Table Of ContentNotes on Morita Equivalence
Andrew Hubery
For more details see for example P.M. Cohn, Algebra Volume 3.
Throughout, A and B will be (associative, unital) rings, and ModA will denote
the category of right A-modules.
Definition. WesaythatAandB areMoritaequivalentprovidedModA∼=ModB.
The first crucial result in this direction is the following lemma.
Lemma 1 (Eilenberg-Watts). The following are equivalent for a functor
F: ModA→ModB.
(1) F has a right adjoint G.
(2) F is right exact and preserves direct sums (coproducts).
(3) F ∼=−⊗ P for some A-B-bimodule P.
A
In this case, G∼=Hom (P,−).
B
Proof. (3) implies (1). Set G:=Hom (P,−).
B
(1) implies (2). We have that F is the left adjoint of G, and so is necessarily
right exact and preserves direct sums.
(2) implies (3). Set P :=F(B). Since F is a functor we have
End (A)→End (F(A))=End (P).
A B B
Thus for each a ∈ A, left multiplication by a is an A-endomorphism λ of A, and
a
henceinducesaB-endomorphismF(λ )ofF(A)=P. Fora∈Aandp∈P weset
a
ap:=F(λ )(p). Then P becomes an A-B-bimodule.
a
We now have for M ∈ModA
M ∼=Hom (A,M)→Hom (F(A),F(M))∼=Hom (P,F(M)).
A B B
This gives a map f : M → Hom (P,F(M)), which is in fact a A-module homo-
M B
morphism. Now, tensor-hom-adjointness gives
Hom (M,Hom (P,F(M)))∼=Hom (M ⊗ P,F(M)),
A B B A
underwhichf inducesaB-modulehomomorphismg : M⊗ P →F(M). Since
M M A
alltheseconstructionsarenatural,weobtainanaturaltransformationg: −⊗ P →
A
F. Since g is an isomorphism, and A is a generator for ModA, it follows that g is
A
a natural isomorphism. (cid:3)
We have used the following general result.
Lemma 2. Let A,B be abelian categories admitting coproducts. Let F,G: A→B
be right exact functors preserving coproducts, and let F →G be a natural transfor-
mation. If F(P)∼=G(P) for some generator P of A, then F ∼=G.
Proof. As usual, we write P(I) for a coproduct. Let M ∈A and take a resolution
P(I) →P(J) →M →0
1
2
Applying F and G, we obtain a diagram with exact rows
F(P)(I) −−−−→ F(P)(J) −−−−→ F(M) −−−−→ 0
y y y
G(P)(I) −−−−→ G(P)(J) −−−−→ G(M) −−−−→ 0
The vertical maps on the left and in the centre are isomorphism, hence the one on
the right is also an isomorphism. (cid:3)
General Form of Morita Equivalences
We now come to the main theorem.
Theorem 3. A and B are Morita equivalent if and only if there exist bimodules
P and Q with bimodule isomorphisms
A B B A
P ⊗ Q∼=A and Q⊗ P ∼=B.
B A
Moreover, in this case, we must have that
(1) P and Q are finitely-generated projective generators,
B A
(2) P ∼=Hom (Q,A) and Q∼=Hom (P,B) as bimodules,
A B
(3) A∼=End (P) and B ∼=End (Q) as algebras.
B A
Proof. Suppose we have an equivalence, given by functors F: ModA → ModB
and G: ModB →ModA.
Since (F,G) is an adjoint pair, the Eilenberg-Watts Lemma applies to give an
A-B-bimodule P and natural isomorphisms
F ∼=−⊗ P and G∼=Hom (P,−).
A B
Similarly, since (G,F) is also an adjoint pair, we obtain a B-A-bimodule Q and
natural isomorphisms
G∼=−⊗ Q and F ∼=Hom (Q,−).
B A
Now, since F and G are (quasi-)inverse functors, we have natural isomorphisms
1∼=GF ∼=−⊗ (P ⊗ Q) and 1∼=FG∼=−⊗ (Q⊗ P).
A B A B
In particular, we have
P ⊗ Q∼=A and Q⊗ P ∼=B,
B A
the first being an isomorphism of A-bimodules, the second an isomorphism of B-
bimodules.
Conversely, suppose that we have bimodules P and Q and bimodule iso-
A B B A
morphisms
P ⊗ Q∼=A and Q⊗ P ∼=B.
B A
Setting F :=−⊗ P and G:=−⊗ Q gives that
A B
FG∼=−⊗ (Q⊗ P)∼=−⊗ B ∼=1 and GF ∼=1.
B A B
Hence F and G are inverses, so induce an equivalence ModA∼=ModB.
We now prove the properties (1), (2) and (3).
Using the two constructions for F, we get
P ∼=F(A)∼=Hom (Q,A),
A
3
which is even an isomorphism of A-B-bimodules. Similarly
B ∼=Q⊗ P ∼=F(Q)∼=End (Q).
A A
This is even an algebra isomorphism B ∼=End (Q).
A
The results for Q and A are dual. This gives (2) and (3).
For (1) observe that since F = Hom (Q,−) admits a right adjoint, it is right
A
exact and preserves direct sums (coproducts). The right exactness implies that Q
is projective, and that it preserves direct sums implies that it is finitely-generated
projective.
To prove the latter, it is enough to consider free modules. For, given any X,
there is a natural transformation
Hom (X,M)(J) →Hom (X,M(J)), (f )7→f where f(x):=(f (x)).
A A j j
This commututes with finite direct sums, so that
Hom (X,M)(J)⊕Hom (Y,M)(J) ∼=Hom (X ⊕Y,M)(J)
A A A
→Hom (X ⊕Y,M(J))∼=Hom (X,M(J))⊕Hom (Y,M(J)).
A A A
As usual we have written M(I) for the direct sum and MI for the direct product.
Now, Q is projective if and only if there exists Q0 with Q⊕Q0 free, and Q
A A
is finitely-generated projective if and only if there exists Q0 with Q⊕Q0 finitely-
generated free.
So, if we can show that the above natural transformation is an isomorphism for
a free module if and only if it is finitely-generated free, we can deduce that it is an
isomorphismforaprojectivemoduleifandonlyifitisfinitely-generatedprojective.
Every free module is isomorphic to one of the form A(I). Moreover, we know
that
Hom (A(I),M)∼=Hom (A,M)I ∼=MI.
A A
So, it is enough to show that the functor M 7→ MI commutes with direct sums if
and only if I is finite.
Now, as above for the functors, regrouping terms gives a natural transformation
(M(J))I →(MI)J, ((m ) ) 7→((m ) ) .
ij j i ij i j
To say that M 7→MI commutes with direct sums says that the image must be in
(MI)(J) for all sets J.
If I is finite, then this is easily seen to be true. On the other hand, if I is
infinite, then take J = I and choose elements m such that m 6= 0 if and only
ij ij
if i = j. Then only a single term in (m ) is non-zero, so (m ) ∈ M(J) and
ij j ij j
((m ) ) ∈ (M(J))I. After regrouping, though, we see that each (m ) ∈ MI is
ij j i ij i
non-zero, and hence ((m ) ) 6∈(MI)(J).
ij i j
Thus M 7→MI commutes with direct sums if and only if I is finite.
We obtain that Q is finitely-generated projective, and similarly that P is
A B
finitely-generated projective.
Lastly, if P ⊕P0 ∼=Bn, then
Qn ∼=Bn⊗ Q∼=(P ⊕P0)⊗ Q∼=(P ⊗ Q)⊕(P0⊗ Q)∼=A⊕(P0⊗ Q).
B B B B B
Thus Qn (cid:16)A, so Q is a generator. Similarly for P . (cid:3)
A B
4
Finding all Morita Equivalences
We now want to show that every finitely-generated projective generator gives
rise to an equivalence.
Let Q be an A-module, and set
A
B :=End (Q) and P :=Hom (Q,A).
A A
Then P and Q arebimodules,andwehavenaturalbimodulehomomorphisms
A B B A
P ⊗ Q→A, f ⊗q 7→f(q),
B
and
Q⊗ P →End (Q)=B, q⊗f 7→[q,f] where [q,f](q0):=qf(q0).
A A
We observe that, given f,g ∈ P = Hom (Q,A) and r ∈ Q, we have [r,f] ∈ B, so
A
g[r,f]∈P. To see how this acts, take q ∈Q. Then
g[r,f]·q :=g([r,f](q))=g(rf(q))=g(r)f(q).
Hence
g[r,f]=g(r)f.
(faithfully-balanced?)
Lemma 4. The natural homomorphism
P ⊗ Q→A, f ⊗q 7→f(q),
B
is an isomorphism if and only if it is surjective, which is if and only if Q is a
generator.
Proof. We observe that the image of the map is the trace of Q in A,
X
tr (A):= Im(f).
Q
f
Hence the map is surjective if and only if Qn (cid:16)A for some n, which is if and only
if Q is a generator.
A
Now suppose that this map is surjective, say with
X X
f ⊗q 7→ f (q )=1 .
i i i i A
i i
To see that it is also injective, let
X X
g ⊗r 7→ g (r )=0.
j j j j
j j
We therefore have
X X X
g ⊗r = g ⊗r f (q )= g ⊗[r ,f ](q )
j j j j i i j j i i
j i,j i,j
X X
= g [r ,f ]⊗q = g (r )f ⊗q =0.
j j i i j j i i
i,j i,j
Hence the map is also injective, hence an isomorphism. (cid:3)
5
Lemma 5 (Dual Basis). The natural homomorphism
Q⊗ P →End (Q)=B, q⊗f 7→[q,f],
A A
is an isomorphism if and only if Q is finitely generated projective.
In fact in this case we have a natural isomorphism of functors
Q⊗ Hom (−,A)∼=Hom (−,Q), q⊗f 7→[q,f].
A A A
Proof. The module Q is finitely generated projective if and only if we have an
A
epimorphism φ: An → Q admitting a retract θ: Q → An; that is, φθ = 1 . If we
Q
let the components of θ be f ∈ Hom (Q,A) = P, and let the components of φ
i A
correspond to elements q ∈Q (via Hom (A,Q)∼=Q)), then
i A
X X
φθ(q)= q f (q)= [q ,f ](q).
i i i i
i i
P
Hence Q is finitely generated projective if and only if there exists q ⊗f ∈
A i i i
P
Q⊗ P such that [q ,f ]=1 , which is if and only if the map Q⊗ P →B is
A i i i Q A
onto.
We now do the same trick as before to see that surjectivity implies injectivity.
For, suppose
X X
r ⊗g 7→ [r ,g ]=0.
j j j j
j j
Then
X X X
r ⊗g = [q ,f ]r ⊗g = q f (r )⊗g
j j i i j j i i j j
j i,j i,j
X X
= q ⊗f (r )g = q ⊗f [r ,g ]=0.
i i j j i i j j
i,j i,j
For the final statement, we know that for all M there is a natural homomor-
A
phism
Q⊗ Hom (M,A)→Hom (M,Q), q⊗f 7→[q,f] where [q,f](m):=qf(m).
A A A
Since Q is finitely generated projective, we can write Q⊕Q0 ∼=An. Then
A
An⊗ Hom (M,A)∼=Hom (M,A)n ∼=Hom (M,An)
A A A A
implies that
Q⊗ Hom (M,Q)∼=Hom (M,Q).
A A A
This yields the natural isomorphism
Q⊗ Hom (−,A)∼=Hom (−,Q)
A A A
as required. (cid:3)
Theorem 6. If Q is a finitely generated projective generator, then Q induces an
A
equivalence
ModA∼=ModEnd (Q), M 7→Hom (Q,M).
A A
Proof. Since Q is a finitely-generated projective generator, we can apply our two
A
lemmas to deduce that
P ⊗ Q∼=A and Q⊗ P ∼=B.
B A
The equivalence of categories now follows from the main theorem. (cid:3)
6
Main Example of Morita Equivalence
This yields the main example of such an equivalence. We take Q = An, a
A
free module. Then B ∼= M (A), corresponding to the usual left action on column
n
vectors. We also see that P = Hom (Q,A) is given by row vectors of length n,
A A
so the bimodule isomorphisms
P ⊗ Q∼=A and Q⊗ P ∼=B
B A
are given by
a a
1 1
(cid:0)b ··· b (cid:1)⊗ .. 7→Xb a and .. ⊗(cid:0)b ··· b (cid:1)7→(a b ).
1 n . i i . 1 n i j
a i a
n n
7
Morita Contexts and Quotient Categories
Can extend these ideas slightly to include quotient categories (P.M. Cohn’s lan-
guage).
Suppose we have F: ModA → B admitting a right adjoint G. We then obtain
an A-B-bimodule P such that
F ∼=−⊗ P and G∼=Hom (P,−).
A B
If moreover G admits a right adjoint H, then we obtain a B-A-bimodule Q such
that
G∼=−⊗ Q and H ∼=Hom (Q,−).
B A
BytheEilenberg-WattsLemma,GadmitsarightadjointifandonlyifP isfinitely
B
generated projective.
It follows that we have a bimodule isomorphism
Q∼=G(A)∼=Hom (P,B)
B
and a natural isomorphism
GF ∼=−⊗ (P ⊗ Q).
A B
We say that ModA is a quotient category of ModB provided that we have
functors F and G as above such that GF ∼= 1. This is if and only if P ⊗ Q ∼= A
B
as bimodules, which then implies that
A∼=P ⊗ Q∼=G(P)∼=End (P).
B B
The analogy is now that we view F as a retract for the functor G.
