Internat. J. Math. & Math. Sci.
VOL. 20 NO. 4 (1997) 825-827
825
SUBRINGS
OFI-RINGS AND S-RINGS
MAMADOUSANGHARE
Drpartement
deMathrmatiquesetInformatiques Facult6 desSciencesetTechniquesUCAD DAKAR (SENEGAL) e-mail sanghare@ucad,refer,sn
(ReceivedMay6,1993and in revised formFebruary 13,1997)
ABSTRACT. Let
R
bea non-commutative associativeringwithunity1:f:
0,aleft R-moduleis saidto satisfy property (I) (resp.(S))
if every injective (resp. surjective) endomorphism of M is an automorphism ofM. ItiswellknownthateveryArtinian(resp. Noetherian)module satisfies property(I) (resp. (S))and that the converse isnot true. AringR
iscalled a leftI-ring (resp. S-ring)ifeveryleft R-module withproperty(I)(resp (S))is Artinian(resp.Noetherian). Itisknown that asubringBof a left I-ring (resp. S-ring)R
isnotingeneralaleft I-ring (resp. S-ring)evenifRis afinitelygenerated B-module, for exampletheringM3 (K)
of3x 3 matrices over a fieldK isaleft I-ring (resp S-ring), whereas itssubringB=
a 07 0
which is a commutativeringwith anon-principal Jacobsonradical
J=K. 1 0
+K.
0 0 000 100
isnotanI-ring (resp. S-ring)(see [4],theorem8). WerecallthatcommutativeI-rings (resp S-tings)are characterized as those whose modules are a direct sum of cyclic modules, these tings are exactly commutative, Artinian, principalidealrings(see[1]). Someclassesofnon-’commutativeI-tingsandS- tings havebeen studied in [2] and
[3].
AringR
is offinite representationtype if it isleftandfight Artinian and has (up to isomorphism) only a finitenumber of finitely generated indecomposable left modules. In the case of commutativetingsorfinite-dimensional algebrasover analgebraically closed field,theclasses of left I-tings, left S-tings and tings offiniterepresentationtype are identical(see [1]and[4])
AringR
is said tobe a ringwith polynomial identity (P. I-ring) ifthere exists apolynomialf(X1,X2,...,X,),
r>
2, inthenon-commutingindeterminatesX1,X2,...,X, overthecenter ,7ofR suchthat oneof themonomialsoff
of highest total degreehas coefficient 1,andf(al,
a2,a)
0for allal,a2,
a
inR. Throughoutthispaperallringsconsidered are associativeringswith unity, and byamoduleM
over aringR
wealwaysunderstandaunitary left R-module. WeuseMR
toemphasizethat
M
is aunitary fight R-module.KEY WORDS AND PHRASES: Left I-ring, left S-ring, ringwithpolynomial idemity, ring offinite representation type.
1991AMSSUBJECTCLASSIFICATION CODES: 16D70, 16P10, 16L60.
826 M. SANOHARE 1. THE MAIN RESULT
THEOREM. LetRbealeftI-ring (resp. S-ring),and
B
be asub-ring ofR
contained in thecenterZ
ofR
SupposethatR
is afinitelygeneratedfiatB-module ThenB
isanI-ring (resp S-ring) To provethistheoremweneedsomeresults.Itiseasytosee that
LEMMA
1.Every
homomorphic image ofaleft I-ring (resp S-ring)is aleft I-ring (resp S-ring) LEMMA 2. LetPx
andP2
be twoprimeidealsof aringR. IfP1
is notcontained inP2
thenHom(R/P, R/P2) {0}
PROOF. Let
f- R/P1 R/P2
be an R-homomorphism, and setf (1 + P1 +
P2, whereER. Letx E
P1 \P2,
and letrbe any elementinR. WehaveP2 f(xr + PI)
xrt+
P_ ThusxRt
P2.
SinceP2
isprime,wehave 6P2,andhencef
0.LEMMA
3. LetR
beaprime ringwithpolynomial identity. IfR
is aleft I-ring(resp.S-ring), thenR
issimpleArtinian.PROOF. Let
R’
bethe totalring of fractions ofR
[5]. Itisknown thatR’
issimpleArtinian[5],so theR-moduleR’
satisfies(I)(resp.(S)).
SinceR
is aleft I-ring (resp. S-ring), thenR’
is an Artinian (resp. Noetherian) R-module and henceR
R.LEMMA
4. LetR
beasemi-prime ringwithpolynomial identity. IfR
isaleft I-ring(resp S-ring), thenR
issemi-simpleArtinian.PROOF. Let
(Pt)te.
be afamily pairwisedistinct minimalprimeidealsofR
such thatteL
By Lemma the quotient tingsR/Pt(g
L)
areleft I-tings (resp. S-rings)withpolynomialidentity Thenitfollows from Lemma3 that the tingsR/Pe( L)
are simpleArtinian, so theleftR-modulesR/Pt(g L)
satisfy(I)
(resp(S)).
FollowingLernma1,HomR(R/Pt, RIPe) {0}
for",
sothe leftR-moduleMtet.R/Pt
satisfies(I) (resp. (S)). SinceR
is aleft I-ring (resp. S-ring), thenMis Artinian. ButR regardedasleft R-moduleisisomorphictoasubmodule of the semi-simpleArtinianleft R-moduleM,
henceR
issemi-simpleA.,’tinian.PROPOSITION5. Let
R
bearingwithpolynomial identity. IfR
is aleft S-ring (resp. I-ring), thenR
isleftArtinian.PROOF. Supposethat
R
is aleft S-ring (resp. I-ring)thenthe quotient ringR/rad(R),
wheretad(R)
is the prime radical ofR,
is a left S-ring (resp. I-ring), so, follqwing Lemma 4, the ringR/rad(R)
issemi-simpleArtinian ThisfactimpliesthatR
issemi-perfect and hencetad(R) J(R),
where
J(R)
is the Jacobson radical of R. Let e be a primitive idempotent of R. Since the endomorphism ringof theR-moduleReisisomorphictothelocal ringere
witha nil Jacobson radicaleJ(R)e,
thenthe R-moduleRe satisfiesproperty(I)
(resp (S)). Itfollowsthat theR-moduleRe
isNoetherian (resp. Artinian). Since
R
regarded as R-module is direct sum of finitely many left R-modules of the formRe,
whereeisaprimitive idempotent ofR,
thenR
is Noetherian. LetP
now be aprime ideal ofR.
Since theprime ringR/P
issimplein virtueof Lemma 3, thenR
isleftArtinian.PROOF OF
THE
MAINTHEOREM. SinceR
is afinitelygeneratedZ-module, thenR
is a ring withpolynomialidentity(see [6]). Soby Proposition 5R
is aleftArtinianring. Thusby[7]
the ringB
is Artinian. Let el ,e, be primitive idempotents of
B
such that B=e,
Bei For everyi,1
<_ <_
n,B,
e,Be,
isalocalArfinian ring. ToshowthatBis aleftI-ring(resp. S-ring)it isenough to show thatforevery i, 1< _<
n,B,
is aleft I-ring(resp. S-ring). WehaveA
$=IAi, whereA,
e,Ae,,
1_< _<
n. Byhypothesis theleftB-module=1 A, A
is flat andfinitelygenerated,so theBi-module
A eAe, - eiAe
(R)BBA
(R),e,Be A
SUBRINGSOFI-RINGSAND S-RINGS 827 isalso flat and finitely generated Since
B,
isan Artinianlocal ring thentheB,-moduleA,
isfaithfully flat(see[8]proposition1,p.44)SupposenowthatB, isnotanI-ring(resp. S-ring) forsome i, 1
_< _<
n Thenby Proposition2of [2],there exists aB,-moduleMofinfinitelength such that, for every integern_>
1, theB,-moduleM satisfies both properties(I) and (S) Following [$] (corollary 2, p. 107), the B,-moduleA, is afree module. LetM’
M(R)B,A,.
SincetheBi-module
M isofinfinitelengthandA,
is afaithfully flat Bi-module, thenM’
is anA,-moduleofinfinitelength. Onthe otherhand,sinceA,
is afree B,-module, thereexists anintegers_>
Isuch thatA, B.
Wehave then theB,-moduleisomorphismM’
M(R)B,A M (R)B,Bt -
M’.Hencethe B,-module
M’
M satisfiesboth properties (I) and (S) and thereforeM’,
regarded as A,-module,satisfiesproperties(I)and(S) Thisfact impliesthatthe homomorphic imageA,
oftheleftI- ring (resp. S-ring)A
is not aleft I-ring(resp.S-ring),in contradiction withLemma1.COROLLARY. Let
R
bealeft I-ring(resp. S-ring). IfR
is afinitelygenerated flat moduleover itscenterZ,
thenZis anI-ring (resp. S-ring).Thefollowing example showsthattheconverseof thetheoremaboveisnot true" LetKbe a field The commutativeringA
K[X,Y]/(X2,XY,
Y2)
isnotanI-ring (resp. S-ring)because its Jacobson radical 3’ KX+
KYis notprincipal(see [1],theorem8). OntheotherhandK
isanI-ring (resp S-ring)andA
is afinite-dimensionalK-vectorspaceACKNOWLEDGEMENT. Theauthor wouldlike to thank thereferee forhisvaluable suggestionsand numerousvery useful remarks aboutthetext.
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