FINITE GENERATION of POWERS of IDEALS 0. Introduction All Rings

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FINITE GENERATION of POWERS of IDEALS 0. Introduction All Rings PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 127, Number 11, Pages 3141{3151 S 0002-9939(99)05199-0 Article electronically published on May 4, 1999 FINITE GENERATION OF POWERS OF IDEALS ROBERT GILMER, WILLIAM HEINZER, AND MOSHE ROITMAN (Communicated by Wolmer V. Vasconcelos) Abstract. Suppose M is a maximal ideal of a commutative integral domain R and that some power M n of M is finitely generated. We show that M is finitely generated in each of the following cases: (i) M is of height one, (ii) R is integrally closed and ht M = 2, (iii) R = K[X; S~] is a monoid domain over a field K,whereS~=S 0 is a cancellative torsion-free monoid such [f g s that ∞ mS = ,andMis the maximal ideal (X : s S). We extend the m=1 ; 2 above results to ideals I of a reduced ring R such that R=I is Noetherian. We T prove that a reduced ring R is Noetherian if each prime ideal of R has a power that is finitely generated. For each d with 3 d , we establish existence of ≤ ≤∞ a d-dimensional integral domain having a nonfinitely generated maximal ideal M of height d such that M 2 is 3-generated. 0. Introduction All rings in this paper are commutative with unity. We consider the following question: Question 0.1. Suppose that some power M n of the maximal ideal M of an integral domain R is finitely generated. Does it follow that M is finitely generated? This question is raised in [3, page 74] and was mentioned in a talk given by the first author at the AMS meeting in Auburn, Alabama in November 1971. It is also listed, for the case of a quasilocal integrally closed domain, as Problem 8 in the questions list on pages 174{176 in the 1973 Notices of the AMS from the problem session organized by Graham Evans at the January 1973 AMS meeting in Dallas. We answer Question 0.1 in the negative in 3. We also consider the following generalization of Question 0.1. § Question 0.2. Suppose I is an ideal of a ring R such that R=I is Noetherian. If some power of I is finitely generated, under what conditions is I finitely generated? In 1 we present positive results on these two questions. Regarding Question 0.1 we§ prove in Theorem 1.24, among other results, that the maximal ideal M is finitely generated under each of the following conditions: (i) R is a reduced ring Received by the editors January 26, 1998. 1991 Mathematics Subject Classification. Primary 13A15, 13E99, 13G05. Key words and phrases. Cohen's theorem, finite generation, maximal ideal, monoid ring, Noe- therian, power of an ideal, Ratliff-Rush closure. The first two authors acknowledge with thanks the hospitality of the mathematics department of the University of North Carolina at Chapel Hill. Partial support of the work of the second author by the National Science Foundation is also gratefully acknowledged. c 1999 American Mathematical Society 3141 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 3142 ROBERT GILMER, WILLIAM HEINZER, AND MOSHE ROITMAN and ht M = 1, (ii) R is an integrally closed domain and ht M = 2. We would like to know the answer to Question 0.1 if R is an integrally closed (or even completely integrally closed) domain, eventually assuming that M is of finite height. Regarding Question 0.2 we prove in Theorem 1.10, among other results, that I is finitely generated if I is the radical of an ideal generated by a nonempty regular sequence. Also in 1 we present for the case of a reduced ring an analogue to Cohen's Theorem [8,§ (3.4)]. We assume in Theorem 1.17 that each prime ideal of a reduced ring R has a power that is finitely generated and conclude that R is Noetherian. Moreover, if R is semiquasilocal, we prove that R is Noetherian if each prime of R of positive height has a finitely generated power (Theorem 1.20). The context in which Question 0.1 was raised in [3] makes it natural to consider the question in a monoid domain over a field. Thus in 2 we prove that the analogue of Question 0.1 for a commutative cancellative semigroup§ S has a positive answer provided m∞=1 mS = (Theorem 2.3). We establish in Proposition 2.1 the con- nection between these∅ two aspects of Question 0.1 (for semigroups and for rings), and this allowsT us to conclude in Theorem 2.5 that if R = K[X; T ]isamonoid ring over a field K,whereT is a cancellative monoid, S = T 0 ; m∞=1 mS = , and M is the maximal ideal (Xs : s S), then M is finitely\{ generated} if some∅ ∈ T power of M is finitely generated. Without the restriction m∞=1 mS = ,weshow in Example 3.2 that the ideal M need not be finitely generated. ∅ T 1. Powers of ideals The main tool for dealing with Questions 0.1 and 0.2 is Theorem 1.9, which states that under the hypothesis of Question 0.2 the ideal I is finitely generated if I is the radical of a principal ideal and cI = (0) for each nonzero c I. The proof of Theorem 1.9 is based on Lemma 1.4. Moreover,6 Lemma 1.8 reduces∈ Question 0.1 to the case of quasilocal rings, and Lemma 1.7 reduces Question 0.2 to the case of radical ideals. Lemma 1.1. Let J I be ideals in a ring R such that In = J n for some positive m ⊆ i m i integer n.ThenI =IJ − for all m n and 0 i m. ≥ ≤ ≤ i m i m m Proof. Since J I, the ideal I J − lies between I and J . Thus it suffices to m m ⊆ n n 1 n+1 n n+1 show I = J . Moreover, I = JI − implies I = JI = J . The lemma follows by induction. Lemma 1.2. If I is an ideal of a ring R such that In is finitely generated for some n 1, then there exists a finitely generated ideal J I such that In = J n. ≥ ⊆ n n Proof. Let I =(a1;::: ;ak). By definition of I , the element aj is a finite sum of n products of n elements of I, and hence belongs to Bj for some finitely generated subideal Bj of I.SetJ= Bj. Remark 1.3. Suppose M isP a maximal ideal of a ring R such that M n is finitely generated for some n 1. Lemmas 1.1 and 1.2 imply that there exists a finitely generated ideal I M≥ such that M is the Ratliff-Rush closure of I in the sense that M is the largest⊆ ideal containing I and having the same high powers as I [10], [6]. In particular, M is integral over I. In connection with generators of powers of an ideal, Eakin and Sathaye prove in [1, page 440] the interesting result that if I is an ideal of a quasilocal ring with infinite residue field and if n and r are positive integers such that In can be License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use FINITE GENERATION OF POWERS OF IDEALS 3143 generated by fewer than n elements, then there exists an r-generated ideal J I r ⊆ such that JIn 1 = In. − Lemma 1.4. Let N K M be modules over a ring R such that N =( k I )M, ⊆ ⊆ j=1 j where Ij is an ideal of R so that R=Ij is Noetherian for 1 j k. Assume that e Q all the R-modules ( I )M for 0 e k are finitely generated≤ ≤ (for e =0we j=1 j ≤ ≤ have e I = R). Then the R-module K is finitely generated. j=1 j Q k k 1 Proof.QConsider the sequence of R-modules: N =( I )M ( − I )M j=1 j ⊆ j=1 j ⊆ I1M M: Each quotient of two successive modules in this sequence is a ··· ⊆ ⊆ Q Q finitely generated module over the Noetherian ring R=Ij for a suitable j,andis therefore a Noetherian R-module. It follows that M=N is a Noetherian R-module, hence K=N, and so also K, is finitely generated. Lemma 1.5. Let I and J be ideals of a ring R such that √I = √J (in particular, we may take J = √I). Then the following two conditions are equivalent: (i) R=I is Noetherian and I is finitely generated, and (ii) R=J is Noetherian and J is finitely generated. Proof. Assume that R=I is Noetherian and that I is finitely generated. Thus √I is finitely generated. For some m 1wehave(√I)m J √I. By Lemma ≥ ⊆ ⊆ 1.4 with Ij = √I for all j, the ideal J is finitely generated. Since the ring R=√J is Noetherian, and since √J is finitely generated, we obtain that all prime ideals of the ring R=J are finitely generated. By Cohen's Theorem [8, (3.4)], R=J is Noetherian. Corollary 1.6. Let I be an ideal of a ring R. Assume that R=I is Noetherian and that In is finitely generated for some n 1.ThenIis finitely generated iff the ring R=In is Noetherian. ≥ Lemma 1.7. Let I be an ideal of a ring R.
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