10 K# R. Goodearl and A# K. Boyle

summand of S A, whence A is infective and so A e 7?(R)«

Therefore 7?(R) = fl(Q).

THEOREM l.l6. [23, Theorem 2.7] If R is a right self-injective

ring, then every finitely generated nonsingular right R-module is

projective and infective.

PROPOSITION 1.17. Let R be a regular, right self-injective

ring, and let A e 7?(R). Then A is a generator in 7&R) if &nd

only if A is faithful.

Proof. First assume that A is a generator, and consider any

nonzero x e R. According to l»l6, xR e 7?(R), hence there must

exist a nonzero map f : A -*xR. Now fA is a nonzero right ideal

2

of R, and (fA) = fA because R is regular. Thus

f(A(fA)) = fA 4 0, whence A(fA) £ 0. Since fA ^ xR, we obtain

Ax 4 = 0. Therefore A is faithful.

Conversely, let A be faithful, and consider any nonzero

B e ?KR). 3y l#l6, all cyclic submodules of B are projective,

hence there exists a nonzero element x e R such that xR is

isomorphic to a submodule of B. Now ax ^ 0 for some a £ A, and

axR is projective by 1.16, from which we see that axR is isomorphic

to a nonzero submodule of B. Thus HocuCaxR^) ^ 0, and

consequently Hom^AjB) ^ 0. Therefore A is a generator in 7KH).

THEOREM 1.18. Let R,S be regular, right self-injective rings.

Then 7lCR) is equivalent to 7((S) if and only if there exists a

faithful nonsingular injective right R-module A such that

EndR(A) =*S.