An Introducton to Morta Theory Matt Booth October 2015 Nov. 2017: made a few revsons. Thanks to Nng Shan for catchng a typo. My man reference for these notes was Chapter II of Bass s book Algebrac K-Theory (1968); you can fnd a more detaled exposton there. 1 Motvaton When we re dong representaton theory we want to study the structure of the category Mod-A for some rng A. So we want to know f and when two dfferent rngs A and B gve us the same category. Wth ths n mnd, two rngs are sad to be Morta equvalent when ther module categores are equvalent. In many cases, we often only care about rngs up to Morta equvalence. If ths s the case, then gven a rng A, we d lke to fnd some partcularly nce representatve of the Morta equvalence class of A. 2 Morta Equvalence Frst some notaton: Let R be a rng. Wrte Mod-R for the catgeory of rght R-modules and R-module homomorphsms. Wrte mod-r for the (full) subcategory of fntely generated R-modules. Wrte Proj-R for the subcategory of projectve modules, and proj-r for the subcategory of fntely generated projectve modules. Two rngs R, S are defned to be Morta equvalent f the categores Mod-R and Mod-S are equvalent. In fact, any such equvalence wll be addtve: ths s a general fact about equvalences between abelan categores, snce the sum of two morphsms f, g : X Y can be recovered as the composton f + g : X X X f g Y Y Y of f g wth the dagonal and fold maps. Note that any property defned categorcally wll be preserved under equvalence: for example f F : Mod-R Mod-S s an equvalence, then F wll take projectve modules to projectve modules and hence nduce an equvalence Proj-R Proj-S. 1
Moreover, an equvalence F as above wl nduce an equvalence between proj-r and proj-s, snce the fntely generated projectve modules are precsely those projectve modules P for whch the functor Hom R (P, ) dstrbutes over drect sums. Example If R s a dvson rng, then R s Morta equvalent to all of ts matrx rngs M n (R). We ll see a proof of a specal case of ths later on, usng quvers. Remark One can show that Z(R) = End [Mod-R,Mod-R] (d), the endomorphsm rng of the dentty functor of Mod-R (sometmes called the centre of Mod-R): frst note that d s just Hom A (A, ) where A s regarded as an A A-bmodule. So by an approprate verson of the Yoneda lemma (we need to be careful about A-lnearty), we get that End [Mod-R,Mod-R] (d) = End A e(a, A) = Z(A), where A e := A A op s the envelopng algebra. So f R and S are Morta equvalent, then Z(R) = Z(S). In partcular two commutatve rngs are Morta equvalent f and only f they are somorphc. So Morta equvalence s only nterestng for noncommutatve rngs! A generator for a category C s an object G such that for any two parallel morphsms f, g : X Y wth f g, then there s some morphsm h : G X such that fh gh. Note that generators are preserved under equvalence. If C s Mod-R, then a generator for C s the same thng as a module G such that every R-module s a quotent of a (possbly nfnte) drect sum of copes of G. A progenerator n Mod-R s a fntely generated projectve generator. Progenerators always exst: as an easy example, R s a progenerator for Mod-R. We care about progenerators because of the followng result: Theorem (Morta). Two rngs R and S are Morta equvalent f and only f there exsts a progenerator P of Mod-R such that S = End R (P ). Note that we re consderng R as a rght R-module. In partcular, R = End R (R). We can easly prove one drecton of ths theorem now: If R and S are Morta equvalent, then take an equvalence F : Mod-R Mod-S. Snce R s a progenerator for Mod-R, F (R) s a progenerator for Mod-S. We have somorphsms R = End R (R) = End S (F (R)). The converse s harder to prove: we need to consder an equvalent charactersaton of Morta equvalence n terms of tensor products of bmodules. 2
3 Bmodules An R S bmodule s an abelan group M whch s both a left R-module and a rght S-module, such that the actons are compatble: (rm)s = r(ms). If M s an R S bmodule and N s an S T bmodule then t makes sense to form the tensor product and get an R T bmodule M S N. Theorem. Two rngs R and S are Morta equvalent f and only f there exsts an S R bmodule P and an R S bmodule Q such that P R Q = S and Q S P = R. Note that these are somorphsms of S S bmodules and R R bmodules respectvely. Proof. To prove the f drecton, suppose we have P and Q such that P R Q = S and Q S P = R. Then settng F := R Q and G := S P, we have GF = ( R Q) S P = R (Q S P ) = R R = d R and smlarly F G = d S. To prove the only f drecton we need a verson of the Elenberg-Watts Theorem, whch tells us that f F : Mod-R Mod-S : G s an equvalence, then there exsts an R S bmodule Q such that F = R Q (and furthermore, Q s a progenerator for Mod-S). So f we have an equvalence as above, applyng Elenberg-Watts twce we get bmodules P and Q such that F = R Q and G = S P. Usng d R = GF and ds = F G t s not hard to check that P R Q = S and Q S P = R. 4 Morta s Theorem Armed wth ths new charactersaton of Morta equvalence we can prove the other half of Morta s Theorem. So suppose we have a progenerator P of Mod-R wth S = End R (P ). The left acton of End R (P ) on P gves us a left acton of S on P makng P nto an S R bmodule. Set Q := Hom R (P, R). Q has a rght acton by S = End R (P ) where we precompose a morphsm P R wth an endomorphsm of P, and a left acton by R = End R (R) where we compose wth an endomorphsm of R. Ths turns Q nto an R S bmodule. If we can prove that P R Q = End R (P ) and Q S P = R then we re done. Defne maps φ : Q S P R and ψ : P R Q End R (P ) by φ(f p) = f(p) and ψ(p f) = pf. The map φ s onto snce P s a generator. The map ψ s onto snce P s a drect summand of R n and so any endomorphsm of P s a sum of endomorphsms factorng through R. Note that we have denttes ψ(x f)(y) = xφ(f y) and g ψ(x f) = φ(g x)f, for x, y P and f, g Q. 3
We just need to prove that φ and ψ are njectve maps. We ll only show that ψ s njectve snce the argument for φ s smlar. Snce ψ s surjectve, frst fnd an element x f wth ψ( x f ) = 1. Let y g be any element of P R Q. Then: y g =,j(y g )ψ(x j f j ) (multplyng by 1 and usng lnearty of ψ) = y φ(g x j )f j,j = y φ(g x j ) f j,j (usng the second dentty) (pullng an element of R through the tensor product) =,j ψ(y g )(x j f j ) (usng the frst dentty) So f ψ( y g ) = 0 then y g = 0, and so ψ s njectve. 5 Induced equvalences between subcategores We know that an equvalence F : Mod-R Mod-S nduces equvalences between the subcategores Proj-R and Proj-S, as well as an equvalence between proj-r and proj-s. What about mod-r and mod-s? It turns out that we have the followng very nce theorem: Theorem (Morta). Mod-R s equvalent to Mod-S f and only f mod-r s equvalent to mod-s. Proof. If F : Mod-R Mod-S s an equvalence, then by the Elenberg-Watts theorem F = R P for some R S bmodule P that s a progenerator for Mod-S. Snce P s a fntely generated S module, F restrcts to a functor mod-r mod-s, and furthermore ths s an equvalence snce the nverse of F restrcts n the same way. Conversely f mod-r and mod-s are equvalent, then we can run the proof at the end of Secton 2 agan to obtan a progenerator P of mod-r such that S = End R (P ). But a progenerator for mod-r s a progenerator for Mod-R, and hence, applyng Morta s Theorem, R and S are Morta equvalent. 4
6 Basc algebras We are nterested n fnte dmensonal algebras over a feld. Therefore for any such algebra A we want to fnd a nce algebra A Morta equvalent to A. Then we can prove results about Mod-A and get results about Mod-A. Here we descrbe how to buld one such A, the basc algebra of A. A = Let A be a fnte dmensonal algebra over a feld. Then A admts a decomposton n =0 e A as rght A-modules, where the e are a complete set of prmtve orthogonal dempotents. We call A basc f for every j, the two A-modules e A and e j A are not somorphc. In general, A need not be basc (see the example below). But there s an easy way to construct a basc algebra from A. Suppose we have a decomposton of A as above. Choose a subset {e a1,..., e am } of {e 1,..., e n } maxmal wth respect to the condton that e a A e aj A whenever j. In partcular every e A s somorphc to some e aj A. Set e := e a1 + + e am, and put A b := eae. Then A b s a basc algebra. Up to somorphsm, A b wll not depend on the choce of {e a1,..., e am }. Call A b the basc algebra assocated to A. Then A and A b are Morta equvalent: one can show that ea s a progenerator for Mod-A, and ts endomorphsm rng s precsely A b. Example Set A = M n (k), and let e be the matrx wth a 1 n poston (, ) and zeroes elsewhere. Then the e are a complete set of prmtve orthogonal dempotents, and for all and j, e A = k n = e j A. So A b = e 1 Ae 1 s a copy of k. Ths provdes an alternate proof that M n (k) s Morta equvalent to k. 5