Internat. J. Math. & Math. Sci.
VOL. 17 NO. 3 (1994) 617-618
617
SEMIPRIME SF-RINGS
WHOSE ESSENTIALLEFT
IDEALSARE TWO-SIDED
ZHANGJULEandDU XHIANNENG Anhui Normal University Wuhu 241000,P.R.of China
(Received
April21,1992)
ABSTRACT. It is provedthat if R is asemiprime ELT-ringand everysimple right R-module is flat thenR is regular. IsRregularif R isasemiprime ELT-ring andeverysimple right R-module isflat?
In
thisnote,wegiveapositiveanswertothe question.KEY WORDS
ANDPHRASES. (Von Neumann) regularring, SF-ring, ELT-ring.
1991AMS SUBJECT CLASSIFICATION
CODE.
16A30.1. INTRODUCTION.
In [1] YueChi Mingproposedthefollowing question: IsR regularif R is asemiprimeELT-
ring and every simple right R-module is flat? In
this note, we give a positive answer to the
question.
All rings considered in this paper are associative with identity, and all modules areunital.
A
ring R is(Von Neumann)
regular provided that for every a R there exists R such thata aba
(see [2]).
R is called astrongly regularring if for each a R,aa2R.
Following[1],
call Rand ELT-ring if every essential left ideal is an ideal of R. We call R a right SF-ring if every simplerightR-moduleisflat
(see [31).
2. MAIN
RESULTS.
Webeginbystatingfollowinglemmaswhich willbe usedinproof ofourmainresult.
LEMMA
1.([4],
p.30,Exercise19)
IfR isasemiprime ring,thenSoc(RR Soc(RR).LEMMA
2.([5],
Corollary8.5)
If R is a semiprime ring,then.every
minimal left (right)ideal isgenerated byanidempotent.
LEMMA
3.([3],
Proposition3.2)
Let Rbealeft (right) SF-ring. IfI isan ideal ofR, then R/Ialso isaleft (right) SF-ring.LEMMA
4.([3],
Theorem4.10)
Let R be a left (right) SF-ring. Ifevery maximal right(left)
idealofR isanideal,thenRstrongly regular.LEMMA
5. If R is a semiprimeELT
and right SF-ring, then R is fully left (right) idempotent.PROOF. From Lemma 1, Soc(RR)=Soc(RR). Now we write instead of Soc(nR).
By
Lemma 2, S is fully left (right) idempotent. Since R is an ELT-ring, and every maximal left idealof R/Sis an image ofamaximal essentialleft idealofR under the natural map v:R--R/S, hence every maximal left ideal of R/S is an ideal.By
Lemma 3, R/S is a right SF-ring. It follows fromLemma
4 that R/S is strongly regular, whence R/S is fullyleft (right) idempotent.Since S isfully left (right)idempotent, thenR isfully left(right)idempotent.
Nowweproveourmainresultwhichgives apositiveanswertothe question raised in
[1].
618 Z. JULE AND D. XIANNENG
THEOREM2.1. IfR isasemiprime
ELT
and rightSF-ring, then Risregular.PROOF. From Lemma5, Ris afully left (right) idempotent ring. IfP is aprimeidealof R, thenit iseasy to know that
RIP
isafully right idempotent ring. Since RisELT,
thisimplies that R/Pis an ELT-ring.By (see [6], Corollary 6),
R/P is regular. Considering that R is fully
idempotent, thus R isaregularring(see [21,Corollary 1.18).
1.18).
ACKNOWLEDGEMENT. This researchwassupportedbyAnhui EducationCouncil of China.
REFERENCES
1.
YUE
CHIMING, R.,
On Von Neumann regular ringsV,
Math. J. Okayama Univ. 22(1980),
151-160.2.
GOODEARL, K.R., Von Neumann RegularRings, Pitman,London, 1979.
3.
REGE, M.B.,
OnVon
Neumann regular rings and SF-rings, Math. Japonica 31 No. 6(1986),
927-936.4.
GOODEARL, K.R.,
Ring Theory: Non-singular Rings and Modules, Dekkar, New York, 1974.5.
FAITH, C.,
Algebra I: Rings,Modules,
andCategories, Springer-Verlag, 1981.6.