Morita equivalence


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

F:ℳR→ℳS   and   G:ℳS→ℳR

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

G⁢F⁢(M)≅RM   and   F⁢G⁢(N)≅SN,

where A≅RB 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.

Proof.

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)∣mi∈M}. Then F⁢(M) becomes a module over Mn⁢(R) if we adopt the standard matrix multiplication m⁢A, where m∈F⁢(M) and A∈Mn⁢(R). If f:M1→M2 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 r∈R and 0 everywhere else. For simplicity we write e:=e⁢(1). Note that e is an idempotentPlanetmathPlanetmath in Mn⁢(R): e=e⁢e, and e commutes with e⁢(r) for any r∈R: e⁢e⁢(r)=e⁢(r)⁢e.

Set G⁢(N)={s⁢e∣s∈N}. For any r∈R, define s⁢e⋅r:=s⁢e⁢e⁢(r)=s⁢e⁢(r)⁢e. Since s⁢e⁢(r)∈N, this multiplication turns G⁢(N) into an R-module. If g:N1→N2 is an Mn⁢(R)-module homomorphism, define G⁢(g):G⁢(N1)→G⁢(N2) by G⁢(g)⁢(s⁢e)=g⁢(s)⁢e. If N1⟶gN2⟶hN3 are Mn⁢(R)-module homomorphisms, then

G⁢(h∘g)⁢(s⁢e)=(h∘g)⁢(s)⁢e=h⁢(g⁢(s))⁢e=G⁢(h)⁢[g⁢(s)⁢e]=G⁢(h)⁢[G⁢(g)⁢s⁢e]=G⁢(h)∘G⁢(g)⁢(s⁢e)

so that G is a covariant functor.

If M is any R-module, then G⁢F⁢(M)={(m1,…,mn)⁢e∣m∈M}={(m1,0,…,0)T∣m∈M}≅M, where mT stands for the transposeMathworldPlanetmath of the row vectorMathworldPlanetmath m∈M into a column vector.

On the other hand, if N is any Mn⁢(R)-module, then F⁢G⁢(N)={(s1⁢e,…,sn⁢e)∣si∈N}. Before proving that F⁢G⁢(N)≅N, let’s do some preliminary work.

Denote ei⁢i by the n×n matrix whose cell (i,i) is 1 and 0 everywhere else. Then each ei⁢i is idempotent, ei⁢i⁢ej⁢j=0 for i≠j, and e11+⋯+en⁢n=1. From this, we see that N=N1⊕⋯⊕Nn, where Ni=N⁢ei⁢i, and Ni≅Nj as Mn⁢(R)-modules. Since N1=N⁢e has an R-module structureMathworldPlanetmath as we had shown earlier, Ni are all R-modules. Let πi:N→Ni be the projection map, ψi:Ni→N be the embedding of Ni into N, and ϕi⁢j:Ni→Nj be the isomorphism from Ni to Nj given by ϕi⁢j⁢(s⁢ei⁢i)=s⁢ej⁢j. All these are Mn⁢(R)-module homomorphisms since ei⁢i⁢A=A⁢ei⁢i.

Now, take any s∈N, then s↦(π1⁢(s),…,πn⁢(s))↦(ϕ11⁢π1⁢(s),…,ϕn⁢1⁢πn⁢(s))∈F⁢G⁢(N) is a homomorphismMathworldPlanetmathPlanetmathPlanetmath α:N→F⁢G⁢(N). Conversely, (s1⁢e,…,sn⁢e)↦(ϕ11⁢(s1⁢e),…,ϕ1⁢n⁢(sn⁢e))↦ψ1⁢(ϕ11⁢(s1⁢e))+⋯+ψn⁢(ϕ1⁢n⁢(sn⁢e))∈N is also a homomorphism β:F⁢G⁢(N)→N. By inspection, α and β are inversesMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of each other, and hence F⁢G⁢(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