There is no analog of the transpose map for infinite matrices

Publicacions Matem`atiques, Vol 41 (1997), 653–657. THERE IS NO ANALOG OF THE TRANSPOSE MAP FOR INFINITE MATRICES Juan Jacobo Sim´ on∗ Abstract In this note we show that there are no ring anti-isomorphism between row finite matrix rings. As a consequence we show that row finite and column finite matrix rings cannot be either isomorphic or Morita equivalent rings. We also show that antiisomorphisms between endomorphism rings of infinitely generated projective modules may exist. 1. Introduction and notation The motivation for this note was to find out the extent to which the transpose for infinite matrices behaves like that for finite matrices. It is well-known that for any commutative ring, R, the transpose map t:Mn(R)→Mn(R) is an anti-automorphism. For infinite matrix rings the analogous transpose map yields an anti-isomorphism t: RFMA(R)→CFMA(R), where RFMA(R) (resp. CFMA(R)) denotes the ring of row-finite (resp. column-finite) matrices, indexed by the set A, having entries in R. Unlike the finite case, this is obviously not an anti-AUTOmorphism of RFMA(R). Thus we are led to ask: for a commutative ring, R, does there exist an anti-automorphism of RFMA(R)? We answer this question in the negative, with a vengeance. In fact, we show in Theorem 2.2 that there are no anti-isomorphisms between rings of the form End(RR(A)) and End(SS(B)) for any rings Rand S(commutative or not), and any infinite sets Aand B. As a corollary we show that for any two rings R and Sthe rings RFMA(R) and CFMB(S) cannot be either isomorphic or Morita equivalent rings. This corollary is well known in the case of commutative domains and A,Bcountable infinite sets, or division rings. A particularly clear proof for commutative domains is given in [2] although ∗Partially supported by DGICYT (PB93-0515-C02-02). 654 J. J. Sim´ on we wish to call to the reader’s attention that Theorem 1.8 (III) does not hold in the generality stated. We establish in Example 2.4 that (normalized) Morita context is the closest relation that may exist between such rings. Finally we show in Example 2.5 that anti isomorphisms (in fact, involutions) between endomorphism rings of countably generated and locally free left projective modules may exist, so that there are modules very close to free that admit anti-isomorphisms. Throughout this paper ring means associative ring with identity and for any ring R,Rop denotes the opposite ring of R[3, Definition 0.1.11]. We denote by R-mod (respectively mod-R) the category of all left (resp. right) unital R-modules. Homomorphisms of modules act opposite scalars. Thus, when fand gare endomorphisms of a left R-module RM, their composition g◦fwill be denoted by fg; in particular End(RM) will be the opposite ring of HomR(M,M) and, for a right R-module NR, End(NR) = HomR(N,N). For a given family of modules {Mα}α∈A, where Ais a set, ⊕AMα will mean the direct sum, while AMαwill mean the direct product. Direct sum and product of copies of a single module M, indexed by A will be written M(A)or ⊕AMand MAor AM, respectively, as usual. For any set A, cardinality will be denoted by |A|, and following most of literature, when Ais finite, say |A|=n∈Nwe will write Mninstead of ⊕AMα=AMα. For any infinite set A, we denote by RFMA (R) (respectively CFMA (R)) the ring of A×Arow-finite matrices (respectively column-finite matrices) over R. It is well-known that there exist ring isomorphisms End(RR(A))∼ =RFMA(R) and End(R(A) R)∼ =CFMA(R). Recall that a ring Rhas (left) SBN (Single Basis Number) when for every n∈N, Rn∼ =R(as left R-modules). Note that, for every ring R, since R(N)∼ =(R(N))nas left and right R-modules then both RFM(R) and CFM(R) have SBN; thus RFM(R)∼ =Mn(RFM(R)) and also CFM(R)∼ =Mn(CFM(R)). 2. Anti isomorphisms We shall see that for any two rings Rand S, the existence of an anti isomorphism between RFMA(R) and RFMB(S) for some sets Aand B implies that either Aor Bmust be finite. To do this, we begin with the following proposition: Proposition 2.1. Let Rbe a ring and Aan infinite set. Then (R(A))Acannot be isomorphic to R(A)as left R-modules. There is no analog of the transpose 655 Proof: It is well-known that there is the first ordinal number αsuch that |α|=|A|. Then for every ordinal β, such that β<α, we must have that |β|<|α|. Since |α|=|A|then we can endow Awith a linear ordering (A, ≤), given by α. By this and properties of αabove, we have that for every a∈A, the set Sa={x∈A|x≤a}satisfies |Sa|<|A|. Now suppose there exists an R-isomorphism ϕ:R(A)A→R(A). Let {ea}a∈(A,≤)be the canonical (well-ordered) basis for R(A)and let {xa}a∈(A,≤)such that xa=ϕ−1(ea). Then, {xa}a∈(A,≤)is a well-ordered basis for R(A)A. Let ρa:R(A)A→R(A) aand ηa:R(A)→Rabe the usual projections. Define, for each a∈A, the set Na=∪b≤aSupp(xbρa), where Supp(x)={a∈A|xηa=0}for x∈R(A). Since Supp(xbρa) is a finite set and |Sa|<|A|then |Na|<|A|and hence Na⊂A(strictly). So, for each a∈Awe may choose ξ(a)∈A\Na; that is, for each a∈A, there exists ξ(a)∈Asuch that xbρaηξ(a)= 0 for all b≤a. Now define x∈R(A)Asuch that, for each a∈A,xρaηξ(a)= 1 and xρaηb=0ifb=ξ(a). We claim that xcannot be a linear combination of {xa}a∈(A,≤). To see this, suppose x=n i=1 rixaiwhere a1<···<a nare elements of Aand r1,... ,r nare elements of R. By definition of xwe must have 1=xρanηξ(an)= n  i=1 rixaiρanηξ(an)=0(asai≤an).A contradiction. Therefore, R(A)Acannot be isomorphic to R(A). Now we can prove our main result. Theorem 2.2. Let Rand Sbe rings, Aand Bsets and δ: End(RR(A))→End(SS(B))a ring anti isomorphism. Then Aand B cannot be infinite sets simultaneously. Proof: Suppose both Aand Bare infinite sets and suppose WLOG |A|≥|B|. Let E= End(Rop(A) Rop ) and F= End(SS(B)). Since both Fand Eare anti isomorphic to End(RR(A)), we have that there is a ring isomorphism σ:E→F. Since |B|≤|A|then (Rop(A) Rop )(B)∼ =Rop(A) Rop and so, there is a natural isomorphism between the functors HomRop ((Rop(A))(B),−)∼ =HomRop (Rop(A),−) 656 J. J. Sim´ on which on applying these isomorphisms to Rop(A)yields that EE∼ =EEB. Using σ−1to endow left E-modules with left F-module structure we have that (by properties of change of rings functors), as left F-modules, FE∼ =FFand FEB∼ =FE; so that FFB∼ =FF. Now, note that since S(B) F(as right F-module) is finitely presented then by [3, Exercise 2.11.9’, p. 337] we have S(B)FBF∼ =BS(B)FF. Now S(B)∼ =S(B) F F∼ =S(B) F  B F∼ = B S(B) F F∼ =S(B)B. But this contradicts Proposition 2.1 and hence Aand Bcannot be infinite sets simultaneously. Corollary 2.3. For any two rings Rand S, and any infinite sets A and B, the rings RFMA(R)and CFMB(S)cannot be isomorphic rings. In fact, RFMA(R)and CFMB(S)cannot be Morita equivalent rings. Proof: The first statement is a direct consequence of Theorem 2.2. For the second statement, take E= RFMA(R) and F= CFMB(S). By [4, Theorem 1] we know that, if Eand Fare Morita equivalent rings then there exists n∈Nsuch that E∼ =Mn(F) as rings; but also Mn(F)∼ =F as rings, because Fhas SBN. So that E∼ =Fwhich is impossible by Theorem 2.2. At present, we have seen that for any two rings Rand Sthe rings RFM(R) and CFM(S) cannot be either isomorphic or Morita equivalent. Another possibly relationship between RFM(R) and CFM(S) would be the existence of a Morita context. We recall from [2, Definition III.1.4, (see p. 51)] that a (left) R-module Mis (left) slender if for every homomorphism θ:RN→M, we have that, θ(en) = 0 for all but finitely many n∈N; where enis the element of RNsuch that Supp(en)={n}and ρn(en) = 1; where ρn:RN→Rnare the usual projections. Example 2.4. The rings R= RFM(Z), S= CFM(Z) and bimodules RMS= HomZ(Z(N),ZN), SNR= HomZ(ZN,Z(N)) make M,Na Morita context. We shall construct a Morita context as in [1, Exercise 22.11]. To do this, note that by [2, Corollary III.2.3] Zis slender so that HomZ(ZN,Z)∼ =⊕NHomZ(Z,Z) as abelian groups, which implies that HomZ(ZN,ZN)∼ =(HomZ(Z,Z)(N))N. Thus End(ZZN)∼ =CFM(Z)as rings. Now by [1, Exercise 22.11(1)] we have that M,Nis a Morita context. There is no analog of the transpose 657 Remark. From Example 2.4 it is easy to construct new examples with noncommutative rings, by using [2, Corollary III.2.11] which, in particular, states that if Ris slender so is RFM(R). Note also that, for any slender ring R,evenifRis commutative, the ring RFM(R) is left slender but not right slender because of the isomorphism of right RFM(R)-modules RFM(R)∼ =RFM(R)N. In view of Theorem 2.2 we ask for modules which are close to infinitely generated free, but admit endomorphism rings with involution. We will give examples of such modules. Example 2.5. There exist a ring Rand a left infinitely generated projective module RP(which satisfies Pn∼ =Pfor all n∈N) such that the ring End(RP) has an involution. Let Sbe any ring and SQ=SS(N). Set T= End(SQ) and U=Top. Setting R=S×U,P=Q×Uwe have that End(RP)∼ =T×Uas rings and note that P∼ =Pnfor every n∈N. Now let δ:T×U→T×Usuch that, for every (a, b)∈T×U, δ(a, b)=(b, a). It is clear that δis an involution. References 1. F. W. Anderson and K. R. Fuller,“Rings and Categories of Modules,” Springer-Verlag, New York, 1974. 2. P. C. Eklof and A. H. Mekler,“Almost Free Modules (Set-Theoretic Methods),” North-Holland, Amsterdam, 1990. 3. L. H. Rowen,“Ring Theory,” vol. 1, Academic Press, Boston, 1988. 4. J. J. Sim´ on, Morita equivalent row finite matrix rings are isomorphic, J. Algebra 173 (1995), 390–393. Departamento de Matem´aticas Universidad de Murcia 30100 Murcia SPAIN e-mail: [email protected] Primera versi´o rebuda el 14 de Novembre de 1996, darrera versi´o rebuda el 20 de Juny de 1997