aa r X i v : . [ m a t h . A C ] S e p CONCH MAXIMAL SUBRINGS
ALBORZ AZARANG
Department of Mathematics, Faculty of Mathematical Sciences and Computer,Shahid Chamran Universityof Ahvaz, Ahvaz-Irana − [email protected] ID: orcid.org/0000-0001-9598-2411 Abstract.
It is shown that if R is a ring, p a prime element of an integral domain D ≤ R with T ∞ n =1 p n D = 0 and p ∈ U ( R ), then R has a conch maximal subring (see [17]). We prove that either aring R has a conch maximal subring or U ( S ) = S ∩ U ( R ) for each subring S of R (i.e., each subringof R is closed with respect to taking inverse, see [30]). In particular, either R has a conch maximalsubring or U ( R ) is integral over the prime subring of R . We observe that if R is an integral domain with | R | = 2 ℵ , then either R has a maximal subring or | Max ( R ) | = 2 ℵ , and in particular if in addition dim ( R ) = 1, then R has a maximal subring. If R ⊆ T be an integral ring extension, Q ∈ Spec ( T ), P := Q ∩ R , then we prove that whenever R has a conch maximal subring S with ( S : R ) = P , then T has a conch maximal subring V such that ( V : T ) = Q and V ∩ R = S . It is shown that if K isan algebraically closed field which is not algebraic over its prime subring and R is affine ring over K ,then for each prime ideal P of R with ht ( P ) ≥ dim ( R ) −
1, there exists a maximal subring S of R with( S : R ) = P . If R is a normal affine integral domain over a field K , then we prove that R is an integrallyclosed maximal subring of a ring T if and only if dim ( R ) = 1 and in particular in this case ( R : T ) = 0. Introduction
Following Faith [17], a subring V of a commutative ring T is called a conch subring if there exists a unit x ∈ T such that x − ∈ V but x / ∈ V and V is maximal respect to this property (i.e., excluding x and in-cluding x − ), V is called x -conch subring or it is said that V conches x in T . In other words, V is a x -conchsubring of a ring T if and only if V is maximal in the set { R | R is a subring of T, Z [ x − ] ⊆ R, x / ∈ R } ,where Z denotes the prime subring of R . But note that clearly, the previous set is nonempty if and onlyif x / ∈ Z [ x − ] which is equivalent to the fact that x is not integral over Z . Krull proved that an integraldomain R is integrally closed if and only if R is the intersection of the conch subrings of K = Q ( R )that contains R , and moreover, each conch subring V of a field K is a chain ring and therefore is avaluation domain (i.e., for each x ∈ K , either x ∈ V or x − ∈ K ), see [25, P.110]. In [17], Faith studiedconch subrings for general commutative rings by the use of Manis valuations rings (or maximal pairs).He proved that if a subring A conches x in a ring Q , then A is integrally closed and ( A, √ x − A ) is amaximal pair of Q ; in particular, A is a valuation ring. Conversely, a maximal pair ( A, P ) comes froma conch subring of Q if and only if P = √ x − A for some unit x ∈ Q ( x − ∈ A ) [Conch Ring Theorem].Moreover, if R is a subring of a ring Q , then the intersection Conch Q ( R ) of the conch subrings of Q containing R is integrally closed, and every element of Conch ∗ Q ( R ) := Conch Q ( R ) ∩ U ( Q ) is integral over R . More interestingly, in [23], Griffin proved that if Q is VNR or has few zero-divisor (i.e., the set of allzero divisor of Q , Zd ( Q ), is a finite union of prime ideals of Q ) and R is integrally closed in Q = Q ( R ),then R is the intersection of valuation subrings of Q . In [17, Section 10], Faith studied conch subringswhich are maximal subrings, i.e., if A conches x in Q , then A is a maximal subring of Q if and only if Q = A [ x ].In this paper, motivated by the above facts and the existence of maximal subring, we are interesting toprove that if R is a ring then either R has a conch maximal subring or each subring of R is closed withrespect to taking inverse, i.e., U ( S ) = S ∩ U ( R ) for each subring S of R , see [30]. In particular, for eachring R either U ( R ) is integral over Z , or R has a conch maximal subring. Date : 5 Jun 2020.2010
Mathematics Subject Classification.
Key words and phrases.
Maximal subring, Conch subring, Integral element, Conductor ideal.
All rings in this note are commutative with 1 = 0. All subrings, ring extensions, homomorphisms andmodules are unital. A proper subring S of a ring R is called a maximal subring if S is maximal withrespect to inclusion in the set of all proper subrings of R . Not every ring possesses maximal subrings (forexample the algebraic closure of a finite field has no maximal subrings, see [7, Remark 1.13]; also see [6,Example 2.6] and [9, Example 3.19] for more examples of rings which have no maximal subrings). A ringwhich possesses a maximal subring is said to be submaximal, see [4], [7] and [9]. If S is a maximal subringof a ring R , then the extension S ⊆ R is called a minimal ring extension (see [18]) or an adjacent extensiontoo (see [13]). Minimal ring extensions first appears in [21], for studying integral domains with a uniqueminimal overring. Next in [18], these extensions are fully studied and some essential facts are obtained.The following result, whose proof could be found in [18] is needed. Before presenting it, let us recall thatwhenever S is a maximal subring of a ring R , then one can easily see that either R is integral over S or S is integrally closed in R . If S ⊆ R is a ring extension, then the ideal ( S : R ) = { x ∈ R | Rx ⊆ S } iscalled the conductor of the extension S ⊆ R . Theorem 1.1.
Let S be a maximal subring of a ring R . Then the following statements hold. (1) ( S : R ) ∈ Spec ( S ) . (2) ( S : R ) ∈ M ax ( S ) if and only if R is integral over S . (3) If S is integrally closed in R , then ( S : R ) ∈ Spec ( R ) . Moreover if x, y ∈ R and xy ∈ S , theneither x ∈ S or y ∈ S . (4) There exist a unique maximal ideal M of S such that S M is a maximal subring of R M and foreach P ∈ Spec ( S ) \ { M } , S P = R P ( M is called the crucial maximal ideal of the extension S ≤ R and ( S : R ) ⊆ M . Moreover, S ≤ R is integral (resp., integrally closed) extension if and only if S M ≤ R M is integral (resp., integrally closed) extension). (5) If S is an integrally closed maximal subring of R , P = ( S : R ) and M be the crucial maximalideal of the extension, then ( RP ) MP is a one dimensional valuation domain. Note that by ( u, u − )-Lemma, one can easily see that if S ⊆ R is an integral ring extension then U ( S ) = S ∩ U ( R ). In particular, if S is a maximal subring of R and S ⊆ R is integral, then U ( S ) = S ∩ U ( R ).On the other hand, if S is a maximal subring of R , then by the second part of (3) of Theorem 1.1 (or bythe use of ( u, u − )-Lemma) for each x ∈ U ( R ), we infer that either x ∈ S or x − ∈ S . Therefore, if S is amaximal subring of R such that there exists a unit x in R such that x ∈ S but x − / ∈ S , we immediatelyconclude that S is integrally closed in R . Now let us prove the following fact for conch subrings by theuse of minimal ring extension. Corollary 1.2.
Let T be a ring and x ∈ U ( T ) . If R is a x -conch subring of T , then x − is contained inexactly one prime (maximal) ideal of R , in other words √ x − R ∈ M ax ( R ) .Proof. Since R is a x -conch subring of T , we immediately conclude that R ⊆ S := R [ x ] is a minimalring extension. Thus by (4) of Theorem 1.1, let M be the crucial maximal ideal of the minimal ringextension R ⊆ S . We prove that M is the only prime ideal of R which contains x − . First note that since R M ⊆ S M is a minimal ring extension ( R M = S M ), we deduce that x / ∈ R M and therefore x − ∈ M .Now for each prime ideal P of R with P = M , we have R P = S P . Therefore x − is a unit in R P , hence x − / ∈ P and we are done. (cid:3) Now, let us sketch a brief outline of this paper. Section 2 is devoted to the existence of integrallyclosed/conch maximal subrings. We prove that if R is a ring, D is an integral domain which is a subringof R , p is a prime element of D with T ∞ i =1 p n D = 0 and p ∈ U ( R ), then R has an integrally closed/conchmaximal subring. In particular, if R is a ring which contains an atomic (or a completely integrally closed)domain D and a prime element of D is invertible in R , then R has an integrally closed/conch maximalsubring. Next, we show that if T is a ring, then either T has an integrally closed/conch maximal subringor U ( R ) = R ∩ U ( T ), for each subring R of T , which is one of the main result in Section 2. This resulthas many consequences. In particular, either a ring T has an integrally closed/conch maximal subring or U ( T ) is integral over Z . It is observed that if T is an integral domain which satisfies ACCP (resp. is aBFD, see [1]), then either T has maximal subring or each subring of T satisfies ACCP (resp. is a BFD).We show that if D is a GCD-domain which is a subring of a ring T , then either T has a maximal subringor U ( T ) ∩ K = U ( D ), where K is the quotient field of D . We prove that a ring T has a conch maximalsubring if and only if Z [ U ( T )] has an integrally closed proper subring. If T is a ring without maximalsubring and Char ( T ) >
0, then each element of the group U ( T ) has finite order. In particular, if inaddition U ( T ) is finitely generated, then U ( T ) is finite. We observe that if T is a residually finite ring ONCH MAXIMAL SUBRINGS 3 with
Char ( T ) = 0 and U ( T ) is finitely generated, then either T has a maximal subring or T = Z . Weprove that if T is a reduced ring with | T | = 2 ℵ , then either T has a maximal subring or | M ax ( T ) | = 2 ℵ .In particular, if T is an integral domain with | T | = 2 ℵ and dim ( T ) = 1, then T has a maximal subring.It is observe that if R = K + Rx , where K is a field which is contained in a ring R , and x / ∈ U ( R ) ∪ Zd ( R ),then R has a maximal subring. Finally in Section 2, we prove that if R is an atomic integral domainwith | R | = 2 ℵ and M is a principal maximal ideal of R , then | R/M | = | R | and in particular R has amaximal subring. Section 3, is devoted to the existence of maximal subring with certain conductor. Wefirst prove that if K is an algebraically closed field which is not algebraic over its prime subring, thenfor each prime ideal P of K [ X ], there exists an integrally closed (conch) maximal subring R of K [ X ]with ( R : K [ X ]) = P . Next, we studied the conductor of integrally closed maximal subrings for integralextensions. In fact, we show that if R ⊆ T is an integral extension, Q ∈ Spec ( T ), P := Q ∩ R , and thereexists an integrally closed maximal subring S of R with U ( R/P ) * S/P ( S : R ) = P , then T has anintegrally closed maximal subring V with ( V : T ) = Q . Conversely, if R ⊆ T is an integral extension andeach maximal ideal of T is conductor of an integrally closed maximal subring of T , then each maximalideal of R is conductor of an integrally closed maximal subring of R . We show that if K is algebraicallyclosed field which is not absolutely algebraic the each prime ideal Q of K [ X , . . . , X n ] with ht ( Q ) ≥ n − K [ X , . . . , X n ]. But for minimal ringextension of K [ X , . . . , X n ], n ≥
2, a prime ideal of K [ X , . . . , X n ], is conductor of a minimal ring ex-tension of K [ X , . . . , X n ] if and only if P is maximal; and for n = 1, each prime ideal of K [ X ] is aconductor of a minimal ring extension of K [ X ]. We prove some results for the zeroness of the conductorof the minimal ring extension of the form R ⊆ R [ u ], where R is a certain ring or u is a certain element of R .Finally, let us recall some notation and definitions. As usual, let Char ( R ), U ( R ), Zd ( R ), N ( R ), J ( R ), M ax ( R ), Spec ( R ) and M in ( R ), denote the characteristic, the set of all units, the set of all zero-divisors,the nil radical ideal, the Jacobson radical ideal, the set of all maximal ideals, the set of all prime idealsand the set of all minimal prime ideals of a ring R , respectively. We also call a ring R , not necessarilynoetherian, semilocal (resp. local) if M ax ( R ) is finite (resp. | M ax ( R ) | = 1). For any ring R , let Z = Z · R = { n · R | n ∈ Z } , be the prime subring of R . We denote the finite field with p n elements,where p is prime and n ∈ N , by F p n . Fields which are algebraic over F p for some prime number p , arecalled absolutely algebraic field. If R is a ring and a ∈ R \ N ( R ), then R a denotes the ring of quotient of R respect to the multiplicatively closed set X = { , a, a , . . . , a n , . . . } . If D is an integral domain, thenwe denote the set of all non-associate irreducible elements of D by Irr ( D ). Also, we denote the set ofall natural prime numbers by P . Suppose that D ⊆ R is an extension of domains. By Zorn’s Lemma,there exists a maximal (with respect to inclusion) subset X of R which is algebraically independent over D . By maximality, R is algebraic over D [ X ] (thus every integral domain is algebraic over a UFD; thiscan be seen by taking D to be the prime subring of R ). If E and F are the quotient fields of D and R , respectively, then X can be shown to be a transcendence basis for F/E (that is, X is maximal withrespect to the property of being algebraically independent over E ). The transcendence degree of F over E is the cardinality of a transcendence basis for F/E (it can be shown that any two transcendence baseshave the same cardinality). We denote the transcendence degree of F over E by tr.deg ( F/E ). We remindthat whenever R ( T is an affine ring extension, i.e., T is finitely generated as a ring over R (in praticular,if T is a finitely generated R -module), then by a natural use of Zorn Lemma, one can easily see that T has a maximal subring S which contains R . If R is a proper subring of T then R is a maximal subring of T if and only if for each x ∈ T \ R , we have R [ x ] = T . Finally we refer the reader to [19], for standarddefinitions of Bezout domains, QR -domains and completely integrally closed domains.2. Existence of Conch Maximal Subring
We begin this section by the following main result which is a generalization of [4, Theorem 2.2].
Theorem 2.1.
Let D be an integral domain with a prime element p such that T ∞ n =1 p n D = 0 . Assumethat D ⊆ R is a ring extension such that p ∈ U ( R ) . Then R has an integrally closed maximal subringwhich contains D and conches p .Proof. We may assume that R is an integral domain which is algebraic over D . To see this, first notethat D \ { } is a multiplicatively closed set in R , therefore R has a prime ideal Q , with Q ∩ D = 0. Inother words, D can be embedded in R/Q , and clearly p ∈ U ( R/Q ) (note that if
S/Q is an integrallyclosed maximal subring of
R/Q which contains the image of D and conches ( p + Q ) − in R/Q , then S is an integrally closed maximal of R which contains D and conches p ). Hence we may suppose that R ALBORZ AZARANG is an integral domain. If R is not algebraic over D , then let X be a transcendence base for R over D .Clearly p is prime in D [ X ] and T ∞ n =1 p n D [ X ] = 0. Hence we can replace D by D [ X ]. Thus assume that D ⊆ R is an algebraic extension of integral domain with quotient fields K ⊆ E . It is clear that E/K isan algebraic extension. Now we claim that D ( p ) [ p ] = K , where K is the quotient field of D . For prooflet x = ab ∈ K , where a, b ∈ D . Since T ∞ n =1 p n D = 0, we conclude that b = p n c , where c / ∈ pD and n ≥
0. Thus ac ∈ D ( p ) and therefore x = a/cp n ∈ D ( p ) [ p ], i.e., K = D ( p ) [ p ]. Next we prove that S := D ( p ) is a maximal subring of K . To see this, let x ∈ K \ S . From T ∞ n =1 p n D = 0 and x / ∈ S , we easily seethat x = ap n b , where a, b / ∈ pD and n ∈ N . Hence we infer that a ∈ S and therefore p n ∈ S [ x ]. Thus p ∈ S [ x ], which immediately by the previous part implies that S [ x ] = K , i.e., S is a maximal subring of K . Thus by [8, Proposition 2.1], we conclude that E has a maximal subring V such that V ∩ K = S .Hence p / ∈ V and therefore we deduce that V is an integrally closed maximal subring of E . Thus in theextension R ⊆ E we have U ( R ) * V . Therefore by [9, Theorem 2.19], we infer that R has an integrallyclosed maximal subring W which contains V ∩ R but p / ∈ W i.e., W is a maximal subring which contains D and conches p in R (note that a conch maximal subring is integrally closed by the comment precedingCorollary 1.2). (cid:3) Corollary 2.2.
Let D be an atomic (or a completely integrally closed) domain and R is a ring extensionof D . If a prime element of D is invertible in R , then R has a conch maximal subring.Proof. Let p be a prime element of D which is invertible in R . We claim that J := T ∞ n =1 ( p n ) = 0. If D iscompletely integrally closed then we are done by [19, Corollary 13.4]. Hence assume that R is atomic and J = 0, then by [24, Ex.5, Sec.1-1], J is a prime ideal of R which is properly contained in M := ( p ). Nowsince J = 0 and R is atomic domain, we immediately conclude that J contains an irreducible element q of R . Thus q ∈ M = ( p ) and therefore q = p which is absurd (see [2] for more interesting result in arbitrarycommutaive rings). Thus in any cases J = 0 and hence we are done by the previous theorem. (cid:3) Fields which have (no) maximal subrings are completely determined in [7]. We need the following insequel.
Corollary 2.3.
Let K be a field which is not absolutely algebraic. Then K has an integrally closed/conchmaximal subring.Proof. If Char ( K ) = 0 then K contains Z and use the previous corollary. If Char ( K ) = 0, then by ourassumption, there exists x ∈ K , which is not algebraic over Z p . Therefore K contains the atomic domain Z p [ x ] and again use the previous corollary. (cid:3) Now we have the following main theorem.
Theorem 2.4.
Let T be a ring. Then either T has a conch maximal subring or U ( R ) = R ∩ U ( T ) foreach subring R of T . In other words either T has a conch maximal subring or each subring of T is closedrespect to taking the inverse.Proof. Suppose that T has a subring R with U ( R ) ( R ∩ U ( T ). Hence there exists x ∈ R \ U ( R ) with x ∈ U ( T ). We show that T has a maximal subring conches x − . First we claim that we may assumethat T is an integral domain. To see this, take X := { xy | y ∈ R } , clearly X is a multiplicativelyclosed subset of R and 0 / ∈ X , for x is not unit in R . Hence there exists a prime ideal P of R such that P ∩ X = ∅ . We may assume that P is a minimal prime ideal of R and therefore x / ∈ P (note, x is nota zero-divisor of R ). Since P is a minimal prime ideal of R , we conclude that there exists a (minimal)prime ideal Q of T such that Q ∩ R = P (see [24, Exercise 1, P. 41]), therefore we can consider R/P as asubring of
T /Q . Now note that the image of x is a unit of T /Q but is not a unit of
R/P ; for otherwisethere exists y ∈ R , such that 1 − xy ∈ P , i.e., 1 + ( − y ) x ∈ P ∩ X which is absurd. Thus the extension R/P ⊆ T /Q is an extension of integral domains and ¯ x := x + P is not a unit in R/P but is a unit in
T /Q .Clearly, if
S/Q is a maximal subring of
T /Q which conches ¯ x − in T /Q , then S is a maximal subring of T conches x − . Thus we may assume T is an integral domain. Now we have two cases:( ) Char ( T ) = p >
0, where p is a prime number (note T is an integral domain). Thus we infer that x is not algebraic over the prime subring of T , for otherwise x is integral over Z p , the prime subring of R (or T ). Therefore by ( u, u − )-Lemma, x ∈ U ( R ) which is absurd. Therefore x is not algebraic over Z p .Hence D := Z p [ x ] is an atomic domain which is contained in T and the prime element x of D is invertiblein T . Thus T has a maximal subring that conches x − by Corollary 2.2.( ) Char ( T ) = 0. If x is not algebraic over Z , then similar to the proof of (1) and considering the atomic ONCH MAXIMAL SUBRINGS 5 domain D := Z [ x ] in T , we conclude that T has a maximal subring conches x − , by Corollary 2.2. Thusassume that x is algebraic over Z . Since x − / ∈ R , we conclude that x − Z [ x ], for Z [ x ] ⊆ R . Thusby [19, Theorem 30.9], dim ( Z [ x ]) = 1 and therefore by [24, Theorem 56], the quotient field of Z [ x ], i.e., Q ( x ) has a valuation V such that xV = V (i.e., x − / ∈ U ( V )). Since Z [ x ] is a one dimensional noetherianintegral domain, then by Krull-Akizuki Theorem ([24, Theorem 93]), we infer that V is a noetherian onedimensional valuation domain. In other words, V is a DVR and therefore V is a UFD. Let M = ( π )denotes the maximal ideal of V . Thus x = vπ for some v ∈ V (note, V is a valuation and x − / ∈ V ).Let K be the quotient field of T , then V ⊆ Q ( x ) ⊆ K . Therefore by Theorem 2.1, K has a maximalsubring W such that V ⊆ W and π / ∈ W . Thus v ∈ W and π ∈ N , where N is the maximal ideal of W .Therefore x ∈ N and thus U ( T ) * W . Therefore by [9, Theorem 2.19], T has a maximal subring thatconches x − . (cid:3) In other words, the previous theorem state that if R ⊆ T is a ring extension and a non invertible element x of R , is invertible in T , then T has a maximal subring. In particular, if a ring T has no maximalsubring, then xR = R implies xT = T , for each subring R of T and x ∈ R . In other words either a ring T has a maximal subring or each principal proper ideal of any subring of T , survives in T . The previoustheorem has several conclusion as follows. Corollary 2.5.
Let T be a ring, then T has a conch maximal subring if and only if U ( T ) is not integralover Z . In particular, if T has no conch maximal subring then Z [ x ] = Z [ x − ] , for each x ∈ U ( T ) .Proof. If T has a subring R which conches x − in T , then by ( u, u − )-Lemma x − is not integral over R and therefore x − is not integral over Z . Conversely, assume that T has no conch maximal subring, thenfor each x ∈ U ( T ), by Theorem 2.4, we conclude that U ( Z [ x ]) = Z [ x ] ∩ U ( T ). Therefore x − ∈ Z [ x ].Thus by ( u, u − )-Lemma, we deduce that x − is integral over Z . Thus U ( T ) is integral over Z . Henceif T has no conch maximal subring, then for each x ∈ U ( T ) we obtain that x − ∈ Z [ x ] and x ∈ Z [ x − ].Therefore Z [ x ] = Z [ x − ]. (cid:3) In other words the previous corollary state that a ring T has a subring S which conches x − in T if andonly if T has a maximal subring which conches x − in T . Corollary 2.6.
Let T be a ring with nonzero characteristic. Then either T has a conch maximal subringor each element of the group U ( T ) has finite order.Proof. Assume that
Char ( T ) = n and therefore Z n is the prime subring of T . Thus if T has no maximalsubring, then for each x ∈ U ( T ), x is integral over Z n . Hence, Z n [ x ] is a finitely generated Z n -module.Therefore Z n [ x ] is finite. This immediately implies that x has finite order in U ( T ). (cid:3) Corollary 2.7.
Let T be a ring without maximal subring and S ⊆ R be subrings of T . Then U ( S ) = S ∩ U ( R ) .Proof. Since T has no maximal subring, then by Theorem 2.4, U ( R ) = R ∩ U ( T ) and U ( S ) = S ∩ U ( T ).Thus U ( S ) = S ∩ U ( T ) = ( S ∩ R ) ∩ U ( T ) = S ∩ ( R ∩ U ( T )) = S ∩ U ( R ). (cid:3) Corollary 2.8.
Let R be a ring and X be a multiplicatively closed subset of regular (i.e., non zero-divisors) of R . Then either R X has a maximal subring or R X = R , i.e., X ⊆ U ( R ) . In particular, if R is an integral domain, then each proper quotient overring of R (i.e., R ( R X ) has a maximal subring.Proof. It suffices to take T = R X in Theorem 2.4. (cid:3) Corollary 2.9.
Let R be an integral domain and X be a multiplicatively closed subset of R which is notcontained in U ( R ) . Then there exists a proper subring S of R X and a ∈ X such that R X = S a . An integral domain R is called a QR -domain, if each overring of R is of the form R X for some multi-plicatively closed subset X of R . It is well-known that each Bezout domain is a QR -domain. Corollary 2.10.
Let R be a QR -domain (which is not absolutely algebraic field). Then each properoverring of R has a maximal subring. Corollary 2.11.
Let R be a ring. Then either Q ( R ) has a maximal subring or Q ( R ) = R .Proof. Let X = R \ Zd ( R ), therefore Q ( R ) = R X . If for some x ∈ X , we have x − / ∈ R , then by Theorem2.4, Q ( R ) has a maximal subring. Hence if Q ( R ) has no maximal subring, then for each x ∈ X weconclude that x − ∈ R and therefore R = Q ( R ). (cid:3) ALBORZ AZARANG
Corollary 2.12.
Let R be a noetherian ring. Then either Q ( R ) has a maximal subring or R is a countableartinian ring with nonzero characteristic which is integral over its prime subring.Proof. Assume that Q ( R ) has no maximal subring. Therefore by Corollary 2.11, R = Q ( R ). Now notethat since R is noetherian ring, then Ass ( o ) is finite and Zd ( R ) = S P ∈ Ass (0) P . This immediately impliesthat Q ( R ) is a semilocal ring. Thus by [9, Proposition 3.13], we infer that R = Q ( R ) is an artinian ringwith nonzero characteristic which is integral over Z . (cid:3) We remind that an integral domain R satisfies the ascending chain condition for principal ideals (ACCP)if there does not exist an infinite strictly ascending chain of principal (integral) ideals of R , see [1]. Corollary 2.13.
Let T be an integral domain satisfies ACCP. Then either T has a conch maximal subringor each subring of T satisfies ACCP. In particular, if T has no maximal subring, then each subring R of T is atomic and if ( a ) ⊃ ( a ) ⊃ · · · is an infinite strictly descending chain of principal ideals of R , then T ∞ i =1 ( a n ) = 0 . Consequently, for each a ∈ R \ U ( R ) we have T ∞ i =1 ( a n ) = 0 Proof.
Assume that T has no maximal subring, then by Theorem 2.4, we conclude that U ( R ) = R ∩ U ( T ).Therefore by [22, Proposition 2.1 and Corollary], R has ACCP. The final part is evident for it is well-known that domains with ACCP are atomic. (cid:3) We remind that an integral domain R is called a bounded factorization domain (BFD) if R is atomic andfor each nonzero nonunit of R there is a bound on the length of factorization into product of irreducibleelements, i.e., for each nonzero nonunit x of R , there exists a positive integer N ( x ) such that whenever x = x · · · x n as a product of irreducible elements of R , then n ≤ N ( x ). One can easily see that a BFDsatisfies ACCP, but the converse does not hold, see [1, Example 2.1]. Also a Noetherian domain or aKrull domain is a BFD, see [1, Proposition 2.2]. Corollary 2.14.
Let T be a BFD. Then either T has a maximal subring or each subring of T is a BFD.Proof. Assume that T has no maximal subring, then by Theorem 2.4, we conclude that U ( R ) = R ∩ U ( T ).Therefore by the comment preceding [1, Proposition 2.6], R is a BFD. (cid:3) Corollary 2.15.
Let R be a GCD-domain and T a ring extension of R . Assume that there exist a, b ∈ R such that gcd ( a, b ) = 1 , ab / ∈ U ( R ) and ab ∈ U ( T ) . Then T has a maximal subring. In particular, if T has no maximal subring, then U ( T ) ∩ K = U ( R ) , where K is the quotient field of R .Proof. Let x = ab . If T has no maximal subring, then x (resp. x − ) is integral over the prime subringof T and therefore x (resp. x − ) is integral over R . Thus x and x − are in R , for a GCD -domain isintegrally closed, i.e., b | a and a | b in R . Therefore a and b are units for gcd ( a, b ) = 1, which is absurd.Thus T has a maximal subring. The final part is evident. (cid:3) Corollary 2.16.
Let T be a ring such that the prime subring Z of T is integrally closed in T . Theneither T has a maximal subring or U ( T ) = U ( Z ) . In particular, if T has no maximal subring, then U ( T ) is finite.Proof. By Theorem 2.4, if T has no maximal subring, then U ( T ) is integral over Z and therefore U ( T ) ⊆ Z ,for Z is integrally closed in T . This immediately implies that U ( T ) = U ( Z ) and therefore U ( T ) isfinite. (cid:3) Proposition 2.17.
Let T be a ring and R := Z [ U ( T )] . The following are equivalent. (1) R has a proper subring which is integrally closed in R . (2) R has an integrally closed maximal subring. (3) T has a maximal subring A and there exists u − ∈ U ( T ) \ A with u ∈ A (i.e., T has a conchmaximal subring). (4) U ( T ) is not integral over Z .Proof. (1) = ⇒ (2) If S is a proper integrally closed subring of R , then clearly U ( T ) is not integral over S .Therefore U ( T ) = U ( R ) is not integral over Z . Thus by Theorem 2.4, R has an integrally closed maximalsubring. (2) = ⇒ (3) If A is an integrally closed maximal subring of R , then U ( R ) = U ( T ) is not integralover A . Therefore U ( T ) is not integral over Z and hence we are done by Theorem 2.4. (3) = ⇒ (4) U ( T ) isnot integral over A and therefore is not integral over Z . (4) = ⇒ (1) By Theorem 2.4, R has an integrallyclosed maximal subring (which is proper). (cid:3) Proposition 2.18.
Let T be a ring without maximal subrings and U ( T ) is finitely generated group. Then ONCH MAXIMAL SUBRINGS 7 (1) If Char ( T ) = n , then Z n [ U ( T )] is finite. In particular, U ( T ) is finite. (2) If Char ( T ) = 0 , then Z [ U ( T )] is a finitely generated Z -module. (3) If Char ( T ) = 0 and T is a residually finite ring, then T = Z .Proof. First note that if U ( T ) = < u , . . . , u n > , then by our assumption and Theorem 2.4, for each i , u − i is integral over Z and therefore u − i ∈ Z [ u i ] ⊆ Z [ u , . . . , u n ]. Therefore Z [ U ( T )] = Z [ u , . . . , u n ] isintegral over Z . Thus Z [ U ( T )] is finitely generated Z -module. In particular, if Char ( R ) = n > Z = Z n ), then Z [ U ( T )] is finite. Hence (1) and (2) hold. For (3), By [4, Theorem 2.29], T = Z [ U ( T )].Therefore by the first part of the proof T = Z [ u , . . . , u n ]. Hence if U ( T ) = U ( Z ), then T is finitelygenerated as a ring over Z and thus T has a maximal subring which is absurd. Therefore U ( T ) = U ( Z )and thus T = Z , hence we are done. (cid:3) A ring R is called clean, if each nonzero element of R is a sum of a unit and an idempotent of R . Nowthe following is in order. Proposition 2.19.
Let R be a clean ring. Then either R has a maximal subring or R has nonzerocharacteristic and is integral over Z .Proof. Assume that R has no maximal subring. Then by Corollary 2.5, U ( R ) is integral over Z . Alsonote that clearly each idempotent of R is integral over Z . Therefore by our assumption each element of R is a sum of two integral element over Z . Hence R is integral over Z . Finally, if Char ( R ) = 0, then dim ( R ) = 1 for R is integral over Z . Let P be prime ideal of R which is not maximal, then R/P isa local integral domain (note, in clean ring each prime ideal is contained in a unique maximal ideal).Therefore by [9, Corollary 2.24],
R/P and therefore R has a maximal subring which is absurd. Hence R has a nonzero characteristic and hence we are done. (cid:3) Example . (1) There exists a ring R which has a maximal subring, R is integral over Z andmoreover U ( R ) is a finitely generated. To see this, let K be a finite field extension of Q ofdegree n ≥ R be the integral closure of Z in K . Then it is well-known that R is a free Z -module with rank n and clearly R (and therefore U ( R ) are integral over Z ) and by Dirichlet’sunit theorem U ( R ) is finitely generated. Finally note that since R = Z is a finitely generatedalgebra, R has a maximal subring (clearly R has no conch maximal subring).(2) There exists a ring R which has a maximal subring, R is not integral (algebraic) over its primesubring and moreover U ( S ) = S ∩ U ( R ) for each subring S of R . To see this, let p be a primenumber and R = Z p [ X ] be the polynomial ring over Z p . Then clearly R has a maximal subring,and R is not algebraic over Z p . Finally note that for each subring S of R , Z p ⊆ S and therefore U ( S ) = U ( R ).(3) For each infinite cardinal number α , there exists a ring R with | R | = | U ( R ) | = α and R has nomaximal subring. In particular, if α > ℵ , then U ( R ) is not finitely generated. To see this notethat if K is a field with zero characteristic and | K | = α , then the idealization R = Z (+) K hasno maximal subring by [8, Example 3.19] (for K has no maximal Z -submodule). It is clear that1(+) K ⊆ U ( R ) and therefore | U ( R ) | = α = | R | .(4) There exists a conch subring of a ring T which is not a maximal subring of T . To see this, let V be a valuation domain with dim ( V ) = n ≥ K . Suppose P and Q are primeideal of V with ht ( Q ) = n and ht ( P ) = n −
1. Let x ∈ Q \ P . Then V conches x − in K but V ( V P ( K , i.e., V is not a maximal subring of K . Proposition 2.21.
Let T be a ring with | T | > ℵ . Let R be the integral closure of Z in T and X be a generator set for the group U ( T ) . Then either T has a maximal subring or | T | = | R | = | X | . Inparticular, if S is an integrally closed subring of T , then U ( T ) = U ( S ) and | S | = | T | .Proof. First note that by [8, Corollary 2.4], either T has a maximal subring or | U ( T ) | = | T | . Now assumethat T has no maximal subring, therefore by Corollary 2.5, U ( T ) is integral over Z . Hence U ( T ) ⊆ R and therefore | R | = | T | . Next, we show | X | = | T | . If X is finite, then U ( T ) is finitely generated andtherefore U ( T ) is countable which is absurd. Hence X is infinite. Similar to the proof of Proposition2.18, Z [ U ( T )] = Z [ X ]. One can easily see that | Z [ X ] | = | X | and therefore | U ( T ) | = | X | . The final partis evident. (cid:3) In the next result we proved that whenever R is a local integral domain which is an integrally closedmaximal subring of a ring T , then R is a conch subring of T . Proposition 2.22.
Let R be an integral domain which is an integrally closed maximal of a ring T . Then ALBORZ AZARANG (1) If R is a local ring, then R is a conch subring of T . (2) If M is the crucial maximal of R ⊆ T , then R M is a conch subring of T M .Proof. First note that by [27, Theorem 10], T is an integral domain. Let y ∈ T \ R . Then by maximalityof R we have R [ y ] = T . Now by a similar proof [19, Lemma 19.14], we conclude that y − ∈ R andtherefore R conch y in T . This proves (1). (2) is trivial by (1) and (4) of Theorem 1.1. (cid:3) In [9, Corollaries 3.6 and 3.7], we prove that if T is a reduced ring with J ( T ) = 0 or | T | > ℵ , then T has a maximal subring. Now the following is in order. Theorem 2.23.
Let T be a reduced ring with | T | = 2 ℵ . Then either T has a maximal subring or | M ax ( T ) | = 2 ℵ and the intersection of each countable family of maximal ideals of T is nonzero.Proof. Assume that T has no maximal subring, then by [9, Corollary 3.6], J ( T ) = 0. Therefore T can beembedded in Q M ∈ Max ( T ) TM . Since T has no maximal subring, then by [7, Propossition 2.6], we concludethat | M ax ( T ) | ≤ ℵ and by [7, Corollary 1.3] for each maximal ideal M of T we have | R/M | ≤ ℵ . Nowif | M ax ( T ) | < ℵ , then we deduce that | T | ≤ | Y M ∈ Max ( T ) TM | ≤ ℵ ℵ which is absurd. Thus we infer that | M ax ( T ) | = 2 ℵ . The proof of the final part is similar. (cid:3) Now we have the following.
Theorem 2.24.
Let T be an integral domain with | T | = 2 ℵ and dim ( T ) = 1 . Then T has a maximalsubring.Proof. First note that if
Char ( T ) = p is a prime number, then T is not algebraic over Z p ; thus thereexists x ∈ T which is not algebraic over the prime subring of T . Now if T has zero characteristic, then wetake D = Z and if Char ( T ) = p > p is a prime number), we take D = Z p [ x ]. In any cases D isa PID with infinitely many non-associate prime elements. Let Irr ( D ) = { q , q , . . . } . We have two cases.If there exists i such that q i ∈ U ( T ), then by Theorem 2.1, T has a maximal subring. Hence supposethat for each i , q i is not invertible in T and therefore T has a maximal ideal M i such that q i ∈ M i . Nowby Theorem 2.23, N := T ∞ i =1 M i = 0. We claim that N ∩ D = 0. To see this, let q ∈ N ∩ D , then foreach i we have q ∈ M i ∩ D = q i D and therefore q = 0. Thus N ∩ X = ∅ , where X := D \ { } is amultiplicatively closed subset in T . Therefore T has a prime ideal P with N ⊆ P and P ∩ X = ∅ . Since dim ( T ) = 1, we conclude that P is a maximal ideal of T . From D ∩ P = 0 we deduce that the field T /P contains a copy of D . Therefore T /P is not absolutely algebraic field and hence by Corollary 2.3,
T /P has a maximal subring. Thus T has a maximal subring and we are done. (cid:3) Proposition 2.25.
Let T be an integral domain with | T | = 2 ℵ and dim ( T ) = 2 . Let H be the set ofall height one prime ideals of T . Then either T has a maximal subring or T P ∈ H P = 0 , | H | ≥ ℵ and | R/P | ≤ ℵ for each P ∈ H .Proof. Assume that T has no maximal subring. Therefore by [9, Corollary 2.24], J ( T ) = 0. Since dim ( T )is finite we conclude that each maximal ideal of T contains a height one prime ideal and clearly eachhight one prime ideal is contained in a maximal ideal of T . Thus T P ∈ H P ⊆ J ( T ) = 0. Hence T embedsin Q P ∈ H R/P . Therefore, if | H | < ℵ , then there exists P ∈ H such that | R/P | ≥ ℵ . Now either R/P is a field or dim ( R/P ) = 1. In the former case
R/P has a maximal subring by Corollary 2.3 and inthe later case
R/P has a maximal subring by Theorem 2.24, which is a contradiction in any cases. Thus | H | ≥ ℵ . (cid:3) Proposition 2.26.
Let K be a field, R a ring extension of K , x ∈ R \ ( U ( R ) ∪ Zd ( R )) . If R = K + Rx ,then R has a maximal subring.Proof. First we show that K + Rx is a proper subring of R . It is clear that K + Rx is a subring of R . Now if R = K + Rx , then there exist a ∈ K and r ∈ R such that x = a + rx . If a = 0, then weconclude that x (1 − rx ) = a ∈ U ( R ) which is absurd. Hence a = 0 and therefore x = rx , and since x is not a zero-divisor we deduce that 1 = rx which is impossible by our assumption. Thus K + Rx isa proper subring of R . Now note that R = K + Rx = K + ( K + Rx ) x = K + Kx + Rx . Therefore( K + Rx )[ x ] = R . Thus R has a maximal subring. (cid:3) ONCH MAXIMAL SUBRINGS 9
Proposition 2.27.
Let R be an uncountable PID, then any ring extension of R has a maximal subring.Proof. Let T be a ring extension of R . If U ( R ) is uncountable then we are done by Theorem 2.4, forin this case U ( R ) is not algebraic over Z and therefore U ( T ) is not integral over Z . Hence assume that U ( R ) is countable. Thus by the proof of [4, Theorem 3.1], we infer that R has a prime elements q suchthat U ( R/ ( q )) is not algebraic over the prime subring of R/ ( q ). Now we have two cases. If qT = T , thenby Theorem 2.4, T has a maximal subring. Otherwise qT is a proper ideal of T and clearly qT ∩ R = qR ,for qR is a maximal ideal of R . Thus T /qT contains a copy of
R/qR . Hence U ( T /qT ) contains a copyof U ( R/qR ). Therefore U ( T /qT ) is not integral over the prime subring of
T /qT . Therefore
T /qT has amaximal subring by Theorem 2.4. Thus T has a maximal subring. (cid:3) Proposition 2.28.
Let R be an atomic (or a completely integrally closed) domain with | R | = 2 ℵ . If M is a principal maximal ideal of R , then | R/M | = | R | and therefore R has a maximal subring.Proof. If M = 0, then R is a field and therefore we are done by Corrollary 2.3. Thus assume that M = ( p ) is a nonzero principal maximal ideal of R . Hence p is a prime element of R . Similar to the proofof Corollary 2.2, we conclude that T ∞ n =1 ( p n ) = 0. Therefore R can be embedded in Q ∞ n =1 R/ ( p n ). Thusthere exists n such that | R/ ( p n ) | = 2 ℵ . Hence by [8, Lemma 2.8], | R/ ( p ) | = 2 ℵ and therefore R/ ( p )has a maximal subring, by Corollary 2.3. Thus R has a maximal subring. (cid:3) Conductor of Integrally Closed Maximal Subring
In this section we are interested to show that if K is an algebraic closed field, which is not algebraic overits prime subfield, and R is affine ring over K , then for each prime ideal P of R with ht ( P ) ≥ dim ( R ) − S of R which is integrally closed in R and ( S : R ) = P .First we begin by the following lemma for arbitrary ring R . Lemma 3.1.
Let R be a maximal subring of a ring T with ( R : T ) ∈ Spec ( T ) \ M ax ( T ) . Then R isintegrally closed in T .Proof. If T is integral over R , then by [20, Theorem 2.8], P := ( R : T ) satisfies exactly one of thefollowing:(1) P is a maximal ideal of T .(2) There exists a maximal ideal M of T such that M ⊆ P ⊆ M . Therefore M = P , for P is prime.(3) There exist distinct maximal ideals M and M of T such that P = M ∩ M . Thus either P = M or P = M , for P is prime.Hence in any case we conclude that P is a maximal ideal of T , which is impossible. Thus R is integrallyclosed in T . (cid:3) We remind that if V is an integral domain with quotient field K = V , then one can easily see that thefollowing are equivalent:(1) V is a maximal subring of K (2) V is a one dimensional valuation domain.(3) V is a real (archimedean) valuation domain.(4) V is a completely integrally closed valuation domain.Now the following is in order. Lemma 3.2.
Let K be a field and T be a ring extension of K . If V is a maximal subring of T which isintegrally closed in T and K * T , then V ∩ K is a maximal subring of K . In particular, V ∩ K is onedimensional valuation domain.Proof. First note that since V is integrally closed in T , then for each u ∈ U ( T ) either u ∈ V or u − ∈ V ,by (3) of Theorem 1.1. Since K * V , we infer that K ∩ V is a proper subring of K . Now for maximalityof K ∩ V in K , it suffices to show that ( K ∩ V )[ α ] = K , for each α ∈ K \ ( K ∩ V ). By the first part of theproof note that α − ∈ V . Since V is a maximal subring of T we conclude that V [ α ] = T . Now suppose β ∈ K , thus β ∈ T = V [ α ], which implies that β = v + v α + · · · + v n α n for some v i ∈ V . Therefore βα − n = v α − n + · · · + v n = v ∈ V , for α − ∈ V . Finally note that since K is a field and α, β ∈ K wehave βα − n = v ∈ K ∩ V and therefore β = vα n ∈ ( K ∩ V )[ α ]. Hence K ∩ V is a non field maximalsubring of K and therefore is a one dimensional valuation domain. (cid:3) Let K be a field, X , . . . , X n are indeterminates over K , T = K [ X , . . . , X n ] and Q ∈ Spec ( T ). Nowa natural and stronger question arises from the existence of integrally closed maximal subrings is asfollows: Does there exists an integrally closed maximal subring of T with conductor Q ? If n = 1 and K is absolutely algebraic field (i.e., K is algebraic over Z p for some prime p ), then K [ X ] has no integrallyclosed maximal subring, by [5, Lemma 4.6] (the only integrally closed subrings of K [ X ] are K and K [ X ]).Therefore the answer to the question is not positive in general. But we show that if K is algebraicallyclosed field which is not algebraic over its prime subring and n − ≤ ht ( Q ) ≤ n , then T has an integrallyclosed maximal subring with conductor Q . We need some observation. Let k be a field, then the Hahnfield (or the field of generalized formal power series) over k with exponents in Q (for general definitionsee [16, Section 2.8]) is denoted by k [[ t Q ]], is the set of all formal power series α = P s ∈ Q α s t s with α s ∈ k and supp ( α ); = { s ∈ Q | α s = 0 } is a well-ordered subset of Q . It is clear that if t is an indeterminate over k , then k ⊆ k [ t ] ⊆ k [[ t Q ]] (also k [[ t ]] ⊆ k (( t )) ⊆ k [[ t Q ]]). A natural valuation on k [[ t Q ]] onto Q , which send α to min ( supp ( α )) is an archimedean valuation and hence its valuation ring V is a maximal subring of k [[ t Q ]]. It is well-known that if k is algebraically closed, then k [[ t Q ]] is algebraically closed and thereforeit contains a copy of algebraic closure of k ( t ), say K . Therefore by Corollary 3.2, K ∩ V is a maximalsubring of K . In [26, Proposition 5.17], the authors characterized exactly maximal subrings of K [ X ]which contain ( K ∩ V )[ X ]. These maximal subrings are integrally closed in K [ X ] and the conductor ofthem is either 0 or is of the form ( X − a ) K [ X ] for some a ∈ K (note K is algebraically closed). Thereforein the later case these maximal subrings are of the form ( V ∩ K ) + ( X − a ) K [ X ]. Now we have thefollowing immediate corollary. Corollary 3.3.
Let K be an algebraically closed field which is not absolutely algebraic. Then K [ X ] hasan integrally closed maximal subring with zero conductor.Proof. By the above observation, it suffices to show that there exists an algebraically closed field k andan indeterminate t over k such that K is equal to the algebraic closure of k ( t ) in k [[ t Q ]]. Let F be theprime subfield of K and S be a transcendence basis for K over F . Since K is not algebraic over F weinfer that S = ∅ . Let t ∈ S and S ′ = S \ { t } , then clearly E = F ( S ′ ) is contained in K , t is not algebraicover E and K is algebraic over E ( t ). Now let k be the algebraic closure of E in K , then it is obvious that t is not algebraic over k , k is algebraically closed and K is algebraic over k ( t ). Thus K is the algebraicclosure of k ( t ). Therefore the algebraic closure of k ( t ) in k [[ t Q ]], say L is isomorphic (as field) to K . Thusby the above observation L [ X ] has an integrally closed maximal subring with zero conductor and hencethe same is true for K [ X ]. (cid:3) We need the following result which is a lying-over property of conductors of integrally closed maximalsubrings in integral extensions.
Theorem 3.4.
Let R ⊆ T be an integral extension of rings, Q ∈ Spec ( T ) and P = Q ∩ R . Assume that R has an integrally closed maximal subring S with ( S : R ) = P and U ( R/P ) * S/P . Then T has anintegrally closed maximal subring V with ( V : T ) = Q .Proof. It is clear that A := R/P ⊆ B := T /Q is an integral extension and C := S/P is an integrallyclosed maximal subring of A with ( C : A ) = 0 and U ( A ) * C . Hence by (3) of Theorem 1.1, assume that α ∈ U ( A ) such that α ∈ C but α − / ∈ C . Thus C [ α − ] = A , by maximality of C . Let D be the integralclosure of C in B , then α − / ∈ D and one can easily see that D [ α − ] = B (similar to the proof of Lemma3.2). Therefore B has a maximal subring E which contains D but α − / ∈ E (see [8, Proposition 2.1]).Clearly, E is integrally closed in B and E ∩ A = C . Let V be a subring of T such that E = V /Q and Q = ( V : T ). Thus P ⊆ Q ∩ R , for Q ⊆ Q . Now let x ∈ Q ∩ R , then x + Q ∈ ( V /Q ) ∩ ( R/P ) =
S/P ,i.e., x ∈ S . Therefore Q ∩ R ⊆ S . Thus P ⊆ Q ∩ R ⊆ ( S : R ) = P which immediately implies that Q ∩ R = P . Now since R ⊆ T is an integral extension (and therefore INC holds) and Q ⊆ Q are primesideals of T with a same contraction in R , we conclude that Q = Q . Therefore ( V : T ) = Q and we aredone. (cid:3) The following is the main result in this section.
Theorem 3.5.
Let K be an algebraically closed field which is not absolutely algebraic and X , . . . , X n be indeterminates over K . Then for each prime ideal Q of T = K [ X , . . . , X n ] with n − ≤ ht ( Q ) ≤ n ,there exists an integrally closed maximal subring S of R with ( S : R ) = Q .Proof. We have two cases. First assume that n = 1. If Q = 0 then we are done by Corollary 3.3, hencesuppose that Q is a maximal ideal of T and therefore T /Q ∼ = K . By Corollary 2.3, T /Q has an integrally
ONCH MAXIMAL SUBRINGS 11 closed maximal subring
V /Q and (
V /Q : T /Q ) = 0. Thus V is an integrally closed maximal subringof T which contains Q and therefore Q = ( V : T ), by maximality of Q . Now suppose that n ≥
2. If ht ( Q ) = n , then Q is a maximal ideal of T and similar to the case n = 1 we are done. Hence assume that ht ( Q ) = n −
1. Thus dim ( TP ) = 1. Therefore by Noether’s Normalization Theorem (see [15, TheoremA1 P.221 or Theorem 13.3]), there exist Y , . . . , Y n in T such that T is integral over R := K [ Y , . . . , Y n ]and P := Q ∩ R = ( Y , . . . , Y n ). Thus T /Q is integral over K [ Y ]. Now by Corollary 3.3, K [ Y ] has anintegrally closed maximal subring with zero conductor which does not contain K (see [5, Lemma 4.6]).Hence T /Q has an integrally closed maximal subring with zero conductor by Theorem 3.4. Thus T hasan integrally closed maximal subring with conductor Q . (cid:3) Corollary 3.6.
Let K be an algebraically closed field which is not absolutely algebraic and R be an affinering over K . Then for each prime ideal Q of R with ht ( Q ) ≥ dim ( R ) − , there exists an integrally closedmaximal subring S of R with ( S : R ) = Q . Proposition 3.7.
Let K be a field which is not absolutely algebraic. The following are equivalent. (1) For each n ≥ , the ring K [ X , . . . , X n ] has an integrally closed maximal subring S with zeroconductor and K * S . (2) For each n ≥ and each prime ideal Q of R = K [ X , . . . , X n ] , there exists an integrally closedmaximal subring S with ( S : R ) = Q and K * S . (3) Each affine ring over K has an integrally closed maximal subring S with zero conductor and K * S .Proof. It suffices to prove (1) = ⇒ (2). Let Q be a nonzero prime ideal of R , and ht ( Q ) = m . Hence1 ≤ m ≤ n . If m = n , then Q is a maximal ideal of R , therefore the field R/Q contains a copy of K . Thus R/Q is not absolutely algebraic field. Therefore by Corollary 2.3,
R/Q has an integrally closed maximalsubring. Hence R has an integrally closed maximal subring S which contains Q . Thus Q ⊆ ( S : R )and therefore ( S : R ) = Q , for Q is a maximal ideal of R . Hence assume that 1 ≤ m ≤ n −
1. Thus d := dim ( RQ ) = n − m ≥
1. Therefore by Noether’s Normalization Theorem (see [15, Theorem A1P.221 or Theorem 13.3]), there exist Y , . . . , Y n in R such that R is integral over T := K [ Y , . . . , Y n ] and P := Q ∩ T = ( Y d +1 , . . . , Y n ). Thus B := R/Q is integral over A := K [ Y , . . . , Y d ]. Now by (1), A hasan integrally closed maximal subring S with zero conductor and K * S . Thus by Theorem 3.4, B hasan integrally closed maximal subring V with zero conductor K * V . Hence R has an integrally closedmaximal subring with conductor Q , and we are done. (cid:3) Let T be a ring, then we denote the set of all integrally closed maximal subrings of T by X i.c ( T ) andalso define Spec ( X i.c ( T )) := { ( S : T ) | S ∈ X i.c ( T ) } . Note that by (3) of Theorem 1.1, Spec ( X i.c ( T )) ⊆ Spec ( T ). Therefore P ∈ Spec ( X i.c ( R )) if and only if T has an integrally closed maximal subring S with( S : T ) = P . Now we have the following. Theorem 3.8.
Let R ⊆ T be an integral extension of rings. If M ax ( T ) ⊆ Spec ( X i.c ( T )) , then M ax ( R ) ⊆ Spec ( X i.c ( R )) .Proof. Let P be a maximal ideal of R , then there exists a maximal ideal Q of T such that Q ∩ R = P .Therefore R/P ⊆ T /Q is an integral extension of fields. Now by our assumption there exists an integrallyclosed maximal subring V of T with ( V : T ) = Q . Hence V /Q is an integrally closed maximal subringof
T /Q . Therefore
R/P * V /Q . Thus by Lemma 3.2,
V /Q ∩ R/P is an integrally closed maximalsubring of
R/P . Hence R has an integrally closed maximal subring W which contains P and therefore( W : R ) = P . (cid:3) Proposition 3.9.
Let R be an integrally closed maximal subring of an integral domain T with U ( T ) * R .If R is noetherian, then ( R : T ) = 0 . In particular, R M is a DVR, where M is the crucial maximal idealof the extension R ⊆ T .Proof. Assume that t − ∈ U ( T ) \ R . Therefore t ∈ R and R [ t ] = T . Now if P = ( R : T ) = 0, then tP = P for t ∈ U ( T ). Therefore t − P = P . This immediately implies that t − is integral over R whichis absurd. The final part is evident by the fact that R is noetherian and (5) of Theorem 1.1. (cid:3) Remark . Note that one can prove the first part of the previous proposition by Krull IntersectionTheorem and the fact that ( R : T ) = T ∞ n =1 Rt n . Corollary 3.11.
Let R be a noetherian one dimensional integral domain. Then for each = a ∈ R \ U ( R ) ,the overring R a = R [ a ] has a maximal subring with zero conductor. In particular, if X is a multiplicativelyclosed set of R which is not contained in U ( R ) , then R X has a maximal subring with zero conductor. Proof.
Let T = R [ a ], then clearly T has a maximal subring S which contains R and a / ∈ S . Therefore S a = T . Now note that by Krull-Akizuki Theorem ([24, Theorem 93]), S is noetherian and thereforeby Krull Intersection Theorem we conclude that ( S : T ) = T ∞ n =1 Sa n = 0. The final part is similar andfollows from Corollary 2.9. (cid:3) Proposition 3.12.
Let R be a normal noetherian integral domain which is an integrally closed maximalsubring of a ring T with crucial maximal ideal M . Then ( R : T ) = 0 and R M is a DVR.Proof. First note that by [27, Theorem 10], T is an integral domain and therefore T is a minimal overrringof R . Hence by [3, Theorem 2.4 and Lemma 2.8], there exists an ideal A of R such that ( R : T ) = T ∞ n =1 A n .Thus by Krull intersection theorem we infer that ( R : T ) = 0 and therefore by (5) of Theorem 1.1, R M is a valuation domain. Since R is noetherian, we conclude that R M is a DVR. (cid:3) Proposition 3.13.
Let R be a completely integrally closed integral domain which is a conch maximalsubring of a ring T . Then ( R : T ) = 0 .Proof. First note that there exists a ∈ R such that T = R [ a ], for R is a conch maximal subring. Nowone can easily see that ( R : T ) = T ∞ n =1 Ra n . Thus by [19, Corollary 13.4], we infer that ( R : T ) = 0 for R is completely integrally closed. (cid:3) Lemma 3.14.
Let R be an integral domain with = ( p ) ∈ Spec ( R ) . Then the following hold. (1) R is a maximal subring of R [ p ] with zero conductor if and only if T ∞ n =1 ( p n ) = 0 and ( p ) ∈ M ax ( R ) . (2) if R is atomic, then ( R : R [ p ]) = 0 . Moreover, R is a maximal subring of R [ p ] if and only if ( p ) ∈ M ax ( R ) .Proof. Assume that T = R [ p ], then one can easily see that ( R : T ) = T ∞ n =1 Rp n . Hence ( R : T ) = 0 ifand only if T ∞ n =1 Rp n = 0. For (1), first suppose that T ∞ n =1 Rp n = 0 and ( p ) ∈ M ax ( R ), then we prove R is a maximal subring of T . Let A be a subring of T which properly contains R . Let x ∈ A \ R , thus wemay assume that x = rp n , where r ∈ R , n ≥ a / ∈ Rp , for T ∞ n =1 Rp n = 0 and x ∈ A ⊆ T but x / ∈ R .Hence rp ∈ A . Since M = ( p ) is a maximal ideal of R , we conclude that ap + br = 1, for some a, b ∈ R .Therefore p = a + b rp ∈ A , and hence T ⊆ A . Thus R is a maximal subring of T . Conversely, assumethat R is a maximal subring of T . Clearly R conches p in T . Therefore if M is the crucial maximal idealof the extension of R ⊆ T , then M is the unique prime ideal of R which contains p , by Corollary 1.2.Therefore M = ( p ). For (2) note that by the proof of Corollary 2.2, T ∞ n =1 Rp n = 0 and therefore we aredone by part (1). (cid:3) Proposition 3.15.
Let R ⊆ T be an integral extension of integral domain with p ∈ R is prime elementin T and pR = R ∩ pT . If R is a maximal subring of R [ p ] with zero conductor, then T is a maximalsubring of T [ p ] with zero conductor.Proof. Since R is a maximal subring of R := R [ p ] with zero conductor, we conclude that 0 = ( R : R ) = T ∞ n =1 Rp n . Let T = T [ p ], we claim that Q := ( T : T ) = T ∞ n =1 T p n = 0. If Q = 0, then Q ∩ R = 0 for T is an integral domain which is integral over R . Let 0 = x ∈ Q ∩ R , then for each n ≥
1, there exists t n ∈ T such that x = p n t n . Thus xp n = t n ∈ R [ p ] is integral over R . Since R is integrally closed in R we deduce that t n ∈ R and therefore x ∈ T ∞ n =1 Rp n = 0, which is absurd. Thus Q = 0 and hence we aredone by Lemma 3.14. (cid:3) Remark . Theorem 3.5 raises a natural question for the dual concept of maximal subrings, i.e.,minimal ring extensions, as follows. Let K be a field and R = K [ X , . . . , X n ], where X , . . . , X n areindependent indeterminate over K . Assume that P ∈ Spec ( R ) is arbitrary prime ideal. Now the naturalquestion arises: Does there exist a minimal ring extension T of R with ( R : T ) = P ? If P ∈ M ax ( R )(for arbitrary ring R ), then the answer to this question is positive by [14, Corollary 2.5], and in this casenote that by (2) of Theorem 1.1, T is integral over R . In fact by [14, Corollary 2.5], for each maximalideal M of R , the idealization T := R (+) RM is a minimal ring extension of R with ( R : T ) = M . Nowsuppose P is a prime ideal of R = K [ X , . . . , X n ] which is not a maximal ideal of R . First note that byLemma 3.1, if there exists a minimal ring extension T of R with ( R : T ) = P , then R is integrally closedin T and ht ( P ) = n −
1, by (5) of Theorem 1.1. Also by [27, Theorem 10], T is an integral domain andtherefore T is an overring of R . Now we have two cases. If n = 1, then P = 0 and one can easily seethat for each irreducible polynomial p ( X ) ∈ R = K [ X ], the overring T := R [ p ( X ) ] is a minimal ring ONCH MAXIMAL SUBRINGS 13 extension of R with ( R : T ) = 0. But if n ≥
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Let R be an affine normal integral domain over a field K . Suppose R is not a field.Then R has a minimal overring if and only if tr.deg ( R/K ) = 1 .Proof.
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