A family of quotients of the Rees algebra
aa r X i v : . [ m a t h . A C ] M a r A family of quotients of the Rees algebra
V. Barucci ∗ , M. D’Anna † , F. Strazzanti ‡ Dedicated to Marco Fontana on occasion of his 65-th birthday
Abstract
A family of quotient rings of the Rees algebra associated to a com-mutative ring is studied. This family generalizes both the classicalconcept of idealization by Nagata and a more recent concept, theamalgamated duplication of a ring. It is shown that several propertiesof the rings of this family do not depend on the particular member.MSC: 20M14; 13H10; 13A30.
Introduction
Let R be a commutative ring and let M be an R -module; the idealization,also called trivial extension, is a classical construction introduced by Nagata(see [15, page 2], [11, Chapter VI, Section 25] and [8]) that produces a newring containing an ideal isomorphic to M . Recently, D’Anna and Fontanaintroduced the so-called amalgamated duplication (see [4], [2], studied also in,e.g., [5], [14] and [1]), that, starting with a ring R and an ideal I , produces anew ring that, if M = I , has many properties coinciding with the idealization(e.g., they have the same Krull dimension and if I is a canonical ideal of alocal Cohen-Macaulay ring R , both of them give a Gorenstein ring). On theother hand, while the idealization is never reduced, the duplication can be ∗ email : [email protected] † email : [email protected] ‡ email : [email protected] t + at + b ∈ R [ t ] and denotingwith R + the Rees algebra associated to the ring R with respect to the ideal I , i.e. R + = L n ≥ I n t n , we study the quotient ring R + / ( I ( t + at + b )),where ( I ( t + at + b )) is the contraction to R + of the ideal generated by t + at + b in R [ t ]. We denote such ring by R ( I ) a,b .In the first section we introduce the family of rings R ( I ) a,b , show thatidealization and duplication are particular cases of them (cf. Proposition1.4) and study several general properties such as Krull dimension, total ringof fractions, integral closure, Noetherianity and spectrum. In Section 2 weassume that R is local; in this case we prove that the rings R ( I ) a,b have thesame Hilbert function and that they are Cohen-Macaulay if and only if I is a maximal Cohen-Macaulay R -module. We conclude this section provingthat, if R is a Noetherian integral domain of positive dimension, there existinfinitely many choices of b such that the ring R ( I ) , − b is an integral domain.Finally in the last section we study the one-dimensional case. If R is local,Noetherian and I a regular ideal we find a formula for the CM type of R ( I ) a,b (cf. Theorem 3.2) and prove that it is Gorenstein if and only if I is a canonicalideal of R . Moreover, we show the connection of the numerical duplicationof a numerical semigroup (see [7]) with R ( I ) , − b , where R is a numericalsemigroup ring or an algebroid branch and b has odd valuation (see Theorems3.4 and 3.6). Let R be a commutative ring with unity and I a proper ideal of R ; let t bean indeterminate. The Rees algebra (also called Blow-up algebra) associatedto R and I is defined as the following graded subring of R [ t ]: R + = M n ≥ I n t n ⊆ R [ t ] . emma 1.1. Let f ( t ) ∈ R [ t ] be a monic polynomial of degree k . Then f ( t ) R [ t ] ∩ R + = { f ( t ) g ( t ); g ( t ) ∈ I k R + } Proof.
Observe first that I k R + = { P ni =0 b i t i ; b i ∈ I k + i } . It is trivial thateach element of the form f ( t ) g ( t ), with g ( t ) ∈ I k R + , is in f ( t ) R [ t ] ∩ R + .Conversely, if g ( t ) ∈ R [ t ] and if f ( t ) g ( t ) ∈ R + , we prove by induction onthe degree of g ( t ), that g ( t ) ∈ I k R + . If the degree of g ( t ) is zero, i.e. g ( t ) = r ∈ R , and if f ( t ) r ∈ R + , then the leading term of f ( t ) r is rt k and r ∈ I k ⊂ I k R + . The inductive step: suppose that the leading term of g ( t )is h n t n ; thus the leading term of f ( t ) g ( t ) is h n t k + n . If f ( t ) g ( t ) ∈ R + , then h n ∈ I k + n and so f ( t ) h n t n ∈ R + . It follows that, if f ( t ) g ( t ) ∈ R + , then f ( t ) g ( t ) − f ( t ) h n t n = f ( t )¯ g ( t ) ∈ R + , where deg(¯ g ( t )) < n = deg( g ( t )). Byinductive hypothesis ¯ g ( t ) ∈ I k R + , hence g ( t ) = ¯ g ( t ) + h n t n ∈ I k R + .We denote the ideal of the previous lemma by ( I k f ( t )). Lemma 1.2.
Let f ( t ) ∈ R [ t ] be a monic polynomial of degree k > . Theneach element of the factor ring R + / ( I k f ( t )) is represented by a unique poly-nomial of R + of degree < k .Proof. The euclidean division of an element g ( t ) of R + by the monic polyno-mial f ( t ) is always possible and gives g ( t ) = f ( t ) q ( t ) + r ( t ), with deg( r ( t )) The ring extensions R ⊆ R + / ( I k f ( t )) ⊆ R [ t ] / ( f ( t )) areboth integral and the three rings have the same Krull dimension.Proof. By the two lemmas above we have the two inclusions. Moreover, theclass of t in R [ t ] / ( f ( t )) is integral over R and over R + / ( I k f ( t )) as well. Itfollows that all the extensions are integral. By a well known theorem onintegral extensions, we get that the three rings have the same dimension.We observe now that, for particular choices of the polynomial f ( t ) above,we get known concepts. 3ecall that the Nagata’s idealization, or simply idealization, of R withrespect to an ideal I of R (that could be defined for any R -module M ) isdefined as the R -module R ⊕ I endowed with the multiplication ( r, i )( s, j ) =( rs, rj + si ) and it is denoted by R ⋉ I .The duplication of R with respect to I is defined as follows: R ✶ I = { ( r, r + i ) | r ∈ R, i ∈ I } ⊂ R × R ;note that R ✶ I ∼ = R ⊕ I endowed with the multiplication ( r, i )( s, j ) =( rs, rj + si + ij ). Proposition 1.4. We have the following isomorphisms of rings: R + / ( I t ) ∼ = R ⋉ I R + / ( I ( t − t )) ∼ = R ✶ I Proof. 1) For each residue class modulo ( I t ), let r + it ∈ R + , with r ∈ R and i ∈ I , be its unique representative; the map α : R + / ( I t ) → R ⋉ I defined setting α ( r + it + ( I t )) = ( r, i ) is an isomorphism of rings: as amatter of fact, α preserves sums and, if r, s ∈ R , i, j ∈ I , we have α (( r + it +( I t ))( s + jt + ( I t ))) = α ( rs + ( rj + si ) t + ijt + ( I t )) = α ( rs + ( rj + si ) t + ( I t )) = ( rs, rj + si ) = ( r, i )( s, j ).2) Similarly to 1), the map β : R + / ( I ( t − t )) → R ✶ I defined setting β ( r + it +( I ( t − t ))) = ( r, r + i ) is an isomorphism of rings. Asfor the product, we have β (( r + it +( I ( t − t )))( s + jt +( I ( t − t )))) = β ( rs +( rj + si ) t + ijt +( I ( t − t ))) = β ( rs +( rj + si + ij ) t + ij ( t − t )+( I ( t − t ))) = β ( rs +( rj + si + ij ) t +( I ( t − t ))) = ( rs, rs + rj + si + ij ) = ( r, r + i )( s, s + j ).The previous proposition makes natural to consider the family R ( I ) a,b = R + / ( I ( t + at + b )), where a, b ∈ R . As R -module R ( I ) a,b ∼ = R ⊕ I andthe natural injection R ֒ → R ( I ) a,b is a ring homomorphism; however 0 ⊕ I in general (if b = 0) is not an ideal of R ( I ) a,b , although this happens foridealization and duplication.Both idealization and duplication can be realized in other cases.4 roposition 1.5. If t + at + b = ( t − α ) , with α ∈ R , then R ( I ) a,b ∼ = R ⋉ I . If t + at + b = ( t − α )( t − β ) , with ( t − α ) and ( t − β ) comaximal idealsof R [ t ] , then R ( I ) a,b ∼ = R ✶ I .Proof. 1) It is enough to consider the automorphism of R [ t ], induced by t t − α .2) By Chinese Remainder Theorem, the map Φ : R [ t ] / ( t + at + b ) → R × R ,defined by Φ( r + st ) = ( r + αs, r + βs ) is an isomorphism of rings, as well asthe map Ψ : R [ t ] / ( t − t ) → R × R , defined by Ψ( r + st ) = ( r, r + s ), so thatΨ − ◦ Φ : R [ t ] / ( t + at + b ) → R [ t ] / ( t − t ) , where (Ψ − ◦ Φ)( r + st ) = ( r + αs ) + ( β − α ) st is also an isomorphism. Ifwe fix an ideal I of R and if we restrict Ψ − ◦ Φ to the subring R ( I ) a,b , i.e.to the elements r + it , with r ∈ R and i ∈ I , we get R ( I ) a,b ∼ = { ( r + αi ) + ( β − α ) it ; r ∈ R, i ∈ I } = { r ′ + ( β − α ) it ; r ′ ∈ R, i ∈ I } ;the last ring is R ✶ J , where J = ( β − α ) I . To finish the proof we show that β − α is invertible. In the authomorphism of R [ t ] induced by t → t + β , theideal ( t − α, t − β ) corresponds to ( t − α + β, t ) = ( β − α, t ) and this last idealis R [ t ] if and only if β − α is invertible. Example 1.6. Let R = Z and t + at + b = t − t + 6 = ( t − t − I of Z , Z ( I ) − , ∼ = Z ✶ I .In this paper we study the family of rings of the form R ( I ) a,b , showingthat many relevant properties are independent by the member of the family.From now on, we denote each element of R ( I ) a,b simply by r + it ( r ∈ R , i ∈ I ). Proposition 1.7. Let Q be the total ring of fractions of R ( I ) a,b . Then eachelement of Q is of the form r + itu , where u is a regular element of R .Proof. Assume that ( s + jt ) is a regular element of R ( I ) a,b and that( r + it ) / ( s + jt ) ∈ Q . Since ( s + jt ) is regular, then x ( s + jt ) = 0, for every x ∈ R \ { } . Hence, xj = 0 implies xs = 0.Consider, now, the element ( ja − s + jt ). To prove the Proposition, it isenough to show that: 5) the product u = ( s + jt )( ja − s + jt ) is a regular element of R ,ii) ( ja − s + jt ) is a regular element of R ( I ) a,b .In fact in this case we can write ( r + it ) / ( s + jt ) = ( r + it )( ja − s − jt ) /u .Observing that − at − t = b ∈ R , we have u = s ( ja − s ) − j b ∈ R .If x ( ja − s + jt ) = 0 (for some x ∈ R \ { } ), then xj = 0, that implies x ( ja − s + jt ) = xs = 0, a contradiction. Hence ( ja − s + jt ) is not killed byany non zero element of R ; it follows that u is regular in R , otherwise therewould exist x ∈ R \ { } such that ux = 0 that implies ( s + jt ) not regular in R ( I ) a,b , since it is killed by ( ja − s + jt ) x = 0. Thus i) is proved.ii): if ( ja − s + jt ) is not regular in R ( I ) a,b , there exists ( h + kt ) = 0such that ( ja − s + jt )( h + kt ) = 0. Hence u ( h + kt ) = ( s + jt )( ja − s + jt )( h + kt ) = 0, so u is not regular in R ( I ) a,b . But then it is not regular in R ; contradiction. Corollary 1.8. Assume that I is a regular ideal; then the rings R ( I ) a,b and R [ t ] / ( t + at + b ) have the same total ring of fractions and the same integralclosure.Proof. Each element of R [ t ] / ( t + at + b ), let’s say r + r t with r, r ∈ R ,is in Q , in fact if i is an element of I regular in R , i is also regular in R [ t ] / ( t + at + b ) and r + r t = ( ir + ir t ) /i ∈ Q . Moreover, if r + r t isregular in R [ t ] / ( t + at + b ), it is also regular in Q . In fact, according toProposition 1.7, an element of Q is of the form ( s + jt ) /u ( s ∈ R , j ∈ I , u ∈ R and regular); if ( r + r t )( s + jt ) /u = 0, then ( s + jt ) /u = 0. It followsthat, if ( r + r t ) / ( s + s t ) is an element of Q ′ , the total ring of fractions of R [ t ] / ( t + at + b ), then r + r t and s + s t belong to Q and s + s t is regularin Q , so ( r + r t ) / ( s + s t ) ∈ Q . On the other hand, if ( r + it ) /u ∈ Q ,with u ∈ R and regular in R , then u is also regular in R [ t ] / ( t + at + b ) and( r + it ) /u ∈ Q ′ .By Corollary 1.8, it follows that the integral closure of R ( I ) a,b contains R [ t ] / ( t + at + b ), where R is the integral closure of R , but it may be strictlylarger. For example, for R = Z and t + at + b = t + 4, we have that Z [ t ] / ( t + 4) is not integrally closed, in fact ( t/ + 1 = 0.Using the chain of inclusions R ⊆ R ( I ) a,b ⊆ R [ t ] / ( t + at + b ) and the factthat these extensions are integral, we can get information on Spec( R ( I ) a,b )with respect to Spec( R ). 6 roposition 1.9. For each prime ideal P of R , there are at most two primeideals of R ( I ) a,b lying over P . Moreover if t + at + b is irreducible on R/ m for any maximal ideal m of R , then there is exactly one prime ideal of R ( I ) a,b lying over P .Proof. Every prime ideal of R ( I ) a,b , lying over P has to be the contractionof a prime ideal of R [ t ] / ( t + at + b ). It is well known (see e.g. [9, Chapter6]) that for every prime ideal P of R , P [ t ] is a prime of R [ t ] lying over P and there exist infinitely many other primes in R [ t ], lying over P , all of themcontaining P [ t ] and with no inclusions among them. In particular, there is abijection between these ideals and the nonzero prime ideals of ( Q ( R/P ))[ t ](here Q ( R/P ) denotes the field of fractions of R/P ); therefore the image of allthese prime ideals J in ( Q ( R/P ))[ t ] is of the form ( f ( t )), for some irreduciblepolynomial f ( t ); hence J = ϕ − P (( f ( t ))), where ϕ P is the composition of thecanonical homomorphisms R [ t ] → ( R/P )[ t ] ֒ → ( Q ( R/P ))[ t ]. Thus the primeideals of R [ t ] / ( t + at + b ) lying over P are of the form J/ ( t + at + b ), with J ⊇ ( t + at + b ). This means that the polynomial f ( t ), corresponding to J , divides the image of t + at + b in Q ( R/P )[ t ]. Hence, if t + at + b isirreducible in Q ( R/P )[ t ], there is only one prime of R [ t ] / ( t + at + b ) lyingover P ; on the other hand, if t + at + b has two distinct irreducible factors in( Q ( R/P ))[ t ], there exist exactly two prime ideals in R [ t ] / ( t + at + b ) lyingover P . Hence there are at most two primes in R ( I ) a,b lying over P and thefirst part of the proposition is proved.Suppose that J/ ( t + at + b )) ∈ Spec( R [ t ] / ( t + at + b )) and J/ ( t + at + b )) ∩ R = P . We know that J = ϕ − P (( f ( t ))), where f ( t ) is an irreduciblefactor of t + at + b in Q ( R/P )[ t ]. If P ′ ∈ Spec( R ), P ′ ⊂ P , then the primeideals of R [ t ] / ( t + at + b ) lying over P ′ correspond to the irreducible factorsof t + at + b in Q ( R/P ′ )[ t ]; since the factorization of t + at + b in Q ( R/P ′ )[ t ]induces a factorization in Q ( R/P )[ t ], f ( t ) is irreducible also in Q ( R/P ′ )[ t ]and we have a prime ideal of R [ t ] / ( t + at + b ) lying over P ′ of the form J ′ / ( t + at + b ), with J ′ = ϕ − P ′ (( f ( t ))) ⊂ J . In particular, if m is a maximalideal of R containing P and t + at + b is irreducible on R/ m , then thereis one and only one prime ideal of R [ t ] / ( t + at + b ) lying over P and thesame happens for R ( I ) a,b because the extension R ( I ) a,b ⊆ R [ t ] / ( t + at + b )is integral. Remark 1.10. 1) Notice that, for particular a and b , the factorization of t + at + b in Q ( R/P )[ t ] may not depend on P . For example, in the case ofthe idealization, the equality t = t · t , implies that there is only one prime7ying over P , both in R [ t ] / ( t ) and in the idealization. As for the case of theduplication, the equality t − t = t · ( t − R [ t ] / ( t − t ) lying over P , namely ( P, t ) and ( P, t − P ⊇ I (see,e.g., [4]).2) By the proof of Proposition 1.9 we see that the extension R ⊆ R [ t ] / ( t + at + b ) and the extension R ⊆ R ( I ) a,b as well fulfill the going down property.In particular a minimal prime of R ( I ) a,b lies over a minimal prime P of R .3) The proof of the previous proposition also implies that a sufficientcondition for R ( I ) a,b to be an integral domain is that R is an integral domainand t + at + b is irreducible in Q ( R )[ t ]. We will see in the next sectionthat, under particular assumptions on R , we can prove the existence of suchpolynomials.We conclude this section characterizing the rings R ( I ) a,b which are Noethe-rian. Proposition 1.11. The following conditions are equivalent: (i) R is a Noetherian ring; (ii) R ( I ) a,b is a Noetherian ring for all a, b ∈ R ; (ii) R ( I ) a,b is a Noetherian ring for some a, b ∈ R .Proof. If R is Noetherian, also the Rees algebra R + is Noetherian; henceit is straightforward that R ( I ) a,b is Noetherian for every a, b ∈ R , being aquotient of a Noetherian ring.Since the condition (iii) is a particular case of (ii), we need to prove onlythat (iii) implies (i). Assume by contradiction that R is not a Noetherian ring;then there exists an ideal J = ( f , f , . . . ) of R that is not finitely generatedand we can assume that f i +1 / ∈ ( f , . . . f i ) for any i . Consider the ideal J R ( I ) a,b of R ( I ) a,b ; by hypothesis, it is finitely generated and its generatorscan be chosen from those of J (regarded as elements of R ( I ) a,b ). Hence we canassume that J R ( I ) a,b = ( f , . . . , f s ). This implies f s +1 = P sk =1 f k ( r k + i k t ),for some r k ∈ R and i k ∈ I , and therefore f s +1 = P sk =1 f k r k ; contradiction. Assume that R is local, with maximal ideal m . Then it is known that both R ✶ I and R ⋉ I are local with maximal ideals m ⊕ I (in the first case under8he isomorphism R ✶ I ∼ = R ⊕ I ). More generally: Proposition 2.1. R is local if and only if R ( I ) a,b is local. In this case themaximal ideal of R ( I ) a,b is m ⊕ I (as R -module).Proof. Let R be local; we claim that all the elements r + it with r / ∈ m are invertible in R ( I ) a,b . As a matter of fact, looking for s + jt such that( r + it )( s + jt ) = 1, we obtain the linear system ( rs − ibj = 1 is + ( r − ia ) j = 0which has determinant δ = r − iar + i b ∈ r + m . Thus δ is invertible in R ;moreover, it is easy to check that if ( s, j ) is the solution of the system, then j ∈ I ; hence s + jt ∈ R ( I ) a,b and it is the inverse of r + it .Conversely, if R ( I ) a,b is local, R has to be local, since R ⊆ R ( I ) a,b is anintegral extension (cf. Proposition 1.3).It is also clear that, if ( R, m ) is local and if we denote by M the maximalideal of R ( I ) a,b , then k = R/ m ∼ = R ( I ) a,b /M . In the sequel, we will alwaysdenote with k the common residue field of R and R ( I ) a,b . Remark 2.2. Since R ( I ) a,b is an R -algebra, every R ( I ) a,b -module N is alsoan R -module and then λ R ( I ) a,b ( N ) ≤ λ R ( N ) (where λ ( ) denote the lengthof a module).If we consider a R ( I ) a,b -module N annihilated by M , we have that, as R -module, N is annihilated by m . Hence it is naturally an R ( I ) a,b /M -vectorspace and an R/ m -vector space; in particular, λ R ( I ) a,b ( N ) = dim k ( N ) = λ R ( N ) (where k = R/ m ∼ = R ( I ) a,b /M ).For a Noetherian local ring ( R, m ), we denote by ν ( I ) the cardinalityof a minimal set of generators of the ideal I . The embedding dimensionof R , ν ( R ), is by definition ν ( m ) and the Hilbert function of R is H ( n ) = λ R ( m n / m n +1 ). Proposition 2.3. Let ( R, m ) be a Noetherian local ring. Then, for every a, b ∈ R the rings R ( I ) a,b have the same Hilbert function. In particular, theyhave the same embedding dimension ν ( R ( I ) a,b ) = ν ( R ) + ν ( I ) and the samemultiplicity. roof. First of all, let us consider M ; we have M = m + m It (and hence, as R -module, it is isomorphic to m ⊕ m I ): in fact, if ( r + it ) and ( s + jt ) are in M , then their product rs − bij +( rj + si − aij ) t ∈ m ⊕ m I . Conversely, pick anelement in m ⊕ m I of the form rs + uit (with r, s, u ∈ m and i ∈ I ); we have rs + uit = rs + u ( it ) ∈ M ; since m ⊕ m I is generated by elements of this formwe have the equality. Arguing similarly for any n ≥ 2, we immediately obtainthat M n = m n + m n − I and, as R -module, it is isomorphic to m n ⊕ m n − I .It follows that, as R -modules, M n /M n +1 ∼ = m n / m n +1 ⊕ I m n − /I m n . Bythe previous remark the length of M n /M n +1 as R ( I ) a,b -module coincides withits dimension as k -vector space and with its length as R -module. The thesisfollows immediately. Remark 2.4. Let R be a Noetherian local ring. By Propositions 1.3 and2.3 we getdim R ( I ) a,b = dim R ≤ ν ( R ) ≤ ν ( R ) + ν ( I ) = ν ( R ( I ) a,b ) . The first inequality is an equality if and only if R is regular and the secondif and only if ν ( I ) = 0, that is equivalent to I = 0, by Nakayama’s lemma.This means that R ( I ) a,b is regular if and only if R is regular and I = 0;clearly if I = 0 one has R ( I ) a,b = R .We want to show that, if R is a local Noetherian integral domain, we canalways find integral domains in the family of rings R ( I ) a,b . The followingproposition was proved in [6] and we publish it with the permission of thesecond author. Proposition 2.5. Let ( R, m ) be a local Noetherian integral domain with dim R ≥ and Q ( R ) its field of fractions. Then for any integer n > ,not multiple of , there exist infinitely many elements b ∈ R such that thepolynomial t n − b is irreducible over Q ( R ) .Proof. We will use the following well-known criterion of irreducibility: if b isnot a p -th power for any prime p | n and b 6∈ − Q ( R ) if 4 | n , then t n − b isirreducible (see [13, Chapter VI, Theorem 9.1]). In particular, if 4 does notdivide n and b is not a d -th power for any integer d > d | n , then t n − b is irreducible.Taking a prime ideal P ⊂ R such that ht P = 1 we have dim R P = 1 and,by the Krull-Akizuki Theorem, its integral closure R P of R P in Q ( R P ) = Q ( R ) is Noetherian (see, e.g. [12, Theorem 4.9.2]), hence it is a Dedekind10ing. So there is at least a discrete valuation v : Q ( R ) ∗ → Z with v (( R P ) M ) = N (with M maximal ideal of R P ). Since R ⊆ R P ⊆ ( R P ) M have the samefield of fractions, it follows that v ( R ) ⊆ N is a semigroup containing twoconsecutive integers; so there exists c > x ∈ N , x ≥ c belongs to v ( R ).In particular, there exist infinitely many elements b ∈ R such that v ( b )is prime to n , so b cannot be a d -th power in Q ( R ) for any d > d | n . Hence we can find infinitely many b ∈ R such that ( t n − b ) ⊂ Q ( R )[ t ] isirreducible. Corollary 2.6. Let R be a local Noetherian integral domain with dim R ≥ ,let Q ( R ) be its field of fractions and let I ⊂ R be an ideal. Then there existinfinitely many elements b ∈ R such that R ( I ) , − b is an integral domain.Proof. By Proposition 2.5 we can find b such that ( t − b ) is irreducible in Q ( R )[ t ]. The thesis now follows by point 3) of Remark 1.10.Now we want to investigate the Cohen-Macaulayness of R ( I ) a,b .Assume that R is a CM ring; we set dim R = depth R = d ; moreover,Ann( R ( I ) a,b ) = (0), hence the dimension of R ( I ) a,b as R -module (i.e., since R ( I ) a,b is a finite R -module, dim( R/ Ann( R ( I ) a,b ))) equals the Krull dimen-sion of R ( I ) a,b . We can assume that d ≥ 1, otherwise both R and R ( I ) a,b aretrivially CM.Given a regular sequence x = x , x , . . . , x d of the ring R , it is not difficultto check that it is an R ( I ) a,b -regular sequence if and only if its image in R ( I ) a,b is a regular sequence of R ( I ) a,b as a ring. Moreover, since x is asystem of parameters of R , then it is a system of parameters of R ( I ) a,b (since R ⊆ R ( I ) a,b is an integral extensions) and x is a system of parameters forthe R -module R ( I ) a,b . Hence, arguing exactly as in [2] we have that R ( I ) a,b is a CM ring if and only if it is a CM R -module.Since R ( I ) a,b ∼ = R ⊕ I as R -module, it follows that depth ( R ⊕ I ) =min { depth I, depth R } = depth I and therefore R ( I ) a,b is a CM R -module ifand only if I is a CM R -module of dimension d (that is if and only if I is amaximal CM R -module).Hence we can state the following: Proposition 2.7. Assume that R is a local CM ring of dimension d . Then R ( I ) a,b is CM if and only if I is a CM R -module of dimension d . In partic-ular, the Cohen-Macaulayness of R ( I ) a,b depends only on the ideal I . emark 2.8. We notice that if I is a canonical ideal of R , since R ( I ) a,b ∼ = R ⊕ I , we can apply a result of Eisenbud (stated and proved in [2]) to getthat R ( I ) a,b is Gorenstein for every a, b ∈ R . We will see that in the one-dimensional case we can determine the CM type of R ( I ) a,b and deduce thatit is a Gorenstein ring if and only if I is a canonical ideal. Remark 2.9. In [3, Corollary 5.8], under the assumption that the ring ( R, m )is a local CM ring with infinite residue field, it has been proved the followingformula about the multiplicity of the duplication: e ( R ✶ I ) = e ( R )+ λ R ( I/IJ )(where J is any minimal reduction of m ); in particular, if dim R = 1, then e ( R ✶ I ) = 2 e ( R ).By Proposition 2.3 we can state that, under the same assumptions, thesame formulas hold for the multiplicity of R ( I ) a,b , for every a, b ∈ R . Assume for all this section that ( R, m ) is a one-dimensional, Noetherian, andlocal ring and I a regular ideal; in this section we determine the CM type of R ( I ) a,b .Since I is regular, it is a maximal Cohen-Macaulay R -module and R isa CM ring; therefore R ( I ) a,b is also CM by Proposition 2.7. In this case thetype of R ( I ) a,b equals the length of ( R ( I ) a,b : M ) /R ( I ) a,b as R ( I ) a,b -module,where M is the maximal ideal of R ( I ) a,b ; so we start studying R ( I ) a,b : M . Lemma 3.1. For any a, b ∈ R , the R ( I ) a,b -module R ( I ) a,b : M is equal to (cid:26) rs + is t ; is ∈ I : m , rs ∈ ( I : I ) ∩ ( R : m ) (cid:27) Proof. Consider a generic element r/s + ( i/s ) t of Q ( R ( I ) a,b ), where r, s ∈ R, i ∈ I and s is regular (cf. Proposition 1.7). It is an element of R ( I ) a,b : M if and only if( r/s + ( i/s ) t )( m + jt ) = rm/s + ( im/s ) t + ( rj/s ) t + ( ij/s ) t = rm/s − ijb/s + ( im/s + rj/s − ija/s ) t is an element of R ( I ) a,b , for any m ∈ m and for any j ∈ I , that is ( rm/s − ijb/s ) ∈ R and ( im/s + rj/s − ija/s ) ∈ I .12uppose that r/s + ( i/s ) t ∈ R ( I ) a,b : M ; in particular, if j = 0 we have rm/s ∈ R and im/s ∈ I , that is r/s ∈ R : m and i/s ∈ I : m . Moreoversince ja ∈ I ⊆ m and i/s ∈ I : m , we have im/s, ija/s ∈ I , hence rj/s ∈ I for any j ∈ I and then r/s ∈ I : I .Conversely, suppose that i/s ∈ I : m and r/s ∈ ( I : I ) ∩ ( R : m ). Then rm/s − ijb/s ∈ R + I = R and im/s + rj/s − ija/s ∈ I + I + I = I ,consequently r/s + ( i/s ) t ∈ R ( I ) a,b : M . Theorem 3.2. The CM type of R ( I ) a,b is t ( R ( I ) a,b ) = λ R (cid:18) ( I : I ) ∩ ( R : m ) R (cid:19) + λ R (cid:18) I : m I (cid:19) ; in particular, it does not depend on a and b .Proof. Consider the homomorphism ϕ of R -modules R ( I ) a,b : M → ( I : I ) ∩ ( R : m ) R × I : m Irs + is t (cid:18) rs + R, is + I (cid:19) . Thanks to the previous lemma, ϕ is well defined and surjective; moreover,its kernel is given by the elements r/s + ( i/s ) t with r/s ∈ R and i/s ∈ I ,that is ker ϕ = R ( I ) a,b ; hence R ( I ) a,b : MR ( I ) a,b ∼ = ( I : I ) ∩ ( R : m ) R × I : m I . Consequently, using Remark 2.2, we have t ( R ( I ) a,b ) = λ R ( I ) a,b (cid:18) R ( I ) a,b : MR ( I ) a,b (cid:19) = λ R (cid:18) R ( I ) a,b : MR ( I ) a,b (cid:19) == λ R (cid:18) ( I : I ) ∩ ( R : m ) R × I : m I (cid:19) == λ R (cid:18) ( I : I ) ∩ ( R : m ) R (cid:19) + λ R (cid:18) I : m I (cid:19) . orollary 3.3. The ring R ( I ) a,b is Gorenstein if and only if I is a canonicalideal of R .Proof. Recall first that a ring is Gorenstein if and only if it has CM type 1.Recall also that I is a canonical ideal of a one-dimensional CM local ring R ,i.e. an ideal I such that I : ( I : J ) = J for each regular ideal J of R , if andonly if λ R (( I : m ) /I ) = 1 (cf. [10], Satz 3.3). Notice that, for any ideal I regular and proper, λ R (( I : m ) /I ) ≥ R ( I ) a,b is Gorenstein ifand only if t ( R ( I ) a,b ) = 1 = 0 + 1; hence, λ R (( I : m ) /I ) = 1, i.e. I is acanonical ideal. Conversely if I is a canonical ideal, then I : I = R and λ R (( I : m ) /I ) = 1; by the same formula we get t ( R ( I ) a,b ) = 0 + 1 = 1, i.e. R ( I ) a,b is Gorenstein.We conclude this section studying two particular cases of one dimensionalrings: numerical semigroup rings and algebroid branches; in both cases weshow the connection with the numerical duplication of a numerical semigroup(see [7]).Recall that a numerical semigroup is a submonoid of N with finite comple-ment in N and it can be expressed in terms of its minimal set of generators, S = h n , . . . , n ν i , with GCD( n , . . . , n ν ) = 1. A semigroup ideal E is a subsetof S such that S + E ⊆ E . We set 2 · E = { s | s ∈ E } . According to [7], thenumerical duplication of S with respect to a semigroup ideal E of S and anodd integer m ∈ S is the numerical semigroup S ✶ m E = 2 · S ∪ (2 · E + m ) . A numerical semigroup ring is a ring R of the form k [[ S ]] = k [[ X n , . . . , X n ν ]],where k is a field and X an indeterminate. Such a ring is a one-dimensional,Noetherian, local integral domain; moreover, it is analytically irreducible (i.e.its integral closure R is a DVR, which is a finite R -module) and in this case R = k [[ X ]], the ring of formal power series. The valuation v induced by k [[ X ]] on k [[ S ]] is given by the order of a formal power series and, if I is anideal of k [[ S ]], v ( I ) = { v ( i ); i ∈ I, i = 0 } is a semigroup ideal of S . Theorem 3.4. Let R = k [[ S ]] be a numerical semigroup ring, let b = X m ∈ R , with m odd, and let I be a proper ideal of R . Then R ( I ) , − b is isomorphicto the semigroup ring k [[ T ]] , where T = S ✶ m v ( I ) . roof. If S = h n , . . . , n ν i , an element of R ( I ) ,b is of the form r ( X ) + i ( X ) t where r ( X ) = r ( X n , . . . , X n ν ) ∈ k [[ S ]] and i ( X ) = i ( X n , . . . , X n ν ) ∈ I .Taking into account that we are factoring out the ideal ( I ( t − X m )), wecan easily check that the map Φ : R ( I ) , − b → k [[ T ]], defined byΦ( r ( X ) + i ( X ) t ) = r ( X ) + i ( X ) X m , is an isomorphism of rings. Example 3.5. If R = k [[ X , X ]], b = X and I = X k [[ X , X ]], then R ( I ) ,b ∼ = k [[ X , X , X ]]. According to Corollary 3.3, we get a Gorensteinring (in fact the semigroup h , , i is symmetric), because the ideal I is acanonical ideal of R .We consider now the case of algebroid branches, i.e. local rings ( R, m ) ofthe form k [[ X , . . . X n ]] /P , where P is a prime ideal of height n − k isalgebraically closed. We have that ( R, m ) is a one-dimensional, Noetherian,complete, local integral domain; moreover, R is analytically irreducible withintegral closure isomorphic to k [[ X ]] and k ⊂ R . If we consider the valuation v induced by k [[ X ]] on R , we get again that v ( R ) = { v ( r ); r ∈ R, r = 0 } isa numerical semigroup and that v ( I ) = { v ( i ); i ∈ I, i = 0 } is a semigroupideal of v ( R ). Theorem 3.6. Let R be an algebroid branch and let I be a proper ideal of R ; let b ∈ R , such that m = v ( b ) is odd. Then R ( I ) , − b an algebroid branchand its value semigroup is v ( R ) ✶ m v ( I ) .Proof. Since v ( b ) is odd, by Proposition 2.5, t − b is irreducible in Q ( R )[ t ]and R ( I ) , − b is an integral domain. Moreover, applying the results of theprevious sections, we know that R ( I ) , − b is local (we will denote by M itsmaximal ideal), Noetherian and one-dimensional. It is not difficult to checkthat the m -adic topology on the R -module R ( I ) , − b coincide with the M -adictopology, hence it is complete. Since R ( I ) , − b contains its residue field k , byCohen structure theorem, it is of the form k [[ Y , . . . , Y l ]] /Q , for some primeideal Q of height l − 1; so it is an algebroid branch.Let V = k [[ Y ]] be the integral closure of R ( I ) , − b in its quotient field Q ( R ( I ) , − b ) = Q ( R )( t ) = k (( Y )). We denote by v ′ the valuation associated15o k [[ Y ]]; in particular v ′ ( Y ) = 1. Since Q ( R ) = k (( X )), we have k (( Y )) = k (( X ))( t ); moreover, t = b implies that 2 v ′ ( t ) = v ′ ( b ) = mv ′ ( X ). In orderto obtain v ′ ( Y ) = 1 it is necessary that v ′ ( t ) = m and v ′ ( X ) = 2.Now, it is straightforward that v ′ ( R ( I ) , − b ) = v ( R ) ✶ m v ( I ). References [1] A. Bagheri, M. Salimi, E. Tavasoli, S. Yassemi, A construction ofquasi-Gorenstein rings , J. Algebra Appl. (2012), no. 1. 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