SET OF PROPER IDEALS
YASUYUKI HIRANO AND HISAYA TSUTSUI Received 5 July 2005
A study of pairs of commutative rings with the same set of prime ideals appears in the literature. In this paper, we investigate pairs of subrings, not necessarily commutative, with a common set of proper ideals.
Throughout, all rings are associative with an identity element but not necessarily com- mutative. During the decade of 1980s, a series of papers appeared in theCanadian Journal of Mathematics[1,2] that had investigated pairs of commutative rings with the same set of prime ideals. In this paper, we consider some generalizations of that study in the non- commutative setting.
Consider the ringR=HomD(V,V), whereV is a vector space over a division ringD with dimD(V)= ℵω0(ω0is the first limit ordinal). LetKbe any subfield of the center ofR, M= {f ∈HomD(V,V)|dimf(V)<ℵω0}, andS=K+M. ThenSandRhave the same set of countably many prime ideals. Further,RandShave the same set of ideals since all of their proper ideals are prime ideals. (See Blair and Tsutsui [4]). We will first observe that examples of this nature are limited in the commutative setting. It is an immediate consequence ofTheorem 1that the only possible pairs of a commutative ringRand its proper unital subringSwith the same set of proper ideals are fields. We will then briefly investigate pairs of subrings with a common ideal. This investigation will yield the fact that a pair of rings has the same set of prime ideals if and only if they have the same set of maximal ideals. Among other things, we consider properties that pass through a pair of rings with a common set of proper ideals.
Hereafter, we reserve the termsubring for aunital subring. Thus, not only a subring inherits its binary operations from its overring, but also it has the same identity element.
We call subringsRandSof a ring (right-)ideally equalif they have the same set of proper (right) ideals, that is,Iis a proper (right) ideal ofRif and only ifIis a proper (right) ideal ofS. Two ringsRandSare calledP- (M-)ideally equalif they have the same set of prime (maximal) ideals.
Theorem1. SupposeS⊆Rare rings which are right-ideally equal. Then eitherS=RorS andRare division rings.
Copyright©2005 Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences 2005:20 (2005) 3261–3268 DOI:10.1155/IJMMS.2005.3261
Proof. IfRandShave two distinct maximal right idealsMandNin common, thenR= M+N=S. Thus, we may assume thatM isthe unique maximal right ideal ofSandR.
If 0=a∈M, thenaR=aS, and for eachr∈Rthere existss∈Ssuch thatar=as, and sor−s∈rR(a), the right annihilator ideal ofainR. SincerR(a)isa proper right ideal ofR,rR(a)⊆Sand sor−s=sfor somes∈S. Thusr=s+s∈SandR=S. If no such nonzeroaexists, thenM=0, andSandRare division rings.
Corollary2. Let RandSbe two distinct subrings of a ring. ThenRandSare division rings if and only ifR,S, andR∩Shave the same set of proper right ideals.
Example 3. LetKbe a field and lettbe an indeterminate. LetT=K(t)[X]/X3, and set α=X+X3. ThenR=K(t)[α2] is a subring ofT, andS=K(t−α)[α2] is also a subring ofT. Sincet·α2=(t−α)·α2, we haveM:=Rα2=Sα2=K(t)α2, and it is easy to see that Mis the only nonzero ideal inRandS. IfR=S, thenα∈R, a contradiction. Hence,Rand Sare distinct subrings of the commutative ringTwhich are ideally equal but they are not fields. Notice that if char(K)=0, thenKα2is an ideal ofR∩S=K+K(t)α2but it is not an ideal ofRnorS. If char(K)=p >0, thenK(tP)α2is an ideal ofR∩S=K(tP) +K(t)α2 that is not an ideal ofRnorS.
We now prove two propositions on a pair of rings with a common ideal.
Proposition4. LetRandSbe subrings of a ring and suppose that they have a common idealI. IfPis a prime ideal ofR, thenP= {a∈S|IaI⊆P}is eitherSor a prime ideal ofS.
Proof. ObviouslyPis an ideal ofS. IfI⊆P, then it is clear thatP=S. SupposeIP.
Then sinceP is prime,I·1·I=I2P. ThusPis a proper ideal ofS. LetA,Bbe two ideals ofSsuch thatAB⊆P. Then (IAI)(IBI) ⊆IABI⊆IPI ⊆P. SinceP is prime, we have eitherIAI⊆P orIBI⊆P. Hence, eitherA⊆PorB⊆P. Therefore, Pis a prime
ideal ofS.
By a primitive idealPof a ringRwe meanR/Phas a faithful irreducible right module.
Proposition5. LetRandSbe subrings of a ring having a common idealI. IfPis a primitive ideal ofR, thenP= {a∈S|IaI⊆P}is eitherSor a primitive ideal ofS.
Proof. Note first that since a primitive ideal is a prime ideal,P=S if and only ifI⊆ P (as was shown in the proof ofProposition 4). SupposeIP and letM be a faithful irreducible rightR/P-module. Then, for any nonzerom∈M, 0=mI⊆M, and somI= M. This implies thatMis an irreducible rightS-module.
We claimMP=0. SinceMP=MIP⊆MI=M, ifMP=0 thenMPI =M=0. How- ever,MPI =MIPI ⊆MP=0, and thusMP=0. Hence,M is a rightS/P-module. Let abe an element ofSsatisfying Ma=0. Then MIaI=0 and soIaI⊆P. Hencea∈P.
Therefore,Mis also a faithful rightS/P-module.
We denote the set of prime ideals of a ringRby Spec(R), the set of maximal ideals of a ringRby Max(R), and the set of primitive ideals of a ringRby Prim(R). The following theorem shows that two ringsRandSareP-ideally equal if and only if they areM-ideally equal.
Theorem6. LetRandSbe subrings of a ring. Then the following statements are equivalent:
(a) Max(S)⊇Max(R), (b) Max(S)⊆Max(R), (c) Spec(S)=Spec(R), (d) Prim(S)=Prim(R).
Proof. Since otherwise the result is trivially true, assume thatR=S. Then, if Max(S)⊇ Max(R),Rhas a unique maximal idealM. LetN be another maximal ideal ofS. Then sinceS=M+N, there existm∈M and n∈N such that 1=m+n. But then n=1− m∈R\M and hence RnR=R. Therefore, M2=MRM=MRnRM⊆N. Since N is a prime ideal ofS, this is a contradiction. ThereforeM is a unique maximal ideal ofS, and hence Max(S)=Max(R)= {M}. This shows the equivalence of the statements (a) and (b). Suppose now that Max(S)=Max(R)= {M}and letP=Mbe a prime ideal of R. Then, byProposition 4,P= {a∈S|MaM⊆P}is a prime ideal ofS. SinceM is the unique maximal ideal ofS, we haveP⊆M, and soPis an ideal ofR. SinceMPM ⊆P, we obtainP⊆P, and therefore P=Pis a prime ideal ofS. Since a primitive ideal is prime, the equivalence of the statements (a), (b), and (d) can be shown similarly by using
Proposition 5.
Corollary7. LetRbe a ring with unique maximal idealMand letS=M+Z(R)where Z(R)is the center ofR. ThenSpec(S)=Spec(R). In this case,Mis also a maximal right and left ideal ofS.
Proof. Clearly,Mis an ideal ofS. Suppose thatM⊂I⊆Sfor a right idealIofS. Then, there existsz∈I withz∈Z(R)\M. HenceR=RzR=zR. Thus there existsr∈Rsuch thatzr=1. But then for anyx∈R,rx=rxzr=zrxr=xr and thereforer∈Z(R)⊆S.
This shows thatMis a maximal ideal and a maximal right ideal ofS. SinceMis the unique maximal ideal of R, we have Max(R)⊆Max(S), and hence, by Theorem 6, Spec(S)=
Spec(R).
A ring is calledfully idempotentif every ideal ofRis idempotent. A ringRis calledvon Neumann regularprovided that for everyx∈Rthere exists y∈Rsuch thatxyx=x. A commutative fully idempotent ring is von Neumann regular. However, the class of fully idempotent rings strictly contains the class of Von Neumann regular rings. The subring S=xA1(k) +k of a Weyl algebraA1(k) is an example of a fully idempotent ring that is not a Von Neumann regular ring.
Corollary8. LetRandSbe fully idempotent subrings of a ring. ThenRandSare ideally equal if and only ifRandSareP-ideally equal.
Proof. ByTheorem 6, ifRandShave the same set of prime ideals, then every ideal ofR andSis contained in the unique maximal idealM. IfIR, thenIS=I(IS)⊆I(MS)= IM⊆I, and similarly,SI⊆I. HenceIS.
We are now in a position to give a few examples.
Example 9(a pair ofP-ideally equalrings which arenot ideally equal). LetR=Q(√3)⊕R andS=Q(√2)⊕Rbe additive abelian groups with multiplication defined by (a,b)(c,d)=
(ac,ad+bc). ThenRandSareM-ideally equal rings with the unique maximal idealM= 0⊕R. HenceRandSareP-ideally equal. LetI=0⊕Q(√2). ThenI is an ideal ofSbut not ofR.
Example 10(a pair of rings with a nonzero common ideal that are notP-ideally equal). Let K be a field and letK[x] andK[y] be two polynomial rings overK. Consider the ring S=K[x]⊕K[y] and its subringR= {(a+x f(x),a)∈S|a∈K, f(x)∈F[x]}. ThenR andShave common idealI= {(x f(x), 0)| f(x)∈K[x]}. ClearlyP= {(0, 0)}is a prime ideal ofR, but it is not a prime ideal ofS.
Example 11(a pair ofP-ideally equalrings which are not fully idempotent). Let ¯Rbe the ring consisting of countable matrices overRof the form
Am 0
a a
0 . ..
, (1)
wherea∈RandAm is an arbitrarym×mmatrix overR, andmis allowed to be any positive integer. Let ¯S=M¯ +FwhereFis a subfield of the center of ¯R, and ¯Mis the subset of ¯Rconsisting of all countable matrices of the form
Am
0 0
. ..
, (2)
whereAmis an arbitrarym×mmatrix overR, andmis allowed to be any positive integer.
Let S=S¯⊕M¯ and R=R¯⊕M¯ be additive abelian groups with multiplication de- fined by (a,b)(c,d)=(ac,ad+bc). Then, S and R are M-ideally equal rings with the unique common maximal ideal M= {(m1,m2)|m1,m2∈M¯} and hence, they areP- ideally equal. However, the idealI= {(0,m)|m∈M¯}is not idempotent.
For a ringT, letS(T) be the set of all subringsSofT with Spec(S)=Spec(T). Note that ifThas more than one maximal ideal, thenS(T)= {T}.
Corollary12. Let T be a ring with unique maximal idealM. ThenS(T)= {p−1(S)| Sisasimple subring ofT/M}wherep:T→T/Mis the canonical epimorphism.
Proof. IfX∈S(T), then obviouslyX=p−1(X/M) andX/Mis a simple subring ofT/M.
Conversely, suppose thatS=X/Mis a simple subring ofT/M. ThenX=p−1(S) is a sub- ring ofTcontainingMas a maximal ideal. SinceMis the unique maximal ideal ofT, we have Max(T)⊆Max(p−1(S)). Hence byTheorem 6,p−1(S)∈S(T).
LetT be a ring with unique maximal idealM. ThenS(T) is a partially ordered set under set-theoretic inclusion. Notice thatS(T/M) is the set of simple subrings ofT/M, andϕ:S(T)→S(T/M) defined byϕ(p−1(S))=S(whereSis a simple subring ofT/M) gives an isomorphism betweenS(T) andS(T/M) byCorollary 12. Hence, ifT/Mis com- mutative, thenS(T/M) is a complete lattice, and so isS(T).
Corollary13. LetRbe a ring all of whose ideals are prime. ThenS(R)forms a complete lattice.
Proof. LetMbe the maximal ideal ofRand letZ(R) be the center ofR. LetT=M+Z(R).
By [4, Theorems 1.3 and 4.1] of Blair and Tsutsui,S(R)=S(T) andT/Mis a field. Hence
S(T/M) is a complete lattice, and so isS(R).
We now investigate properties that pass through a pair of rings with a common set of proper ideals.
For a ringR, letB(R) denote its prime radical and letJ(R) denote its Jacobson radical.
Proposition14. LetRandSbe subrings of a ring with a common idealI.
(a)IfRis a semiprime ring and ifrS(I)=S(I)=0, thenSis a semiprime ring.
(b)IfRis a semiprimitive ring and ifrS(I)=S(I)=0, thenSis a semiprimitive ring.
Proof. (a) LetPbe a prime ideal ofR. Then byProposition 4,IB(S)I⊆IPI ⊆P. Hence, IB(S)I⊆B(R)=0. But then sincerS(I)=S(I)=0, we haveB(S)=0.
(b) LetPbe a primitive ideal ofR. Then, byProposition 5,IJ(S)I⊆IPI ⊆P. Hence, IJ(S)I⊆J(R)=0. But then, sincerS(I)=S(I)=0, we haveJ(S)=0.
LetR⊆Sbe rings with a common idealI, and letPbe a prime ideal ofRwithIP.
Then “lying over” holds, that is, there exists a prime idealQinSsuch thatQ∩R=P.
(See, e.g., Rowen [6].)
Proposition15. LetR⊆Sbe rings with a common idealI. IfPis a prime ideal ofSwith IP, thenP∩Ris a prime ideal ofR.
Proof. LetA,Bbe two ideals ofRsuch thatAB⊆P∩R. Then (SA)(ISB)⊆SAIB⊆SAB⊆ SP=P. SincePis prime,SA⊆PorISB⊆P.
IfSA⊆P, thenA⊆P∩R. Suppose thatISB⊆P. SinceIP and sinceP is prime,
B⊆SB⊆P. HenceB⊆P∩R.
Proposition16. LetR⊆Sbe rings with a common idealI. Then ifSis a semiprime ring and ifIis an essential ideal ofS, thenRis a semiprime ring.
Proof. LetP is be a prime ideal ofS. IfI⊆P, thenB(R)∩I⊆P. On the other hand, if IP, then byProposition 15,P∩Ris a prime ideal of R, and henceB(R)∩I⊆B(R)⊆ P∩R⊆P. Therefore,B(R)∩I⊆B(S)=0. But then, sinceR⊆SandIis an essential ideal
ofS, we haveB(R)=0.
Note that the “primitive versions” of Propositions15and16are also valid.
A fully idempotent ring is in particular, a semiprime ring.
Proposition17. LetRandSbeM-ideally equal rings with the common maximal idealM.
IfRis fully idempotent, then so isS, and in this case they are ideally equal.
Proof. LetI be a proper ideal ofS. ThenIR⊆RIR=(RIR)2=RI(RIR)⊆RI(RMR)= RIM⊆RIand similarly,RI⊆IR. HenceIR=RIis an ideal ofR. But thenIR=(IR)2= I(RIR)⊆IM⊆I and therefore,I is an ideal ofR. This shows thatSis fully idempotent and the second claim now follows byTheorem 6andCorollary 8.
Every right ideal of a von Neumann regular is idempotent. A ring all of whose right ideals are idempotent is called afully right idempotent ring and it has received some at- tention in the literature.
Proposition18. LetRandSbeM-ideally equal rings with the common maximal idealM.
Then ifRis fully right idempotent, then so isS, and in this case they are ideally equal.
Proof. LetI be a proper right ideal of Sanda∈I. If a /∈M, thenSaS=S, and hence a∈aS=aSaS⊆I2. On the other hand, sinceRis fully right idempotent, ifa∈M,a∈ aR=aRaR. Now, sinceRis in particular fully idempotent, byProposition 17,RaRis an ideal ofS. SinceRaRis the smallest ideal ofRcontainingaandSaSis the smallest ideal of Scontaininga, we haveSaS=RaR. Hencea∈aR=aRaR=aSaS⊆I2. The next natural question is whether or not the “regularity” passes throughM-ideally equal rings.
Proposition19. LetRandSbeM-ideally equal rings with the common maximal ideal M. IfRis von Neumann regular, thenSis von Neumann regular if and only ifS/Mis von Neumann regular.
Proof. IfR is von Neumann regular, thenM is von Neumann regular as a ring by [5, Lemma 1.3] of Goodearl. By the same lemma, ifM andS/M are both von Neumann
regular, then so isS.
Example 20. LetW denote thenth Weyl algebra over a field of characteristic zero. It is well known thatWis a simple Noetherian domain, and henceWis an Ore domain. Let Ddenote the field of fraction ofW. LetRbe the set of countable matrices overDof the form
Am 0
a a
0 . ..
, (3)
wherea∈DandAm∈Mm(D), the ring ofm×mmatrices overD, andmis allowed to be any positive integer. LetSbe the same set of matrices excepta∈W. ThenRandSare M-ideally equal rings with the maximal idealMthat consists of countable matrices of the form
Am 0
0 0
0 . ..
. (4)
Any factor ring of a von Neumann regular ring is von Neumann regular, and a Noetherian von Neumann regular ring is semisimple Artinian. Therefore,Sis not a von Neumann regular ring sinceS/MWis a simple Noetherian but not an Artinian ring. The fact that Ris a von Neumann regular ring follows immediately by the definition of a von Neumann regular ring sinceMm(D) is von Neumann regular and 0=a∈Dis invertible.
The center of a simple ring is a field and there are simple rings that are not right Noetherian. Thus, in the noncommutative setting, neither the descending nor ascending chain condition passes through a pair of ideally equal rings.
The most natural generalization of commutative rings is the class of PI-rings, that is, rings that satisfy a polynomial identity.
Proposition21. LetR⊂SbeM-ideally equal rings with the common maximal idealM. If Ssatisfies a polynomial identity, thenRis right Noetherian if and only ifSis right Noetherian andS/Mis finitely generated rightR/M-module.
Proof. LetMbe the maximal ideal. SinceRandSare PI-rings,Mis the Jacobson radical of the two rings. Suppose thatRis right Noetherian. ThenMis a finitely generatedR- module. Following the proof of [1, Lemma 3.27] of Anderson and Dobbs, one can show that Sis a finitely generated right R-module. Sis then right Noetherian and S/Mis a finitely generated rightR/M-module.
Conversely, assume thatS is right Noetherian andS/M is a finitely generated right R/M-module. SinceR/Mis a simple Artinian ring,R/Mis a Noetherian rightR-module.
Hence, to show thatRis right Noetherian, it suffices to show thatMRis Noetherian. Let M1⊆M2⊆...be an ascending chain ofR-submodules ofMR. ThenM1M⊆M2M⊆...
is an ascending chain ofS-submodules ofMS. SinceMSis right Noetherian, there exists a positive integernsuch thatMnM=Mn+1M=....Consider next the chainMnS/MnM⊆ Mn+1S/MnM⊆....This is an ascending chain ofS-submodules of (M/MnM)S. Thus there exists a positive integerm > nsuch thatMmS/MnM=Mm+1S/MnM=....SinceMmS/MnM is a finitely generated rightS/M-module and sinceS/Mis a simple Artinian ring, it has finite length as a rightS/M-module. SinceS/Mis a finitely generated rightR/M-module, MmS/MnMhas finite length as a rightR/M-module. Since, for eachk > m,Mk/MnMis an R/M-submodule ofMmS/MnM, there exists a positive integert > msuch thatMt/MnM=
Mt+1/MnM=....But thenMt=Mt+1=....
Example 22. Let R=Q⊕Rand S=R⊕Rbe additive abelian groups with multipli- cation defined by (a,b)(c,d)=(ac,ad+bc). ThenRand Sare commutativeM-ideally equal rings with the unique maximal idealM=0⊕R.M=0⊕Ris the only nonzero proper ideal ofSand henceSis Noetherian. On the other hand, 0⊕Q⊂0⊕Q(√2)⊂ 0⊕Q(√2,√3)⊂0⊕Q(√2,√3,√5)⊂...is an ascending chain of ideals inRthat does not terminate.
Acknowledgment
We wish to thank the referee for a very helpful critique of the manuscript.
References
[1] D. F. Anderson and D. E. Dobbs,Pairs of rings with the same prime ideals, Canad. J. Math.32 (1980), no. 2, 362–384.
[2] ,Pairs of rings with the same prime ideals. II, Canad. J. Math.40(1988), no. 6, 1399–
1409.
[3] F. W. Anderson and K. R. Fuller,Rings and Categories of Modules, 2nd ed., Graduate Texts in Mathematics, vol. 13, Springer, New York, 1992.
[4] W. D. Blair and H. Tsutsui,Fully prime rings, Comm. Algebra22(1994), no. 13, 5389–5400.
[5] K. R. Goodearl,Von Neumann Regular Rings, Monographs and Studies in Mathematics, vol. 4, Pitman, Massachusetts, 1979.
[6] L. H. Rowen,Ring Theory. Vol. I, Pure and Applied Mathematics, vol. 127, Academic Press, Massachusetts, 1988.
Yasuyuki Hirano: Department of Mathematics, Okayama University, Okayama 700-8530, Japan E-mail address:[email protected]
Hisaya Tsutsui: Department of Mathematics, Embry-Riddle Aeronautical University, Prescott, AZ 86301, USA; Department of Mathematics, Millersville University, Millersville, PA 17551, USA
E-mail address:[email protected]
