Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 40 (1999), No. 2, 437-458.

On the Cohen-Macaulayness of the Coordinate Ring

of Certain Projective Monomial Curves

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 n_{1}, …, n_{e} be a sequence of positive integers with gcd(n_{1}, …, n_{e}) = 1 and n_{1} < n_{2} < ⋯ < n_{e}. Let A′ be the coordinate ring of the projective monomial curve in the projective e-space ℙ_{K}^{e} defined parametrically by Z_{0} = X^{n_{e}}, …, Z_{i} = X^{n_{e}−n_{i}}Y^{n_{i}}, …, Z_{e} = Y^{n_{e}} where n_{0} := 0. In this paper under some assumptions, we discuss when exactly the graded ring A′ is Cohen-Macaulay and we give numerical criterion for this in terms of the standard basis of the semigroup generated by n_{1}, …, n_{e} in the case when some e − 1 terms of n_{1}, …, n_{e} form an arithmetic sequence. Our special assumptions are satisfied in the case e = 3, in particular, for the class of monomial projective space curves, we get a criterion for arithmetically Cohen-Macaulayness.

1. Introduction

In this article we assume that A′ is the coordinate ring of a projective monomial curve in the projective e-space ℙ_{K}^{e} defined parametrically by Z_{0} = X^{n_{e}}, …, Z_{i} = X^{n_{e}−n_{i}}Y^{n_{i}}, …, Z_{e} = Y^{n_{e}} where n_{0} := 0 and n_{1} < ⋯ < n_{e} is a sequence of positive integers with gcd(n_{1}, …, n_{e}) = 1. In special case when the sequence n_{1}, …, n_{e} is an arithmetic sequence then A′ is always Cohen-Macaulay (see [3; Proposition (1.2)]). Cavaliere and Niesi (see [1; Lemma (3.4)] and [2; Lemma (4.3)]) proved that A′ is Cohen-Macaulay if and only if Z_{0}, Z_{e} is an A′-regular sequence. However, in concrete examples to check that Z_{0}, Z_{e} is an A′-regular sequence is tedious and requires long computation.

∗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} 1999 Heldermann Verlag

Our aim of this paper is to give simple necessary and sufficient conditions for A′ to be Cohen-Macaulay and which are easy to check. We consider the case when the sequence m_{0} := n_{1}, …, m_{e−1} := n_{e} is an almost arithmetic sequence, that is, m_{0} < ⋯ < m_{e−2} is an arithmetic sequence and n := m_{e−1} is arbitrary. We prove that (in most cases) 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} ℕm_{i} +ℕn. 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]). In addition to these non-negative integers we define four more non-negative integers z′, μ′, λ′ and w′. Given integers m_{0}, …, m_{e−2}, n, it is easy to find the non-negative integers λ, μ, ν, u, v, z, w z′, μ′, λ′, w′ and then we only have to check essentially only one inequality for 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 4.8, 5.8 and 6.6 we prove the following results:

• In the case n < m_{0} :

(1) If w = 0 then A′ is Cohen-Macaulay.

(2) If w ≠ 0 then A′ is Cohen-Macaulay if and only if ν ≥ q_{u−z} + 1 + (v − w) .

• In the case m_{0} < n < m_{p} :

(1) If z′ ≥ p then A′ is Cohen-Macaulay if and only if ν ≥ q_{u−z} + 1 + (v − w) and v ≥ μ′ + q_{z′} + 1 .

(2) If 1 ≤ z′ < p then :

(a) If A′ is Cohen-Macaulay then v + 1 ≥ mindeg ((μ′ + 1)m_{0}) + 1 .

(b) If v ≥ μ′ + 1 then A′ is Cohen-Macaulay.

(3) If z′ = 0 then A′ is Cohen-Macaulay if and only if v + w′ ≥ μ − λ′ + q_{u} + 1 .

• In the case m_{p} < n :

(1) If w′ ≠ 0 then A′ is Cohen-Macaulay if and only if q_{u+k} + 1 ≥ mindeg ((λ′ − ε_{k})m_{0} + g_{k}) + w′ for every k ∈ [0, p − r_{u}].

(2) If w′ = 0 then the following statements are equivalent :

(i) A′ is Cohen-Macaulay.

(ii) mindeg(h) = deg (δ(h)) for every h ∈ Γ.

(iii) For every j ∈ ℕ with a(g_{j}) ≥ 1 and b(g_{j}) ≥ 1, there exist unique integers s, t ∈ ℕ^{+} such that j = i(g_{j}) + tz + su and sλ′′ ≤ tv′′, where λ′′ := λ′ − q_{u} − 1 and v′′ := μ + q_{z} + 1 − v ,

where δ(h), a(g_{j}), i(g_{j}) and b(g_{j}) are as in the Definition 6.2.

Since any three integers are in almost arithmetic sequence with n < m_{0} < m_{1}, the Theorem 4.8 determines all monomial projective space curves which are arithmetically Cohen-Macaulay. We also give many examples to illustrate.

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

2. Numerical criterion for Cohen-Macaulayness

In this section we give a numerical criterion for the Cohen-Macaulayness of the coordinate ring of a projective monomial curve.

2.1. Notations. Let ℤ (respectively ℕ, ℕ^{+}) denote the set of all (respectively non-negative, positive) 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 n_{1}, …, n_{e} be a sequence of positive integers with n_{1} < ⋯ < n_{e} and gcd(n_{1}, …, n_{e}) = 1. We assume that n_{1}, …, n_{e} is a minimal set of generators for the semigroup Γ := ∑_{i=1}^{e} ℕn_{i}.

• Let A′ := K[Y^{n_{e}}, X^{n_{1}}Y^{n_{e}−n_{1}}, …, X^{n_{i}}Y^{n_{e}−n_{i}}, …, X^{n_{e}}]. Let n_{0} := 0 and for 0 ≤ i ≤ e, let Z_{i} := X^{n_{i}}Y^{n_{e}−n_{i}}.

• Let E := (ℕ)^{e}.

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

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

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

• For α = ∑_{i=1}^{e} a_{i}e_{i} ∈ E(h), we put deg(α) := ∑_{i=1}^{e} 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 minimal elements with respect to the order ≤_{deg}. Let min(E(h)) denote the set of all minimal elements in E(h). Note that all elements of min(E(h)) have the same degree, therefore this degree we shall denote by mindeg(h).

2.2. Lemma. With the notations as above,

(1) Let (a_{0}, a_{1}, …, a_{e}), (b_{0}, b_{1}, …, b_{e}) ∈ (ℕ)^{e+1}. Then Z_{0}^{a_{0}}Z_{1}^{a_{1}} ⋯ Z_{e}^{a_{e}} = Z_{0}^{b_{0}}Z_{1}^{b_{1}} ⋯ Z_{e}^{b_{e}} in A′ if and only if ∑_{i=0}^{e} a_{i} = ∑_{i=0}^{e} b_{i} and ∑_{i=0}^{e} a_{i}n_{i} = ∑_{i=0}^{e} b_{i}n_{i}.

(2) If Z_{1}^{a_{1}} ⋯ Z_{e}^{a_{e}} = Z_{0}^{b_{0}}Z_{1}^{b_{1}} ⋯ Z_{e}^{b_{e}} with b_{0} > 0 then ∑_{i=1}^{e} a_{i} > ∑_{i=1}^{e} b_{i}.

(3) Z_{1}^{a_{1}} ⋯ Z_{e}^{a_{e}} ∉ Z_{0}A′ if and only if mindeg (∑_{i=1}^{e} n_{i}a_{i}) = ∑_{i=1}^{e} a_{i}.

Proof. (1) is clear and (2) follows from (1).

(3) A monomial Z_{1}^{a_{1}} ⋯ Z_{e}^{a_{e}} ∈ Z_{0}A′ if and only if there exists (b_{0}, b_{1}, …, b_{e}) ∈ (ℕ)^{e+1} such that b_{0} ≠ 0 and Z_{1}^{a_{1}} ⋯ Z_{e}^{a_{e}} = Z_{0}^{b_{0}}Z_{1}^{b_{1}} ⋯ Z_{e}^{b_{e}}. Therefore the assertion follows from (1) and (2). □

2.3. Proposition. The following statements are equivalent:

(i) A′ is Cohen-Macaulay.

(ii) mindeg(h + tn_{e}) = mindeg(h) + t for every h ∈ Γ and t ∈ ℕ.

Proof. Since A′ is Cohen-Macaulay if and only if Z_{e}^{t} is a non-zero divisor in A′/Z_{0}A′ for every t ≥ 1 by [2; Lemma (4.3)], the equivalence follows from 2.2 (3). □

3. Key-Lemma and its consequences

We continue using the notations introduced in 2.1, except that the sequence n_{1}, …, n_{e} will be denoted by m_{0}, …, m_{e−1}. This is convenient for using the same notations and results from the papers [4], [5] and [6].

3.1. Standard basis. The set S_{m_{0}} := {z ∈ Γ | z − m_{0} ∉ Γ} is called the standard basis or Apéry 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}.

3.2. 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 Γ′ := ∑_{i=0}^{p} ℕm_{i} and Γ = Γ′ + ℕn. Let S := S_{m_{0}} be the standard basis of Γ with respect to m_{0}. For t ∈ ℕ, let q_{t} ∈ ℤ, r_{t} ∈ [1, p] , g_{t} ∈ Γ′ and γ_{t} ∈ E(g_{t}) be defined by the equations t = q_{t}p + r_{t} , g_{t} = q_{t}m_{p} + m_{r_{t}} and γ_{t} = q_{t}e_{p} + e_{r_{t}} respectively.

The Key-Lemma from [6; §3] (see also [4; §2]) is the following:

3.3. Key-Lemma. With the notations as in 3.2,

(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 ∈ ℕ | 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 ∈ ℕ and (s, b) ∈ U.

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

Proof. See [6 ; §3]. □

3.4. Definition. Let z′ := max{i ∈ ℕ | vn = am_{0} + g_{i} for some a ∈ ℕ} and let λ′ := min{a ∈ ℕ^{+} | g_{u} = am_{0} + bn for some b ∈ ℕ}.

3.5. Lemma.

(1) z′ ≥ z and there is a unique integer μ′ ∈ ℕ such that vn = μ′m_{0} + g_{z′}. In particular, if z′ = 0 then z = 0 and μ′ = μ.

(2) λ′ ≤ λ and there is a unique integer w′ ∈ ℕ such that g_{u} = λ′m_{0} + w′n.

(3) If z′ > z then z′ ≥ u. Moreover, if w ≠ 0 then z′ = z and μ′ = μ and if w = 0 then z′ = su + z for some s ∈ ℕ.

(4) If 1 ≤ z′ < p then z = z′.

(5) If w′ > w then w′ ≥ v. Moreover, if z ≠ 0 then w′ = w and if z = 0 then w′ = tv + w for some t ∈ ℕ.

Proof. (1) and (2) clear from the definitions.

(3) If z′ ≠ z then from g_{z} ≡ g_{z′}(mod m_{0}) and the Key-Lemma, we have z′ ≥ u. Now, from the equations g_{u} = λm_{0} + wn and εm_{0} + g_{z′} = g_{u} + g_{z′−u} with ε ∈ {0, 1}, we get μm_{0} + g_{z} = vn = μ′m_{0} + g_{z′} = (μ′ + λ − ε)m_{0} + g_{z′−u} + wn. If w ≠ 0 this contradicts the definition of v. Let s ∈ ℤ^{+}, j ∈ [0, u − 1] be the unique integers defined by the equation z′ = su + j. If w = 0 then g_{u} ≡ 0(mod m_{0}) and so g_{z} ≡ g_{z′} ≡ g_{j}(mod m_{0}). Therefore z = j by the Key-Lemma.

(4) follows from z′ = su + z and u > p.

(5) If w′ ≠ w then from wn ≡ w′n(mod m_{0}) and the Key-Lemma, we have w′ ≥ v. Now, from the equations vn = μm_{0} +g_{z}, g_{u} = λ′m_{0} +w′n and εm_{0} +g_{u} = g_{z} +g_{u−z} with ε ∈ {0, 1}, we get g_{u−z} = (λ′ + μ + ε)m_{0} + (w′ − v)n ∉ S. If z ≠ 0 this contradicts the definition of u (see Key-Lemma). Let t ∈ ℤ^{+}, b ∈ [0, v − 1] be the unique integers defined by the equation w′ = tv + b. If z = 0 then wn ≡ g_{u} = (λ′ + tμ)m_{0} + bn ≡ bn(mod m_{0}) and so w = b by the Key-Lemma. □

In the sequel the symbols u, v, w, z, λ, μ, ν, U, W and z′, μ′, λ′, w′ have the meaning assigned to them by the above Key-Lemma, 3.4 and 3.5. Sometimes giving reference of the Key-Lemma 3.3 is omitted.

3.6. Lemma. Suppose that z′ = 0. Then

(1) μ ≥ λ′.

(2) If w = 0 and λ > λ′ then λ ≥ λ′ + μ.

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

(4) (v + w′)e_{p+1} ∈ E ((μ − λ′)m_{0} + g_{u}).

Proof. Since z′ = 0, z = 0 and μ′ = μ by 3.5 (1).

(1) From vn = μm_{0} and v ≥ 1, we have μ > 0. If λ′ = μ + a with a ∈ ℕ^{+} then a < λ′. Therefore from the equations vn = μm_{0} and g_{u} = λ′m_{0} + w′n, we get g_{u} = am_{0} + (v + w′)n a contradiction to the definition of λ′.

(2) Since w = 0 and λ > λ′, w′ > 0 and λm_{0} = g_{u} = λ′m_{0} + w′n and so w′ ≥ v by the definition of v. Therefore (λ − λ′)m_{0} = w′n ≥ vn = μm_{0} and so λ − λ′ ≥ μ.

(3) If n < m_{p} and λ′ + w′ < q_{u} + 1 then q_{u}m_{p} + m_{r_{u}} = g_{u} = λ′m_{0} + w′n < (λ′ + w′)m_{p} ≤ q_{u}m_{p} < g_{u}, a contradiction.

(4) Follows from (1) and the equations vn = μm_{0} and g_{u} = λ′m_{0} + w′n. □

3.7. Proposition.

(1) mindeg(μ′m_{0}) = min{v, μ′}.

(2) If z′ ≠ 0 then mindeg(μ′m_{0} + m_{r_{z′}}) = min{v, μ′ + 1}.

(3) Suppose that z′ = 0. Then

(a) mindeg(λ′m_{0}) ∈ {v, λ′, q_{u} + 1}.

(b) mindeg ((μ − λ′)m_{0} + m_{r_{u}}) = (μ − λ′) + 1.

Proof. Let j ∈ {0, r_{z′}} and let α be a minimal element in E(μ′m_{0} + g_{j}). Then, in view of [4; Lemma 2.6 (3)] we may assume that α = ae_{0} + γ_{i} + be_{p+1}. From vn = μ′m_{0} + g_{z′} and am_{0} + g_{i} + bn = μ′m_{0} + g_{j}, we have

(3.7.a)  vn = \{ \begin{matrix}(a + ε)m_{0} + g_{i+z′} + bn, & if j = 0, \\ am_{0} + g_{z′−j+i} + bn & if j = r_{z′}. \end{matrix}

with ε ∈ {0, 1}.

(1) For j = 0 from the equation (3.7.a) and the definitions of v and z′, it follows that either b = v and a = 0 = i, or b = 0, i = 0 and a = μ′.

(2) Since z′ ≠ 0, z′ − r_{z′} ≥ 0. Therefore for j = r_{z′} from the equation (3.7.a) and the definitions of v and z′, it follows that either b = v, a = 0, i = 0, z′ = r_{z′} or b = 0, i = r_{z′} and a = μ′.

(3) Since z′ = 0, z = 0, μ′ = μ and μ ≥ λ′ by 3.6 (1).

(a) Let α be a minimal element in E(λ′m_{0}). Then, in view of [4; Lemma 2.6 (3)] we may assume that α = ae_{0} + γ_{i} + be_{p+1}. If b ≠ 0 then b = v by definition of v and a = 0 = i. Since g_{u} ≥ λ′m_{0} = am_{0} + g_{i} + bn, i ≤ u. Therefore, if b = 0 then either i = 0 and a = λ′, or i = u and a = 0 by the Key-Lemma.

(b) Let α be a minimal element in E ((μ − λ′)m_{0} + m_{r_{u}}). Then, in view of [4; Lemma 2.6 (3)] we may assume that α = ae_{0} + γ_{i} + be_{p+1}. From vn = μm_{0} and (μ − λ′)m_{0} + m_{r_{u}} = am_{0} + g_{i} + bn, it follows that b < v and μ − λ′ ≥ a. Suppose that μ − λ′ > a. Then (μ − λ′ − a)m_{0} + m_{r_{u}} = g_{i} + bn and so i ≥ u, since (r_{u}, 0) ∈ U and z = 0. Therefore (v − b)n = (λ′ + a − ε)m_{0} + g_{i−r_{u}} where ε ∈ {0, 1} be such that εm_{0} + g_{i} = g_{r_{u}} + g_{i−r_{u}}. Now, from the definition of v, it follows that b = 0 and (μ − λ′ − a + ε)m_{0} = g_{i−r_{u}}. Therefore vn = μm_{0} = (λ′ + a − ε)m_{0} + g_{i−r_{u}}. Now, from the definition of z′, it follows that i = r_{u}, a contradiction, since i ≥ u > r_{u}. Therefore μ − λ′ = a. □

3.8. Lemma. Let s, i ∈ ℕ. Then there exists a unique integer b_{s} ∈ [0, s − 1] such that sg_{i} = b_{s}m_{0} + g_{si} and q_{si} = sq_{i} + (s − 1 − b_{s}).

Proof. Use [4; Lemma 2.6] and induction on s. □

3.9. Lemma. Suppose that w = 0. Then for every i ∈ ℕ there exists j ∈ [0, u − 1] such that g_{i} ≡ g_{j}(mod m_{0}).

Proof. Let i ∈ ℕ and let k ∈ ℕ, j ∈ [0, u − 1] be the unique integers defined by the equation i = ku + j. Now, since w = 0 and g_{u} ≡ 0(mod m_{0}), we have g_{i} = g_{ku+j} ≡ kg_{u} + g_{j} ≡ g_{j}(mod m_{0}). □

3.10. Lemma. Let a, a′ ∈ ℕ and b, b′ ∈ [0, v − 1] be such that a′e_{0} + b′e_{p+1} ∈ E(am_{0} + bn). Then a = a′ and b = b′.

Proof. Since bn ≡ am_{0} + bn = a′m_{0} + b′n ≡ b′n(mod m_{0}) and (0, b), (0, b′) ∈ U, we have b = b′ by the Key-Lemma. □

3.11. Lemma. Let a, a′ ∈ [0, μ − 1] and b, b′ ∈ ℕ be such that am_{0} + bn = a′m_{0} + b′n. Then a = a′ and b = b′.

Proof. If b < b′ then a > a′ and (a − a′)m_{0} = (b′ − b)n. Therefore b′ − b ≥ v by the definition of v and so (a − a′)m_{0} ≥ vn ≥ μm_{0}, a contradiction. □

3.12. Lemma. Suppose that m_{0} < n < m_{p} and z′ = 0. Let c, c′, b, b′ ∈ ℕ and i ∈ [1, u − 1] be such that c′m_{0} + b′n = cm_{0} + g_{i} + bn. Then c < c′, b′ < b and c′ − c > q_{i} + 1 + b − b′.

Proof. Since z′ = 0, z = 0 and μ′ = μ by 3.5 (1). Suppose that c ≥ c′. Then from vn = μm_{0} and c′m_{0} + b′n = cm_{0} + g_{i} + bn, we get b′_{0}n ≡ g_{i} + b_{0}n(mod m_{0}) for some b′_{0}, b_{0} ∈ [0, v − 1]. Now, since z = 0, both (0, b′_{0}), (i, b_{0}) ∈ U and so (0, b′_{0}) = (i, b_{0}), a contradiction. Therefore c < c′ and so b′ < b, since (i, 0) ∈ U. Now, from m_{0} < n < m_{p} and (c′ − c)m_{0} = g_{i} + (b − b′)n, we have c′ − c > q_{i} + 1 + b − b′. □

4. The case n < m_{p}

Let m_{0}, …, m_{p}, n be an almost arithmetic sequence of positive integers with gcd(m_{0}, …, m_{p},

n) = 1 and m_{0} < ⋯ < m_{p}. Let Γ be the semigroup generated by m_{0}, …, m_{p}, n and let A′ := K[Y^{n_{e}}, …, X^{n_{i}}Y^{n_{e}−n_{i}}, …, X^{n_{e}}]. We assume that m_{0}, …, m_{p}, n is a minimal set of generators for the semigroup Γ.

We continue using the same notations as in Section 3. In particular, remember the Key-Lemma 3.3 and sometimes giving reference of the Key-Lemma is omitted.

In this section we assume that n < m_{p}.

4.1. Lemma. Suppose that either w = 0 or ν ≥ q_{u−z} +1+(v−w). Let a, i ∈ ℕ , b ∈ [0, v−1] be such that γ_{i} + be_{p+1} ∈ E(am_{0}). Then a ≥ q_{i} + 1 + b.

Proof. If either n > m_{0} or b = 0 then the assertion is clear.

First suppose that w = 0. Then by 3.9 there exist j ∈ [0, u − 1] and a′ ∈ ℕ such that g_{i} = a′m_{0} + g_{j}. Now, from the equation am_{0} = a′m_{0} + g_{j} + bn and (j, b) ∈ U (since w = 0), we get j = 0 = b and a = a′ by the Key-Lemma. Now, from the equation am_{0} = g_{i}, we have q_{i} + 1 ≤ a.

Now, assume that n < m_{0}, b ≠ 0 and ν ≥ q_{u−z} + 1 + (v − w).

Since g_{i} + bn = am_{0} ≡ 0(mod m_{0}), we have (i, b) ∉ U and so i ≥ u − z, since b < v. We consider the two cases b < v − w and b ≥ v − w separately.

Case b < v − w: In this case from the equations νm_{0} = g_{u−z} + (v − w)n, am_{0} = g_{i} + bn and b ≠ 0, we get (a + ε − ν)m_{0} + (v − w − b)n = g_{i−u+z} where ε ∈ {0, 1} be such that εm_{0} + g_{i} = g_{u−z} + g_{i−u+z}. Therefore a + ε − ν + (v − w − b) ≥ q_{i−u+z} + 1, since n, m_{0} < m_{p}.

Case b ≥ v − w: In this case we have (a + ε − ν)m_{0} = g_{i−u+z} + (b − v + w)n where ε ∈ {0, 1} be such that εm_{0} + g_{i} = g_{u−z} + g_{i−u+z}. Now, since b − v + w < b, it follows by the case b < v − w and induction on b that a + ε − ν ≥ q_{i−u+z} + 1 + (b − v + w). In any case, since ν ≥ q_{u−z} + 1 + (v − w) by assumption and ε + q_{i} = q_{i−u+z} + q_{u−z} + 1 by [4; Lemma 2.6 (2)], we have a ≥ b + q_{i} + 1. □

4.2. Lemma. Suppose that v ≥ μ′ + q_{z′} + 1. Then for every h ∈ Γ and α ∈ E(h), there exist a, i ∈ ℕ and b ∈ [0, v − 1] such that ae_{0} + γ_{i} + be_{p+1} ∈ E(h) and a + q_{i} + 1 + b ≤ deg(α).

Proof. By [4; Lemma 2.6 (3)] there exist a′, b′, j ∈ ℕ such that deg(α) = a′+q_{j}+1+b′. If b′ < v then there is nothing to prove. Suppose that b′ ≥ v. Then, since εe_{0} + γ_{j+z′} = γ_{j} + γ_{z′} with ε ∈ {0, 1}, we have deg(β) ≤ deg(α), where β := (a′ + μ′ + ε)e_{0} + γ_{j+z′} + (b′ − v)e_{p+1} ∈ E(h). If b′ − v ≥ v then repeating we get b ∈ [0, v − 1] such that ae_{0} + γ_{i} + be_{p+1} ∈ E(h) and a + q_{i} + 1 + b = deg(ae_{0} + γ_{i} + be_{p+1}) ≤ deg(α) for some a, i ∈ ℕ. □

4.3. Lemma. Suppose that m_{0} < n < m_{p} and z′ = 0. Then for every h ∈ Γ and α ∈ E(h), there exist a ∈ [0, μ − 1] and i, b ∈ ℕ such that a + q_{i} + 1 + b ≤ deg(α).

Proof. Since z′ = 0, z = 0 and μ′ = μ. By [4; Lemma 2.6 (3)] there exist a′, b′, j ∈ ℕ such that deg(α) = a′ + q_{j} + 1 + b′. Let a_{1} ∈ ℕ and a_{0} ∈ [0, μ − 1] be the unique integers defind by the equation a′ = a_{1}μ + a_{0}. Since n < m_{p} and vn = μm_{0}, we have v < μ. Now, the assertion is clear by taking a := a_{0}, b := (a_{1}v + b′) and i := j. □

4.4. Definition. For h ∈ Γ, let

F(h) := \{ \begin{matrix}\{ae_{0} + γ_{i} + be_{p+1} ∈ E(h) | b ∈ [0, v − 1]\}, & if either n < m_{0} or z′ ≠ 0, \\ \{ae_{0} + γ_{i} + be_{p+1} ∈ E(h) | a ∈ [0, μ − 1]\}, & \begin{aligned}if & m_{0} < n < m_{p} \\ & and z′ = 0. \end{aligned} \end{matrix}

and let

i(h) := max{i ∈ ℕ | ae_{0} + γ_{i} + be_{p+1} ∈ F(h)}.

4.5. Proposition. Let h ∈ Γ. Then

(1) The set F(h) is finite and non-empty. In particular, i(h) exists.

(2) There exists a unique pair (a(h), b(h)) such that

a(h)e_{0} + γ_{i(h)} + b(h)e_{p+1} ∈ F(h) .

(3) Let δ(h) := a(h)e_{0}+γ_{i(h)}+b(h)e_{p+1} ∈ F(h). Then i(h+m_{p}) = i(h)+p and δ(h+m_{p}) = δ(h) + γ_{p}.

Proof. (1) Since F(h) ⊆ E(h) and E(h) is finite, F(h) is finite. Further, F(h) ≠ ∅ by

3.3,

if either n < m_{0}
or z′ ≠ 0,
4.3, \begin{aligned}if & m_{0} < n < m_{p} \\ & and z′ = 0. \end{aligned}

(2) is immediate from \{ \begin{matrix}3.10, & if either n < m_{0} or z′ ≠ 0, \\ 3.11, & \begin{aligned}if & m_{0} < n < m_{p} \\ & and z′ = 0. \end{aligned} \end{matrix}

(3) Let α = ae_{0} +γ_{i(h)} +be_{p+1} ∈ F(h). Then α +γ_{p} ∈ F(h+m_{p}) and so i(h+m_{p}) ≥ i(h)+p. Let β = a′e_{0}+γ_{i′}+be_{p+1} ∈ F(h+m_{p}). If i′ ≥ i(h)+p then α′ := β−γ_{p} = a′e_{0}+γ_{i′−p}+b′e_{p+1} ∈ F(h) and so i′ − p ≤ i(h) by definition of i(h). □

4.6. Proposition. Suppose that v ≥ μ′ + q_{z′} + 1 and either w = 0 or ν ≥ q_{u−z} + 1 + v − w. Let h ∈ Γ and α′ := a′e_{0} + γ_{i′} + b′e_{p+1} ∈ E(h). Then

(1) deg (δ(h)) ≤ deg(α′), (2) mindeg(h) = deg (δ(h)) ,

(3) mindeg(h + tm_{p}) = mindeg(h) + t for every t ∈ ℕ.

Proof. (1) Since v ≥ μ′ + q_{z′} + 1, we have z′ ≠ 0 if m_{0} < n. In view of 4.2 we may assume that b′ ∈ [0, v − 1]. Then α′ ∈ F(h). Put i := i(h), a := a(h) and b := b(h). Then i ≥ i′ by definition of i(h). If i = i′ then a′ = a and b′ = b by 4.5 (2) and there is nothing to prove. Therefore assume that i > i′. Then am_{0} + g_{i−i′} + bn = (a′ + ε)m_{0} + b′n where ε ∈ {0, 1} be such that εm_{0} + g_{i} = g_{i′} + g_{i−i′}. Now, since (0, b′) ∈ U, a < a′ + ε and so

(4.6.a)  g_{i−i′} + bn = (a′ + ε − a)m_{0} + b′n .

We consider the two cases b < b′ and b ≥ b′ separately.

Case b < b′: In this case from the equation (4.6.a), we get

(a′ + ε − a)m_{0} + (b′ − b)n = g_{i−i′}

with a′ + ε − a > 0 and so i − i′ ≥ u > p. Therefore, since n < m_{p}, we have

(4.6.b)  q_{i−i′} + 1 ≤ (a′ + ε − a) + (b′ − b) .

Case b ≥ b′: In this case from the equation (4.6.a), we get (a′ + ε − a)m_{0} = g_{i−i′} + (b − b′)n. Therefore by 4.1 we have

(4.6.c)  q_{i−i′} + 1 + (b − b′) ≤ (a′ + ε − a) .

Now, in any case, use [4; Lemma 2.6 (2)] to conclude that deg (δ(h)) = a + q_{i} + 1 + b ≤ a′ + q_{i′} + 1 + b′ = deg(α′).

(2) Follows from [4; Lemma 2.6 (3)] and (1).

(3) Since δ (h + m_{p}) = δ (h) + γ_{p}, the assertion follows from (2) by induction on t. □

4.7. Proposition. Suppose that m_{0} < n < m_{p} and z′ = 0. Let α′ := a′e_{0}+γ_{i′}+b′e_{p+1} ∈ E(h), h ∈ Γ. Then:

(1) If i(h) − i′ ∉ ℕu then deg (δ(h)) < deg(α′).

(2) If i(h) − i′ ∈ ℕu and v + w′ ≥ μ − λ′ + q_{u} + 1 then deg (δ(h)) ≤ deg(α′).

(3) If v + w′ ≥ μ − λ′ + q_{u} + 1 then mindeg(h) = deg (δ(h)).

(4) If v + w′ ≥ μ − λ′ + q_{u} + 1 then mindeg(h + tm_{p}) = mindeg(h) + t for every t ∈ ℕ.

Proof. Since z′ = 0, z = 0 and μ′ = μ by 3.5 (1). In view of 4.3 we may assume that a′ < μ. Then α′ ∈ F(h). Put i := i(h), a := a(h) and b := b(h). Then i ≥ i′ by definition of i(h). If i = i′ then a′ = a and b′ = b by 4.5 (2) and there is nothing to prove. Therefore assume that i > i′. Then a′m_{0} + g_{i′} + b′n = h = am_{0} + g_{i} + bn and so

(4.7.a)  (a′ + ε)m_{0} + b′n = am_{0} + g_{i−i′} + bn

where ε ∈ {0, 1} be such that

(4.7.b)  εm_{0} + g_{i} = g_{i′} + g_{i−i′} .

Let b_{1}, b′_{1} ∈ ℕ and b_{0}, b′_{0} ∈ [0, v − 1] be the unique integers defined by the equations b = b_{1}v + b_{0} and b′ = b′_{1}v + b′_{0}. Then from the equations vn = μm_{0} and (4.7.a), we get b′_{0}n ≡ g_{i−i′} + b_{0}n(mod m_{0}). Now, since (0, b′_{0}) ∈ U and z = 0, it follows from the Key-Lemma that i − i′ ≥ u. Put j := i−i′ and let j_{1} ∈ ℕ and j_{0} ∈ [0, u−1] be the unique integers defined by the equation j = j_{1}u + j_{0}. Then j_{1}g_{u} + g_{j_{0}} = (b_{j_{1}} + ε′)m_{0} + g_{j} where b_{j_{1}} ∈ [0, j_{1} − 1] and ε′ ∈ {0, 1} with

(4.7.c)  ε′m_{0} + g_{j} = g_{j_{0}} + g_{j_{1}u} .

With this from the equations (4.7.a) and g_{u} = λ′m_{0} + w′n, we get

(4.7.d)  (a′ + ε + ε′ + b_{j_{1}})m_{0} + b′n = (a + j_{1}λ′)m_{0} + g_{j_{0}} + (b + j_{1}w′)n .

(1) Since i − i′ ∉ ℕu, we have j_{0} ≠ 0. Therefore from the equation (4.7.d) and 3.12, we get (a′ + ε + ε′ + b_{j_{1}}) − (a + j_{1}λ′) > q_{j_{0}} + 1 + (b + j_{1}w′) − b′. Now, by using the equations (4.7.b), (4.7.c), 3.6 (3), 3.8 (applied to s = j_{1} and i = u) and [4; Lemma 2.6 (2)], it is easy to see that deg(α′) = a′ + b′ + q_{i′} + 1 > a + b + q_{i} + 1 = deg(δ(h)).

(2) Since i − i′ ∈ ℕu, we have j_{0} = 0, ε′ = 0 and the equation (4.7.d) is

(4.7.e)  c′m_{0} + b′n = cm_{0} + (b + j_{1}w′)n

where

(4.7.f)  c′ := (a′ + ε + b_{j_{1}}) and c := (a + j_{1}λ′) .

We consider two cases c′ ≥ c and c′ < c separately.

Case c′ ≥ c: From the equation (4.7.e), we have c′−c ≥ (b+j_{1}w′)−b′ and so c′+b′ ≥ c+b+j_{1}w′. Therefore by 3.6 (3), [4; Lemma 2.6 (2)] and 3.8 (applied to s = j_{1} and i = u), we have deg (δ(h)) = a + q_{i} + 1 + b ≤ a′ + q_{i′} + 1 + b′ = deg(α′).

Case c′ < c: Let c_{1}, c′_{1} ∈ ℕ and c_{0}, c′_{0} ∈ [0, μ − 1] be the unique integers defined by the equations

(4.7.g)  c = c_{1}μ + c_{0} and c′ = c′_{1}μ + c′_{0} .

Then from the equations vn = μm_{0} and (4.7.e), we get c′_{0}m_{0} + (c′_{1}v + b′)n = c_{0}m_{0} + (c_{1}v + b + j_{1}w′)n. Therefore by 3.11 we have

(4.7.h)  c′_{0} = c_{0} and c′_{1}v + b′ = c_{1}v + b + j_{1}w′

and so from (4.7.f), (4.7.g) and (4.7.h), we have

(4.7.i)  a = (c_{1} − c′_{1})μ + a′ + ε + b_{j_{1}} − j_{1}λ′ and b = b′ − (c_{1} − c′_{1})v − j_{1}w′ .

Further, since λ′ ≤ μ by 3.6 (1) and a ≤ μ, we have (c_{1} − c′_{1})μ = c − c′ ≤ c = a + j_{1}λ′ ≤ a + j_{1}μ < (j_{1} + 1)μ and so

(4.7.j)  (c_{1} − c′_{1}) ≤ j_{1} .

Now, from the equations (4.7.i), (4.7.j), (4.7.b), (4.7.c), [4; Lemma 2.6 (2)], 3.8 (applied to s = j_{1} and i = u) and the assumption v + w′ ≥ μ − λ′ + q_{u} + 1, it is easy to see that deg(α′) = a′ + b′ + q_{i′} + 1 ≥ a + b + q_{i} + 1 = deg(δ(h)).

(3) is immediate from (1) and (2).

(4) Since δ (h + m_{p}) = δ (h) + γ_{p}, the assertion follows from (3) by induction on t. □

4.8. Theorem.

(1) If n < m_{0} and w = 0 then A′ is Cohen-Macaulay.

(2) In each of the following cases

(a) n < m_{0} and w ≠ 0,

(b) m_{0} < n < m_{p} and z′ ≥ p,

A′ is Cohen-Macaulay if and only if ν ≥ q_{u−z} + 1 + (v − w) and v ≥ μ′ + q_{z′} + 1.

(3) Suppose that m_{0} < n < m_{p} and 1 ≤ z′ < p. Then

(a) If A′ is Cohen-Macaulay then v + 1 ≥ mindeg ((μ′ + 1)m_{0}) + 1.

(b) If v ≥ μ′ + 1 then A′ is Cohen-Macaulay.

(4) If m_{0} < n < m_{p} and z′ = 0 then A′ is Cohen-Macaulay if and only if v + w′ ≥ μ − λ′ + q_{u} + 1.

Proof. (1) Since n < m_{0} and vn = μ′m_{0} +g_{z′}, v ≥ μ′ +q_{z′} +1. Therefore the assertion follows from 4.6 (3) and 2.3.

(2) First assume that n < m_{0} and w ≠ 0. Then v ≥ μ′+q_{z′}+1. If ν ≥ q_{u−z}+1+(v−w) then A′ is Cohen-Macaulay by 4.6 (3) and 2.3. Conversely, suppose that A′ is Cohen-Macaulay. Then, since (q_{u−z} + 1)e_{p} + (v − w)e_{p+1} ∈ E ((ν − 1)m_{0} + m_{p−r_{u−z}}) and mindeg ((v − w)n) = v − w (by definition of v), by 2.3 we have

(ν − 1) + 1 ≥ mindeg ((q_{u−z} + 1)m_{p} + (v − w)n)
= mindeg ((v − w)n) + q_{u−z} + 1 = (v − w) + q_{u−z} + 1.

Now, assume that m_{0} < n < m_{p} and z′ ≥ p. Then ν ≥ q_{u−z}+1+(v−w). If v ≥ μ′+q_{z′}+1 then A′ is Cohen-Macaulay by 4.6 (3) and 2.3. Conversely, suppose that A′ is Cohen-Macaulay. Then from the equation vn = μ′m_{0} + g_{z′} = μ′m_{0} + m_{r_{z′}} + q_{z′}m_{p} and 2.3, we get

v ≥ \{ \begin{matrix}mindeg(μ′m_{0} + m_{r_{z′}}) + q_{z′} & if z′ > p, \\ mindeg(μ′m_{0}) + 1 & if z′ = p. \end{matrix}

Therefore v ≥ μ′ + q_{z′} + 1 by 3.7 (1) and 3.7 (2).

(3) (a) Follows from vn + m_{p−z′} = μ′m_{0} + m_{r_{z′}} + m_{p−z′} = (μ′ + 1)m_{0} + m_{p} and 2.3.

(b) Since 1 ≤ z′ < p, q_{z′} = 0 and so v ≥ μ′ + 1 = μ′ + q_{z′} + 1 by assumption. Moreover, ν ≥ q_{u−z} + 1 + (v − w), since m_{0} < n < m_{p}. Therefore A′ is Cohen-Macaulay by 4.6 (3) and 2.3.

(4) Since m_{0} < n < m_{p} and νm_{0} = g_{u−z} + (v − w)n, ν ≥ q_{u−z} + 1 + (v − w). Further, since z′ = 0, z = 0 and μ′ = μ by 3.5 (1). If v + w′ ≥ μ − λ′ + q_{u} + 1 then A′ is Cohen-Macaulay by 4.7 (3) and 2.3. Conversely, suppose that A′ is Cohen-Macaulay. Then from the equation (v + w′)n = (μ − λ′)m_{0} + g_{u} and 2.3, we get v + w′ ≥ mindeg ((μ − λ′)m_{0} + m_{r_{u}}) + q_{u} = μ − λ′ + q_{u} + 1 by 3.7 (3)(b). □

4.9. Remark. The Theorem 4.8 gives a numerical criterion for the Cohen-Macaulayness of A′ in the following cases:

(1) n < m_{0} (see 4.8 (1) and 4.8 (2)).

(2) m_{0} < n < m_{p} and either z′ ≥ p or z′ = 0 (see 4.8 (2) and 4.8 (4)).

In the case m_{0} < n < m_{p} and 1 ≤ z′ < p partial results are given in 4.8 (3). These results are not complete, the necessary condition given in 4.8 (3)(a) is not sufficient and the sufficient condition given in 4.8 (3)(b) is not necessary, as is demonstrated in Section 7 by examples (see 7.3 (3) and 7.3 (4)).

4.10. Lemma. Suppose that m_{0} < n < m_{p}. Let ε ∈ {0, 1} be such that εm_{0}+g_{u+z′} = g_{u}+g_{z′}. Then:

(1) (v + w′)n = (μ′ − λ′ + ε)m_{0} + g_{u+z′}.

(2) If (μ′ − λ′ + ε) ≤ 0 then (v + w′) − (μ′ − λ′ + ε) ≥ q_{u+z′} + 1.

(3) If 0 ≤ (μ′ − λ′ + ε) ≤ μ′ then mindeg ((μ′ − λ′ + ε)m_{0}) = (μ′ − λ′ + ε).

(4) If (μ′ − λ′ + ε) ≥ 0 and A′ is Cohen-Macaulay then

(v + w′) ≥ mindeg ((μ′ − λ′ + ε + m_{r_{u+z′}})) + q_{u+z′} .

Proof. (1) Follows from the equations vn = μ′m_{0} + g_{z′} and g_{u} = λ′m_{0} + w′n.

(2) Since n < m_{p} and u + z′ ≥ u > p, the assertion follows from (1).

(3) Let α be a minimal element in E ((μ′ − λ′ + ε)m_{0}). By [4; Lemma 2.6 (3)] we may assume that α = ae_{0} + γ_{i} + be_{p+1}. From the equations vn = μ′m_{0} + g_{z′}, g_{u} = λ′m_{0} + w′n and am_{0} + g_{i} + bn = (μ′ − λ′ + ε)m_{0}, we have

(4.10.a)  vn = (λ′ − ε + a + ε′)m_{0} + g_{i+z′} + bn

where ε′ ∈ {0, 1} be such that ε′m_{0} + g_{i+z′} = g_{i} + g_{z′}. Now, since (μ′ − λ′ + ε) < μ′, from the definitions of v, z′ and the equation (4.10.a), it is easy to see that b = 0, i = 0, ε′ = 0 and a = (μ′ − λ′ + ε). Therefore mindeg ((μ′ − λ′ + ε)m_{0}) = deg(α) = a = (μ′ − λ′ + ε).

(4) In view of (2) we may assume that (μ′ − λ′ + ε) ≥ 1. Now, since m_{0} < n < m_{p}, the assertion follows from the equation (v + w′)n = (μ′ − λ′ + ε)m_{0} + g_{u+z′}, 2.3 and (3). □

4.11. Question. Suppose that m_{0} < n < m_{p} and z′ ∈ [1, p − 1]. Let ε ∈ {0, 1} be such that εm_{0} + g_{u+z′} = g_{u} + g_{z′}. Then:

(a) If μ′ − λ′ + ε ≤ 0 is then A′ always Cohen-Macaulay?

(b) If μ′ − λ′ + ε ≥ 0 and (v + w′) ≥ mindeg ((μ′ − λ′ + ε + m_{r_{u+z′}})) + q_{u+z′} is then A′ always Cohen-Macaulay?

5. The case m_{p} < n and w′ ≠ 0

In this section we continue using the same notations as in Section 3 and assume that m_{p} < n.

5.1. Notations. We fix the following notations:

For k ∈ [0, p − r_{u}], let ε_{k} := \{ \begin{matrix}0 & if k = 0, \\ 1 & if k ≠ 0, \end{matrix} and δ_{k} := a_{k}e_{0} + γ_{u_{k}} + b_{k}e_{p+1} be a fixed minimal

element in E ((λ′ − ε_{k})m_{0} + g_{k}). In particular, deg(δ_{k}) = mindeg ((λ′ − ε_{k})m_{0} + g_{k})).

5.2. Lemma. δ_{k} + w′e_{p+1} ∈ E(g_{u+k}) for every k ∈ [0, p − r_{u}]. Moreover, if w′ ≠ 0 then u_{k} < u + k for every k ∈ [0, p − r_{u}].

Proof. Clear from the equations g_{u+k} = (λ′ − ε_{k})m_{0} + g_{k} + w′n, ∂(δ_{k}) = (λ′ − ε_{k})m_{0} + g_{k} and δ_{k} = a_{k}e_{0} + γ_{u_{k}} + b_{k}e_{p+1}. □

5.3. q_{u+k} + 1 ≥ mindeg((λ′ − ε_{k})m_{0} + g_{k}) + w′ for every k ∈ [0, p − r_{u}].

5.4. Lemma. Suppose that w′ ≠ 0 and 5.3 holds. Then for every h ∈ Γ and α ∈ E(h), there exist a, b ∈ ℕ and i ∈ [0, u − 1] such that ae_{0} + γ_{i} + be_{p+1} ∈ E(h) and a + q_{i} + 1 + b ≤ deg(α).

Proof. By [4; Lemma 2.6 (3)] there exist a′, b′, j ∈ ℕ such that deg(α) = a′ + q_{j} + 1 + b′. If j < u there is nothing to prove. Therefore assume that j ≥ u. Let k := max{0, r_{j} − r_{u}}. Then k ∈ [0, p − r_{u}] and it is easy (by using Key-Lemma) to see that g_{j} = g_{u+k} + g_{j−u−k}. Let

β := (a′ + a_{k} + ε)e_{0} + γ_{j−u−k+u_{k}} + (b′ + b_{k} + w′)e_{p+1}

where ε ∈ {0, 1} be such that εe_{0} + γ_{j−u−k+u_{k}} = γ_{u_{k}} + γ_{j−u−k}. Using 5.2 and the assumption 5.3, it is easy to see that β ∈ E(h), j−u−k+u_{k} < j and deg(β) ≤ deg(α). If j−u−k+u_{k} < u there is nothing to prove. If j − u − k + u_{k} ≥ u then repeating we get i ∈ [0, u − 1] such that ae_{0} + γ_{i} + be_{p+1} ∈ E(h) and a + q_{i} + 1 + b = deg(ae_{0} + γ_{i} + be_{p+1}) ≤ deg(α). □

5.5. Definition. For h ∈ Γ, let

G(h) := {ae_{0} + γ_{i} + be_{p+1} ∈ E(h) | i ∈ [0, u − 1]}

and let

b(h) := max{b ∈ ℕ | ae_{0} + γ_{i} + be_{p+1} ∈ G(h)} .

S. Molinelli, D. P. Patil, G. Tamone: On the Cohen-Macaulayness … 451

5.6. Proposition. Let h ∈ Γ. Then:

(1) The set G(h) is a finite and non-empty. In particular, b(h) exists.

(2) There exists a unique pair (a(h), i(h)) such that a(h)e_{0} + γ_{i(h)} + b(h)e_{p+1} ∈ G(h) .

(3) Let δ(h) := a(h)e_{0} + γ_{i(h)} + b(h)e_{p+1} ∈ G(h). Then b(h + n) = b(h) + 1 and δ(h + n) = δ(h) + e_{p+1}.

Proof. (1) Since G(h) ⊆ E(h) and E(h) is finite, G(h) is finite. Further, G(h) ≠ ∅ by the Key-Lemma.

(2) is immediate from the Key-Lemma.

(3) Let α = ae_{0}+γ_{i}+b(h)e_{p+1} ∈ G(h). Then α+e_{p+1} ∈ G(h+n) and so b(h+n) ≥ b(h)+1. Let β = a′e_{0}+γ_{i′}+b′e_{p+1} ∈ G(h+n). If b′ ≥ b(h)+1 then α′ := β−e_{p+1} = a′e_{0}+γ_{i′}+(b′−1)e_{p+1} ∈ G(h) and so b′ − 1 ≤ b(h) by definition of b(h). □

5.7. Proposition. Suppose that w′ ≠ 0 and 5.3 holds. Let α′ := a′e_{0} + γ_{i′} + b′e_{p+1} ∈ E(h), h ∈ Γ. Then

(1) deg (δ(h)) ≤ deg(α′),

(2) mindeg(h) = deg (δ(h)),

(3) mindeg(h + tn) = mindeg(h) + t for every t ∈ ℕ.

Proof. (1) In view of 5.4 we may assume that i′ ∈ [0, u − 1]. Then α′ ∈ G(h). Put a := a(h), i := i(h) and b := b(h). Then b ≥ b′ by definition of b(h). If b = b′ then i = i′ and a = a′ by 5.6 (2) and there is nothing to prove. Therefore assume that b > b′. Then a′m_{0} + g_{i′} = am_{0} + g_{i} + (b − b′)n. We consider the cases i′ ≤ i and i′ > i separately. Case i ≤ i′: In this case (a′ − a − ε)m_{0} + g_{i′−i} = (b − b′)n where ε ∈ {0, 1} be such that εm_{0} + g_{i′} = g_{i} + g_{i′−i}. Therefore a′ − a − ε + q_{i′−i} + 1 ≥ (b − b′), since n > m_{p} > m_{0}. Case i > i′: In this case (a′ − a + ε)m_{0} = g_{i−i′} + (b − b′)n where ε ∈ {0, 1} be such that εm_{0} + g_{i} = g_{i′} + g_{i−i′}. Therefore a′ − a + ε ≥ q_{i−i′} + 1 + (b − b′), since n > m_{p} > m_{0}. Now, in any case, use [4; Lemma 2.6 (2)] to conclude that deg (δ(h)) = a + q_{i} + 1 + b ≤ a′ + q_{i′} + 1 + b′ = deg(α′).

(2) Follows from [4; Lemma 2.6 (3)] and (1).

(3) Since δ(h + n) = δ(h) + e_{p+1}, the assertion follows from (2) by induction on t. □

5.8. Theorem. Suppose that m_{p} < n and w′ ≠ 0. Then A′ is Cohen-Macaulay if and only if 5.3 holds.

Proof. If 5.3 holds then A′ is Cohen-Macaulay by 5.7 (3) and 2.3. Conversely, suppose that A′ is Cohen-Macaulay. Then by 2.3, we have

q_{u+k} + 1 = mindeg(g_{u+k}) = mindeg ((λ′ − ε_{k})m_{0} + g_{k} + w′n)
= mindeg ((λ′ − ε_{k})m_{0} + g_{k}) + w′

for every k ∈ [0, p − r_{u}]. □

6. The case m_{p} < n and w′ = 0

In this section we continue using the same notations as in Section 3 and assume that m_{p} < n and w′ = 0. Therefore g_{u} = λ′m_{0}.

6.1. Lemma. For every h ∈ Γ and α ∈ E(h), there exist a ∈ [0, λ′ − 1] and i, b ∈ ℕ such that ae_{0} + γ_{i} + be_{p+1} ∈ E(h) and a + q_{i} + 1 + b ≤ deg(α).

Proof. Let h ∈ Γ and let α ∈ E(h). Then by [4; Lemma 2.6 (3)] there exist a′, j, b′ ∈ ℕ such that a′e_{0} + γ_{j} + b′e_{p+1} ∈ E(h) and deg(α) = a′ + q_{j} + 1 + b′. If a′ < λ′ then there is nothing to prove. Suppose that a′ ≥ λ′. Put β := (a′ − λ′ + ε)e_{0} + γ_{u+j} + b′e_{p+1} where ε ∈ {0, 1} be such that εm_{0} + g_{u+j} = g_{u} + g_{j}. Then β ∈ E(h). Further, since g_{u} = λ′m_{0}, we have λ′ ≥ q_{u} + 1 ≥ 2 and so deg(β) ≤ deg(α) and a′ − λ′ + ε < a′. If a′ − λ′ + ε ≥ λ′ then repeating we get a ∈ [0, λ′ − 1] such that ae_{0} + γ_{i} + be_{p+1} ∈ E(h) and a + q_{i} + 1 + b ≤ deg(α) for some i, b ∈ ℕ. □

6.2. Definition. For h ∈ Γ, let

G(h) := {(a, i, b) ∈ ℕ^{3} | a ∈ [0, λ′ − 1], ae_{0} + γ_{i} + be_{p+1} ∈ E(h)} .

Then G(h) ≠ ∅ by 6.1. Order G(h) by the lexicographic order from right to left and let δ(h) := a(h)e_{0} + γ_{i(h)} + b(h)e_{p+1} denote the maximal element in G(h).

6.3. Lemma. Let h ∈ Γ and let δ(h) be as in the Definition 6.2. Put a_{0} := a(h), i_{0} := i(h) and b_{0} := b(h).

(1) If a ∈ [0, λ′ − 1] and a′e_{0} + γ_{i′} + b′e_{p+1} ∈ E(am_{0}) with a′ ∈ [0, λ′ − 1] then either a′ = a, i′ = 0 = b′, or b′ > 0.

(2) a_{0} ∈ [0, λ′ − 1] and mindeg(am_{0}) = a for every a ∈ [0, a_{0}].

(3) If a ∈ [0, a_{0}] then mindeg(am_{0} + g_{i_{0}}) = a + q_{i_{0}} + 1.

Proof. (1) We need only to prove that if b′ = 0 then a′ = a. Suppose that a′ < a. Then i′ > 0 and (a − a′)m_{0} = g_{i′}. Therefore i′ ≥ u by the Key-Lemma and so (a − a′)m_{0} ≥ g_{u} = λ′m_{0}. Therefore a − a′ ≥ λ′, a contradiction to the assumption a < λ′.

(2) Follows easily from the definition of δ(h) and the part (1).

(3) It is enough to prove the assertion for a = a_{0}. Let α = a′e_{0} + γ_{i′} + b′e_{p+1} be a minimal element in E(a_{0}m_{0} + g_{i_{0}}). Then a′ < λ′ by 6.1 and b′ = 0, i′ ≤ i_{0} by the definition of δ(h).

Therefore a_{0} ≤ a′ and from the equation a_{0}m_{0}+g_{i_{0}} = a′m_{0}+g_{i′}, we get (a′−a_{0}+ε)m_{0} = g_{i_{0}−i′} where ε ∈ {0, 1} be such that εm_{0} + g_{i_{0}} = g_{i′} + g_{i_{0}−i′}. Therefore, if i_{0} > i′ then a′ − a_{0} + ε > q_{i_{0}−i′} + 1 and so deg(α) > deg(a_{0}e_{0} + g_{i_{0}}) ≥ deg(α) which is absurd. This proves that i′ = i_{0} and a′ = a_{0}. □

6.4. Proposition. Let h ∈ Γ and let δ(h) be as in the Definition 6.2. Put a_{0} := a(h), i_{0} := i(h) and b_{0} := b(h). Let α = ae_{0} + γ_{i} + be_{p+1} ∈ E(h) with a ∈ [0, λ′ − 1] be such that either b = b_{0}, or b < b_{0} and a_{0} ≤ a. Then deg (δ(h)) ≤ deg(α).

Proof. We have

(6.4.a)  a_{0}m_{0} + g_{i_{0}} + b_{0}n = h = am_{0} + g_{i} + bn .

First suppose that b = b_{0}. Then from the equation (6.4.a) and the definition of δ(h), we have either i = i_{0} and a = a_{0}, or i < i_{0}. Therefore from the equation (6.4.a), we get

(6.4.b)  g_{i_{0}−i} = (a − a_{0} + ε)m_{0}

where ε ∈ {0, 1} be such that εm_{0} + g_{i_{0}} = g_{i} + g_{i_{0}−i}. Now, from n > m_{p} > m_{0}, the equation (6.4.b) and [4; Lemma 2.6 (2)], it is easy to see that deg (δ(h)) = a_{0} + q_{i_{0}} + 1 + b_{0} ≤ a + q_{i} + 1 + b = deg(α).

Now, suppose that b < b_{0} and a ≥ a_{0}. Then from the equation (6.4.a) we have

(6.4.c)  g_{i_{0}} + (b_{0} − b)n = (a − a_{0})m_{0} + g_{i} .

We consider then two cases i_{0} ≥ i and i_{0} < i separately.

Case i_{0} ≥ i: In this case from the equation (6.4.c) we have

(6.4.d)  g_{i_{0}−i} + (b_{0} − b)n = (a − a_{0} + ε)m_{0}

where ε ∈ {0, 1} be such that εm_{0} + g_{i_{0}} = g_{i} + g_{i_{0}−i}. Therefore, since n > m_{p} > m_{0}, we get

(6.4.e)  q_{i_{0}−i} + 1 + (b_{0} − b) ≤ (a − a_{0} + ε) .

Case i_{0} < i: In this case from the equation (6.4.c) we have

(6.4.f)  εm_{0} + (b_{0} − b)n = (a − a_{0})m_{0} + g_{i−i_{0}}

where ε ∈ {0, 1} be such that εm_{0} + g_{i} = g_{i_{0}} + g_{i−i_{0}}. Therefore, since n > m_{p} > m_{0}, from the equation (6.4.f), it is easy to see that

(6.4.g)  b_{0} − b ≤ (a − a_{0} − ε) + q_{i−i_{0}} + 1 .

Now, in any case, use [4; Lemma 2.6 (2)] and equations (6.4.e), (6.4.g), to conclude that deg (δ(h)) = a_{0} + q_{i_{0}} + 1 + b_{0} ≤ a + q_{i} + 1 + b = deg(α). □

6.5. Lemma. Let h ∈ Γ and let δ(h) be as in the Definition 6.2. Put a_{0} := a(h), i_{0} := i(h) and b_{0} := b(h). Let α = ae_{0} + γ_{i} + be_{p+1} ∈ E(h) be such that b < b_{0} and a < a_{0}. Then there exist unique integers s, t ∈ ℕ^{+} such that:

(1) b_{0} − b = tv and i − i_{0} = tz + su. In particular, (a_{0} − a)m_{0} + g_{i_{0}} + tvn = g_{i}.

(2) deg (δ(h)) = deg(α) + (sλ′′ − tv′′), where λ′′ := λ′ − q_{u} − 1 and v′′ := μ + q_{z} + 1 − v.

Proof. (1) From the equation a_{0}m_{0} + g_{i_{0}} + b_{0}n = h = am_{0} + g_{i} + bn and b < b_{0}, a < a_{0}, it follows that i_{0} < i and

(6.5.a)  (a_{0} − a + ε)m_{0} + (b_{0} − b)n = g_{i−i_{0}}

where with ε ∈ {0, 1} be such that

(6.5.b)  εm_{0} + g_{i} = g_{i_{0}} + g_{i−i_{0}} .

Let t ∈ ℕ and v′ ∈ [0, v−1] be the unique integers defined by the equation b_{0}−b = tv+v′. Then from the equations tvn = tg_{z} + tμm_{0},

(6.5.c)  tg_{z} = g_{tz} + c_{t}m_{0}

with c_{t} ∈ [0, t − 1] and (6.5.a), we get

(6.5.d)  (a_{0} − a + ε + ε′ + c_{t} + tμ)m_{0} + v′n = g_{i−i_{0}−tz}

where ε′ ∈ {0, 1} be such that

(6.5.e)  ε′m_{0} + g_{i−i_{0}} = g_{tz} + g_{i−i_{0}−tz}

and a_{0} − a + ε + ε′ + c_{t} + tμ ≥ a_{0} − a > 0. Therefore from the equation (6.5.d), it follows that i − i_{0} − tz ≥ u. Let s ∈ ℕ^{+} and u′ ∈ [0, u − 1] be the unique integers defined by the equation i − i_{0} − tz = su + u′. Then, since w′ = 0, we have g_{i−i_{0}−tz} ≡ sg_{u} + g_{u′} ≡ g_{u′} and so from the equation (6.5.d), it follows that v′n ≡ g_{u′}. Therefore, since both (u′, 0) and (0, v′) ∈ U, we have u′ = 0 = v′ by the Key-Lemma. This proves that b_{0} − b = tv and i − i_{0} − tz = su.

(2) From n > m_{p} > m_{0}, g_{u} = λ′m_{0} and vn = μm_{0} + g_{z}, it follows that λ′′ , v′′ ∈ ℕ^{+}. From the equation (6.5.d) we have

(6.5.f)  \begin{aligned}(a_{0} − a + ε + ε′ + c_{t} + tμ)m_{0} + v′n & = g_{i−i_{0}−tz} \\ & = g_{su} \\ & = sg_{u} − c_{s}m_{0} \\ & = (sλ′ − c_{s})m_{0}. \end{aligned}

where

(6.5.g)  c_{s}m_{0} + g_{su} = sg_{u}

with c_{s} ∈ [0, s − 1] and v′ = 0 by (1). Therefore

(6.5.h)  sλ′ = a_{0} − a + ε + ε′ + tμ + c_{t} + c_{s} .

Now, since

(6.5.i)  sλ′′ − tv′′ = sλ′ − sq_{u} − s − tμ − tq_{z} − t + tv ,

from (6.5.e), (6.5.b), (6.5.g), (6.5.c), (6.5.h), (6.5.i), [4; Lemma 2.6 (2)] and [4; Lemma 3.8], it is easy to see that deg (δ(h)) = a_{0} + q_{i_{0}} + 1 + b_{0} = a + q_{i} + 1 + b + (sλ′′ − tv′′) = deg(α) + (sλ′′ − tv′′). □

6.6. Theorem. Suppose that m_{0} < m_{p} < n and w′ = 0. Then the following statements are equivalent :

(i) A′ is Cohen-Macaulay.

(ii) mindeg(h) = deg (δ(h)) for every h ∈ Γ.

(iii) For every j ∈ ℕ with a(g_{j}) ≥ 1 and b(g_{j}) ≥ 1, there exist unique integers s, t ∈ ℕ^{+} such that j = i(g_{j}) + tz + su and sλ′′ ≤ tv′′, where λ′′ := λ′ − q_{u} − 1 and v′′ := μ + q_{z} + 1 − v.

Proof. (ii) ⇒ (i): Let h ∈ Γ and let c ∈ ℕ. Then δ(h + cn) = δ(h) + ce_{p+1} by the definition of δ(h + cn). Therefore by (ii) we have mindeg(h + cn) = deg (δ(h + cn)) = deg (δ(h)) + c = mindeg(h) + c. Therefore A′ is Cohen-Macaulay by 2.3.

(i) ⇒ (ii): Suppose that A′ is Cohen-Macaulay. Let h ∈ Γ and let δ(h) be as in the Definition 6.2. Put a_{0} := a(h), i_{0} := i(h) and b_{0} := b(h). Let α ∈ E(h) be a minimal element in E(h). In view of 6.1, we may assume that α = ae_{0} + γ_{i} + be_{p+1} with a ∈ [0, λ′ − 1] and i, b ∈ ℕ. Then b_{0} ≥ b by the definition of δ(h). We consider the following two cases separately:

• either b = b_{0}, or b < b_{0} and a ≥ a_{0};

• b < b_{0} and a < a_{0}.

Case {either b = b_{0}, or b < b_{0} and a ≥ a_{0}}: In this case by the minimality of α in E(h), the definition of δ(h) and 6.4, it follows that

(6.6.a)  deg (δ(h)) = deg(α) = mindeg(h) .

This proves (ii).

Case {b < b_{0} and a < a_{0}}: In this case by 6.5 there exist unique integers s, t ∈ ℕ^{+} such that b_{0} − b = tv, i − i_{0} = tz + su and

(6.6.b)  (a_{0} − a)m_{0} + g_{i_{0}} + tvn = g_{i} .

Now, since A′ is Cohen-Macaulay, t ≥ 1 and n > m_{p} from the equation (6.6.b), 2.3 and 6.3 (3), it follows that

q_{i} + 1 = deg(γ_{i}) ≥ mindeg(g_{i}) = mindeg ((a_{0} − a)m_{0} + g_{i_{0}} + tvn)
= mindeg ((a_{0} − a)m_{0} + g_{i_{0}}) + tv
= (a_{0} − a) + q_{i_{0}} + 1 + tv .

This proves that deg (δ(h)) = a_{0} + q_{i_{0}} + 1 + b_{0} ≤ a + q_{i} + 1 + b = deg(α) and so deg (δ(h)) = deg(α) = mindeg(h) by the minimality of α in E(h).

(ii) ⇒ (iii): Let j ∈ ℕ be such that a(g_{j}) ≥ 1 and b(g_{j}) ≥ 1. Put a′ := a(g_{j}), i′ := i(g_{j}) and b′ := b(g_{j}). Then applying 6.5 to h = g_{j} and α = γ_{j}, there exist integers s, t ∈ ℕ^{+} such that b′ = tv, j − i′ = tz + su and

(6.6.c)  deg (δ(g_{j})) = deg(γ_{j}) + (sλ′′ − tv′′)

where λ′′ := λ′ − q_{u} − 1 and v′′ := μ + q_{z} + 1 − v. Now, since mindeg(g_{j}) = deg (δ(g_{j})) by (ii) and γ_{j} ∈ E(g_{j}), it follows from the equation (6.6.c) that sλ′′ ≤ tv′′. This proves (iii).

(iii) ⇒ (ii): Let h ∈ Γ and let δ(h) be a s in 6.2. Put a_{0} := a(h), i_{0} := i(h) and b_{0} := b(h). Let α ∈ E(h) be a minimal element in E(h). In view of 6.1, we may assume that α = ae_{0} + γ_{i} + be_{p+1} with a ∈ [0, λ′ − 1] and i, b ∈ ℕ. Then b_{0} ≥ b by the definition of δ(h). If either b = b_{0}, or b < b_{0} and a ≥ a_{0} then (ii) is proved in (6.6.a). Now, assume that b < b_{0} and a < a_{0}. Then (6.6.b) holds and by the definitions of δ(g_{i}) and t, we have b(g_{i}) ≥ tv. Further, from the definition of δ(h), it follows that b(g_{i}) = b_{0} − b = tv and a(g_{i}) ≥ a_{0} − a. Therefore by (iii) we have sλ′′ ≤ tv′′ where λ′′ := λ′ − q_{u} − 1 and v′′ := μ + q_{z} + 1 − v. Therefore deg (δ(h)) ≤ deg(α) by 6.5 (2) and so deg (δ(h)) ≤ deg(α) = mindeg(h) by the minimality of α in E(h). □

7. Examples

7.1. Example. (see [3; Proposition (1.2)]) Let m_{0}, …, m_{p}, m_{p+1} be an arithmetic sequence and let n := m_{p+1}. Then u = p + 1, r_{u} = p, λ = λ′ = 1, w = w′ = 1 and so q_{u} + 1 = λ′ + w′. Therefore A′ is Cohen-Macaulay by 5.8.

7.2. Examples. These are examples with n < m_{0}.

(1) Let n = 7, m_{0} = 10 and m_{1} = 15. Then u = 2, λ = 3, w = 0, v = 5, μ = 2, z = 1 and so by 4.8 (1) A′ is Cohen-Macaulay.

(2) Let n = 7, m_{0} = 8 and m_{1} = 10. Then u = 3, λ = 2, w = 2, v = 4, μ = μ′ = 1, z = z′ = 2 and ν = 3. Therefore ν = 3 = q_{u−z} + 1 + (v − w) and so by 4.8 (2) A′ is Cohen-Macaulay.

(3) Let n = 13, m_{0} = 16 and m_{1} = 21. Then u = 2, λ = 1, w = 2, v = 9, μ = 6, z = 1 and ν = 7. Therefore ν = 7 < 8 = q_{u−z} + 1 + (v − w) and so by 4.8 (2) A′ is not Cohen-Macaulay.

7.3. Examples. These are examples with m_{0} < n < m_{p}.

(1) Let m_{0} = 7, m_{1} = 10 and n = 9. Then v = 3, z = z′ = 2 > p = 1, μ = μ′ = 1 and so v = 3 = μ′ + q_{z′} + 1. Therefore A′ is Cohen-Macaulay by 4.8 (2).

(2) Let m_{0} = 7, m_{1} = 12 and n = 10. Then v = 4, z = z′ = 1 = p, μ = μ′ = 4 and so v = 4 < 5 = μ′ + q_{z′} + 1. Therefore A′ is not Cohen-Macaulay by 4.8 (2).

(3) Let m_{0} = 7, m_{1} = 16, m_{2} = 25 and n = 17. Then p = 2, v = 3, z = z′ = 1 < p, μ = μ′ = 5 and from (μ′ + 1)m_{0} = 6m_{0} = m_{2} + n, we have mindeg ((μ′ + 1)m_{0}) = 2 and so v + 1 = 4 > 3 = mindeg ((μ′ + 1)m_{0}) + 1. For h := 4m_{0} + 2m_{1} = 5m_{0} + m_{2}, we have mindeg(h) = 6 and from h + m_{2} = 5n, we have mindeg(h + m_{2}) = 5. Therefore mindeg(h + m_{2}) = 5 ≠ 7 = mindeg(h) + 1 and so A′ is not Cohen-Macaulay by 2.3. This example shows that the converse of 4.8 (3)(a) is not true.

(4) Let m_{0} = 7, m_{1} = 12, m_{2} = 17 and n = 11. Then p = 2, v = 3, z = z′ = 1 < p, μ = μ′ = 3 and from (μ′ + 1)m_{0} = 4m_{0} = m_{2} + n, we have mindeg ((μ′ + 1)m_{0}) = 2 and so v + 1 = 4 > 3 = mindeg ((μ′ + 1)m_{0}) + 1. In this example v = 3 < 4 = μ′ + 1 and by CoCoA A′ is Cohen-Macaulay. Therefore the converse of 4.8 (3)(b) is not true.

(5) Let m_{0} = 10, m_{1} = 19, m_{2} = 28, m_{3} = 37 and n = 35. Then p = 3, u = 5, λ = λ′ = 3, w = w′ = 1, q_{u} = 1, r_{u} = 2, v = 2, z = z′ = 0, μ = μ′ = 7 and so v + w′ = 3 < 6 = μ − λ′ + q_{u} + 1. Therefore A′ is not Cohen-Macaulay by 4.8 (4).

7.4. Examples. These are examples with m_{p} < n.

(1) Let m_{0} = 10, m_{1} = 15 and n = 17. Then u = 2, w = w′ = 0, λ = λ′ = 3, v = 5, z = 1, μ = 7 and A′ is Cohen-Macaulay by CoCoA.

(2) Let m_{0} = 15, m_{1} = 25 and n = 26. Then u = 3, w = w′ = 0, λ = λ′ = 5, v = 5, z = 1, μ = 7, z′ = 4, μ′ = 2 and A′ is not Cohen-Macaulay by 6.6. In fact, we have δ(g_{7}) = 3e_{0} + 5e_{p+1} and so a(g_{7}) ≥ 1, b(g_{7}) ≥ 1, i(g_{7}) = 0 and 7 = i(g_{7}) + tz + su with t = 1, s = 3. Further, sλ′′ = 2(5−q_{3}−1) = 4 > 3 = 7+q_{1}+1−5 = v′′ = tv′′. Therefore the condition (iii) of 6.6 does not hold for j = 7 and so A′ is not Cohen-Macaulay.

(3) Let m_{0} = 15, m_{1} = 18 and n = 20. Then p = 1, u = 5, r_{u} = 1, q_{u} = 4, λ = 6, w = 0, λ′ = 2, w′ = 3 and so q_{u} +1 = 5 = mindeg(λ′m_{0})+w′. Therefore A′ is Cohen-Macaulay by 5.8.

(4) Let m_{0} = 17, m_{1} = 22, m_{2} = 27 and n = 38. Then p = 2, u = 11, q_{u} = 5, r_{u} = 1 < p, λ = λ′ = 7, w = w′ = 1. Further, from λ′m_{0} = g_{6} + n and (λ′ − 1)m_{0} + g_{1} = 6m_{0} + m_{1}, we have mindeg(λ′m_{0}) = 4 and mindeg ((λ′ − 1) + g_{1}) = 7. Therefore q_{u} + 1 = 6 > 5 = mindeg(λ′m_{0}) + w′ and q_{u+1} + 1 = 6 < 8 = mindeg ((λ′ − 1) + g_{1}) + w′. Therefore A′ is not Cohen-Macaulay by 5.8.

References

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[2] Cavaliere, M. P.; Niesi, G.: On monomial curves and Cohen-Macaulay type. Manuscr. Math. 42 (1983), 147–159.

[3] Molinelli, S.; Tamone, G.: On the Hilbert function of certain rings of monomial curves. J. Pure Appl. Algebra 101 (1995), 191–206.

[4] Molinelli, S.; Patil, D. P.; Tamone, G.: On the Cohen-Macaulayness of the associated graded ring of certain monomial curves. Beiträge Algebra Geom. 39 (1998), 433–446.

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Received February 16, 1998