Morita equivalence

Let R be a ring. Write R for the categoryMathworldPlanetmath of right modules over R. Two rings R and S are said to be Morita equivalent if R and S are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath as categories ( What this means is: we have two functorsMathworldPlanetmath

F:RS   and   G:SR

such that for any right R-module M and any right S-module N, we have

GF(M)RM   and   FG(N)SN,

where ARB means that there is an R-module isomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath between A and B.

Example. Any ring R with 1 is Morita equivalent to any matrix ring Mn(R) over it.


Assume n>1. For convenience, we will also say a module to mean a right module.

Let M be an R-module. Set F(M)={(m1,,mn)miM}. Then F(M) becomes a module over Mn(R) if we adopt the standard matrix multiplication mA, where mF(M) and AMn(R). If f:M1M2 is an R-module homomorphismMathworldPlanetmath. Set F(f):F(M1)F(M2) by F(f)(m1,,mn)=(f(m1),,f(mn))F(M2). Then F is a covariant functor by inspection.

Next, let N be an Mn(R)-module. Write e(r) as the n×n matrix whose cell (1,1) is rR and 0 everywhere else. For simplicity we write e:=e(1). Note that e is an idempotentPlanetmathPlanetmath in Mn(R): e=ee, and e commutes with e(r) for any rR: ee(r)=e(r)e.

Set G(N)={sesN}. For any rR, define ser:=see(r)=se(r)e. Since se(r)N, this multiplication turns G(N) into an R-module. If g:N1N2 is an Mn(R)-module homomorphism, define G(g):G(N1)G(N2) by G(g)(se)=g(s)e. If N1gN2hN3 are Mn(R)-module homomorphisms, then


so that G is a covariant functor.

If M is any R-module, then GF(M)={(m1,,mn)emM}={(m1,0,,0)TmM}M, where mT stands for the transposeMathworldPlanetmath of the row vectorMathworldPlanetmath mM into a column vector.

On the other hand, if N is any Mn(R)-module, then FG(N)={(s1e,,sne)siN}. Before proving that FG(N)N, let’s do some preliminary work.

Denote eii by the n×n matrix whose cell (i,i) is 1 and 0 everywhere else. Then each eii is idempotent, eiiejj=0 for ij, and e11++enn=1. From this, we see that N=N1Nn, where Ni=Neii, and NiNj as Mn(R)-modules. Since N1=Ne has an R-module structureMathworldPlanetmath as we had shown earlier, Ni are all R-modules. Let πi:NNi be the projection map, ψi:NiN be the embedding of Ni into N, and ϕij:NiNj be the isomorphism from Ni to Nj given by ϕij(seii)=sejj. All these are Mn(R)-module homomorphisms since eiiA=Aeii.

Now, take any sN, then s(π1(s),,πn(s))(ϕ11π1(s),,ϕn1πn(s))FG(N) is a homomorphismMathworldPlanetmathPlanetmathPlanetmath α:NFG(N). Conversely, (s1e,,sne)(ϕ11(s1e),,ϕ1n(sne))ψ1(ϕ11(s1e))++ψn(ϕ1n(sne))N is also a homomorphism β:FG(N)N. By inspection, α and β are inversesMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of each other, and hence FG(N)N. ∎

Remark. A property P in the class of all rings is said to be Morita invariant if, whenever R has property P and S is Morita equivalent to R, then S has property P as well. By the example above, it is clear that commutativity is not a Morita invariant property.

Title Morita equivalence
Canonical name MoritaEquivalence
Date of creation 2013-03-22 16:38:49
Last modified on 2013-03-22 16:38:49
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 6
Author CWoo (3771)
Entry type Definition
Classification msc 16D90
Defines Morita equivalent
Defines Morita invariance
Defines Morita invariant