THERE IS NO ANALOG OF THE TRANSPOSE MAP FOR INFINITE MATRICES

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.


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 : M n (R) → M n (R) is an anti-automorphism.For infinite matrix rings the analogous transpose map yields an anti-isomorphism t : RFM A (R) → CFM A (R), where RFM A (R) (resp.CFM A (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 RFM A (R).
Thus we are led to ask: for a commutative ring, R, does there exist an anti-automorphism of RFM A (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( R R (A) ) and End( S S (B) ) for any rings R and S (commutative or not), and any infinite sets A and B. As a corollary we show that for any two rings R and S the rings RFM A (R) and CFM B (S) cannot be either isomorphic or Morita equivalent rings.This corollary is well known in the case of commutative domains and A, B countable infinite sets, or division rings.A particularly clear proof for commutative domains is given in [2] although 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, R op 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 f and g are endomorphisms of a left R-module R M , their composition g • f will be denoted by fg; in particular End( R M ) will be the opposite ring of Hom R (M, M ) and, for a right For a given family of modules {M α } α∈A , where A is a set, ⊕ A M α will mean the direct sum, while A M α will mean the direct product.Direct sum and product of copies of a single module M , indexed by A will be written M (A) or ⊕ A M and M A or A M , respectively, as usual.For any set A, cardinality will be denoted by |A|, and following most of literature, when A is finite, say For any infinite set A, we denote by RFM A (R) (respectively CFM A (R)) the ring of A × A row-finite matrices (respectively column-finite matrices) over R. It is well-known that there exist ring isomorphisms End( R R (A) ) ∼ = RFM A (R) and End(R Recall that a ring R has (left) SBN (Single Basis Number) when for every n ∈ N, R n ∼ = R (as left R-modules).Note that, for every ring R, since R (N) ∼ = (R (N) ) n as left and right R-modules then both RFM(R) and CFM(R) have SBN; thus RFM(R) ∼ = M n (RFM(R)) and also CFM(R) ∼ = M n (CFM(R)).

Anti isomorphisms
We shall see that for any two rings R and S, the existence of an anti isomorphism between RFM A (R) and RFM B (S) for some sets A and B implies that either A or B must be finite.To do this, we begin with the following proposition: Proposition 2.1.Let R be a ring and A an infinite set.Then (R (A) ) A cannot be isomorphic to R (A) as left R-modules.
Proof: It is well-known that there is the first ordinal number α such 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 A with a linear ordering (A, ≤), given by α.By this and properties of α above, we have that for every a ∈ A, the set Now suppose there exists an R-isomorphism ϕ : R (A) A → R (A) .Let {e a } a∈(A,≤) be the canonical (well-ordered) basis for R (A) and let {x a } a∈(A,≤) such that x a = ϕ −1 (e a ).Then, {x a } a∈(A,≤) is a well-ordered basis for R (A) A .Let ρ a : R (A) A → R Now define x ∈ R (A) A such that, for each a ∈ A, xρ a η ξ(a) = 1 and xρ a η b = 0 if b = ξ(a).We claim that x cannot be a linear combination of {x a } a∈(A,≤) .
To see this, suppose x = n i=1 r i x ai where a 1 < • • • < a n are elements of A and r 1 , . . ., r n are elements of R.
By definition of x we must have Therefore, R (A) A cannot be isomorphic to R (A) .Now we can prove our main result.
Theorem 2.2.Let R and S be rings, A and B sets and δ : End( R R (A) ) → End( S S (B) ) a ring anti isomorphism.Then A and B cannot be infinite sets simultaneously.
Proof: Suppose both A and B are infinite sets and suppose WLOG |A| ≥ |B|.Let E = End(R op(A) R op ) and F = End( S S (B) ).Since both F and E are anti isomorphic to End( R R (A) ), we have that there is a ring isomorphism σ : and so, there is a natural isomorphism between the functors Using σ −1 to endow left E-modules with left F -module structure we have that (by properties of change of rings functors), as left F -modules, F (as right F -module) is finitely presented then by [3, Exercise 2.11.9', p. 337] we have S (B) But this contradicts Proposition 2.1 and hence A and B cannot be infinite sets simultaneously.
Corollary 2.3.For any two rings R and S, and any infinite sets A and B, the rings RFM A (R) and CFM B (S) cannot be isomorphic rings.In fact, RFM A (R) and CFM B (S) cannot be Morita equivalent rings.
Proof: The first statement is a direct consequence of Theorem 2.2.For the second statement, take E = RFM A (R) and F = CFM B (S).By [4, Theorem 1] we know that, if E and F are Morita equivalent rings then there exists n ∈ N such that E ∼ = M n (F ) as rings; but also M n (F ) ∼ = F as rings, because F has SBN.So that E ∼ = F which is impossible by Theorem 2.2.
At present, we have seen that for any two rings R and S the 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 M is (left) slender if for every homomorphism θ : R N → M , we have that, θ(e n ) = 0 for all but finitely many n ∈ N; where e n is the element of R N such that Supp(e n ) = {n} and ρ n (e n ) = 1; where ρ n : R N → R n are the usual projections.
Example 2.4.The rings R = RFM(Z), S = CFM(Z) and bimodules We shall construct a Morita context as in [1,Exercise 22.11].To do this, note that by [2, Corollary III.2.3] Z is slender so that Hom Z (Z N , Z) ∼ = ⊕ N Hom Z (Z, Z) as abelian groups, which implies that Hom Note also that, for any slender ring R, even if R is 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 R and a left infinitely generated projective module R P (which satisfies P n ∼ = P for all n ∈ N) such that the ring End( R P ) has an involution.
Let S be any ring and S Q = S S (N) .Set T = End( S Q) and U = T op .Setting R = S × U , P = Q × U we have that End( R P ) ∼ = T × U as rings and note that P ∼ = P n for every n ∈ N. Now let δ : T × U → T × U such that, for every (a, b) ∈ T × U , δ(a, b) = (b, a).It is clear that δ is an involution.
and η a : R (A) → R a be the usual projections.Define, for each a ∈ A, the setN a = ∪ b≤a Supp(x b ρ a ), where Supp(x) = {a ∈ A | xη a = 0} for x ∈ R (A) .Since Supp(x b ρ a ) is a finite set and |S a | < |A| then |N a | < |A| and hence N a ⊂ A (strictly).So, for each a ∈ A we may choose ξ(a) ∈ A\N a ; that is, for each a ∈ A, there exists ξ(a) ∈ A such that x b ρ a η ξ(a) = 0 for all b ≤ a.
as rings.Now by [1, Exercise 22.11(1)] we have that M, N is a Morita context.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 R is slender so is RFM(R).