information for you to conclude that the ring is not left noetherian. In
particular, there is no noetherian ring B with A ∼ B.

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References[H]: Hernstein – Noncommutative Rings.
[AF]: Anderson, Fuller – Rings and Categories of Modules.

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