Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 39 (1998), No. 2, 433-446.

On the Cohen-Macaulayness of the Associated Graded Ring of Certain Monomial Curves

To the memory of Professor Wolfgang Vogel

S. Molinelli  D. P. Patil^{∗}  G. Tamone

Dipartimento di Matematica, Università di Genova via Dodecaneso 35, I-16146 Genova, Italy e-mail: tamone@dima.unige.it molinell@.dima.unige.it

Department of Mathematics, Indian Institute of Science, Bangalore 560 012, India e-mail: patil@math.iisc.ernet.in

Abstract. Let K be a field and let m_{0}, …, m_{e−1} be a sequence of positive integers. Let C be an algebroid monomial curve in the affine e-space 𝔸_{K}^{e} defined parametrically by X_{0} = T^{m_{0}}, …, X_{e−1} = T^{m_{e−1}} and let A be the coordinate ring of C. In this paper we discuss when exactly the associated graded ring gr_{𝔪}(A) is Cohen-Macaulay. If some e − 1 terms of m_{0}, …, m_{e−1} form an arithmetic sequence then we give necessary and sufficient conditions for the Cohen-Macaulayness of the associated graded ring gr_{𝔪}(A).

1. Introduction

It is important and often difficult to compute the Hilbert function of a local ring (A, 𝔪), which is a good measure of singularity of (A, 𝔪) at 𝔪. However, if the associated graded ring gr_{𝔪}(A) is Cohen-Macaulay then the computation of the Hilbert function of A can be reduced to the computation of the Hilbert function of an Artinian local ring, which is simple. Therefore, the question whether the associated graded ring gr_{𝔪}(A) is Cohen-Macaulay is important and considered by many authors. For example, in [8], [9] and [10], Sally proved that if either emdim(A) = mult(A) or mult(A) − 2 ≤ emdim(A) and A is Gorenstein then gr_{𝔪}(A) is Cohen-Macaulay, where emdim(A) and mult(A) denote the embedding dimension and multiplicity of A respectively. There are examples (see (5.1)) such that gr_{𝔪}(A) is Cohen-Macaulay with A non-Gorenstein and mult(A) >> emdim(A).

∗During this work the second author was supported by Consiglio Nazionale delle Ricerche, Italy. This work was done during June 1994 and July 1996, while the second author was visiting Genova University, Genova, Italy, whose hospitality is gratefully acknowledged.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 1998 Heldermann Verlag

In this article we assume that A is the coordinate ring of an algebroid monomial curve in the affine e-space 𝔸_{K}^{e} defined parametrically by X_{0} = T^{m_{0}}, …, X_{e−1} = T^{m_{e−1}} with 0 < m_{0} < m_{1} < ⋯ < m_{e−1} and gcd(m_{0}, …, m_{e−1}) = 1. In special case when the sequence m_{0}, …, m_{e−1} is an arithmetic sequence then gr_{𝔪}(A) is always Cohen-Macaulay (see [3 ; Proposition (1.1)]). In the case of monomial space curves, that is, e = 3, Robbiano and Valla (see [7; Section 3]) gave necessary and sufficient conditions for gr_{𝔪}(A) to be Cohen-Macaulay. In [2] Herzog also considered the case of monomial space curves. However, in concrete examples (particularly if mult(A) is large) to check these conditions is tedious and requires long computation.

Our aim of this paper is to give simple necessary and sufficient condition for gr_{𝔪}(A) to be Cohen-Macaulay and which is easy to check. We consider the case when the sequence m_{0}, …, m_{e−1} is an almost arithmetic sequence, that is, m_{0} < ⋯ < m_{p}, p := e−2 is an arithmetic sequence and n := m_{e−1} is arbitrary. We prove that (in most cases) gr_{𝔪}(A) is Cohen-Macaulay if and only if exactly one inequality between two positive integers holds. This inequality involves the non-negative integers λ, μ, ν, u, v, z, w which were defined in [6], using the explicit description of the standard basis S_{m_{0}} of the semigroup Γ := ∑_{i=0}^{e−2} ℤ^{+}m_{i} + ℤ^{+}n. The explicit description of S_{m_{0}} and the properties of the non-negative integers λ, μ, ν, u, v, z, w were used to find explicit minimal sets of generators for the derivation module and the relation ideal of A (see [5] and [6]). Further in [4] the explicit set of generators for the relation ideal of A and properties of the non-negative integers λ, μ, ν, u, v, z, w were used to prove that these curves are set-theoretic complete intersections. Given integers m_{0}, …, m_{p}, n, it is easy to find the non-negative integers λ, μ, ν, u, v, z, w and then we only have to check only one inequality for gr_{𝔪}(A) to be Cohen-Macaulay.

For an integer t ∈ ℤ^{+}, let q_{t} ∈ ℤ, r_{t} ∈ {1, 2, …, p} be the unique integers defined by the equation t = q_{t}p + r_{t}. Then, in 3.4 and 4.4 we prove the following results:

Case I: m_{p} < n : gr_{𝔪}(A) is Cohen-Macaulay if and only if λ + w ≥ q_{u} + 1 and v ≤ q_{z} + 1 + μ.

Case II: m_{0} < n, m_{p} : if μ ≠ 0 then gr_{𝔪}(A) is Cohen-Macaulay if and only if λ + w ≥ q_{u} + 1 and v ≤ q_{z} + 1 + μ.

Case III: n < m_{0} :

(1) if either z = 0 or ν ≤ q_{u−z} + 1 + v − w then gr_{𝔪}(A) is Cohen-Macaulay.

(2) if μ ≠ 0 and z ≠ 0 then gr_{𝔪}(A) is Cohen-Macaulay if and only if ν ≤ q_{u−z}+1+(v−w).

Therefore for the case of monomial space curves, we can determine when exactly (see the Case I above) the associated graded ring is Cohen-Macaulay, since any three integers are in almost arithmetic sequence with m_{0} < m_{1} < n. This is much more easier to check than, for instance, the necessary and sufficient conditions given by Robbiano and Valla. We also give many examples to illustrate.

For many values of m_{0}, …, m_{p}, n, we used the Computer algebra program CoCoA to test whether or not the associated graded ring gr_{𝔪}(A) is Cohen-Macaulay. This helped us to find the right inequalities.

2. Key-Lemma

In this section we recall the explicit description of the standard basis from [6; §3]. Our proof depends mainly on this description and so we call this description as a Key-Lemma.

We fix the following notations throughout this paper.

2.1. Notations. Let ℤ (respectively ℤ^{+}) denote the set of all (respectively non-negative) integers. For a, b ∈ ℤ, let [a, b] := {i ∈ ℤ | a ≤ i ≤ b}. Unless mentioned otherwise, the symbols a, b, c, d, e, i, j, k, l, m, n, p, q, r, s, t, u, v, w, z denote integers.

Let m_{0}, …, m_{e−1} be a sequence of positive integers with m_{0} < ⋯ < m_{e−1} and gcd(m_{0}, …, m_{e−1}) = 1. We assume that m_{0}, …, m_{e−1} is a minimal set of generators for the semigroup Γ := ∑_{i=0}^{e−1} ℤ^{+}m_{i}.

− Let E := (ℤ^{+})^{e}.

− For i ∈ [0, e − 1], we put e_{i} := (δ_{ij})_{0≤j≤e−1}, where δ_{ij} denote the Kronecker delta.

− For α = ∑_{i=0}^{e−1} a_{i}e_{i}, let ∂(α) := ∑_{i=0}^{e−1} a_{i}m_{i}.

− For h ∈ Γ, let E(h) := {α ∈ E | ∂(α) = h}.

− For α = ∑_{i=0}^{e−1} a_{i}e_{i} ∈ E(h), we put deg(α) := ∑_{i=0}^{e−1} a_{i}.

− For α, β ∈ E(h), we write α ≤_{deg} β if deg(α) ≤ deg(β).

Then ≤_{deg} is an order on E(h) and since E(h) is a finite set, E(h) has maximal elements with respect to the order ≤_{deg}. Let max(E(h)) denote the set of all maximal elements in E(h). Note that all elements of max(E(h)) have the same degree, therefore this degree we shall denote by maxdeg(h).

2.2. Proposition. Let A := K[[T^{m_{0}}, …, T^{m_{e−1}}]] and let 𝔪 be the maximal ideal of A. Let τ_{0}, … τ_{e−1} denote the images of T^{m_{0}}, …, T^{m_{e−1}} in gr_{𝔪}(A) respectively. Then

(1) gr_{𝔪}(A) = K[τ_{0}, …, τ_{e−1}] and a monomial τ_{0}^{a_{0}} ⋯ τ_{e−1}^{a_{e−1}} is non-zero in gr_{𝔪}(A) if and only if ∑_{i=0}^{e−1} a_{i} = maxdeg (∑_{i=0}^{e−1} a_{i}m_{i}).

(2) For a positive integer t and a non-zero monomial τ_{0}^{a_{0}} ⋯ τ_{e−1}^{a_{e−1}} in gr_{𝔪}(A), the monomial τ_{0}^{t} (τ_{0}^{a_{0}} ⋯ τ_{e−1}^{a_{e−1}}) is non-zero in gr_{𝔪}(A) if and only if maxdeg(∑_{i=0}^{e−1} a_{i}m_{i}+tm_{0}) = maxdeg(∑_{i=0}^{e−1} a_{i}m_{i}) + t.

(3) The following statements are equivalent:

(i) gr_{𝔪}(A) is Cohen-Macaulay.

(ii) maxdeg(h + tm_{0}) = maxdeg(h) + t for every h ∈ Γ and t ∈ ℤ^{+}.

Proof. (1) is immediate from definition and (2) follows from (1).

(3) Since gr_{𝔪}(A) is Cohen-Macaulay if and only if τ_{0}^{t} is a non-zero divisor in gr_{𝔪}(A) for every t ≥ 1 by [1; Theorem 7], the equivalence follows from (2). □

To give necessary and sufficient conditions so that the condition (ii) of (3) holds, we shall use the explicit description of the standard basis from [6; §3], which we recall below.

2.3. Standard basis. The set S_{m_{0}} := {z ∈ Γ | z − m_{0} ∉ Γ} is called the standard basis or Apèry set of Γ with respect to m_{0}. It is clear that S_{m_{0}} depends on Γ and m_{0}, but for simplicity we put S := S_{m_{0}}. It is easy to see that S = {s_{0} := 0, s_{1}, …, s_{m_{0}−1}}, where s_{1}, …, s_{m_{0}−1} ∈ Γ are positive integers with the following properties:

(i) s_{i} ≡ i (mod m_{0}) for every i ∈ [0, m_{0} − 1].

(ii) If z ∈ Γ then z ≡ i (mod m_{0}) for a unique i ∈ [0, m_{0} − 1] and z ≥ s_{i}.

2.4. Explicit description of the standard basis of a semigroup generated by almost arithmetic sequence. Put p := e−2 and assume that m_{0}, …, m_{p} is an arithmetic sequence, that is, there is a positive integer d with m_{i} = m_{0} +id for all 0 ≤ i ≤ p. Let n be an arbitrary positive integer with gcd(m_{0}, d, n) = 1. Let A := K[[T^{m_{0}}, …, T^{m_{p}}, T^{n}]], Γ′ := ∑_{i=0}^{p} ℤ^{+}m_{i} and Γ = Γ′ + ℤ^{+}n. Let S := S_{m_{0}} be the standard basis of Γ with respect to m_{0}. For t ∈ ℤ^{+}, let q_{t} ∈ ℤ, r_{t} ∈ [1, p] and g_{t} ∈ Γ′ be defined by t = q_{t}p + r_{t} and g_{t} = q_{t}m_{p} + m_{r_{t}}.

The Key-Lemma from [6; §3] (see also [5; §1]) is the following:

2.5. Key-Lemma. With the notations as above

(1) g_{s} + g_{t} = εm_{0} + g_{s+t} with ε = 1 or 0 according as r_{s} + r_{t} ≤ p or r_{s} + r_{t} > p.

(2) Let u := min{t ∈ ℤ^{+} | g_{t} ∉ S} and v := min{b ≥ 1 | bn ∈ Γ′}. Then there exist unique integers w ∈ [0, v − 1], z ∈ [0, u − 1], λ ≥ 1, μ ≥ 0 such that

(i) g_{u} = λm_{0} + wn ;

(ii) vn = μm_{0} + g_{z} ;

(iii) g_{u−z} + (v − w)n = νm_{0}. Moreover, ν = λ + μ + ε where ε = 1 or 0 according as r_{u−z} < r_{u} or r_{u−z} ≥ r_{u}.

(3) Let V := [0, u − 1] × [0, v − 1], W := [u − z, u − 1] × [v − w, v − 1] and U := V \ W. Then S = {g_{s} + bn | (s, b) ∈ U}. In particular, if (s, b), (t, c) ∈ U with g_{s} + bn ≡ g_{t} + cn (mod m_{0}) then (s, b) = (t, c).

(4) Every element of Γ can be expressed uniquely in the form am_{0} + g_{s} + bn with a ∈ ℤ^{+} and (s, b) ∈ U.

(5) The map E → (ℤ^{+})^{2} defined by ∑_{i=0}^{p+1} a_{i}e_{i} ↦ (∑_{i=0}^{p} a_{i}m_{i}, a_{p+1}n) is a bijection between S and U.

Proof. See [6; §3]. □

In the sequel the symbols u, v, w, z, λ, μ, ν, U and W have the meaning assigned to them by the above Key-Lemma.

In addition, below we prove few technical lemmas which are consequences of the above Key-Lemma and are needed for the proof of the main result.

2.6. Lemma.

(1) deg(γ_{j}) = q_{j} + 1 for every j ∈ ℤ^{+}.

(2) For i, j ∈ ℤ^{+}, there is ε ∈ {0, 1} such that γ_{i}+γ_{j} = εe_{0}+γ_{i+j} and q_{i}+q_{j} +1 = ε+q_{i+j}.

(3) For h ∈ Γ and α ∈ E(h), there exist a, b, j ∈ ℤ^{+} such that deg(α) = a + q_{j} + 1 + b.

Proof. Clear from the definitions and the Key-Lemma 2.5. □

2.7. Lemma.

(1) If n > m_{p} then v < μ + q_{z} + 1.

(2) If n > m_{0} then ν > q_{u−z} + 1 + v − w.

(3) If n < m_{p} then λ + w ≥ q_{u} + 1.

Proof. (1) and (2) are clear from (2)(ii) and (2)(iii) of the Key-Lemma 2.5 respectively.

(3) If λ + w < q_{u} + 1 then g_{u} = q_{u}m_{p} + m_{r_{u}} ≥ (λ + w)m_{p} + m_{r_{u}} > λm_{0} + wn = g_{u}, a contradiction. □

2.8. Lemma. Suppose that λ + w ≥ q_{u} + 1. Then (1) If j ∈ ℤ^{+} , j ≥ u then q_{j} ≤ λ − ε + q_{j−u} + w with ε ∈ {0, 1}.

(2) For every h ∈ Γ and α ∈ E(h), there exist a, b ∈ ℤ^{+} and i ∈ [0, u − 1] such that ae_{0} + γ_{i} + be_{p+1} ∈ E(h) and deg(α) ≤ a + q_{i} + 1 + b.

(3) Let h ∈ Γ and let α be a maximal element in E(h). Then there exist a, b ∈ ℤ^{+} and i ∈ [0, u − 1] such that ae_{0} + γ_{i} + be_{p+1} ∈ E(h) and deg(α) = deg(ae_{0} + γ_{i} + be_{p+1}).

Proof. (1) Since γ_{j} + εe_{0} = γ_{u} + γ_{j−u} with ε ∈ {0, 1} and by assumption deg(γ_{u}) = q_{u} + 1 ≤ λ + w, we have q_{j} = q_{u} + q_{j−u} + 1 − ε ≤ λ + w + q_{j−u} − ε.

(2) By 2.6 (3) there exist a′, b′, j ∈ ℤ^{+} such that deg(α) = a′ + q_{j} + 1 + b′ = deg(a′e_{0} + γ_{j} + b′e_{p+1}). If j < u then there is nothing to prove. If j ≥ u then by the above part (1), we have β := (a′ + λ − ε)e_{0} + γ_{j−u} + (w + b′)e_{p+1} ∈ E(h) where γ_{j} + εe_{0} = γ_{u} + γ_{j−u} with ε ∈ {0, 1} and deg(α) ≤ deg(β). If j − u ≥ u then repeating we get i ∈ [0, u − 1] such that deg(α) ≤ deg(ae_{0} + γ_{i} + be_{p+1}) = a + q_{i} + 1 + b for some a, b ∈ ℤ^{+}.

(3) By the part (2) above there exist a, b ∈ ℤ^{+} and i ∈ [0, u − 1] such that β := ae_{0} + γ_{i} + be_{p+1} ∈ E(h) and deg(α) ≤ deg(β). Now by the maximality of α in E(h), it follows that deg(α) = deg(β). □

2.9. Lemma. If ae_{0} + γ_{i} ∈ E(bn) for some a, b ∈ ℤ^{+} and i ∈ [1, u − 1] then z ≠ 0.

Proof. Suppose that z = 0. Then vn = μm_{0} ≡ 0 (mod m_{0}). Let b_{1} ∈ ℤ^{+} and b_{0} ∈ [0, v − 1] be the unique integers defined by the equation b = b_{1}v + b_{0}. Then g_{i} ≡ bn = b_{1}vn + b_{0}n ≡ b_{0}n (mod m_{0}). This is a contradiction, since i ≠ 0 and both (i, 0), (0, b_{0}) belong to U. □

2.10. Lemma. Suppose that z = 0. If γ_{i} + be_{p+1} ∈ E(am_{0}) for some i ∈ [0, u − 1] and a, b ∈ ℤ^{+} then i = 0.

Proof. Let b_{1} ∈ ℤ^{+} and b_{0} ∈ [0, v − 1] be the unique integers defined by the equation b = b_{1}v + b_{0}. Then from μm_{0} = vn, it follows that g_{i} + b_{0}n ≡ 0 (mod m_{0}). Therefore, since (i, b_{0}) ∈ U, we must have (i, b_{0}) = (0, 0). □

2.11. Lemma. If t ∈ [0, v − 1] then E(tn) = {te_{p+1}}. In particular, maxdeg(tn) = t for every t ∈ [0, v − 1].

Proof. Let t ∈ [0, v −1] and let α ∈ E(tn). Since ∂(α) = tn ∈ S, it follows that α = γ_{i} +be_{p+1} for some i ∈ ℤ^{+} and b ∈ [0, t]. Now, from (t − b)n = g_{i} and the definition of v, it follows that i = 0 and t = b. In particular, α = te_{p+1}. □

2.12. Lemma. Let j ∈ [0, u − 1] and let

I_{j} := {(i, k) ∈ (ℤ^{+})^{2} | j = i + kz and g_{i} + kg_{z} = g_{j}} .

Then

E(g_{j}) = \{ \begin{matrix}\{γ_{j}\} & if μ ≠ 0, \\ \{γ_{i} + kve_{p+1} | (i, k) ∈ I_{j}\} & if μ = 0. \end{matrix}

In particular, (1) if j < z then E(g_{j}) = {γ_{j}} and maxdeg(g_{j}) = q_{j} + 1.

(2) E(g_{z}) = \{ \begin{matrix}\{γ_{z}\}, & if μ ≠ 0, \\ \{γ_{z}, ve_{p+1}\}, & if μ = 0 \end{matrix} and

maxdeg(g_{z}) = \{ \begin{matrix}\{q_{z} + 1\}, & if μ ≠ 0, \\ max\{q_{z} + 1, v\}, & if μ = 0 \end{matrix} .

(3) if μ ≠ 0 and z ≠ 0 then E(g_{u−z}) = {γ_{u−z}} and maxdeg(g_{u−z}) = q_{u−z} + 1.

Proof. Let α ∈ E(g_{j}). Since ∂(α) = g_{j} ∈ S, it follows that α = γ_{i} + be_{p+1} with i ≤ j and b ∈ ℤ^{+}. If b = 0 then i = j and there is nothing to prove, since (j, 0) ∈ I_{j}. Therefore assume that b ≥ 1 and i < j. Then g_{j−i} ≡ bn (mod m_{0}). Therefore, since (j − i, 0) ∈ U, we have b ≥ v. Let b_{0} ∈ [0, v − 1] and b_{1} ∈ ℤ^{+} be the unique integers defined by the equation b = b_{1}v + b_{0}. Then g_{j} = (a + b_{1}μ)m_{0} + g_{i+b_{1}z} + b_{0}n ∈ S, where g_{i} + b_{1}g_{z} = am_{0} + g_{i+b_{1}z}, a ∈ ℤ^{+}. Therefore g_{j−(i+b_{1}z)} = (a + b_{1}μ + ε)m_{0} + b_{0}n, where g_{j} + εm_{0} = g_{j−(i+b_{1}z)} + g_{i+b_{1}z}. Now, since both (j − (i + b_{1}z), 0) ∈ U and (0, b_{0}) ∈ U, it follows that a = 0 = ε, μ = 0, b_{0} = 0 and j − (i + b_{1}z) = 0 and so g_{j} = g_{i} + g_{b_{1}z} = g_{i} + b_{1}g_{z}. Therefore (i, b_{1}) ∈ I_{j} and α = γ_{i} + b_{1}ve_{p+1}. □

3. The case n > m_{0}

Let m_{0}, …, m_{p}, n be an almost arithmetic sequence of positive integers with gcd(m_{0}, …, m_{p}, n) = 1. Let A := K[[T^{m_{0}}, …, T^{m_{p}}, T^{n}]] and let 𝔪 be the maximal ideal of A. Suppose that m_{0}, …, m_{p}, n is a minimal set of generators for the semigroup Γ := ∑_{i=0}^{p} ℤ^{+}m_{i} + ℤ^{+}n.

We continue using the same notations as in Section 2. In particular, remember the Key-Lemma. Sometimes giving reference of the Key-Lemma 2.5 is omitted.

In this section assume that n > m_{0}.

3.1. Proposition. Suppose that v ≤ μ + q_{z} + 1. Then (1) Let a, b, i ∈ ℤ^{+} be such that γ_{i} + be_{p+1} ∈ E(am_{0}). Then a ≥ q_{i} + 1 + b.

(2) Let a, b ∈ ℤ^{+} , i ∈ [0, u − 1] be such that ae_{0} + γ_{i} ∈ E(bn). Then a + q_{i} + 1 ≥ b.

(3) Let a, a′, b, b′ ∈ ℤ^{+} and i, i′ ∈ [0, u − 1] be such that α′ = a′e_{0} + γ_{i′} + b′e_{p+1} ∈ E(am_{0} + g_{i} + bn). If b ≤ b′ then deg(α) ≥ deg(α′), where α := ae_{0} + γ_{i} + be_{p+1}

Proof. (1) This is clear from n > m_{0} and am_{0} = g_{i} + bn.

(2) Since am_{0} + g_{i} = bn, it follows from the definition of v (see the Key-Lemma) that b ≥ v. Therefore

(3.1.a)  am_{0} + g_{i} = μm_{0} + g_{z} + (b − v)n.

and so a ≥ μ, since (i, 0) ∈ U. We now consider two cases i ≤ z and i > z separately.

Case i ≤ z : In this case from the equation (3.1.a) we get

(a − μ + ε)m_{0} = g_{z−i} + (b − v)n

with γ_{z} + εe_{0} = γ_{i} + γ_{z−i}, ε ∈ {0, 1} and so a − μ + ε > q_{z−i} + 1 + b − v, since n > m_{0}. Therefore a + q_{i} + 1 ≥ q_{i} + 1 + q_{z−i} + 1 − ε + μ − v + b = q_{z} + 1 + μ − v + b ≥ b.

Case i > z : In this case, since i − z < u, we have (i − z, 0) ∈ U and so from the equation (3.1.a) we get

(a − μ − ε)m_{0} + g_{i−z} = (b − v)n

with γ_{i} +εe_{0} = γ_{i−z} +γ_{z}, ε ∈ {0, 1} and i−z ≥ 1. Now, since z ≠ 0 by 2.9, it follows the case i ≤ z and by induction on i that a−μ−ε+q_{i−z}+1 ≥ b−v. Therefore a−μ−ε+q_{i−z}+1+q_{z}+1 ≥ b − v + q_{z} + 1. This proves that a + q_{i} + 1 ≥ b, since v ≤ μ + q_{z} + 1 by assumption.

(3) Since (i, 0) ∈ U, b ≤ b′ and am_{0} + g_{i} + bn = a′m_{0} + g_{i′} + b′n, it follows that a′ ≤ a and so (a − a′)m_{0} + g_{i} = g_{i′} + (b′ − b)n. Therefore

{εm_{0} + (a − a′)m_{0} = g_{i′−i} + (b′ − b)n if i ≤ i′ \atop (a − a′)m_{0} + g_{i−i′} = εm_{0} + (b′ − b)n if i > i′}

with

{γ_{i′} + εe_{0} = γ_{i} + γ_{i′−i} if i ≤ i′, \atop γ_{i} + εe_{0} = γ_{i′} + γ_{i−i′} if i > i′, ε ∈ \{0, 1\}}

and so

(3.1.b)  ε + a − a′ ≥ q_{i′−i} + 1 + (b′ − b) by the part (1) above,
(3.1.c)  (a − a′) + q_{i−i′} + 1 ≥ ε + (b′ − b) by the part (2) above.

Now by using 2.6 (2), we get

ε + q_{i} + 1 + a + b ≥ a′ + q_{i} + 1 + q_{i′−i} + 1 + b′
= a′ + ε + q_{i′} + 1 + b′  if i ≤ i′,
q_{i} + 1 + ε + a + b = q_{i′} + 1 + q_{i−i′} + 1 + a + b
≥ a′ + q_{i′} + 1 + ε + b′  if i > i′.

In both the cases, this proves that deg(α) ≥ deg(α′). □

3.2. Proposition. Suppose that λ + w ≥ q_{u} + 1 and v ≤ μ + q_{z} + 1. Let h ∈ Γ and let a ∈ ℤ^{+}, (i, b) ∈ U be such that h = am_{0} + g_{i} + bn. Then maxdeg(h) = a + q_{i} + 1 + b.

Proof. Let α′ ∈ E(h) be an element of maximal degree. Then deg(α′) = maxdeg(h) and in view of 2.8 (3), we may assume that α′ = a′e_{0} +γ_{i′} +b′e_{p+1} with a′, b′ ∈ ℤ^{+} and i′ ∈ [0, u−1]. Put α := ae_{0} + γ_{i} + be_{p+1}. Then α ∈ E(h) and deg(α) ≤ deg(α′) by the maximality of α′. Since (i, b) ∈ U and am_{0} + g_{i} + bn = h = a′m_{0} + g_{i′} + b′n, we have a′ ≤ a. Further, b′ ≥ b. For, if b′ < b then (i, b − b′) ∈ U and g_{i′} ≡ g_{i} + (b − b′)n, which contradicts the fact that (i′, 0) ∈ U. Then deg(α) ≥ deg(α′) by 3.1 (3). Therefore deg(α) = deg(α′) and so α must be a maximal element in E(h). In particular, maxdeg(h) = deg(α) = a + q_{i} + 1 + b.

3.3. Corollary. Suppose that λ + w ≥ q_{u} + 1 and v ≤ μ + q_{z} + 1. Let h ∈ Γ and let t ∈ ℤ^{+}. Then maxdeg(h + tm_{0}) = maxdeg(h) + t.

Proof. Let a ∈ ℤ^{+} and (i, b) ∈ U be such that h = am_{0} +g_{i} +bn. Then h+tm_{0} = (a+t)m_{0}+ g_{i} + bn. Therefore by 3.2 we have maxdeg(h + tm_{0}) = a + t + q_{i} + 1 + b = maxdeg(h) + t. □

3.4. Theorem. If either m_{p} < n, or m_{0} < n < m_{p} and μ ≠ 0 then the following statements are equivalent :

(i) gr_{𝔪}(A) is Cohen-Macaulay.

(ii) maxdeg(h + tm_{0}) = maxdeg(h) + t for all h ∈ Γ and for all t ∈ ℤ^{+}.

(iii) λ + w ≥ q_{u} + 1 and v ≤ μ + q_{z} + 1.

Proof. (i) ⟺ (ii) : is proved in the proposition 2.2.

(ii) ⇒ (iii) : From the equation g_{u} = λm_{0}+wn, (ii) and 2.11, it follows that q_{u}+1 = deg(γ_{u}) ≤ λ + maxdeg(wn) = λ + w. Similarly from the equation vn = μm_{0} + g_{z}, (ii), 2.12 and either m_{p} < n, or m_{0} < n < m_{p} and μ ≠ 0, it follows that v = deg(vn) ≤ μ+maxdeg(g_{z}) = μ+q_{z}+1.

(iii) ⇒ (ii) : is proved in the corollary 3.3. □

3.5. Proposition. Suppose that m_{0} < n < m_{p} and μ = 0. Let z_{1} ∈ ℤ^{+} and z_{0} ∈ [0, z − 1] be the unique integers defined by the equation u − z = z_{1}z + z_{0}. Then there exists an integer ν_{1} ≥ −1 such that (ν + ν_{1})m_{0} + g_{z−z_{0}} = (z_{1} + 1)vn + (v − w)n.

Proof. Since g_{u−z} + bm_{0} = g_{z_{0}} + z_{1}g_{z} = g_{z_{0}} + z_{1}vn for some b ∈ ℤ^{+}, from the equation νm_{0} + g_{z} = g_{u−z} + vn + (v − w)n, it follows that (ν + b − ε)m_{0} + g_{z−z_{0}} = (z_{1} + 1)vn + (v − w)n with ε ∈ {0, 1}. □

3.6. Corollary. Suppose that m_{0} < n < m_{p} and μ = 0. Let z_{1} ∈ ℤ^{+}, z_{0} ∈ [0, z − 1] and ν_{1} ≥ −1 be as in 3.5. If z_{0} ≠ 0 and gr_{𝔪}(A) is Cohen-Macaulay then ν + ν_{1} + q_{z−z_{0}} + 1 ≥ (z_{1} + 1)v + (v − w).

Proof. Since m_{0} < n and 0 ≠ z_{0} < z, by 2.12 and 2.2 we have

ν + ν_{1} + q_{z−z_{0}} + 1 = ν + ν_{1} + maxdeg(g_{z−z_{0}})
= maxdeg ((ν + ν_{1})m_{0} + g_{z−z_{0}}) ≥ (z_{1} + 1)v + (v − w) .  □

3.7. Questions. Suppose that m_{0} < n < m_{p} and μ = 0. Let z_{1} ∈ ℤ^{+}, z_{0} ∈ [0, z − 1] and ν_{1} ≥ −1 be as in 3.5. Then

(1) In the case z_{0} = 0, is gr_{𝔪}(A) always Cohen-Macaulay?

(2) Is the converse of 3.6 true? That is, in the case m_{0} < n < m_{p}, μ = 0, z_{0} ≠ 0 and ν + ν_{1} + q_{z−z_{0}} + 1 ≥ (z_{1} + 1)v + (v − w) where z_{0}, z_{1}, ν_{1} are as in 3.5, is gr_{𝔪}(A) Cohen-Macaulay?

3.8. Remark. If p = 1, m_{0} < n < m_{1} and z_{0} = 0 then in [7; Remark 3 of § 3], it is proved that the question 3.7 (1) has an affirmative answer. In all the examples 5.3 the converse of 3.6 holds.

4. The case n < m_{0}

In this section, we continue using the same notations as in Section 3 and assume that n < m_{0}. Therefore λ + w ≥ q_{u} + 1 by 2.7 (3).

4.1. Lemma. Suppose that z = 0. Then μ ≠ 0 and β := λ_{0}e_{0} + λ_{1}ve_{p+1} is a maximal element in E(λm_{0}), where λ_{1} ∈ ℤ^{+} and λ_{0} ∈ [0, μ − 1] be the unique integers defined by the equation λ = λ_{1}μ + λ_{0}. In particular, maxdeg(λm_{0}) = λ_{1}v + λ_{0}.

Proof. Since z = 0, clearly μ ≠ 0 by 2.5 (2)(ii) and ∂(β) = λ_{0}m_{0} + λ_{1}vn = λ_{0}m_{0} + λ_{1}μm_{0} = λm_{0} and so β ∈ E(λm_{0}). Let α ∈ E(λm_{0}) be an element of maximal degree. Then maxdeg(λm_{0}) = deg(α) and in view of 2.8 (3), we may assume that α = ae_{0} +γ_{i} +be_{p+1} with a, b ∈ ℤ^{+} and i ∈ [0, u − 1]. Let b_{0} ∈ [0, v − 1] and b_{1} ∈ ℤ^{+} be the unique integers defined by the equation b = b_{1}v +b_{0}. Then, since b_{1}vn = b_{1}μm_{0}, we have 0 ≡ ∂(α) ≡ g_{i} +b_{0}n (mod m_{0}). Now, (i, b_{0}) ∈ U (see the Key-Lemma 2.5), since z = 0 and so (i, b_{0}) = (0, 0). This proves that α = ae_{0} + b_{1}ve_{p+1}. If b_{1} > λ_{1} then vn ≤ (b_{1} − λ_{1})vn = (λ_{0} − a)m_{0} < μm_{0} = vn, a contradiction. Therefore b_{1} ≤ λ_{1} and (λ_{1} − b_{1})vn = (a − λ_{0})m_{0} and so (λ_{1} − b_{1})v ≥ a − λ_{0}, since n < m_{0}. This shows, by the maximality of α in E(λm_{0}) that we must have λ_{1} = b_{1} and a = λ_{0}. In particular, β = α is a maximal element in E(λm_{0}). □

4.2. Proposition. Suppose that either z = 0 or ν ≤ q_{u−z} + 1 + (v − w).

(1) Let a, b, i ∈ ℤ^{+} be such that ae_{0} + γ_{i} ∈ E(bn). Then a + q_{i} + 1 ≤ b.

(2) Let a, b ∈ ℤ^{+} , i ∈ [0, u − 1] be such that γ_{i} + be_{p+1} ∈ E(am_{0}). Then a ≤ q_{i} + 1 + b.

(3) Let a, a′, b, b′ ∈ ℤ^{+} and i, i′ ∈ [0, u − 1] be such that α′ = a′e_{0} + γ_{i′} + b′e_{p+1} ∈ E(am_{0} + g_{i} + bn). If b ≤ b′ then deg(α) ≤ deg(α′), where α := ae_{0} + γ_{i} + be_{p+1}

Proof. (1) Clear from n < m_{0} and am_{0} + g_{i} = bn.

(2) If z = 0 then i = 0 by 2.10 and so a ≤ b, since n < m_{0}. Now, assume that ν ≤ q_{u−z} + 1 + v − w. Since g_{i} + bn = am_{0} ≡ 0 (mod m_{0}), we have (i, b) ∉ U and so we have two cases i < u − z and i ≥ u − z separately:

Case i < u − z : In this case b ≥ v. Let j ≥ 1 be an integer with i + j = u − z. Then g_{j} + g_{i} = g_{u−z} + εm_{0} with ε ∈ {0, 1}. Since νm_{0} = g_{u−z} + (v − w)n, we have am_{0} + g_{j} = (ν + ε)m_{0} + (b − v + w)n and so a ≥ ν + ε, since (j, 0) ∈ U. Now, from (a − ν − ε)m_{0} + g_{j} = (b − v + w)n we get a − ν − ε + q_{j} + 1 < b − v + w. Therefore a < b − q_{j} − 1 − v + w + ν + ε = b + q_{i} + 1 − (q_{u−z} + 1 + v − w − ν) ≤ b + q_{i} + 1, since q_{u−z} + ε = q_{i} + q_{j} + 1 by 2.6 (2).

Case i ≥ u−z : In this case we have b ≥ v−w and am_{0} = g_{i}+bn ≥ g_{u−z}+(v−w)n = νm_{0} and so a ≥ ν. Now, since g_{i} + εm_{0} = g_{u−z} + g_{i−u+z} with ε ∈ {0, 1} and νm_{0} = g_{u−z} + (v − w)n, we get (a − ν + ε)m_{0} = g_{i−u+z} + (b − v + w)n. Now, since u − z ≠ 0, it follows by the case i < u − z and induction on i that a − ν + ε ≤ q_{i−u+z} + 1 + (b − v + w) and so a ≤ b + q_{i} + 1 + (ν − (q_{u−z} + 1 + v − w)) ≤ b + q_{i} + 1, since q_{u−z} + q_{i−u+z} + 1 = ε + q_{i} by 2.6 (2).

(3) Since (i, 0) ∈ U, b ≤ b′ and am_{0} + g_{i} + bn = a′m_{0} + g_{i′} + b′n, it follows that a′ ≤ a and so (a − a′)m_{0} + g_{i} = g_{i′} + (b′ − b)n. Therefore

εm_{0} + (a − a′)m_{0} = g_{i′−i} + (b′ − b)n if i ≤ i′
(a − a′)m_{0} + g_{i−i′} = εm_{0} + (b′ − b)n if i > i′

with

γ_{i′} + εe_{0} = γ_{i} + γ_{i′−i} if i ≤ i′,
γ_{i} + εe_{0} = γ_{i′} + γ_{i−i′}  if i > i′, ε ∈ {0, 1}

and so

(4.2.a)  ε + a − a′ ≤ q_{i′−i} + 1 + (b′ − b) by the part (2) above,
(4.2.b)  (a − a′) + q_{i−i′} + 1 < ε + (b′ − b) by the part (1) above.

Now by using 2.6 (2), we get

ε + q_{i} + 1 + a + b ≤ a′ + q_{i} + 1 + q_{i′−i} + 1 + b′
= a′ + ε + q_{i′} + 1 + b′  if i ≤ i′,
q_{i} + 1 + ε + a + b = q_{i′} + 1 + q_{i−i′} + 1 + a + b
≥ a′ + q_{i′} + 1 + ε + b′  if i > i′.

In both the cases, this proves that deg(α) ≤ deg(α′). □

4.3. Corollary. Suppose that either z = 0 or ν ≤ q_{u−z} + 1 + v − w.

(1) Let a, b ∈ ℤ^{+}, i ∈ [0, u − 1] be such that α = ae_{0} + γ_{i} + be_{p+1} is a maximal element in E(am_{0} + g_{i} + bn). Then α + te_{p+1} is also maximal element in E(am_{0} + g_{i} + bn + tn) for every t ∈ ℤ^{+}.

(2) maxdeg(h + tn) = maxdeg(h) + t for every h ∈ Γ and t ∈ ℤ^{+}.

Proof. (1) Put h := am_{0} + g_{i} + bn. It is enough to prove the assertion for t = 1. Let β ∈ E(h + n) be a maximal element. In view of 2.8 (3) we may assume that β = a′e_{0} + γ_{i′} + b′e_{p+1} with a′, b′ ∈ ℤ^{+} and i′ ∈ [0, u − 1]. If b′ ≤ b + 1 then deg(β) ≤ deg(α + e_{p+1}) by 4.2 (3). If b′ > b + 1 then α and β′ = β − e_{p+1} ∈ E(h) and deg(β′) ≤ deg(α) by the maximality of α in E(h) and so deg(β) = deg(β′) + 1 ≤ deg(α) + 1 = deg(α + e_{p+1}). In any case we have deg(α + e_{p+1}) = deg(β) by the maximality of β in E(h + n). In particular α + e_{p+1} is also maximal in E(h + n).

(2) Let h ∈ Γ. Then by 2.8 (3) there is a maximal element in E(h) of the form ae_{0} +γ_{i} +be_{p+1} with a, b ∈ ℤ^{+} and i ∈ [0, u − 1] and the assertion follows from (1). □

4.4. Theorem. Suppose that n < m_{0}.

(1) If either z = 0 or ν ≤ q_{u−z} + 1 + v − w then gr_{𝔪}(A) is Cohen-Macaulay.

(2) If μ ≠ 0 and z ≠ 0 then the following statements are equivalent :

(i) gr_{𝔪}(A) is Cohen-Macaulay.

(ii) maxdeg(h + tn) = maxdeg(h) + t for all h ∈ Γ and for all t ∈ ℤ^{+}.

(iii) ν ≤ q_{u−z} + 1 + v − w.

Proof. (1) is immediate from 4.3 (2) and 2.2.

(2) (i) ⟺ (ii) : is proved in the proposition 2.2.

(ii) ⇒ (iii) : From νm_{0} = g_{u−z} + (v − w)n, (ii) and 2.12, it follows that ν ≤ maxdeg(νm_{0}) ≤ maxdeg(g_{u−z}) + (v − w) = q_{u−z} + 1 + v − w.

(iii) ⇒ (ii) : is proved in the corollary 4.3 (2). □

4.5. Proposition. Suppose that n < m_{0} and μ = 0. Let I_{u−z} be as in 2.12, z_{0} := min{i ∈ ℤ^{+} | (i, k) ∈ I_{u−z} for some k ∈ ℤ^{+}} and let z_{1} ∈ ℤ^{+} be such that (z_{0}, z_{1}) ∈ I_{u−z}. Then

(1) E(g_{z_{0}}) = {γ_{z_{0}}}. In particular, maxdeg(g_{z_{0}}) = q_{z_{0}} + 1.

(2) If gr_{𝔪}(A) is Cohen-Macaulay then ν ≤ q_{z_{0}} + 1 + z_{1}v + (v − w).

Proof. (1) is immediate from 2.12.

(2) Since n < m_{0}, by 2.12, 2.2 and νm_{0} = g_{z_{0}} + z_{1}vn + (v − w)n, we have ν ≤ maxdeg(g_{z_{0}}) + z_{1}v + (v − w) = q_{z_{0}} + 1 + z_{1}v + (v − w). □

4.6. Question. Is the converse of 4.5 (2) true? That is, in the case n < m_{0}, μ = 0 and ν ≤ q_{z_{0}} + 1 + z_{1}v + (v − w) where z_{0}, z_{1} ∈ ℤ^{+} are as in (4.5), is gr_{𝔪}(A) Cohen-Macaulay?

4.7. Remark. In all the examples 5.4 the converse of 4.5 (2) holds.

5. Examples

5.1. Example. Let m_{0}, …, m_{p}, m_{p+1} be an arithmetic sequence and let n := m_{p+1}. Then u = p + 1, λ = 1, w = 1 and so λ + w = 2 = q_{u} + 1. Therefore gr_{𝔪}(A) is Cohen-Macaulay by 3.4.

5.2. Examples. Let m_{0}, m_{1}, n be positive integers with gcd(m_{0}, m_{1}, n) = 1. Suppose that m_{0}, m_{1}, n is a minimal set of generators for the semigroup Γ := ℤ^{+}m_{0} + ℤ^{+}m_{1} + ℤ^{+}n. Then

(1) If m_{0} = 3 and 3 < m_{1} < n then it is easy to see that u = 2 = v and either (λ, w) = (1, 1) or λ ≥ 2. Therefore λ + w ≥ 2 ≥ u = q_{u} + 1 and so by gr_{𝔪}(A) is Cohen-Macaulay by 3.4.

(2) If m_{0} = 4 and 4 < m_{1} < n and the semigroup Γ is symmetric then it is easy to see that 1 < u < 4, 1 < v ≤ 3 and one of the following cases occurs

(a) u = 2, v = 2 and w = 0 and so λ ≥ 2.

(b) u = 2, v = 2 and w = 1.

In any case we have λ + w ≥ 2 = u = q_{u} + 1 and so gr_{𝔪}(A) is Cohen-Macaulay by 3.4.

(3) If m_{0} = 4, m_{1} = 5 and n = 11 then the semigroup Γ is not symmetric and u = 3, λ = 1, w = 1, v = 2 and so q_{u} + 1 = u = 3 > 2 = λ + w. Therefore gr_{𝔪}(A) is not Cohen-Macaulay by 3.4.

(4) If m_{0} = 5 and 5 < m_{1} < n and the semigroup Γ is symmetric then it is easy to see that 1 < u < 5, 1 < v ≤ 3, w ≠ 0 and one of the following cases occurs

(a) u = 2, v = 3 and w = 1.

(b) u = 3, v = 3 and w = 2.

(c) u = 3, v = 2, w = 1, z = 1, μ = 0 and so λ = ν ≥ 2.

(d) u = 4, v = 2, w = 1, z = 3, μ = 0 and λ = ν ≥ 3.

In any case we have λ + w ≥ u = q_{u} + 1 and so gr_{𝔪}(A) is Cohen-Macaulay by 3.4.

(5) If m_{0} = 5, m_{1} = 8 and n = 27 then the semigroup Γ is not symmetric and u = 4, λ = 1, w = 1, v = 2 and so q_{u} + 1 = u = 4 > 2 = λ + w. Therefore gr_{𝔪}(A) is not Cohen-Macaulay by 3.4.

(6) If m_{0} = 6, m_{1} = 7 and n = 15 then the semigroup Γ is symmetric and u = 3, λ = 1, w = 1, v = 2 and so q_{u} + 1 = u = 3 > 2 = λ + w. Therefore gr_{𝔪}(A) is not Cohen-Macaulay by 3.4.

(7) (Sally). Let m_{0} be any integer ≥ 4, m_{1} = m_{0} + 1 and let n := (m_{0} − 1)m_{0} − 1. Then from the equations (m_{0} − 1)m_{1} = m_{0} + n, 2n = m_{0}(m_{0} − 1) + (m_{0} − 2)m_{1} and m_{0}^{2} = m_{1} + n, it follows that u = m_{0} − 1, λ = 1, w = 1, v = 2, μ = m_{0} − 1, z = m_{0} − 2 and ν = m_{0}. Therefore, since λ + w = 2 < m_{0} − 1 = u = q_{u} + 1, gr_{𝔪}(A) is not Cohen-Macaulay by 3.4.

(8) Let m_{0} = 7, m_{1} = 8 and n = 20. Then u = 6, λ = 4, w = 1, v = 2, μ = 0, z = 5, ν = 4 and so λ + w = 5 < 6 = u = q_{u} + 1. Therefore gr_{𝔪}(A) is not Cohen-Macaulay by 3.4.

(9) Let m_{0} = 30, m_{1} = 39 and n = 50. Then u = 10, w = 0, λ = 13, v = 3, z = 0, μ = 5, ν = 18 and so λ + w = 13 > 10 = u = q_{u} + 1. Therefore gr_{𝔪}(A) is Cohen-Macaulay by 3.4.

5.3. Examples with m_{0} < n < m_{p} and μ = 0.

(1) Let m_{0} = 5, m_{1} = 9 and n = 6. In this case, since the semigroup Γ generated by 5, 6, 9 is symmetric, gr_{𝔪}(A) is Cohen-Macaulay by the example 5.2 (4). Further, u = 3, λ = 3, w = 2, v = 3, μ = 0, z = 2, ν = 3, z_{1} = 0, z_{0} = 1, ν_{1} = 0 (see 3.5 for notations) and ν + ν_{1} + q_{z−z_{0}} + 1 = ν + 1 = 4 = (z_{1} + 1)v + v − w.

(2) Let m_{0} = 7, m_{1} = 20 and n = 8. Then u = 3, λ = 4, w = 4, v = 5, μ = 0, z = 2, ν = 4, z_{1} = 0, z_{0} = 1, ν_{1} = 0 (see 3.5 for notations), ν + ν_{1} + q_{z−z_{0}} + 1 = 5 < 6 = (z_{1} + 1)v + (v − w) and gr_{𝔪}(A) is not Cohen-Macaulay by the example 5.2 (8).

5.4. Examples with n < m_{0} and μ = 0.

(1) Let m_{0} = 39, m_{1} = 50 and n = 30. Then u = 9, w = 2, λ = 10, v = 5, z = 3, μ = 0, ν = 10, z_{0} = 0, z_{1} = 2 (see 4.5 for notations) and so q_{u−z} + 1 + (v − w) = 9 < ν = 10 < 13 = q_{z_{0}} + 1 + z_{1}v + v − w and gr_{𝔪}(A) is Cohen-Macaulay by the example 5.1 (9).

(2) Let m_{0} = 15, m_{1} = 17, m_{2} = 19 and n = 12. Then u = 6, w = 1, λ = 3, v = 3, z = 3, μ = 0, ν = 4, z_{0} = 0, z_{1} = 1 (see 4.5 for notations) and so q_{u−z} + 1 + (v − w) = ν = 4 < 5 = q_{z_{0}} + 1 + z_{1}v + v − w and gr_{𝔪}(A) is Cohen-Macaulay by 4.4 (1).

(3) Let m_{0} = 27, m_{1} = 35, m_{2} = 43 and n = 26. Then u = 10, w = 1, λ = 7, v = 3, z = 3, μ = 0, ν = 8, z_{0} = 4, z_{1} = 1 (see (4.5) for notations), ν = 8 > 7 = q_{z_{0}} +1+z_{1}v +v −w and gr_{𝔪}(A) is not Cohen-Macaulay by 4.5 (2).

5.5. Examples with n < m_{0} and z = 0. Let p be any positive integer, n := 2(p + 1) and let m_{0} := 3(p + 1).

(1) Let m_{i} := 2m_{0}+i for 1≤i≤p. Let Γ be the semigroup generated by n, m_{0}, m_{1}, …, m_{p}. Then from the equations: g_{p+1} = 3m_{0} + 2n, 3n = 2m_{0}, 5m_{0} = g_{p+1} + n, it follows that u = p + 1, λ = 3, w = 2, v = 3, μ = 2, z = 0 and ν = 5. Therefore gr_{𝔪}(A) is Cohen-Macaulay by 4.4 (1), but in this case ν = 5 > 3 = q_{u−z} + 1 + v − w.

(2) Let m_{i} := 2m_{0} +3i for 1 ≤ i ≤ p. Let Γ be the semigroup generated by n, m_{0}, m_{1}, …, m_{p}. Then from the equations: g_{p+1} = 5m_{0}, 3n = 2m_{0}, 7m_{0} = g_{p+1} + 3n, it follows that u = p + 1, λ = 5, w = 0, v = 3, μ = 2, z = 0 and ν = 7. Therefore gr_{𝔪}(A) is Cohen-Macaulay by 4.4 (1), but in this case ν = 7 > 5 = q_{u−z} + 1 + v − w.

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Received September 5, 1997