Hilbert Coefficients and Buchsbaumness of Associated Graded Rings

Shiro Goto and Koji Nishida^{∗}

Abstract Let A be a Noetherian local ring with the maximal ideal 𝔪 and I an 𝔪-primary ideal. The purpose of this paper is to generalize Northcott’s inequality on Hilbert coefficients of I given in [8], without assuming that A is a Cohen-Macaulay ring. We will investigate when our inequality turns into an equality. It is related to the Buchsbaumness of the associated graded ring of I.

AMS classification codes: 13D40, 13H10, 13H15 Keywords: Hilbert coefficients, standard ideal, Buchsbaum ring

1 Introduction

Let A be a d-dimensional Noetherian local ring with the maximal ideal 𝔪 and I an 𝔪-primary ideal of A. Then there exist integers e_{0}(I), e_{1}(I), ⋯, e_{d}(I) such that

ℓ_{A}(A/I^{n+1}) = e_{0}(I)\binom{n + d}{d} − e_{1}(I)\binom{n + d − 1}{d − 1} + ⋯ + (−1)^{d}e_{d}(I)

for n ≫ 0. These integers are called the Hilbert coefficients of I and a lot of results are known on them in the case where A is a Cohen-Macaulay ring. For example, as was proved by Northcott [8], we always have e_{0}(I) − ℓ_{A}(A/I) ≤ e_{1}(I). Moreover, provided A/𝔪 is infinite, Huneke and Ooishi [6, 9] proved that e_{0}(I) − ℓ_{A}(A/I) = e_{1}(I) if and only if I^{2} = QI for some (any) minimal reduction Q of I, and when this is the case, by [11], the associated graded ring G(I) = ⨁_{n≥0} I^{n}/I^{n+1} is a Cohen-Macaulay ring. The purpose of this paper is to extend their results without assuming that A is a Cohen-Macaulay ring.

Suppose that I contains a parameter ideal Q as a reduction. Then, from Northcott’s inequality, one can easily deduce that e_{0}(I)−ℓ_{A}(A/I) ≤ e_{1}(I)−e_{1}(Q) (See 3.1). Assuming that Q is a standard ideal in the sense of [10, Definition 19 of Appendix], we will investigate when the equality e_{0}(I) − ℓ_{A}(A/I) = e_{1}(I) − e_{1}(Q) holds. In order to state our result, let us fix some notation. For an ideal 𝔮 of A which is minimally generated by a_{1}, ⋯, a_{s}, we set

∗The authors are supported by the Grant-in-Aid for Scientific Reserches in Japan (C(2), No. 13640044, No. 11640011).

Σ(𝔮) = 𝔮 + ∑_{i=1}^{s}[(a_{1}, ⋯, a_{i−1}, a_{i+1}, ⋯, a_{s}) :_{A} a_{i}] .

It is easy to see that Σ(𝔮) does not depend on the choice of the minimal system of generators. For a module M over a ring R, we denote by H_{𝔞}^{i}(M) the i-th local cohomology module of M with respect to 𝔞. In particular, we set W = H_{𝔪}^{0}(A). Then we have the following.

Theorem 1.1 Suppose that I contains a standard parameter ideal Q as a reduction. Then e_{0}(I) − ℓ_{A}(A/I) = e_{1}(I) − e_{1}(Q) if and only if I^{2} ⊆ QI + W and Σ(Q) ⊆ I.

If the length of H_{𝔪}^{i}(A), which is denoted by h^{i}(A), is finite for any 0 ≤ i < d, we have that

−e_{1}(Q) ≤ ∑_{i=0}^{d−1} \binom{d − 2}{i − 1}h^{i}(A)

with equality when Q is a standard ideal (See 2.4). Therefore, as a consequence of 1.1 and [3], we get the next result.

Corollary 1.2 If A is a quasi-Buchsbaum ring, then

sup_{\sqrt{I}=𝔪}\{e_{0}(I) − ℓ_{A}(A/I) − e_{1}(I)\} = ∑_{i=0}^{d−1} \binom{d − 2}{i − 1}h^{i}(A) .

Moreover, assuming that A is a Buchsbaum ring or a slightly different condition, for ideals I which enjoy the property stated in 1.1, we will study the Buchsbaumness of G(I) together with I(G(I)) and a(G(I)), where I(∗) and a(∗) denote the I-invariant (cf. [10, p. 254]) and a-invariant (cf. [4]) respectively.

Theorem 1.3 Suppose that either (i) A is a Buchsbaum ring or (ii) A is a quasi-Buchsbaum ring and I ⊆ 𝔪^{2}. If I contains a parameter ideal Q such that I^{2} ⊆ QI + W and Σ(Q) ⊆ I, then G(I) is a Buchsbaum ring with I(G(I)) = I(A) and a(G(I)) ≤ 2 − d.

Throughout this paper (A, 𝔪) denotes a commutative Noetherian local ring with d = dim A > 0 and I an 𝔪-primary ideal of A. The Rees algebra R(𝔞) of an ideal 𝔞 of a ring R is the subring R[It] of R[t], where t is an indeterminate. The associated graded ring G(𝔞) is the quotient ring R(𝔞)/𝔞R(𝔞). For f ∈ R(𝔞), we denote it’s image in G(𝔞) by \overline{f}.

2 Preliminaries

We begin with the following result of one dimensional case.

Lemma 2.1 Let d = 1. If I contains a parameter ideal Q as a reduction, then we have that e_{0}(I) − ℓ_{A}(A/I) ≤ e_{1}(I) + ℓ_{A}(I ∩ W) with equality if and only if I^{2} ⊆ QI + W.

Proof. Let B = A/W. Then B is a Cohen-Macaulay ring with dim B = 1 and QB is a parameter ideal of B contained in IB as a reduction. Hence, by Northcott’s inequality and the result of Huneke and Ooishi stated in Introduction, we have that e_{0}(IB)−ℓ_{B}(B/IB) ≤ e_{1}(IB) with equality if and only if I^{2}B = QIB. On the other hand, as ℓ_{B}(B/I^{n+1}B) = ℓ_{A}(A/I^{n+1}) − ℓ_{A}(W) for n ≫ 0, we have e_{0}(IB) = e_{0}(I) and e_{1}(IB) = e_{1}(I) + ℓ_{A}(W). Moreover, ℓ_{B}(B/IB) = ℓ_{A}(A/I) − ℓ_{A}(W) + ℓ_{A}(I ∩ W). Therefore we get the required assertion as I^{2}B = QIB if and only if I^{2} ⊆ QI + W.

When we investigate higher dimensional case, we reduce the dimension using a superficial element (cf. [7, Section 22]), and the next result, which may be well known, plays a key role.

Lemma 2.2 Let d ≥ 2 and a be a superficial element of I. We set B = A/aA. Then dim B = d − 1 and

e_{i}(IB) = \{ \begin{matrix}e_{i}(I) if 0 ≤ i < d − 1 \\ e_{d−1}(I) + (−1)^{d−1}ℓ_{A}(0 :_{A} a) if i = d − 1 . \end{matrix}

Proof. Let n ≫ 0. Then I^{n+1} ∩ aA = aI^{n} and I^{n} ∩ (0 :_{A} a) = 0. Hence we have an exact sequence

0 ⟶ 0 :_{A} a ⟶ A/I^{n} ⟶^{a} (aA + I^{n+1})/I^{n+1} ⟶ 0 ,

so that

ℓ_{B}(B/I^{n+1}B)
= ℓ_{A}(A/I^{n+1}) − ℓ_{A}(A/I^{n}) + ℓ_{A}(0 :_{A} a)
= ∑_{i=0}^{d}(−1)^{i}e_{i}(I)\binom{n + d − i}{d − i} − ∑_{i=0}^{d}(−1)^{i}e_{i}(I)\binom{n − 1 + d − i}{d − i} + ℓ_{A}(0 :_{A} a)
= ∑_{i=0}^{d−2}(−1)^{i}e_{i}(I)\binom{n + d − 1 − i}{d − 1 − i} + (−1)^{d−1}\{e_{d−1}(I) + (−1)^{d−1}ℓ_{A}(0 :_{A} a)\} .

Thus we get the required assertion.

Lemma 2.3 Suppose that A/𝔪 is infinite and J is a reduction of I. Then there exists an element a ∈ J which is superficial for both of I and J. Moreover, for such element a ∈ J, setting B = A/aA, we have e_{1}(I) − e_{1}(J) = e_{1}(IB) − e_{1}(JB) provided d ≥ 2.

Proof. By taking a general linear form in G(J)/𝔪G(J), we see the existence of a ∈ J satisfying the required condition. If d ≥ 3, we get the equality since e_{1}(IB) = e_{1}(I) and e_{1}(JB) = e_{1}(J). Even if d = 2, we have

e_{1}(IB) − e_{1}(JB) = {e_{1}(I) − ℓ_{A}(0 :_{A} a)} − {e_{1}(J) − ℓ_{A}(0 :_{A} a)}
= e_{1}(I) − e_{1}(J) .

Lemma 2.4 Let Q be a parameter ideal of A. We have the following statements provided h^{i}(A) is finite for any 0 ≤ i < d.

(1) Let d = 1. Then −e_{1}(Q) = h^{0}(A).

(2) Let d ≥ 2. Then we have that

−e_{1}(Q) ≤ ∑_{i=1}^{d−1} \binom{d − 2}{i − 1}h^{i}(A)

with equality if Q is a standard ideal.

Proof. Let d = 1. Then, taking n ≫ 0 such that W = 0 :_{A} Q^{n} and ℓ_{A}(A/Q^{n}) = e_{0}(Q)·n− e_{1}(Q), we see that −e_{1}(Q) = ℓ_{A}(W) since e_{0}(Q) · n = e_{0}(Q^{n}) = ℓ_{A}(A/Q^{n}) − ℓ_{A}(0 :_{A} Q^{n}). Thus we get the assertion (1).

Next we assume that d ≥ 2. Moreover, in order to prove the assertion (2), we may assume that A/𝔪 is infinite. Then we can choose a ∈ Q\𝔪Q which is a superficial element of Q. Let B = A/aA and 0 ≤ i < d − 1. Considering the exact sequence

0 ⟶ 0 :_{A} a ⟶ A ⟶^{a} A ⟶ B ⟶ 0 ,

we get the exact sequence

(#) H_{𝔪}^{i}(A) ⟶^{a} H_{𝔪}^{i}(A) ⟶ H_{𝔪}^{i}(B) ⟶ H_{𝔪}^{i+1}(A) ⟶^{a} H_{𝔪}^{i+1}(A) .

Hence it follows that h^{i}(B) ≤ h^{i}(A) + h^{i+1}(A) with equality when Q is a standard ideal.

Let d = 2. Then −e_{1}(Q) = −{e_{1}(QB) + ℓ_{A}(0 :_{A} a)} = h^{0}(B) − ℓ_{A}(0 :_{A} a). Because the exact sequence (#) implies h^{0}(B) ≤ ℓ_{A}(0 :_{W} a) + h^{1}(A), we have −e_{1}(Q) ≤ h^{1}(A).

Furthermore, if Q is standard, then h^{0}(B) = h^{0}(A) + h^{1}(A) and 0 :_{A} a = W, so that −e_{1}(Q) = h^{1}(A).

Let d ≥ 3. Then e_{1}(QB) = e_{1}(Q). Hence we can easily verify the assertion (2) by induction on d.

3 General case

As a result in general case, we give the following assertion, which is a generalization of Northcott’s inequality.

Theorem 3.1 If I contains a parameter ideal Q as a reduction, then e_{0}(I) − ℓ_{A}(A/I) ≤ e_{1}(I) − e_{1}(Q).

Proof. We prove by induction on d. If d = 1, the assertion follows from 2.1 and 2.4. Suppose that d ≥ 2. We may assume that A/𝔪 is infinite, so that there exists a ∈ Q\𝔪Q which is superficial for both of I and Q. Then, setting B = A/aA, we have

e_{0}(I) − ℓ_{A}(A/I) = e_{0}(IB) − ℓ_{B}(B/IB) by 2.2
≤ e_{1}(IB) − e_{1}(QB) by the inductive hypothesis
= e_{1}(I) − e_{1}(Q) by 2.3 .

Thus we get the required inequality.

The next result gives a sufficient condition under which the inequality of 3.1 turns into an equality in the case where I = 𝔪.

Proposition 3.2 Let Q be a parameter ideal which is a reduction of 𝔪. If there exists an ideal V of A such that dim_{A} V < d and 𝔪^{2} ⊆ Q𝔪 + V, then e_{0}(𝔪) − 1 = e_{1}(𝔪) − e_{1}(Q).

Proof. We prove by induction on d. If d = 1, then V ⊆ W ⊆ 𝔪, so that by 2.1 we have e_{0}(𝔪) − 1 = e_{1}(𝔪) + ℓ_{A}(W), which yields the required equality since −e_{1}(Q) = ℓ_{A}(W) by 2.4. Suppose that d ≥ 2. As we may assume that A/𝔪 is infinite, it is possible to take an element a ∈ Q \ 𝔪Q such that dim_{A} V/aV < d − 1 and a is a superficial element for both of 𝔪 and Q. Let B = A/aA. Then dim_{B} V B < dim B as V B is a homomorphic image of V/aV, so that by the inductive hypothesis we have e_{0}(𝔪B) − 1 = e_{1}(𝔪B) − e_{1}(QB), from which the required equality follows since e_{0}(𝔪B) = e_{0}(𝔪) and e_{1}(𝔪B) − e_{1}(QB) = e_{1}(𝔪) − e_{1}(Q).

Corollary 3.3 Let Q be a parameter ideal which is a reduction of 𝔪. Then e_{0}(𝔪) = 1 if and only if e_{1}(𝔪) = e_{1}(Q).

Proof. Because 0 ≤ e_{0}(𝔪) − 1 = e_{0}(𝔪) − ℓ_{A}(A/𝔪) ≤ e_{1}(𝔪) − e_{1}(Q), we get e_{0}(𝔪) = 1 if e_{1}(𝔪) = e_{1}(Q). In order to prove the converse implication, we may assume that A is complete. Now suppose that e_{0}(𝔪) = 1. Let 𝔞(𝔭) be the 𝔭-primary component of a primary decomposition of 0. We set V = ⋂_{𝔭∈Assh A} 𝔞(𝔭), where Assh A denotes the set of associated primes of A whose coheight is d, and B = A/V. Then dim_{A} V < d and e_{0}(𝔪B) = e_{0}(𝔪) = 1, which implies that B is a regular local ring. Hence we have 𝔪 = Q + V, so that 𝔪^{2} ⊆ Q𝔪 + V. Therefore, by 3.2 it follows that e_{1}(𝔪) = e_{1}(Q).

4 The case where Q is a standard ideal

Lemma 4.1 Let d ≥ 2 and Q = (a_{1}, a_{2}, ⋯, a_{d}) be a standard parameter ideal of A. We set a = a_{1}, b = a_{d}, J = (a_{1}, a_{2}, ⋯, a_{d−1}) and K = (a_{2}, a_{3}, ⋯, a_{d}). Then we have the following.

(1) aJ :_{A} b^{2} = aJ :_{A} b.

(2) aJ ∩ bA ⊆ aJI provided Σ(Q) ⊆ I.

(3) I^{2} ⊆ QI + W provided Σ(Q) ⊆ I, I^{2} ⊆ JI + [bA :_{A} a] and I^{2} ⊆ KI + [aA :_{A} b].

Proof. (1) Let us take any x ∈ aJ :_{A} b^{2} and write b^{2}x = ay, with y ∈ J. Then, as y ∈ [b^{2}A :_{A} a] ∩ (b^{2}, a_{1}, ⋯, a_{d−1}), there exists z ∈ A such that y = b^{2}z. Here we notice that bz ∈ J since z ∈ J :_{A} b^{2} = J :_{A} b. On the other hand, as b^{2}x = ab^{2}z, we have bx − abz ∈ [0 :_{A} b] ∩ bA = 0, so that bx = a · bz ∈ aJ. Thus we get aJ :_{A} b^{2} ⊆ aJ :_{A} b and the converse inclusion is obvious.

(2) Let us take any ξ ∈ aJ ∩ bA and write ξ = ay = bz, with y ∈ J and z ∈ A. Moreover, we write y = a_{1}y_{1} + ⋯ + a_{d−1}y_{d−1}, with y_{1}, ⋯, y_{d−1} ∈ A. It is enough to show y_{i} ∈ I for any 1 ≤ i ≤ d − 1. However, as y_{1} ∈ K :_{A} a^{2} = K :_{A} a ⊆ Σ(Q) ⊆ I, we may consider only the case that d ≥ 3 and 2 ≤ i ≤ d − 1. Because ay_{1} ∈ K, we can express ay_{1} = a_{2}z_{2}+⋯+a_{d}z_{d}, with z_{2}, ⋯, z_{d} ∈ A. Then z_{i} ∈ (a_{1}, ⋯, a_{i−1}, a_{i+1}, ⋯, a_{d}) :_{A} a_{i} ⊆ I for any 2 ≤ i ≤ d. On the other hand, as

bz = a(a_{2}z_{2} + ⋯ + a_{d}z_{d}) + aa_{2}y_{2} + ⋯ + aa_{d−1}y_{d−1}
= aa_{2}(y_{2} + z_{2}) + ⋯ + aa_{d−1}(y_{d−1} + z_{d−1}) + aa_{d}z_{d} ,

it follows that

\begin{aligned}y_{i} + z_{i} & ∈ (a_{2}, ⋯, a_{i−1}, a_{i+1}, ⋯, a_{d}) :_{A} aa_{i} \\ & ⊆ (a_{2}, ⋯, a_{i−1}, a_{i+1}, ⋯, a_{d}) :_{A} a_{i} ⊆ I \end{aligned}

for 2 ≤ i ≤ d − 1, and hence we get y_{i} ∈ I.

(3) It is enough to show [aA :_{A} b] ∩ I^{2} ⊆ JI + W. Let us take any x ∈ [aA :_{A} b] ∩ I^{2}. Then, bx = ay for some y ∈ A, and ax = aξ + bz for some ξ ∈ JI and z ∈ A. From these equalities we get a^{2}y = abξ + b^{2}z. Hence z ∈ aJ :_{A} b^{2} = aJ :_{A} b, so that bz = aη for some η ∈ JI. Then it follows that ax = aξ + aη, which implies x − ξ − η ∈ 0 :_{A} a = W. Thus we have x ∈ JI + W and the proof is completed.

Proof of Theorem 1.1. We prove by induction on d. By 2.1 and 2.4 we get the assertion when d = 1. Suppose that d ≥ 2. As we may assume that A/𝔪 is infinite, it is possible to choose a minimal system of generators a_{1}, ⋯, a_{d} of Q such that a_{1} and a_{d} are superficial for both of I and Q. We set a = a_{1}, b = a_{d}, B = A/aA, J = (a_{1}, ⋯, a_{d−1}), K = (a_{2}, ⋯, a_{d}) and Q_{i} = (a_{1}, ⋯, a_{i−1}, a_{i+1}, ⋯, a_{d}) for 1 ≤ i ≤ d. Because e_{0}(I) = e_{0}(IB), ℓ_{A}(A/I) = ℓ_{B}(B/IB) and e_{1}(I) − e_{1}(Q) = e_{1}(IB) − e_{1}(KB), by the inductive hypothesis we have e_{0}(I) − ℓ_{A}(A/I) = e_{1}(I) − e_{1}(Q) if and only if I^{2}B ⊆ KIB + H_{𝔪}^{0}(B) and Σ(KB) ⊆ IB, which holds if I^{2} ⊆ QI + W and Σ(Q) ⊆ I since W B ⊆ H_{𝔪}^{0}(B) and Σ(KB) ⊆ Σ(Q)B. Now we assume that e_{0}(I) − ℓ_{A}(A/I) = e_{1}(I) − e_{1}(Q). Then it follows that I^{2} ⊆ KI + [aA :_{A} b] and Q_{i} :_{A} a_{i} ⊆ I for 2 ≤ i ≤ d. Moreover, by passing A/bA we get I^{2} ⊆ JI + [bA :_{A} a] and Q_{i} :_{A} a_{i} ⊆ I for 1 ≤ i ≤ d − 1. Therefore, as Σ(Q) ⊆ I, we have I^{2} ⊆ QI + W by 4.1 and the proof is completed.

Proof of Corollary 1.2. We may assume that A/𝔪 is infinite. Then any ideal of A has a minimal reduction, so that by 2.4 and 3.1 we have

e_{0}(I) − ℓ_{A}(A/I) − e_{1}(I) ≤ ∑_{i=1}^{d−2} \binom{d − 2}{i − 1}h^{i}(A)

for any 𝔪-primary ideal I. Hence it is enough to find an 𝔪-primary ideal for which the equality holds. Let x_{1}, ⋯, x_{d} be an sop for A contained in 𝔪^{2} and n_{1}, ⋯, n_{d} be integers not less than 2. We set Q = (x_{1}^{n_{1}}, ⋯, x_{d}^{n_{d}}) and I = Q :_{A} 𝔪. Then Q is a standard parameter ideal by [10, Proposition 2.1] and I^{2} = QI by [3]. Because we obviously have Σ(Q) ⊆ I, by 1.1 and 2.4 it follows that

e_{0}(I) − ℓ_{A}(A/I) − e_{1}(I) = ∑_{i=1}^{d−2} \binom{d − 2}{i − 1}h^{i}(A) ,

and the proof is completed. Example 4.2 Let R = k[[X, Y, Z, W]] be the formal power series ring with variables X, Y, Z and W over an infinite field k. Let 𝔞 = (X^{2}, Y)R, 𝔟 = (Z, W)R and A = R/𝔞 ∩ 𝔟. Let x, y, z and w respectively denote the images of X, Y, Z and W in A. We set Q = (x − z, y − w)A and 𝔪 = (x, y, z, w)A. Then we have the following assertion.

(1) dim A = 2, depth A = 1, h^{1}(A) = 2 and A is not a quasi-Buchsbaum ring.

(2) 𝔪^{3} = Q𝔪^{2}, but 𝔪^{2} ≠ Q𝔪.

(3) If V is an ideal of A with dim_{A} V < 2, then V = 0, so that 𝔪^{2} ⊈ Q𝔪 + V.

(4) e_{0}(𝔪) = 3, e_{1}(𝔪) = 1 and e_{1}(Q) = −1, so that e_{0}(𝔪) − ℓ_{A}(A/𝔪) = e_{1}(𝔪) − e_{1}(Q). Proof. From the exact sequence 0 ⟶ A ⟶ R/𝔞 ⊕ R/𝔟 ⟶ R/𝔞 + 𝔟 ⟶ 0, we get the assertion (1). One can directly check the assertion (2). Because dim A/𝔭 = 2 for any 𝔭 ∈ Ass A, we have the assertion (3). The associated graded ring G(𝔪) of 𝔪 is isomorphic to

k[X, Y, Z, W]/(X^{2}, Y) ∩ (Z, W) ,

so that we have the exact sequence

0 ⟶ G(𝔪) ⟶ k[X, Z, W]/(X^{2}) ⊕ k[X, Y] ⟶ k[X]/(X^{2}) ⟶ 0 .

This implies that the Poincaré series P(G(𝔪), λ) of G(𝔪) is

\frac{1 + λ}{(1 − λ)^{2}} + \frac{1}{(1 − λ)^{2}} − (1 + λ) ,

from which it follows that

ℓ_{A}(A/𝔪^{n+1}) = \frac{3}{2}n^{2} + \frac{7}{2}n

for n ≥ 2. Hence e_{0}(𝔪) = 3 and e_{1}(𝔪) = 1. Because k is infinite, there exists μ ∈ k such that c = (x − z) + μ(y − w) is a superficial element of Q. Let B = A/cA. Then e_{1}(Q) = e_{1}(QB) = −h^{0}(B) and the exact sequence 0 ⟶ A ⟶^{c} A ⟶ B ⟶ 0 yields the exact sequence

0 ⟶ H_{𝔪}^{0}(B) ⟶ H_{𝔪}^{1}(A) ⟶^{c} H_{𝔪}^{1}(A) .

Because H_{𝔪}^{1}(A) ≅ R/𝔞 + 𝔟 ≅ k[[X]]/(X^{2}) and (X − Z) + μ(Y − W) ≡ X mod 𝔞 + 𝔟, we have H_{𝔪}^{0}(B) ≅ [(X^{2}) :_{k[[X]]} X]/(X^{2}) = (X)/(X^{2}). Thus we get e_{1}(Q) = −1 and the proof is completed.

5 Buchsbaumness of G(I)

Throughout this section we assume that I contains a parameter ideal Q = (a_{1}, ⋯, a_{d}) as a reduction. We set R = R(I) and G = G(I). The graded maximal ideal of G is denoted by M. Furthermore, we set f_{i} = a_{i}t ∈ R for 1 ≤ i ≤ d. For certain elements x_{1}, ⋯, x_{n} of a ring S and an S-module L, we denote by e(x_{1}, ⋯, x_{n}; L) the multiplicity symbol of x_{1}, ⋯, x_{n} with respect to L (cf. [10, p. 24]).

Lemma 5.1 e(f_{1}^{n_{1}}, ⋯, f_{d}^{n_{d}}; G_{M}) = e(a_{1}^{n_{1}}, ⋯, a_{d}^{n_{d}}; A) for any n_{1}, ⋯, n_{d} > 0.

Proof. Let G_{+} be the ideal of G generated by homogeneous elements of positive degree. As (f_{1}, ⋯, f_{d})G is a reduction of G_{+}, we have e(f_{1}, ⋯, f_{d}; G_{M}) = e_{0}((G_{+})_{M}). On the other hand, as ℓ_{G_{M}} (G/(G_{+})^{n}) = ℓ_{A}(A/I^{n}) for any n > 0, we have e_{0}((G_{+})_{M}) = e_{0}(I). Hence it follows that e(f_{1}, ⋯, f_{d}; G_{M}) = e(a_{1}, ⋯, a_{d}; A). Therefore, for any n_{1}, ⋯, n_{d} > 0

\begin{aligned}e(f_{1}^{n_{1}}, ⋯, f_{d}^{n_{d}}; G_{M}) & = n_{1}n_{2} ⋯ n_{d} · e(f_{1}, ⋯, f_{d}; G_{M}) \\ & = n_{1}n_{2} ⋯ n_{d} · e(a_{1}, ⋯, a_{d}; A) = e(a_{1}^{n_{1}}, ⋯, a_{d}^{n_{d}}; A) . \end{aligned}

Thus we get the required equality.

In the rest of this section, we furthermore assume that Q is a standard ideal such that I^{2} ⊆ QI + W, I^{3} ⊆ Q and Σ(Q) ⊆ I.

Lemma 5.2 Let n_{1}, ⋯, n_{d} be positive integers. Then

(a_{1}^{n_{1}}, ⋯, a_{i}^{n_{i}}) ∩ I^{n} = ∑_{j=1}^{i} a_{j}^{n_{j}}I^{n−n_{j}}

for any n ∈ ℤ and 1 ≤ i ≤ d. Hence we have

G/(f_{1}^{n_{1}}, ⋯, f_{i}^{n_{i}})G ≅ G(IB) ,

where B = A/(a_{1}^{n_{1}}, ⋯, a_{i}^{n_{i}}).

Proof. We may assume that n > n_{j} for any 1 ≤ j ≤ i. Let x ∈ (a_{1}^{n_{1}}, ⋯, a_{i}^{n_{i}}) ∩ I^{n}. Then, as x ∈ Q ∩ (Q^{n−1}I + W) = Q^{n−1}I, we can express

x = ∑_{λ∈Λ} y_{λ}a^{λ} (y_{λ} ∈ I) ,

where Λ is the set of λ = (λ_{1}, ⋯, λ_{d}) ∈ ℤ^{d} such that λ_{1} + ⋯ + λ_{d} = n − 1 and a^{λ} = a_{1}^{λ_{1}}a_{2}^{λ_{2}} ⋯ a_{d}^{λ_{d}}. On the other hand, as

x ∈ (a_{1}^{n_{1}}, ⋯, a_{i}^{n_{i}}) ∩ Q^{n−1} = ∑_{j=1}^{i} a_{j}^{n_{j}}Q^{n−1−n_{j}} ,

we can write

x = ∑_{γ∈Γ} z_{γ}a^{γ} (z_{γ} ∈ A) ,

where Γ = {γ ∈ Λ | γ_{j} ≥ n_{j} for some 1 ≤ j ≤ i}. It is enough to show that z_{γ} ∈ I for any γ ∈ Γ.

Let B = A[T_{1}, ⋯, T_{d}] be the polynomial ring with variables T_{1}, ⋯, T_{d} over A and φ : B ⟶ R(Q) be the homomorphism of A-algebras such that φ(T_{j}) = f_{j} for 1 ≤ j ≤ d. Because a_{1}, ⋯, a_{d} is a d-sequence, ker φ is generated by homogeneous elements of degree one (cf. [5]), so that ker φ ⊆ IB as Σ(Q) ⊆ I. Now we set

f = ∑_{λ∈Λ\Γ} y_{λ}T^{λ} + ∑_{γ∈Γ}(y_{γ} − z_{γ})T^{γ} .

Then f ∈ ker φ. Hence we get z_{γ} ∈ I for any γ ∈ Γ.

Lemma 5.3 We have

(1) [0 :_{G} f_{1}]_{n} = \{\overline{wt^{n}} | w ∈ W ∩ I^{n}\},

(2) 0 :_{G} f_{1} = [0 :_{G} f_{1}]_{1} ⊕ [0 :_{G} f_{1}]_{2},

(3) ℓ_{G_{M}} (0 :_{G} f_{1}) = ℓ_{A}(W), and hence depth G > 0 if depth A > 0.

Proof. (1) Let x ∈ I^{n} and \overline{xt^{n}} ∈ 0 :_{G} f_{1}. Then a_{1}x ∈ I^{n+2}, so that by 5.2 we have a_{1}x = a_{1}y for some y ∈ I^{n+1}, which implies x ∈ I^{n+1} + W since x − y ∈ 0 :_{A} a_{1} = W. Hence \overline{xt^{n}} = \overline{wt^{n}} for some w ∈ W ∩ I^{n}. Thus we get [0 :_{G} f_{1}]_{n} ⊆ \{\overline{wt^{n}} | w ∈ W ∩ I^{n}\}, and the converse inclusion is obvious.

(2) This follow from the assertion (1) as W ∩ I^{n} ⊆ W ∩ Q = 0 for n ≥ 3.

(3) We get this assertion since [0 :_{G} f_{1}]_{1} ≅ W/W ∩ I^{2} and [0 :_{G} f_{1}]_{2} ≅ W ∩ I^{2}.

Lemma 5.4 f_{1}, ⋯, f_{d} is a standard system of parameters for G_{M}. In particular, it follows that H_{M}^{0}(G) = 0 :_{G} f_{1}, so that h^{0}(G_{M}) = h^{0}(A). Moreover, we have I(G_{M}) = I(A).

Proof. By 5.2 we have G/(f_{1}, ⋯, f_{d})G ≅ G(I/Q), so that

ℓ_{G_{M}} (G/(f_{1}, ⋯, f_{d})G) = ℓ_{A}(A/Q) .

Similarly, setting 𝔞 = (a_{1}^{2}, ⋯, a_{d}^{2}), we have

ℓ_{G_{M}} (G/(f_{1}^{2}, ⋯, f_{d}^{2})G) = ℓ_{A}(A/𝔞) .

Then, using 5.1 and that a_{1}, ⋯, a_{d} is a standard system of parameters for A, we get

ℓ_{G_{M}} (G/(f_{1}, ⋯, f_{d})G) − e(f_{1}, ⋯, f_{d}; G_{M})
= ℓ_{A}(A/Q) − e(a_{1}, ⋯, a_{d}; A)
= ℓ_{A}(A/𝔞) − e(a_{1}^{2}, ⋯, a_{d}^{2}; A)
= ℓ_{G_{M}} (G/(f_{1}^{2}, ⋯, f_{d}^{2})G) − e(f_{1}^{2}, ⋯, f_{d}^{2}; G_{M}) .

Therefore by [10, Theorem and Definition 17 in Appendix], we have the required assertion.

Lemma 5.5 We have the following.

(1) If 0 < i < d, then H_{M}^{i}(G) is concentrated in degree 1 − i.

(2) a(G) ≤ 2 − d.

Proof. We prove by induction on d. Let d = 1. In this case, the assertion (1) insists nothing. In order to prove the assertion (2), let us consider the exact sequence

0 ⟶ H_{M}^{0}(G)(−1) ⟶ G(−1) ⟶^{f_{1}} G ⟶ G/f_{1}G ⟶ 0 .

This sequence yields the exact sequence

H_{M}^{0}(G/f_{1}G) ⟶ H_{M}^{1}(G)(−1) ⟶^{f_{1}} H_{M}^{1}(G) ⟶ 0 ,

which implies [H_{M}^{1}(G)]_{n−1} ≅ [H_{M}^{1}(G)]_{n} for n ≥ 3 since [G/f_{1}G]_{n} ≅ I^{n}/QI^{n−1} + I^{n+1} = 0 for n ≥ 3. Hence we get [H_{M}^{1}(G)]_{n} = 0 for n ≥ 2, so that a(G) ≤ 1.

Now we assume that d ≥ 2. Let B = A/W. Then the kernel of the graded homomorphism G ⟶ G(IB) of A-algebras induced from the canonical surjection A ⟶ B has finite length, so that we have H_{M}^{i}(G) ≅ H_{M}^{i}(G(IB)) for i > 0. On the other hand, QB is a standard parameter ideal of B such that I^{2}B = QIB and Σ(QB) ⊆ IB. Hence by 5.3 and 5.4 we have that f_{1} is G(IB)-regular and f_{1} · H_{M}^{i}(G(IB)) = 0 for any 0 ≤ i < d.

Furthermore, setting C = B/a_{1}B, we have G(IB)/f_{1}G(IB) ≅ G(IC) by 5.2. Therefore we get the exact sequence

0 ⟶ G(IB)(−1) ⟶^{f_{1}} G(IB) ⟶ G(IC) ⟶ 0 ,

from which we see that H_{M}^{i}(G(IB)) ↪ H_{M}^{i}(G(IC)) for 0 ≤ i < d and H_{M}^{d−1}(G(IB)) is a homomorphic image of H_{M}^{d−2}(G(IC))(1). Because QC = (a_{2}, ⋯, a_{d})C is a standard parameter ideal of C such that I^{2}C = QIC and Σ(QC) ⊆ IC, the inductive hypothesis insists that H_{M}^{i}(G(IC)) = [H_{M}^{i}(G(IC))]_{1−i} for any 0 ≤ i < d − 1 and a(G(IC)) ≤ 3 − d. Now the assertion (1) can be verified easily. In order to see the assertion (2), let us consider the exact sequence

H_{M}^{d−1}(G(IC)) ⟶ H_{M}^{d}(G(IB))(−1) ⟶^{f_{1}} H_{M}^{d}(G(IB)) ⟶ 0 .

If n > 3 − d, then [H_{M}^{d−1}(G(IC))]_{n} = 0, so that [H_{M}^{d}(G(IB))]_{n−1} ≅ [H_{M}^{d}(G(IB))]_{n}. Hence we have [H_{M}^{d}(G)]_{n} ≅ [H_{M}^{d}(G(IB))]_{n} = 0 for any n ≥ 3 − d. Therefore we get the assertion (2) and the proof is completed.

Lemma 5.6 Suppose that a_{1}, ⋯, a_{d} form a weak sequence (cf. [10, Definition 1.1]) in any order. We arbitrary take x_{i} ∈ 𝔪 for 1 ≤ i ≤ d and set ξ_{i} = x_{i} − a_{i}t. Then

(ξ_{1}, ⋯, ξ_{d})G ∩ H_{M}^{0}(G) = 0 .

Proof. Let us take any φ ∈ (ξ_{1}, ⋯, ξ_{d})G ∩ H_{M}^{0}(G). As H_{M}^{0}(G) = 0 :_{G} f_{1} by 5.4, we can express φ = \overline{w_{1}t + w_{2}t^{2}}, with w_{j} ∈ W ∩ I^{j} for j = 1, 2. We would like to show that w_{j} ∈ I^{j+1} for j = 1, 2. For that, we write φ = ∑_{i=1}^{d} \overline{ξ_{i}} · \overline{η_{i}}, with η_{i} ∈ R for 1 ≤ i ≤ d. Taking N ≫ 0, we can express η_{i} = ∑_{j=1}^{N} η_{ij}t^{j} (η_{ij} ∈ I^{j}) for 1 ≤ i ≤ d. Our assumption implies 𝔪W = 0, so that 𝔪I^{2} ⊆ 𝔪QI. Hence I^{j} ⊆ Q for j ≥ 3. Then, by 5.1 we have η_{ij} ∈ QI^{j−1} for j ≥ 3. Furthermore, we can choose η_{i2} in QI since ξ_{i} ∈ 𝔪A[t], I^{2} ⊆ QI + W and 𝔪W = 0. Because

w_{1}t + w_{2}t^{2} ≡ ∑_{i=1}^{d} ξ_{i}η_{i} mod IR ,

we get the following congruence equations:

0 ≡ ∑_{i=1}^{d} x_{i}η_{i0} mod I,
w_{1} ≡ ∑_{i=1}^{d}(x_{i}η_{i1} − a_{i}η_{i0}) mod I^{2} ,
w_{2} ≡ ∑_{i=1}^{d}(x_{i}η_{i2} − a_{i}η_{i1}) mod I^{3} ,
0 ≡ ∑_{i=1}^{d}(x_{i}η_{ij} − a_{i}η_{i,j−1}) mod I^{j+1} for 3 ≤ j ≤ N and
0 ≡ ∑_{i=1}^{d} a_{i}η_{iN} mod I^{N+2} .

The third equation implies w_{2} ∈ Q, so that w_{2} = 0. Hence it is enough to show w_{1} ∈ I^{2}. We need the following.

Claim There exist elements y_{αβ}^{(j)} ∈ I^{j} for any 1 ≤ j ≤ N and 1 ≤ α < β ≤ d such that

∑_{i=1}^{d} a_{i}(η_{ij} + ∑_{α<i} x_{α}y_{αi}^{(j)} − ∑_{i<β} x_{β}y_{iβ}^{(j)}) ∈ I^{j+2} .

If this is true, setting

v_{i} = η_{i1} + ∑_{α<i} x_{α}y_{αi}^{(1)} − ∑_{i<β} x_{β}y_{iβ}^{(1)} ,

we have ∑_{i=1}^{d} a_{i}v_{i} ∈ I^{3} = QI^{2}. Hence there exist v′_{i} ∈ I^{2} for 1 ≤ i ≤ d such that ∑_{i=1}^{d} a_{i}(v_{i} − v′_{i}) = 0. Then, for any 1 ≤ i ≤ d we get

v_{i} − v′_{i} ∈ (a_{1}, ⋯, a_{i−1}, a_{i+1}, ⋯, a_{d}) :_{A} a_{i}
= (a_{1}, ⋯, a_{i−1}, a_{i+1}, ⋯, a_{d}) :_{A} 𝔪 ,

so that x_{i}(v_{i} − v′_{i}) ∈ Q, which implies x_{i}v_{i} ∈ Q as x_{i}v′_{i} ∈ 𝔪I^{2} ⊆ Q. On the other hand, we have ∑_{i=1}^{d} x_{i}v_{i} = ∑_{i=1}^{d} x_{i}η_{i1}, so that q ∈ Q, where q = ∑_{i=1}^{d}(x_{i}η_{i1} − a_{i}η_{i0}). Because w_{1} − q ∈ I^{2}, we have w_{1} − q = q′ + w′ for some q′ ∈ QI and w′ ∈ W. Then, as w_{1} − w′ = q + q′ ∈ Q ∩ W = 0, we get w_{1} ∈ I^{2}.

Proof of Claim. We prove by descending induction on j. First, we set y_{αβ}^{(N)} = 0 for any 1 ≤ α < β ≤ d. Next, we assume that 2 ≤ j ≤ N and we have the required elements y_{αβ}^{(j)}. Of course, y_{αβ}^{(j)} ∈ QI^{j−1} if j ≥ 3. However, even if j = 2 we can choose y_{αβ}^{(j)} in QI^{j−1} since I^{2} ⊆ QI + W and 𝔪W = 0. Now we set

v_{ij} = η_{ij} + ∑_{α<i} x_{α}y_{αi}^{(j)} − ∑_{i<β} x_{β}y_{iβ}^{(j)} .

Let K_{•} = K_{•}(f_{1}, ⋯, f_{d}; G) be the Koszul complex with the differential maps ∂_{p} : K_{p} ⟶ K_{p−1} and let T_{1}, T_{2}, ⋯, T_{d} be the free bases of K_{1}. We set

σ = ∑_{i=1}^{d} \overline{v_{ij}t^{j}} · T_{i} .

Then σ ∈ (f_{1}, ⋯, f_{d})K_{1} as v_{ij} ∈ QI^{j−1} for any 1 ≤ i ≤ d. On the other hand,

∂_{1}(σ) = ∑_{i=1}^{d} f_{i} · \overline{v_{ij}t^{j}} = \overline{(∑_{i=1}^{d} a_{i}v_{ij})t^{j+1}} = 0

in G, so that σ ∈ Z_{1}(K_{•}). Because f_{1}, ⋯, f_{d} is a d-sequence on G, we have

(f_{1}, ⋯, f_{d})K_{1} ∩ Z_{1}(K_{•}) = B_{1}(K_{•}) .

As a consequence, it follows that there exist elements y_{αβ}^{(j−1)} ∈ I^{j−1} for any 1 ≤ α < β ≤ d such that

∂_{2}(∑_{α<β} \overline{y_{αβ}^{(j−1)}t^{j−1}} · T_{α} ∧ T_{β}) = σ .

The left hand side is equal to

∑_{i=1}^{d} \overline{(∑_{α<i} a_{α}y_{αi}^{(j−1)} − ∑_{i<β} a_{β}y_{iβ}^{(j−1)})t^{j}} · T_{i} ,

so that we have

v_{ij} ≡ ∑_{α<i} a_{α}y_{αi}^{(j−1)} − ∑_{i<β} a_{β}y_{iβ}^{(j−1)} mod I^{j+1}

for any 1 ≤ i ≤ d. This implies

∑_{i=1}^{d} x_{i}v_{ij} ≡ ∑_{α<β} a_{α}x_{β}y_{αβ}^{(j−1)} − ∑_{α<β} x_{α}a_{β}y_{αβ}^{(j−1)} mod I^{j+1} .

On the other hand,

∑_{i=1}^{d} x_{i}v_{ij} = ∑_{i=1}^{d} x_{i}η_{ij} ≡ ∑_{i=1}^{d} a_{i}η_{i,j−1} mod I^{j+1} .

Therefore we get

∑_{i=1}^{d} a_{i}(η_{i,j−1} + ∑_{α<i} x_{α}y_{αi}^{(j−1)} − ∑_{i<β} x_{β}y_{iβ}^{(j−1)}) ∈ I^{j+1}

and the proof is completed.

Proof of Theorem 1.3. Only the Buchsbaumness of G is left to show. We prove by induction on d. Because H_{M}^{0}(G) = \{\overline{w_{1}t + w_{2}t^{2}} | w_{1} ∈ W, w_{2} ∈ W ∩ I^{2}\} and 𝔪W = 0, we have M · H_{M}^{0}(G) = 0. Hence G is a Buchsbaum ring if d = 1.

Suppose that d ≥ 2. Let B = A/W and C = B/a_{1}B. Then C and IC inherits the assumption on A and I in 1.3 (cf. Proof of 5.5). Therefore the inductive hypothesis implies that G(IC) is a Buchsbaum ring, so that G(IB) is also a Buchsbaum ring since G(IB)/f_{1}G(IB) ≅ G(IC), f_{1} is G(IB)-regular and f_{1} · H_{M}^{i}(G(IB)) = 0 for any i < d (cf. [10, Proposition 2.19]). Furthermore, it is easy to see that the kernel of the graded homomorphism G ⟶ G(IB) coincides with H_{M}^{0}(G). Thus we get that G/H_{M}^{0}(G) is a Buchsbaum ring.

Let V = 𝔪 + It ⊆ R. Because we may assume that A/𝔪 is infinite, we can choose a system of generators ξ_{1}, ⋯, ξ_{ℓ} of V such that {ξ_{i}}_{i∈Λ} form an sop for G_{M} for any subset Λ ⊆ {1, 2, ⋯, ℓ} with d-elements. In order to prove the Buchsbaumness of G, it is enough to show that

({ξ_{i}}_{i∈Λ})G ∩ H_{M}^{0}(G) = 0

for any Λ stated above (cf. [10, Proposition 2.22]). Let Λ = {i_{1} < i_{2} < ⋯ < i_{d}} and ξ_{i_{k}} = x_{k} − b_{k}t (x_{k} ∈ 𝔪, b_{k} ∈ I) for 1 ≤ k ≤ d. Because (b_{1}t, ⋯, b_{d}t)G + 𝔪G coincides with the M-primary ideal (ξ_{i_{1}}, ⋯, ξ_{i_{d}})G + 𝔪G, we have that b_{1}t, ⋯, b_{d}t is an sop for G/𝔪G. Hence Q′ = (b_{1}, ⋯, b_{d}) is a reduction of I. Then, by our assumption that (i) A is a Buchsbaum ring or (ii) A is a quasi-Buchsbaum ring and I ⊆ 𝔪^{2}, we have that Q′ is a standard parameter ideal of A, and hence by 1.1 we get I^{2} ⊆ Q′I + W and Σ(Q′) ⊆ I. Therefore, by 5.6 we have (ξ_{i_{1}}, ⋯, ξ_{i_{d}}) ∩ H_{M}^{0}(G) = 0 and the proof is completed.

The next example insists that the assumption of 1.3 that I ⊆ 𝔪^{2} is necessary when A is a quasi-Buchsbaum ring but not a Buchsbaum ring.

Example 5.7 Let F = k[[X, Y, Z, W]] be the formal power series ring with variables X, Y, Z and W over a field k. Let 𝔞 = (X, Y)F ∩ (Z, W)F ∩ (X^{2}, Y, Z^{2}, W)F and A = F/𝔞. Let x, y, z and w respectively denote the images of X, Y, Z and W in A. We set 𝔪 = (x, y, z, w)A, a = x − z, b = y − w and Q = (a, b)A. Then we have the following.

(1) A is a 2-dimensional quasi-Buchsbaum ring but not a Buchsbaum ring.

(2) Q is a standard parameter ideal of A such that 𝔪^{2} = Q𝔪 + W. We obviously have Σ(Q) ⊆ 𝔪.

(3) G(𝔪) is not a Buchsbaum ring.

Proof. Let 𝔫 = (X, Y, Z, W)F and 𝔟 = (X, Y)F ∩ (Z, W)F. Then we have the exact sequence 0 ⟶ F/𝔟 ⟶ F/(X, Y)F ⊕ F/(Z, W)F ⟶ F/𝔫 ⟶ 0, which implies that F/𝔟 is a 2-dimensional Buchsbaum ring such that depth F/𝔟 = 1, H_{𝔫}^{1}(F/𝔟) ≅ k and e_{0}(𝔫/𝔟) = 2. Because 𝔟 = 𝔞 + XZF and XZ𝔫 ⊆ 𝔞, considering the exact sequence 0 ⟶ 𝔟/𝔞 ⟶ A ⟶ F/𝔟 ⟶ 0, we get

W = H_{𝔪}^{0}(A) = 𝔟/𝔞 = xzA ≅ k ,
H_{𝔪}^{1}(A) ≅ H_{𝔫}^{1}(F/𝔟) ≅ k ,
e_{0}(𝔪) = e_{0}(𝔫/𝔟) = 2 .

Hence A is a 2-dimensional quasi-Buchsbaum ring with I(A) = h^{0}(A) + h^{1}(A) = 2.

On the other hand, It is easy to see that A/Q ≅ k[[X, Y]]/(X^{3}, XY, Y^{2}) and Q is a reduction of 𝔪. Then ℓ_{A}(A/Q) = 4 and e(a, b; A) = e_{0}(𝔪) = 2, so that ℓ_{A}(A/Q) − e(a, b; A) = I(A), which implies that Q is a standard ideal of A. Because F/𝔟 is a Buchsbaum ring with e_{0}(𝔫/𝔟) = 2 and depth F/𝔟 > 0, by [1] and [2] it follows that F/𝔟 has maximal embedding dimension, so that we have 𝔫^{2} = (X − Z, Y − W)𝔫 + 𝔟. Hence we get 𝔪^{2} = Q𝔪 + W.

Let a′ = x − w and b′ = y − z. Then A/(a′, b′)A ≅ k[[X, Y]]/(X^{2}, XY, Y^{2}) and (a′, b′)A is a reduction of 𝔪. Hence ℓ_{A}(A/(a′, b′)A) = 3 and e(a′, b′; A) = 2, so that ℓ_{A}(A/(a′, b′)A) − e(a′, b′; A) ≠ I(A). Therefore a′, b′ is not a standard sop for A, which implies that A is not a Buchsbaum ring. Then G(𝔪) is also not a Buchsbaum ring since

G(𝔪) ≅ S/{(X, Y)S ∩ (Z, W)S ∩ (X^{2}, Y, Z^{2}, W)S} ,

where S = k[X, Y, Z, W], and the proof is completed.

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DEPARTMENT OF MATHEMATICS SCHOOL OF SCIENCE AND TECHNOLOGY MEIJI UNIVERSITY 214-8571, JAPAN e-mail: goto@math.meiji.ac.jp

DEPARTMENT OF MATHEMATICS AND INFORMATICS SCHOOL OF SCIENCE AND TECHNOLOGY CHIBA UNIVERSITY 263-8522, JAPAN e-mail: nishida@math.s.chiba-u.ac.jp