NEW ZEALAND JOURNAL OF MATHEMATICS Volume 25 (1996), 195-197

A N E L E M E N T A R Y PR OO F OF A C H A R A C T E R IZ A T IO N OF SE M IP E R FE C T R IN G S

W .K . N i c h o l s o n

(Received September 1995)

Abstract. An elementary proof is given of a recent characterization of semiper­fect rings as the rings with no infinite orthogonal family of idempotents in which every element is the sum of an idempotent and a unit.

A ring R is called clean if every element is the sum of a unit and an idempotent, and R is called I-finite if it contains no infinite orthogonal family of idempotents (equivalently if the idempotents in R satisfy the ACC with respect to the partial order e < f if and only if e £ f R f ). If we write J for the Jacobson radical of R, then R is called semiperfect if R/J is semisimple artinian and idempotents can be lifted modulo J. Recently Camillo and Yu [1] obtained the following new characterization of semiperfect rings:

Theorem 1. A ring R is semiperfect if and only if it is I-finite and clean.

Their proof involves a number of deep results about exchange rings and, because of the importance of the theorem, we give a proof in this note which uses only elementary ring theory. Three preliminary lemmas, each of interest in itself, will be needed in the proof. Throughout the paper all rings are associative with unity, and all modules are unital.

Lemma 2. If the module r M has finite composition length, then end(M) is clean.

Proof. If a € end(M), Fitting’s Lemma shows (writing a to the right of its ar­gument) that there exists an integer n > 0 such that M = M an © ker(an) where M a n = M a n + 1 = . . . and ker(an) = ker(an+1) = . . . . Then M an and ker(an) are cn-invariant submodules, a = ct\Man is a unit in end(Man) and rj = a|ker(an) is nilpotent in end(ker(an)). Thus a is represented as

a 0 a 0 0 o'0 rj — 0 -1 + r) + 0 1

which is the sum of a unit and an idempotent. □

Corollary 3. If A is a division ring, the ring Mn[A] of all n x n matrices over A is clean.

Lemma 4. A ring R is clean if and only if R/J is clean and idempotents lift modulo J.

1991 AMS Mathematics Subject Classification: Primary 16L30; Secondary 16U60. Key words and phrases: Semiperfect ring, Clean ring.Research supported by NSERC Grant 8075.

196 W .K . NICHOLSON

Proof. If R is clean, R/J is clean because any image of R is clean. Let x 2 — x G J, and write x = e + u where e2 = e and u is a unit. Then

u[x — u~l { 1 — e)u] = ue + u2 — u + eu = x 2 — x G J

and it follows that x — u-1 (l — e)u G J. Thus idempotents lift modulo J.Conversely, if the conditions hold and x G R, let x = e + u in R/J where e2 = e

and u is a unit. By hypothesis we may assume that e2 = e. Then x — e is a unit in R/J so 1 — (x — e)v = a € J for some v 6 R. Hence (x — e)v = 1 — a is a unit in R, so x — e has a right inverse in R. Similarly x — e has a left inverse, and we have proved that R is clean. □

Remark 5. The proof shows that, if R is clean, then idempotents lift modulo any ideal of R in place of J. In fact idempotents lift modulo every one-sided ideal, and this is known to characterize the exchange rings [2].

Lemma 6 . If R is a clean ring and L J is a left ideal, then L contains a nonzero idempotent.

Proof. It suffices to show that if x £ J then Rx contains a nonzero idempotent. Suppose, on the contrary, that e2 = e G Rx implies e = 0. If a G R, write y = ax and let y = e + u where e2 = e and u is a unit. As before, u[y — u-1 (l — e)u] = y2 — y. If we write / = u-1 (l — e)u, this shows that y — f = u~l (y2 — y), so f 2 = / G Ry C Rx. Thus / = 0 by assumption, whence

1 = 1 - / = 1 - y + u~l {y2 - y) G R(l - y) = R{ 1 - ax).

As a e R was arbitrary, this shows that x G J, a contradiction. □

Proof of the Theorem. Assume that R is semiperfect. If ei, e2, . •. are orthogo­nal idempotents in R , then the same is true of eT,ej , . . . in R /J , whence en = en+i = . . . = 0 for some n. Since J contains no nonzero idempotents, we obtain en = en+i = . . . = 0, which proves that R is I-finite. The Wedderburn-Artin theorem gives

R/J = Mni [Ai] x M „2 [A2| x . . . x Mnm [Am]

where each A j is a division ring. Thus it follows from Corollary 3 that R/J is clean. Hence R is clean by Lemma 4.

Conversely, if R is clean then idempotents lift modulo J by Lemma 4, so it remains to prove that R/J is semisimple artinian. If L/J ^ 0 is a left ideal of R/J, we show that L/J is a summand of R/J. As R is I-finite, choose (using Lemma 6) e maximal in {e G R \ 0 ^ e2 = e G L}. Then L + R( 1 — e) = R, so it suffices to show that L fl P(1 — e) C J. If not we can choose 0 / / 2 = / g L ( 1 i?(l — e) by Lemma 6. Then f e = 0 so g = e + / — e f is an idempotent in L and e < g. Hence e = g by the maximality of e, whence / = e f. But then f = f 2 = f { e f ) = 0, a contradiction.

AN ELEMENTARY PROOF OF A CHARACTERIZATION OF SEMIPERFECT RINGS 197

Remark 7. A ring R is called potent if idempotents can be lifted modulo J and every left (equivalently right) ideal L J contains a nonzero idempotent. The proof of the Theorem shows that every I-finite potent ring is semiperfect. Conversely, semiperfect rings are I-finite (by the proof of the Theorem) and potent (by the Theorem and Lemma 6). Thus the semiperfect rings are just the I-finite, potent rings. Indeed, we have shown the following useful fact: If R is semiperfect and L J is a left ideal, there exists e2 = e G R such that L = Re © (L fl R(1 — e)) and L fl R( 1 — e) C J. In other words, R(1 — e) is a projective cover of R/L.

References

1. V.P. Camillo and Hua-Ping Yu, Exchange rings, units and idempotents Comm, in Alg. 2 2 (1994), 4737-4749.

2 . W.K. Nicholson, Lifting idempotents and exchanqe rinqs, Trans. A.M.S. 229 (1977), 269-278.

W .K . NicholsonDepartment of Mathematics and StatisticsUniversity of CalgaryCalgaryAlbertaC A N A D A T2N 1N4 wknichol@acs.ucalgary.ca