NON-COHEN-MACAULAY PROJECTIVE MONOMIAL CURVES WITH POSITIVE h-VECTOR

VICTORIA E. DE QUEHEN AND LESLIE G. ROBERTS

Abstract. We find an infinite family of projective monomial curves all of which have h-vector with no negative values and are not Cohen-Macaulay.

Let 𝒮 = {m_{0}, …, m_{p+1}} be a sequence of integers with 0 < m_{0} < m_{1} < ⋯ < m_{p+1}, gcd({m_{i}}) = 1 and Γ the monoid generated by the elements of 𝒮. Let k be a field and R = 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}}]. Then R is the homogeneous coordinate ring of the projective monomial curve C = Proj(R), and is ℕ-graded by assigning degree 1 to its algebra generators. The degree of the curve C is m_{p+1}. The Hilbert function H_{R} of R is defined by H_{R}(n) = dim_{k} R_{n}. The difference sequence ΔH_{R} is defined by ΔH_{R}(0) = 1 and ΔH_{R}(n) = H_{R}(n) − H_{R}(n − 1) for n ≥ 1. The h-vector of R (or C) is defined to be the second difference sequence Δ_{R}^{2} := Δ(ΔH_{R}). If R is Cohen-Macaulay then it is immediate that Δ_{R}^{2} has no negative values (being the Hilbert function of R/(u^{m_{p+1}}, t^{m_{p+1}})). One might ask about the converse. Question 1. If R is the homogeneous coordinate ring of a projective monomial curve and Δ^{2}H_{R} has no negative values, is R Cohen-Macaulay?

Counterexamples to Question 1 appear to be unknown (although a non-monomial counterexample is given in [3, (b)page 513]). In this note we give an infinite class of counterexamples. Our proof is based on the algorithm for computing ΔH_{R} which we describe in the next paragraph. More details can be found in [2] or [1]. We learned about the problem from Dilip Patil.

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}) .

2000 Mathematics Subject Classification. Primary 14H45.

This paper was supported by an NSERC Undegraduate Student Research Award of the first author and the NSERC operating grant of the second author, and is based on the summer 2002 research project of the first author.

The set 𝔐_{i} is the set obtained from the set Θ_{i} by removing those integers which have occurred in Θ_{j}, for some 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. In [1] it was shown that ΔH_{R}(n) = #𝔐_{n} (by defining a ring gr(𝒮) isomorphic to R/u^{m_{p+1}} which has Hilbert function H(n) = #𝔐_{n}, # denoting cardinality). Furthermore R is Cohen-Macaulay if and only if addition of m_{p+1} gives a map 𝔐_{i} → 𝔐_{i+1} for all i ≥ 0.

For example, if 𝒮 = {1, 3, 4} we get Macaulay’s non-Cohen-Macaulay example R = k[u^{4}, tu^{3}, t^{3}u, t^{4}]. Here 𝔐_{0} = {0}, 𝔐_{1} = {1, 3, 4}, 𝔐_{2} = {2, 5, 6, 7, 8}, 𝔐_{3} = {9, 10, 11, 12}. The sequence ΔH_{R}(n) = #𝔐_{n} starts out 1, 3, 5, 4 → and the h-vector Δ^{2}H_{R} is 1, 2, 2, −1, 0 →. The quickest way to see that R is not Cohen-Macaulay is to observe that Δ^{2}H_{R}(3) = −1. But one might also observe that 4 + 𝔐_{2} = 4 + {2, 5, 6, 7, 8} = {6, 9, 10, 11, 12} ⊈ 𝔐_{3} = {9, 10, 11, 12}. The simple test using the h-vector seems to work in all familiar examples, but the results of this paper show that it does not suffice in an infinite number of cases.

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. Order corresponds to degree in the grading of R so the two words can be used interchangeably.

Notation 2. Throughout {a, b, c, …} is a set of integers whose elements are a, b, c, . . .. If S and T are two sets of integers then S + T is {s + t|s ∈ S, t ∈ T }. If S is a set and n is an integer then n + S is {n + s|s ∈ S}.

NON-COHEN-MACAULAY CURVES WITH POSITIVE h-VECTOR 3

The examples: We did a systematic search for counterexamples to Question 1 using Mathematica programs based on the algorithm described above. We found no counterexamples with p = 1 (projective monomial curves in ℙ^{3}) or more generally for 𝒮 an almost arithmetic progression (all but one element of 𝒮 in an arithmetic progression). (If 𝒮 is an arithmetic progression, then R is Cohen-Macaulay, so no counterexample is possible.) If p = 2, out of 91390 = \binom{40}{4} cases tested (all cases up to degree 40, of which 5619 are ruled out by not having greatest common divisor 1) we found 230 counterexamples, the smallest degree being 20. If p = 3 the smallest degree of a counterexample is 12, for example 𝒮 = {1, 2, 5, 8, 12}. For p = 4, 5, 6, 7, 8 the smallest degree of counterexample is respectively 14,16,18,20,22. (Does this pattern continue? It seems unlikely. In any event enough further cases were tested to show that 12 is the lowest degree of a counterexample.) If p = 2 one of the degree 20 examples is 𝒮 = {5, 9, 11, 20}. This is the first of our proposed infinite family of counterexamples, sets 𝒮_{x} of the form {−1 + 6x, 3 + 6x, 5 + 6x, 8 + 12x} for x an integer greater than or equal to 1. Curiously, of the 230 counterexamples with p = 2 of degrees between 20 and 40, there are no counterexamples with degrees 21,25,26,27, and only 2 of degree 33 (compared with 14 of degree 34 and 26 of degree 32), so the existence of counterexamples seems to be a bit delicate. Having the sum of the second and third coordinates equal to the fourth seems to be part of what makes our family work. However out of our original 230 counterexamples only 30 have this feature. A further 30 have the sum of the first and second coordinates equal to the fourth. These are obtainable from the previous 30 by interchanging t and u in the corresponding ring R. There are two examples with the sum of the first and third coordinates equal to the fourth.

For the rest of this paper let 𝒮 = 𝒮_{x} as defined above. Then p = 2 and m_{p+1} = m_{3} = 8 + 12x. The sets 𝔐_{i} decompose naturally into subsets according to the maximum number of copies of m_{p+1} in a minimal 𝒮expression of an integer n ∈ Γ. Based on this idea we define sets (i, j) where 0 ≤ j ≤ min(i, 2x + 3) as follows:

(1) If 0 ≤ i ≤ 2x + 1 then (a) (i, 0) = {(8 + 12x)i}

(b) (i, 1) = (8 + 12x)i − (9 + 6x) + {0, 4, 6}

(c) (i, 2) = (8 + 12x)i − 2(9 + 6x) + {0, 4, 6, 8, 12}

(d) (i, j) = (8 + 12x)i − j(9 + 6x) + {0, 4, 6, 8, 12, 6j} for 3 ≤

j ≤ i.

(2) If i = 2x + 2 then (a) If 0 ≤ j ≤ 2x + 1 then (i, j) is defined as in case 1.

(b) (2x + 2, 2x + 2) = (8 + 12x − 9 − 6x)i + {4, 6, 8, 12, 6j} (same as (i, i) in case 1 except that the smallest element is omitted).

(3) If i ≥ 2x + 3 then (a) If 0 ≤ j ≤ 2x + 1 then (i, j) is defined as in cases 1 and 2.

(b) If j = 2x+2 then (i, j) = (8+12x)i−j(9+6x)+{6, 8, 12, 6j} (the two smallest elements omitted from previous formulae).

(c) (i, 2x + 3) = {(8 + 12x)i − (9 + 6x)j + 12} (a singleton)

We sometimes refer to the (i, j) as blocks because (i, j) is (almost) a consecutive set if the elements of {∪_{j}(i, j)} are listed by increasing magnitude. (This is illustrated by Example 5 below.) The definitions of the (i, j) were formulated by studying the 𝔐_{i} for various values of x, but so far they are just sets defined as above. We relate them to the 𝔐_{i} by the following lemma.

Lemma 3. The sets (i, j) are pairwise disjoint, and 𝔐_{i} = ∪_{j}(i, j).

Assuming Lemma 3 we can prove our main result.

Theorem 4. For all integers x ≥ 1, 𝒮_{x} = {−1 + 6x, 3 + 6x, 5 + 6x, 8 + 12x} gives a counterexample to Question 1. More explicitly, the corresponding homogeneous coordinate ring R has h-vector 1 3 5 6 ⋯ 6 5 0 (with 6 in degrees 3 ≤ n ≤ 2x + 1) and is not Cohen-Macaulay.

Proof. From the definitions m_{p+1} + (i, j) = (i + 1, j), with the one exception i = j = 2x + 2. Given Lemma 3 this means (with one exception) that Δ^{2}H_{R}(n) is the number of elements in the new block (n, n) that appears in degree n. (0, 0) contains 1 element (namely 0). (1, 1) = {−1 + 6x, 3 + 6x, 5 + 6x} contains 3 elements, and (2, 2) contains 5 elements. For 3 ≤ n ≤ 2x + 1, (n, n) contains 6 elements (and since x ≥ 1 the interval 3 ≤ n ≤ 2x + 1 is non-empty). The block (2x+2, 2x+2) contains 5 elements. The block (2x+3, 2x+2) contains 4 elements so in passing from (2x+2, 2x+2) to (2x+3, 2x+2) one element disappears. But it is replaced by the one element in (2x + 3, 2x + 3) so Δ^{2}H_{R}(2x+3) = 0 and the h-vector of R is 1 3 5 6 ⋯ 6 5 0 → (with 6 in degrees 3 ≤ n ≤ 2x+1). In particular Δ^{2}H_{R}(n) has no negative values. But R is not Cohen-Macaulay because adding m_{p+1} to (2x + 2, 2x + 2) is not a map into (2x + 3, 2x + 2). In the language of [1, (1.6)], Γ has one unstable element (8 + 12x)(2x + 2) − (9 + 6x)(2x + 2) + 4 = (−1 + 6x)(2x + 2) + 4 in degree 2x + 2. □

The proof of Lemma 3 now proceeds in steps. Steps 1-3 show that the blocks (i, j) are pairwise disjoint. Step 4 shows that if n ∈ (i, j) then ord_{𝒮}(n) ≤ i. Steps 5 and 6 are the technical ingredients for showing in step 7 that the reverse implication holds, i.e. if ord_{𝒮}(n) = i then n ∈ (i′, j′) with i′ ≤ i, and j′ an integer such that 0 ≤ j′ ≤ min(i, 2x + 3). Lemma 3 is then proved in a brief final paragraph. Step 1. First arrange the (i, j) in columns where, as we go down one step in each column, i is increased by 1 and j is increased by 2. Thus if 0 ≤ i ≤ 2x + 3 column i has (i, i) at the bottom and for i = 2ℓ even, (ℓ, 0) at the top, for i = 2ℓ + 1 odd, (ℓ + 1, 1) at the top. Column i contains ℓ + 1 elements if 0 ≤ i ≤ 2x + 3. If i > 2x + 3 and i = 2ℓ is even then (ℓ, 0) is at the top of column i and (ℓ + x + 1, 2x + 2) is at the bottom. If i > 2x + 3 and i = 2ℓ + 1 is odd then (ℓ + 1, 1) is at the top of column i and (ℓ + x + 2, 2x + 3) is at the bottom of column i. If i > 2x + 3 then column i contains x + 2 elements.

Example 5. To help the reader visualize this set up we illustrate it with 𝒮 = {5, 9, 11, 20} corresponding to x = 1. The columns have been shifted down as we proceed to the right, so that each row contains elements with the same value of i. Thus, after establishing Lemma 3, the rows turn out to be the 𝔐_{i}. The diagram is given twice, once with the names (i, j), and a second time with the actual sets. The columns begin at the left with column 0 and continue indefinitely to the right, with columns up to 8 shown. Because of space limitations the second diagram is split in two, with columns 0-4 in the first part and 5-8 in the second, the asterisk indicating where the two halves are fitted together. The blocks with the largest interval of their column are indicated in bold type (see step 3). Diagram of indices (i, j).

(0, 0)

(1, 1) (1, 0)
(2, 2) (2, 1) (2, 0)
(3, 3) (3, 2)  (3, 1) (3, 0)
(4, 4)∗ ∗(4, 3) (4, 2) (4, 1) (4, 0)
(5, 5) (5, 4) (5, 3) (5, 2)
(6, 5) (6, 4)

Corresponding diagram of the sets (i, j).

{0}

{5, 9, 11}  {20}
{10, 14, 16, 18, 22} {25, 29, 31} {40}
{15, 19, 21, 23, 27, 33} {30, 34, 36, 38, 42}
{24, 26, 28, 32, 44}∗
{45, 49, 51}  {60}
∗{35, 39, 41, 43, 47, 53} {50, 54, 56, 58, 62} {65, 69, 71} {80}
{37}  {46, 48, 52, 64} {55, 59, 61, 63, 67, 73} {70, 74, 76, 78, 82}
{57}  {66, 68, 72, 84}

Step 2. The sets (i, j) in each column are pairwise disjoint. To see this it is convenient to replace the set (i, j) by the (possibly larger) set (i, j)′ where if j = 0, 1 then (i, j)′ = (i, j) and if j ≥ 2 then (i, j)′ = (8+12x)i−j(9+6x)+{0, 4, 6, 8, 12, 6j}. (If j = 2 the element 6j coincides with one of the other elements. To form the set (i, 2)′ first use this formula, then discard the duplicate.) As i increases by 1 and j increases by 2 the integer (8+12x)i−j(9+6x) decreases by 10 and the 6j term (always the largest) increases by 12 (which after subtracting 10 is a net increase of 2). Since the integers 0,4,6,8,12 are in distinct congruence classes mod 10 the sets (i, j)′ corresponding to a particular column are pairwise disjoint, hence a fortiori so are the sets (i, j) in that column.

Step 3. The sets (i, j) are pairwise disjoint. Let [i, j] be the set of all integers n such that a ≤ n ≤ b, where a is the smallest integer of (i, j) and b is the largest. ([i, j] will be referred to as the interval of (i, j).) From the definition of the (i, j) and the discussion in step 2 we see that as we go down one step in a column the intervals [i, j] get larger with the upper bound increasing by 2 and the lower bound decreasing by 10 (or 6 in the bottom step of column 2x + 2, 4 in the bottom step of columns i, i even with i > 2x + 2) except that the last interval [ℓ + x + 2, 2x + 3] in column i (i = 2ℓ + 1 > 2x + 2) is a singleton which is contained in the previous interval [ℓ + x + 1, 2x + 1] (the smallest element of the latter being two less than the singleton). Notice that the parity of all integers in (i, j) is that of j, so all integers in a block of column i are even if i is even and odd if i is odd. Therefore to show that the (i, j) are disjoint it suffices to observe that the largest intervals in columns of the same parity do not overlap. This is illustrated by the following table. LI(i) indicates the largest interval in column i. The gap column is the smallest element of the largest interval in column i minus the largest element in the largest interval of column i − 2.

\begin{matrix}i & LI(i − 2) & LI(i) & gap \\ 2 ≤ i ≤ 2x + 1 & [i − 2, i − 2] & [i, i] & 12x − 6i + 10 \\ 2x + 2 & [2x, 2x] & [2x + 2, 2x + 2] & 2 \\ 2x + 3 & [2x + 1, 2x + 1] & [2x + 2, 2x + 1] & 2 \\ i = 2ℓ ≥ 2x + 4 & [ℓ + x, 2x + 2] & [ℓ + x + 1, 2x + 2] & 2 \\ i = 2ℓ + 1 > 2x + 4 & [ℓ + x, 2x + 1] & [ℓ + x + 1, 2x + 1] & 2 \end{matrix}

The gap in the first row is greater than 0 since in this row i ≤ 2x+1. Since the gaps are all positive step 3 follows.

The calculations are all straightforward using the above description of the largest interval in each column, and the definitions of the (i, j). As one example we will work through the i = 2ℓ ≥ 2x+4 row. If i = 2ℓ is even and i ≥ 2x + 4 then the largest interval in column i − 2 is [ℓ + x, 2x+2] and the largest interval in column i is [ℓ+x+1, 2x+2]. These intervals respectively have largest value (8 + 12x)(ℓ + x) − (2x + 2)(9 + 6x)+6(2x+2) and smallest value (8+12x)(ℓ+x+1)−(9+6x)(2x+2)+6 with difference (8+12x)(ℓ+x+1)−(9+6x)(2x+2)+6−(8+12x)(ℓ+ x) + (9 + 6x)(2x + 2) − 6(2x + 2) = (8 + 12x) + 6 − 6(2x + 2) = 2. As an illustration, if x = 1, i = 6, this yields ℓ = 3, blocks (5, 4), (4, 4) and difference 46 − 44 = 2 in Example 5. Step 4. Every element in (i, j) is the sum of i elements of 𝒮. This is obvious for i = 0, 1. If i > 1 then it is immediate from the definitions that (a) if j < i then (i, j) ⊆ 8 + 12x + (i − 1, j) (recall that 8 + 12x = m_{p+1} = m_{3}) (b) if i ≥ 1, (i, i) ⊆ (1, 1) + (i − 1, i − 1). The claim of this step then follows by induction.

Step 5. 8+12x+(i, j) = (i+1, j) with the exception that (8+12x)(2x+ 2) − (9 + 6x)(2x + 2) + 4 ∈ (2x + 2, 2x + 2) but (8 + 12x)(2x + 3) − (9 + 6x)(2x + 2) + 4 = (8 + 12x)(2x + 2) − (9 + 6x)(2x + 2) + 6(2x + 2) ∈ (2x + 2, 2x + 2) (this is the unstable element referred to in the proof of Theorem 4. In Example 5, 20+24=44).

Step 6. For 1 ≤ i ≤ 2x + 3, (i, i) + (1, 1) ⊆ ∪_{i′≤i+1}(i′, j′). Recall that (1, 1) = (8+12x)−(9+6x)+{0, 4, 6}. We check the claim for successive i. First of all (1, 1) + (1, 1) = 2(8 + 12x) − 2(9 + 6x) + {0, 4, 6, 8, 10, 12}. The only element in (1, 1) + (1, 1) that is not in (2, 2) is 2(8 + 12x) − 2(9 + 6x) + 10, which equals 8 + 12x ∈ (1, 0).

Similarly (1, 1)+(2, 2) = (8+12x)−(9+6x)+{0, 4, 6}+2(8+12x)− 2(9+6x)+{0, 4, 6, 8, 12} = 3(8+12x)−3(9+6x)+{0, 4, 6, 8, 10, 12, 14, 16, 18}. The only elements in (1, 1) + (2, 2) but not in (3, 3) are 3(8 + 12x) − 3(9 + 6x) + 10 = 2(8 + 12x) − (9 + 6x) ∈ (2, 1), 3(8 + 12x) − 3(9 + 6x) + 14 = 2(8 + 12x) − (9 + 6x) + 4 ∈ (2, 1), and 3(8 + 12x) − 3(9 + 6x) + 16 = 2(8 + 12x) − (9 + 6x) + 6 ∈ (2, 1).

Now suppose that 3 ≤ i ≤ 2x. Then (1, 1) + (i, i) = (8 + 12x) − (9 + 6x)+{0, 4, 6}+(8+12x)i−(9+6x)i+{0, 4, 6, 8, 12, 6i} = (8+12x)(i+ 1)−(9+6x)(i+1)+{0, 4, 6, 8, 10, 12, 14, 16, 18, 6i, 6i+4, 6(i+1)}. The elements not in (i + 1, i + 1) come from the 10, 14, 16, 18, 6i, 6i + 4 in the brackets. We have (8 + 12x)(i + 1) − (9 + 6x)(i + 1) + 10 = (8+12x)i−(9+6x)(i−1) ∈ (i, i−1), (8+12x)(i+1)−(9+6x)(i+1)+14 = (8 + 12x)i − (9 + 6x)(i − 1) + 4 ∈ (i, i − 1), (8 + 12x)(i + 1) − (9 + 6x)(i + 1) + 16 = (8 + 12x)i − (9 + 6x)(i − 1) + 6 ∈ (i, i − 1), and (8+12x)(i+1)−(9+6x)(i+1)+18 = (8+12x)i−(9+6x)(i−1)+8 ∈ (i, i − 1). If i = 3 then (8 + 12x)(i + 1) − (9 + 6x)(i + 1) + 6i is the same as the 18 case just covered. If i = 4 then (8 + 12x)(i + 1) − (9 + 6x)(i + 1) + 6i = (8 + 12x)(i − 1) − (9 + 6x)(i − 3) + 4 ∈ (3, 1). If i > 4 then (8+12x)(i+1)−(9+6x)(i+1)+6i = (8+12x)(i−2)−(9+6x)(i−5)+ 6(i−5) ∈ (i−2, i−5). Finally (8+12x)(i+1)−(9+6x)(i+1)+6i+4 = (8 + 12x)i − (9 + 6x)(i − 1) + 6(i − 1) ∈ (i, i − 1).

Now consider i = 2x + 1. Then (1, 1) + (2x + 1, 2x + 1) = (8 + 12x) − (9 + 6x) + {0, 4, 6} + (8 + 12x)(2x + 1) − (9 + 6x)(2x + 1) + {0, 4, 6, 8, 12, 6(2x + 1)} = (8 + 12x)(2x + 2) − (9 + 6x)(2x + 2) + {0, 4, 6, 8, 10, 12, 14, 16, 18, 6(2x + 1), 6(2x + 1) + 4, 6(2x + 2)}. The elements of (1, 1)+(2x+1, 2x+1) not in (2x+2, 2x+2) come from the 0, 10, 14, 16, 18, 6(2x + 1), 6(2x + 1) + 4. The 10, 14, 16, 18, 6(2x + 1) + 4 are handled in the same way as the previous case, yielding an element of (2x + 1, 2x). If x = 1 then the 6(2x + 1) case coincides with the 18 case. If x > 1 then (8 + 12x)(2x + 2) − (9 + 6x)(2x + 2) + 6(2x + 1) = (8 + 12x)(2x − 1) − (9 + 6x)(2x − 4) + 6(2x − 4) ∈ (2x − 1, 2x − 4).

Finally the 0 case. (8+12x)(2x+2)−(9+6x)(2x+2) = (8+12x)(2x− 1) − (9 + 6x)(2x − 2) + 6(2x − 2) ∈ (2x − 1, 2x − 2).

Now consider i = 2x + 2. Then (1, 1) + (2x + 2, 2x + 2) = (8 + 12x) − (9+6x)+{0, 4, 6}+(8+12x)(2x+2)−(9+6x)(2x+2)+{4, 6, 8, 12, 6(2x+ 2)} = (8+12x)(2x+3)−(9+6x)(2x+3)+{4, 6, 8, 10, 12, 14, 16, 18, 6(2x+ 2), 6(2x+2)+4, 6(2x+3)}. Of these elements only (8+12x)(2x+3)−(9+ 6x)(2x+3)+12 ∈ (2x+3, 2x+3) so we have to say where all the other elements are. Of the integers in brackets, 10,14,16,18, and 6(2x+2)+4 all yield an element of (2x+2, 2x+1) as in the discussion of the 3 ≤ i ≤ 2x case, replacing i by 2x + 2 in the argument for 10, 14, 16, 18, 6i + 4. If x = 1, (8 + 12x)(2x + 3) − (9 + 6x)(2x + 3) + 6(2x + 2) = 49 ∈ (3, 1). If x > 1, (8 + 12x)(2x + 3) − (9 + 6x)(2x + 3) + 6(2x + 2) ∈ (2x, 2x − 3) by the argument for 6i in the 3 ≤ i ≤ 2x case, replacing i by 2x + 2. And (8 + 12x)(2x + 3) − (9 + 6x)(2x + 3) + 6(2x + 3) = (8 + 12x)(2x + 3) − (9 + 6x)(2x + 1) ∈ (2x + 3, 2x + 1) leaving only 4,6,8. We have (8+12x)(2x+3)−(9+6x)(2x+3)+4 = (8+12x)(2x−1)−(9+6x)(2x− 3) + 6(2x − 3) ∈ (2x − 1, 2x − 3), (8 + 12x)(2x + 3) − (9 + 6x)(2x + 3) + 6 = (8 + 12x)2x − (9 + 6x)(2x − 1) + 6(2x − 1) ∈ (2x, 2x − 1), and (8+12x)(2x+3)−(9+6x)(2x+3)+8 = (8+12x)(2x+1)−(9+6x)(2x+ 1)+6(2x+1) ∈ (2x+1, 2x+1). If x = 1 we need a separate calculation for (8 + 12x)(2x + 3) − (9 + 6x)(2x + 3) + 4 = 29 = 9 + 20 ∈ (2, 1).

Now consider i = 2x + 3. We have (1, 1) + (2x + 3, 2x + 3) = (8+12x)−(9+6x)+{0, 4, 6}+(8+12x)(2x+3)−(9+6x)(2x+3)+12 = (8 + 12x)(2x + 4) − (9 + 6x)(2x + 4) + {12, 16, 18}. But (2x + 4, 2x + 4) is empty so we have to find where all three elements are. As in the 3 ≤ i ≤ 2x case we have (8 + 12x)(2x + 4) − (9 + 6x)(2x + 4) + 16 = (8 + 12x)(2x + 3) − (9 + 6x)(2x + 2) + 6 ∈ (2x + 3, 2x + 2) and (8+12x)(2x+4)−(9+6x)(2x+4)+18 = (8+12x)(2x+3)−(9+6x)(2x+ 2)+8 ∈ (2x+3, 2x+2). For the 12 case we have (8+12x)(2x+4)−(9+ 6x)(2x+4)+12 = (8+12x)(2x+1)−(9+6x)(2x)+6(2x) ∈ (2x+1, 2x).

This completes the proof of step 6.

Step 7. Every element n ∈ Γ with ord_{𝒮}(n) = i occurs in ∪_{i′≤i}(i′, j′). To see this, suppose we have such an n. Choose an 𝒮-expression for n of length i with as many as possible copies of 8 + 12x, say n = a_{1}+a_{2}+⋯+a_{i−j}+j(8+12x) with all a_{ℓ} ∈ (1, 1). If j = i then n ∈ (i, 0). If ∑_{ℓ=1}^{i−j} a_{ℓ} ∈ (i − j, i − j) step 7 follows from step 5. Otherwise suppose there is some integer k, 2 ≤ k ≤ i − j such that ∑_{ℓ=1}^{k} a_{ℓ} ∉ (k, k). (k = 1 is impossible since a_{1} ∈ (1, 1).) Pick the smallest such k. Then by step 6, ∑_{ℓ=1}^{k} a_{ℓ} ∈ (k′, j′) with either k′ < k or k′ = k and j′ < k. If k′ < k then by step 4 we have an 𝒮-expression for n of length less than i. If k′ = k and j′ < k then by step 5, n can be rewritten as the sum of i elements of 𝒮 with more than j copies of 8 + 12x, both of which yield a contradiction. This completes the proof of step 7.

We can now prove Lemma 3. Disjointness of the (i, j) was established in step 3. Step 7 says that 𝔐_{i} ⊆ ∪_{i′≤i}(i′, j). Suppose x ∈ 𝔐_{i}. Then x ∈ (i′, j) for unique i′, j. If i′ < i then by step 4, x ∈ 𝔐_{i′′} for i′′ ≤ i′ (hence i′′ < i). This is a contradiction, and Lemma 3 follows. Hence also does Theorem 4.

References

[1] Dilip Patil and Leslie G. Roberts. Hilbert functions of monoial curves. To appear in JPAA.

[2] Leslie G. Roberts. Certain projective curves with unusual Hilbert function. JPAA, 104:303–311, 1995.

[3] Leslie G. Roberts and Balwant Singh. Seminormality and Cohomology of Projective Varieties. J.Algebra, 103:500–519, 1986.

Department of Mathematics and Statistics, Queen’s University, Kingston, Ontario, Canada, K7L 3N6

  E-mail address: 9qved@qlink.queensu.ca

Department of Mathematics and Statistics, Queen’s University, Kingston, Ontario, Canada, K7L 3N6

  E-mail address: robertsl@mast.queensu.ca