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
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.
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