NONSINGULAR INJECTIVE MODULES 9

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,

IX

[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

e

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