QUASI-SOCLE IDEALS IN GORENSTEIN NUMERICAL SEMIGROUP RINGS SHIRO GOTO, SATORU KIMURA, AND NAOYUKI MATSUOKA 8 0 0 2 Abstract. Quasi-socle ideals, that is the ideals I of the form I = Q :mq in Goren- n stein numerical semigroup rings over fields are explored, where Q is a parameter a ideal, and m is the maximal ideal in the base local ring, and q ≥ 1 is an integer. J The problems of when I is integral over Q and of when the associated graded ring 7 G(I)= In/In+1 of I is Cohen-Macaulay are studied. The problems are rather 1 Ln≥0 wild; examples are given. ] C A 1. Introduction . h This paper aims at a study of the Polini-Ulrich Conjecture 1.1 ([PU]) of one- t a m dimensional case. We shall explore Gorenstein numerical semigroup rings over fields [ as the test case. Before stating our own result, let us explain the reason why we are 2 interested in the conjecture of the special case. See Section 2 for the statement of the v 6 main Theorem 2.1 of this paper. 8 3 Let A be a Cohen-Macaulay local ring with the maximal ideal m and d = dimA > 0. 1 Let Q = (a ,a ,··· ,a ) be a parameter ideal in A and let q > 0 be a positive integer. . 1 2 d 0 1 Then we put I = Q : mq and refer to those ideals as quasi-socle ideals in A. 7 0 The study of socle ideals Q : m dates back to the research of L. Burch [B], where she : v explored socle ideals of finite projective dimension and gave a very nice characterization i X of regular local rings (cf. [GH, Theorem 1.1]). More recently, A. Corso and C. Polini r a [CP1, CP2] showed, with the interaction to linkage theory of ideals, that if A is a Cohen-Macaulay local ring which is not regular, one has the equality I2 = QI for every parameter ideal Q in A, where I = Q : m. Subsequently, the first author and H. Sakurai [GSa1, GSa2, GSa3] showed the equality I2 = QI could hold true, where I = Q : m, for numerous parameter ideals Q in A, even though the base rings A are not necessarily Cohen-Macaulay. However, a more important thing is the following. If J is an equimultiple Cohen-Macaulay ideal of reduction number one, the associated Key words and phrases: Quasi-socle ideal, Numerical semigroup, Gorenstein local ring, associated graded ring, Rees algebra,integral closure, multiplicity. 2000 Mathematics Subject Classification: 13H10, 13A30, 13B22,13H15. 1 2 SHIROGOTO, SATORUKIMURA,ANDNAOYUKIMATSUOKA gradedringG(J) = Jn/Jn+1 isaCohen-Macaulay ringand, soistheReesalgebra Ln≥0 R(J) = Jn, provided ht J ≥ 2. One also knows the number and degrees of the Ln≥0 A defining equations of R(J), so that one can understand fairly explicitly the process of desingularization of SpecA along the subscheme V(J). This observation motivated the ingenious research of C. Polini and B. Ulrich [PU], where they posed, among many important results, the following conjecture. Conjecture 1.1 ([PU]). Let (A,m) be a Cohen-Macaulay local ring with dimA ≥ 2. Assume that dimA ≥ 3 when A is regular. Let q ≥ 2 be an integer and let Q be a parameter ideal in A such that Q ⊆ mq. Then Q : mq ⊆ mq. This conjecture was recently settled by H.-J. Wang [W], whose beautiful theorem says: Theorem 1.2 ([W]). Let (A,m) be a Cohen-Macaulay local ring with d = dimA ≥ 2. Let q ≥ 1 be an integer and Q a parameter ideal in A. Assume that Q ⊆ mq and put I = Q : mq. Then I ⊆ mq, mqI = mqQ, and I2 = QI, provided that A is not regular if d ≥ 2 and that q ≥ 2 if d ≥ 3. Added to it, the very recent research of S. Goto, N. Matsuoka, and R. Takahashi [GMT] reported a different approach to the Polini-Ulrich conjecture and proved the following. Theorem 1.3 ([GMT]). Let (A,m) be a Gorenstein local ring with d = dimA > 0 and e0(A) ≥ 3, where e0(A) denotes the multiplicity of A with respect to m. Let Q m m be a parameter ideal in A and put I = Q : m2. Then m2I = m2Q, I3 = QI2, and G(I) = In/In+1 is a Cohen-Macaulay ring, so that R(I) = In is also Ln≥0 Ln≥0 Cohen-Macaulay, provided d ≥ 3. The researches [W] and [GMT] were independently performed and their methods of proof are totally different from each other’s. Unfortunately, the technique of [GMT] can not go beyond the restrictions that A is a Gorenstein ring, q = 2, and e0(A) ≥ 3 m and however, despite these restrictions, the result [GMT, Theorem 1.1] holds true even in the case where dimA = 1, while Wang’s result says nothing about the case where QUASI-SOCLE IDEALS IN GORENSTEIN NUMERICAL SEMIGROUP RINGS 3 dimA = 1. As is suggested in [GMT], the one-dimensional case is rather different from higher-dimensional cases and much more complicated to control. It seems natural to ask how one can modify the Polini-Ulrich conjecture, so that it covers also the one-dimensional case. This question has motivated the present re- search. We then decided to explore Gorenstein numerical semigroup rings over fields, as the starting point of our investigations, because they are typical one-dimensional Cohen-Macaulay local rings and because higher-dimensional phenomena are often re- alized, with primitive forms, in those rings of dimension one. We expect, with further investigations, a generalization of the results in this paper to higher-dimensional cases and a possible modification of the Polini-Ulrich conjecture, as well. Letusexplainhowthispaperisorganized. Thestatement ofthemainresult Theorem 2.1 and its proof will be found in Section 2. Theorem 2.1 gives a generalization of [GMT, Theorem 1.1] in the case where the base rings are numerical semigroup rings. As an application of Theorem 2.1 we will explore in Section 3 numerical semigroup rings A = k[[ta,ta+1]] (a > 1) over fields k, where t is an indeterminate. We will give a criterion fortheideal I = (ts) : mq tobeintegral over theparameter ideal(ts) inA(here q > 0 is an integer and 0 < s ∈ H = ha,a+1i, the numerical semigroup generated by a,a+1). The problem of when the ring G(I) is Cohen-Macaulay is answered in certain special cases. We agree with the observation in [GMT] that the one-dimensional case is wild. To confirm this, we will note two examples in Section 4. 2. The main result and the proof Let 0 < a < a < ··· < a (ℓ ≥ 1) be integers with GCD(a ,a ,··· ,a ) = 1. We 1 2 ℓ 1 2 ℓ put ℓ H = ha1,a2,··· ,aℓi = {Xαiai | 0 ≤ αi ∈ Z}. i=1 Then, because GCD(a ,a ,··· ,a ) = 1, H ∋ n for all n ∈ Z with n ≫ 0. We put 1 2 ℓ c(H) = min{m ∈ Z | H ∋ n for all integers n ≥ m}, the conductor of H. Let V = k[[t]] be the formal power series ring over a field k. We put A = k[[H]] = k[[ta1,ta2,··· ,taℓ]] ⊆ V. Let m = (ta1,ta2,··· ,taℓ) be the maximal ideal in A. Then A is a Cohen-Macaulay local ring with dimA = 1 and e0(A) = a , where e0(A) denotes the multiplicity of A m 1 m with respect to the maximal ideal m. The ring V is a module-finite birational extension 4 SHIROGOTO, SATORUKIMURA,ANDNAOYUKIMATSUOKA of A. Hence A = V, where A denotes the normalization. We say that the numerical semigroup H is symmetric, if for every n ∈ Z, n ∈ H ⇐⇒ α−n 6∈ H, where α = c(H)−1 denotes the Frobenius number of H. This condition is equivalent to saying that A = k[[H]] is a Gorenstein ring ([HK]). With this notation we are interested in the problem of when the results of [W] and [GMT] hold true and our result is summarized into the following. Theorem 2.1. Suppose that A = k[[H]] is a Gorenstein ring. Let q > 0 be an integer and assume that the following two conditions (C ) and (C ) are satisfied for q, where 1 2 c = c(H): (C ) tn ∈ mq for all integers n ≥ c; 1 (C ) Let n ∈ H. Then n < a (q −1), if tn 6∈ mq−1. 2 1 Let 0 < s ∈ H. Let Q = (ts) and I = Q : mq. Then the following assertions hold true. (1) mqI = mqQ and Q∩I2 = QI. (2) I2 = QI, if s ≥ c. (3) I3 = QI2 and the associated graded ring G(I) = In/In+1 is Cohen- Ln≥0 Macaulay, if s ≥ a (q−1). 1 Before going ahead, let us note a few remarks on Theorem 2.1. Remark 2.2. (1) Conditions (C ) and (C ) in Theorem 2.1 are naturally satisfied if 1 2 a ≥ 2 and q = 1. We will later show that they are satisfied also in the following two 1 cases. (i) A = k[[H]] is a Gorenstein ring, a ≥ 3, and q = 2. 1 (ii) ℓ = 2, a > 1, a = a +1, and 0 < q < a . 1 2 1 1 (2)InTheorem2.1theringG(I)isnotnecessarilyCohen-Macaulayandthereduction number r (I) = min{0 ≤ n ∈ Z | In+1 = QIn} Q of I with respect to Q can go up, unless s ≥ a (q −1). See Theorem 3.8 and Example 1 4.1. (3) Unless condition (C ) is satisfied, Theorem 2.1 (3) does not hold true in general, 2 although condition (C ) is satisfied (and hence I is integral over Q; cf. Lemma 2.4). 1 See Example 4.2. QUASI-SOCLE IDEALS IN GORENSTEIN NUMERICAL SEMIGROUP RINGS 5 The rest of this section is devoted to the proof of Theorem 2.1. Let us restate our setting. Setting. Let 0 < a < a < ··· < a (ℓ ≥ 1) be integers with GCD(a ,a ,··· ,a ) = 1, 1 2 ℓ 1 2 ℓ H = ha | 1 ≤ i ≤ ℓi, c = c(H), a = a = min [H \{0}], i 1 k a field, V = k[[t]] the formal power series ring over k, A = k[[H]] = k[[ta1,ta2,··· ,taℓ]] ⊆ V, and m = (ta1,ta2,··· ,taℓ) the maximal ideal in A. We begin with the following. Lemma 2.3. Suppose that a ≥ 3 and let α ≥ a−1 be an integer. Let Λ = {n ∈ Z | 0 ≤ n ≤ α} and assume that for every n ∈ Λ, n ∈ H ⇔ α−n ∈/ H. Then α = c−1, so that H is symmetric. Proof. Let 1 ≤ m < a be an integer. Then m ∈/ H and so α−m ∈ H, whence α+n ∈ H for all 1 ≤ n ≤ a − 1. Therefore, since α 6∈ H, to see that α = c − 1, it suffices to show α + a ∈ H. Assume α + a 6∈ H and put Γ = {n ∈ Z | 0 ≤ n ≤ α + a}. Let ∆ = {n ∈ Z | α+1 ≤ n ≤ α+a−1} and let ϕ : Γ∩H → Γ\H, n 7→ α+a−n. Then, since α+a 6∈ H and ∆ ⊆ H, we see Γ∩H = (Λ∩H)∪∆ and Γ\H = (Λ\H)∪{α+a}. Therefore, because the map ϕ is injective and ♯(Λ∩H) = ♯(Λ\H) , we have a−1 = ♯∆ ≤ 1, whence a ≤ 2, which is impossible. Thus α+a ∈ H so that α = c−1. (cid:3) Let q > 0 be an integer and let 0 < s ∈ H. We put Q = (ts) and I = Q : mq. Then I = (tn | n ∈ H,tn ∈ I), which is a monomial ideal in A. Let Q denote the integral closure of Q. We then have Q = tsV ∩A. Lemma 2.4. Suppose tn ∈ mq for all n ∈ Z such that n ≥ c. Then aq ≤ c and I ⊆ Q. Proof. We have aq ≤ c, since tc ∈ mq ⊆ taqV (recall m ⊆ taV, since a = min [H \{0}]). Let n ∈ H and assume tn ∈ I. We want to show n ≥ s. Assume the contrary and we see (s+c−1)−n = (c−1)+(s−n) ≥ c 6 SHIROGOTO, SATORUKIMURA,ANDNAOYUKIMATSUOKA becauses > n,whencet(s+c−1)−n ∈ mq byassumption. Therefore, sincetn ∈ I = Q : mq, we get ts+c−1 = t(s+c−1)−ntn ∈ Q = (ts) whence tc−1 ∈ A = k[[H]], which is impossible (recall c = c(H)). Thus tn ∈ tsV, so that I ⊆ tsV ∩A = Q as is claimed. (cid:3) The following result shows that condition (C ) in Theorem 2.1 is satisfied, if A = 1 k[[H]] is a Gorenstein ring, a ≥ 3, and q = 2. Proposition 2.5. Suppose that A is a Gorenstein ring and let a ≥ 3. Then tn ∈ m2 for all n ∈ Z such that n ≥ c. Hence (ts) : m2 ⊆ (ts) for all 0 < s ∈ H. Proof. We may assume that H is minimally generated by {a } . Hence ℓ ≥ 2 and i 1≤i≤ℓ H 6= ha | 1 ≤ j ≤ ℓ,j 6= ii for all 1 ≤ i ≤ ℓ. We have c ≥ a ≥ 3, since 0 < c ∈ H. j Notice that c > a. In fact, assume that c = a. Then H ∋ n for all integers n ≥ a. Therefore, because a + a − a ≥ a for all 1 ≤ i,j ≤ ℓ, we have m2 = tam, so that i j ℓ (A/taA) ≤ 2, since A is a Gorenstein local ring. This is however impossible, because A ℓ (A/taA) = e0(A) = a ≥ 3, where e0(A) denotes the multiplicity of A with respect A m m to m. Hence c > a. Let n ≥ c be an integer and assume that tn ∈/ m2. Then n = a for some 1 ≤ i ≤ ℓ. i We have i > 1, since c > a. Let K = ha | 1 ≤ j ≤ ℓ,j 6= ii. j Then a ∈/ K. We have GCD(a | 1 ≤ j ≤ ℓ,j 6= i) = 1. In fact, let 1 ≤ m < a be an i j integer. Then m 6∈ H but a +m ∈ H, since a ≥ c. We write i i ℓ ai +m = Xcjaj j=1 with 0 ≤ c ∈ Z. Then c = 0, because m 6∈ H. Therefore a +m ∈ K for all 1 ≤ m < a. j i i Hence a +1,a +2 ∈ K, because a ≥ 3. Thus GCD(a | 1 ≤ j ≤ ℓ,j 6= i) = 1. i i j We now apply Lemma 2.3 to the numerical semigroup K. Let α = c − 1 and let 0 ≤ m ≤ α be an integer. Then, since 0 ≤ m < c ≤ a , we have m ∈ K = i ha | 1 ≤ j ≤ ℓ,j 6= ii, once m ∈ H (recall that a = a < a < ··· < a ). Suppose j 1 2 ℓ now that α−m 6∈ K. Then α−m 6∈ H as 0 ≤ α−m ≤ α, whence m ∈ H because the numerical semigroup H is symmetric, so that we have m ∈ K. Conversely, if m ∈ K, then m ∈ H, whence α−m ∈/ H so that α−m 6∈ K. Consequently, because α ≥ a and a = min [K \ {0}], by Lemma 2.3 we get c(K) = α+ 1 = c. Hence a ∈ K, because i QUASI-SOCLE IDEALS IN GORENSTEIN NUMERICAL SEMIGROUP RINGS 7 a ≥ c. This is impossible. Thus tn ∈ m2 for all integers n ≥ c. The second assertion i (cid:3) follows from Lemma 2.4. We are now ready to prove Theorem 2.1. Proof of Theorem 2.1. (1) We will show that mq−1I ⊆ mqQ : m. We put Λ = {(α ,α ,··· ,α ) ∈ Zℓ | α ≥ 0 for all 1 ≤ i ≤ ℓ and ℓ α = q − 1}. Let 1 2 ℓ i Pi=1 i ℓ α = (α ,α ,··· ,α ) ∈ Λ and let n ∈ H such that tn ∈ I. Let ϕ = tPi=1αiai·tn. 1 2 ℓ Then ϕ ∈ mq−1I ⊆ Q : m = (ts)+(ts+c−1), where the equality Q : m = (ts)+(ts+c−1) follows from the fact that A is a Gorenstein ring (notice that ts+c−1 6∈ Q = (ts) but tm·ts+c−1 = ts·tm+c−1 ∈ Q = (ts) for every 0 < m ∈ H, because c = c(H) is the conductor of H). Consequently ϕ ∈ (ts) or ϕ ∈ (ts+c−1), since ϕ is a monomial in t. Because tm·ts+c−1 = tm+(c−1)·ts ∈ mqQ for all 0 < m ∈ H (use condition (C ); notice that m+(c−1) ≥ c), we have m·ts+c−1 ⊆ mqQ. 1 Hence mϕ ⊆ mqQ if ϕ ∈ (ts+c−1). Suppose that ϕ ∈ (ts) = Q and write ℓ Xαiai +n = h+s i=1 with h ∈ H. Then, since n ≥ s by Lemma 2.4, we get ℓ h = Xαiai +(n−s) i=1 ℓ ≥ Xαiai i=1 ℓ ≥ a·Xαi = a(q −1), i=1 so that we have th ∈ mq−1 by condition (C ). Hence ϕ = tPℓi=1αiai+n = th·ts ∈ mq−1Q 2 and so mϕ ⊆ mqQ. Thus mq−1I ⊆ mqQ : m, whence mqI = m(mq−1I) ⊆ mqQ. Let us show Q∩I2 = QI. Since mqI = mqQ, we have mqIn = mqQn for all n ∈ Z. Let x ∈ Q∩I2 and write x = tsy with y ∈ A. Then for all α ∈ mq, we have ts·αy = αx ∈ mqI2 ⊆ Q2 = (t2s). Henceαy ∈ Q = (ts)sothatwehavey ∈ Q : mq = I. Thusx ∈ QI whence Q∩I2 = QI. 8 SHIROGOTO, SATORUKIMURA,ANDNAOYUKIMATSUOKA (2) It suffices to show I2 ⊆ Q. Let m,n ∈ H such that tm,tn ∈ I. Then m,n ≥ s ≥ c by Lemma 2.4. We get m+n−s ∈ H, since m+n−s = m+(n−s) ≥ c. Therefore tmtn = tm+n−sts ∈ Q, whence I2 ⊆ Q. (3) We may assume that I2 6= QI. Hence I2 6⊆ Q, because Q ∩ I2 = QI. We have I ⊆ mq−1 by condition (C ), since s ≥ a(q − 1) and I ⊆ Q ⊆ tsV. Then, since 2 I2 ⊆ mq−1I, we get Q ( Q+I2 ⊆ Q : m = Q+(ts+c−1). Therefore, since ℓ ([Q : m]/Q) = 1 (recall that A is a Gorenstein ring), we have A Q+I2 = Q : m = Q+(ts+c−1), whence ts+c−1 ∈ I2 because ts+c−1 6∈ Q. Consequently I2 = (Q∩I2)+(ts+c−1) = QI +(ts+c−1) because Q∩I2 = QI, whence I3 = QI2 +I·ts+c−1. Let uscheck thatI·ts+c−1 ⊆ QI2. Let n ∈ H andassume thattn ∈ I. We willshow that tnts+c−1 ∈ QI2. Wemayassumethatn > s. Leth = (n+s+c−1)−2s = (n−s)+(c−1). Then h ∈ H since h ≥ c. Therefore αth·t2s = α·tnts+c−1 ∈ mqI3 ⊆ Q3 = (t3s) for all α ∈ mq and so αth ∈ Q. Consequently, th ∈ Q : mq = I, whence tnts+c−1 = t2sth ∈ Q2I ⊆ QI2. Thus I·ts+c−1 ⊆ QI2 so that I3 = QI2. Since I3 = QI2 and Q∩I2 = QI, we get Q∩Ii+1 = QIi for all i ∈ Z, whence G(I) is a Cohen-Macaulay (cid:3) ring. Combining Proposition 2.5 and Theorem 2.1, we readily get [GMT, Theorem 1.1] in the case where the base rings are numerical semigroup rings. Notice that condition (C ) is automatically satisfied for q = 2. 2 Corollary 2.6 (cf. [GMT, Theorem 1.1]). Suppose that A = k[[H]] is a Gorenstein ring and that a ≥ 3. Let 0 < s ∈ H and put I = Q : m2, where Q = (ts). Then the following assertions hold true. (1) m2I = m2Q and I3 = QI2. (2) G(I) = In/In+1 is a Cohen-Macaulay ring. Ln≥0 (3) I2 = QI, if s ≥ c. QUASI-SOCLE IDEALS IN GORENSTEIN NUMERICAL SEMIGROUP RINGS 9 3. The case where H = ha,a+1i In this section let H = ha,a+1i with a ≥ 2. Applying Theorem 2.1, we shall explore the numerical semigroup H = ha,a+1i. Let c = a(a − 1), that is the conductor of H. Similarly as in Section 2, let k be a field and A = k[[H]] = k[[ta,ta+1]] ⊆ V, where V = k[[t]] is the formal power series ring over k. We denote by m = (ta,ta+1) the maximal ideal in A. Let 0 < s ∈ H, Q = (ts), and I = Q : mq with q > 0 an integer. We study the problems of when I is integral over Q and of when the associated graded ring G(I) = In/In+1 is a Cohen-Macaulay ring. Ln≥0 Let us begin with the following. Lemma 3.1. The following assertions hold true. (1) Let ℓ,i ≥ 0 be integers. Then aℓ+i ∈ H, if i ≤ ℓ. The converse is also ture, if i < a. (2) mℓ = (taℓ+i | 0 ≤ i ≤ ℓ) = (tn | n ∈ H, n ≥ aℓ) for all integers ℓ ≥ 0. Proof. (1) If i ≤ ℓ, then certainly aℓ + i = a(ℓ − i) + (a + 1)i ∈ H. Suppose that aℓ + i ∈ H and i < a. We write aℓ + i = αa + β(a + 1) with 0 ≤ α,β ∈ Z. Then β = a[ℓ −(α+β)]+i and so, letting m = ℓ−(α+β), we see m ≥ 0, because β ≥ 0 and i < a. Hence ℓ ≥ α+β ≥ β = am+i ≥ i. Thus i ≤ ℓ. (2) Let ℓ ≥ 0 be an integer. Then since a(ℓ−i)+(a+1)i = aℓ+i for all 0 ≤ i ≤ ℓ, we get (♯) mℓ = (ta,ta+1)ℓ = (taℓ+i | 0 ≤ i ≤ ℓ). To see mℓ ⊇ (tn | n ∈ H, n ≥ aℓ), let n ∈ H such that n ≥ aℓ. We write n = ap + i with p ≥ ℓ and 0 ≤ i < a. Then p ≥ i by assertion (1), so that tn = tap+i ∈ mp by equality (♯). Hence tn ∈ mℓ, because p ≥ ℓ. Thus mℓ = (tn | n ∈ H,n ≥ aℓ). (cid:3) Proposition 3.2. Conditions (C ) and (C ) in Theorem 2.1 are satisfied for q if and 1 2 only if q < a. Proof. Assume that q < a and let n ≥ c be an integer. Then n ≥ aq, since q < a and c = a(a−1). Hence tn ∈ mq by Lemma 3.1 (2). Let n ∈ H and assume that tn 6∈ mq−1. 10 SHIROGOTO, SATORUKIMURA,ANDNAOYUKIMATSUOKA We then have again by Lemma 3.1 (2) that n < a(q − 1). Thus conditions (C ) and 1 (C ) in Theorem 2.1 are satisfied. See Lemma 2.4 for the only if part. (cid:3) 2 The question of when I is integral over Q is now answered in the following way. Theorem 3.3. The following three conditions are equivalent to each other. (1) I ⊆ Q. (2) mqI = mqQ. (3) q < a. Proof. (2) ⇒ (1) This is clear and well known ([NR]). (3) ⇒ (2) This follows from Proposition 3.2. See Theorem 2.1. (1) ⇒ (3) Assume q ≥ a. We will check that s−a 6∈ H. Suppose s−a ∈ H and let n ∈ H with n ≥ aq. Then n−a ≥ aq −a ≥ a2 −a = c, whence (n+s−a)−s = n−a ∈ H, so that tnts−a = t(n+s−a)−sts ∈ Q. Because s−a ∈ H and mq = (tn | n ∈ H,n ≥ aq) by Lemma 3.1 (2), we get ts−a ∈ Q : mq = I ⊆ Q ⊆ tsV by assumption (1), which is impossible. Thus s − a 6∈ H whence s > a. We write s = aℓ+r with ℓ ≥ 1 and 0 ≤ r < a. Then r > ℓ−1 by Lemma 3.1 (1) since s−a = a(ℓ−1)+r ∈/ H, while r ≤ ℓ by Lemma 3.1 (1) since 0 ≤ r < a and s = aℓ+r ∈ H. Thus r = ℓ so that s = (a+1)ℓ. Hence ℓ < a because s−a < c (= a(a−1)). Let n ∈ H with n ≥ aq. Then aℓ+n−s = n−ℓ ≥ aq −ℓ ≥ a2 −(a−1) = c+1, whence aℓ+n−s ∈ H, so that tntaℓ = taℓ+n−sts ∈ Q for all n ∈ H with n ≥ aq. Thus taℓ ∈ Q : mq = I since mq = (tn | n ∈ H,n ≥ aq) by Lemma 3.1 (2). Consequently taℓ ∈ Q ⊆ tsV by assumption (1), so that aℓ ≥ s = (a+1)ℓ, which is impossible because ℓ ≥ 1. Thus q < a as is claimed. (cid:3) Corollary 3.4. Assume that q < a. Then the following assertions hold true. (1) I2 = QI, if s ≥ aq. (2) I3 = QI2 and G(I) is a Cohen-Macaulay ring, if s ≥ a(q −1). Proof. Since q < a, conditions (C ) and (C ) in Theorem 2.1 are satisfied (Proposition 1 2 3.2). Hence Q∩I2 = QI by Theorem 2.1 (1). Therefore, to see assertion (1), it suffices to show that I2 ⊆ Q. Let n ∈ H with tn ∈ I. Then, since tn ∈ Q ⊆ tsV by Theorem