Proof. Let H be the intersection of all right ideals in
L*(R~), which is easily seen to be a two-sided ideal of R. If A
is any nonsingular right R-module, then the right annihilator of
any element of A belongs to L*(R ), whence AH = 0. In addition,
[4, Proposition l.ll] says that R/H is a right nonsingular ring,
and that any right (R/H)-module is nonsingular as an (R/H)-module
if and only if it is nonsingular as an R-module. Thus the category
of nonsingular right R-modules coincides with the category of
nonsingular right (R/H)-modules.
Any A E 7?(R) is clearly injective as an (R/H)-module, whence
A c 7KR/H). Thus 7?(R) C r(R/H). Now consider any A
WH/H), and
set B = E(A^). Since A is a nonsingular essential right
R-submodule of B, B must be nonsingular, whence B is a right
(R/H)-module. However, -^p/tr ^s injective and thus is a direct
summand of B, whence A = B and consequently A e 7?(R). Therefore
«R) = WR/H).
We may now replace R by R/H; i.e., we may assume, without
loss of generality, that R is a right nonsingular ring.
Using the notation of [4], set Q = S R, which is a regular,
right self-injective ring by [4, Proposition 1.16]. Given any
A e 7?(R), we see that S°A = A, hence A is a right Q-module.
Then A is a nonsingular Q-module by [4, Proposition 1.10], and the
remarks following [4, Theorem 1.14] show that A is injective,
whence A e ??(Q). In addition, we see from the remarks following
[4, Theorem 1.9] that Hom^AjB) = - Honu(A,B) for all A,B e 7J(R)f
and consequently 7?(R) c ??(Q). Conversely, consider any A e 7J(Q),
and note from [4, Proposition 1.10] that A_ is nonsingular. In
view of the remarks preceding [4, Proposition 1.10], A must be a
Q-submodule of S A. Since A is injective, it is thus a direct
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