Hilbert functions of monomial curves∗

D. P. Patil^{1} and L. G. Roberts^{2}

1 Department of Mathematics, Indian Institute of Science, Bangalore 560 012, India. E-mail : patil@math.iisc.ernet.in 2 Department of Mathematics and Statistics, Queen’s University, Kingston, Ontario, Canada, K7L 3N6. E-mail : robertsl@mast.QueensU.CA

Abstract We study the Hilbert function of certain projective monomial curves. We determine which of our curves are Cohen-Macaulay, and find the Cohen-Macaulay type of those that are Cohen-Macaulay.

MSC: 14M05

§0 Introduction.

We discuss the Hilbert function of the homogeneous co-ordinate ring of certain projective monomial curves, namely those that arise from the almost arithmetic progression {d, m_{0}, …, m_{p+1}} where m_{i} = m + id for 0 ≤ i ≤ p + 1, p ≥ 0, d ≥ 1, m ≥ 2, gcd(d, m) = 1 and from the set {d, m_{0}, …, m_{p+1}, n} for d = 1, m_{i} as above, and n > m_{p+1}. We determine which of our curves are Cohen-Macaulay, and discuss the Cohen-Macaulay type of some of those that are Cohen-Macaulay. Our basic tool is an algorithmic approach from [7, pages 304-306]. In the next section we recall how this can be used to compute Hilbert functions of projective monomial curves, and in section 5 we adapt this method to the computation of Cohen-Macaulay type. In addition to the use of results from [7] some other novel features of our presentation are our way of representing elements of the monoid corresponding to the curve (3.1) (which turns out here to be an effective alternative to the standard basis of the monoid, which has often been used, for example, in [1] and [4]), and the explicit partition of monoid elements into stable and unstable (1.6). We found it interesting that for our arithmetic progressions the Hilbert functions of the stable and unstable monoid elements are both easily described, but in one case (Theorem 4.9) they combine to produce a deceptively simple Hilbert function, whereas in another case (Theorem 4.11) they combine to produce a seemingly complicated result that we were not able to formulate correctly using only experimental calculations.

Patil-Roberts : Hilbert functions of monomial curves Page 2

The algorithm of [7] is completely elementary, based solely on the definition of homogenization of an ideal, not requiring any of the semigroup theory of [2] (which was used in [1]). Nonetheless our algorithm seems quite efficient. In addition to using the algorithm theoretically we have programmed it (in Mathematica) and used our programs to formulate conjectures, to test our theorems, and to verify our examples. We have not tried programming the approach of [1]. All of our examples have also been checked with Macaulay 2 [3] applied to the homogeneous coordinate ring of our curve, except that our Macaulay 2 program for Cohen-Macaulay type failed on Example 6.5. Hopefully there is a better way to program it. The methods of the present paper even work by hand on this example.

∗This work was partially supported by the NSERC grant of the second author. The first author thank Department of Mathematics and Statistics, Queen’s University, Kingston, Canada for its hospitality.

Our results provide a generous supply of examples some of which are Gorenstein and level, topics of considerable current interest. The techniques should be applicable to other curves and other ring properties (such as Buchsbaum).

We would like to thank the referee for suggesting a way to better organize the results now found in Theorem 4.11.

(0.1) Notation. If A = ⊕_{i≥0}A_{i} is a graded k-algebra (k a field) then the Hilbert function of A is defined by H_{A}(i) = dim_{k}(A_{i}). If A is the homogeneous co-ordinate ring of a projective curve C we may also write H_{C} := H_{A}. Throughout, ℕ will denote the non-negative integers {0, 1, 2, …}. By ‘arithmetic sequence’ we will mean a finite set of integers which are consecutive terms in an arithmetic progression.

§1 Recall the Construction.

Let 𝒮′ = {m_{0}, …, m_{p+1}} be a sequence of integers with 0 < m_{0} < m_{1} < ⋯ < m_{p+1} and gcd({m_{i}}) = 1. Let Γ′ be the monoid generated by 𝒮′.

Let Θ′_{i} be the set of all sums (repetitions allowed) of i elements of 𝒮′ and let

𝔐′_{i} := Θ′_{i} \ (∪_{j<i}Θ′_{j}) .

The set 𝔐′_{i} is the set obtained from the set Θ′_{i} by removing those integers which have occured in earlier Θ′_{j}, j < i. Starting with 𝔐′_{0} = Θ′_{0} = {0}, 𝔐′_{1} = Θ′_{1} = 𝒮′, and noting that Θ′_{i+1} is the set of all sums of an element of 𝒮′ with an element of Θ′_{i}, one can find the sets 𝔐′_{i} recursively, thereby obtaining an effective algorithm for computing the Hilbert functions of the rings described below.

If n ∈ Γ′ then there exists a unique integer i such that n ∈ 𝔐′_{i} and we write ord_{𝒮}′(n) := i. An 𝒮′-expression for n is a way of writing n as the sum of elements of 𝒮′ and ord_{𝒮}′(n) is the smallest cardinality of an 𝒮′-expression of n.

(1.1) The ring gr(𝒮′). Let gr(𝒮′)_{i} be the k-vector space with basis {t^{n} | n ∈ 𝔐′_{i}} and let gr(𝒮′) = ⊕_{i≥0}gr(𝒮′)_{i}.

Define a multiplication in gr(𝒮′) by

t^{a} · t^{b} := \{ \begin{matrix}t^{a+b}, & if ord_{𝒮}′(a + b) = ord_{𝒮}′(a) + ord_{𝒮}′(b), \\ 0, & otherwise. \end{matrix}

It is clear that with this multiplication gr(𝒮′) is a ring, with H_{gr(𝒮′)}(i) = #𝔐_{i}.

(1.2) Geometric interpretation. Let φ : S := k[X_{0}, …, X_{p+1}] → k[t] be the kalgebra homomorphism defined by X_{i} ↦ t^{m_{i}} and let I := Ker(φ). Let J be the homogenisation of I with respect to the variable W, and let R := k[W, X_{0}, …, X_{p+1}]. Then

gr(𝒮′) ≅ R/(J, W) .

The graded k-algebra R/J is the homogeneous co-ordinate ring of the projective closure C of the affine curve S/I ≅ the k-subalgebra of k[t] generated by {t^{n} | n ∈ Γ′}. Note that R/J ≅ k[u^{m_{p+1}}, t^{m_{0}}u^{m_{p+1}−m_{0}}, …, t^{m_{p}}u^{m_{p+1}−m_{p}}, t^{m_{p+1}}] and C is the projective monomial curve with generic point (u^{m_{p+1}}, t^{m_{0}}u^{m_{p+1}−m_{0}}, …, t^{m_{p}}u^{m_{p+1}−m_{p}}, t^{m_{p+1}}).

(1.3) It is well known that gr(𝒮′) is Cohen-Macaulay if and only if t^{m_{p+1}} is a nonzero divisor in gr(𝒮′). Indeed {t^{m_{p+1}}} is a homogeneous systems of parameters and a graded k-algebra is Cohen-Macaulay if and only if every homogeneous system of parameters is regular. Furthermore, for all 𝒮′, if i is sufficiently large, then t^{m_{p+1}} : gr(𝒮′)_{i} → gr(𝒮′)_{i+1} is an isomorphism (between vector spaces of dimension m_{p+1}). Also R/J is an integral domain (and hence W is always a non-zero divisor in R/J ). Therefore in view of (1.2), R/J is Cohen-Macaulay (of dimension 2) if and only if gr(𝒮′) is Cohen-Macaulay (of dimension 1). In terms of Hilbert functions we have H_{C} = H_{R/J} and the difference sequence of H_{C} is H_{gr(𝒮′)} (with H_{gr(𝒮′)}(i) = #𝔐′_{i}). Furthermore the h-vector of R/J (i.e. Δ^{2}H_{R/J}) is ΔH_{gr(𝒮′)}. (As usual for any Hilbert function H, ΔH is defined by ΔH(0) = 1 and ΔH(i) = H(i) − H(i − 1) for i > 0, and Δ^{2}H = Δ(ΔH).)

More precisely we have the following, which gives a test for stopping our algorithm.

(1.4) Lemma. If t^{m_{p+1}} : gr(𝒮′)_{i} → gr(𝒮′)_{i+1} is onto (equivalently every element of 𝔐′_{i+1} is the sum of m_{p+1} and an element of 𝔐′_{i}) and #𝔐′_{i+1} = m_{p+1} then t^{m_{p+1}} : gr(𝒮′)_{j} → gr(𝒮′)_{j+1} is an isomorphism for j ≥ i + 1. Equivalently adding m_{p+1} gives a bijection from 𝔐′_{j} to 𝔐′_{j+1} for j ≥ i + 1.

Proof: If t^{m_{p+1}} : gr(𝒮′)_{i} → gr(𝒮′)_{i+1} is onto then (gr(𝒮′)/t^{m_{p+1}}gr(𝒮′))_{i+1} = 0. Since gr(𝒮′)/t^{m_{p+1}}gr(𝒮′) is generated by elements of degree one we have (gr(𝒮′)/t^{m_{p}+1}gr(𝒮′))_{j} = 0 for j ≥ i+1 from which it follows that t^{m_{p+1}} : gr(𝒮′)_{j} → gr(𝒮′)_{j+1} is onto for j ≥ i. Since gr(𝒮′)_{i+1} has k-dimension m_{p+1} and dim_{k} gr(𝒮′)_{j} = m_{p+1} for j sufficiently large the lemma follows.

(1.5) Note that gr(𝒮′) depends on the actual set 𝒮′, and not just on the monoid Γ′. Indeed if m_{0} = 1, then gr(𝒮′) may still be quite interesting, even though in this case Γ′ = ℕ. For example 𝒮′ = {1, 2, 3} gives the rational normal cubic in ℙ^{3} and 𝒮′ = {1, 3, 4} gives Macaulay’s example of a non-normal non-Cohen-Macaulay rational quartic in ℙ^{3}. However, repetitions of m_{i} would not change gr(𝒮′) and so there is no loss of generality in assuming that m_{0} < m_{1} < ⋯ < m_{p+1}.

(1.6) Motivated by (1.4) we define an element n ∈ Γ′ to be u n s t a b l e if there exists a ∈ ℕ such that

ord_{𝒮}′ (am_{p+1} + n) < a + ord_{𝒮′} (n) .

If no such a exists then we say that n is s t a b l e . (Equivalently n is unstable if and only if the element t^{n} ∈ gr(𝒮′) is killed by some power of t^{m_{p+1}}). Then from (1.4)(1) we have that gr(𝒮′) is Cohen-Macaulay if and only if 𝒮′ has no unstable elements.

(1.7) From [7] we have that for any 𝒮′ = {m_{0}, m_{1}, …, m_{p+1}} then if n is sufficiently large and 𝒮 = {m_{0}, m_{1}, …, m_{p+1}, n} then gr(𝒮) is Cohen-Macaulay. We will investigate the Cohen-Macaulay type of gr(𝒮) for certain fixed 𝒮′ as a function of n.

§2 The case of arithmetic progression.

As a first illustration of our methods we consider the case where the m_{i} form an arithmetic sequence. Let m, d be two relatively prime positive integers and m_{i} := m + id, 0 ≤ i ≤ p + 1, p ≥ 0. Let 𝒮′ = {m_{0}, m_{1}, …, m_{p+1}}. Then Θ′_{i} = {im_{0}, im_{0} + d, …, im_{p+1}} (a set with differences d). Note that #Θ′_{i} = i(p + 1) + 1. The following simple observation is very useful, and will be often be used without explicit mention.

(2.1) If Θ′_{i} ∩ Θ′_{j} ≠ ∅ then im_{0} ≡ jm_{0} (mod d). Since gcd(m, d) = 1 it follows that i ≡ j (mod d).

Let n = im_{0} + jd ∈ Θ′_{i}. If n ≤ (i − d)m_{p+1} then (noting that n ≥ (i − d)m_{0} since n ≥ im_{0}) we have n ∈ Θ′_{i−d} so ord_{𝒮′} (n) < i and hence n ∉ 𝔐′_{i}. If n > (i − d)m_{p+1} then n ∉ Θ′_{j} for any j < i so n ∈ 𝔐′_{i}. Thus 𝔐′_{i} = {n, n + d, …, im_{p+1}} (differences d) where n = Max(im_{0}, (i − d)m_{p+1} + d), and #𝔐′_{i} = Min(i(p + 1) + 1, m_{p+1}). In other words 𝔐′_{i} is either all of Θ′_{i} or is the largest m_{p+1} elements of Θ′_{i}. From this it follows that x → x + m_{p+1} is an inclusion from 𝔐′_{i−1} to 𝔐′_{i}. Hence, by (1.3), we have the following result.

(2.2) Theorem. Let m and d be relatively prime integers and let 𝒮′ be the arithmetic sequence {m_{0}, m_{1}, …, m_{p+1}} where m_{i} = m + id, 0 ≤ i ≤ p + 1. Then H_{gr(𝒮′)}(i) = min(i(p + 1) + 1, m_{p+1}) and gr(𝒮′) is Cohen-Macaulay.

This result is [5, Theorem 1.2, and Corollary 1.10] where it has a longer proof. The Cohen-Macaulay type in the arithmetic sequence case can also be determined by our methods, with the result that for gr(𝒮′) as in Theorem 2.2 we have that gr(𝒮′)/t^{m_{p+1}}gr(𝒮′) is level and gr(𝒮′) has Cohen-Macaulay type m − 1 (mod p + 1) if (p + 1) ∤ (m − 1) and p + 1 if (p + 1)|(m − 1). See the beginning of section 5 for a recall of level algebras and Cohen-Macaulay type. The argument is similar to the proof of Theorem 5.2, but simpler. These results have also been obtained in [5, Proposition 1.12 and Corollary 1.10].

§3 An almost arithmetic progression case.

After the arithmetic progression, the next most tractable case seems to be when all but one of the m_{i} lie in an arithmetic progression. This almost arithmetic progression case has been studied by various authors (for example [4] and the references therein, where the main tool has been the standard basis of the monoid, or [6] which studies ideal generators in the affine case). For the general almost arithmetic progression there seem to be too many possibilities to get a nice answer for the Hilbert function, so we study a special almost arithmetic progression which is more interesting than the arithmetic progression, but still manageable. Let 𝒮′ = {m, m_{1}, …, m_{p+1}} and 𝒮 = {d, m, m_{1}, …, m_{p+1}}, where m_{i} = m + id (0 ≤ i ≤ p + 1), d ≥ 1, m > 1, gcd(d, m) = 1, p ≥ 0. Let Γ be the monoid generated by 𝒮, Θ_{i} be the set of all sums (repetitions allowed) of i elements of 𝒮 and let 𝔐_{i} := Θ_{i} \ (∪_{j<i}Θ_{j}) , so that Γ, Θ and 𝔐 are constructed from 𝒮 in the same way that Γ′, Θ′ and 𝔐′ are constructed from 𝒮′. In this section we determine the 𝒮-order of all elements of Γ, and in the next we will count the number of elements of Γ with given order, so as to describe the Hilbert function of gr(𝒮). The result has a different flavour in the cases 1 ≤ d < m and d > m (in the latter case m < d = m_{1} − m < m_{1}).

If n ∈ Γ there is not a unique way of writing n as a sum of elements of 𝒮. We have found it convenient to avoid non-uniqueness by writing integers n in the form

(3.1) n = id + cm_{p+1} with 0 ≤ i ≤ m_{p+1} − 1, c ∈ ℤ.

This way we achieve uniqueness, but at the cost of requiring c negative for a finite number of n ∈ Γ. These n will be described in the course of our arguments. Throughout this section, whenever we write n = id + cm_{p+1}, i and c will be as in (3.1).

A key observation is the following:

(3.2) If n = id + cm_{p+1} ∈ Γ then in any 𝒮-expression n = i′d + ∑ c_{i}m_{i} for n we have that the number ∑ c_{i} of m’s (counting repetition) is congruent to c (mod d).

The proof is easy, and similar to that given in (2.1). Now we go through various cases, covering all integers n ≥ 0.

(3.3) If 0 ≤ i ≤ d < m and c < 0 then n ∉ Γ.

Proof: Suppose n = id+cm_{p+1} with 0 ≤ i ≤ d < m and c < 0. Then n = id+cm_{p+1} < dm + cm = (d + c)m (id < dm because i ≤ d and d < m, cm_{p+1} < cm because c < 0 and m < m_{p+1}). If n ∈ Γ then by (3.2) any 𝒮-expression for n must contain jd + c m’s. We must have j ≥ 1 so that jd + c ≥ 0. Furthermore n ≥ (jd + c)m. Then n < (d + c)m ≤ (jd + c)m ≤ n, which is a contradiction. Thus n ∉ Γ, as claimed.

(3.4) If n = id + cm_{p+1} with c ≥ 0 and 0 ≤ i ≤ d then ord_{𝒮}(n) = i + c and n is a stable element of Γ.

Proof: Here it is obvious that n ∈ Γ. Since n = id + cm_{p+1} is an 𝒮-expression for n we have ord_{𝒮}(n) ≤ i + c. By (3.2) any other 𝒮-expression for n will be of the form n = i′d + x where i′ ≥ 0 and x ∈ Θ′_{c+jd} (necessarily with c + jd ≥ 0). If j ≤ 0 write n = id + cm_{p+1} = (i − jm_{p+1})d + (c + jd)m_{p+1}. Clearly x ≤ (c + jd)m_{p+1} so i′ ≥ i − jm_{p+1}. Then i′ + c + jd ≥ i − jm_{p+1} + c + jd = i + c − j(m_{p+1} − d) ≥ i + c (since j ≤ 0 and m_{p+1} > d). If j > 0 then i′ + c + jd ≥ c + jd ≥ c + d ≥ c + i. In all cases our alternate 𝒮-expression for n is of cardinality greater than or equal to i + c so ord_{𝒮}(n) = i + c. If c′ > 0 and n′ = n + c′m_{p+1} then n′ = id + (c + c′)m_{p+1} and by the above ord_{𝒮}(n′) = i + c + c′, that is, n is stable.

(3.5) Suppose d > m and n = id + cm_{p+1} with c < 0, 0 ≤ i < d. Then n ∈ Γ if and only if m − (p + 1)c ≤ i < d. Furthermore ord_{𝒮}(n) = c + d, and n is an unstable element of Γ.

Proof: Suppose n ∈ Γ. Since c < 0 we cannot have n = cm. By (3.2) we must have n ≥ (d + c)m. Substituting the value of n we obtain id + (m + (p + 1)d)c ≥ (d + c)m. Cancelling cm and dividing by d yields i ≥ m − (p + 1)c, which, given the assumption that i < d, yields the ‘only if’ implication. From the inequalities d > i ≥ m − (p + 1)c we obtain that c + d > 0. Note also that n = id + cm_{p+1} < (c + d)m_{p+1} so n ∈ Θ′_{c+d}, which implies ord_{𝒮}(n) ≤ c + d. By (3.2) ord_{𝒮}(n) ≡ c + d (mod d) and since c < 0 it follows that ord_{𝒮}(n) = c + d. The ‘if’ implication holds because the above calculations show that n = id + cm_{p+1} and d > i ≥ m − (p + 1)c imply c + d > 0 and (c + d)m ≤ n < (c + d)m_{p+1} so n ∈ Θ′_{c+d} ⊂ Γ. Finally n is unstable since n + |c|m_{p+1} = n − cm_{p+1} = id so ord_{𝒮}(n + |c|m_{p+1}) ≤ i < c + d + |c| = ord_{𝒮}(n) + |c|.

(3.6) Suppose n = id + cm_{p+1} with d ≤ i < m_{p+1} and n ≥ (c + d)m. Then n is a stable element of Γ and ord_{𝒮}(n) = d + c.

Proof: We have (c + d)m ≤ n < (c + d)m_{p+1} (the second inequality holding since i < m_{p+1}) so c + d > 0 (which should be noted because c < 0 is possible here). Thus n ∈ Θ′_{c+d} and ord_{𝒮}(n) ≤ (c + d). By (3.2) every 𝒮-expression for n is of the form i′d + x where i′ ≥ 0 and x ∈ Θ′_{c+jd}. If j ≥ 1 this expression is of cardinality at least c + d. The 𝒮-expression of this type with j ≤ 0 of smallest cardinality is clearly the given expression n = id + cm_{p+1} (provided c ≥ 0. If c < 0 then Θ′_{c+jd} is empty for j ≤ 0 so we don’t have to worry about the case j ≤ 0) and by the assumption on i we have i + c ≥ d + c so ord_{𝒮}(n) = d + c. Finally n is stable by an argument similar to that given in (3.4).

(3.7) Suppose n = id + cm_{p+1} with c < 0 and n < (c + d)m. Then n ∉ Γ.

Proof: By (3.2) if n ∈ Γ we must have n = i′d+x with i′ ≥ 0 and x ∈ Θ′_{c+jd}, c+jd ≥ 0. Since c < 0 we must have j ≥ 1. But this is not possible if n < (c + d)m.

(3.8) Suppose n = id + cm_{p+1} with c ≥ 0, d < i < m_{p+1} and n < (c + d)m. Then n is an unstable element of Γ and ord_{𝒮}(n) = i + c. For a given value of c ≥ 0, the conditions on n and i are equivalent to d < i < m − (p + 1)c (so that this case can only occur for d < m).

Proof: Since n < (c + d)m, by (3.2) we cannot have an 𝒮-expression for n with more than c m_{i}’s. Using fewer m’s will lead to a larger cardinality of the 𝒮-expression because d < m_{p+1}. More formally suppose also that n = i′d+x where x ∈ Θ′_{c−jd}, j ≥ 1. We need to show that i′ +c−jd > i+c. (Note that since j ≥ 1 we must have i′ > i. In particular, i′ ≠ i.) From id + cm_{p+1} = i′d + x we have cm_{p+1} − x = (i′ − i)d < (i′ − i)m_{p+1} from which it follows that (c + i − i′)m_{p+1} < x. Since x ∈ Θ′_{c−jd} we have x ≤ (c − jd)m_{p+1}. Now we have (c + i − i′)m_{p+1} < (c − jd)m_{p+1} from which the desired inequality i′ + c − jd > i + c follows. Therefore ord_{𝒮}(n) = i + c. Since m < m_{p+1} we have n + jm_{p+1} ≥ (c + j + d)m for j sufficiently large. By (3.6) we have ord_{𝒮}(n + jm_{p+1}) = c + j + d < c + j + i = j + i + c = j + ord_{𝒮}(n). Therefore n is unstable.

The condition n < (c + d)m is equivalent to id + c(m + (p + 1)d) < (c + d)m which after cancelling cm from both sides of the inequality and rearranging is seen to be equivalent to i < m − (p + 1)c. Given the assumption that d < i, the last inequality is equivalent to d < i < m − (p + 1)c as claimed.

§4 Examples and description of the Hilbert function.

Let 𝒮 = {d, m_{0}, …, m_{p+1}) as introduced in the previous section, where d, m, p and 𝔐_{j} retain their previous meaning. We are interested in the Hilbert function H(j) = dim_{k}(gr(𝒮)_{j}) = #𝔐_{j}, which is the difference sequence of the Hilbert function of the corresponding projective monomial curve C. Since we will be mostly working directly with 𝔐_{j} (rather than the homogeneous co-ordinate ring of C) j will be referred to as the order of an integer in 𝔐_{j}.

(4.1) First consider the stable elements of type (3.4). These are of the form n = id + cm_{p+1} for c ≥ 0, 0 ≤ i ≤ d and have order j = i + c. For a given j ≥ 0 these integers are of the form n = id + (j − i)m_{p+1} for 0 ≤ i ≤ min(j, d). They form an arithmetic sequence with differences m_{p+1} − d = m_{p} and largest element jm_{p+1} (corresponding to i = 0). If j ≤ d the smallest element in the sequence is jd (corresponding to i = j) and there are j + 1 of them. If j ≥ d the smallest element is d^{2} + (j − d)m_{p+1} (corresponding to i = d) and there are d + 1 of them.

(4.2) Now consider the stable integers of the form (3.6). These are of the form n = id + cm_{p+1} for d ≤ i < m_{p+1}, n ≥ (c + d)m and have order j = d + c. For a given order j the condition n ≥ (c+d)m is equivalent to i ≥ (d−j)(p+1)+m, so the integers n of order j are of the form n = id + (j − d)m_{p+1} for max(d, m_{p+1} − j(p + 1)) ≤ i < m_{p+1}. They form an arithmetic sequence with largest element jm_{p+1} − d for i = m_{p+1} − 1 and differences d. (If j = 0 there are no such integers, if j = 1 we get m = m_{0}, m_{1}, …, m_{p}. The progression is extended by jm_{p+1}, which we have counted in (4.1).) If m_{p+1} − j(p + 1) ≥ d there are j(p + 1) integers of type 3.6, with the smallest being jm. If m_{p+1} − j(p + 1) ≤ d there are m_{p+1} − d = m_{p} integers of type 3.6, with the smallest being n = d^{2} + (j − d)m_{p+1}.

(4.3) Note that if j ≥ d and m_{p+1} −j(p+1) ≤ d (the last inequality being equivalent to j ≥ (m_{p+1}−d)/(p+1)) then (4.1) and (4.2) contain the same smallest element (this is the first place less than jm_{p+1} that the progressions intersect since gcd(d, m) = 1) so when describing the Hilbert function we have to take care to avoid double counting. This arises from i = d being allowed in both (3.4) and (3.6). However either d or (m_{p+1} − d)/(p + 1) could be the larger so we cannot make an arbitrary choice and put i = d in one place but not the other.

(4.4) Now consider the unstable elements of the type (3.5). These occur only for d > m and are of the form n = id+cm_{p+1} with c < 0, m−(p+1)c ≤ i < d. They have order c + d. For a given order j they are the integers of the form n = id + (j − d)m_{p+1} with j < d and m_{p+1} − j(p + 1) = m + (p + 1)(d − j) ≤ i < d. They continue the progression of (4.2) so that (4.4) and (4.2) together in order j < d consist of the integers n = id + (j − d)m_{p+1} with m_{p+1} − j(p + 1) ≤ i < m_{p+1}, making a total of j(p + 1) integers in all. This set is vacuous for m < d ≤ m + p + 1 because then even c = −1 yields an empty set of possible i values, and non-empty for d > m + p + 1.

(4.5) Now we describe the unstable integers of case (3.8). These occur only for d < m and are of the form n = id + cm_{p+1} where c ≥ 0, d < i < m − (p + 1)c and have order i + c. It is convenient to arrange these integers in a diagram whose rows correspond to c = 0, c = 1, … respectively. This set is empty if d = m − 1 because then even the c = 0 row is empty, and non-empty if d < m − 1.

(d + 1)d  (d + 2)d  …  (m − 1)d
(d + 1)d + m_{p+1} (d + 2)d + m_{p+1} … (m − 1 − (p + 1))d + m_{p+1}
⋮

The first row is the integers id, (d < i ≤ m−1) arranged in increasing order. Each successive row is obtained from the preceding by adding m_{p+1} in each co-ordinate and dropping the last p+1 entries. The i^{th} row contains m−d−1−(i−1)(p+1) elements so there are

σ = ⌈\frac{m − d − 1}{p + 1}⌉

rows and the last row contains τ = m−d−1−(p+1)(σ −1) elements (1 ≤ τ ≤ p+1).

The integers of a given 𝒮-order d + 1, d + 2, d + 3, …, are the upward diagonals {(d + 1)d}, {(d + 1)d + m_{p+1}, (d + 2)d}, … starting in the upper left hand corner.

(4.6) Remark. If we keep d and m the same in (4.4), and vary p, then the set of unstable elements for d, m, p is contained in the set of unstable elements for d, m, p−1. The largest number of unstable elements occurs for p = 0, and is 0 if p becomes large enough. In contrast to this if we keep d and m fixed in (4.5) and increase p then the first row of our table stays the same, and a greater number p + 1 of elements are dropped in passing to the next row. If p is large enough we are left only with the first row, which remains no matter how large p becomes.

Now we give a couple of examples to illustrate the above considerations. In each case the entries of type (4.1) are in bold type, those of type (4.2) are in normal type, those of both are doubly underlined, and the unstable elements are singly underlined.

(4.7) Example. Let 𝒮 = {4, 11, 15, 19}, so that d = 4, m = 11, p = 1, σ = ⌈6/2⌉ = 3 and τ = 11 − 4 − 1 − 3 = 3. In the following table row j is the set 𝔐_{j}, 1 ≤ j ≤ 11. Subsequent rows are obtained by adding 19 = m_{p+1} to the preceding row.

{4,11,15,19}
{8, 22, 23, 26, 30, 34, 38}
{12, 27, 33, 37, 41, 42, 45, 49, 53, 57}
{16, 31, 44, 46, 48, 52, 56, 60, 61, 64, 68, 72, 76}
\{\underline{20}, 35, 50, 55, 59, 63, 65, 67, 71, 75, 79, 80, 83, 87, 91, 95\}
\{\underline{24}, \underline{39}, 54, 66, 69, 70, 74, 78, 82, 84, 86, 90, 94, 98, 99, 102, 106, 110, 114\}

\{\underline{28}, \underline{43}, \underline{58}, 73, 77, 81, 85, 88, 89, 93, 97, 101, 103, 105, 109, 113, 117, 118, 121, 125, 129, 133\} \{\underline{32}, \underline{47}, \underline{62}, \underline{\underline{92}}, 96, 100, 104, 107, 108, 112, 116, 120, 122, 124, 128, 132, 136, 137, 140, 144, 148, 152\} \{\underline{36}, \underline{51}, \underline{\underline{111}}, 115, 119, 123, 126, 127, 131, 135, 139, 141, 143, 147, 151, 155, 156, 159, 163, 167, 171\} \{\underline{40}, \underline{\underline{130}}, 134, 138, 142, 145, 146, 150, 154, 158, 160, 162, 166, 170, 174, 175, 178, 182, 186, 190\} \{\underline{\underline{149}}, 153, 157, 161, 164, 165, 169, 173, 177, 179, 181, 185, 189, 193, 194, 197, 201, 205, 209\}

The table of unstable elements given by (4.5) is

20 24 28 32 36 40
39 43 47 51
58 62

yielding elements 20 of order d + 1 = 5, {24, 39} of order 6, {28, 43, 58} of order 7, {32, 47, 62} of order 8, {36,51} of order 9, and 40 of order 10, with no elements in order m = 11, as obtained above by direct calculation.

(4.8) Example. Let 𝒮 = {8, 3, 11, 19}, so that d = 8, m = 3, p = 1. In the following table row j is the set 𝔐_{j}, 1 ≤ j ≤ 8. Subsequent rows are obtained by adding 19 = m_{p+1} to the preceding row.

{3,8,11,19}
{6,14,16,22,27,30,38}
{9,17,24,25,33,35,41,46,49,57}
{12,20,28,32,36,43,44,52,54,60,65,68,76}
{15,23,31,39,40,47,51,55,62,63,71,73,79,84,87,95}
\{\underline{18},26,34,42,48,50,58,59,66,70,74,81,82,90,92,98,103,106,114\}

\{\underline{21},\underline{29},\underline{37},45,53,56,61,67,69,77,78,85,89,93,100,101,109,111,117,122,125,133\} \{\underline{\underline{64}},72,75,80,86,88,96,97,104,108,112,119,120,128,130,136,141,144,152\}

Now we are ready to assemble the above results into an explicit description of the Hilbert function H. First we have

(4.9) Theorem. Let 𝒮 = {d, m_{0}, …, m_{p+1}} where gcd(m, d) = 1, m_{i} = m + id for 0 ≤ i ≤ p + 1, m > 1. Suppose d > m. Then H(j) = 1 + (p + 2)j for 0 ≤ j ≤ d − 1 and H(j) = m_{p+1} for j ≥ d.

Proof: If j < d there are j + 1 integers of order j given by (4.1) and j(p + 1) by (4.4) (some of which may be unstable) making a total of 1 + j(p + 2) as claimed. If j ≥ d then we obtain d+1 integers of order j from (4.1). We have m_{p+1} −j(p+1) ≤ m_{p+1} − d(j + 1) = m < d so from (4.2) we obtain m_{p} making a total of d + 1 + m_{p} − 1 = m_{p+1} as claimed. (We subtract 1 because of the double counting discussed in (4.3). In any case these elements are all stable and there cannot be more than m_{p+1} stable elements in any given degree.)

(4.10) Now we consider the case d < m, which is a bit more difficult. Here we have stable elements of types (4.1) and (4.2) and unstable elements, if present, are of type (4.5). If 0 ≤ j ≤ d there are j + 1 stable elements of type (4.1) of order j after which there are d + 1 of them. There are {j(p + 1)} stable elements of type (4.2) in order j so long as j(p + 1) ≤ m_{p} after which there are m_{p} of them. The smallest order in which m_{p} is attained is ρ := ⌈m_{p}/(p + 1)⌉ = ⌈(m + pd)/(p + 1)⌉. Note that d + σ = ⌈(m + pd − 1)/(p + 1)⌉ so ρ = d + σ + 1 if (m + dp − 1)/(p + 1) is an integer (equivalently (m − d − 1)/(p + 1) is an integer, equivalently τ = p + 1) and ρ = d + σ otherwise. The correction term of (4.3) is applied when the two types of stable element have both reached their maximal cardinality, that is, if and only if j ≥ ρ so the number of stable elements is the sum of the numbers of elements of the two types if 0 ≤ j ≤ ρ − 1 and m_{p+1} if j ≥ ρ.

By the discussion in (4.5) these unstable elements have Hilbert function

1 2 3 … σ − 1 σ σ … σ σ − 1 … σ − 1 σ − 2 … σ − 2 … 1 1 … 1 0

starting with 1 in order d + 1,with increments of 1 until reaching σ in order d + σ staying at σ until order d + σ + τ − 1 (τ times in all), then decreasing with each of the values σ − j, 1 ≤ j ≤ σ − 1 occurring p times, so that σ − 1 first occurs in order d + σ + τ, σ − 2 first occurs in order d + σ + τ + p, …, 0 first occurs in order d + σ + τ + (σ − 1)p = m. On the interval d + 1 ≤ j ≤ ρ − 1 (≤ d + σ) the unstable elements compensate for the stabilization of the type (4.1) elements, and althogether H(j) = 1 + j(p + 2) for 0 ≤ j ≤ ρ − 1. On the interval ρ ≤ j ≤ m − 1 the unstable elements of order j are most easily counted by working backwards from 0 in order m and 1 in order m − 1 as ⋯ 2 2 1 1 ⋯ 1 1 0 with each value occuring p times (except perhaps the left most, which can occur fewer times). This pattern continues all the way down to ρ whether we have ρ = d+σ or ρ = d+σ +1. (However it is crucial that when τ = p + 1 we have ρ = d + σ + 1 so that the smallest element in the bottom row of the table of unstable elements is of order ρ − 1.) The number of unstable elements in order j, ρ ≤ j ≤ m − 1, is thus seen to be ⌈(m − j)/p⌉. If j ≥ ρ there are m_{p+1} stable elements, so H(j) = m_{p+1} + ⌈(m − j)/p⌉ for ρ ≤ j ≤ m − 1. If j ≥ m there are only the m_{p+1} stable elements, so H(j) = m_{p+1} for j ≥ m. These results can be summarised by

(4.11) Theorem. Let 𝒮 = {d, m_{0}, …, m_{p+1}} where gcd(m, d) = 1, m_{i} = m + id for 0 ≤ i ≤ p + 1, m > 1. Suppose d < m and let ρ := ⌈m_{p}/(p + 1)⌉. Then

H(j) = \{ \begin{matrix}j(p + 2) + 1 & for 0 ≤ j < ρ \\ m_{p+1} + ⌈\frac{m−j}{p} ⌉ & for ρ ≤ j < m \\ m_{p+1} & for m ≤ j \end{matrix}

Note that (if d < m) there are no unstable elements of type (4.5) if and only if d = m − 1, in which case ρ = m and the second case in Theorem 4.11 is empty. Furthermore if p = 0 then also ρ = m so the second case is again empty. However one might prefer to require p > 0 in Theorem 4.11 so as to avoid the appearance of dividing by 0. If p = 0 we can interchange d and m and apply Theorem 4.9.

(4.12) Remark. Clearly if gr(𝒮) is Cohen-Macaulay (i.e. d = m − 1 or m < d ≤ m + p + 1) then the h-vector ΔH of R/J contains no negative entries. Conversely suppose that gr(𝒮) is not Cohen-Macaulay. If d > m + p + 1 then by Theorem 4.9 ΔH(d) = m + p + 1 − d < 0. If d < m − 1 and p = 0 we can interchange d and m to obtain the case p = 0 and d > m + 1 = m + p + 1 and again conclude by Theorem 4.9 that ΔH has a negative value. If d < m − 1 and p > 0 an easy calculation shows that ρ < m and then by Theorem 4.11 we have ΔH(m) = −⌈1/p⌉ = −1. Thus in all cases the h-vector contains no negative values if and only if R/J is Cohen-Macaulay. Recently Victoria De Quehen, an NSERC undergraduate summer research student of the second author, has found many projective monomial curves whose homogeneous co-ordinate ring is not Cohen-Macaulay and whose h-vector contains no negative values. However extensive searching has found no such curves arising from an almost arithmetic progression.

(4.13) Examples. Here we look at our two examples.

(i) In Example 4.7 we have p = 1, d = 4, m = 11, ρ = 8. Theorem 4.11 predicts H(j) = 1 + 3j for 0 ≤ j ≤ 7, H(j) = m_{p+1} = 19 for j ≥ 11, H(10) = 20, H(9) = 21 and H(8) = 22, i.e. H is the sequence 1 4 7 10 13 16 19 22 22 21 20 19 →, which is indeed that observed in the Example.

(ii) In Example 4.8 we have m = 3, d = 8, p = 1 which is covered by Theorem 4.9. This predicts Hilbert function H(j) = 1 + 3j, 0 ≤ j ≤ d − 1 = 7 and H(j) = m_{p+1} = 19 for j ≥ d = 8, i.e H is the sequence 1 4 7 10 13 16 19 22 19 →, which is indeed what was observed in the Example.

It is easy to find examples of all the cases of our theorems. We have checked many with our computer programs, providing evidence for the correctness of our results.

§5 The Cohen-Macaulay type of 𝒮.

For preliminary discussion in this section 𝒮 can be an arbitrary set {n_{0}, n_{1}, …, n_{p+1}} with 1 ≤ n_{0} < n_{1} < ⋯ < n_{p+1}. Our first observation is that the discussion of section 1 gives an effective algorithm for deciding whether or not gr(𝒮) is Cohen-Macaulay. By (1.3) gr(𝒮) is Cohen-Macaulay if and only if x → x + n_{p+1} maps 𝔐_{j} injectively into 𝔐_{j+1}. By Lemma 1.4 this involves a finite amount of checking and we can decide when we are done.

Now we discuss how to determine the Cohen-Macaulay type of gr(𝒮), in the case where gr(𝒮) is Cohen-Macaulay. First a definition.

(5.1) Definition. Let 𝒮 = {n_{0}, n_{1}, …, n_{p+1}} with 1 ≤ n_{0} < n_{1} < ⋯ < n_{p+1} and suppose that t^{n_{p+1}} is a non-zero-divisor in gr(𝒮). Let 𝒮′ = {n_{0}, n_{1}, …, n_{p}} and Γ the submonoid of ℕ generated by 𝒮. Let x ∈ Γ\(n_{p+1} + Γ). Then we will say that x is a maximal element of Γ\(n_{p+1} + Γ) if for all j ∈ 𝒮′ we have either x + j ∈ n_{p+1} + Γ or ord_{𝒮}(x + j) < ord_{𝒮}(x) + ord_{𝒮}(j).

Note that if x ∈ Γ\(n_{p+1} +Γ) then ord_{𝒮}(x) = ord_{𝒮′} (x). Definition 5.1 has been set up so that x is maximal if and only if t^{x} is a non-zero element of gr(𝒮)/t^{n_{p+1}}gr(𝒮) which annihilates the homogeneous maximal ideal of gr(𝒮)/t^{n_{p+1}}gr(𝒮). Since Γ\(n_{p+1} + Γ) is finite the maximal elements are readily determined.

Recall that the Cohen-Macaulay type of gr(𝒮) is the dimension of the socle of the Artinian ring gr(𝒮)/t^{n_{p+1}}gr(𝒮). The socle of gr(𝒮)/t^{n_{p+1}}gr(𝒮) is the annihilator of the graded maximal ideal of gr(𝒮)/t^{n_{p+1}}gr(𝒮) and has k-basis {t^{M}|M is a maximal element of Γ\(n_{p+1} + Γ)}.

Now we return to the situation 𝒮 = {d, m_{0}, …, m_{p+1}}, and Γ the monoid generated by 𝒮, as in the previous two sections. By (1.3) gr(𝒮) is Cohen-Macaulay if and only if there are no unstable elements in Γ. If d < m then by (4.5) gr(𝒮) is Cohen-Macaulay if and only if d = m − 1. If d > m then by (4.4) gr(𝒮) is Cohen-Macaulay if and only if m < d ≤ m + p + 1. In this section we determine the Cohen-Macaulay type of gr(𝒮) in these cases.

As usual H(j) will denote dim_{k}(gr(𝒮)_{j}) = #𝔐_{j}. Since we are in the Cohen-Macaulay case, t^{m_{p+1}} is a non-zero divisor in gr(𝒮), and gr(𝒮)/t^{m_{p+1}}gr(𝒮) has basis {t^{n}|n ∈ Γ\(m_{p+1} + Γ)}, with the degree j part having basis {t^{n}|n ∈ 𝔑_{j} := 𝔐_{j}\(m_{p+1} + 𝔐_{j−1})}. Furthermore ΔH(j) = dim_{k}(gr(𝒮)/t^{m_{p+1}}gr(𝒮))_{j} = #𝔑_{j}. Recall also that an Artinian ℕ-graded k-algebra is level if the socle is concentrated in one degree. Our result then is

(5.2) Theorem. Let 𝒮 = {d, m_{0}, …, m_{p+1}} where gcd(m, d) = 1, m_{i} = m + id for 0 ≤ i ≤ p+1, d = m−1 or m < d ≤ m+p+1 (so that we are in the Cohen-Macaulay case) and Γ the monoid generated by 𝒮. Then the k-algebra gr(𝒮)/t^{m_{p+1}}gr(𝒮) is level (with socle in degree d if d = m − 1 or m < d < m + p + 1 and degree d − 1 if d = m + p + 1), and the Cohen-Macaulay type of gr(𝒮) is p + 2 if d = m − 1, m_{p+1} − 1 − (p + 2)(d − 1) if m < d < m + p + 1, and p + 2 if d = m + p + 1. The Cohen-Macaulay type of gr(𝒮) is one (i.e. gr(𝒮) is Gorenstein) if and only if d = m + p.

Proof: If d = m−1 by Theorem 4.11 ΔH(j) = p+2 for 1 ≤ j ≤ d. If m < d ≤ m+p+1 by Theorem 4.9 ΔH(j) = p+2 for 1 ≤ j ≤ d−1 and ΔH(d) = m_{p+1}−1−(p+2)(d−1). ΔH(j) = 0 for all other j > 0. Noting that m_{p+1} − 1 − (p + 2)(d − 1) = m − d + p + 1 (which is 0 if d = m + p + 1 and non-zero if m < d < m + p + 1) the claimed Cohen-Macaulay type will follow if we show that gr(𝒮)/t^{m_{p+1}}gr(𝒮) is level. This is equivalent to asserting that the maximal elements of Γ\(m_{p+1}+Γ) occur only in the largest order for which ΔH is non-zero. We will first analyse the case ΔH(j) = ΔH(j − 1) = p + 2 (i.e. 2 ≤ j ≤ d if d = m − 1 and 2 ≤ j ≤ d − 1 if d > m). We can give a common analysis for both the cases d < m and d > m. Because ΔH still has its maximal value of p + 2 the discussion of (4.1)(4.2) shows that 𝔑_{j−1} consists of {(j − 1)d} from (4.1) and {(j − 1)m, (j − 1)m + d, …, (j − 1)m + pd} from (4.2). Similarly 𝔑_{j} consists of {jd} from (4.1) and {jm, jm + d, …, jm + pd} from (4.2). Since (j − 1)d + d = jd and (j − 1)m + id + m = jm + id, 𝔑_{j−1} contains no maximal elements.

It remains only to show that 𝔑_{d−1} contains no maximal elements of Γ\(m_{p+1}+Γ) if m < d < m+p+1. In this case 𝔑_{d−1} = {(d−1)d, (d−1)m, (d−1)m+d, …, (d−1)m+ pd} as described above (except for the first term (d−1)d, this is an arithmetic sequence with differences d, containing p + 1 elements) and 𝔑_{d} = {d^{2}, d^{2} + d, …, d(m + p)} (an arithmetic sequence with differences d containing m + p + 1 − d = p + 1 − (d − m) elements). Since (d − 1)d + d = d^{2} the element (d − 1)d of 𝔑_{d−1} is not maximal. Since d^{2} − (d − 1)m = d(d − m) + m ∈ 𝒮 the smallest p + 1 − (d − m) elements of 𝔑_{d−1}\{(d − 1)d} differ from the corresponding element of 𝔑_{d} by an element of 𝒮 and hence are not maximal. Since d(m + p) − (d − 1)m − pd = m the remaining d − m elements of 𝔑_{d−1} differ from d(m + p) ∈ 𝔑_{d} by an element of 𝒮 and hence are not maximal either, completing the proof.

The following example may help the reader understand the above theorem.

(5.3) Example. Let 𝒮 = {5, 4, 9, 14, 19}, so that d = 5, m = 4 and p = 2. The j^{th} row of the following table is 𝔐_{j}, j ≥ 0. The normal type entries in row j are 19 + 𝔐_{j−1}, i.e. 19 more than the corresponding entries of the previous row. The remaining entries in row j make up 𝔑_{j} and are in bold.

{0}
{4, 5, 9, 14, 19}
{8, 10, 13, 18, 23, 24, 28, 33, 38}
{12, 15, 17, 22, 27, 29, 32, 37, 42, 43, 47, 52, 57}
{16, 20, 21, 26, 31, 34, 36, 41, 46, 48, 51, 56, 61, 62, 66, 71, 76}
{25, 30, 35, 39, 40, 45, 50, 53, 55, 60, 65, 67, 70, 75, 80, 81, 85, 90, 95}
{44, 49, 54, 58, 59, 64, 69, 72, 74, 79, 84, 86, 89, 94, 99, 100, 104, 109, 114}

For example 𝔑_{3} = {12, 15, 17, 22}, 𝔑_{4} = {16, 20, 21, 26} and 15+5 = 20, 12+4 = 16, 17+4 = 21, 22+4 = 26 so none of the elements of 𝔑_{3} is maximal in Γ\(m_{p+1}+Γ). Similarly 𝔑_{5} = {25, 30} and 20 + 5 = 25, 16 + 9 = 25, 21 + 9 = 30, 26 + 4 = 30 so none of the elements in 𝔑_{4} is maximal in Γ\(m_{p+1} +Γ). The algebra gr(𝒮)/t^{m_{p+1}}gr(𝒮) is level, with its socle having k-basis t^{25}, t^{30}, both of order 5, and the Cohen-Macaulay type is 2 = 19 − 1 − 4 · 4, as claimed by the Theorem.

§6 A different type of example.

Let 𝒮′ = {1, m, m+1, …, m+p+1} (so this is the previous type of example where d = 1, and m and p have their usual meanings), and let 𝒮 = {1, m, m+1, …, m+p+1, n} for some n > m + p + 1. As usual Γ is the submonoid of ℕ generated by 𝒮 and Γ′ is the submonoid generated by 𝒮′. (Of course, since d = 1, Γ = Γ′ = ℕ but the notation Γ and Γ′ is still useful because elements of ℕ will have different orders as elements of Γ or Γ′). In this section we determine a necessary and sufficient condition for 𝒮 to be Cohen-Macaulay, and find the Cohen-Macaulay type of those that are Cohen-Macaulay. This work is motivated by [7] where such examples are used to give a counterexample to the Hibi conjecture. First of all, by (1.3) and [7] Theorem 3 we have

(6.1) Theorem. Let 𝒮′ and 𝒮 be as above. Then

(1) If n is sufficiently large then gr(𝒮) is Cohen-Macaulay.

(2) If gr(𝒮) is not Cohen-Macaulay then there is an integer a such that ord_{𝒮′} (a) ≥ ord_{𝒮′} (a + n).

Now we have

(6.2) Theorem. Let 𝒮′ and 𝒮 be as defined above. Then gr(𝒮) is Cohen-Macaulay if and only if n > (m − 1)(m + p).

Proof: If m = 2 the hypothesis n > (m − 1)(m + p) = p + 2 is satisfied by any n that is greater than m + p + 1 so gr(𝒮) should always be Cohen-Macaulay. Indeed if m = 2 then 𝔐_{j} = {i|(j−1)(m+p+1)+1 ≤ i ≤ j(m+p+1)} (p+3 consecutive integers) from which it is clear that no integer a as in 6.1(2) exists. So suppose that m > 2. Then by (3.8) ord_{𝒮′} (m−1) = m−1 (actually, this assertion is obvious). Hence the smallest and largest elements of 𝔐′_{m−1} are respectively m−1 and (m−1)(m+p+1), with difference (m−1)(m+p). By (4.5) there are no unstable elements of gr(𝒮′) in order j ≥ m, so if j ≥ m then by (4.2) 𝔐′_{j} consists of the integers (j −1)(m+p+1)+1, …, j(m+p+1) (differences 1, the (4.1) elements being (j − 1)(m + p + 1) + 1 and j(m + p + 1)). Consequently if 0 ≤ x ≤ (m − 1)(m + p + 1) then ord_{𝒮′} (x) ≤ m − 1. Now suppose that m + p + 1 < n ≤ (m − 1)(m + p). Then m − 1 + n ≤ m − 1 + (m − 1)(m + p) = (m − 1)(m + p + 1) so ord_{𝒮}(m − 1 + n) ≤ ord_{𝒮′} (m − 1 + n) ≤ m − 1 < ord_{𝒮}(m − 1) + ord_{𝒮}(n) = m − 1 + 1 = m so t^{m−1} · t^{n} = 0 in gr(𝒮), so the latter cannot be Cohen-Macaulay. On the other hand if n > (m − 1)(m + p) we claim that there is no integer a as in Theorem 6.1(2). For if ord_{𝒮′} (a) < m − 1 then ord_{𝒮′} (a + n) > ord_{𝒮′} (a) because the largest integer of order ord_{𝒮′} (a) is at most (m − 2)(m + p + 1) and a + n ≥ n > (m − 1)(m + p) > (m − 2)(m + p + 1). If ord_{𝒮′} (a) ≥ m − 1 then ord_{𝒮′} (a + n) > ord_{𝒮′} (a) by the above description of 𝔐_{j} for j ≥ m − 1 and the way the integer (m − 1)(m + p) was chosen. It now follows from 6.1(2) that gr(𝒮) is Cohen-Macaulay. This completes the proof.

Now we find the Cohen-Macaulay type of gr(𝒮) in the Cohen-Macaulay case n > (m − 1)(m + p). We have Γ = Γ′ = ℕ, so Γ\(n + Γ) = [0, n − 1] := {i|0 ≤ i ≤ n − 1} (with x ∈ [0, n − 1] having order ord_{𝒮}(x) = ord_{𝒮′} (x)). This leads to the description of gr(𝒮)/t^{n}gr(𝒮) in [7,Theorem 3(e)]. Namely gr(𝒮)/t^{n}gr(𝒮) is isomorphic to the ring gr′(𝒮) obtained from gr(𝒮′) by setting to 0 all t^{j} with j ≥ n. (It is important for this observation that d = 1. If d > 1 then t^{j} is non-zero in gr(𝒮)/t^{n}gr(𝒮) for some j > n so the analysis would be more complicated if d > 1.) The Cohen-Macaulay type of gr(𝒮) will then be the number of maximal elements in [0, n − 1] (as defined in Definition 5.1) which we find in the following theorem.

(6.3) Theorem. Let 𝒮′ and 𝒮 be as defined at the beginning of this section, with n > (m − 1)(m + p) if m > 2 and n > m + p + 1 if m = 2. Then the Cohen-Macaulay type of gr(𝒮) is σ + ℓ where σ = ⌈\frac{m−2}{p+1}⌉ (the σ associated to 𝒮′) and

ℓ = \{ \begin{matrix}m + p + 1, & if n ≡ 1 (mod m + p + 1); \\ m − 1, & if n ≡ i (mod m + p + 1) where 2 ≤ i ≤ m; \\ i − 1, & if n ≡ i (mod m + p + 1) where m < i ≤ m + p + 1. \end{matrix}

Proof: We continue the discussion begun before the statement of the Theorem. Note that ord_{𝒮′} (n) ≥ m − 1 (≥ m if m = 2) and that gr′(𝒮) is equal to gr(𝒮′) in orders less than ord_{𝒮′} (n). Furthermore the unstable elements of Γ′ are all in [0, n − 1]. The maximal elements of [0, n−1] are of two types, those that are unstable elements of Γ′, and those that are stable elements of Γ′. There is one unstable maximal element from each row in the discussion of (4.5), namely the largest element of each row, hence the term σ. Let 𝔖′_{j} denote the stable elements of order j in Γ′. Then it follows from the discussion in (4.1) and (4.2) that 𝔖′_{j} = {1 + (j − 1)(m + p + 1)} ∪ {i|Max(jm, 1 + (j − 1)(m+p+1)) ≤ i ≤ j(m+p+1)} (with 1+(j −1)(m+p+1) being the unique element of type (4.1) other than j(m+p+1)). If j ≥ m−1 then 1+(j −1)(m+p+1) ≥ jm−1 so 𝔖′_{j} = {i|1 + (j − 1)(m + p + 1) ≤ i ≤ j(m + p + 1)} (m + p + 1 elements in all), and if m > 2 then 𝔖′_{m−2} ⊆ {i|(m − 3)(m + p + 1) + 1 ≤ i ≤ (m − 2)(m + p + 1)} (with equality except if p = 0 or m = 3, in which cases only (m − 3)(m + p + 1) + 2 is missing from 𝔖′_{m−2}). Now let j = ord_{𝒮′} (n) (so that j ≥ m − 1, with equality if m > 2 and n has its smallest possible value (m − 1)(m + p) + 1). The maximal elements of [0, n − 1] arising from stable elements are all elements of 𝔖′_{j} that are less than n, and all elements of 𝔖′_{j−1} that are not ‘covered’ by an element of 𝔖′_{j} that survives in [0, n − 1]. Elements of lower order are clearly not maximal since if i < j − 1 then addition of m + p + 1 maps 𝔖′_{i} into 𝔖′_{i+1}. (We will say that x is covered by y if y − x ∈ 𝒮′. Then if x, y < n with ord_{𝒮}(y) = 1 + ord_{𝒮}(x) then x is not a maximal element of [0, n − 1]). We now count these elements, first assuming that 𝔖′_{j−1} = {i|(j − 2)(m + p + 1) + 1 ≤ i ≤ (j − 1)(m + p + 1)} (which is the case except possibly if m > 2 and j = m−1). Suppose that n ≡ 1 (mod m + p + 1). Then there are no elements of 𝔖′_{j} that are less than n, so none of the elements of 𝔖′_{j−1} are covered by an element of order j in [0, n−1]. There are m+p+1 elements in 𝔖′_{j−1} so ℓ = m + p + 1, which is the first case of ℓ. Now suppose that n ≡ 2 (mod m + p + 1). This yields one maximal element of order j in [0, n − 1], namely (j − 1)(m + p + 1) + 1. This covers the largest element (j − 1)(m + p + 1) of 𝔖′_{j−1} since 1 ∈ 𝒮. Similarly the p+2 consecutive integers (j−1)(m+p+1)+1−(m+p+1), …, (j−1)(m+p+1)+1−m at the beginning of 𝔖′_{j−1} are covered by (j − 1)(m + p + 1) + 1. This yields p + 3 elements that are covered, and hence m + p + 1 − (p + 3) = m − 2 that are not. Althogether ℓ = (m − 2) + 1 = m − 1. If we increase n by 1 then we gain one more maximal element of [0, n − 1] of order j and cover one more element of [0, n − 1] so ℓ remains at m − 1. This process continues until all of 𝔖′_{j−1} is covered, which occurs when (j − 1)(m + p + 1) + i − m = (j − 1)(m + p + 1), i.e. when i = m. This yields the second case of ℓ. Finally if n ≡ i where m < i ≤ m + p + 1 then all of 𝔖′_{j−1} is covered, so [0, n − 1] has only the i − 1 maximal elements of order j, which is the third case of ℓ. To conclude, notice that if j = m − 1 then the smallest element of {i|(m−3)(m+p+1)+1 ≤ i ≤ (m−2)(m+p+1)} not covered by an element of order m − 1 in [0, n − 1] (assuming that there are any such elements) is n − m and n − m > (m−1)(m+p)−m ≥ Max((m−2)m, 1+(m−3)(m+p+1)) so by the above description of 𝔖′_{m−2} all uncovered elements of {i|(m−3)(m+p+1)+1 ≤ i ≤ (m−2)(m+p+1)} are contained in 𝔖′_{m−2}. Thus the value of ℓ remains as found above even if 𝔖′_{m−2} is properly contained in {i|(m − 3)(m + p + 1) + 1 ≤ i ≤ (m − 2)(m + p + 1)}. This completes the proof.

(6.4) The basis elements of the socle of gr(𝒮)/t^{n}gr(𝒮) are in the following orders:

(1) The largest unstable element of the rows are in orders m−1, m−1−p, m− 1 − 2p, …, m − 1 − (σ − 1)p (all in order m − 1 if p = 0, no such elements if σ = 0).

(2) If n ≡ 1 (mod m + p + 1) there are m + p + 1 basis elements in order ord_{𝒮′} (n) − 1.

(3) If n ≡ i (mod m + p + 1) where 2 ≤ i ≤ m then there are i−1 basis elements in order ord_{𝒮′} (n) and and m − i in order ord_{𝒮′} (n) − 1.

(4) If n ≡ i (mod m + p + 1) where m < i ≤ m + p + 1 then there are i − 1 basis elements in order ord_{𝒮′} (n).

Note that for gr(𝒮)/t^{n}gr(𝒮) to be level we need either

(i) σ = 0 (equivalently m = 2) and n is arbitrary. gr(𝒮)/t^{n}gr(𝒮) is Gorenstein if in addition n ≡ 2 (mod m + p + 1). or

(ii) (a) σ = 1 or p = 0 and (b) ord_{𝒮}(n) = m − 1, n ≡ i (mod m + p + 1), m ≤ i ≤ m + p + 1 or ord_{𝒮}(n) = m, n ≡ 1 (mod m + p + 1).

The following example may help the reader understand the above results.

(6.5) Example. Let 𝒮′ = {1, 7, 8, 9} so that m = 7, p = 1. Then (m − 1)(m + p) = 48 so for 𝒮 = {1, 7, 8, 9, n} then gr(𝒮) is Cohen-Macaulay if and only if n ≥ 49. Because of the size of n we will only illustrate Γ′. In the following table row j (j ≥ 0) gives the elements of order j in Γ′, with integers greater than or equal to 49 in normal type (so that [0, 48] = Γ\(49 + Γ) is the bold entries).

{0}
{1, 7, 8, 9}
{2, 10, 14, 15, 16, 17, 18}
{3, 11, 19, 21, 22, 23, 24, 25, 26, 27}
{4, 12, 20, 28, 29, 30, 31, 32, 33, 34, 35, 36}
{5, 13, 37, 38, 39, 40, 41, 42, 43, 44, 45}
{6, 46, 47, 48, 49, 50, 51, 52, 53, 54}
{55, 56, 57, 58, 59, 60, 61, 62, 63}

Here the unstable elements are {2}, {3 11}, {4 12 20}, {5 13} {6} (grouped by order) so σ = 3 with the unstable maximal elements being 6 13 20, and 46 covers 37 and 45, 47 covers 38, 48 covers 39, 40, 41. The stable elements of order 5 that are not covered by stable elements of order 6 are 42, 43, 44, so the Cohen-Macaulay type of gr(𝒮) is 9, and the socle of gr(𝒮)/t^{49}gr(𝒮) has basis t^{6}, t^{13}, t^{20}, t^{42}, t^{43}, t^{44}, t^{46}, t^{47}, t^{48}. In this case we have m + p + 1 = 9 and n ≡ 4 (mod 9). Also ord_{𝒮′} (49) = 6 = m − 1 so we are in agreement with case 3 of (6.4).

If we take n ≥ 49 the Cohen-Macaulay types of gr(𝒮) (obtained by direct calculation) are successively 9 9 9 9 10 11 12 9 9 9 9 9 9 10 11 12 ..., continuing with period 9 (in agreement with the cases of Theorem 6.3). We note that the smallest n which makes gr(𝒮) Cohen-Macaulay (namely (m − 1)(m + p) + 1) is congruent to p + 3 (mod m + p + 1) so we can start with either case 2 or case 3 of Theorem 6.3 (but clearly not case 1).

In the situation of Theorem 6.3 the h-vector of gr(𝒮) is the “truncation” of the Hilbert function of gr(𝒮′) at n, i.e. ΔH_{gr(𝒮)}(i) = H_{gr(𝒮′)}(i) so long as ∑_{j≤i} H_{gr(𝒮′)}(j) ≤ n and the final non-zero value of ΔH_{gr(𝒮)} is chosen so that ∑_{j≥0} ΔH_{gr(𝒮)}(j) = n. This amounts to deleting all integers greater than or equal to n in the tables of the type in the previous two examples. Thus in Example 6.5 Theorem 4.11 yields 1 4 7 10 12 11 10 9 → for H_{𝒮′} and truncating this at 49 yields 1 4 7 10 12 11 4 0 which is indeed what we observed for the h-vector of gr(𝒮) in Example 6.5.

(6.6) Remark. There should be similar results if d > 1. However the situation is more complicated: gr(𝒮)/t^{n}gr(𝒮) is still of cardinality n but no longer has such a simple description, and the sets 𝔖′_{j} no longer consist of consecutive integers for large j. There appears to be a different bound for Cohen-Macaulayness in each congruence class (mod d). We will leave such things for possible future investigation.

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