aa r X i v : . [ m a t h . A C ] S e p Maximal non valuative domains
Rahul Kumar & Atul Gaur Department of MathematicsUniversity of Delhi, Delhi, India.E-Mail: [email protected]; [email protected]
Abstract
The notion of maximal non valuative domain is introduced and char-acterized. An integral domain R is called a maximal non valuative do-main if R is not a valuative domain but every proper overring of R isa valuative domain. Maximal non valuative domains have at most fourmaximal ideals. Various properties of maximal non valuative domainsare discussed. Conditions are given under which pseudo-valuation do-mains and maximal non pseudo-valuation domains are maximal nonvaluative domains. Mathematics Subject Classification:
Primary 13G05, 13B02, Secondary13B22, 13B30, 13A15.
Keywords:
Maximal non valuative domain, valuative domain, valuation do-main, pseudo-valuation domain, B´ezout domain.
Our work is motivated by [3]. An integral domain R with the quotient fieldqf( R ) is said to be a valuative domain, see [3], if for each x ∈ qf( R ), either R ⊆ R [ x ] or R ⊆ R [ x − ] has no intermediate ring. In this paper, we introducethe concept of maximal non valuative domains. Let R ⊂ T be an extension ofintegral domains. Then we say that R is a maximal non valuative subring of T if R is not a valuative domain but each subring of T containing R properly is avaluative domain. Moreover, if T = qf( R ), then R is said to be a maximal nonvaluative domain. In this paper, we discuss various properties of maximal nonvaluative domain and characterize the same in terms of B´ezout domain. All The author was supported by the SRF grant from UGC India, Sr. No. 2061440976. The author was supported by the MATRICS grant from DST-SERB India, No.MTR/2018/000707.
Rahul Kumar and Atul Gaur rings considered below are integral domains. By an overring of R , we mean asubring of the quotient field of R containing R . A ring with a unique maximalideal is called a local ring. The symbol ⊆ is used for inclusion and ⊂ is usedfor proper inclusion. Throughout this paper, qf( R ) denotes the quotient fieldof an integral domain R , R ′ denotes the integral closure of R in qf( R ). For aring R , dim( R ) denotes the Krull dimension of R .In this paper, we show that if R is a maximal non valuative subring of T ,then T is an overring of R , see Theorem 2.1. If R is a maximal non valuativedomain, then R ′ is a Pr¨ufer domain, see Corollary 2.3. Moreover, if R isnot integrally closed, then R has at most three maximal ideals, the set of non-maximal prime ideals of R is linearly ordered by inclusion, and there is at mostone maximal ideals of R that does not contain all non-maximal prime ideals of R , see Theorem 2.7; and if R is integrally closed, then R has at least two and atmost four maximal ideals, there are exactly two non-maximal prime ideals thatare not comparable in case R has exactly two maximal ideals, otherwise the setof non-maximal prime ideals of R is linearly ordered by inclusion, and there areat most two maximal ideals of R that do not contain all non-maximal primeideals of R , see Proposition 2.8. We characterize integrally closed maximalnon valuative domains in terms of B´ezout domains, see Theorem 2.9. Finally,we characterize local non integrally closed maximal non valuative domain R .We also discuss the cases where either R is a pseudo-valuation domain or amaximal non pseudo-valuation subring of R ′ .For any ring R , Spec( R ) denotes the set of all prime ideals of R ; Max( R )denotes the set of all maximal ideals of R . As usual, | X | denotes the cardinalityof a set X . A ring extension R ⊆ T is said to be residually algebraic if for any prime ideal Q of T , T /Q is algebraic over R/ ( Q ∩ R ), see [4, 1]. Moreover, if R ⊆ S isresidually algebraic, for any subring S of T containing R , then ( R, T ) is saidto be a residually algebraic pair, see [1].In our first theorem, we list some properties of an extension of integraldomains in which every intermediate ring is a valuative domain.
Theorem 2.1.
Let R ⊂ T be a ring extension of integral domains. If eachsubring of T properly containing R is a valuative domain, then the followinghold:(i) R ⊂ T is an algebraic extension.(ii) ( R, T ) is a residually algebraic pair. (iii) If R is not a field, then T is an overring of R .Proof. (i) If possible, suppose that R ⊂ T is not an algebraic extension.Then there exists an element t ∈ T \ R which is transcendental over R .Take S = R [ t , t ]. Then S is a subring of T containing R properly.Therefore, S is a valuative domain. Clearly, t ∈ qf( S ). Thus, either S ⊆ S [ t ] or S ⊆ S [ t − ] has no intermediate ring. Now, note that S ⊂ R [ t + t , t , t ] ⊂ S [ t ] = R [ t ] and S ⊂ R [ t − + t , t , t ] ⊂ S [ t − ] = R [ t − , t , t ], which is a contradiction. Hence, R ⊂ T is an algebraicextension.(ii) Let S be a subring of T properly containing R and Q be a prime idealof S . If possible, suppose that R/ ( Q ∩ R ) ⊂ S/Q is not an algebraicextension. Then there exists an element ¯ t ∈ S/Q that is not algebraicover R/ ( Q ∩ R ). Take S ′ = ( R/Q ∩ R )[¯ t ]. Then S ′ = S ′′ / ( Q ∩ S ′′ )for some subring S ′′ of S containing R properly. Therefore, S ′′ is avaluative domain. Thus, by [3, Theorem 2.2(i)], S ′′ has at most threemaximal ideals and hence S ′ has at most three maximal ideals, which isa contradiction.(iii) Let K = qf( R ). If possible, suppose that T K . Choose t ∈ T \ K .Then t is algebraic over R , by part (i). Therefore, α = tr is integral over R for some non zero r ∈ R . Clearly, α / ∈ K . Let n = [ K ( α ) : K ]. Then { , α, α , . . . , α n − } is a basis of K ( α ) over K . Let β be any non zero,non unit of R . Take S = R + αβ R + α β R + · · · + α n − β R . Then S is a subring of T properly containing R . Therefore, S is a valuativedomain. Note that qf( S ) = K ( α ). Now, if αβ − ∈ S , then αβ − = r + αβ r + α β r + · · · + α n − β r n − for some r , r , . . . , r n − ∈ R . Itfollows that β − = β r ∈ R , a contradiction. Also, if α − β ∈ S , then α − β = r + αβ r + α β r + · · · + α n − β r n − for some r , r , . . . , r n − ∈ R . It follows that β = r α + α β r + α β r + · · · + α n β r n − . Let α n = P n − i =0 α i x i for some x , x , . . . , x n − ∈ R . Consequently, we have β = r α + α β r + α β r + · · · + α n − β r n − + β r n − (cid:16) n − X i =0 α i x i (cid:17) It follows that β = β r n − x , that is, 1 = β r n − x , a contradiction.Thus, αβ − , α − β ∈ K ( α ) \ S . Since S is a valuative domain, either S ⊂ S [ αβ − ] or S ⊂ S [ α − β ] has no intermediate ring. Assume that S ⊂ S [ αβ − ] has no intermediate ring. It follows that either S = S [ α ]or S [ α ] = S [ αβ − ]. Now, if S = S [ α ], then α ∈ S . Therefore, α = r + αβ r + α β r + · · · + α n − β r n − for some r , r , . . . , r n − ∈ R .Consequently, we have 1 = β r , which is a contradiction. Thus, we Rahul Kumar and Atul Gaur have S [ α ] = S [ αβ − ], that is, αβ − ∈ S [ α ]. This gives αβ − = (cid:16) n − X i =0 α i β r i (cid:17) + (cid:16) n − X i =0 α i β r i (cid:17) α + · · · + (cid:16) n − X i =0 α i β r ( n − i (cid:17) α n − Now, using α n = P n − i =0 α i x i , we get β − = β r ∈ R for some non zero r ∈ R , a contradiction. Thus, we may assume that S ⊂ S [ α − β ] hasno intermediate ring. It follows that either S = S [ α − β ] or S [ α − β ] = S [ α − β ]. If former holds, then α − β ∈ S . Therefore, α − β = r + αβ r + α β r + · · · + α n − β r n − for some r , r , . . . , r n − ∈ R . Thisgives β = ( r + αβ r + α β r + · · · + α n − β r n − ) α Consequently, we have β = β r for some non zero r ∈ R , that is, 1 = βr ,a contradiction. Finally, we assume that S [ α − β ] = S [ α − β ], that is, α − β ∈ S [ α − β ]. This gives α − β = (cid:16) n − X i =0 α i β r i (cid:17) + (cid:16) n − X i =0 α i β r i (cid:17) α − β + · · · + (cid:16) n − X i =0 α i β r mi (cid:17) α − m β m , that is, α m − β = (cid:16) n − X i =0 α i β r i (cid:17) α m + (cid:16) n − X i =0 α i β r i (cid:17) α m − β + · · · + (cid:16) n − X i =0 α i β r mi (cid:17) β m On comparing the coefficient of α m − , we conclude that β is a multipleof β , which is again a contradiction. Therefore, T ⊆ K .We now define the maximal non valuative subrings of an integral domainformally. Definition 2.2.
Let R be a proper subring of an integral domain T . Then R is said to be a maximal non valuative subring of T if R is not a valuativedomain but every subring of T properly containing R is a valuative domain.A domain R is said to be a maximal non valuative domain if R is a maximalnon valuative subring of its quotient field qf( R ).The next corollary shows that the integral closure of maximal non valuativedomain is a Pr¨ufer domain. Corollary 2.3.
Let R be a maximal non valuative domain. Then R ′ is aPr¨ufer domain. Moreover, if R ′ is local, then R ′ is a valuation domain.Proof. Note that ( R, qf( R )) is a residually algebraic pair, by Theorem 2 . R is called an i-domain if for each overring T of R , thecanonical contraction map Spec( T ) → Spec( R ) is injective, see [10]. The nextcorollary is a direct consequence of Corollary 2.3 and [10, Corollary 2.15]. Corollary 2.4.
Let R be a maximal non valuative domain. If R ′ is local,then R is an i-domain. Now, in the next proposition we discuss the impact of localization on max-imal non valuative subrings of a domain.
Proposition 2.5.
Let R be a maximal non valuative subring of an integraldomain T and N be a multiplicatively closed subset of R . Then either N − R is a valuative domain or N − R is a maximal non valuative subring of N − T .Proof. If N − R is valuative, then we are done. Now, assume that N − R isnot valuative. Let S ′ be a subring of N − T containing N − R properly. Then S ′ = N − S , for some subring S of T properly containing R . Now, by [3,Proposition 2.4], S ′ is valuative as S is valuative by assumption. Thus, N − R is a maximal non valuative subring of N − T .As a consequence of Proposition 2.5, the requirement of ring to be local inCorollary 2.4 can be dropped. Corollary 2.6.
Let R be a maximal non valuative domain. If R is integrallyclosed, then R is an i-domain.Proof. As i-domain is a local property, it is enough to show that R P is ani-domain for all prime ideals P of R . Let P be a prime ideal of R . Thenby Proposition 2.5, either R P is a valuative domain or R P is a maximal nonvaluative domain. If R P is a valuative domain, then R P is an i-domain, by [3,Corollary 3.3]. Otherwise R P is a maximal non valuative domain. As R P is alocal integrally closed domain, the result now follows from Corollary 2.4.In the next theorem, we list some properties of maximal non valuativedomains which are not integrally closed. Theorem 2.7.
Let R be a maximal non valuative domain. If R is notintegrally closed, then the following statements hold:(i) | Max ( R ) | ≤ .(ii) The set of non-maximal prime ideals of R is linearly ordered by inclusion.(iii) There is at most one maximal ideal of R that does not contain all non-maximal prime ideals of R . Rahul Kumar and Atul Gaur
Proof.
Note that R is a maximal non valuative subring of R ′ . In particular, R ′ is a valuative domain. Consequently, the statements (i), (ii), and (iii) holdfor R ′ , by [3, Theorem 2.2]. As R ⊂ R ′ is an integral extension of domains, weconclude that the statements (i), (ii), and (iii) hold for R as well.We now present some properties of maximal non valuative domains whichare integrally closed. Proposition 2.8.
Let R be a maximal non valuative domain. If R is inte-grally closed, then the following statements hold:(i) ≤ | Max ( R ) | ≤ .(ii) If | Max ( R ) | = 2 , then there are exactly two non-maximal prime idealsof R that are not comparable. Otherwise, the set of non-maximal primeideals of R is linearly ordered by inclusion.(iii) There are at most two maximal ideals of R that do not contain all non-maximal prime ideals of R .Proof. (i) Note that if R has more than four maximal ideals, then R has aproper overring with four maximal ideals, which is a contradiction by [3,Theorem 2.2(i)]. It follows that | Max( R ) | ≤
4. Now, if R is local, then R is a valuation domain, by Corollary 2.3, a contradiction.(ii) First, assume that M and N are the only maximal ideals of R . Then R M and R N are valuative as R is maximal non valuative. It follows that R M and R N are valuation domains, by [3, Proposition 3.1]. Thus, R is a B´ezout domain with exactly two maximal ideals. Since R is nota valuative domain, M and N do not contains each non-maximal primeideal of R , by [3, Theorem 3.7]. Consequently, there are at least two non-maximal prime ideals of R that are not comparable. Since R M and R N are valuation domains, there are exactly two non-maximal prime idealsof R that are not comparable.Now, let M , M and M be any three maximal ideals of R . Then R M i is avaluative domain for i = 1 , ,
3. It follows that R M i is a valuation domainfor i = 1 , ,
3, by [3, Proposition 3.1]. If possible, suppose that P and Q are any two incomparable non-maximal prime ideals of R . Without lossof generality, we may assume that P ⊂ M but P M and Q ⊂ M but Q M . Since R is maximal non valuative, R M ∩ R M is a valuativedomain, which is a contradiction, by [3, Theorem 3.7]. Thus, the set ofnon-maximal prime ideals of R is linearly ordered by inclusion.(iii) If | Max( R ) | = 2, then nothing to prove. Now, assume that R has exactlythree maximal ideals, say M , M , and M . Let P i be a non-maximalprime ideal of R such that P i M i for i = 1 , ,
3. Also by part (ii), wemay assume that P ⊆ P ⊆ P . Then P is a non-maximal prime idealof R that is not contained in any maximal ideal of R , a contradiction.Finally, assume that R has exactly four maximal ideals, say M , M , M ,and M . Let P i be a non-maximal prime ideal of R such that P i M i for i = 1 , ,
3. Again, by part (ii), we may assume that P ⊆ P ⊆ P . Then P ⊂ M . Note that S = R M ∩ R M ∩ R M is a valuative domain. Thus,by [3, Theorem 2.2], at most one maximal ideal of S does not containeach non-maximal prime ideal of S , a contradiction. Therefore, there areat most two maximal ideals of R that do not contain all non-maximalprime ideals of R .In the next theorem, we present a necessary and sufficient condition for anintegrally closed domain to be a maximal non valuative domain. Theorem 2.9.
Let R be an integrally closed domain. Then the followingstatements are equivalent:(1) R is a maximal non valuative domain.(2) Exactly one of the following holds:(i) R is a B ´ ezout domain with exactly two maximal ideals and exactlytwo non-maximal prime ideals of R are not comparable.(ii) R is a B ´ ezout domain with exactly three maximal ideals andexactly two maximal ideals of R do not contain exactly one non-maximalprime ideal of R whereas the third maximal ideal of R contains all non-maximal prime ideal of R .(iii) R is a B ´ ezout domain with exactly four maximal ideals and atmost one maximal ideal of R does not contain all non-maximal primeideals of R .Proof. (1) ⇒ (2) Note that R is a Pr¨ufer domain, by Corollary 2.3. Also, byProposition 2.8(i), we have 2 ≤ | Max( R ) | ≤
4. It follows that R is a B´ezoutdomain. Now, assume that | Max( R ) | = 2, then (i) follows from Proposition2.8(ii). Also, if | Max( R ) | = 3, then exactly two maximal ideals of R do notcontain at least one non-maximal prime ideal of R , by Proposition 2.8(iii)and [3, Theorem 3.7]. Now, assume that M, N and U are maximal idealsof R . If possible, assume that P, Q are non-maximal prime ideals of R suchthat only U contains both of them. Since R U is a valuation domain, either P ⊂ Q or Q ⊂ P . Without loss of generality, assume that P ⊂ Q . Then by[3, Theorem 3.7], R M ∩ R N ∩ R Q is not a valuative domain, a contradictionas R is maximal non valuative. Thus, M and N do not contain exactly onenon-maximal prime ideal of R . Finally, assume that | Max( R ) | = 4. Then by Rahul Kumar and Atul Gaur
Proposition 2.8(iii), there are at most two maximal ideals of R that do notcontain all non-maximal prime ideals of R . If possible, assume that there areexactly two maximal ideals of R that do not contain all non-maximal primeideals of R . Let M , M , M and M be maximal ideals of R , where M and M do not contain all non-maximal prime ideals of R . Let P , P be non-maximalprime ideals of R such that P M and P M . Moreover, by Proposition2.8(ii), we may assume that P ⊆ P . Then P M . Note that P ⊂ M .Since R is a maximal non valuative domain, S = R M ∩ R M ∩ R M is a valuativedomain, which is a contradiction, by [3, Theorem 3.7].(2) ⇒ (1) Suppose (i) holds. Then R is not valuative, by [3, Theorem 2.2].Now, let M, N be maximal ideals of R and P, Q be incomparable non-maximalprime ideals of R . Since R is a Pr¨ufer domain, R M , R N are valuation domains.Thus, we may assume that P ⊂ M , P N and Q ⊂ N , Q M . Now, by [3,Corollary 3.9], it is enough to show that R M ∩ R U and R V ∩ R N are valuativedomains for some arbitrary non-maximal prime ideals U, V of R . First, weclaim that R M ∩ R U is a valuative domain. If U ⊂ M , then we are done.Therefore, we may assume that U M and so U ⊂ N . It follows that P U .Now if U ⊂ P , then again we are done. Otherwise U = Q , by assumption.Then by [3, Theorem 3.7], R M ∩ R Q is a valuative domain. Similarly, we canprove that R V ∩ R N is a valuative domain.Now, assume that (ii) holds. Then R is not valuative, by [3, Theorem 3.7].Let S be a proper overring of R . Then S = ∩ P ∈ X R P for some subset X ofSpec( R ). Let M, N , and U be maximal ideals of R where M and N do notcontain exactly one non-maximal prime ideal of R , whereas U contains allnon-maximal prime ideals of R . Now, the following cases arise:Case (i): Let M, N, U ∈ X . Then S = R , a contradiction.Case (ii): Let U ∈ X . Then S = R U is a valuation domain (and so isvaluative) as R is a B´ezout domain.Case (iii): Let M ∈ X but N, U / ∈ X . Then R M ∩ R U is a subring of S contains R properly. Note that R M ∩ R U is valuative, by [3, Theorem 3.7].It follows that S is valuative, by [3, Corollary 3.9]. Similarly, if N ∈ X but M, U / ∈ X , then we are done.Case (iv): Let M, N ∈ X but U / ∈ X . Then S = R M ∩ R N ∩ ( ∩ P ∈ X R P ),where X = X \ { M, N } . Without loss of generality, we may assume that X does not contains any prime ideal that is either contained in M or N . By[3, Corollary 3.9], we may assume that X is non empty. We claim that X is a singleton set. If possible, suppose that P, Q ∈ X . By assumption, wemay assume that P ⊂ M, P N and Q ⊂ N, Q M . Since P, Q ⊂ U and R U is a valuation domain, P and Q are comparable, which contradicts ourassumption. Thus, we may assume that X = { P } . Now, by [3, Theorem 3.7], S = R M ∩ R N ∩ R P is a valuative domain.Case (v): Let M X, N X , and U X . Then S contains a valuationdomain R U properly and hence S is a valuation domain.Finally, assume that (iii) holds. Then again by [3, Theorem 3.7], R is notvaluative. Let S be a proper overring of R . Then S = ∩ P ∈ X R P for somesubset X of Spec( R ). Now, the following cases arise:Case (i): Let all four maximal ideals be in X . Then R = S , which is acontradiction.Case (ii): Let there be more than one maximal ideals in X . Then S is aB´ezout domain with at most three maximal ideals and at most one maximalideal of S does not contain all non-maximal prime ideals of S . Thus, S isvaluative, by [3, Theorem 3.7].Case (iii): Let there be at most one maximal ideal in X .Subcase (i): Let all the maximal ideals of R contain all non-maximal primeideals of R . Then S ′ = R M is a subring of S contains R properly, where M is amaximal ideal of R . Thus, S ′ is a valuation domain and hence S is a valuationdomain.Subcase (ii): Let N be the maximal ideal that does not contain all non-maximal prime ideals of R . If N / ∈ X , then again S is a valuation domain.Now, assume that N ∈ X . Then take S ′′ = R M ∩ R N , where M is a maximalideal of R other than N . Note that S ′′ is a subring of S contains R properly.Now, S ′′ is valuative, by [3, Theorem 3.7]. Thus, by [3, Corollary 3.9], S is avaluative domain. Hence, R is a maximal non valuative domain. Corollary 2.10.
Let R be a finite dimensional B ´ ezout domain. Assumethat dim ( R ) = n . Then the following statements hold:(1) If R is a maximal non valuative domain, then | Spec ( R ) | = n + | Max ( R ) | ,where ≤ | Max ( R ) | ≤ .(2) R is a maximal non valuative domain if and only if exactly one of thefollowing holds:(i) | Spec ( R ) | = n + 2 , R has exactly two maximal ideals with height n , and exactly two non-maximal prime ideals of R are not comparable.(ii) | Spec ( R ) | = n +3 , R has exactly three maximal ideals with exactlytwo maximal ideals of R do not contain exactly one non-maximal primeideal of R , and exactly one maximal ideal have height n .(iii) | Spec ( R ) | = n + 4 , R has exactly four maximal ideals, and atleast three of these maximal ideals have height n .Proof. By Proposition 2.8 and Theorem 2.9, the result holds.A ring extension R ⊂ T is said to be a minimal extension or R is said tobe a maximal subring of R , if there is no ring between R and T , see [5 , R ⊂ T is a pointwise minimal extension, if R ⊂ R [ x ] is minimal0 Rahul Kumar and Atul Gaur for each x ∈ T \ R , see [3]. Our next theorem gives a necessary and sufficientcondition for a domain R to be a maximal non valuative subring of R ′ provided R ′ is local. Theorem 2.11.
Let R be a ring such that R ′ is local. Then R is a maximalnon valuative subring of R ′ if and only if the following statements hold:(i) R ′ is a valuation domain.(ii) R ⊂ R ′ is not a pointwise minimal extension.(iii) for each ring S such that R ⊂ S ⊆ R ′ , we have either S = R ′ or S ⊂ R ′ is a pointwise minimal extension.Proof. Let R be a maximal non valuative subring of R ′ . Then R ′ is a valuativedomain and so is a valuation domain, by [3, Proposition 3.1]. Also, R ⊂ R ′ isnot a pointwise minimal extension, by [3, Proposition 5.1]. Let S be a ring suchthat R ⊂ S ⊆ R ′ . Then S is a valuative domain. Thus, by [3, Proposition 5.1],either S = R ′ or S ⊂ R ′ is a pointwise minimal extension.Conversely, assume that (i), (ii), and (iii) hold. Then R is not a valuativedomain by (ii) and [3, Proposition 5.1]. Also, every proper overring of R contained in R ′ is a valuative domain, by (i), (iii), and [3, Proposition 5.1].Recall from [7] that a domain R is said to be a pseudo-valuation domainif for any prime ideal P of R and any x, y in the quotient field of R suchthat xy ∈ P , then either x ∈ P or y ∈ P . Every PVD R admits a canonicallyassociated valuation overring V , in which every prime ideal of R is also a primeideal of V and both R and V are local domains with the same maximal ideal,see [7, Theorem 2.7]. In the next theorem, we give several equivalent conditionsfor a pseudo-valuation domain to be a maximal non valuative domain. Theorem 2.12.
Let ( R, M ) be a pseudo-valuation domain, with canonicallyassociated valuation overring ( V, M ) . Assume that K := R/M , L := R ′ /M ,and F := V /M . Then the following conditions are equivalent:(i) R is a maximal non valuative subring of V ; (ii) R is a maximal non valuative domain;(iii) R ′ = V , R ⊂ R ′ is not a pointwise minimal extension, and for each ring S such that R ⊂ S ⊆ R ′ , we have either S = R ′ or S ⊂ R ′ is a pointwiseminimal extension;(iv) L = F , K ⊂ L is not a pointwise minimal extension, and for each ring S such that K ⊂ S ⊆ L , we have either S = L or S ⊂ L is a pointwiseminimal extension. Proof.
Note that (ii) ⇒ (i) holds trivially by definition. For (i) ⇒ (ii), assumethat (i) holds. Let T be a proper overring of R . Then either T ⊆ V or V ⊂ T , by [2, Lemma 1.3]. If T ⊆ V , then T is valuative. If V ⊂ T , then T is a valuation domain. Thus, (ii) holds. Note that (i) ⇔ (iii) follows fromTheorem 2.11. Finally it is easy to see (iii) ⇔ (iv).After pseudo-valuation domain the natural question is when maximal nonpseudo-valuation ring is a maximal non valuative domain. This we address inthe next theorem. Recall from [8] that a maximal non pseudo-valuation subringof a domain S is a proper subring R of S that is not a pseudo-valuation ringbut each subring of S properly containing R is pseudo-valuation. Moreover,a ring T is called the unique minimal overring of R in S if R ⊂ T and anyintermediate ring A between R and S not equal to R contains T , see [8]. Theorem 2.13.
Let R be a maximal non pseudo-valuation subring of R ′ .Then the following are equivalent:(i) R is a maximal non valuative subring of R ′ .(ii) R is not valuative, R ′ is a valuation domain, and R has a unique minimaloverring S in R ′ that is valuative.Proof. Let R be a maximal non valuative subring of R ′ . Then R ′ is valuation,by Theorem 2.11. It follows that R has a unique minimal overring, say S in R ′ ,by [8, Theorem 6]. Note that S is valuative as R is a maximal non valuativesubring of R ′ .Conversely, assume that (ii) holds. Let T be a proper overring of R in R ′ . Then S ⊆ T , by assumption. Since R is a maximal non pseudo-valuationsubring of R ′ , S is a pseudo-valuation domain. It follows that T is valuative,by [3, Corollary 5.4]. Thus, R is a maximal non valuative subring of R ′ .A proper overring T of a domain R is called the unique minimal overringof R if any proper overring of R contains T , see [6]. Theorem 2.14.
Let R be a maximal non pseudo-valuation subring of qf ( R ) .If R is local, then the following are equivalent:(i) R is a maximal non valuative domain.(ii) R is not valuative and R has a unique minimal overring S that is avaluative pseudo-valuation domain with associated valuation overring R ′ .Proof. Let R be a maximal non valuative domain. Then R is not integrallyclosed, by Proposition 2.8. Now, (ii) follows from [8, Theorem 4].Conversely, assume that (ii) holds. Let T be a proper overring of R . Then S ⊆ T , by assumption. It follows that T is valuative, by [3, Corollary 5.4].Thus, R is a maximal non valuative domain.2 Rahul Kumar and Atul Gaur