if S is infinite there is no good relation between property A(n) for R and property A(n) for the connected component rings of R (Examples 5.16 and 5.21). §5 consists of eight counterexamples, four of which have been mentioned above. Example 5.1 is a noncommutative ring which is not a left A(n) ring for any «. Three sufficient conditions for a ring R to be an A(l) ring are: (1) R is an A(0) ring; (2) R is a C-ring, meaning that any exact sequence 0->M-> F-> N ^0 of finitely generated 7?-modules with M flat and F free splits ; and (3) R is a 73-ring, meaning that for any index set 7, R' is a submodule of a flat module. Examples 5.2, 5.3, and 5.6 show that these three conditions are logically independent and hence that none of them are necessary for R to be an A(\) ring. Our notation is basically that of [3]; in addition we use "f.g." for finitely generated and "f.p." for finitely presented. ls is the identity function on a set S; » stands for all kinds of isomorphism; Si; is the Kronecker delta; Zis the ring of integers, and N is the set of positive integers. 2. General results. This section includes all those statements about A(n) rings which we know to be valid for rings which are not necessarily commutative. Propositions 2.1 and 2.2, as well as the essential ideas in the proofs of 2.3 and 2.4 are all due to M. Auslander. Proposition 2.3 and Theorem 2.4 were first proved by S. Endo [8] for the special case when R is commutative and T is a ring of quotients of R with respect to a multiplicative system of regular elements. 2.1. Proposition. Let R be a ring, M a left R-module. For each set I, put M' = \~\iei M, and let a(M): R1 g) M->■ M1 be the canonical homomorphism. Then (i) M is f.g. o a(M) is surjective,for all sets 7(l). (ii) M is f.p. o c(M) is bijective,for all sets I. Proof. First suppose o(M): RM ® M'-> MM is surjective. Then 3/j,..., fkeRM, and xx,.. .,xke M such (hat F-> M-> 0 be exact with E and F free left modules. Note that A" is an exact functor of TV, o is functorial, and o(H) is bijective whenever H is f.g. and free. If M is f.g., we take F f.g., so o(F) bijective implies o(M) surjective. If M is f.p., we take E and F f.g., so o(E), a(F) bijective imply o(M) bijective. Finally, if o(M) is bijective for all sets 7, we may take F f.g., and let AT=ker (u). Then a(M), o(F) bijective imply o(K) surjective, so K is f.g., so M is f.p. Q.E.D. 2.2. Proposition. If for each set I, R1 is a submodule of aflat right module, then R is a left A(\) ring. Proof. Suppose u : M-*■ E is an injective morphism of f.g. left modules, with M flat and E free. Let 7 be a set and select an injective morphism v: R'-> N, (') The related assertion in [3, §1, Exercise 2a] is incorrect as worded.