On the Complete Integral Closure of the Rees Algebra

8/3/2019 On the Complete Integral Closure of the Rees Algebra

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ON THE COMPLETE INTEGRAL CLOSUREOF THE REES ALGEBRA

STEFANIA GABELLI AND ANNA GUERRIERI

1. Introduction and background.

In this work we analyze the complete integral closure of graded domains con-structed using and manipulating some specific ideal filtrations in a domain R.Among the objects that encode information on the asymptotic behavior of theideal J, the Rees algebra of J,

is the chief example of blowup algebra, where by this name one refers to thosealgebraic structures that are loosely related, when R is Noetherian, to the conceptof blowing up a variety along a subvariety (see [26]). A wealth of work has beenproduced on the integral closure of R(J) in its field of quotients in a Noetheriansetting. Much less is known on the complete integral closure of R(J) when R isnot necessarily Noetherian; the difficulties naturally arising because of the erraticbehavior of the complete integral closure.

Let R be a an integral domain and K the quotient field of R. A nonzero elementx of K is said to be almost integral on R if there is a nonzero element d of R suchthat dx" E R, for i?: O. This is equivalent to saying that all the powers of x belongto a finite R-module, [6, Chap. V, 1, n.4], or that they generate a fractionary idealof R, [7, Lemma 3.1]. The complete integral closure of R in K, denoted R , is theset of the elements of K almost integral over R and one says that R is completelyintegrally closed if R = R. Every element integral over R is almost integral overR and the converse holds in the Noetherian case. Thus, if R' denotes the integralclosure of R in K, one has R ~ R' ~ Rand R' = R in the Noetherian case. It iswell known that R is always integrally closed, see [25, page 172].

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The complete integral closure however is not a proper closure operation. Manyexamples have been given to show that the complete integral closure of a domainmight not be completely integrally closed itself, see [9] and [10]. In [23], Sheldonprovides an example of a two-dimensional Priifer domain whose complete integralclosure is not a completely integrally closed domain. In [14], Hill constructs anexample where iterating the process of taking the complete integral closure doesnot lead to a completely integrally closed domain in a finite number of steps.

Among valuation rings, the only completely integrally closed rings are the rank-one valuation rings, see [6, Chap. VI, 4, Proposition 9] and [19]. Any intersectionof completely integrally closed domains with the same quotient field is completelyintegrally closed. Hence, Krull domains, which are locally finite intersection ofdiscrete valuation rings of rank one (DVR) in their quotient field, are a class ofcompletely integrally closed domains that are not necessarily Noetherian, see [6,Chap. VII, 3, Definition 3]. Intersections of rank-one valuation rings, however, donot exhaust the class of completely integrally closed domains as proved in [23]. It isfurther evidence of anomalous behavior the fact that a localization of a completelyintegrally closed domain is not necessarily completely integrally closed; an exampleis given by the ring of entire functions over the complex field [11].

In this work we examine the structure of the complete integral closure of R(J)and we give sufficient and necessary conditions on the ring R and the ideal J forR(J) being completely integrally closed or Krull. As in the Noetherian case, itturns out that the complete integral closure of R(J) is deeply connected with thestructure of other blowup algebras like the extended Rees algebra and the associatedgraded ring. Namely

R(J)e = R(J)[Cl] = tBnEzrtn ~ R[t, Cll,g(J) = tBn~o(In / In+l) ~ R(J)/ JR(J) ~ R(J)e /ClR(J)e,

where In = R for all n : S O. Barshay proves in [5, Theorem 5] that if J is an idealsuch that g(J) is reduced, then R(J) is integrally closed in R[t]. In particular, ifR is integrally closed and g(J) is reduced, then R(J) is integrally closed. In thisway, when R is Noetherian, Barshay's result gives a very simple sufficient conditionto guarantee the complete integral closure of R(J). When R is not Noetherian thesituation is more complicated and it is not difficult to give an example of a domain


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and an ideal such that the associated graded ring is reduced, R(J) is integrallyclosed but not completely integrally closed (see Example 3.8).

After having studied in Section 2 the structure of the complete integral closureof a graded domain, in Section 3 we prove that if (;(J) is a domain, then R(J) iscompletely integrally closed (Krull) if and only if R is completely integrally closed(Krull) and n n ~ O In = 0; see Theorem 3.6 and Corollary 3.16. Further, if JR(J)can be written as an irredundant finite intersection of strongly primary ideals (e.g.(;(J) reduced with finitely many minimal primes) and PI, ... ,Pn are the associatedprimes of JR(J), then R(J) is completely integrally closed (Krull) if and only if Ris completely integrally closed (Krull) and R(J)Pi is completely integrally closed(Krull) for each i= 1,... ,n , if and only if R is completely integrally closed (Krull)and R(J)Pi is a DVR for each i= 1, ... ,n; see Theorem 3.13. Besides, the conditionthat JR(J) is as an irredundant finite intersection of strongly primary ideals is alsonecessary for R(J) being a Krull domain; see Corollary 3.15.

In Section 4 we concentrate on the Rees algebra of a prime ideal p of R. Weprove that if (; (pRp) is a domain, then the complete integral closure of R(p) iscontained in the symbolic algebra of p; see Proposition 4.1. When R p is a DVRand (;(p) is reduced with only finitely many minimal primes, we obtain that (;(p)is a domain if and only if R(p) is completely integrally closed and the prime idealq = Q n R has height at most one, for any height-one prime ideal Q of R(p); seeTheorem 4.6.

We thank William Heinzer for the E-mail conversations regarding this topic.

2. The complete integral closure of a graded domain

In this section we examine the general structure of the complete integral closureof graded domains and in particular of the Rees algebra of an ideal.

Let A = EB,EI'A, be a domain graded by a torsionless additive monoid randlet Q(A) be its quotient field. We denote by H = U , E I ' A, the set of homogeneouselements of A. Then H is a multiplicative set of A and AH is graded by the group< r >. That is AH = EB aE< I '> (AH)an where


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A H is called the homogeneous quotient field of A , even though in general it is nota field. It is clear that A H is contained in Q(A) and that (AH)O is a field containingthe quotient field of A o. We say that an element of Q(A) is homogeneous of degreea if it is contained in (A H )a.It has been proved in [3, Proposition 2.1] that, regardless to A , A H is always

completely integrally closed.Next result guarantees that the complete integral closure of a graded structure

is still graded. The same assertion can be found in [27, Proposition 4.17], we givethe proof for completeness.

Proof: Since A H is completely integrally closed, then A ~ A H . Whence anyelement x E A is a sum of finitely many homogeneous elements Xaj of Q(A). Wehave to show that Xaj E A for each j. Let d = d'l +...+ d,t E A be such thatdx" E A for each n ?: O . It is then clear that d,l (xa1)n E A and so Xa1 E A . Hencex - Xa1 E A and, by induction, Xaj E A for each j. D

The analogous statement for the integral closure is in [28, Vol II, Theorem 11].Next result, a direct consequence of Lemma 2.1, appears also in [3, Proposition 5.2].

Lemma 2.2: If A = EB,EI'A, is a graded domain, the following are equivalent:(1) A is completely integrally closed in Q(A);(2) A is completely integrally closed in AH;(3) A o is completely integrally closed in (AH)O and each homogeneous element

of Q(A) of nonzero degree which is almost integral over A is already in A.

Observe that, if x is an homogeneous element of Q(A) that is almost integralover A , then there is a nonzero homogeneous element h of A such that lux" E A forn?: O .

Now assume that r= N, the natural integers. In this case, let y be any nonzeroelement of Q(A) of degree 1. If x E (AH)z, z E Z, then x/yZ E (AH)o. Hence(AH)z = (AH)OYz and A H = (AH)o[y,y-l] . Therefore Q(A) = (AH)O(Y). More-over y is trascendental over the field (AH)O [28, Vol. II, page 157]. In particular,


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5if A is either R [tJ, R [t, t-l], R(J) or R(J)e, the homogeneous quotient field of A isK[t, e] and the quotient field of A is K(t), where K is the quotient field of R .

Next results make use of the following observation: A domain R is completelyintegrally closed if and only if the polynomial ring R[XI, ... ,Xn] is completelyintergrally closed, see [6, Chap. V, 4, Proposition 14] and [9, Corollary 3].

Proposition 2.3: Let R be a dom ain w ith quotien t field K , then the follow inga re e qu iv ale n t:

(1 ) R is com pletely in tegrally closed in K ;( 2 ) R[t] is com pletely in tegrally closed in K (t);( 3 ) R[t , e] is com pletely in tegrally closed in K (t).

Proof: The equivalence of (1) and (2) is proved in [6, Chap. V, 4, Proposition 14].We now prove that (1) implies (3). By Lemma 2.2, we just need to show that

each homogeneous element of K(t) which is almost integral over R [t, t-l] is alreadyin R [t, t-l]. Let (a/b)tC be an element of K(t) almost integral over R [t, t-l]. Thusthere is a nonzero element dt " E R[t , e] such that d(a/b)nte+n c E R[t , e] for alln ?: O. Therefore d(a/b)n E R for all n ?: O. This concludes the argument since nowwe have that alb is almost integral over R and therefore it is in R by hypothesis.

To prove that (3) implies (1) just observe that R = R[t , el] nK. D

Proposition 2.4: R(J) ~ R[t] .Proof: Let K be the quotient field of R . Since K[t] is completely integrally closed,then R(ij ~ K[t]. Let (a/b)tm E K[t] be almost integral over R(J) and let dtk ER(J), d i- 0, be such that dtk((a/b)tm)n = d(a/b)ntk+ mn E R(J) for each n ?: O.Then d(a/b)n E R , that is (a/b) E R . Hence R(J) ~ R[t] . D

Recall that an integral domain R is called a K rull dom ain if there exists afamily {lI~hEI of discrete valuations on the quotient field of R , Q (R ), such thatthe intersection of the rings of the 1I~ is R and, for all nonzero x E Q(R) , the set ofindices i E J such that 1I~(x ) i- 0 is finite.It is well known that if F is a field and {Rih:O:::;i:o:::;ns a finite family of subrings


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of F which are Krull domains, then n u; is still a Krull domain; see [6].l:Si:Sn

Proposition 2.5: If R(I) is completely integrally closed (respectively a Krulldomain) then R is completely integrally closed (respectively a Krull domain).Proof: It is enough to observe that R = R(I) n K. DProposition 2.6: The following conditions are equivalent:

(1) R(I) is completely integrally closed in K(t);( 2 ) R(I)e is completely integrally closed in K(t);( 3 ) R is completely integrally closed in K and R(I) is completely integrally

closed in R[t].

Proof: Proposition 2.4 and Proposition 2.5 guarantee that (1) implies (3).To prove that (3) implies (2) observe that, by Proposition 2.2, we just need

to consider the homogeneous elements. Let (alb)tC E K(t) be an element almostintegral on R(I)e. There is di" E R(I)e, d E R - {O}, such that d(alb)nte+cn ER(I)e for n ?: O. In particular alb is almost integral over R and therefore itbelongs to R. Let alb = a E R. Thus (alb)tC = oi" E R[t, ell. If e : S 0, thenat" E R(I)e. If e > 0, then at" E R[t] and if we show that at" is almost integralover R(I) we are done by hypothesis. Let n' be the least nonnegative integer forwhich f = e + en' > O. Let E = idtf with iany nonzero element in t=', Thus E isan element of R(I). Since dan E t+=. we get ido" E Icn'+e+cn = u+ for alln ?: O. Thus E(atC)n = idtf antnC is an element of R(I) for all n as wanted.

Eventually, (2) implies (1) since R(I) = R(I)e n K[t] is the intersection withinK(t) of two completely integrally closed domains. D

3. The complete integral closure of the Rees algebra

In this section we give necessary and sufficient conditions for the Rees algebraof an ideal to be completely integrally closed or Krull. Because of Proposition 2.6we start by studying when the Rees algebra is completely integrally closed in R[t].As in the Noetherian case, the complete integral closure of R(I) depends on theproperties of the associated graded ring (;(1).


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The remark below gathers some well known properties of (;(1).

Remark 3.1: For any ideal I the order function Vt : R --+ Z~o U{oo] is definedas v[(a) = sup {n E Z I a E In}. Here v[(a) = 00 if and only if a belongs to In for alln ?: O . It is well known that v[(a + b) ?: min {v[(a), v[(b)}, with equality holdingif v[(a) i- v[(b), and v[(ab) ?: v[(a) + v[(b). Furthermore, v[(ab) = v[(a) + v[(b)for all the elements a and b if and only if (; (I ) is a domain. In this case, the idealI and the ideal Po = n n ~ O In are prime ideals. Thus R/ Po is a domain and theorder function t/t induces a discrete valuation on R / Po centered in 1/ P o. It followsthat, if (;(1) is a domain and n n ~ O In = (0), then Vt is a discrete valuation on Rcentered in I. The condition v[(ad) = dv[(a) for all a E A, a > 0 is equivalent tothe weaker condition that (;(1) is reduced.

Given a E In - In+1, denote by a* the image of a in (;(1), in other words, theclass of a in In /In+l. This is called the leading form of a. If a belongs to all thepowers of I, set a* = O .

Remark 3.2: The complete integral closure of R(I) in R[t] is linked to the natureof the zero divisors of (;(1). If there are an homogeneous zero divisor a* E (;(1) andan homogeneous element i* E (0 :(}(I) a*), such that (iaj)*a* = 0 for all j?: 1, thenR(I) is not completely integrally closed in R[t]. In fact, assuming a E Im_Im+l andiE Il_Il+1, we have (itl)(atm+l)j = iajtmj+l+j. One can proceed by induction onj to prove that iaj is in Imj+l+j for all j?: 1. Let j= 1. By hypothesis 1*a* = 0,thus ia E Im+l+1 as wanted. Let j> 1 and assume iaj E r=+. This means thatiaj E I" - Ir+1 with r ?: mj + I + j. By hypothesis, (iaj + Ir+1 )(a + Im+l) = 0,thus iaj+1 E Ir+m+l ~ Imj+l+j+m+l = Im(j+l)+l+(j+l) as wanted. In conclusionatm+1 is an element of R[t] - R(I) which is completely integral on R(I).

Proposition 3.3: If R(I) is completely integrally closed in R[tJ, then one hasn n ~ O In = O .Proof: If it were not so, there would be a nonzero element d E In for all n ?: O .This would imply that dti E Iiti E R(I) for all i?: 0 while t ~ R(I). ConsequentlyR(I) would not be completely integrally closed in R[t]. D


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Remark 3.4: IfR is a domain, I ~ R an ideal and p a prime ideal containing Isuch that Rp is a Noetherian ring, then the intersection of all the powers of I iszero. In fact if x belongs to In for all n ?: 0, the image of x vanishes after localizingat p. It follows that x itself must be zero. Conditions that force the intersection ofall the powers of an ideal to be equal to zero in the general case have been studiedby D. D. Anderson, [1].

Proposition 3.5: If (;(1) is a domain and I is an ideal of R with n n ~ O In = 0,then R(I) is completely integrally closed in R[t].Proof: If it were not so, there would be a nonzero element atm E R[t] - R(I) andan element itl i- 0 in R(I) such that

for all j ?: O . Being a different from zero, there would be an integer n such thata E In - In+1 with n < m since atm ~ R(I). Being itl E R(I) different from zerotoo, iE Il+h - Il+h+1 for some h ?: O . Moreover lIJ(ah+1) = (h+ l)lIJ(a) = (h+ l)nsince (;(1) is a domain. One thus would have iah+1 E Il+m(h+l) with m ?: n + 1.Having Il+m(h+l) ~ Il+(n+l)(h+l) = Il+n(h+l)+h+l, the product

would be equal to zero in (; (I ) contradicting the hypothesis that (; (I ) is a do-main. D

Theorem 3.6: If (;(1) is a domain, then R(I) is completely integrally closed ifand only if R is completely integrally closed and n n ~ O In = O.Proof: Use Proposition 2.6 and Proposition 3.3. D

Remark 3.7: A fractional divisorial ideal of a domain R is an ideal of type(R:K J) for some fractional ideal J of R. Hence a fractional ideal I of R is divisorialif and only if I = (R:K (R:K I)). If R is completely integrally closed and p is adivisorial prime, then ti; is a DVR [8, Theorem 2.1], thus nn~opn = O . It followsthat, if p is a divisorial prime of a completely integrally closed domain Rand (;(p)


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is a domain, then R(p) is completely integrally closed. Observe that Rp a DVRforces R(pRp) to be completely integrally closed as well.

The condition that (;(J) be a domain does not force the intersection of all thepowers of J to be equal to zero, as it is shown in the following example.

Example 3.8: Let R be a valutation domain of dimension one that is not aDVR and let m be its maximal ideal. R is a completely integrally closed domainwith m = m 2. Thus nn~O m " = m i- O . Since R(m) = R + tm[t] = R + m[t] andm R ( m ) = m[t] , the associated graded ring of m is

(;(m) ~ R(m) /mR(m) = R(m)/m[t ] = Rim.Note that R(m) is integrally closed but not completely integrally closed. In fact itscomplete integral closure is R[t ] .

It is simple to provide an example of an ideal J in a completely integrally closeddomain R for which R(J) is completely integrally closed and (;(J) is not evenreduced. Consider, for example, a principal ideal J = a R in a completely integrallyclosed domain R . Since R(J) = R[a t ] is a polynomial ring over R , it is completelyintegrally closed, however (;(J) is not a domain if J is not prime and it is not reducedif a = b2

In what follows we study the case when the associated graded ring is not a do-main. To do so we are led to impose conditions on the structure of the extendedideal JR(J). In particular we study the case in which JR(J) has a primary decom-position.

Definition 3.9: The prime ideals of a domain R maximal among the ideals of zero-divisors modulo an ideal J are called the maximal primes of J. Max(J) denotes theset of maximal primes of J.

Ifa E R - { O } , as in the proof of [17, Theorem 53], one has

R=Ra nPEMax (aR)


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Proposition 3.10: If R is completely integrally closed (respectively a Krull do-main) and R(I)p is completely integrally closed (respectively a Krull domain) foreach maximal prime P of IR(I), then R(I) is completely integrally closed (respec-tively a Krull domain).Proof: By Proposition 2.6 it is enough to show that R(I)e is completely integrallyclosed. Since R(I)jIR(I) ~ R(I)e jc1R(I)e, P is a maximal prime of IR(I)if and only if Q = (P, t-1) is a maximal prime of t-lR(I)e. Observe also thatR(I)p ~ R(I)'Q. Since

R(I)e = R(I)~_l n {R(I)'Q},QEMax (t-ln(I)e)

by Proposition 2.3, if R is completely integrally closed (respectively Krull), thenR(I)~_l = R[t, t-1] is completely integrally closed. Thus, if R(I)p is completelyintegrally closed for all P, then R(I)e is completely integrally closed being theintersection of completely integrally closed domains with the same quotient field.The same proof shows that if R is a Krull domain and R( I) p is Krull for eachmaximal prime P of IR(I), then R(I)e is Krull and consequently R(I) = R(I)e nR[t] is Krull as well. D

Definition 3.11: An ideal of a ring R is said to be strongly primary if it is primaryand contains a power of its radical.

Next lemma is a direct consequence of well known results. We give the proof forcompleteness.

Lemma 3.12: Let R be a domain and I an ideal such that I = Ll n..nLn is anirredundant finite intersection of strongly primary ideals. Then v'Li = Pi = (I:R xdfor some Xi E R - Pi and the the maximal primes of I are the ideals which aremaximal among the Pi.Proof: By [4, Theorem 4.5], each Pi is the radical of (I:R x) for any x not in Li.Going modulo I one may assume 1=0. Thus Ann (x ) is contained in Pi for any xnot in Li, Since L, is strongly primary it contains a power of its radical, thus thereis an integer m for which pr is contained in Ls, Let Ii = n#Lj. Then we have


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JiPr = O . Let n be the least non-negative integer such that JiPr = 0 and let x bea nonzero element of t.tr>. Since PiX = 0, for such an x we have Pi ~ Ann (x),and hence Pi = Ann (x). Finally, since the set of zero-divisors modulo J is theunion of PI, ,Pn [4, Proposition 4.7], each ideal of zero-divisors is contained insome Pi. D

Observe that, with the notation of the previous Lemma, if J = aR is principal,then the ideals Pi are divisorial. In fact Pi = (aR:Rxd = (R:K (l,Xia-I)).

Theorem 3.13: Assume that JR(J) can be written as an irredundant finiteintersection of strongly primary ideals (e.g. (;(J) reduced with finitely many minimalprimes) and let PI, ... ,Pn be the associated primes of JR(J). Then the followingare equivalent:

(1 ) R(J) is completely integrally closed (respectively Krull);( 2 ) R is completely integrally closed (respectively Krull) and R(J)Pi is com-

pletely integrally closed (respectively Krull) for each i= 1, ... ,n;(3) R is completely integrally closed (respectively Krull) and R(J)Pi is a DVR

for each i= 1, ... ,n.

Proof: Assume that R(J) is completely integrally closed (respectively Krull). ThenR is completely integrally closed (respectively Krull) by Proposition 2.5. SinceR(J) / JR(J) ~ R(J)e /t-IR(J)e, the ideals Qi = (Pi, t-l) are the primes associatedwith the principal ideal cIR(J)e. Hence they are divisorial by Lemma 3.12. SinceR(J)e is completely integrally closed by Proposition 2.6, then R(J)'Qi ~ R(J)Pi isa DVR for each i= 1, ... ,n by [8, Theorem 2.1]. This proves that (1) implies (3).

That (3) implies (2) is clear.Eventually, that (2) implies (1) follows from Proposition 3.10. D

A simple consequence of the previous result and Theorem 3.6 is that, if (;(J) isa domain and R is completely integrally closed, then n n ~ O In = 0 if and only ifR(J)p is a DVR where P is the prime ideal JR(J).

Corollary 3.14: Let R be a completely integrally closed domain (respectivelyKrull), J an ideal of R such that (;(J) is reduced with only finitely many minimal


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primes, PI, ... ,Pn Let P i be the lift to R of the contraction of Pi to R/ I. If y(IpJis reduced and RPi is Noetherian for each i = 1, ... ,n, then R(I) is completelyintegrally closed (respectively Krull).

Proof: By [5], the hypothesis RPi integrally closed and y(IpJ reduced for each i=1, ... ,n, implies that R(IpJ is integrally closed for each i= 1,... ,n. Since R(I)piis isomorphic to a localization ofR(IpJ, which is Noetherian, then R(I) Pi is normal,hence completely integrally closed. The proof now follows from Theorem 3.13. D

Note that, if R(I) is Krull, the contractions to R of the associated primes ofIR(I) are exactly the associated primes of In for all n 0, see [24, Lemma 2.3].Corollary 3.15: Let R be a Krull domain, then the following are equivalent:

(1 ) R(I)e is Krull;( 2 ) R(I) is Krull;( 3 ) R(I) is completely integrally closed and IR(I) is an irredundant finite in-

tersection of strongly primary ideals;(4) IR(I) = pind n ... n ps(ns) and R(I)pi is a DVR for each i.

Proof: The fact that (4) implies (1) and (1) implies (2) follows by the proof ofProposition 3.10.

Since the ideal R(I)+ = tJJn?_IIntn is a height-one prime ideal of R(I), then,if R(I) is Krull, R(I)+ is divisorial. It follows that also the ideal IR(I) =tJJn?_IIn+ltn = CIR(I)+ is divisorial. Hence IR(I) is of type pind n .. . n pins)for some height-one prime ideals PI, ... , P; of R(I) and R(I)pi is a DVR for eachi= 1, ... ,s. This proves that (2) implies (4).

Theorem 3.13 proves that (4) implies (3).At last (3) implies (2) since by Theorem 3.13, if R(I) is completely integrally

closed and IR(I) is an irredundant finite intersection of strongly primary ideals,then R(I)pi is a DVR for each prime Pi associated to I. Again by Theorem 3.13,if R is Krull, then R(I) is Krull. D

In the Noetherian case the equivalence between (2) and (4) has been proved in[13, Proposition 2.13].


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Corollary 3.16: If g(I) is a domain, then R(I) is Krull if and only if R is Krulland n n ~ O In = o .Proof: It follows by Proposition 2.5, Theorem 3.6 and Corollary 3.15. D

Therefore, if R is the integral closure of a Noetherian domain and g (I ) is adomain, then R(I) is a Krull domain. In fact, in this case, for every ideal I, theintersection of all the powers of Iis zero, see [1, Theorem 5].

Remark 3.17: A Mori domain is a domain with the ascending chain conditionon divisorial ideals. It is well known that a Krull domain is a completely integrallyclosed Mori domain. Thus we are led to ask if an analogue of Corollary 3.15 holdsfor Mori domains. The difficulty lies in the fact that if R is Mori, the polynomialring R [ t ] need not be Mori [22]. However, assuming that R [ t ] is a Mori domain,by using the same argument given in the Krull case, we can conclude that R(I)is Mori if and only if IR(I) has finitely many maximal primes Pi and R(I)pi isMori for each i. In fact, like in the Krull case, a height-one prime ideal of a Moridomain is divisorial. Hence, if R(I) is Mori, IR(I) is divisorial and it has finitelymany maximal primes [15, Corollary 2.2]. Further, localizations of Mori domainsare Mori and finite intersections of Mori domains are Mori, see [20].

We do not know whether the fact that R(I) is Mori forces R [ t ] to be Mori.

4. The Rees algebra of a prime ideal

In this section we analyze the complete integral closure of the Rees algebra of aprime ideal p of R in relation with the structure of the associated graded ring andthe symbolic algebra of p. The symbolic algebra of p is the Rees algebra associatedto the filtration given by the symbolic powers of p , namely

S ( p ) = E B n ~ O p ( n ) .

Recall that the n-th symbolic power of a prime ideal p in a ring R is defined tobe the ideal p"Rp n R. If p" has a primary decomposition, p e n ) is its p-prirnarycomponent.


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Proposition 4.1: Let R be a completely in tegrally closed domain , p a prim eideal of R such that n n ~ O pen) = (0 ) {e.g. R p Noetherian} and g(pR p) is a d om ain .Then the com plete in tegral closure ofR (p) is con tain ed in S(p).

Proof: By Proposition 2.4, R(p) is contained in R[t] . Let at" be a nonzero homo-geneous element of R[t] almost integral over R(p) . We show that a E p(n) , thusshowing that at" belongs to S(p).

Since at" is almost integral over R(p) , there is a non zero element bt": E R(p)such that bai E pm+ni for all i ?: O . If n = 0, then bai E p m ~ R for alli ?: O . Since R is completely integrally closed it follows that a belongs to R.Assume now that n ?: 1. Let v be the order function of pR p Since a o f - 0 andn n ~ O p"R p = 0, we may assume that v(a/1) = I is finite. Since g(pRp) is a domain,we have that v((b/1)(a i /1)) = v(b/1) + iv(a/1) = v(b/1) + il . On the other hand,v( (b /1)( ai /1)) ?: m+in , because bai E pm+ni . Thus v(b/1) ?: m+i(n-l). It followsthat b/1 E p m+ (n -l)i R p for all i. If (n - I ) > 0, then b/1 E n i ~ O p n+ (n -l)i R p = 0,contradicting the fact that b o f - O . It follows that I ?: n and then a E p(l) ~ p(n ). D

It is well known that, under the same hypotheses, the integral closure of theRees algebra of p , Ra(p) , is contained in the symbolic algebra. In fact, if g(pRp) isa domain, the powers p"R p are integrally closed since they are valuation ideals ofthe p-adic valuation of R p.

Remark 4.2: Observe that if Rand R(pRp) are both completely integrally closed,the symbolic algebra of p over R is completely integrally closed since S(p) =R(pRp) n R[t] , and R[t] is completely integrally closed. (Recall that when R pis a DVR then R(pRp) is completely integrally closed). Similarly, if R is Krull,then S (p ) is Krull if and only if R(pRp) is completely integrally closed; see [18] or[24, Lemma 2.5] in the Noetherian case.

In conclusion, if R is completely integrally closed and R p is a DVR, we have

R (p) ~ R a(P) ~ R (p) ~ S(p) = S(p).

It is therefore natural to ask:


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Question 4.3: Let R be completely integrally closed and R p a DVR. Is R(p)completely integrally closed? Under which conditions is R(p) equal to S(p) ?

Recalling the results in [9], one may observe that R(p) = S(p) if the conductorof S(p) in R(p) is nonzero, i.e. if S(p) is contained in a finite R (p )-submodule ofthe quotient field, see [9, Lemma 5]. Again by [9, Corollary 6], R(p) is completelyintegrally closed if the conductor of R(p) in R(p) is nonzero.

Obvious, but restrictive, cases in which R(p) = S(p) are those when either R(p)or Ra(P) coincide with S(p).

Remark 4.4: IfS = R - p , let S* denote the multiplicatively closed set in g (p ]whose elements are the leading forms of elements in S. It is clear that R (pR p) ~R(p) Q9R ti; ~ R(p)s and that g(pR p) ~ g(p) Q9R u; ~ g(p)s*. The kernel of thenatural map from g(p) to g(pRp) is EBn~O((pn n p(n+l))/pn+l). Notice that theelements of this kernel are zero-divisors in g (p). Let

Each prime ideal Q of R(p) containing pR(p) such that Q n R = p , contains P.In fact Q n S = 0 and going modulo pR(p) one obtains a prime Q* of g(p) thatinverting S* remains prime. Since Q* is equal to the contraction of its extensionvia S*, it is obvious that the kernel of the canonical homomorphism in containedin Q*. Therefore P ~ Q.

Furthermore, if g(pRp) is a domain, then P is the only minimal prime idealcontaining pR(p) such that P n R = p. In fact, being the contraction of a prime, itis prime and any other prime of that type must contain it.

Consequently, if g(pRp) is a domain, R(p) = S(p) if and only if P = pR(p) ifand only if pR(p) is a prime ideal of R(p) if and only if g(p) is a domain; see [21,Corollary 1.2].

In general, even assuming R p a DVR, R(p) can be different from the symbolicalgebra as shown in the following example.

Example 4.5: Let K be a field and R the 2-dimensional normal domain R =K [lx, y, zl]/(x2 - yz). Consider the height-one prime ideal p = (x , y )R . Clearly


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R p is a DVR. Actually, p is an almost complete intersection and it is generatedby a d-sequence, see [16]. The symmetric and the Rees algebra of p are the sameand R(p) is isomorphic to R [U, V ]/(yU - xV, xU - zV ) while g(p) is isomorphicto K [z, U , V ]/(zV ) which is not a domain. Therefore R(p) does not coincide withS(p) since this happens if and only if g(p) is a domain; see Remark 4.4. HoweverR(p) is integrally closed (hence completely integrally closed since R is Noetherian)because g (p ) is reduced, see [5].

The last observation of Remark 4.4 can be improved when R p is a DVR.

Theorem 4.6: Let R be completely integrally closed and p a prime ideal of Rsuch that R p is a D VR and g(p) is reduced with finitely many minimal primes.Then the following are equivalent:(1) g(p) is a domain;(2 ) R(p) is completely integrally closed and, for any height-one prime ideal Q

of R(p), the prime ideal q = Q n R has height at most one;(3 ) R(p) is completely integrally closed and, for any associated prime of pR(p),

the prime ideal q = Q n R has height at most one.Proof: Since R p is a DVR, g(pRp) is a domain. Thus, by Remark 4.4, P =pR(pRp)nR(p) is the only minimal prime ideal containing pR(p) such that pn R =p. Since, by hypothesis, pR(p) has a primary decomposition given by finitely manyprimes which are minimal over it and P is formed by zero-divisors, we deduceby Lemma 3.12 that P appears in this decomposition. Therefore we may writepR(p) = P n Ql n ... n Qs where P is the only one with P n R = p. Hence,qi = Qi n R is prime in R of height at least 2, for each i= 1, ... ,s. This showsthat (1) and (3) are equivalent. In fact, if g (p ) is a domain we have pR(p) = Pand R(p) is completely integrally closed by Theorem 3.6. Viceversa, if (3) holds,clearly the Qi'S do not appear in the primary decomposition of pR(p). ThereforepR(p) = P and g(p) is a domain.

It is clear that (2) implies (3). To see that (3) implies (2), recall that, byTheorem 3.13, if R(p) is completely integrally closed, then all the primes idealsassociated to pR(p) have height one. Besides, if Q is a prime ideal of R(p) notcontaining pR(p) and q = Q n R i- 0, then q is a prime ideal of R not containing


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p and R (P )R -q ~ R (pR q) ~ R q[x]. Thus Q is the only prime contracting to q andthe ideals Q and q have the same height. D

Definition 4.7: One says that an extension A ~ B of Krull rings with the propertythat P n A has height at most one, for any height-one prime ideal Q of B, satisfiescondition PDE (Pas d'eclatement) or NBD (No blowing up), see [7].

Remark 4.8: Recall that the divisorial prime ideals of a Krull domain are exactlythe height-one primes (see Remark 3.7 for the definition of divisorial ideal). Denoteby C(A) the divisor class group of the Krull domain A, that is the quotient groupD(A)/P(A), where D(A) is the free abelian group generated by the height-oneprimes of A and P(A) is the group of principal ideals. Condition PDE implies thatthe map 'lj; : D(A) --+ D(B) defined by I f--t (B: (B: IB)) is a homomorphism; see[2, Proposition 5.1].If R(I) is Krull, C(R(I)) is generated by the classes of the height-one primes

containing some nonzero constant and the map i f ; : C(R) --+ C(R(I)) inducedby ' lj; is always injective with a left inverse induced by the contraction, because'lj;(I) n R = (R (I): (R (I): IR (I))) n R = I, for any divisorial ideal I of R ; see [2,Theorem 6.4].

Theorem 4.9: Let R be a Krull domain and p a divisorial prime of R such thatg(p) is reduced with only finitely many minimal primes. Then the following areequivalent:(1) g (p ) is a domain;(2) R(p ) is Krull and the extension R ~ R (p ) satisfies condition PDE;(3) R(p ) is Krull and the map i f ; : C(R) --+ C(R(p)) is an isomorphism.

Proof: (2) implies (1) by Theorem 4.6. Besides, if (1) holds, then R(p ) is Krull byCorollary 3.16 and the inclusion R ~ R (p ) satisfies condition PDE by Theorem 4.6once again.

To prove that (2) implies (3), it is enough to show that ' lj; is surjective. Noticethat, via (1), g(p) is a domain and then pR(p ) = P is prime. Since pR(p ) is divi-sorial, ' l j;(p) = ( R (p ) : ( R (p ) :p R ( p )) ) = pR(p) . On the other hand, any height-one


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prime Q of R(p) different from P and containing some nonzero constant contractsto a height-one prime q of R and is the only prime of R(p) contracting to q. SinceD(R(p)) is generated by the height-one primes, ' l j I(q) = (R (p ): (R (p ): q R(p ))) = Q.It follows that 'l jI is surjective.

Finally, (3) implies (1). In fact, if i f ; is bijective, then 'l jI is surjective and theremust be a divisorial ideal J of R such that 'ljI(J) = P = pR(pRp) n R(p). SinceJ = 'ljI(J) n R = P n R = p , then P = ' l jI(p) = pR(p) and pR(p) is prime. D

Among the works that, in the Noetherian case, have investigated the relationshipamong the class groups of R, R(I) and 5(1), we point out [12], [13] and [24].

The previous results motivates the following question.

Question 4.10: Let R be a completely integrally closed domain and let p be aprime ideal of R such that R p is a DV R and g(p) is reduced with only finitely manyminimal primes. Is R(p) completely integrally closed?

References

[1] D. D. Anderson, The Krull intersection Theorem, Pacific J. Math. 57(1975), 11-14.[2] D. F. Anderson, Graded Krull domains, Comm. Alg. 7 (1979), 79-106.[3] D. D. Anderson and D. F. Anderson, Divisibility properties of graded do-

mains, Can. J. Math. 34 (1982), 196-215 .[4] M. F. Atiya and 1. G. Macdonald, Introduction to Commutative Algebra,

Addison-Wesley, (1969).[5] J. Barshay, Graded algebras of powers of ideals generated by A -sequences,

J. Algebra 25 (1973), 90-99.[6] N. Bourbaki, Algebre Commutative, Chap.1-7, Hermann, (1964).[7] R.M. Fossum, The divisor Class group of a Krull domain, Springer-Verlag,

(1973).[8] S. Gabelli, Completely Integrally Closed Domains and t-ideals Bollettino

UMI3-B (1989), 327-342.[9] R. G. Gilmer and W. Heinzer, On the complete integral closure of an

integral domain, J. Austral. Math. Soc. 6 (1966), 351-361.[10] W. Heinzer, Some remarks on complete integral closure, J. Austral. Math.

Soc. 9 (1969), 310-314.


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[11] M. Henriksen, On the ideal structure of the ring of the entire functions,Pacific J. Math. 2 (1952), 179-184.

[12] J. Herzog and W. V. Vasconcelos, On the divisor class group of Rees alge-bras, J. Algebra 93, 182-188 (1985).

[13] J. Herzog, A. Simis and W. V. Vasconcelos, Arithmetic of normal Reesalgebras, J. Algebra 143 (1991), 299-294.[14] P. Hill, On the complete integral closure of a domain, Proc. Amer. Math.

Soc 36 (1) (1972), 26-30.[15] E. G. Houston, T. G. Lucas and T.M. Viswanathan, Primary decomposi-

tion of divisorial ideals in Mori domains, J. Algebra 117 (1988), 327-342.[16] C. Huneke, Symbolic powers of prime ideals and special graded algebras,

Comm. Alg. 9 (4) (1981), 339-366.[17] 1. Kaplansky, Commutative rings, Univ. of Chicago Press, Chicago, (1974).[18] D. Katz and L. J. Ratliff Jr, On the symbolic Rees ring of a primary ideal,Comm. Alg. 14 (1986), 959-970.[19] W. Krull, Allgemeine Bewertungstheorie, J. reine angew. Math. 167

(1931), 160-196.[20] N. Raillard (Dessagnes), Sur les anneaux de Mori, C. R. Acad. Sci. Paris

Ser A 280 (1975), 1571-1573.[21] L. Robbiano and G. Valla, Primary powers of a prime ideal, Pac. J. Math.

63 (1976), 491-498.[22] M. Roitman, On polynomial extensions of Mori domains over countable

fields, J. Pure Appl. Algebra 64 (1990), 315-328.[23] P. B. Sheldon, Two counterexamples involving complete integral closure in

finite-dimensional Priifer domains, J. Algebra 27 (1973), 462-474.[24] A. Simis and N. V. Trung, The Divisor Class Group of Ordinary and Sym-

bolic Blow-ups, Math. Z. 198 (1988), 479-491.[25] B.L. van der Waerden, Algebra II, Springer-Verlag, New York, (1991).[26] W. V. Vasconcelos, Arithmetic of Blowup Algebras, LNMS Lecture Notes

Series 195, Cambridge University Press, (1994).[27] L. J. Wallace, Graded Mori rings, PhD. Thesis, University of California at

Riverside, (1998).[28] O. Zarisky and P. Samuel, Commutative Algebra, Van Nostrand, (1960).

Universite di Roma III - Dipartimento di Matematica,Largo S. Leonardo Murialdo,l ; 00100 Roma, Italy

E-mail: [email protected] dell'Aquila - Dipartimento di Matematica,Via Vetoio, Loc. Coppito; 67100 L'Aquila, Italy
mailto:[email protected]:[email protected]


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E-mail: [email protected]

1991 Mathematics Subject Classification. 13A30, 13C05, 13E05, 13H15Key words and phrases. Complete integral closure, Rees algebra, associated graded ring,

Krull rings.
mailto:[email protected]:[email protected]

On the Complete Integral Closure of the Rees Algebra

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