A note on pairs of rings with same prime ideals
aa r X i v : . [ m a t h . A C ] M a y A note on pairs of rings with same prime ideals
Rahul Kumar & Atul Gaur Department of MathematicsUniversity of Delhi, Delhi, India.E-Mail: [email protected]; [email protected]
Abstract
We study the ring extensions R ⊆ T having the same set of primeideals provided N il ( R ) is a divided prime ideal. Some conditions aregiven under which no such T exist properly containing R . Using ideal-ization theory, the examples are also discussed to strengthen the results. Mathematics Subject Classification:
Primary 13B99; Secondary 13A15,13A18.
Keywords: φ -PVR, φ -chained ring, overring. Throughout this paper, all rings are commutative with nonzero identity andif R is a subring of T , then the identity element of R and T is same. By anoverring of R , we mean a subring of the total quotient ring of R containing R . By a local ring, we mean a ring with unique maximal ideal. The symbol ⊆ is used for inclusion, while ⊂ is used for proper inclusion. We use T ( R ) todenote the total quotient ring of R , R ′ to denote the integral closure of a ring R in T ( R ), N il ( R ) to denote the set of nilpotent elements of R , and Z ( R ) todenote the set of zero-divisors of R . If I and J are R -submodules of T ( R ), then( I : J ) = { x ∈ T ( R ) : xJ ⊆ I } . Badawi defined the divided prime ideal of R in [5], as a prime ideal which is comparable to every ideal of R and a ring R issaid to be divided if each prime ideal of R is divided, see [5]. In [2], Andersonand Badawi introduced the notion H to be the set of all rings R such that N il ( R ) is a divided prime ideal and named these rings as φ -rings. They used The author was supported by a grant from UGC India, Sr. No. 2061440976. The author was supported by the MATRICS grant from DST-SERB, No.MTR/2018/000707.
Rahul Kumar and Atul Gaur H to denote the subset of H such that N il ( R ) = Z ( R ). For further study onthese rings, see [2], [3], [6], [7], [8], [9], [10], [11], [14], [12], [13].In this paper, we will focus on the ring extensions in class H having thesame set of prime ideals which was extensively studied by Anderson and Dobbsfor integral domains, see [1]. Naturally, one may think to study these ringextensions for some larger class. Note that if R is an integral domain, then R isin H . This motivates us to study the ring extensions in class H having the sameset of prime ideals. Note that if R ∈ H , then there is a ring homomorphism φ from T ( R ) to R Nil ( R ) given by φ ( r/s ) = r/s , for all r ∈ R and s ∈ R \ Z ( R ).Moreover, the restriction of φ to R is also a ring homomorphism given by φ ( r ) = r/
1, for all r ∈ R , see [6]. Note that if R ∈ H , then φ ( R ) = R .In this paper, we discuss examples using this idealization theory. In [20],Nagata defined a new class of rings, namely idealization of a module. If R is aring and M is an R -module, then the idealization R (+) M is the ring definedas follows: Its additive structure is that of the abelian group R ⊕ M , andits multiplication is defined by ( r , m ) ( r , m ) := ( r r , r m + r m ), for all r , r ∈ R and m , m ∈ M . It will be convenient to view R as a subring of R (+) M via the canonical injective ring homomorphism that sends r to ( r, The ring extensions R ⊆ T for which Spec( R ) = Spec( T ) are studied byAnderson and Dobbs in [1]. They discussed various type of domains R forwhich there exist domains T satisfying Spec( R ) = Spec( T ), and also givenconditions on R for which no such T exists. In Lemma 3 .
1, they proved thatif R ⊆ T is a ring extension with a common nonzero ideal I which containsa non-zerodivisor of R , then T is contained in the total quotient ring of R ,and indeed T ⊆ ( I : I ). The result does not seems to be correct. The proofgoes wrong if for any nonzero t ∈ T and for any non-zerodivisor y ∈ I (of R ), ty = 0. In fact, there is a class of ring extensions for which ty = 0 and hencethereby proving that [1, Lemma 3.1] is not correct. For example, consider R = Z and T = Z (+) Z / Z . Then I = 2 Z is a common nonzero ideal withevery nonzero element of I is a non-zerodivisor of R . But T T ( R ). Notethat this is true for all domains R with nonzero maximal ideal M . In thiscase, take T = R (+) R/M . Then M is a common nonzero ideal but T T ( R ).Note that if ty = 0, then the proof of [1, Lemma 3.1] will work. Thus, if wetake y to be a non-zerodivisor of T , then [1, Lemma 3.1] is correct with thesame proof. We now state the modified result. Lemma 2.1.
Let R ⊆ T be a ring extension with a common nonzero ideal I .If I contains a non-zerodivisor of T , then T is contained in the total quotientring of R , and indeed T ⊆ ( I : I ) . Also, in [1, Remark 3.4(a)], Anderson and Dobbs concluded the following:Let R ⊂ T be a ring extension such that Spec( R ) = Spec( T ) and M be theunique maximal ideal of both R and T . If M contains a non-zerodivisor of R ,then T ⊆ ( M : M ).They mentioned that the same proof of [1, Lemma 3.1] will work to provethe above result. As [1, Lemma 3.1] has been modified, some argument isneeded to establish the above result. Note that M contains a non-zerodivisorof T . Otherwise M = Z ( T ) which gives M = Z ( R ) as Spec( R ) = Spec( T ).This contradicts that M contains a non-zerodivisor of R . Thus, the resultfollows by Lemma 2 . R ⊂ S be an extension of integral domains. Then R is a said to be apseudo-valuation subring (PV, for short) in S , see [4], if for each x ∈ S \ R and non-unit a of R , we have x − a ∈ R . Note that PV in a domain is ageneralization of PVD. We now continue to investigate the properties of ringextensions having the same set of prime ideals initiated in [1]. Here is our firstresult. Theorem 2.2.
Let R ⊆ T ⊂ U be an extension of integral domains suchthat Spec ( R ) = Spec ( T ) . Then R is a PV in U if and only if T is a PV in U .Proof. Assume that R is a PV in U . Take x ∈ U \ T and a non-unit y in T . Then y is also a non-unit in R as Spec( R ) = Spec( T ). It follows that x − y ∈ R . Thus, T is a PV in U . Conversely, suppose that T is a PV in U . Take x ∈ U \ R and y a non-unit in R . Then y is also a non-unit in T as Spec( R ) = Spec( T ). If x / ∈ T , then x − y ∈ T which is not a unit in T asotherwise x ∈ T . It follows that x − y ∈ R and we are done. On the otherhand, if x ∈ T , then x is a unit in T as Spec( R ) = Spec( T ). It follows that x − y is a non-unit in T and so x − y ∈ R . Thus, R is a PV in U .Note that the above proposition generalizes [1, Proposition 2.2] to PV ina domain. In Proposition 2 .
4, we generalize the same to φ -PVR as φ -PVRgeneralizes PVD. Recall from [6] that a ring R ∈ H is said to be a φ -pseudo-valuation ring ( φ -PVR, for short), if each prime ideal P of R is φ -stronglyprime, that is, if xy ∈ φ ( P ) for x, y ∈ R Nil ( R ) , then either x ∈ φ ( P ) or y ∈ φ ( P ). First, we need the following lemma. Lemma 2.3.
Let R be a ring in H . If T is a ring containing R such that Spec ( R ) = Spec ( T ) , then T ∈ H .Proof. Clearly,
N il ( T ) = N il ( R ) is a prime ideal of T . It remains to showthat N il ( R ) is divided in T . Let I be a proper ideal of T . Then I is an idealof R as Spec( R ) = Spec( T ). Thus, N il ( R ) is divided in T . Proposition 2.4.
Let R ⊆ T be a ring extension such that R ∈ H and Spec ( R ) = Spec ( T ) . Then R is a φ -PVR if and only if T is a φ -PVR. Rahul Kumar and Atul Gaur
Proof.
Note that
N il ( R ) is a prime ideal of both R and T . It follows thatSpec( R/N il ( R )) = Spec( T /N il ( R )). Now, if R is a φ -PVR, then R/N il ( R )is a PVD, by [9, Proposition 2.9]. It follows that T /N il ( R ) is a PVD, by[1, Proposition 2.2]. Also, T ∈ H , by Lemma 2 .
3. Thus, by another appeal to[9, Proposition 2.9], T is a φ -PVR. The converse is similar.A domain R is called a going-down domain if R ⊆ T satisfies the going-down property for each overring T of R , see [15]; and a ring R is called agoing-down ring if R/P is a going-down domain for all prime ideals P of R , see[16]. The next theorem generalizes [1, Corollary A.4] and [1, Proposition A.1]to rings in H . Theorem 2.5.
Let R ∈ H . If T is a ring containing R properly such that Spec ( R ) = Spec ( T ) , then(i) R is not a going down ring.(ii) R is divided if and only if T is divided.Proof. To prove ( i ), note that N il ( R ) is a prime ideal of both R and T . Itfollows that Spec( R/N il ( R )) = Spec( T /N il ( R )). Now, if R is a going downring, then R/N il ( R ) is a going down domain, which is a contradiction by[1, Corollary A.4].For ( ii ), let R be divided and P be any prime ideal of T . Then P is adivided prime ideal of R . Let I be a proper ideal of T . Then I is a properideal of R and so is comparable with P . Thus, T is divided. Conversely,assume that T is divided. Then T /N il ( R ) is divided and so R/N il ( R ) isdivided, by [1, Proposition A.1]. Since N il ( R ) is divided, we conclude that R is divided.A ring R in H is said to be a φ -chained ring, if for each x ∈ R Nil ( R ) \ φ ( R ),we have x − ∈ φ ( R ), see [8]. As a direct consequence of [8, Proposition 3.3],[8, Corollary 3.4], and [1, Proposition 3.8], we observe the generalization of[1, Proposition 2.5] to rings in H . Proposition 2.6.
Let R ∈ H be a local ring with maximal ideal M . If M contains a non-zerodivisor, then the following are equivalent:(1) R is a φ -PVR;(2) ( M : M ) is a φ -chained overring of R with maximal ideal M ;(3) R has a φ -chained overring V such that Spec ( R ) = Spec ( V ) . The next corollary is a companion to Theorem 2 . . H . Corollary 2.7.
Let R ∈ H be a local ring such that the maximal idealcontains a non-zerodivisor of R . Then R is a φ -chained ring if and only if R is both a going down ring and a φ -PVR.Proof. Let R be a φ -chained ring. Then it is a φ -PVR trivially. Also, R/N il ( R )is a valuation domain, by [2, Theorem 2.7]. It follows that R/N il ( R ) is a goingdown domain. Consequently, by [16, Proposition 2.1(a)], R is going down.Conversely, assume that R is both a going down ring and a φ -PVR. Thenby Proposition 2 .
6, it follows that R has a φ -chained overring V such thatSpec( R ) = Spec( V ). Thus, R = V , by part ( i ) of Theorem 2 . Proposition 2.8.
Let R ⊆ T be rings such that Spec ( R ) = Spec ( T ) . Then R P = T T \ P for each non-maximal prime ideal P of R . The next proposition generalizes [1, Corollary 3.20(2)] to rings in H . Proposition 2.9.
Let R ⊆ T be a ring extension such that T ∈ H and R is local with maximal ideal M . Assume that M is a finitely generated ideal of R . Then Spec ( R ) = Spec ( T ) if and only if M ∈ Spec ( T ) .Proof. The “only if” assertion is trivial. For the converse part, assume that M ∈ Spec( T ). Then N il ( T ) = N il ( R ). Also, R/N il ( R ) is local with finitelygenerated maximal ideal M/N il ( R ) and T /N il ( R ) is a domain. It follows thatSpec( R/N il ( R )) = Spec( T /N il ( R )), by [1, Corollary 3.20]. Thus, the resultholds. Remark 2.10.
Let R be a Pr¨ufer domain which is not a field. Then thereis no ring T which contains R properly such that Spec( R ) = Spec( T ), see[1, Remark 3.17(b)]. Note that this can be extended to φ -Pr¨ufer rings also (Aring R ∈ H is said to be a φ -Pr¨ufer ring, see [2], if φ ( R ) is a Pr¨ufer ring). Moreprecisely, if R is a φ -Pr¨ufer ring such that N il ( R ) is not a maximal ideal of R and T is a ring containing R such that Spec( R ) = Spec( T ), then R = T . Tosee this, first note that N il ( R ) = N il ( T ). It follows that both R/N il ( R ) and T /N il ( R ) have the same set of prime ideals. As by [2, Theorem 2.6], R/N il ( R )is a Pr¨ufer domain, we have R/N il ( R ) = T /N il ( R ) and so R = T .In the next proposition, we investigate the overrings of R ∈ H havingthe same set of prime ideals. Moreover, it can be seen as a generalization of[1, Corollary 3.21]. Rahul Kumar and Atul Gaur
Proposition 2.11.
Let R ∈ H be a local ring with maximal ideal M suchthat R ′ is a Pr¨ufer ring. Let T be an overring of R . Then Spec ( R ) = Spec ( T ) if and only if M ∈ Spec ( T ) .Proof. It is easy to see that both R ′ and T are in H . Moreover, by [6], N il ( T ) = N il ( R ′ ) = N il ( R ) and T ( R/N il ( R )) = T ( R ) /N il ( R ). It followsthat T /N il ( R ) is an overring of R/N il ( R ). Moreover, by [3, Lemma 2.8],( R/N il ( R )) ′ = R ′ /N il ( R ). Now, by [2, Corollary 2.10], ( R/N il ( R )) ′ is aPr¨ufer domain. Consequently, Spec( R/N il ( R )) = Spec( T /N il ( R )) if and onlyif M/N il ( R ) ∈ Spec(
T /N il ( R )), by [1, Corollary 3.21]. Hence, the result fol-lows. Remark 2.12.
In [1], Anderson and Dobbs raised the following question:given a ring R , how does one can obtain all overrings T of R which satisfiesSpec( R ) = Spec( T ). They answered the question in [1, Theorem 3.5] for do-mains. However, observe that the proof of [1, Theorem 3.5] works for ringsalso.We now discuss an example that gives a class of ring extensions having thesame set of prime ideals. Example 2.13.
Let B be a ring of the form K + M , where K is a field and M is a maximal ideal of B . Assume that D is a subring of K , and A = D + M .Set U = A (+) T ( B ) and V = B (+) T ( B ). Then Spec( U ) = Spec( V ) if andonly if V is local and D is a field. To see this, note that if Spec( U ) = Spec( V ),then Spec( A ) = Spec( B ), by [18, Theorem 25.1(3)]. It follows that B is localand D is a field, by [1, Corollary 3.11]. Consequently, V is local, by anotherappeal to [18, Theorem 25.1(3)]. Conversely, assume that V is local and D is a field. Then B is local, by [18, Theorem 25.1(3)]. It follows that Spec( A )= Spec( B ), by [1, Corollary 3.11]. Consequently, by another application of[18, Theorem 25.1(3)], Spec( U ) = Spec( V ).The next proposition gives a class of rings R in H which does not admitany ring T properly containing R and having the same set of prime ideals. Fordomains, it was observed in [1, pg. 372]. Proposition 2.14.
Let R ∈ H be an integrally closed local ring such that R = T ( R ) . If the maximal ideal M of R is finitely generated, then there is noring T properly containing R such that Spec ( R ) = Spec ( T ) .Proof. If possible, suppose there exists a ring T properly containing R suchthat Spec( R ) = Spec( T ). Then we have N il ( T ) = N il ( R ) and Spec( R/N il ( R ))= Spec( T /N il ( R )). Moreover, by [6] and [3, Lemma 2.8], we conclude that T ( R/N il ( R )) = T ( R ) /N il ( R ) and ( R/N il ( R )) ′ = R ′ /N il ( R ), respectively.By hypothesis, it follows that R/N il ( R ) is an integrally closed local domainwhich is not a field. Moreover, the maximal ideal M/N il ( R ) of R/N il ( R ) isfinitely generated. Consequently, we have a contradiction, by last paragraphof [1, pg. 372].The next example shows that the condition on M of being finitely generatedcan not be removed. Example 2.15.
Let F ⊂ K be a purely transcendental field extension.Take a valuation domain B = K + M with unique nonzero maximal ideal M .Set R = A (+) T ( B ) and T = B (+) T ( B ) where A = F + M . Then it is easyto see that Z ( R ) = N il ( R ) = N il ( T ) = { } (+) T ( B ). Let (0 , x ) ∈ N il ( R )and ( y, z ) ∈ R \ N il ( R ). Then y = 0 and (0 , x ) = ( y, z )(0 , x/y ). It followsthat R ∈ H . Since A is an integrally closed domain, R is an integrallyclosed ring, by [18, Theorem 25.6]. Also, by [18, Corollary 25.5(3)], R = T ( R ).Since M is not finitely generated, M (+) T ( B ) is not finitely generated. Notethat M (+) T ( B ) is the unique maximal ideal of R , by [18, Theorem 25.1(3)].However, by Example 2 .
13, Spec( R ) = Spec( T ).Let C ( R ) denotes the complete integral closure of R in T ( R ). In thenext theorem, we show that the ring extensions in H having the same setof prime ideals will have the same complete integral closure. This generalizes[1, Proposition 3.15] for rings in H . Theorem 2.16.
Let R ∈ H be such that R = T ( R ) . If T is a ringcontaining R such that Spec ( R ) = Spec ( T ) , then C ( R ) = C ( T ) .Proof. It is easy to conclude that
N il ( R ) = N il ( T ), Z ( T ) = Z ( R ) andSpec( R/N il ( R )) = Spec( T /N il ( R )). Also, T ( R/N il ( R )) = T ( R ) /N il ( R ),by [6], and so R/N il ( R ) is not a field. By [1, Proposition 3.15], it followsthat C ( R/N il ( R )) = C ( T /N il ( R )). Now, by Lemma 2 . T ∈ H . Conse-quently, we have C ( R ) /N il ( R ) = C ( T ) /N il ( R ), by [3, Lemma 2.8]. Thus, C ( R ) = C ( T ).The following corollary is an immediate consequence of Theorem 2 .
16 when R = C ( R ). Corollary 2.17.
Let R ∈ H be a completely integrally closed ring suchthat R = T ( R ) . Then there is no ring T properly containing R such that Spec ( R ) = Spec ( T ) . In [1, Lemma 2.4], Anderson and Dobbs proved the next result for localdomains which are not fields. Here, we generalize the same for rings. In fact,the proof follows mutatis mutandis from the proof of [1, Lemma 2.4]. Notethat an ideal I of a ring R is said to be divisorial if I = ( R : ( R : I )). Rahul Kumar and Atul Gaur
Theorem 2.18.
Let R be a local ring with maximal ideal M . If M containsa non-zerodivisor of R , then the following hold:(i) ( R : M ) = ( M : M ) if and only if M is not a principal ideal of R .Moreover, if M = Rr for some r ∈ R , then ( R : M ) = Rr − and ( M : M ) = R .(ii) If M is not a divisorial ideal of R , then ( R : M ) = ( M : M ) . The next proposition generalizes [1, Proposition 3.23].
Proposition 2.19.
Let R be a local ring with maximal ideal M such that M contains a non-zerodivisor of R . Then ( M : M ) = R if and only if M is anon-principal divisorial ideal of R .Proof. The “only if” assertion follows mutatis mutandis from the proof of[1, Proposition 3.23] and “if” assertion follows from Theorem 2 . .
19 andLemma 2 . Corollary 2.20.
Let R be a local ring such that the maximal ideal of R contains a non-zerodivisor of R . Assume that the maximal ideal of R is eitherprincipal or a non-divisorial ideal of R . Then there is no ring T properlycontaining R such that Spec ( R ) = Spec ( T ) . It is easy to see that [1, Corollary 3.30] is not true for fields as if R is afield, then R ′ is not of rank one. However, if we exempt R to be a field, then R ′ is a discrete (rank 1) valuation domain. In Proposition 2 .
21, we generalizethe same to rings in H . First, we recall some definitions. An ideal of a ring R is said to be a nonnil ideal if I N il ( R ). A ring R ∈ H is called a nonnil-Noetherian ring if every nonnil ideal of R is finitely generated, see [10]. Recallfrom [3] that a ring R ∈ H is said to be a discrete φ -chained ring if R is a φ -chained ring with at most one nonnil prime ideal and every nonnil ideal of R is a principal ideal. Proposition 2.21.
Let R ∈ H . If R is a nonnil-Noetherian φ -PVR, thendim( R ) ≤ and R ′ is a discrete φ -chained ring.Proof. Note that (
R/N il ( R )) ′ = R ′ /N il ( R ), by [3, Lemma 2.8]. Also, by[10, Theorem 2.2] and [9, Proposition 2.9], R/N il ( R ) is a Noetherian PVD.It follows that dim ( R ) ≤ R ′ /N il ( R ) is a discrete valuation domain, by[1, Corollary 3.30]. Thus, the result holds, by [3, Lemma 2.9].Recall from [19] that a ring R is said to be a weakly finite-conductor ring if Ra ∩ Rb is a finitely generated ideal for all a, b ∈ R . We end this paper with thegeneralization of [1, Corollary A.3] to rings in H . Note that [1, Corollary A.3]fails to hold when R ⊂ T is a field extension such that T is not a finitelygenerated R -module. Therefore, the exemption of R to be a field is requiredin the statement of [1, Corollary A.3]. Proposition 2.22.
Let R ⊂ T be a ring extension such that R ∈ H is aweakly finite-conductor ring, R = T ( R ) , and Spec ( R ) = Spec ( T ) . Let M bethe maximal ideal of R . Then both M and T are finitely generated R -modules.Proof. Note that
R/N il ( R ) is not a field as T ( R/N il ( R )) = T ( R ) /N il ( R ),by [6]. Also, it is easy to see that R/N il ( R ) is a finite-conductor domainwith maximal ideal M/N il ( R ). Thus, M/N il ( R ) and T /N il ( R ) are finitelygenerated R/N il ( R )-module, by [1, Corollary A.3]. Since R = T ( R ), M is anonnil ideal of R . Now, the proof follows mutatis mutandis from the proof of[2, Lemma 2.4]. References [1] D. F. Anderson and D. E. Dobbs, Pairs of rings with the same primeideals, Can. J. Math., 32 (1980) 362-384.[2] D. F. Anderson, A. Badawi, On φ -Pr¨ u fer rings and φ -Bezout rings,Houston J. Math., 30 (2004) 331-343.[3] D. F. Anderson, A. Badawi, On φ -Dedekind rings and φ -Krull rings,Houston J. Math., 31(4) (2005) 1007-1022.[4] A. Ayache, O. Echi, Valuation and pseudovaluation subrings of an in-tegral domain, Comm. Algebra, 34(7) (2006) 2467-2483.[5] A. Badawi, On divided commutative rings, Comm. Algebra 27(3) (1999)1465-1474.[6] A. Badawi, On φ -pseudo-valuation rings, Lecture Notes Pure Appl.Math., Marcel Dekker, New York/Basel, 205 (1999) 101-110.[7] A. Badawi, On φ -pseudo-valuation rings II, Houston J. Math. 26(3)(2000) 473-480.[8] A. Badawi, On φ -chained rings and φ -pseudo-valuation rings, HoustonJ. Math., 27 (2001) 725-736.0 Rahul Kumar and Atul Gaur [9] A. Badawi, On divided rings and φ -pseudo-valuation rings, Interna-tional J. of Commutative Rings (IJCR), Nova Science/New York, 1(2002) 51-60.[10] A. Badawi, On Nonnil-Noetherian rings, Comm. Algebra 31(4) (2003)1669-1677.[11] A. Badawi, Factoring nonnil ideals into prime and invertible ideals,Bull. London Math. Soc. 37(5) (2005) 665-672.[12] A. Badawi and T. G. Lucas, Rings with prime nilradical, Arithmeticalproperties of commutative rings and monoids, Chapman & Hall/CRC241 (2005) 198-212.[13] A. Badawi and T. G. Lucas, On φ -Mori rings, Houston J. Math. 32(1)(2006) 1-32.[14] A. Badawi and A. Jaballah, Some finiteness conditions on the set ofoverrings of a φφ