R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
G IUSEPPE B ACCELLA
Generalized V -rings and von Neumann regular rings
Rendiconti del Seminario Matematico della Università di Padova, tome 72 (1984), p. 117-133
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V-Rings
and Von Neumann Regular Rings.
GIUSEPPE
BACCELLA (*)
. Introduction and notations.
Throughout
thepresent
paper allrings
are associative with 0 ~ 1 and all modules areunitary.
Given aring .1~,
we shall denote with the Jacobson radical of l~. S will be a choosen set ofrepresentatives
of all
simple right R-modules,
while P is the subset of S of therepresenta-
tives of all
simple projective right
R-modules. Given aright
R-moduleM,
we shall write for theinjective envelope
of.~;
the notation(resp.
will mean that N is an B-submodule(resp.
an essential
.R-submodule)
of For agiven
subset A cS,
we shalldenote
by SocA
theA-homogeneous conponent
of the socle Soc ~11 ofif;
we shall write instead of(M),
for agiven
U E S.For every
right (resp. left)
R-moduleM, rR(M) (resp.
will bethe annihilator in .R of M.
A
ring B
is aright V-ring (resp. right GV-ring)
if allsimple (resp.
all
simple singular) right
R-modules areinjective.
We say that R isfully right idempotent (FBI,
forshort)
if a = a2 for everyright
ideala of .R or,
equivalently,
ifR(-R/t))
is flat for every two-sidedideal ~
of 1~. Several authors studied the connections between the above
concepts
and theconcept
of a(Von Neumann) regular ring (see [2],
[7], [13], [14], [16], [17], [18], [19], [20]).
Thepresent
paper is intended (*) Indirizzo dell’A.: Istituto Matematico, Universitadell’Aquila,
Via Roma, 33 - 67100L’Aquila (Italia).
Lavoro finanziato con i fondi concessi dal Ministero della Pubblica Istru- zione.
to
give
further contributions to thestudy
of the above classes ofrings (it
is well knownthat,
in the commutative case, all these coincide with the class ofregular rings).
In the first section we concentrate our attention to the class of those
rings R
for whichSoc .RR
isprojective
andR/80c ,RR
isfully right idempotent;
we call themGFRI rings (generalized .I’,RI rings).
This class
properly
includes the class ofright GV-rings and,
of course, the class of .F.RIrings.
We prove, amongothers,
that a GFRIring
has a
regular centre;
moreover, if SocRR
nJ(rR)
=0,
thenGFBI implies
FRI(actually,
we shall prove alsothat,
under the samehypo- thesis,
ifR/Soc RR
isregular,
then R isregular).
In sect. 2right GY-rings
are characterized
by
means of the behavior of theprojective components
of the socle of each
right
module. We also consider the class ofright SI-rings
introducedby
K.R. Goodearl in[8] (~
is aright SI-ring provided
eachsingular right
R-module isinjective;
.T~ is then aright GV-ring).
We prove that R is aregular right S1-ring iff R
is semi-prime
and issemisimple; thus,
forregular rings,
thecondition is
right/left symmetrical.
In sect. 3 we consider the class of
regular rings
which arestrongly regular
modulo theirsocle;
we call them ASRrings.
R. Yue ChiMing
considered in
[19,20]
some classes ofrings, namely
MELT leftsemiprime
MELT leftp-Y-rings,
MELTregular rings
andsemiprime
MELT
rings
whosesimple right
modules are flat(MELT
means thatall maximal essential left ideals are
two-sided).
We prove that all these classes coincide with the class of ASRrings. Finally
we provethat a
ring R
is aself-injective
ASRring
iff .R= S’ X S" X T,
whereS’ is a
semisimple ring,
S" is a directproduct
of divisionrings
and Tis a
strongly regular self-injective ring
with zero socle. Thisgives
a refinement of
[20,
Theor.27].
In the last section we discuss some
examples.
1.
Rings
which arefully right idempotents
modulo theright
socle.Throughout
this section .~ will be agiven ring.
Recall that if f is a two-sided ideal of.R,
then thefollowing
conditions areeqivalent:
(a)
isflat, (~)
for each x Ef, x
Exf, (c)
for each module andsubmodule
N ~ .M, (d)
for eachsimple
moduleSRIF, E(SR/t)
=E(SR) (the proof
of theequivalence
between(d)
and theother conditions can be found in
[2,
Theor.1.1]).
We shall makeextensive use of the
following result,
which wasoriginally proved
in
[3, Prop. 1.7,
Coroll. 1.5 and1.11] :
PROPOSITION 1.1..h’or every
ring
J? one hasSocp (.RR)
:=:(Soc RR)2.
Moreover, if
’~ is a two-sided ideal contained in Soc then thef oltowing
conditions are
equivalent:
(1)
f2 --- f.(2) R(-R/f)
isflat.
(3)
There is a subset A c P such that f =SOCA (.RR).
If
these conditionshold,
thenfor
each moduleMR
one hasSocA (31)
_ .M~.It is well known that if B is
semiprime,
then SocRR
=Soc RR
is
projective
on both sides.Moreover,
if n is a minimalright ideal,
then xt = eR for an
idempotent
e E R andSocn (RR)
= ReR is a mi-nimal two-sided ideal.
LEMMA 1.2. Let C be the centre
of
thering
1?. Given P E P and setf =
Socp (RR)’ tor
every a E C thefollowing
conditions areequivalent:
PROOF.
(1)
~(2).
There is anidempotent e
E .R such that eR and t == We have then = aBeR ReaR and henceCaR =,4-- 0,
from which eaR = eR. It follows that at = = f.(2) => (3):
it is obvious.(3)
~(1). Being flat,
we have0 ~ 1 = ~ n aR =
_=af.
PROPOSITION 1.3. Let PEP ==
Soc,
Then t isgenerated by a
centralidempotent i f
andonly i f
f n Cen(R)
# 0.PROOF. The
« only
if »part
is clear.Suppose
that 0 ~ a e t r’1n Cen
(R).
Since isflat,
we have 0 0 aR == aR n t = at andthen, by
Lemma1.2, t
= at aRc t,
so that t = We claim that~ n
J(R)
=0,
from which we will conclude that f isgenerated by
acentral
idempotent, according
to[3,
Theor.2.7]
andbeing f, finitely generated.
From the above andProp.
1.1 we have aR = f = f2 == aRaR =
a2.R,
therefore a = a2b for some b E .R.Suppose
thatx E f f1
J(R)
and let y E .R be such that x = ay. Then we have y = ay ==
a2by
=ba2y
=baay
= bax =0,
because ba = ab E f. Thus x = 0.PROPOSITION 1.4. Given a
ring R,
thefollowing
conditions areequivalent :
(1) R/(Soc Rn)2 is
aI’.RI-ring.
tor
eachsimple singular right
R-module S.(3)
SocI~R
isprojective
andi f
a is aright
idealo f
R such thattor
someA c P,
then a = a2.PROOF.
First,
observe that asimple right
R-module issingular
if and
only
if it is annihilatedby Soc RB. According
toProp. 1.1 r
B(R/(Soc RB)2)
is flat andhence, given
asimple singular right
R-moduleS,
theR/(Soc RR)2-injective envelope
of S is the same asE(SR).
Thusthe
equivalence
of conditions(1)
and(2)
is a consequence of[2,.
Theor.
3.2].
(2) => (3).
We shall prove thatE(SR/SOCRR)
=E(S)
for each sin-gular simple right
R-moduleS,
from which it will follow thatB(R/Soc J?R)
is flat and hence
Soc .RR
will beprojective by Prop.
1.1.Indeed,
ifSR
issimple
andsingular,
thenS(Soc RR)
= 0 and therefore(2) implies
that
(E(SR))
= 0.Taking
into account of[15, Prop. 2.27]
we
get
Now,
assume that a is aright
ideal of .R such that a r’1 Soc.RR
==
SocA (Rn)
for some subset ~1 c P and let x E a.By
theequivalence
of conditions
(1)
and(2),
there exist 0153l,...,0153rq 9 andz E
(Soc Rn)2
= SocRR
such that x = xlyl+ ... -~- -~-
z. This im-plies
that z ESoc RR
=SocA (Rn).
Since the latter isidempotent
we infer that z E a2 and we conclude that x E a2.
(3)
~(1):
it is clear.If .R satisfies the
equivalent
conditions ofProp. 1.4,
then we call1~ a
(generalized .F’.RI-ring) . Thus R
is GFRIexactly
when Soc
RR
isprojective
andR/Soc RR
is aPROPOSITION 1.5.
I f
R is aGFRI-ring,
then the centre Co f
R isa
regular ring.
PROOF. Assume that R is GFRI. We claim
that,
for every a EC,
there exists b E .Z~ such that a = a2b. Then the
regularity
of C willfollow
by
the sameargument
as in theproof
of[13,
Lemma2.3(d)].
Given a E
C,
for every P E P we have aR nSocp (.RR)
=a(Socp
by
the flatness of It follows from Lemma 1.2 that eitherSocp (RR)
c aR orSocp (RR)
n aR = 0. Inasmuch as SocRR
isprojective by Prop. 1.4,
we inferthat,
in any case,Soc RR
==
SocA (RR)
for some subset .A c P(possibly
A =0). Thus, according
to
Prop. 1.4,
aR =aRaB
= a2.R and hence a = a2b for some b E .1~.A module
MR
is said to bep-injective
if for everyprincipal right
ideal a of
1~,
allB-homomorphisms
from a to M extends to 1~.Equiva- lently,
.M isp-injective
ifExt’ M) = 0
for each x E .R.p-injec-
tive
modules, expecially simple p-injective modules,
were studiedextensively by
R. Yue ChiMing,
who also gave thefollowing
defini-tion : the
ring
R is aright p-Y-ring (resp. p-Y’-ring)
if everysimple (resp. simple singular) right
.R-module isp-injective (see [16, 17,
18, 19]).
If R is aright p-V-ring,
then is aFRI-ring (see [16,
Lemma
1]) ;
moreover .R isregular
iff eachright
R-module isp-injec-
tive
(see [17,
Lemma2]).
PROPOSITION 1.6. Given a
ring R,
thefollowing
conditions areequivatent :
( 1 )
.R is aright p - Y’-ring.
(2) Soc RR is projective
andR/Soc .RR
is aright p-Y-ring.
Thus a
right p-V’-ring
is GSRI.PROOF.
(1) =>(2).
If n is ap-injective
minimalright
ideal of.R,
then n is
generated by an idempotent.
This shows that Soc.RR
isprojec-
tive. Since the
simple singular right
.R-modnles areexactly
thesimple right R/Soc RR-modules,
it iseasily
seen that if R is aright p-Y’-ring,
then
R/Soc RR
is aright p- Y-ring.
(2) => (1).
Assume that(2)
holds and set Given asimple singular right
.R-moduleS,
we have Sf = 0.Being (B/Soc RR)
flat,
for each x E .R we have(we
areusing
the fact that,R/x,R -~- ~
^~+
and xR+ -~- f/t
is aprincipal right
ideal ofRjt) .
Thus R is aright
PROPOSITION 1.7. be a two-sided ideal
of
thering
.R containedin
Soc En
and suppose that f r1J(R)
= 0. Then thefollowing
are true:(i)
R is andonly if Rlf
is FRI.(ii)
R isregular if
andR If
isregular.
PROOF. In both statements
(i)
and(ii)
the« only
if »part
is tri-vial,
so we must prove the « if »part. Firstly, by
thehypothesis
and[3, Prop. 2.1],
both and(Rlf)n
are flat and f =SocA (RR)
forsome subset A c P.
(i).
Assume thatR/f
is We claim that for eachsimple right
.R-module S theequality
=rR(E(SR)) holds,
from which it will follow that R is
by [2,
Theor.3.1]. Suppose
that and
set ~
=Since ~
r1J(R)
=0, taking
intoaccount of
[4,
Theor.1.3]
weget rR(S)
= = c cc
rR(S),
whencerR(S)
=rn(E(Sn)).
Assume that S E S - A. Then ,S~ = 0and, being flat,
we have that =E(SR).
SinceRlf
is.F’.RI,
then rB/t(S)
=rR/r(E(SRlt))
=rR/t(E(SB))
and weinfer, again,
=rn(E(Sn).
(ii). Suppose
thatRlf
isregular.
In order to prove that .I~ isregular
we must prove that every module
M
is flat. Inasmuch asfR
isprojec- tive,
then it isflat; thus, given
any moduleMR,
weget
the exactsequence
According
toProp. 1.1,
we have Mt =SocA (~j
and hence isflat, being projective.
FurthermoreM/l£t
is flat as aright R/f-module,
because is
regular. Being flat,
it follows from[1, Prop. 10]
that is flat.
Finally,
from the exact sequence(1)
we concludethat
MR
is fiat.As in
[4],
we call thering
.Rright weakly semiprime (right
w. s.for
short)
if(Soc RR)2
r1J(R)
= 0. Thus theforegoing proposition
allows us to state the
following corollary,
the first statement of whichgeneralizes [18, Prop. 6], according
toProp.
1.6.COROLLARY 1.8.
Suppose
that .R isright
w. s.. Then thefollowing
are true :
(i)
andonl y i f
R is GFRI.(ii)
R is andRf(Soc Rll)2 is regular.
2.
Applications
togeneralized V-rings.
In
[14,
Theor.3.3]
aright GV-ring
.Z~ is characterized in terms of the behavior of the Jacobson radical of eachright
R-module. We want togive
an alternativedescription
of ,1~ which involves theright
socle of ,R. It is clear that a
right GV-ring
with zeroright
socle is aright V-ring.
PROPOSITION 2.1. For every
ring
R thef ollowing
conditions areequivalenti:
(1)
is aright V-ring.
(2) If
M is aright
then every essential submoduleof
_
M is an intersection
of
maximal submodulesof
M.PROOF.
(1)
~(2).
Let M be aright R-modnle
and let .L be anessential submodule of if. Since
RR)
cSoc(.M),
thenM(Soc RB)
cc .L. If
R/Soc RB
is aright V-ring,
then it follows from[13,
Theor.2.1]
that as a
right R/Soc BB-submodule
ofis an intersection of maximal
R/Soc RR-submodules
ofThis is
enough
to conclude that L is an intersection of maximal sub- modules of M.(2) => (1).
Let S be asimple right R/Soc BR-module,
let a be aright
ideal of.R,
withSocRR c a
anda/Soc RR
essential inR/Soc Rll,
and let
f :
be a non-zeroR/Soc RR-homomorphism.
Weclaim that if ~: a
a/Soc RR
is the canonicalepimorphism
andb = then b « aR. If
not, then, by
the definition ofb,
thereis a minimal
right
ideal n ofR
such that a == b(D
n, in contradiction with the fact that SocRB
c b. Inasmuch asa/Soc Rll
is essential inR/Soc Ell’
a is essential in tR and hence b is essential inSince b #
afrom
(2)
it follows that there is a maximalright
ideal m of .R such that ’b em andaim. Being b
maximal in a, we have that b = a r1 mand therefore
Ker(f)
=b/Soc RR
=(a
nm)/Soc .RR
=(a/Soc
r1(mj
Soc From this and
by
the « 3 X 3 » lemma it follows thatt
extendsto an
R/Soc RR-homomorphism
fromR/Soc .RR
to S. This shows thatS RISoc RR
isinjective..
THEOREM 2.2. Given a
ring .R,
thef ollowing
conditions areequivalent : ( 1 )
.R is aright GV-ring.
(2)
SocRR
isprojective
andR/Soc RR
is aright V-ring.
(3) If
M is aright
thenZ(M)
r1M(Soc RR)
= 0 andevery essential submodule
o f
M is an intersectionof
maximalsubmodules.
PROOF.
(1)
«(2).
Theproof
is similar to theproof
ofProp.
1.6.(2)
«(3).
Thisequivalence
is a consequence ofProp. 2.1, taking
into account that if Soc
RR
isprojective, then,
for each moduleMR,
we have
M(Soc .RR)
=Socp(.M~) by Prop. 1.1,
whence r1M(Soc RR)
= 0. ·Since a
right GV-ring
isGFRI,
from Coroll. 4.8 we obtain the follow-ing
two results(recall
that if .R is aFRI-ring,
thenJ(R)
=0) .
PROPOSITION 2.3..F’or a
right GV-ring
.R thefollowing
conditionsare
equivalent:
( 1 )
.R is(2)
.R isright
w.s..(3) J(.R)
= 0. oPROPOSITION 2.4. A
ring
.R is aregular right GV-ring if
andonly if
= 0 andR/Soc .RR
is aregular V-ring..
Combining Prop.
2.3 with[2,
Theor.4.5],
we obtain thefollowing
theorem which
generalize [14,
Theor.4.8] (see
also[11, Prop. 1.3]).
THEOREM 2.5. Given a
ring .R,
thefollowing
conditions areequivalent : (1)
R is asemiprime right GV-ring
withfinite right (resp. left)
Goldie dimension.
(2 )
,R is asemiprime right GV-ring
and R is aright (resp. left)
Goldie
ring.
(3) 1? its isomorphic
to a directproduct o f f initely
manysimple right V-rings
which areright (resp. left)
Goldierings..
We
point
out that the condition(2)
of[14,
Theor.4.8]
is not con-gruent
with the others stated there.Indeed,
if S is asemisimple ring
and T is a
simple right
NoetherianV-ring
which is not Artinian(see [6]) ,
then S X T is asemiprime right
Noetherianright GV-ring and, obviously,
it is neithersimple
nor Artinian.However,
in view of ourTheor.
2.5,
that condition should bereplaced
with thefollowing:
is
isomorphic
to a directproduct
offinitely
manysimple right
Noetherian
right V-rings)>.
A
special
class ofright GV-rings
is that ofright SI-rings
introducedby
Goodearl in[8,
Ch.III] ;
these are characterizedby
the fact that everysingular right
module isinjective.
LEMMA 2.6. A
right SI-ring
with essentialright
soel eif
andonly i f
Soc.RR is projective
andR/Soc RR
issemisimple.
PROOF. The
« only
if »part
is astraightforward
consequence of the definition of aright SI-ring. Conversely,
assume that Soc1~R
is
projective
andR/Soc RR
issemisimple.
IfhCR
issingular,
thenM(Soc RR
=0,
so M is aninjective right Rjsoc RR-module. Being RR)
flat(see Prop. 1.1),
it follows that.MR
isinjective..
Goodearl
proved
thefollowing decomposition
theorem forright SI-rings (see [8,
Theor.3.11]):
THEOREM 2.7..~ is a
right SI-ring if
andonly if
Rdecomposes
as :where
(see
above Lemma2.6)
and eachRi
is asimple right SI-ring
which is Moritaequivalent
to a domain.The most difficult
(and longest) steep
in Goodearl’sproof
of thistheorem is to find the
decomposition (2),
where S has essentialright
socle and each
,Rz
issimple.
We wish to exhibit here asimpler
andshorter
proof
of this fact.PROOF OF THE DECOMPOSITION
(2).
Assume that R is aright SI-ring
and set f = Soc
RR. According
to[8, Prop. 3.6], Rlf
is aright
Noethe-rian, right V-ring
andthen, being FBI,
from[13,
Lemma3.1]
== where each
R,
is asimple right Noetherian, right V-ring.
Let us write= hi/f,
for a suitable two-sidedideal hi
s ofB,
and setIi
=1,,(f) n 1),
1(1 c i c m). By Prop.
1.1 we have1,
~1 f ==
l~f
= 0 and henceII,
= 0. There are elements a1, ... , ame,B
with thefollowing properties:
(a) aiE1)"
(b)
alai - ai Et,
(c)
E t wheneveri =1= j, (d)
air - raz E t for each r EBuy
Given i E
11,
... ,m~,
if1,
=0,
then f a(~ t)R
andconversely. Assume
that
1,
# 0. Inasmuch ax= Ri
is asimple ring and l1 n f
=0, then ~ f ;
hence there areunique ei
E 9f
suchthat a,
_= 6i
+ fix.
We claim
that ei
is a centralidempotent
andet~ = li . Indeed,
it is clear
that
=e~ +
so that =e~ -
e~+ f ~ - f s,
whence
e2i
= Given r E.R,
we have f 3 air - rai = eir - re=+ f ir
-- therefore esr - rei =
0, being
two-sided ideals and f = 0.Thus 6i
is a centralidempotent. Finally,
since0 ~ 6¡R
c1,
andli
gzis a
simple ring,
we conclude that6¡B
=li
and the canonical map
R -+ R/f
induces anisomorphism Ri.
We may now assume that there is an
integer n,
with0 c n ~ m,
such that iff and we have the central
idempotents
61’ ... , en such that every
ei,R
is asimple ring. Taking
into account of(c),
if1 c i ~ j ~ n,
= 6iej+ -~- tiei +
thereforee,e; = 0
(remember
thatlif
= 0 =ffi).
Thus the areorthogonal.
If we set
f =1- e1- ... - en,
thenf
is a centralidempotent and, by
theabove, fR
=+
...+ 1)m
and We conclude that.R decomposes
as in(2 ),
where S =fR
is aright SI-ring
with essentialright
socle and eachR~
=e;R
is asimple right Noetherian, right
SI-ring.
Since a
right SI-ring
hasprojective right
socle and aring
withessential, projective right
socle isright non-singular,
we deduce fromTheor. 2.7 the
following corollary:
COROLLARY 2.8. A
right SI-ring
isright non-singular.
The
following
resultslightly improves [8,
Coroll.3.7].
PROPOSITION 2.9. Given a
ring R,
thefollowing
conditions areequivalent :
(1 )
R is aright ,SI-ring,
Soc.RR v ~R
andJ(R)
= 0.(2 )
R issemiprime
andR/Soc RR
issemisimple.
(3)
R is aregular left
PROOF.
(1)
«(2).
It is a consequence of Lemma 2.6.(2)
~(3).
Inasmuch as R issemiprime,
then SocRn
=(Soc BR)2 Soc,,R
and Coroll.1.8implies
that 1~ isregular. Combining Prop , I ,I
with
[2, Prop. 1.4],
we see thatSocR
«R~ implies
Soc sR -aRB.
Thus R is a left
SI-ring by
Lemma 2.6.(3)
~(2).
If R is aregular
leftSI-ring,
thenR/Soc
R is a leftNoetherian
regular ring by [8, Prop. 3.6]; hence R/Soc
=R is semi-simple..
3.
Regular rings
which arestrongly regular
modulo their socle.Let R be a
given ring.
We call R anASR-ring (almost strongly regular ring)
if .R isregular
andR/Soc
R isstrongly regular.
In thefollowing
theorem we characterize theASR-rings
among thoserings
for which all essential maximal
right
ideals are two-sided(it
is clearthat an
ASR-ring
has thisproperty).
THEOREM 3.1. Assume that every essential maximal
right
idealo f
the
ring :R
is Then thefollowing
conditions areequivalent
to the condition that R is ASR:
( 1 )
R isright
W.8. and allsimple left
R-module8 areflat.
(2)
R isright
W.8. andRj(Soc
isstrongly regular.
(3)
R isright
w.s. and is aright p-Y’-ring.
(4)
R isright
w,s, and is aright (5)
R isregular.
( 6 )
R i8If,
inaddition,
everyprojective simple right R-module
P isRlrR (P)- injective,
then the above conditions areequivalent
to :(7) right V-ring.
Finally, if
all maximalright
idealsof
rR aretwo-sided,
then all the above conditions areequivalent
to thefollowing
one :(8)
.R isstrongly regular.
PROOF.
(1) =>(2).
This is a consequence of[2,
Theor.3.7],
sincethe
hypothesis
of the theoremimplies
that all maximalright
idealsof
RI(Soc RR)2
are two-sided.(2) ~ (5).
It is a consequence of Coroll. 1.8.(5) =>(6).
It is clear.(6) => (4).
If R is.P.RI,
then R issemiprime
and hence Soc B =thus
BISOC RR
is aright V-ring by [2,
Theor.3.7]
and .Ris a
right GV-ring by
Theor. 2.2.(4) => (3).
It is clear.(3) ~ (2). According
toProp. 1.6,
if .R is aright p-V’-ring,
thenSoc RR = (Soc RR)2
andRI80c
is aright p - V-ring.
ThusR/Soc BR
is
FBI by [16,
Lemma1]
and(2)
follows from[2,
Theor.3.7].
(5) => (1).
It is clear.Now, by
virtue of theequivalence
between(5)
and(2),
we seethat
(5)
isequivalent
to the fact that I~ is A~SR.Assume that every
simple projective right R-module
P isR/rR(P)- injective
and that conditions(1) through (6)
are fulfilled. Then is flatby (5)
and hencePR
isinjective.
Thus(4) implies (7).
Conversely (7) implies (4),
because aright V-ring
issemiprime.
The last
part
of the theorem isagain
a consequence of[2,
Theor.3.7]. ·
REMARK 3.2.
Rings
with theproperty
that all essential maximal left ideals are two-sided were named MELTrings by
Yue ChiMing
in
[19, 20];
he also consideredsemiprime
MELT leftp- Y’-rings, semiprime
MELT leftp- Y-rings,
MELTregular rings
andsemiprime
MELT
rings
whosesimple right
modules are flat. In view of the fore-going theorem,
we see that all the aboveconcepts
coincide with theconcept
of an ASRring. Consequently,
Theor. 3.1improves [18
Theor.
2]
and[19, Prop.
9 and10].
REMARK 3.3. If .R is an
prime ring,
then Soc 1~ ~0;
otherwise.R would be a
prime strongly regular ring, namely
a divisionring,
and hence Soc .R = .R : a contradiction. Thus
Soc B
v .Rand,
inparti- cular,
.I~ isprimitive.
Recall that .l~ is a
right full
linearring
if R isisomorphic
to theendomorphism ring
of aright
vector space over some divisionring.
The
following
lemmaprovides
ageneralization
of[19,
Theor. 11equivalence
of conditions(1)
and(4)].
LEMMA 3.4. Let R be a
prime regular ring
with maximalright
quo-tient
ring Q. I f Q is ASR,
then R is Artinian.PROOF.
By
the above Remark 3.3 thisimplies that Q
is a
right
full linearring.
It follows from thehypothesis
thatQ
isArtinian,
so 1~ has finiteright
Goldie dimension. We conclude that .~is
simple Artinian, being regular.
For A S.R
prime right V-rings
we have thefollowing
result:PROPOSITION 3.5.
Suppose
that .R is aprime
ASRring.
T hen Ris a
right V-ring i f
andonly i f
there is aright full linear ring Q
such thatSoc
Q
c R cQ. It it is
the case, then R is also aleft V-ring i f
andonly i f
Artinian.PROOF. Let us write f = Soc
According
to Remark3.3, f
isessential in
.I~;
moreover there is anidempotent e
e R such that f = BeR and eR is theunique (up
to anisomorphism) simple projective right
.R-module. It is well known that
Q
= EndfR,
thereforeQf
= ~. SinceE(eRR)
=eQ,
it is clear that isinjective
iff ’~ =fQ, namely
iffSoc
Q c
.R. The firstpart
of ourproposition
is now a consequence of Theor. 3.1.If .R is a
right
and leftV-ring,
then R has faithfulinjective, projec-
tive
simple right
and leftmodule,
hence R is Artinianby [12,
Coroll.2.2]. N
If R is an
ring,
then every essentialright
ideal is two-sided.Right selfinjective rings
with the latterproperty
were studiedby
K.
Byrd
in[5],
under the name ofright q-rings.
We aregoing
toprove a structure theorem for
right selfinjective
ASRrings
whichis a decisive
improvement
of[20,
Theor.27].
Since theserings
aremerely right q-rings
with zero Jacobsonradical,
our theorem couldbe
proved by using Byrd’s
structure theorem[5,
Theor.6]. However,
we exhibit here a more direct
proof.
THEOREM 3.6. Let R be a
right selfinjective ASR ring.
Then Rdecomposes as R
=S’ X
S" XT,
where S’ is asemisimple ring,
S" is adirect
product of
divisionrings
and T is aright (and
henceleft) selfinjec- tive, strongly regular ring
with zero socle. Thus .R is aright
andleft
V-ring
with bounded indexof nilpotence (see [10,
page71]) .
PROOF. Let us write f = Soc R.
Being
.I~right selfinjective,
thereis an
idempotent
such that eR = We claim that e is central.Indeed,
sinceRjf
isstrongly regular,
for each r E ,R there isx E f such that er = re
-~-
x, hence re E eR. Thus Be c e,R and therefore= eR.
Being R semiprime,
it follows that e is central. It is nowclear that T =
(1- e).R
is aright (and
henceleft) selfinjective strongly regular ring
with zero socle.The
ring
S = eR isregular, right selfinjective
and Soc ~S = t isessential in S. As it is well known
(see
e.g.[9,
Theor.3.28,
page92]),
there is a
family (Si)iEI
ofright
full linearrings
such thatFrom the
hypothesis, each Si
isASR,
so it is of the formMn. (Di),
where
Di
is a divisionring and ni
apositive integer.
We claim that3~i = 1 for all but a finite number of indices i E I.
Assume,
on thecontrary,
that there is an infinite subset J c I with ni > 1 for all i E J.For every
i E J,
choose two non zeroidempotents Vi E Si
suchthat uivi = vi and viui = 0 and consider a, b E S with
Then a,
b areidempotents and, taking
into account that all idem-potents
in arecentral,
weget
b = ab -4el
a contradiction. We conclude that S
= ~S’ >C S",
where ~S’ is a semi-simple ring
and S" is a directproduct
of divisionrings.
4.
Examples.
,,- ,- .. ~
a)
Let D be a divisionring,
letV~
be an infinite dimensional vector space overD,
setQ
=End YD
and f = SocQ.
Let S be anystrongly
regular subring
ofQ
and set A = S+ f,
thesubring
ofQ generated
by
t and ~. It is clear that Soc A = t ~ A and A isprime.
SinceAll
= ,S+ f/f
~s/s
isstrongly regular
and t2 =f,
it followsfrom Theor. 3.1 that A is ASR and is
a right
and leftG V-ring. According
to
Prop. 3.5,
.A. is aright V-ring
which is not a leftV-ring.
Inparticu- lar,
if T is thesubring of Q
of all scalar transformations ofYD,
thenB = T
+
f is thering
described in[10, example 6.19,
page68].
Notethat B is a
right (and left) SI-ring,
sinceBft
= D.The
following examples
envolve formaltriangular
matrixrings.
For the basic
properties
of theserings
see[9, Chap. 4].
b)
Let~’,
~’ be twosemisimple rings
and let M be a faithful(T, S)-bimodule.
Then is aright
and leftSI-ring
whichis neither a
right
nor a leftV-ring,
sincec)
Let A and f be as inexample a)
and consider thering
Then Soc and Soc are both
projective.
SinceRfsoc RB I"J R/SocR
B - A and the latter is aright V-ring (recall
thatA/f
isstrongly regular)
which is not a leftV-ring,
it follows from Theor. 2.2 that R is aright GV-ring
and isnot a left
GV-ring.
SinceJ(R) = (0 0) # f 0 0
’ .R is not aright
gV-ring.
Observe that R is
right
and leftnonsingular
and is neither aright
nor a left
8I-ring.
d)
Let be the domain described in[6,
page78].
is asimple
non Artinianright
and leftV-ring
which is aZ2-algebra.
Ifwe consider the
ring .
I then SocRR
= 0 andSOCR
I~ _is
idempotent. Since gz RM
is aV-ring,
it fol-lows from Theor. 2.2 that l~ is a left
GV-ring.
It cannot be aright GV-ring,
otherwise it would be aright V-ring,
in contradiction with the fact that 0.e)
Let B and f be as inexample a)
and consider thering
Then Soc
/f 0B
and
Soc
Since
BA,
R is not aright GV-ring, although
R/Soc RR
r-v D x D is aV-ring.
On the other hand we havethus .R is a left
SI-ring by
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