Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 41 (2000), No. 1, 67-84.

Kähler Differentials of Affine Monomial Curves^{1}

S. Geller  Alok Kumar Maloo  D. P. Patil  L. G. Roberts

Department of Mathematics, Texas A & M University College Station, Texas 77802, USA, e-mail: geller@math.tamu.edu

Department of Mathematics, Indian Institute of Technology Kanpur 208 016, UP, India, e-mail: akmaloo@iitk.ernet.in

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

Department of Math. and Stat., Queen’s University Kingston, Ontario, Canada, K7L 3N6, e-mail: robertsl@mast.QueensU.CA

0. Introduction

For a field k and a k-algebra C let Ω_{C/k}^{1} denote the module of Kähler differentials, d_{C} : C → Ω_{C/k}^{1} the universal derivation, Ω_{C/k}^{q} = Λ_{C}^{q}(Ω_{C/k}^{1}) for q ≥ 0 and Ω_{C/k}^{∗} = ⊕_{q≥0}Ω_{C/k}^{q}. For simplicity of notation, we will write d instead of d_{C}. For a submonoid Γ of ℤ^{+} = {0, 1, 2, …}, let k[Γ] be the monoid ring of Γ. In this paper we describe Ω_{k[Γ]/k}^{∗} in the case where characteristic k = 0 and Γ is generated by an arithmetic progression.

For fixed integers m, p, d with m ≥ 2, 1 ≤ p ≤ m − 1, d ≥ 1 , gcd(m, d) = 1 we define m_{i} = m + id, (i ∈ ℤ^{+}, 0 ≤ i ≤ p), S = {m_{0}, m_{1}, m_{2}, …, m_{p}} and Γ to be the monoid generated by the arithmetic progression S. We will write A for k[Γ] = k[t^{m_{0}}, t^{m_{1}}, …, t^{m_{p}}]. The ring A is graded by weight if we set wt(t) = 1 and grade the A-modules Ω_{A/k}^{q} by in addition setting wt(dt) = 1. It should be clear from the context whether d denotes the differential or the common difference of the arithmetic progression. The n-th weight piece of Ω_{A/k}^{q} will be denoted by (Ω_{A/k}^{q})_{n}.

1This work was partially supported by the NSERC grant of the fourth author. The first and third authors thank Department of Mathematics and Statistics, Queen’s University, Kingston, Canada for its hospitality. The second author thanks Department of Mathematics, Indian Institute of Science, Bangalore for its hospitality.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2000 Heldermann Verlag

This work is motivated in part by the relation between the Ω_{k[Γ]/k}^{q} and K-Theory, as discussed for example in [6; appendix]. It is hoped that the results obtained here will at least partially survive in subrings of k[t] that are not generated by monomials, for example as considered in [2]. We expect our calculation to yield interesting invariants of the singularity of k[Γ] at the origin or of the monoid Γ itself.

In Section 3 we compute dim_{k}(Ω_{A/k}^{1})_{n} in terms of partitions of n as the sum of integers from the set S. For general Γ, dim_{k}(Ω_{k[Γ]/k}^{1})_{n} is likely too fine an invariant to be easily interpreted. However in the arithmetic progression case we are able to give a combinatorial interpretation in terms of certain subsets of S that are introduced in Section 2. This was surprising to us and fails if either the characteristic of k is non-zero or S is not an arithmetic progression. In Section 4 we determine the index of regularity of Ω_{A/k}^{1} and in Section 5 we describe an interesting relationship between the Hilbert functions of Ω_{A/k}^{1} and Ω_{A′/k}^{1} where A′ = k[t^{m}, t^{m+1}, …, t^{m+p}]. Another motivation for this paper was to explain the symmetry of the Hilbert function of Ω_{A/k}^{1}/dA proved in [6] in the case where p = m − 1 and d = 1. In Section 5 we generalize this result to the case p = m − 1, d > 1. Our improved methods suggest that the symmetry is an accident. In Section 6 we completely describe Ω_{A/k}^{q} for q ≥ 3, generalizing the result of [7] to the case p ≤ m − 1. This permits us to determine the Hilbert function of Ω_{A/k}^{2} in terms of Ω_{A/k}^{1} and the Gaussian polynomials. In each section examples are given to illustrate the results.

1. Presentation of Ω_{A/k}^{1}

In addition to the notation fixed in the introduction, let φ : R = k[X_{0}, …, X_{p}] ⟶ A be the k-algebra homomorphism defined by X_{i} ↦ t^{m_{i}}, 0 ≤ i ≤ p, and let 𝔭 be the kernel of φ. For 0 ≤ i ≤ p, define x_{i} = t^{m_{i}}, deg(X_{i}) = 1 and wt(X_{i}) = m_{i}. Then 𝔭 is homogeneous if R is graded by weight, but not if R is graded by degree.

Recall (from [3; (4.13)]) the presentation for Ω_{A/k}^{∗} given by Ω_{A/k}^{∗} ≅ Ω_{R/k}^{∗}/(𝔭, d𝔭), where (𝔭, d𝔭) is the 2-sided ideal of Ω_{R/k}^{∗} generated by {F, dF | F ∈ 𝔭}.

For a = (a_{0}, …, a_{p}) ∈ (ℤ^{+})^{p+1}, let X^{a} = ∏_{i=0}^{p} X_{i}^{a_{i}}. The image of a monomial X^{a} ∈ R of weight n = ∑_{i=0}^{p} a_{i}m_{i} in A is t^{n}. Therefore the n-th weight piece 𝔭_{n} of 𝔭 is generated by binomials {X^{a} − X^{b} | ∑_{i=0}^{p} a_{i}m_{i} = ∑_{i=0}^{p} b_{i}m_{i} = n} as a k-vector space. (An explicit minimial set of ideal generators for 𝔭 is given in [5; (4.5)]).

It is more convenient to consider Ω_{A/k}^{1}/dA rather than Ω_{A/k}^{1} as we then have a simpler presentation. But we should be careful to remember that Ω_{A/k}^{1}/dA is only a k-vector space, not an A-module.

1.1. Theorem. Let m, p, d be positive integers with gcd(m, d) = 1 and let A = k[t^{m_{0}}, t^{m_{1}}, t^{m_{2}}, …, t^{m_{p}}]. For any non-negative integer n, let V_{n} be the k-vector space with basis {t^{n−m_{i}}dt^{m_{i}} | i ∈ [0, p], t^{n−m_{i}} ∈ A, n − m_{i} ≠ 0}, and W_{n} the subspace of V_{n} spanned by {∑_{i=0}^{p} a_{i}t^{n−m_{i}}dt^{m_{i}} | ∑_{i=0}^{p} a_{i}m_{i} = n, a_{i} ≥ 0}. Then (Ω_{A/k}^{1}/dA)_{n} ≅ V_{n}/W_{n}.

Proof. Since Ω_{R/k}^{1} is the free R-module with basis dX_{0}, …, dX_{p}, we have that Ω_{R/k}^{1} ⊗_{R} A = Ω_{R/k}^{1}/(𝔭Ω_{R/k}^{1}) is a free A-module with basis dx_{0}, …, dx_{p}. We know that 𝔭 is generated by {X^{a} − X^{b} | ∑_{i=0}^{p} a_{i}m_{i} = ∑_{i=0}^{p} b_{i}m_{i} = n} and dt^{n} = d((t^{m_{0}})^{a_{0}} … (t^{m_{p}})^{a_{p}}) = ∑_{i=0}^{p} a_{i}(t^{m_{0}})^{a_{0}} … (t^{m_{i}})^{a_{i}−1} … (t^{m_{p}})^{a_{p}} dt^{m_{i}} for (a_{0}, …, a_{p}) with ∑_{i=0}^{p} a_{i}m_{i} = n. Since (Ω_{A/k}^{1}/dA)_{n} is (Ω_{R/k}^{1}/(𝔭Ω_{R/k}^{1}, d𝔭))_{n} modulo dt^{n}, the assertion follows. □

1.2. Remark. For large n, dim_{k}(Ω_{A/k}^{1})_{n} = 1 and dim_{k}(Ω_{A/k}^{q})_{n} = 0 for all q ≥ 2. If i, j ∈ Γ, j > 0 then t^{i}dt^{j} ≠ 0 in Ω_{A/k}^{1}.

Proof. Since the quotient field of A is k(t) and Ω_{A/k}^{q} ⊗_{A} k(t) ≅ Ω_{k[t]/k}^{q} ⊗_{A} k(t) ≅ Ω_{k(t)/k}^{q} for all q ≥ 0, the assertions follow. □

Before translating 1.1 into explicit results about Ω_{A/k}^{1}, we will make some remarks about partitions.

2. Partitions

In addition to the notation previously introduced, for non-negative integers a, b, let [a, b, d] denote {a + id | i ∈ ℤ^{+}, a + id ≤ b}, Floor[a] the largest integer less than or equal to a, and Ceiling[a] the smallest integer greater than or equal to a. We will write [a, b] instead of [a, b, 1]. Note that in this notation the set S = [m, m + pd, d].

For integers n , r, let S(n, r) be the set of all integers in S that can occur in a partition of n as a sum of r elements of S, that is,

S(n, r) = {m_{j} | n = ∑_{i=0}^{p} a_{i}m_{i} with ∑_{i=0}^{p} a_{i} = r and a_{j} ≠ 0} .

Let S(n) = ∪_{r≥1}S(n, r). Further, the cardinality |S(n, r)| of S(n, r) will be denoted by s(n, r) and P (n) = {r ∈ ℤ^{+} | S(n, r) ≠ ∅}.

For example, if m = 6, p = 2, d = 1, then S = {6, 7, 8}, S(14, 2) = {6, 7, 8} since we have 14 = 7 + 7 = 6 + 8, S(18, 3) = {6} since 18=6+6+6 is the only way to write 18 as the sum of three elements of S, and S(17, 3) = ∅.

The set {m_{j} | 0 ≤ j ≤ p and a_{j} ≠ 0} is called the support of the partition n = ∑_{i=0}^{p} a_{i}m_{i} of n, and the integer 1 + (l − s) is called the spread of this partition, where m_{l} and m_{s} are the largest and smallest element in the support respectively.

For example, if S = {5, 7, 9, 11} then the partition 5 + 5 + 9 + 11 of 30 has support {5, 9, 11} and spread 4. Note that if S(n, r) ≠ ∅ then n ≡ rm mod d. A partition of n with spread 1 is of the form n = rx, x ∈ S.

2.1. Lemma. Let n, r be non-negative integers.

(a) For x ∈ S, n−x lies in the monoid generated by S if and only if x ∈ S(n). In particular V_{n} has basis {t^{n−m_{i}}dt^{m_{i}} | m_{i} ∈ S(n)}.

(b) S(n, 2) ∩ S(n, r) = ∅ for r > 2.

(c) If S(n, r) ≠ ∅ then n has a partition as the sum of r elements of S with either spread 1 or 2, but not both.

(d) There are exactly Ceiling[s(n, 2)/2] partitions of n as the sum of two elements of S. More precisely, n = m_{s} + m_{l} = m_{s+1} + m_{l−1} = · · · · · · = m_{s+j} + m_{l−j}, where m_{l} and m_{s} are the largest and smallest elements of S(n, 2) respectively and j =(Ceiling[\frac{s(n,2)}{2}]−1).

(e) If r ≥ 3 then for i = 3, …, s(n, r) there is a partition of n as the sum of r elements of S with spread i. Furthermore these partitions can be chosen to have “nested” supports (see Example 2.2 below).

(f) For fixed r, {n | S(n, r) ≠ ∅} = [rm, rm_{p}, d], which is a set of consecutive elements in the congruence class of rm mod d. Moreover, if n = rm + jd, 0 ≤ j ≤ rp then

S(n, r) = \{ \begin{matrix}[m, m_{j}, d] & if 0 ≤ j ≤ p − 1 \\ S & if p ≤ j ≤ (r − 1)p \\ [m_{j−(r−1)p}, m_{p}, d] & if (r − 1)p + 1 ≤ j ≤ rp. \end{matrix}

Explicitly, S(n, r) = [max(m, n − (r − 1)m_{p}), min(m_{p}, n − (r − 1)m), d], which is a set of consecutive elements of S containing either m or m_{p} (or both).

(g) For fixed n, P (n) is a set of consecutive elements in a congruence class mod d.

(h) If S(n, r) ≠ ∅ and S(n, r + d) ≠ ∅ then m ∈ S(n, r + d) and m_{p} ∈ S(n, r). If S(n, r), S(n, r + d) and S(n, r + 2d) are all non-empty then S(n, r + d) = S.

Proof. (a) Straightforward from the definitions.

(b) Immediate since S is the minimal set of generators of Γ.

(c) Since S(n, r) ≠ ∅ we have n = rm + id where 0 ≤ i ≤ rp. Write i = αr + β with 0 ≤ β ≤ r − 1. Then, β = 0 ⟺ n = rm_{α} ⟺ n has a partition with spread 1. If β > 0 then α ≤ p − 1 so m_{α}, m_{α+1} ∈ S. Therefore n = (r − β)m_{α} + βm_{α+1} expresses n as the sum of r elements of S with spread 2.

(d) is straightforward to verify.

(e) If r ≥ 3 we can increase the spread of the partition in the proof of (c) 1 at a time (by 2 when going from spread 1 to spread 3) by replacements of the type m_{i} + m_{j} ↦ m_{i−1} + m_{j+1}, 1 ≤ i ≤ j ≤ p − 1 which proves (e).

(f) is straightforward to verify.

(g) From (f) we have S(n, r) ≠ ∅ ⟺ n ≡ rm mod d and rm ≤ n ≤ r(m + pd) ⟺ r ≡ m′n (mod d) and n/(m + pd) ≤ r ≤ n/m, where m′ is an inverse mod d of m.

(h) follows from (f). □

2.2. Example. Let m = 5, d = 2 and p = 3. Then S = {5, 7, 9, 11} and for n = 32, r = 4, we have n = 7+7+9+9 = 7+7+7+11 = 5+7+9+11. The supports of these partitions are {7, 9}, {7, 11}, {5, 7, 9, 11} respectively. By “nested” we mean that [7, 9] ⊂ [7, 11] ⊂ [5, 11], i.e. the smallest intervals containing the supports are nested.

2.3. Lemma. Let n and r be non-negative integers and let n = ∑_{i=0}^{p} c_{i}(m+id) be a partition of n as the sum of r elements of S. Let c = (c_{0}, …, c_{p}) and b_{r} = (b_{r,0}, …, b_{r,p}) where b_{r,i} = rm_{i} − n. Then b_{r} · c = 0. Furthermore if m_{s} and m_{l} are the smallest, respectively largest, elements of S(n, r) then b_{r,s} < 0 and b_{r,l} > 0.

Proof. We have n = ∑_{i=0}^{p} c_{i}m_{i}, c_{i} ≥ 0 for all 0 ≤ i ≤ p and ∑_{i=0}^{p} c_{i} = r. Therefore 0 = rn − rn = r (∑_{i=0}^{p} c_{i}m_{i}) − n (∑_{i=0}^{p} c_{i}) = ∑_{i=0}^{p}(rm_{i} − n)c_{i} = b_{r} · c. The second part follows from the fact that the co-ordinates of b_{r} are in an increasing arithmetic progression and the c have non-negative co-ordinates. □

The notation S(n, r), s(n, r), P (n) introduced in this section will be used later without explicit reference.

3. The Hilbert function of Ω_{A/k}^{1}

In this section we use the results on partitions in Section 2 to determine the Hilbert function of Ω_{A/k}^{1}.

3.1. Definition. For n, r positive integers let V_{n,r} be the k-vector space with basis {t^{n−m_{i}}dt^{m_{i}} | m_{i} ∈ S(n, r)} and let W_{n,r} be the k-subspace of V_{n,r} spanned by {∑_{i=0}^{p} a_{i}t^{n−m_{i}}dt^{m_{i}} | a_{i} ≥ 0, ∑_{i=0}^{p} a_{i} = r, ∑_{i=0}^{p} a_{i}m_{i} = n}.

Note that V_{n} = ∑_{r} V_{n,r} and W_{n} = ∑_{r} W_{n,r} where V_{n} and W_{n} are as in 1.1.

3.2. Lemma. Let n, r be positive integers and let b_{r} = (rm_{0} − n, …, rm_{p} − n) as in 2.3.

(a) W_{n,r} ⊆ b_{r}^{⊥} = {∑_{m_{i}∈S(n,r)} λ_{i}t^{n−m_{i}}dt^{m_{i}} ∈ V_{n,r} | ∑ λ_{i}b_{r,i} = 0}.

(b) V_{n,1}/W_{n,1} = 0.

(c) dim_{k} V_{n,2}/W_{n,2} = Floor[s(n, 2)/2] with relations

{t^{m_{j}}dt^{m_{i}} + t^{m_{i}}dt^{m_{j}} = 0 | n = m_{i} + m_{j}, 0 ≤ i ≤ j ≤ p} .

(d) Suppose that r ≥ 3. If m_{i} ∈ S(n, r) and b_{r,i} ≠ 0 then

{b_{r,i}t^{n−m_{j}}dt^{m_{j}} − b_{r,j}t^{n−m_{i}}dt^{m_{i}} | m_{j} ∈ S(n, r), j ≠ i}

is a basis of W_{n,r}. In particular, equality holds in (a) and

dim_{k} V_{n,r}/W_{n,r} = \{ \begin{matrix}0 & if s(n, r) = 1 \\ 1 & if s(n, r) > 1 \end{matrix}

Moreover, if s(n, r) > 1 then t^{n−m_{i}}dt^{m_{i}} is a basis for V_{n,r}/W_{n,r} provided b_{r,i} ≠ 0 (in particular, by 2.3, m_{i} can be either the largest or the smallest element of S(n, r)).

Proof. (a) and (b) are immediate from 2.3 and 2.1(a). (c) Follows from 2.1(d).

(d) If s(n, r) = 1 then dim_{k} V_{n,r} = 1 and W_{n,r} ≠ 0, so V_{n,r}/W_{n,r} = 0. If s(n, r) > 1 then dim_{k} V_{n,r}/W_{n,r} ≤ 1 by the “nested” part of 2.1(d) and dim_{k} V_{n,r}/W_{n,r} ≥ 1 by (a) since b_{r} has at most one co-ordinate equal to 0. Now, the equality in (a) follows. □

Note that 3.2 permits effective computation in Ω_{A/k}^{1}/dA, not just the computation of Hilbert functions. It is not difficult to lift this computation to Ω_{A/k}^{1} and, using the results of Section 6, to Ω_{A/k}^{2}. However we will not pursue the latter.

3.3. Lemma. Let n, r be positive integers. Then

(a) If P (n) = {r} then (Ω_{A/k}^{1}/dA)_{n} = V_{n,r}/W_{n,r}.

(b) If S(n, r) ≠ ∅, S(n, r + d) ≠ ∅, and S(n, r) ∩ S(n, r + d) = ∅, then (Ω_{A/k}^{1}/dA)_{n} = V_{n,r}/W_{n,r} ⊕ V_{n,r+d}/W_{n,r+d}.

(c) If s(n, r) ≥ 2, s(n, r + d) ≥ 2 and |S(n, r) ∩ S(n, r + d)|=1, then dim_{k}(Ω_{A/k}^{1}/dA)_{n} =1.

(d) If |S(n, r)∩S(n, r+d)|=1 and either s(n, r)=1 or s(n, r+d)=1, then (Ω_{A/k}^{1}/dA)_{n} =0.

(e) If |S(n, r) ∩ S(n, r + d)| ≥ 2 then (Ω_{A/k}^{1}/dA)_{n} = 0. In particular, if |P (n)| ≥ 3 then

(Ω_{A/k}^{1}/dA)_{n} = 0.

Proof. (a) Follows from 1.1, since V_{n} = V_{n,r}, W_{n} = W_{n,r}.

(b) V_{n} =V_{n,r} ⊕ V_{n,r+d} and W_{n} =W_{n,r} ⊕ W_{n,r+d} since S(n, s)=∅ for s≠r, r + d by 2.1(h). The assertion now follows from 1.1.

(c) Since |S(n, r) ∩ S(n, r + d)| = 1 we have r ≥ 3 by 2.1(b) and S(n, s) = ∅ if s ≠ r, r + d by 2.1(h). Suppose S(n, r) ∩ S(n, r + d) = m_{i}. By 2.1(h), m_{i} is the smallest element of S(n, r) and the largest of S(n, r + d). By 2.3, b_{r,i} ≠ 0 and b_{r+d,i} ≠ 0. Now (c) follows from 3.2(d) and 1.1.

(d) If |S(n, r) ∩ S(n, r + d)| = 1 and s(n, r) = 1 then S(n, r) = {m_{p}} and S(n, r + d) = S by 2.1(h) and 2.1(f). By 3.2(d) V_{n,r+d}/W_{n,r+d} is one-dimensional with basis t^{n−m_{p}}dt^{m_{p}}. Furthermore the canonical map V_{n,r+d}/W_{n,r+d} → (Ω_{A/k}^{1}/dA)_{n} is onto. Since S(n, r) = {m_{p}} we have n = rm_{p} from which it follows that t^{n−m_{p}}dt^{m_{p}} ∈ W_{n,r} and hence t^{n−m_{p}}dt^{m_{p}} = 0 in (Ω_{A/k}^{1}/dA)_{n}. Therefore (Ω_{A/k}^{1}/dA)_{n} = 0 as claimed. If |S(n, r) ∩ S(n, r + d)| = 1 and s(n, r + d) = 1 then S(n, r) = S and S(n, r + d) = {m}. From this it follows similarly that (Ω_{A/k}^{1}/dA)_{n} = 0.

(e) Since |S(n, r) ∩ S(n, r + d)| > 1 we have r ≥ 3 by 2.1(b). By 3.2(d) W_{n,r} = b_{r}^{⊥}, W_{n,r+d} = b_{r+d}^{⊥}. Suppose that m_{i}, m_{j} ∈ S(n, r) ∩ S(n, r + d), i ≠ j. Then b_{r,j}t^{n−m_{i}}dt^{m_{i}}−b_{r,i}t^{n−m_{j}}dt^{m_{j}} ∈ W_{n,r} and b_{r+d,j}t^{n−m_{i}}dt^{m_{i}} − b_{r+d,i}t^{n−m_{j}}dt^{m_{j}} ∈ W_{n,r+d}. Since the determinant of the coefficient matrix is (m_{i} − m_{j})nd ≠ 0, by 1.1, t^{n−m_{i}}dt^{m_{i}} = t^{n−m_{j}}dt^{m_{j}} = 0 in Ω_{A/k}^{1}/dA. Thus (Ω_{A/k}^{1}/dA)_{n} = 0 by 3.2(d) and 2.1(h). Finally if |P (n)| ≥ 3 then either |S(n, r) ∩ S(n, r + d)| ≥ 2 for some r, in which case we have just proved that (Ω_{A/k}^{1}/dA)_{n} = 0, or s(n, r) = 1, S(n, r + d) = S and s(n, r + 2d) = 1 for some r. In this case (Ω_{A/k}^{1}/dA)_{n} = 0 by 3.3(d). □

The S(n, r) are described explicitly in 2.1. These sets completely determine the Hilbert function of Ω_{A/k}^{1}:

3.4. Theorem. Let m, p, d be integers with m ≥ 2, 1 ≤ p ≤ m−1, d ≥ 1, and gcd(m, d) = 1. Let A = k[t^{m}, t^{m+d}, …, t^{m+pd}] and let H : ℤ^{+} → ℤ^{+}, where H(n) = dim_{k}(Ω_{A/k}^{1})_{n}, be the Hilbert function of Ω_{A/k}^{1}. Then for any n ∈ ℤ^{+} we have:

(a) H(0) = 0 and H(n) = \{ \begin{matrix}0 & if S(n) = ∅ \\ 1 + dim_{k}(Ω_{A/k}^{1}/dA)_{n} & if S(n) ≠ ∅ \end{matrix}.

(b) Suppose that P (n) = {r}. Then

H(n) = \{ \begin{matrix}1 & if s(n, r) = 1 \\ 1 + Floor[s(n, 2)/2] & if r = 2 and s(n, 2) > 1 \\ 2 & if r > 2 and s(n, r) > 1 \end{matrix} .

(c) Suppose that there are distinct integers r_{1}, r_{2} such that ∅ ≠ S(n, r_{1}) ⊆ S(n, r_{2}). Then H(n) = 1. In particular, if |P (n)| ≥ 3 then H(n) = 1.

(d) Suppose that |P (n)| = 2. Then there is an integer r ≥ 2 such that S(n, r) ≠ ∅ and S(n, r + d) ≠ ∅.

(d1) Suppose that S(n, r) ∩ S(n, r + d) = ∅ (which is necessarily the case if r = 2). Then if r = 2

H(n) = \{ \begin{matrix}1 + Floor[s(n, 2)/2] & if s(n, 2 + d) = 1 \\ 2 + Floor[s(n, 2)/2] & if s(n, 2 + d) > 1 \end{matrix}

and if r > 2

H(n) = \{ \begin{matrix}1 & if s(n, r) = 1, s(n, r + d) = 1 \\ 2 & if exactly one of s(n, r), s(n, r + d) is 1 \\ 3 & if s(n, r) > 1, s(n, r + d) > 1 \end{matrix} .

(d2) Suppose that |S(n, r) ∩ S(n, r + d)| = 1 (necessarily r > 2). Then

H(n) = \{ \begin{matrix}1 & if one of s(n, r), s(n, r + d) is 1 \\ 2 & if both s(n, r) > 1, s(n, r + d) > 1. \end{matrix}

(d3) Suppose that |S(n, r) ∩ S(n, r + d)| > 1. Then H(n) = 1.

Proof. (a) The final assertion of (a) follows from 1.2.

(b) Follows from 3.2(c), 3.2(d), and 3.3(a).

(c) Follows from 3.3(d) and 3.3(e).

(d) (d1) follows from 3.3(b), 3.2(c), and 3.2(d); (d2) follows from 3.3(c) and 3.3(d); (d3) follows from 3.3(e). □

3.5. Examples. The following examples show that all the cases in 3.4 can occur. For use in the table we number consecutively the cases in each part of 3.4. For example, 3.4(d1)(case 3) means (among the cases of 3.4(d1)) r > 2 and s(n, r) = 1 = s(n, r + d). This case can occur only if p = 1 by 2.1(h). The hardest to find is 3.4(d2)(case 2).

(m, p, d) S  n P (n)  r S(n, r)  H(n) Reason
(4, 3, 1) {4, 5, 6, 7} 4 {1}  1 {4}  1 3.4(b)(case 1)
11 {2}  2 {4, 5, 6, 7} 3 3.4(b)(case 2)
15 {3}  3 {4, 5, 6, 7} 2 3.4(b)(case 3)
12 {2, 3} 2 {5, 6, 7}  2 3.4(d1)(case 1)
3 {4}
13 {2, 3} 2 {6, 7}  3 3.4(d1)(case 2)
3 {4, 5}
16 {3, 4} 3 {4, 5, 6, 7} 1 3.4(d2)(case 1)
4 {4}
17 {3, 4} 3 {4, 5, 6, 7} 1 3.4(c) or 3.4(d3)
4 {4, 5}
20 {3, 4, 5} 3 {6, 7}  1
4 {4, 5, 6, 7}
5 {4}  3.4(c)
(3, 1, 1) {3, 4} 12 {3, 4} 3 {4} 1 3.4(d1)(case 3)
4 {3}
(9, 4, 1) {9, …, 13} 36 {3, 4} 3 {10, …, 13} 2 3.4(d1)(case 4)
4 {9}
37 {3, 4} 3 {11, 12, 13} 3 3.4(d1)(case 5)
4 {9, 10}
(12, 6, 1) {12, …, 18} 49 {3, 4} 3 {13, …, 18} 2 3.4(d2)(case 2)
4 {12, 13}

We conclude this section with the following interesting application of 3.2 and 3.3.

3.6. Proposition. Let Γ^{1} ⊆ Γ the monoid generated by S^{1} = {m_{s}, m_{s+1}, …, m_{l}} ⊆ S and let A^{1} = k[Γ^{1}]. Then, for all n, the induced homomorphism of k-vector spaces ω_{n} : (Ω_{A^{1}/k}^{1}/dA^{1})_{n} → (Ω_{A/k}^{1}/dA)_{n} is either injective or surjective.

Proof. The notation of Section 2 and Section 3 will be applied to A^{1} by applying ^{1} to the corresponding notation for A. Then S^{1}(n, r) ⊆ S(n, r) for all r, n so P^{1}(n) ⊆ P (n). If |P (n)| ≥ 3 then (Ω_{A/k}^{1}/dA)_{n} = 0 by 3.3(e) so ω_{n} is surjective. From V_{n,r}^{1} ∩ W_{n,r} = W_{n,r}^{1} we have an inclusion V_{n,r}^{1}/W_{n,r}^{1} ↪ V_{n,r}/W_{n,r}. It follows from this that if |P (n)| = 1 or if |P (n)| = 2, say P (n) = {r, r + d}, and S(n, r) ∩ S(n, r + d) = ∅ (which is the case if r = 2 by 2.1(b)), then ω_{n} is injective by 3.3(b). The only case remaining is P (n) = {r, r + d}, r ≥ 3 and S(n, r)∩S(n, r+d) ≠ ∅. If |S(n, r)∩S(n, r+d)| > 1 or if one of S(n, r) or S(n, r+d) is a singleton, then (Ω_{A/k}^{1}/dA)_{n} = 0 by 3.2(d),(e) and ω_{n} is surjective. If |S(n, r) ∩ S(n, r + d)| = 1, s(n, r) > 1, and s(n, r + d) > 1 then dim_{k}(Ω_{A/k}^{1}/dA)_{n} = 1 by 3.3(c). In this case if (Ω_{A^{1}/k}^{1}/dA^{1})_{n} = 0 then ω_{n} is an injection; otherwise by 1.1, Ω_{A^{1}/k}^{1}/dA^{1} contains non-zero elements of the form t^{n−m_{i}}dt^{m_{i}}, m_{i} ∈ S^{1}. Any such element maps to a non-zero element in (Ω_{A/k}^{1}/dA)_{n} by 3.2(d) and 3.3(c) so ω_{n} is surjective. □

3.7. Corollary. For fixed non-negative integers m, d and n, with gcd(m, d) = 1, the function p ↦ dim_{k}(Ω_{A/k}^{1}/dA)_{n} is a unimodal function of p on the interval [1, m − 1].

Proof. Recall that A depends on parameters m, p and d. As p ranges from 1 to m − 1 we get an increasing sequence of monoids, to which we apply 3.6 in weight n. The assertion is then immediate from the fact that a composition of k-vector space homomorphisms V_{1} → V_{2} → V_{3} cannot be injective or surjective if dim_{k}(V_{2})<min{dim_{k} V_{1}, dim_{k} V_{3}}. □

4. The index of regularity of Ω_{A/k}^{1}

The least integer N such that dim_{k} A_{n} = 1 for all n ≥ N is called the conductor or index of regularity of A. It follows from [4; Ch. 3, (5.24)] that if we define integers q ≥ 1, 1 ≤ r ≤ p by the equation m = qp + r, then the index of regularity of A is qm_{p} + m_{r−1} − m + 1 if r > 1 and qm_{p} − m + 1 if r = 1. Similarly, we define the index of regularity of Ω_{A/k}^{1} to be the least integer n_{0} such that dim_{k}(Ω_{A/k}^{1})_{n} = 1 for all n ≥ n_{0}, which exists by 1.2. In this section we shall determine n_{0} explicitly.

Let ℓ be an integer with 2 ≤ ℓ ≤ d + 1. Then

((ℓ + id)m − d) − ((ℓ + (i − 2)d)m_{p}) = 2dm − d (1 + (ℓ + (i − 2)d)p)

is positive if i = 0 and decreases as i increases. Therefore we can make the following definition.

4.1. Definitions. For 2 ≤ ℓ ≤ d + 1, let i_{ℓ} be the largest integer i (necessarily ≥ 0) such that the inequality (ℓ + (i − 2)d) m_{p} < (ℓ + id)m − d holds. Further, let n_{ℓ} = (ℓ + i_{ℓ}d)m − d and r_{ℓ} = ℓ + (i_{ℓ} − 1)d.

These definitions are motivated by the following considerations: S(n_{ℓ}, r_{ℓ} + d) = ∅ (by the definition of r_{ℓ}) and S(n_{ℓ}, r_{ℓ} − d) = ∅ (by the inequality in 4.1).

4.2. Lemma. For 2 ≤ ℓ ≤ d + 1, we have i_{ℓ} = Ceiling [\frac{2m−1−ℓp}{pd}] + 1.

Proof. The inequality in 4.1 ⟺ i < \frac{2m−1}{pd} + 2 − \frac{ℓ}{d} ⟺ i < \frac{2m−1−ℓp}{pd} + 2. Since i < α ⟺ i ≤ Ceiling[α] − 1 (for i an integer and α real), the assertion follows. □

4.3. Remarks. For 2 ≤ ℓ ≤ d + 1, let i_{ℓ} and n_{ℓ} be as in 4.1.

(i) i_{ℓ} is a non-increasing function of ℓ and takes at most two values which are consecutive integers.

(ii) If i_{ℓ} = i_{ℓ+1} then n_{ℓ+1} = n_{ℓ} + m.

(iii) Suppose that there is an integer l such that 2 ≤ l ≤ d and i_{l} = i_{l+1} + 1. Then n_{l} = n_{l+1} + (d − 1)m and max{n_{ℓ} | 2 ≤ ℓ ≤ d + 1} = n_{l}. Otherwise, max{n_{ℓ} | 2 ≤ ℓ ≤ d + 1} = n_{d+1}.

(iv) Let l be any integer. Define i_{l} = Ceiling [\frac{2m−1−lp}{pd}] + 1, n_{l} = (l + i_{l}d)m − d and r_{l} = l + (i_{l} − 1)d. Then n_{l} and r_{l} (but not i_{l}) depend only on the congruence class of l modulo d. We chose ℓ with 2 ≤ ℓ ≤ d + 1 for definiteness in our proofs.

4.4. Theorem. Let ℓ be an integer, 2 ≤ ℓ ≤ d + 1, let i_{ℓ} and n_{ℓ} be as in 4.1. Then dim_{k}(Ω_{A/k}^{1})_{n_{ℓ}} = 2 and dim_{k}(Ω_{A/k}^{1})_{n} = 1 for all n > n_{ℓ} with n ≡ mℓ (mod d). Moreover, n_{0} = max{n_{ℓ} | 2 ≤ ℓ ≤ d + 1} + 1 is the index of regularity for Ω_{A/k}^{1}.

Proof. Since r_{ℓ} ≥ 3 by 4.5(c) below, the assertions follow from 4.6 and 3.4. □

4.5. Lemma. Let ℓ be an integer with 2 ≤ ℓ ≤ d + 1 and let i_{ℓ} and r_{ℓ} be as in 4.1. Then for an integer i ≥ i_{ℓ}, we have :

(a) m − 1 ≤ (ℓ + id − 1)p.

(b) 2m − 1 ≤ (ℓ + (i − 1)d)p.

(c) i_{ℓ} ≥ 1 and r_{ℓ} ≥ ℓ. Moreover, i_{2} ≥ 2 and r_{2} ≥ 2 + d.

(d) If either (m, p) ≠ (2, 1) or i > i_{ℓ} then m < (ℓ + (i − 1)d − 1)p.

(e) If (m, p) = (2, 1) then S((ℓ + id)m, ℓ + (i − 1)d) = S.

Proof. (a) follows from (b), since d ≥ 1. To prove (b) we may assume i = i_{ℓ}. Then by substituting the value of i_{ℓ} from 4.2 we get (i_{ℓ} − 1)pd = Ceiling[\frac{2m−1−ℓp}{pd}] pd ≥ 2m − 1 − ℓp, from which (b) follows.

(c) If i_{ℓ} = 0 then from (b) we have 2m − 1 ≤ (ℓ − d)p ≤ p which contradicts the assumption p ≤ m − 1. So i_{ℓ} ≥ 1. If i_{2} = 1 then from (b) we have 2m − 1 ≤ 2p which again contradicts p ≤ m − 1. Lastly, r_{ℓ} ≥ ℓ and r_{2} ≥ 2 + d by the definition of r_{ℓ}.

(d) Note that p ≤ m − 1 by assumption. If p < m − 1 then (b) ⇒ (d) and if p = m − 1,

(b) ⇒ m ≤ (ℓ + (i − 1)d − 1)p. Further, if p = m − 1 and (m, p) ≠ (2, 1) then p does not divide 2m − 1 − ℓp (since p > 1), and so \frac{2m−1−ℓp}{pd} is not an integer. Therefore m < (ℓ + (i_{ℓ} − 1)d − 1) p ≤ (ℓ + (i − 1)d − 1) p.

(e) Since (m, p) = (2, 1), by (4.2) we have i_{ℓ} = 2 if ℓ = 2 and i_{ℓ} = 1 if ℓ ≥ 3. Let r = ℓ + id and n = (ℓ + id)m. Then (r − d − 1)p = r − d − 1 ≥ ℓ + (i_{ℓ} − 1)d − 1 ≥ 2 = m. Since n = rm = (r − d)m + md, it follows from 2.1(f) that S(n, r − d) = S. □

4.6. Lemma. Let ℓ be an integer 2 ≤ ℓ ≤ d + 1 and let i_{ℓ}, r_{ℓ}, n_{ℓ} be as in 4.1. Then

(a) P (n_{ℓ}) = {r_{ℓ}} and S(n_{ℓ}, r_{ℓ}) = S. In particular, s(n_{ℓ}, r_{ℓ}) > 1.

(b) Let n be an integer with n ≡ mℓ (mod d) and n > n_{ℓ}. Then there exists an integer r such that either |S(n, r − d) ∩ S(n, r)| > 1 or one of S(n, r − d), S(n, r) has cardinality 1 and the other equals S.

Proof. (a) Since n_{ℓ} = r_{ℓ}m + (m − 1)d and p ≤ m − 1 ≤ (r_{ℓ} − 1)p = (ℓ + (i_{ℓ} − 1)d − 1)p by 4.5(b), we have S(n_{ℓ}, r_{ℓ}) = S by 2.1(f). From the definition of i_{ℓ} it now follows that P (n_{ℓ}) = {r_{ℓ}}.

(b) Since n > n_{ℓ} and n ≡ mℓ (mod d), we have that n = (ℓ + id)m + jd with i ≥ i_{ℓ} and 0 ≤ j ≤ m − 1. For fixed i ≥ i_{ℓ}, put r = ℓ + id. Then n = rm + jd and, since 0 ≤ j ≤ m−1 ≤ (ℓ+id)p = rp by 4.5(a), S(n, r) ≠ ∅ by 2.1(f). Furthermore, n = rm+jd = (r − d)m + (j + m)d and 0 ≤ j + m ≤ 2m − 1 ≤ (ℓ + (i − 1)d)p = (r − d)p by 4.5(b), whence S(n, r − d) ≠ ∅ by 2.1(f). We shall prove that this r is the desired one by considering the two cases 0 ≤ j ≤ p − 1 and p ≤ j ≤ m − 1 separately. Case 1: 0 ≤ j ≤ p − 1. In this case we have S(n, r) = [m, m_{j}, d] by 2.1(f). If S(n, r − d) = S then we are done. Therefore assume S(n, r − d) ≠ S. Then (m, p) ≠ (2, 1) by 4.5(e). Since n = (r − d)m + (j + m)d, by 2.1(f) we have (r − d − 1)p + 1 ≤ j + m ≤ (r − d)p and S(n, r − d) = [m_{k}, m_{p}, d] with k = j + m − (r − d − 1)p. By 4.5(d) j − k = (r − d − 1)p − m = (ℓ + (i − 1)d − 1)p − m > 0, so |S(n, r − d) ∩ S(n, r)| > 1. Case 2: p ≤ j ≤ m − 1. In this case S(n, r) = S by 4.5(a) and 2.1(f), so we are done. □

4.7. Examples. The index of regularity n_{0} can be any congruence class modulo d. This and Remarks 4.3 are illustrated by the examples in the following tables.

\begin{aligned}(m, p, d) & ℓ & i_{ℓ} & r_{ℓ} & n_{ℓ} & n_{0} \\ (7, 5, 3) & 2 & 2 & 5 & 53 & 54 \\ & 3 & 1 & 3 & 39 \\ & 4 & 1 & 4 & 46 \\ (7, 4, 3) & 2 & 2 & 5 & 53 \\ & 3 & 2 & 6 & 60 & 61 \\ & 4 & 1 & 4 & 46 \\ (7, 3, 3) & 2 & 2 & 5 & 53 \\ & 3 & 2 & 6 & 60 \\ & 4 & 2 & 7 & 67 & 68 \end{aligned}  \begin{aligned}(m, p, d) & ℓ & i_{ℓ} & r_{ℓ} & n_{ℓ} & n_{0} \\ (7, 5, 2) & 2 & 2 & 4 & 40 & 41 \\ & 3 & 1 & 3 & 33 \\ (7, 4, 2) & 2 & 2 & 4 & 40 \\ & 3 & 2 & 5 & 47 & 48 \end{aligned}

The entire Hilbert function of two of the above cases is given in the table below. The box indicates the index of regularity.

(m, p, d) Hilbert function of Ω_{A/k}^{1}

(7, 5, 3) \begin{matrix}n & 0-13 & 14-27 \\ H(n) & 0,0,0,0,0,0,0,1,0,0,1,0,0,1, & 1,0,1,2,0,1,2,1,1,3,2,0,3,2, \\ n & 28-41 & 42,. . . . . . . . . . . . .,53, 54,55 \\ H(n) & 1,4,2,2,3,2,2,3,2,2,3,2,2,3, & 1,2,2,1,2,2,1,1,2,1,1,2,1,1, \end{matrix}

(m, p, d) Hilbert function of Ω_{A/k}^{1}

(7, 5, 2) \begin{matrix}n & 0-13 & 14-27 \\ H(n) & 0,0,0,0,0,0,0,1,0,1,0,1,0,1, & 1,1,2,1,2,0,3,1,3,2,4,2,3,2, \\ n & 28, . . . . . . . . . . . . ,40, 41 & 42-55 \\ H(n) & 3,2,3,2,3,2,2,1,2,1,2,1,2,1, & 1,1,1,1,1,1,1,1,1,1,1,1,1,1, \end{matrix}

5. Symmetry and comparison of the case d = 1 with d > 1

As usual let A = k[t^{m}, t^{m+d}, …, t^{m+pd}], and define A′ = k[t^{m}, t^{m+d′}, … t^{m+pd′}] for 1 ≤ d′ < d. The notation of Section 2 and Section 3 will be applied to A′ by applying′ to the corresponding notation for A. For example S′ = [m, m + pd′, d′].

In this situation we investigate a curious relationship between the Hilbert functions of Ω_{A/k}^{1}/dA and Ω_{A′/k}^{1}/dA′ when p is not “too small” compared to m and show that, if p = m−1, then the Hilbert function of Ω_{A/k}^{1}/dA is symmetric, a generalization of [6; (8.5)]. We were initially interested in the case d′ = 1 but essentially the same proof works for d′ ≥ 1.

Let h(n) = dim_{k}(Ω_{A/k}^{1}/dA)_{n} and h′(n) = dim_{k}(Ω_{A′/k}^{1}/dA′)_{n}. Given two arithmetic progressions a, a + d, …, a + sd and a′, a′ + d′, …, a + sd′, we will say that h and h′ coincide on [a, a + sd, d] and [a′, a′ + sd′, d′] if h(a + id) = h′(a′ + id′) for i = 1, 2, …, s.

5.1. Theorem. Let I = [2m + d, 2m + (2m − 1)d, d], I′ = [2m + d′, 2m + (2m − 1)d′, d′], J = [2m + d, 2m + (2m − 1)d] and J′ = [2m + d′, 2m + (2m − 1)d′].

(a) If m/(2 + d′) ≤ p ≤ m − 1 then h and h′ coincide on the intervals I and I′ respectively.

(b) If (2m − 1)/3 ≤ p ≤ m − 1 then h(n) = 0 if n ∉ J and h′(n′) = 0 if n′ ∉ J′.

(c) If m = p + 1 then h is symmetric on the interval J and 0 outside J.

Proof. (a) Define σ : I′ → I by 2m + jd′ ↦ 2m + jd for 1 ≤ j ≤ 2m − 1. Then we need only show that h(σ(n′)) = h′(n′) for n′ ∈ I′. If n ∈ I then P (n) = {r|S(n, r) ≠ ∅} ⊆ {2, 2 + d} (cf. 2.1(g)), and if n′ ∈ I′ then P′(n′) ⊆ {2, 2 + d′}. In view of 2.1(b) and 3.4 it suffices to show that

(i) if n′ ∈ I′ then τ(S′(n′, 2)) = S(σ(n′), 2) where τ is the bijection τ : S′ → S defined by m + jd′ ↦ m + jd, 0 ≤ j ≤ p, and

(ii) σ({n′ ∈ I′ | s′(n′, 2 + d′) > 1}) = {n ∈ I | s(n, 2 + d) > 1}.

(i) follows from the equality σ(a + b) = τ(a) + τ(b) for all a, b ∈ S′. For (ii) note that {n | s(n, 2 + d′) > 1} = [(2 + d′)m + d′, (2 + d′)(m + pd′)−d′, d′] and {n | s(n, 2 + d) > 1} = [(2+d)m+d, (2+d)(m+pd)−d, d] by 2.1(f). Since σ((2+d′)m+d′) = (2+d)m+d the proof will be complete if (2+d′)(m+pd′)−d′ ≥ 2m+(2m−1)d′ and (2+d)(m+pd)−d ≥ 2m+(2m−1)d. These inequalities are equivalent to m/(2 + d′) ≤ p and m/(2 + d) ≤ p respectively. The second follows from the first, which is our hypothesis.

(b) Note that J and J′ are the entire intervals containing the arithmetic progressions I and I′ respectively. It is enough to prove the result for h (since the result for h′ is obtained by putting d = d′). For n < 2m + d = the left endpoint of J, s(n, r) ≤ 1 so h(n) = 0 by 3.4. If ℓ ≥ 3 then 2m − 1 − ℓp ≤ 2m − 1 − 3p ≤ 0 (by hypothesis). Therefore, by 4.2, i_{ℓ} ≤ 1, and by 4.5(c), i_{ℓ} ≥ 1, so i_{ℓ} = 1 for ℓ ≥ 3, whence r_{ℓ} = ℓ and n_{ℓ} = (ℓ + d)m − d. Furthermore i_{2} = 2 by 4.5(c) and 4.3(i), so r_{2} = 2 + d, n_{2} = (2 + 2d)m − d. Thus max{n_{ℓ} | 2 ≤ ℓ ≤ d + 1} = n_{2} which is the right end point of J. The assertion follows from 4.4.

(c) The midpoint of the interval J is 2m + md. Let ρ : J → J be the reflection n ↦ 2(2m + md) − n. We have to show that h(ρ(n)) = h(n) for n ∈ J which follows from the observations below:

(i) Let I_{2} = I and for 3 ≤ ℓ ≤ d + 1, let I_{ℓ} = [ℓm + d, ℓm + (m − 1)d, d]. Then ∪_{ℓ=2}^{d+1}I_{ℓ} is a disjoint union and h = 0 outside ∪_{ℓ=2}^{d+1}I_{ℓ}.

(ii) ρ(I_{2}) = I_{2} and h(ρ(n)) = h(n) for n ∈ I_{2}.

(iii) For ℓ ≥ 3, ρ(I_{ℓ}) = I_{d+4−ℓ} and h(n) = 1 for all n ∈ I_{ℓ}.

Proof of (i): The disjointness follows from the fact that gcd(m, d) = 1. From the proof of (b) we have r_{2} = 2 + d, n_{2} = (2 + 2d)m − d, and, for ℓ ≥ 3, r_{ℓ} = ℓ and n_{ℓ} = ℓm + (m − 1)d. Therefore, since n_{ℓ} is the right hand end point of I_{ℓ} for 2 ≤ ℓ ≤ d + 1, we have, by 4.4, h(n) = 0 if n ≡ ℓm (mod d) and n > n_{ℓ}. If n ≡ ℓm (mod d) and n < ℓm + d, then S(n, ℓ) is either {m} or ∅, and so h(n) = 0 by 3.4. Proof of (ii): In view of (a), we may assume that d = 1. This case was proved in [6; (8.5)] (see 5.2 for further discussion). Proof of (iii): From ρ(ℓm+d) = (d+4−ℓ)m+(m−1)d and ρ(ℓm+(m−1)d) = (d+4−ℓ)m+d the first part follows. If n ∈ I_{ℓ} then n < (ℓ + d)m so P (n) = {ℓ} by 2.1(g), s(n, ℓ) > 1 by 2.1(f), and h(n) = 1 for all n ∈ I_{ℓ}, 3 ≤ ℓ ≤ d + 1 by 3.4. □

5.2. Examples. We illustrate 5.1 with several examples (each with d′ = 1).

(1) Let (m, p) = (7, 5). Then h′(n) is displayed in the following table.

ℓ n  h′(n) ↓
2 [15,27] 1, 1, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 1,

If d = 3 then h(n) is displayed in the rows of the following table according as n varies in the congruence class ≡ ℓm (mod d), 2 ≤ ℓ ≤ 4 = d + 1.

\begin{aligned}ℓ & n \\ 2 & [17,53,3] & 1, & -, & -, & 1, & -, & -, & 2, \\ 3 & [18,51,3] & -, & 0, & -, & -, & 0, & -, & -, \\ 4 & [19,52,3] & -, & -, & 0, & -, & -, & 0, & -, \\ & [17,53] & 1, & 0, & 0, & 1, & 0, & 0, & 2, \end{aligned}\begin{aligned}& & & & & & h(n) \\ -, & -, & 2, & -, & -, & 3, & -, & -, & 2, & -, \\ 1, & -, & -, & 1, & -, & -, & 1, & -, & -, & 1, \\ -, & 0, & -, & -, & 0, & -, & -, & 1, & -, & -, \\ 1, & 0, & 2, & 1, & 0, & 3, & 1, & 1, & 2, & 1, \end{aligned}\begin{matrix}↓ \\ -, & 2, & -, & -, & 2, & -, & -, & 2, & -, & -, \\ -, & -, & 1, & -, & -, & 1, & -, & -, & 0, & -, \\ 1, & -, & -, & 1, & -, & -, & 1, & -, & -, & 1, \\ 1, & 2, & 1, & 1, & 2, & 1, & 1, & 2, & 0, & 1, \end{matrix}1,-, -, 1,-, -, 1,-, -, 1, -, 0,-, -, 0,-, -, 0,-, -, -, -, 1,-, -, 0,-, -, 0,-, 1,0,1,1,0,0,1,0,0,1,

Note that h′(n) from the first table is same as the ℓ = 2 row of the second table, illustrating 5.1(a). In this example h′ is not symmetric (about the midpoint 21 of I′ = [15, 27] which is indicated by the down-arrow ↓).

(2) Let (m, p) = (7, 6). Then, as in the example (1), h′(n) and h(n) for d = 3 are displayed in the following two tables:

ℓ n  h′(n) ↓
2 [15,27] 1, 1, 2, 2, 3, 3, 3, 3, 3, 2, 2, 1, 1,

\begin{aligned}ℓ & n \\ 2 & [17,53,3] & 1, & -, & -, & 1, & -, & -, & 2, \\ 3 & [18,51,3] & -, & 0, & -, & -, & 0, & -, & -, \\ 4 & [19,52,3] & -, & -, & 0, & -, & -, & 0, & -, \\ & [17,53] & 1, & 0, & 0, & 1, & 0, & 0, & 2, \end{aligned}\begin{aligned}& & & & & & h(n) \\ -, & -, & 2, & -, & -, & 3, & -, & -, & 3, & -, \\ 1, & -, & -, & 1, & -, & -, & 1, & -, & -, & 1, \\ -, & 0, & -, & -, & 0, & -, & -, & 1, & -, & -, \\ 1, & 0, & 2, & 1, & 0, & 3, & 1, & 1, & 3, & 1, \end{aligned}\begin{matrix}↓ \\ -, & 3, & -, & -, & 3, & -, & -, & 3, & -, & -, \\ -, & -, & 1, & -, & -, & 1, & -, & -, & 0, & -, \\ 1, & -, & -, & 1, & -, & -, & 1, & -, & -, & 1, \\ 1, & 3, & 1, & 1, & 3, & 1, & 1, & 3, & 0, & 1, \end{matrix}2,-, -, 2,-, -, 1,-, -, 1, -, 0,-, -, 0,-, -, 0,-, -, -, -, 1,-, -, 0,-, -, 0,-, 2,0,1,2,0,0,1,0,0,1,

Again h′(n) from the first table is same as the ℓ = 2 row of the second table, illustrating 5.1(a). This time h′ is symmetric (about the midpoint 21 of I′ = [15, 27] which is indicated by the down-arrow ↓), illustrating 5.1(c). In the above two examples the rows corresponding to ℓ = 3 are equal and so are the rows ℓ = 4. Under the reflection about the midpoint 35 of J = [17, 53] the rows ℓ = 3 and ℓ = 4 are interchanged, illustrating 5.1(c) part (iii) in the proof.

(3) In the above two examples the row ℓ = 2 contains contributions to h(n) from V_{n,r}/W_{n,r}, r = 2, 2 + d. To illustrate this, in view of 5.1(a), it suffices to consider d = 1. For (m, p, d) = (7, 5, 1) and (7, 6, 1), this is displayed in the following tables respectively.

r  n  dim_{k}(V_{n,r}/W_{n,r})
2  [15,27] 1, 1, 2, 2, 3, 2, 2, 1, 1, 0, 0, 0, 0
3  [15,27] 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1
h′(n) [15,27] 1, 1, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 1
r  n  dim_{k}(V_{n,r}/W_{n,r})
2  [15,27] 1, 1, 2, 2, 3, 3, 3, 2, 2, 1, 1, 0, 0
3  [15,27] 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1
h′(n) [15,27] 1, 1, 2, 2, 3, 3, 3, 3, 3, 2, 2, 1, 1

In each case the contributions for r = 2 and r = 3 are individually symmetric, but the sum is symmetric only in the case p = 6. More generally, for any m the sum is symmetric if and only if p = m − 1. This symmetry seems to be an accident.

5.3. Remark. Even though the Hilbert function h of Ω_{A/k}^{1}/dA is symmetric, the Hilbert function H of Ω_{A/k}^{1} need not be symmetric. This is because the equality H(n) = h(n) + 1 of 3.4(a) holds only when A_{n} ≠ 0. An example is (m, p, d) = (7, 6, 3).

6. Ω_{A/k}^{q} for q ≥ 3

The natural homomorphism π : A → A/𝔪^{2} of k-algebras induces a homomorphism of Amodules Ω_{A/k}^{q} → Ω_{(A/𝔪^{2})/k}^{q} for all q ≥ 0. In this section we show that if q ≥ 3 then this map is an isomorphism. More generally we have

6.1. Theorem. Let K be a commutative ring in which 2 is invertible and let R = K[X_{0}, …, X_{p}] be the polynomial ring. Let 𝔞 be an ideal in R with 𝔞 ⊂ (X_{0}, …, X_{p})^{2} and X_{α}X_{β}−X_{γ}X_{δ} ∈ 𝔞 whenever α + β = γ + δ, 0 ≤ α, β, γ, δ ≤ p. Let C = R/𝔞 = K[x_{0}, …, x_{p}] and let 𝔪 be the maximal ideal of C generated by x_{0}, …, x_{p}. Then for q ≥ 0 there is an isomorphism of Kmodules Ω_{(C/𝔪^{2})/K}^{q} ⟶^{≅} Λ^{q}(𝔪/𝔪^{2})⊕Λ^{q+1}(𝔪/𝔪^{2}) and for q ≥ 3, the K-algebra homomorphism π : C ⟶ C/𝔪^{2} induces an isomorphism π_{q}^{∗} : Ω_{C/K}^{q} ⟶ Ω_{(C/𝔪^{2})/K}^{q} of C-modules.

Proof. Note that C/𝔪^{2} is isomorphic to K ⊕𝔪/𝔪^{2} as a K-algebra, where the multiplication in K ⊕ 𝔪/𝔪^{2} is given by (a, v) · (b, w) = (ab, aw + bv), a, b ∈ K and v, w ∈ 𝔪/𝔪^{2}. Let P_{q}([0, p]) be the set of subsets of {0, 1, …, p} of cardinality q and, for I ∈ P_{q}([0, p]), let min(I) denote the smallest element of I. For I ∈ P_{q}([0, p]), we write x_{I} = x_{i_{1}} ∧ ⋯ ∧ x_{i_{q}} (∈ Λ^{q}(𝔪/𝔪^{2})) and dx_{I} = dx_{i_{1}} ∧ ⋯ ∧ dx_{i_{q}} (∈ Ω_{(C/𝔪^{2})/K}^{q}), where i_{1} < i_{2} < ⋯ < i_{q} are the elements of I. If K is a field of characteristic ≠ 2, then it was proved in [7; Theorem 1] that for all q ≥ 0 there is an isomorphism θ : Ω_{(C/𝔪^{2})/K}^{q} → Λ^{q}(𝔪/𝔪^{2}) ⊕ Λ^{q+1}(𝔪/𝔪^{2}) of K-modules such that θ(dx_{I}) = x_{I} and θ(x_{i_{0}}dx_{I}) = x_{i_{0}} ∧ x_{I}. The same proof works in the case of any commutative ring K in which 2 is invertible. Since Ω_{C/K}^{q} is generated by {dx_{I} | I ∈ P_{q}([0, p])} as a C-module, it follows from 6.2 below that, for q ≥ 3, {dx_{I} | I ∈ P_{q}([0, p])}∪{x_{i_{0}}dx_{I} | I ∈ P([0, p]) and 0 ≤ i_{0} < min(I)} generates Ω_{C/K}^{q} as a K-module. Under the map θ ∘ π_{q}^{∗} this generating set is mapped onto the K-basis {x_{I} | I ∈ P_{q}([0, p])} ∪ {x_{i_{0}}x_{I} | I ∈ P_{q}([0, p]) and 0 ≤ i_{0} < min(I)} of Λ^{q}(𝔪/𝔪^{2}) ⊕ Λ^{q+1}(𝔪/𝔪^{2}). Therefore the map θ ∘ π_{q}^{∗} is an isomorphism and hence so also is π_{q}^{∗}. □

6.2. Lemma. With the same notation and assumptions as in 6.1, for integers 0 ≤ i, j, α, β, γ, δ ≤ p we have in Ω_{C/K}^{3}:

(a) If α < β < γ then x_{α}dx_{α} ∧ dx_{β} ∧ dx_{γ} = x_{β}dx_{α} ∧ dx_{β} ∧ dx_{γ} = x_{γ}dx_{α} ∧ dx_{β} ∧ dx_{γ} = 0.

(b) x_{α}dx_{β} ∧ dx_{γ} ∧ dx_{δ} = −x_{β}dx_{α} ∧ dx_{γ} ∧ dx_{δ} = x_{γ}dx_{α} ∧ dx_{β} ∧ dx_{δ} = −x_{δ}dx_{α} ∧ dx_{β} ∧ dx_{γ}.

(c) x_{i}x_{j}dx_{α} ∧ dx_{β} ∧ dx_{γ} = 0.

Proof. (a) The assertion follows by using induction on γ − α and the same arguments as in [7; (1.2)].

(b) In view of (a), we may assume that α, β, γ, δ are distinct and α < β < γ < δ. The assertion follows by using the same arguments as in [7; (1.3)].

(c) In view of (a) and (b) we may assume that i ≤ j < α < β < γ. Then by (b) we have x_{i}x_{j}dx_{α} ∧ dx_{β} ∧ dx_{γ} = −x_{i}x_{α}dx_{j} ∧ dx_{β} ∧ dx_{γ} = −x_{α}x_{i}dx_{j} ∧ dx_{β} ∧ dx_{γ} = x_{α}x_{j}dx_{i} ∧ dx_{β} ∧ dx_{γ} and x_{i}x_{j}dx_{α} ∧ dx_{β} ∧ dx_{γ} = x_{j}x_{i}dx_{α} ∧ dx_{β} ∧ dx_{γ} = −x_{j}x_{α}dx_{i} ∧ dx_{β} ∧ dx_{γ}. Therefore 2x_{i}x_{j}dx_{α} ∧ dx_{β} ∧ dx_{γ} = 0 and so x_{i}x_{j}dx_{α} ∧ dx_{β} ∧ dx_{γ} = 0, since 2 is invertible in K. □

6.3. Corollary. Let m, p, d be positive integers with 1 ≤ p ≤ m − 1, gcd(m, d) = 1 and let m_{i} = m + id, 0 ≤ i ≤ p. Let k be a field of characteristic ≠ 2, A = k[t^{m_{0}}, …, t^{m_{p}}], and let 𝔪 be the maximal ideal of A generated by t^{m_{0}}, …, t^{m_{p}}. Then, for q ≥ 0 there is an isomorphism of k-vector spaces Ω_{(A/𝔪^{2})/k}^{q} ⟶^{≅} Λ^{q}(𝔪/𝔪^{2}) ⊕ Λ^{q+1}(𝔪/𝔪^{2}), and, for q ≥ 3, the k-algebra homomorphism π : A ⟶ A/𝔪^{2} induces an isomorphism Ω_{A/k}^{q} ⟶ Ω_{(A/𝔪^{2})/k}^{q} of A-modules.

Proof. We use the notation of Section 1. Since 𝔭 ⊂ (X_{0}, …, X_{p})^{2} and X_{α}X_{β} − X_{γ}X_{δ} ∈ 𝔭 whenever α + β = γ + δ, 0 ≤ α, β, γ, δ ≤ p, the corollary follows from 6.1. □

6.4. Remark. One can apply 6.1 in situations other than monomial curves, for example, in the case of the coordinate axes, i.e., C = K[X_{0}, …, X_{p}]/𝔞 where 𝔞 is the ideal generated by {X_{i}X_{j} | 0 ≤ i < j ≤ p}.

6.5. The deRham Complex. Recall that the deRham complex

0 → A →^{d} Ω_{A/k}^{1} ⟶^{d} Ω_{A/k}^{2} ⟶^{d} ⋯ ⟶^{d} Ω_{A/k}^{q} ⟶^{d} Ω_{A/k}^{q+1} ⟶^{d} ⋯

of A/k is a complex of graded k-vector spaces (with degree of the differential d equal to 0) and is exact (characteristic of k is 0 is needed here) at each q ≥ 1 by [6; Theorem (1.2)]. Combining this with the isomorphism in 6.3 we get a complex of graded k-vector spaces

0 → A ⟶^{d} Ω_{A/k}^{1} ⟶^{d} Ω_{A/k}^{2} ⟶^{d} Λ^{3}(𝔪/𝔪^{2}) ⊕ Λ^{4}(𝔪/𝔪^{2}) ⟶^{d} Λ^{4}(𝔪/𝔪^{2}) ⊕ Λ^{5}(𝔪/𝔪^{2}) ⋯,

where d maps Λ^{q}(𝔪/𝔪^{2}) isomorphically onto Λ^{q}(𝔪/𝔪^{2}) for q ≥ 4. Therefore we have a short exact sequence of graded k-vector spaces

(6.6)  0 → Ω_{A/k}^{1}/dA ⟶^{d} Ω_{A/k}^{2} ⟶^{d} Λ^{3}(𝔪/𝔪^{2}) ⟶ 0.

From the short exact sequence (6.6) we can determine the Hilbert function of Ω_{A/k}^{2} if we know that for Ω_{A/k}^{1}/dA and Λ^{3}(𝔪/𝔪^{2}). The Hilbert function of Ω_{A/k}^{1}/dA was determined in 3.4 and the determination of the Hilbert function of Λ^{3}(𝔪/𝔪^{2}) is described below.

6.7. Lemma. With notation and assumptions as in 6.3, for non-negative integers q, n with q ≤ p + 1 , and f = q(m + \frac{(q−1)}{2} d) we have

(a) dim_{k} (Λ^{q}(𝔪/𝔪^{2}))_{n} is the coefficient of T^{(n−f)/d} in the Gaussian polynomial

(6.7.1)  [ \begin{matrix}p+1 \\ q \end{matrix} ]_{T} = \frac{(T^{p+1} − 1)(T^{p} − 1) ⋯ (T^{p−q+2} − 1)}{(T^{q} − 1)(T^{q−1} − 1) ⋯ (T − 1)} .

(b) The largest weight n for which dim_{k} (Λ^{q}(𝔪/𝔪^{2}))_{n} ≠ 0 is ∑_{i=p−q+1}^{p} m_{i} and the smallest weight is ∑_{i=0}^{q−1} m_{i}.

(c) The largest weight n for which (Ω_{A/k}^{2})_{n} ≠ 0 is max{3m + 3d(p − 1), n_{0} − 1} where n_{0} is as in 4.4.

Proof. (a) The k-vector space 𝔪/𝔪^{2} is graded with basis elements of weights m = m_{0}, m_{1}, …, m_{p}, respectively and {x_{I} | I ∈ P_{q}([0, p]) (see the proof of 6.1 for this notation) is a kbasis of Λ^{q}(𝔪/𝔪^{2}). Therefore dim_{k}(Λ^{q}𝔪/𝔪^{2})_{n} is the number of partitions of n as the sum of q distinct elements of {m_{0}, …, m_{p}}. As noted in [7], by subtracting m_{0} from the first integer, m_{1} from the second, …, m_{q−1} from the q-th, we see that this is the number of partitions of (n − f)/d as the sum of q non-negative integers, each at most (p + 1 − q), where f = ∑_{i=0}^{q−1} m_{i} = q(m + \frac{(q−1)}{2} d). Now the assertion follows from [1; pp. 33, 35].

(b) The elements x_{p−q+1} ∧ x_{p−q+2} ∧ ⋯ ∧ x_{p} and x_{0} ∧ x_{1} ∧ ⋯ ∧ x_{q−1} ∈ Λ^{q}(𝔪/𝔪^{2}) are of largest and smallest weight respectively.

(c) From (b), 3m + 3d(p − 1) is the maximum n such that (Λ^{3}(𝔪/𝔪^{2}))_{n} ≠ 0 and by definition n_{0} − 1 is the maximum n such that dim_{k}(Ω_{A/k}^{1}/dA)_{n} ≠ 0. Therefore the assertion follows from (6.6). □

The examples (m, p, d) = (7, 5, 3) and (7, 4, 3) show that either of the two numbers in 6.7(c) can be the maximum, these pairs are (57, 53) and (48, 60) respectively.

The Gaussian polynomials have been much studied, with quite a large literature. The following is useful for the discussion of the Hilbert function of Ω_{A/k}^{q}.

6.8. Remark. Let p and q be any non-negative integers with q ≤ p. Then the non-zero coefficients of [ \begin{matrix}p \\ q \end{matrix} ]_{T} are symmetric and unimodal. The first part is easy and the second part

is well-known but deep. For the latter see [8]^{1}.

6.9. Example. For (m, p, d) = (7, 5, 3), the following table illustrates the use of (6.6) to compute the Hilbert function of Ω_{A/k}^{2}, where h(n) = dim_{k}(Ω_{A/k}^{1}/dA)_{n}, H_{2}(n) = dim_{k}(Ω_{A/k}^{2})_{n} and λ_{3}(n) = dim_{k}(Λ^{3}(𝔪/𝔪^{2}))_{n}. The row of h(n) is from 5.2. The row of λ_{3}(n) is computed using 6.7 as follows:

[ \begin{matrix}6 \\ 3 \end{matrix} ]_{T} = \frac{(T^{6} − 1)(T^{5} − 1)(T^{4} − 1)}{(T^{3} − 1)(T^{2} − 1)(T − 1)} = 1 + T + 2T^{2} + 3T^{3} + 3T^{4} + 3T^{5} + 3T^{6} + 2T^{7} + T^{8} + T^{9}

and the coefficient of T^{i} in [ \begin{matrix}6 \\ 3 \end{matrix} ]_{T} is λ_{3}(3i + 30), 0 ≤ i ≤ 9. The H_{2}(n) row is the sum of the

other two rows by (6.6).

\begin{matrix}n & 16-58 \\ h(n) & 0,1,0,0,1,0,0,2,1,0,2,1,0,3,1,1,2,1,1,2,1,1,2,1,1,2,0,1,1,0,1,1,0,0,1,0,0,1,0,0,0,0,0 \\ H_{2}(n) & 0,1,0,0,1,0,0,2,1,0,2,1,0,3,2,1,2,2,1,2,3,1,2,4,1,2,3,1,1,3,1,1,3,0,1,2,0,1,1,0,0,1,0 \\ λ_{3}(n) & 0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,2,0,0,3,0,0,3,0,0,3,0,0,3,0,0,2,0,0,1,0,0,1,0 \end{matrix}

It is clear from (6.7.1) that the Gaussian polynomial is a product of the cyclotomic polynomials Φ_{r}. However, the question remains as to which cyclotomic polynomials occur. Because [ \begin{matrix}p \\ q \end{matrix} ]_{T} = [ \begin{matrix}p \\ p−q \end{matrix} ]_{T}, we can assume that q ≤ p/2.

6.10. Theorem. Let Φ_{r} be the cyclotomic polynomial of primitive r-th roots of 1. Let p, q be positive integers with q ≤ p/2. Then [ \begin{matrix}p \\ q \end{matrix} ]_{T} is the product (without repetition) of the Φ_{r}

such that r satisfies one of the following two conditions:

(i) r > q and r divides one of the integers p, p − 1, …, p − q + 1 .

(ii) 2 ≤ r < q and if p ≡ s′_{r} (mod r), 0 ≤ s′_{r} < r, q ≡ s_{r} (mod r), 0 ≤ s_{r} < r, then 0 ≤ s′_{r} ≤ s_{r} − 1 .

Proof. For 1 ≤ i ≤ q, let F_{i}(T) = (T^{p−i+1} − 1) and G_{i} = (T^{i} − 1). Then F (T) = F_{1} ⋯ F_{q} is the numerator of (6.7.1) and G(T) = G_{1} ⋯ G_{q} is the denominator of (6.7.1). Note that, since q ≤ p/2, we have F_{i} ≠ G_{j} for all 1 ≤ i, j ≤ q. Throughout the following proof we use the well-known fact:

(6.10.1)  Φ_{r} is irreducible and divides X^{n} − 1 if and only if r divides n .

Case 1: r > q: In this case Φ_{r} does not divide G(T) and divides at most one of F_{1}, …, F_{q}. Now, the assertion follows from (6.10.1).

1Thanks to Dom de Caen for telling us about this reference.

Case 2: r ≤ q: Since X − 1 divides F_{i} and G_{i} for all 1 ≤ i ≤ q, Φ_{1} does not divide [ \begin{matrix}p \\ q \end{matrix} ]_{T}.

Write p = a′r + s′_{r} and q = ar + s with a′, a ∈ ℤ^{+}, 0 ≤ s′_{r}, s_{r} < r. Then Floor[q/r] = a and by (6.10.1) we have

(6.10.2)  |{G_{i} | Φ_{r} divides G_{i}, 1 ≤ i ≤ q}| = |{jr | 1 ≤ j ≤ a}| = a.

Since p − q + 1 = (a′ − a)r + (s′_{r} − s_{r}) + 1 ≤ (a′ − a + 1)r, we have

\{p − i + 1 | r divides p − i + 1, 1 ≤ i ≤ q\} = \{ \begin{matrix}\{(a′ − j)r | 0 ≤ j ≤ a − 1\}, & if s′_{r} ≥ s_{r}, \\ \{(a′ − j)r | 0 ≤ j ≤ a\}, & if s′_{r} < s_{r}, \end{matrix}

and so by (6.10.1), we have

(6.10.3)  |\{F_{i} | Φ_{r} divides F_{i}, 1 ≤ i ≤ q\}| = \{ \begin{matrix}a, & if s′_{r} ≥ s_{r}, \\ a + 1, & if s′_{r} < s_{r}. \end{matrix}

From (6.10.2) and (6.10.3) it follows that Φ_{r} divides [ \begin{matrix}p \\ q \end{matrix} ]_{T} if and only if 0 ≤ s′_{r} < s_{r}. □

6.11. Example. To illustrate 6.10 consider q =10 and p=24. Then we have {r |r satisfies (i) of 6.10}={11, 12, [15, 24]}. For r ∈[2, 9] we successively have (p, q)=(24, 10)≡{(0, 0), (0, 1), (0, 2), (4, 0), (0, 4), (3, 3), (0, 2), (6, 1)} (mod r). These are the pairs (s′_{r}, s_{r}) for r ∈ [2, 9] and so \{r | r satisfies (ii) of 6.10 \} = \{3, 4, 6, 8\}. Therefore [ \begin{matrix}24 \\ 10 \end{matrix} ]_{T} = Φ_{3}Φ_{4}Φ_{6}Φ_{8}Φ_{11}Φ_{12} ∏_{i=15}^{24} Φ_{i}

by 6.10.

References

[1] Andrews, G. E.: The Theory of Partitions. Addison-Wesley 1976.

[2] Cortinas, C.; Geller, S. C.; Weibel, C.: The Artinian Berger Conjecture. To appear in Math. Zeitschr.

[3] Kunz, E.: Kähler Differentials. Advanced Lectures Mathematics, Vieweg, Braunschweig-Wiesbaden 1986.

[4] Patil, D. P.: Generators for the Derivation Modules and the Defining Ideals of Certain Monomial Curves. Ph.D. Thesis, University of Bombay 1989.

[5] Patil, D. P.: Minimal Sets of Generators for the Relation Ideals of Certain Monomial Curves. Manuscr. Math. 80 (1993), 239–248.

[6] Roberts, L. G.: Kähler differentials and HC_{1} of certain graded k-algebras. In: J.F.Jardine and V.P.Snaith, eds., Algebraic K-Theory: Connections with Geometry and Topology, NATO ASI series, Series C, 279, Kluwer Academic Publishers, Dordrecht 1989, 389–424.

[7] Roberts, L. G.: Kähler Differentials of Certain Cusps. Proceedings of the Great Lakes K-Theory Conference, Fields Institute Communications 16 (1997), 279–284.

[8] Zielberger, D.: Kathy O’Hara’s Constructive Proof of the Unimodality of the Gaussian Polynomials. Amer. Math. Monthly 96 (1989), 590–602.

Received July 6, 1998