REGULAR SEQUENCES OF SYMMETRIC POLYNOMIALS

ALDO CONCA, CHRISTIAN KRATTENTHALER^{†}, JUNZO WATANABE

Abstract. A set of n homogeneous polynomials in n variables is a regular sequence if the associated polynomial system has only the obvious solution (0, 0, …, 0). Denote by p_{k}(n) the power sum symmetric polynomial in n variables x_{1}^{k} +x_{2}^{k} + ⋯+x_{n}^{k}. The interpretation of the q-analogue of the binomial coefficient as Hilbert function leads us to discover that n consecutive power sums in n variables form a regular sequence. We consider then the following problem: describe the subsets A ⊂ ℕ^{∗} of cardinality n such that the set of polynomials p_{a}(n) with a ∈ A is a regular sequence. We prove that a necessary condition is that n! divides the product of the degrees of the elements of A. To find an easily verifiable sufficient condition turns out to be surprisingly difficult already for n = 3. Given positive integers a < b < c with gcd(a, b, c) = 1, we conjecture that p_{a}(3), p_{b}(3), p_{c}(3) is a regular sequence if and only if abc ≡ 0 (mod 6). We provide evidence for the conjecture by proving it in several special instances.

1. Introduction

For d, n ∈ ℕ let P_{d,n}(q) be the q-analogue of the binomial coefficient (also called Gaußian polynomial)

P_{d,n}(q) := [\begin{matrix}n + d \\ n \end{matrix}]_{q} = \frac{[n + d]_{q}!}{[n]_{q}! [d]_{q}!}  (1.1)

where

[j]_{q}! = ∏_{i=1}^{j} \frac{1 − q^{i}}{1 − q} = ∏_{i=1}^{j}(1 + q + ⋯ + q^{i−1}).

It is well known that P_{d,n}(q) ∈ ℤ[q]. Furthermore for a finite field F, P_{d,n}(q) evaluated at q = |F | gives exactly the number of F -subspaces of F^{n+d} of dimension n, [S, 1.3.18]. It is also known that the coefficient of q^{k} in P_{d,n}(q) is the number of partitions of k with at most n parts and parts bounded by d, see [S, 1.3.19]. This interpretation leads to the following construction. Consider the polynomial ring R = ℂ[x_{1}, x_{2}, …, x_{n}] with the usual action of the permutation group G = S_{n}. The monomial complete intersection I = (x_{1}^{d+1}, x_{2}^{d+1}, …, x_{n}^{d+1}) is fixed by G, so that G acts as well on the quotient ring A = R/I. The homogeneous part A_{k} of degree k of A has a ℂ-basis consisting of the monomials in x_{1}^{a_{1}}x_{2}^{a_{2}} ⋯ x_{n}^{a_{n}} with

2000 Mathematics Subject Classification. Primary 05E05; Secondary 13P10 11C08. Key words and phrases. symmetric polynomials, regular sequences. †Research partially supported by the Austrian Science Foundation FWF, grant S9607-N13,

in the framework of the National Research Network “Analytic Combinatorics and Probabilistic Number Theory”.

2 ALDO CONCA, CHRISTIAN KRATTENTHALER^{†}, JUNZO WATANABE

a_{1} + a_{2} + ⋯ + a_{n} = k and a_{i} ≤ d for all i. For a partition λ : λ_{1} ≥ λ_{2} ≥ ⋯ ≥ λ_{n} denote by m_{λ} the monomial symmetric polynomial

m_{λ} = x_{1}^{λ_{1}}x_{2}^{λ_{2}} ⋯ x_{n}^{λ_{n}} + symmetric terms,

where “symmetric terms” means the sum of all terms that have to be added to complete the monomial x_{1}^{λ_{1}}x_{2}^{λ_{2}} ⋯ x_{n}^{λ_{n}} to a symmetric polynomial in x_{1}, x_{2}, …, x_{n}. The invariant ring A^{G} in degree k has a ℂ-basis consisting of the elements m_{λ} where λ is a partition of k in at most n parts with parts smaller than d + 1. In other words, P_{d,n}(q) is the Hilbert series of A^{G}. In characteristic 0 the extraction of the invariant submodule is an exact functor. So we have A^{G} = R^{G}/I^{G}, where R^{G} = ℂ[e_{1}, e_{2}, …, e_{n}] and e_{i} is the elementary symmetric polynomial of degree i. Since the e_{i}’s are algebraically independent, the Hilbert series of R^{G} is 1/ ∏_{i=1}^{n}(1 − q^{i}). Since we know that P_{d,n}(q) is the Hilbert series of A^{G}, we see that A^{G} has the Hilbert series of a complete intersection in R^{G} defined by elements of degree d+1, d+2, …, d+n. These considerations suggest that the ideal I^{G} = I ∩ R^{G} might be generated by

p_{d+1}, p_{d+2}, …, p_{d+n}

where p_{i} = x_{1}^{i} + x_{2}^{i} + ⋯ + x_{n}^{i} is the power sum of degree i and that they form a regular sequence. Since the inclusion (p_{d+1}, p_{d+2}, …, p_{d+n}) ⊆ I^{G} is obvious, to prove the equality it is enough to prove that p_{d+1}, p_{d+2}, …, p_{d+n} form a regular sequence in R^{G} or, which is the same, in R. We will see that this is indeed the case, see Proposition 2.9. We give two (simple) proofs of Proposition 2.9; one is based on Newton’s relations and the other on Vandermonde’s determinant. Denote by h_{i} the complete symmetric polynomial of degree i, that is, the sum of all the monomials of degree i in x_{1}, …, x_{n}. More generally, we are led to consider regular sequences of symmetric polynomials. In particular regular sequences of power sums p_{i} and regular sequences of complete symmetric polynomials h_{i}. Even for n = 3 these questions turn out to be difficult. For indices a, b, c with gcd(a, b, c) = 1 we conjecture that p_{a}, p_{b} and p_{c} form a regular sequence exactly when 6 | abc. We are able to verify this conjecture in a few cases in which the property under investigation is translated into the non-vanishing of a rational number which appears as a coefficient in the relevant expressions or on the irreducibility over the rationals of certain polynomials obtained by elimination, see Theorem 2.11, Proposition 2.13 and Remark 2.14. Other interesting algebraic aspects of ideals and algebras of power sums are investigated by Lascoux and Pragacz [LP] and by Dvornicich and Zannier [DZ]. We thank Riccardo Biagioli, Francesco Brenti, Alain Lascoux and Michael Filaseta for useful suggestions and comments.

2. Regular sequences of symmetric polynomials

Set R_{n} = ℂ[x_{1}, x_{2}, …, x_{n}]. Following Macdonald, we denote by e_{k}(n), p_{k}(n) and h_{k}(n) the elementary symmetric polynomial, the power sum and the complete symmetric polynomial of degree k in n-variables, respectively. When n is clear from the context or irrelevant we will simply denote them by e_{k}, p_{k} and h_{k}, respectively. For a subset A ⊂ ℕ^{∗} we set

p_{A}(n) = {p_{a}(n) : a ∈ A} and h_{A}(n) = {h_{a}(n) : a ∈ A}.

Question 2.1. For which subsets A ⊂ ℕ^{∗} with |A| = n is the set of polynomials p_{A}(n) (respectively h_{A}(n)) a regular sequence in R_{n}? That is, when is (0, …, 0) the only common zero of the equations p_{a}(n) = 0 (respectively h_{a}(n) = 0), a ∈ A?

In the following we list several auxiliary observations.

Lemma 2.2. Let A ⊂ ℕ^{∗} with |A| = n. Set d = gcd(A) and A′ = {a/d : a ∈ A}. Then p_{A}(n) is a regular sequence if and only if p_{A′}(n) is a regular sequence.

Proof. Note that if (z_{1}, z_{2}, …, z_{n}) is a common zero of the p_{a}(n)’s with a ∈ A then (z_{1}^{d}, z_{2}^{d}, …, z_{n}^{d}) is a common zero of the p_{a}(n)’s with a ∈ A′ and that if (z_{1}, z_{2}, …, z_{n}) is a common zero of the p_{a}(n)’s with a ∈ A′ then (z_{1}^{1/d}, z_{2}^{1/d}, …, z_{n}^{1/d}) is a common zero of the p_{a}(n)’s with a ∈ A where z_{i}^{1/d} denotes any d-th root of z_{i}. □

Remark 2.3. For a set A of cardinality n the condition that p_{A}(n) is a regular sequence can be rephrased in terms of a resultant: the resultant of the set must be non-zero. More directly, one can express the same fact by imposing the condition that the ideal generated by p_{A}(n) contains all the forms of degree

(∑_{a∈A} a) − n + 1.

This boils down to the evaluation of the rank of a {0, 1}-matrix of large size. However, we have not been able to compute the resultant or to evaluate the rank of the matrix efficiently.

Lemma 2.4. Let A ⊂ ℕ^{∗} with |A| = n. Set d = gcd(A) and A′ = {a/d : a ∈ A}. Then we have:

(a) The equations p_{a}(2) = 0, a ∈ A, have a non-trivial common zero in ℂ^{2} if and only if the elements in A′ are all odd.

(b) The equations h_{a}(2) = 0, a ∈ A, have a non-trivial common zero in ℂ^{2} if and only if gcd({a + 1 : a ∈ A}) ≠ 1.

Proof. (a) As in the proof of the lemma above, the polynomials p_{a}(2) with a ∈ A have a non-trivial common zero if and only if the polynomials p_{a}(2) with a ∈ A′ have a non-trivial common zero. So we may assume that A = A′. If all the elements of A are odd then (1, −1) is a non-trivial common zero. Conversely, if there is a non-trivial common zero, then we may assume this zero to be (x, 1). In this case, we have x^{a} = −1 for all a ∈ A. Since gcd(A) = 1 we can find a linear combination of the elements in A = {a_{1}, a_{2}, …, a_{n}} of the type ∑ j_{k}a_{k} = 1 with j_{k} ∈ ℤ. Hence we have x^{j_{k}a_{k}} = (−1)^{j_{k}} and thus x = (−1)^{∑ j_{k}}. Therefore we have either x = 1, which is impossible, or x = −1 which implies that all the elements of A are odd.

(b) Denote by (*) the condition: the equations h_{a}(2) = 0 with a ∈ A have a non-trivial common zero in ℂ^{2}. So (*) holds if and only if the equations have a common zero of type (c, 1). Obviously c ≠ 1, so we can multiply each h_{a}(2) by c − 1. We obtain that (*) holds if and only if the equations x^{a+1} − 1 = 0 with a ∈ A have a common root ≠ 1. In other words, gcd({x^{a+1} − 1 : a ∈ A}) ≠ x − 1. But gcd({x^{a+1} − 1 : a ∈ A}) = x^{t} − 1 with t = gcd({a + 1 : a ∈ A}), and we obtain the desired characterization. □

This answers Question 2.1 for n = 2. Namely, Lemma 2.4 implies the following fact.

Lemma 2.5. Let a, b ∈ ℕ \ {0} and d = gcd(a, b). Then we have:

(1) (p_{a}(2), p_{b}(2)) is a regular sequence if and only if a/d or b/d is even.

(2) (h_{a}(2), h_{b}(2)) is a regular sequence if and only if gcd(a + 1, b + 1) = 1.

Here is another fact.

Lemma 2.6. (1) If a ≢ 0 (mod c) and u is a primitive c-th root of unity then

(1, u, u^{2}, …, u^{c−1}, 0, …, 0) ∈ ℂ^{n}

is a zero of p_{a}(n) in ℂ^{n} for all n ≥ c.

(2) Let A ⊂ ℕ^{∗} with |A| = n and 1 ≤ c ≤ n. Set β_{c} = |{a ∈ A : a ≡ 0 (mod c)}|. If β_{c} < ⌊n/c⌋ then p_{A}(n) is not a regular sequence. Here ⌊x⌋ = max{m ∈ ℤ : m ≤ x} is the standard floor function.

Proof. (1) Evaluating p_{a}(n) at (1, u, u^{2}, …, u^{c−1}, 0, …, 0) yields ∑_{i=1}^{c} u^{(i−1)a}. Multiplying the sum by u^{a} − 1, we obtain u^{ca} − 1 which is 0. Since u is a primitive c-th root of unity and a ≢ 0 (mod c), we have u^{a}−1 ≠ 0. It follows that ∑_{i=1}^{c} u^{(i−1)a} = 0.

(2) Write n = qc + r with 0 ≤ r < c so that q = ⌊n/c⌋. Let u be a primitive c-th root of unity. Let (y_{1}, y_{2}, …, y_{q}) ∈ ℂ^{q} and consider the element ỹ ∈ ℂ^{n} obtained by concatenating y_{i}(1, u, …, u^{c−1}) for i = 1, 2, …, q and by adding the zero vector of length r at the end. By (1) we know that every such ỹ is a zero of the p_{a}(n)’s with a ≢ 0 (mod c). If we also impose that ỹ is a zero of the p_{a}(n)’s with a ∈ A and a ≡ 0 (mod c) we obtain β_{c} homogeneous equations in q variables. If β_{c} < q then there exists a non-zero common solution to that system of equations. □

We list another auxiliary result.

Lemma 2.7. (1) Let c > 2 and a+2 ≡ 0 or 1 (mod c). Let u_{1}, u_{2}, u_{3} be distinct c-th roots of unity. Then p = (u_{1}, u_{2}, u_{3}, 0, …, 0) ∈ ℂ^{n} is a zero of h_{a}(n) in ℂ^{n} for all n ≥ 3.

(2) Let A ⊂ ℕ^{∗} with |A| = n and 2 < c. Assume that for all a ∈ A one has a + 2 ≡ 0 or 1 (mod c). Then h_{A}(n) is not a regular sequence.

Proof. (1) We may assume n = 3. To show that h_{a}(3) evaluated at p = (u_{1}, u_{2}, u_{3}) is 0 we first multiply h_{a}(3) with Δ = (x_{1} − x_{2})(x_{1} − x_{3})(x_{2} − x_{3}). A simple calculation shows that

h_{a}(3)Δ = x_{1}^{a+2}(x_{2} − x_{3}) + x_{2}^{a+2}(x_{3} − x_{1}) + x_{3}^{a+2}(x_{1} − x_{2}).

Since, by assumption, we have u_{i}^{a+2} = 1 or u_{i}^{a+2} = u_{i} for all i, this expression vanishes at p.

(2) It follows immediately from (1) that p is a common zero of the equations h_{a}(n). So h_{A}(n) is not a regular sequence. □

For the general case of n homogeneous symmetric polynomials in R_{n}, there holds the following criterion.

Lemma 2.8. Let f_{1}, f_{2}, …, f_{n} be a regular sequence of homogeneous symmetric polynomials in R_{n}. Then n! divides (deg f_{1})(deg f_{2}) ⋯ (deg f_{n}).

Proof. For obvious reasons, f_{1}, f_{2}, …, f_{n} is also a regular sequence in the ring of symmetric polynomials ℂ[e_{1}, e_{2}, …, e_{n}]. Then the Hilbert series of

ℂ[e_{1}, e_{2}, …, e_{n}]/(f_{1}, f_{2}, …, f_{n})

is ∏_{i=1}^{n}(1−q^{deg f_{i}})/(1−q^{i}), and it must be a polynomial with integral coefficients. If we take the limit q → 1, we obtain ( ∏_{i=1}^{n} deg f_{i})/n!, which must be an integer. □

Next we prove that power sums, respectively complete symmetric polynomials, with consecutive indices form regular sequences.

Proposition 2.9. Let A ⊂ ℕ^{∗} be a set of n consecutive elements. Then both p_{A}(n) and h_{A}(n) are regular sequences in R_{n}.

Proof. We present two proofs for power sums and (sketch) two proofs for complete symmetric polynomials. Let a be the minimum of A. We first prove the assertion for the power sums.

Proof (1) for power sums: We argue by induction on a and n. If n = 1 then the assertion is obvious. If a = 1, then any common zero of p_{1}, p_{2}, …, p_{n} is also a common zero of e_{1}, e_{2}, …, e_{n} because p_{1}, p_{2}, …, p_{n} also generate the algebra of symmetric polynomials. But, obviously, the only common zero of the elementary symmetric polynomials is (0, …, 0).

Now assume n > 1 and a > 1. For all h ∈ ℕ we have Newton’s identity

∑_{k=0}^{n}(−1)^{k}e_{n−k}p_{k+h} = 0,

with the convention that p_{0} = n and e_{0} = 1. For h = a − 1 we have

e_{n}p_{a−1} = ∑_{k=1}^{n}(−1)^{k+1}e_{n−k}p_{k+a−1}.

If z = (z_{1}, z_{2}, …, z_{n}) is a common zero of p_{a}, p_{a+1}, …, p_{a+n−1} then it is also a zero of e_{n}p_{a−1}. So z is either a zero of p_{a−1}, and we conclude by induction on a that z = (0, …, 0) or z is a zero of e_{n}. In the second case, one of the coordinates of z is 0 and we conclude by induction on n that z = (0, …, 0).

Proof (2) for power sums: Let z = (z_{1}, z_{2}, …, z_{n}) ∈ ℂ^{n} be a solution of the polynomial system associated to p_{A}(n). We have to prove that z = 0. Let c be the cardinality of the set {z_{i} : i = 1, …, n}. We may assume that z_{1}, z_{2}, …, z_{c} are the distinct values among z_{1}, z_{2}, …, z_{n}, and for i = 1, …, c set m_{i} = |{j : z_{j} = z_{i}}|, so that m_{i} > 0. Hence p_{k}(z_{1}, …, z_{n}) = m_{1}z_{1}^{k} + ⋯ + m_{c}z_{c}^{k}. By assumption, p_{a+j}(z_{1}, …, z_{n}) = 0 for j = 0, …, n − 1. Let V be the Vandermonde matrix (z_{j}^{i}) with i = 0, …, c − 1 and j = 1, …, c. By construction, det V ≠ 0 and the column vector (m_{1}z_{1}^{a}, m_{2}z_{2}^{a}, …, m_{c}z_{c}^{a}) is in the kernel of V. It follows that (m_{1}z_{1}^{a}, m_{2}z_{2}^{a}, …, m_{c}z_{c}^{a}) = 0, that is, c = 1 and z_{1} = 0. This shows that z = 0.

Proof (1) works also for complete symmetric polynomials by replacing Newton’s identity with the corresponding identity for the complete symmetric polynomials. For a second proof for complete symmetric polynomials, one observes that the initial ideal with respect to any term order with x_{1} > x_{2} > ⋯ > x_{n} of the ideal generated by h_{a}(n), h_{a+1}(n), …, h_{a+n−1}(n) contains (and hence is equal to) (x_{1}^{a}, x_{2}^{a+1}, …, x_{n}^{a+n−1}). □

The above results (and many computer calculations) suggest the following conjecture.

Conjecture 2.10. Let A = {a, b, c}, where a, b, c are positive integers with a < b < c and gcd(A) = 1. Then p_{A}(3) is a regular sequence in R_{3} if and only if abc ≡ 0 (mod 6).

The “only if” part has been proved in Lemma 2.8. In direction of the “if” part, we are able to offer the following partial result.

Theorem 2.11. Conjecture 2.10 is true if A either contains 1 and n with 2 ≤ n ≤ 7, or if it contains 2 and 3.

Proof. For simplicity, let us denote by p_{k} and e_{k} the corresponding symmetric polynomials in 3 variables. In the following considerations we will use the basic fact that e_{1}, e_{2}, e_{3} are algebraically independent generators for the algebra of symmetric polynomials in x_{1}, x_{2}, x_{3}.

Assume first that A = {1, n, m} with 2 ≤ n ≤ 7, m ≠ n, and 6 | nm. Formulas expressing the power sum p_{h} in terms of the elementary symmetric polynomials are well-known, see [M, Ex. 20, p. 33]. We will use the fact that every monomial e_{1}^{β_{1}}e_{2}^{β_{2}}e_{3}^{β_{3}} with β_{1} + 2β_{2} + 3β_{3} = h appears in the expression of p_{h} with a non-zero coefficient. The cases n = 2, 3, 4, 5, 7 are easy since in those cases p_{n} is a monomial in e_{2} and e_{3} mod (e_{1}). For instance, p_{4} = ue_{2}^{2} mod (e_{1}) and p_{5} = ue_{2}e_{3} mod (e_{1}), where u stands for a non-zero integer. This is enough to show that

\sqrt{(p_{1}, p_{n})} = \begin{cases}\sqrt{(e_{1}, e_{2})} & if n = 2, 4, \\ \sqrt{(e_{1}, e_{3})} & if n = 3, \\ \sqrt{(e_{1}, e_{2}e_{3})} & if n = 5, 7, \end{cases}

where \sqrt{I} denotes the radical of the ideal I. In the cases n = 2 or n = 4, we have m = 3v , hence p_{m} = ue_{3}^{v} mod (e_{1}, e_{2}) for some non-zero integer u. This implies that \sqrt{(p_{1}, p_{n}, p_{m})} = \sqrt{(e_{1}, e_{2}, e_{3})}. One concludes in a similar manner in the cases n = 3 and n = 5, 7.

The proof for n = 6 is more complicated since p_{6} is not a monomial mod (e_{1}). Indeed, p_{6} = −2e_{2}^{3} + 3e_{3}^{2} mod (e_{1}). However, reducing p_{m} mod (e_{1}) and p_{6}, that is, replacing e_{1} with 0 and e_{3}^{2} with (2/3)e_{2}^{3} we obtain

p_{m} = \begin{cases}a_{m}e_{2}^{h} mod (p_{1}, p_{6}) & if m = 2h, \\ a_{m}e_{2}^{h−1}e_{3} mod (p_{1}, p_{6}) & if m = 2h + 1, \end{cases}

where a_{m} is an integer. The assertion that we have to prove is equivalent to the non-vanishing of coefficient a_{m}. The integer a_{m} can be computed using the formula expressing p_{m} in terms of the e_{i}’s, see [M, Ex. 20, p. 33]. Explicitly, we have

a_{m} = \begin{cases}m ∑_{b=0}^{⌊h/3⌋} \frac{(−1)^{h−b}}{h − b} \binom{h − b}{2b}(2/3)^{b} & if m = 2h, \\ −m ∑_{b=0}^{⌊h/3⌋} \frac{(−1)^{h−b}}{h − b} \binom{h − b}{2b + 1}(2/3)^{b} & if m = 2h + 1. \end{cases}

To show that a_{m} ≠ 0 for m ≠ 1 and 6, we consider

f_{m}(x) = \begin{cases}m ∑_{b=0}^{⌊h/3⌋} \frac{(−1)^{h−b}}{h − b} \binom{h − b}{2b}x^{b} & if m = 2h, \\ −m ∑_{b=0}^{⌊h/3⌋} \frac{(−1)^{h−b}}{h − b} \binom{h − b}{2b + 1}x^{b} & if m = 2h + 1. \end{cases}

Since

\frac{2h}{h − b}\binom{h − b}{2b} = 2\binom{h − b}{2b} − \binom{h − b − 1}{2b − 1}

and

\frac{2h + 1}{h − b}\binom{h − b}{2b + 1} = 2\binom{h − b}{2b + 1} − \binom{h − b − 1}{2b},

the polynomials f_{m}(x) are in ℤ[x]. We have to show that 2/3 is not a root of f_{m}(x) for m ≠ 1, 6. If m < 8, then f_{m}(x) is a non-zero constant. So we may assume m ≥ 8. If m is odd, then the coefficient of the term of degree 0 in f_{m}(x) is odd. If m is even and m ≢ 0 (mod 6) and m ≢ 10 (mod 18) then the leading coefficient of of f_{m}(x) is not divisible by 3. This is enough to conclude that 2/3 is not a root of f_{m}(x) in this case. If m ≡ 0 (mod 6) or m ≡ 10 (mod 18) one needs a more sophisticated analysis of the 3-adic valuation of the other coefficients of f_{m}(x). The argument is given in the appendix.

Finally, assume that A contains 2 and 3, say A = {2, 3, d} for some d > 3. Since p_{1}, p_{2}, p_{3} generates the algebra of symmetric polynomials in x_{1}, x_{2}, x_{3}, we have p_{d} = c_{d}p_{1}^{d} mod (p_{2}, p_{3}) for a uniquely determined rational number c_{d}. The statement we have to prove is equivalent to the non-vanishing of c_{d}. Since e_{2} = \frac{1}{2}p_{1}^{2} and e_{3} = \frac{1}{6}p_{1}^{3} mod (p_{2}, p_{3}), from Newton’s equation p_{d} = e_{1}p_{d−1} − e_{2}p_{d−2} + e_{3}p_{d−3} we obtain that

c_{d} = c_{d−1} − \frac{1}{2}c_{d−2} + \frac{1}{6}c_{d−3} for d > 3, with c_{1} = 1, c_{2} = 0, c_{3} = 0.

Solving the linear recurrence, see [S, 4.1.1], we get that

c_{d} = α^{d} + β^{d} + β̄^{d},

where α ∈ ℝ, 0 < α < 1, and β, β̄ ∈ ℂ are the roots of the polynomial

x^{3} − x^{2} + \frac{1}{2}x − \frac{1}{6} = 0.

We show that c_{d} > 0 for d > 3. To this end, it is enough to show that α^{d} > 2|β|^{d} for d > 3, equivalently, (α/|β|)^{d} > 2. Hence it is enough to prove the statement for d = 4. With the help of a computer algebra program, we find that (α/|β|)^{4} = 2.17 . . .. □

In order to generalize part of Theorem 2.11 we need the following lemma.

Lemma 2.12. Let A = {a, b, c}, where gcd(A) = 1 and abc ≡ 0 (mod 3). Then the zero-set of the polynomial system associated to p_{A}(3) intersects {(z_{1}, z_{2}, z_{3}) ∈ ℂ^{3} : |z_{1}| = |z_{2}| = |z_{3}|} only in (0, 0, 0).

Proof. By contradiction, assume (z_{1}, z_{2}, z_{3}) ∈ ℂ^{3} is a solution of the polynomial system associated to p_{A}(3) and |z_{1}| = |z_{2}| = |z_{3}| ≠ 0. Dividing by z_{3}, we may assume that z_{3} = 1 and |z_{1}| = |z_{2}| = 1. Note that the only complex numbers w_{1}, w_{2} satisfying |w_{1}| = |w_{2}| = 1 and w_{1} + w_{2} + 1 = 0 are the two primitive third roots of unity. Hence z_{1}^{a}, z_{1}^{b} and z_{1}^{c} are primitive third roots of 1. Since gcd(a, b, c) = 1, there exist integers α, β, γ such that aα + bβ + cγ = 1. It follows that z_{1} = z_{1}^{aα+bβ+cγ} = (z_{1}^{a})^{α}(z_{1}^{b})^{β}(z_{1}^{c})^{γ} and hence z_{1} itself is a third root of 1. But one among a, b, c, say a, is divisible by 3. Hence z_{1}^{a} = 1 which is a contradiction. □

We can now state, and prove, the announced partial generalization of Theorem 2.11.

Proposition 2.13. Conjecture 2.10 holds if A contains a and at with t ∈ {2, 3, 4, 5, 7}.

Proof. Let ρ be a primitive third root of 1. We claim that for t ∈ {2, 3, 4, 5, 7} the zero-set of p_{1}(3) and p_{t}(3) consists (up to multiples and permutations) of at most (1, ρ, ρ^{2}) (if t ≢ 0 mod 3)) and (1, −1, 0) (if t is odd). The assertion follows from the fact that, for these values of t, p_{t}(3) is a monomial in e_{2}(3) and e_{3}(3) mod (e_{1}(3)). From Lemma 2.5 it follows that every non-zero point z = (z_{1}, z_{2}, z_{3}) ∈ ℂ^{3} in the zero set of p_{A}(3) satisfies z_{1}z_{2}z_{3} ≠ 0. Hence, according to the claim above, (z_{1}^{a}, z_{2}^{a}, z_{3}^{a}) equals (1, ρ, ρ^{2}) (up to multiples and permutations). Hence |z_{1}| = |z_{2}| = |z_{3}| and this contradicts Lemma 2.12. □

Remark 2.14. For b ∈ ℕ, b > 1, set f_{b}(x) = 1 + x^{b} + (−1)^{b}(x + 1)^{b}. Note that f_{b}(x) = p_{b}(x, 1, −1 − x). Let A = {1, a, b} with ab ≡ 0 (mod 6). Conjecture 2.10 for A is equivalent to gcd(f_{a}(x), f_{b}(x)) = 1. We expect f_{b}(x) to be irreducible in ℚ[x] up to the factor x and cyclotomic factors 1 + x or 1 + x + x^{2} which are present or not depending on b mod 6. In particular, we expect f_{b}(x) to be irreducible ℚ[x] if b ≡ 0 (mod 6). A simple computation shows that Eisenstein’s criterion applies to f_{b}(x + 1) with respect to p = 3 for b of the form 3^{u}(3^{v} + 1) with u > 0 and v ≥ 0. These considerations imply that, if b is of that form, then f_{b} is irreducible (over ℚ). Hence, Conjecture 2.10 holds for A = {1, a, b} where a < b and b = 3^{u}(3^{v} + 1) with u > 0 and v ≥ 0.

For n > 3 the condition ( ∏_{i∈A} i) ≡ 0 (mod n!) does not imply that p_{A}(n) is a regular sequence. For n = 4, computer experiments suggest the following conjecture.

Conjecture 2.15. Let A ⊂ ℕ^{∗} with |A| = 4, say A = {a_{1}, a_{2}, a_{3}, a_{4}}, and assume gcd(A) = 1. Then p_{A}(4) is a regular sequence if and only if A satisfies the following conditions:

(1) At least two of the a_{i}’s are even, at least one is a multiple of 3, and at least one is a multiple of 4.

(2) If E is the set of the even elements in A and d = gcd(E) then the set {a/d : a ∈ E} contains an even number.

(3) A does not contain a subset of the form {d, 2d, 5d}.

Remark 2.16. (a) Condition (1) is obviously stronger than 4! divides a_{1}a_{2}a_{3}a_{4}. For instance, the set {1, 3, 5, 8} does not satisfy (1) and the product of its elements is divisible by 4!.

(b) The conditions (2) and (3) are independent. For example, the set {1, 3, 4, 12} satisfies (1) but not (2) and {1, 2, 5, 12} satisfies (1) and (2) but not (3).

(c) We can prove that conditions (1), (2) and (3) are necessary. Indeed, assume p_{A}(4) is a regular sequence. Then (1) holds by Lemma 2.6. To get (2), consider the point P = (x, −x, y, −y) ∈ ℂ^{4}. Obviously P is a solution of the equation p_{k}(4) = 0 with k odd. If k is even, then p_{k}(4) evaluated at P is 2p_{k}(2) evaluated at (x, y). Hence we see that the only common root of p_{k}(2) with k ∈ E is (0, 0). So, by Lemma 2.4, at least one element of {a/d : a ∈ E} is even. To show that (3) is necessary, note that, by degree reasons, p_{5}(4) is in the ideal (p_{1}(4), p_{2}(4)). Replacing x_{i} by x_{i}^{d}, we see that p_{5d}(4) is in the ideal (p_{d}(4), p_{2d}(4)). So (3) follows.

For complete symmetric polynomials in three variables we formulate the following conjecture.

Conjecture 2.17. Let A = {a, b, c} with a < b < c. Then h_{A}(3) is a regular sequence if and only if the following conditions are satisfied:

(1) abc ≡ 0 (mod 6).

(2) gcd(a + 1, b + 1, c + 1) = 1.

(3) For all t ∈ ℕ with t > 2 there exist d ∈ A such that d + 2 ≢ 0, 1 (mod t).

Again, the “only if” part follows from the considerations above. For instance, if (2) does not hold then the three polynomials have a common non-zero solution of the form (x, 1, 0).

Appendix: Non-vanishing of the coefficient a_{m}

We want to prove that

∑_{b=0}^{⌊h/3⌋} \frac{(−1)^{h−b}}{h − b} \binom{h − b}{2b} (\frac{2}{3})^{b}  (A.1)

is non-zero except for h = 3.

We may assume from now on that h > 3. The idea is a 3-adic analysis of the summands. All of them are rational numbers. If h ≢ 3, 5 (mod 9), then we shall show that the 3-adic valuation of the last summand (the summand for b = ⌊h/3⌋ is smaller than the 3-adic valuations of all other summands. In this situation, the sum cannot be zero. Similarly, if h ≡ 3, 5 (mod 9), we shall show that the 3-adic valuations of the summands for b ≤ ⌊\frac{h}{3}⌋ − 2 are all larger than the 3-adic valuation of the sum of the two summands for b = ⌊\frac{h}{3}⌋ and b = ⌊\frac{h}{3}⌋ − 1. Here, summands with b ≤ ⌊\frac{h}{3}⌋ − 2 do indeed exist (since we assumed that h > 3 which, together with h ≡ 3, 5 (mod 9), implies that h must be at least 3 + 9 = 12), and therefore the sum cannot be zero.

We write v_{3}(\frac{r}{s}) for the 3-adic valuation of the rational number \frac{r}{s} which, by definition, is 3^{a−b}, where 3^{a} is the largest power of 3 dividing r, and where 3^{b} is the largest power of 3 dividing s.

We shall use the following (well-known and easy to prove) fact: the 3-adic valuation of a binomial coefficient \binom{m}{n}, v_{3} (\binom{m+n}{n}), is equal to the number of carries which occur during the addition of the numbers m and n in ternary notation. From now on, whenever we speak of “carries during addition of two numbers”, we always mean the addition of these two numbers when written in ternary notation.

Case 1: h ≡ 0 (mod 3). Let h = 3k, k > 1. In (A.1) replace b by k − b to obtain the sum

∑_{b=0}^{k} \frac{(−1)^{b}}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b} .  (A.2)

Let first k ≢ 1 (mod 3) (i.e., h ≢ 3 (mod 9)).

For the summand in (A.2) for b = 0, we have

v_{3} (\frac{1}{2k} (\frac{2}{3})^{k}) ≤ −k.

We claim that we have

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) > −k  (A.3)

for all b ≥ 1.

To prove the claim we assume that v_{3}(2k + b) = e and that 3^{s−1} ≤ 3b < 3^{s}. Clearly e ≥ 0 and s ≥ 2. Using these conventions, we have

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b})

= b − k − e + #(carries during addition of 3b and 2k − 2b). (A.4)

Since b ≥ 1, the inequality (A.3) holds for e = 0. From now on we assume that e > 0.

If s > e > 0, then from (A.4) we get

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) ≥ 3^{s−2} − k − e
≥ −k + 3^{e−1} − e
≥ −k.

This proves (A.3) for e > 1, since in this case we have actually 3^{e−1} > e, and thus > −k in the last line of the inequality chain. It also proves (A.3) for e = 1 and s > 2, since in this case we have 3^{s−2} > 3^{e−1}, and thus we have > −k + 3^{e−1} − e in the inequality chain. The only case which is left open is e = 1 and s = 2.

Let now s ≤ e. Let us visualize the numbers 2k + b and 3b in ternary notation,

(2k + b)_{3} = … \underbrace{0 . . . . . . 0}_{e},
(3b)_{3} =  \underbrace{… 0}_{s} .

When we add 2k − 2b to 3b, then we get 2k + b. Since (2k + b)_{3} has 0’s as s-th, (s + 1)-st, …, e-th digit (from the right), we must have at least e − s + 1 carries when adding 2k − 2b and 3b. From (A.4) we then have

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) ≥ 3^{s−2} − k − e + (e − s + 1)
≥ −k + 3^{s−2} − s + 1
≥ −k.

The last inequality is in fact strict if s > 2, proving the claim (A.3) in this case.

In summary, except for s = 2 and e > 0, the claim (A.3) is proved. However, if s = 2, then b = 1 or b = 2. If b = 2, then we have b > 3^{s−2}. If we use this in the first line of the inequality chains, then both of them become strict inequalities. Therefore the only case left is b = 1 and e > 0. However, if b = 1, then e = v_{3}(2k + b) = v_{3}(2k + 1) = 0 since we assumed that k ≢ 1 (mod 3). This is absurd, and hence the claim is established completely.

Now we address the (more complicated) case that k ≡ 1 (mod 3) (i.e., h ≡ 3 (mod 9)). In this case we combine the summands in (A.2) for b = 0 and b = 1:

\frac{1}{2k} (\frac{2}{3})^{k} − \frac{1}{2k + 1}\binom{2k + 1}{3} (\frac{2}{3})^{k−1} = −\frac{2^{k−1}}{3^{k}} \frac{(k − 1)(2k^{2} + k + 1)}{k}.

The 3-adic valuation of this expression is

v_{3}(k − 1) − k.

Let us write f = v_{3}(k − 1), and, as before, v_{3}(2k + b) = e and 3^{s−1} ≤ 3b < 3^{s}. Since k ≡ 1 (mod), we know that f ≥ 1. We claim that for b ≥ 2 we have

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) > −k + f.  (A.5)

Since the notation is as before, we may again use Equation (A.4) for the computation of the 3-adic valuation on the left-hand side.

For convenience, we visualize the involved numbers,

(2k)_{3} = … \underbrace{0 . . . . . . . . . 02}_{f}
(2k + b)_{3} = . . . . . . . . . \underbrace{0 … 00}_{e}
(b)_{3} =  \underbrace{… \underbrace{2 … 21}_{min\{e,f\}}}_{s−1}

(3b)_{3} =  \underbrace{… \underbrace{22..210}_{min\{e,f\}+1}}_{s}

For later use, we note that we must have

b ≥ 3^{min{e,f}} − 2,  (A.6)

and if s − 1 > e even

b ≥ 3^{s−2} + 3^{min{e,f}} − 2.  (A.7)

Another general observation is that, if e > 0, then the number of carries when adding 3b and 2k − 2b must be at least

e + χ(f ≥ e) · max{0, f − s + χ(s ≠ e + 1)},  (A.8)

because there must be carries when adding up the second, third,…, (e + 1)-st digits (from the right), and because (2k + b)_{3} has 0’s as s-th, (s + 1)-st, …, f-th digit. Here, χ(A) = 1 if A is true and χ(A) = 0 otherwise. The truth value χ(s ≠ e + 1) occurs because if s = e + 1 we cannot count the carry at the (s = e + 1)-st digit twice.

(a) e ≥ f ≥ 1. From (A.4), (A.6), and (A.8), we get

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) ≥ (3^{f} − 2) − k − e + e
≥ −k + f.

Moreover, if f > 1 then the last inequality is strict, proving the claim (A.5) in this case. The only case which remains open is f = 1. If s > 2, we could use (A.7) instead of (A.6), in which case the claim would also follow. Thus, we are left with considering the case s = 2 and f = 1. In that case, we would have b = 1, a contradiction.

(b) f ≥ s = e + 1 ≥ 2. From (A.4), (A.6), and (A.8), we get

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) ≥ 3^{e} − 2 − k − e + (e + f − s)
≥ −k + f + 3^{s−1} − s − 2
> −k + f,

as long as s > 2. On the other hand, if s = 2 and, hence, e = 1, then we would have b = 1, a contradiction.

(c) f ≥ s > e + 1 ≥ 2. From (A.4), (A.7), and (A.8), we get

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) ≥ 3^{s−2} + 3^{e} − 2 − k − e + (e + f − s + 1)
≥ 3^{s−2} + 3^{e} − 1 − k + f − s
> (s − 1) + 1 − k + f − s
> −k + f,

since s > 2 and e ≥ 1.

(d) s > f > e ≥ 1. From (A.4), (A.7), and (A.8), we get

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) ≥ 3^{s−2} + 3^{e} − 2 − k − e + e
≥ 3^{f−1} + 3^{e} − 2 − k
> −k + f,

since f > 1 and e ≥ 1.

(e) f > e = 0. Our previous visualization of the involved numbers then collapses to

(2k)_{3} = … \underbrace{0 . . . . . . . . . 02}_{f}
(2k + b)_{3} = . . . . . . . . . . . . . . . .
(b)_{3} =  \underbrace{. . . . . . . . .}_{s−1}
(3b)_{3} =  \underbrace{. . . . . . . . . 0}_{s}

Instead of (A.6) or (A.7), we have now b ≥ 3^{s−2}. The number of carries when adding 3b and 2k − 2b must still be at least max{0, f − s + 1}. Hence, from (A.4) we get

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) = 3^{s−2} − k + max\{0, f − s + 1\}.  (A.9)

If f < s, then we obtain from (A.9) that

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) ≥ 3^{f−1} − k
≥ −k + f,

since f ≥ 1. This inequality chain is in fact strict as long as s > f + 1 or f > 1. Thus, the only case which is open is f = 1 and s = 2. In that case, we must necessarily have b = 2. So, instead of b ≥ 3^{s−2} to obtain (A.9), we could have used b > 1 = 3^{s−2}, again leading to a strict inequality.

If f ≥ s, then we obtain from (A.9) that

v_{3} (\frac{1}{2k + b}\binom{2k + b}{3b} (\frac{2}{3})^{k−b}) ≥ 3^{s−2} − k + f − s + 1
≥ −k + f,

since s ≥ 2. This inequality chain is in fact strict whenever s ≥ 3. If s = 2 then the same argument as in the previous paragraph leads also to a strict inequality.

This completes the verification of the claim (A.5).

Case 2: h ≡ 1 (mod 3). Let h = 3k + 1, k ≥ 1. In (A.1) replace b by k − b to obtain the sum

∑_{b=0}^{k} \frac{(−1)^{b+1}}{2k + b + 1}\binom{2k + b + 1}{3b + 1} (\frac{2}{3})^{k−b} .  (A.10)

For the summand in (A.10) for b = 0, we have

v_{3} (−\frac{1}{2k + 1}(2k + 1) (\frac{2}{3})^{k}) = −k.

We claim that we have

v_{3} (\frac{1}{2k + b + 1}\binom{2k + b + 1}{3b + 1} (\frac{2}{3})^{k−b}) > −k

for all b ≥ 1. To see this we argue as in Case 1. Everything works in complete analogy. However, what makes things less complicated here is the fact that the first digit (from the right) of 3b + 1 is a 1. Therefore there is an additional carry when adding 3b+1 and 2k −2b (in comparison to the addition of 3b and 2k −2b in Case 1; namely when adding the first digits), and this implies that the complications that we had in Case 1 when k ≡ 1 (mod 3) do not arise here.

Case 3: h ≡ 2 (mod 3). Let h = 3k + 2, k ≥ 1. In (A.1) replace b by k − b to obtain the sum

∑_{b=0}^{k} \frac{(−1)^{b}}{2k + b + 2}\binom{2k + b + 2}{3b + 2} (\frac{2}{3})^{k−b} .  (A.11)

For the summand in (A.11) for b = 0, we have

v_{3} (\frac{1}{2k + 2} \frac{(2k + 2)(2k + 1)}{2} (\frac{2}{3})^{k}) = −k + v_{3}(2k + 1).

Let us write f = v_{3}(2k + 1). Thus,

v_{3} (\frac{1}{2k + 2} \frac{(2k + 2)(2k + 1)}{2} (\frac{2}{3})^{k}) = −k + f.  (A.12)

Similarly to earlier, we assume that e = v_{3}(2k + b + 2) and that 3^{s−1} ≤ 3b + 3 < 3^{s}. Then the analogue of (A.4) is

v_{3} (\frac{1}{2k + b + 2}\binom{2k + b + 2}{3b + 2} (\frac{2}{3})^{k−b})

= b − k − e + #(carries during addition of 3b + 2 and 2k − 2b). (A.13)

For convenience, we visualize the involved numbers,

(2k + 1)_{3} = … \underbrace{0 . . . . . . . . . 00}_{f}
(2k + b + 2)_{3} = . . . . . . . . . \underbrace{0 … 00}_{e}
(b + 1)_{3} =  \underbrace{… \underbrace{0 … 00}_{min\{e,f\}}}_{s−1}

(3b + 2)_{3} =  \underbrace{… \underbrace{22..222}_{min\{e,f\}+1}}_{s}

The visualization of (3b + 2)_{3} has to be taken with a grain of salt because, if b = 3^{s−2} − 1, then (3b + 2)_{3} has only s − 1 digits.

For later use, we note that we must have s − 1 > min{e, f} and

b ≥ 3^{s−2} − 1 + χ(e = 0).  (A.14)

Moreover, if s = 2, then b + 1 has just one digit and therefore necessarily b = 1. Another general observation is that the number of carries when adding 3b and 2k−2b must be at least

e + χ(f ≥ e) · (χ(e > 0) + max{0, f − s + 1}),  (A.15)

because there must be carries when adding up the first, second, …, e-th digits (from the right), because if f ≥ e > 0 there must also occur a carry when adding up the (e + 1)-st digits, and because (2k + b + 2)_{3} has 0’s as s-th, (s + 1)-st, …, f-th digit.

If we use (A.14) and (A.15) in (A.13), then we obtain the inequality

v_{3} (\frac{1}{2k + b + 2}\binom{2k + b + 2}{3b + 2} (\frac{2}{3})^{k−b})

≥ 3^{s−2} − 1 + χ(e = 0) − k − e + (e + χ(f ≥ e) · (χ(e > 0) + max{0, f − s + 1}))

≥ −k + 3^{s−2} − 1 + χ(f ≥ e) + χ(f ≥ e) · max{0, f − s + 1}.  (A.16)

Let first k ≢ 1 (mod 3) (i.e., h ≢ 5 (mod 9)) or, equivalently, f = 0. If we do the according simplifications in (A.16), then we arrive at

v_{3} (\frac{1}{2k + b + 2}\binom{2k + b + 2}{3b + 2} (\frac{2}{3})^{k−b}) ≥ −k + 3^{s−2} − 1 ≥ −k.

If s > 3 the last inequality is in fact a strict inequality. If s = 2 then necessarily b = 1. In that case we could have used b > 0 = 3^{s−2} − 1 instead of (A.14), which would also lead to a strict inequality. If we compare this with (A.12), then it follows that our sum cannot vanish.

Now let k ≡ 1 (mod 3) (i.e., h ≡ 5 (mod 9)). Equivalently, f ≥ 1. We combine the summands in (A.11) for b = 0 and b = 1:

\frac{1}{2k + 2}\binom{2k + 2}{2} (\frac{2}{3})^{k} − \frac{1}{2k + 3}\binom{2k + 3}{5} (\frac{2}{3})^{k−1}

= −\frac{2^{k−2}}{5 · 3^{k}} (2k + 1)(2k^{3} + k^{2} − k − 10).

The 3-adic valuation of this expression is

v_{3}(2k + 1) − k = −k + f.

We claim that we have

v_{3} (\frac{1}{2k + b + 2}\binom{2k + b + 2}{3b + 2} (\frac{2}{3})^{k−b}) > −k + f

for all b ≥ 2. It should be noted that b ≥ 2 implies s ≥ 3.

(a) f ≥ s ≥ 3. Since s − 1 > min{e, f}, we must have f > e. Thus, from (A.16), we get

v_{3} (\frac{1}{2k + b + 2}\binom{2k + b + 2}{3b + 2} (\frac{2}{3})^{k−b}) ≥ −k + 3^{s−2} + (f − s + 1)
> −k + f,

since s ≥ 3.

(b) s > f ≥ 0. From (A.16), we get

v_{3} (\frac{1}{2k + b + 2}\binom{2k + b + 2}{3b + 2} (\frac{2}{3})^{k−b}) ≥ −k + 3^{s−2} − 1

≥ −k + max{1, 3^{f−1}} − 1
≥ −k + f,

as long as f ≥ 1. If f ≥ 3, the inequality chain is in fact strict. If f = 0 or f = 1 then, because of s ≥ 3, we could have used 3^{s−2} > 1 to obtain that the 3-adic valuation in question must be at least −k + f = −k. If f = 2 and s > 3, then we could have used the estimation 3^{s−2} > 3^{f−1} instead. The only remaining case is f = 2 and s = 3. Since we must have s − 1 > min{e, f}, ths only options for e are e = 0 or e = 1. If e = 0 then, using (A.14) and (A.15) in (A.13), we obtain

v_{3} (\frac{1}{2k + b + 2}\binom{2k + b + 2}{3b + 2} (\frac{2}{3})^{k−b}) = 3^{s−2} − k > −k + 2 = −k + f.

Finally, if e = 1, then from the visualization of (3b + 2)_{3} we see that there must be at least 2 carries when adding 3b + 2 and 2k − 2b (namely when adding the first and second digits). If we use this together with (A.14) in (A.13), the we arrive at

v_{3} (\frac{1}{2k + b + 2}\binom{2k + b + 2}{3b + 2} (\frac{2}{3})^{k−b}) = 3^{s−2} − 1 − k − 1 + 2 > −k + 2 = −k + f.

This completes the proof of our claim.

References

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[LP] Lascoux, A., Pragacz, P., Jacobians of symmetric polynomials, Ann. Comb. 6 (2002), no. 2, 169–172.

[M] Macdonald, I. G., Symmetric functions and Hall polynomials, second edition, Oxford Mathematical Monographs. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1995.

[S] Stanley, R. Enumerative combinatorics, Vol. 1, Cambridge Studies in Advanced Mathematics,

49. Cambridge University Press, Cambridge, 1997.

Dipartimento di Matematica, Università di Genova, Via Dodecaneso, 35, I-16146

Genova, Italy. WWW: http://www.dima.unige.it/~conca/

Fakultät für Mathematik, Universität Wien, Nordbergstraße 15, A-1090 Vienna,

Austria. WWW: http://www.mat.univie.ac.at/~kratt.

Department of Mathematics, Tokai University, Hiratsuka 259-1292, Japan