Regular Sequences of Symmetric Polynomials

Regular Sequences of Symmetric Polynomials

ALDOCONCA(*) - CHRISTIANKRATTENTHALER(**)^y- JUNZOWATANABE(***)

ABSTRACT- A set ofnhomogeneous polynomials innvariables is a regular sequence if the associated polynomial system has only the obvious solution (0;0;. . .;0).

Denote by pk(n) the power sum symmetric polynomial in n variables

x^k₁‡x^k₂‡. . .‡x_n^k. The interpretation of theq-analogue of the binomial coeffi-

cient as Hilbert function leads us to discover thatnconsecutive power sums inn variables form a regular sequence. We consider then the following problem:

describe the subsetsANof cardinalitynsuch that the set of polynomials pa(n) witha2Ais a regular sequence. We prove that a necessary condition is thatn!divides the product of the elements ofA. To find an easily verifiable sufficient condition turns out to be surprisingly difficult already for nˆ3.

Given positive integers a5b5c with gcd(a;b;c)ˆ1, we conjecture that pa(3);pb(3);pc(3) is a regular sequence if and only if abc0(mod6). We provide evidence for the conjecture by proving it in several special instances.

1. Introduction.

For d;n2N let Pd;n(q) be theq-analogue of the binomial coefficient (also called Gaussian polynomial)

P_d;n(q):ˆ n‡d n

qˆ [n‡d]_q! [n]_q![d]_q! (1:1)

(*) Indirizzo dell'A.: Dipartimento di Matematica, UniversitaÁ di Genova, Via Dodecaneso, 35, I-16146 Genova, Italy.

WWW: http://www.dima.unige.it/Äconca/

(**) Indirizzo dell'A.: FakultaÈt fuÈr Mathematik, UniversitaÈt Wien, Nordberg- strasse 15, A-1090 Vienna, Austria.

WWW: http://www.mat.univie.ac.at/Äkratt.

(***) Indirizzo dell'A.: Department of Mathematics, Tokai University, Hiratsu- ka 259-1292, Japan

yResearch 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''.

2000 Mathematics Subject Classification. Primary 05E05; Secondary 13P10 11C08.

where

[j]_q!ˆY^j

iˆ1

1 qⁱ 1 q ˆY^j

iˆ1

(1‡q‡. . .‡q^{i 1}):

It is well known thatPd;n(q)2Z[q]. Furthermore for a finite fieldF,Pd;n(q) evaluated at qˆ jFjgives exactly the number of F-subspaces of F^n‡d of dimensionn, [S, 1.3.18]. It is also known that the coefficient ofq^kinP_d;n(q) is the number of partitions ofkwith at mostnparts and parts bounded by d, [S, 1.3.19]. This interpretation leads to the following construction.

Consider the polynomial ringRˆC[x1;x2;. . .;xn] with the usual action of the permutation group GˆSn. The monomial complete intersection Iˆ(x^d‡1₁ ;x^d‡1₂ ;. . .;x^d‡1_n ) is fixed byG, so thatGacts as well on the quotient ringAˆR=I. The homogeneous partA_k of degreekof Ahas aC-basis consisting of the monomials inx^a₁¹x^a₂² x^a_nⁿwitha₁‡a₂‡. . .‡a_n ˆkand aid for all i. For a partition l:l1l2. . .ln denote byml the monomial symmetric polynomial

m_lˆx^l₁¹x^l₂² x^l_nⁿ‡symmetric terms;

where ``symmetric terms'' means the sum of all terms that have to be added to complete the monomial x^l₁¹x^l₂² x^l_nⁿ to a symmetric polynomial in

x1;x2;. . .;xn. The invariant ringA^Gin degreekhas aC-basis consisting of

the elementsm_lwherelis a partition ofkin at mostnparts with parts smaller thand‡1. In other words,P_d;n(q) is the Hilbert series ofA^G. In characteristic 0 the extraction of the invariant submodule is an exact functor. So we haveA^GˆR^G=I^G, whereR^GˆC[e1;e2;. . .;en] andeiis the elementary symmetric polynomial of degree i. Since the e_i's are algeb- raically independent, the Hilbert series ofR^Gis 1 Qⁿ

iˆ1(1 qⁱ). Since we know thatP_d;n(q) is the Hilbert series ofA^G, we see thatA^Ghas 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^GˆI\R^Gmight be generated by

p_d‡1;p_d‡2;. . .;p_d‡n

wherep_iˆxⁱ₁‡xⁱ₂‡ ‡xⁱ_n is the power sum of degreeiand that they form a regular sequence. Since the inclusion (p_d‡1;p_d‡2;. . .;p_d‡n)I^Gis obvious, to prove the equality it is enough to prove thatpd‡1;pd‡2;. . .;pd‡n

form a regular sequence inR^Gor, which is the same, inR. W e 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 byhithe complete symmetric poly- nomial of degree i, that is, the sum of all the monomials of degree i in

x₁;. . .;xn. More generally, we are led to consider regular sequences of

symmetric polynomials. In particular regular sequences of power sumsp_i and regular sequences of complete symmetric polynomials h_i. Even for nˆ3 these questions turn out to be difficult. For indices a;b;c with gcd(a;b;c)ˆ1 we conjecture thatpa;pb andpc form a regular sequence exactly when 6jabc. We are able to verify this conjecture in a few cases in which the property under investigation is translated into the non-vanish- ing of a rational number which appears as a coefficient in the relevant expressions or on the irreducibility over the rationals of certain poly- nomials 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 Dvor- nicich 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_nˆC[x₁;x₂;. . .;x_n]. Following Macdonald, we denote by e_k(n),

pk(n) andhk(n) the elementary symmetric polynomial, the power sum and the complete symmetric polynomial of degree k in n-variables, respec- tively. Whennis clear from the context or irrelevant we will simply denote them bye_k;p_k andh_k, respectively. For a subsetAN we set

p_A(n)ˆ fp_a(n):a2Ag and h_A(n)ˆ fh_a(n):a2Ag:

QUESTION 2.1. For which subsets AN with jAj ˆn is the set of polynomials p_A(n)(respectivelyh_A(n)) a regular sequence in R_n? That is, when is (0;. . .;0) the onlycommon zero of the equations pa(n)ˆ0 (respectivelyha(n)ˆ0), a2A?

In the following we list several auxiliary observations.

LEMMA 2.2. Let AN with jAj ˆn. Set dˆgcd(A) and A⁰ˆ fa=d:a2Ag. Then pA(n) is a regular sequence if and onlyif p_A⁰(n) is a regular sequence.

PROOF. Note that if (z₁;z₂;. . .;z_n) is a common zero of the p_a(n)'s witha2Athen (z^d₁;z^d₂;. . .;z^d_n) is a common zero of thepa(n)'s witha2A⁰

and that if (z1;z2;. . .;zn) is a common zero of thepa(n)'s witha2A⁰then (z^1=d₁ ;z^1=d₂ ;. . .;z^1=dn ) is a common zero of thepa(n)'s witha2Awherez^1=d_i

denotes any d-th root ofz_i. p

REMARK2.3. For a setAof cardinalitynthe condition thatpA(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

X

a2A

a

!

n‡1:

This boils down to the evaluation of the rank of af0;1g-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 AN with jAj ˆn. Set dˆgcd(A) and A⁰

ˆ fa=d:a2Ag. Then we have:

(a) The equations p_a(2)ˆ0, a2A, have a non-trivial common zero inC²if and onlyif the elements in A⁰are all odd.

(b) The equations ha(2)ˆ0, a2A, have a non-trivial common zero inC²if and onlyifgcd(fa‡1:a2Ag)ˆ1.

PROOF. (a) As in the proof of the lemma above, the polynomials pa(2) with a2A have a non-trivial common zero if and only if the polynomials pa(2) with a2A⁰ have a non-trivial common zero. So we may assume thatAˆA⁰. If all the elements ofAare 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^aˆ 1 for all a2A. Since gcd(A)ˆ1 we can find a linear combina- tion of the elements in Aˆ fa1;a2;. . .;ang of the type P

jkakˆ1 with jk2Z. Hence we have x^jkakˆ( 1)^jk and thus xˆ( 1)^Sjk. Therefore we have eitherxˆ1, which is impossible, orxˆ 1 which implies that all the elements of A are odd.

(b) Denote by (*) the condition: the equationsh_a(2)ˆ0 witha2Ahave a non-trivial common zero inC². So (*) holds if and only if the equations have a common zero of type (c;1). Obviouslyc6ˆ1, so we can multiply each ha(2) by c 1. We obtain that (*) holds if and only if the equations x^a‡1 1ˆ0 with a2A have a common root 6ˆ1. In other words,

gcd(fx^a‡1 1:a2Ag)6ˆx 1. But gcd(fx^a‡1 1:a2Ag)ˆx^t 1 with tˆgcd(fa‡1:a2Ag), and we obtain the desired characterization. p This answers Question 2.1 for nˆ2. Namely, Lemma 2.4 implies the following fact.

LEMMA2.5. Let a;b2Nn f0gand dˆgcd(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) (ha(2);hb(2)) is a regular sequence if and onlyif gcd(a‡1;b‡1)ˆ1.

Here is another fact.

LEMMA2.6.

(1) If a60 (mod c)and u is a primitive c-th root of unitythen (1;u;u²;. . .;u^{c 1};0;. . .;0)2Cⁿ

is a zero of p_a(n)inCⁿ for all nc.

(2) Let AN with jAj ˆn and 1cn. Set b_c

ˆ jfa2A:a0 (mod c)gj. If b_c5bn=cc then pA(n) is not a regular sequence. Here bxc ˆmaxfm2Z:mxg is the standard floor func- tion.

PROOF. (1) Evaluating pa(n) at (1;u;u²;. . .;u^{c 1};0;. . .;0) yields P^c

iˆ1u^{(i 1)a}. Multiplying the sum byu^a 1, we obtainu^ca 1 which is 0. Since uis a primitivec-th root of unity anda60 (modc), we haveu^a 16ˆ0. It follows thatP^c

iˆ1u^{(i 1)a}ˆ0.

(2) Writenˆqc‡rwith 0r5cso thatqˆ bn=cc. Letube a primitive c-th root of unity. Let (y₁;y₂;. . .;y_q)2C^qand consider the element~y2Cⁿ obtained by concatenatingy_i(1;u;. . .;u^{c 1}) foriˆ1;2;. . .;qand by adding the zero vector of lengthrat the end. By (1) we know that every suchy~is a zero of thepa(n)'s witha60 (modc). If we also impose that~yis a zero of the p_a(n)'s witha2Aanda0 (modc) we obtainb_chomogeneous equations in q variables. Ifb_c5q then there exists a non-zero common solution to that

system of equations. p

We list another auxiliary result.

LEMMA2.7.

(1) Let c>2 and a‡20 or1 (modc). Let u1;u2;u3 be distinct c-th roots of unity. Then pˆ(u1;u2;u3;0;. . .;0)2Cⁿis a zero of ha(n)in Cⁿ for all n3.

(2) Let AN with jAj ˆn and 25c. Assume that for all a2A one has a‡20or1 (modc). Then h_A(n)is not a regular sequence.

PROOF. (1) We may assume nˆ3. To show that ha(3) evaluated at pˆ(u₁;u₂;u₃) is 0 we first multiply h_a(3) with Dˆ(x₁ x₂) (x₁ x₃) (x₂ x₃). A simple calculation shows that

ha(3)Dˆx^a‡2₁ (x2 x3)‡x^a‡2₂ (x3 x1)‡x^a‡2₃ (x1 x2):

Since, by assumption, we haveu^a‡2_i ˆ1 oru^a‡2_i ˆu_ifor alli, this expres- sion vanishes atp.

(2) It follows immediately from (1) that p is a common zero of the equationsha(n). SohA(n) is not a regular sequence. p For the general case ofnhomogeneous symmetric polynomials inRn, there holds the following criterion.

LEMMA 2.8. Let f₁;f₂;. . .;f_n be a regular sequence of homogeneous symmetric polynomials in R_n. Then n!divides(degf₁)(degf₂) (degf_n).

PROOF. For obvious reasons,f₁;f₂;. . .;f_nis also a regular sequence in the ring of symmetric polynomialsC[e1;e2;. . .;en]. Then the Hilbert series

of C[e₁;e₂;. . .;e_n]=(f₁;f₂;. . .;f_n)

isQⁿ

iˆ1(1 q^degfi)=(1 qⁱ), and it must be a polynomial with integral coeffi- cients. If we take the limitq!1, we obtain Qⁿ

iˆ1degf_i

n!, which must be

an integer. p

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

PROPOSITION2.9. Let ANbe 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. Letabe the minimum ofA. We first prove the assertion for the power sums.

Proof (1) for power sums: We argue by induction on a and n. If nˆ1 then the assertion is obvious. If aˆ1, then any common zero of p₁;p₂;. . .;pn is also a common zero ofe₁;e₂;. . .;en becausep₁;p₂;. . .;pn

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 h2N we have New-

ton's identity

Xⁿ

kˆ0

( 1)^ke_{n k}p_k‡hˆ0;

with the convention that p0ˆn and e0ˆ1. For hˆa 1 we have e_np_{a 1}ˆXⁿ

kˆ1

( 1)^k‡1e_{n k}p_{k‡a 1}:

Ifzˆ(z1;z2;. . .;zn) is a common zero ofpa;pa‡1;. . .;pa‡n 1then it is also a zero ofenpa 1. Sozis either a zero ofpa 1, and we conclude by induction ona

that zˆ(0;. . .;0) or zis a zero ofen. In the second case, one of the co-

ordinates ofzis 0 and we conclude by induction onnthatzˆ(0;. . .;0).

Proof (2) for power sums: Let zˆ(z₁;z₂;. . .;z_n)2Cⁿ be a solution of the polynomial system associated to pA(n). We have to prove that zˆ0. Let c be the cardinality of the set fzi:iˆ1;. . .;ng. W e may assume that z1;z2;. . .;zc are the distinct values among z1;z2;. . .;zn, and for iˆ1;. . .;c set m_iˆ jfj:z_jˆz_igj, so that m_i>0. Hence p_k(z₁;. . .;z_n) ˆm₁z^k₁‡. . .‡m_cz^k_c. By assumption, p_a‡j(z₁;. . .;z_n)ˆ0

for jˆ0;. . .;n 1. Let V be the Vandermonde matrix (zⁱ_j) with

iˆ0;. . .;c 1 and jˆ1;. . .;c. By construction, detV 6ˆ0 and the

column vector (m1z^a₁;m2z^a₂;. . .;mcz^a_c) is in the kernel of V. It follows that (m₁z^a₁;m₂z^a₂;. . .;m_cz^a_c)ˆ0, that is, cˆ1 and z₁ˆ0. This shows that zˆ0.

Proof (1) works also for complete symmetric polynomials by re- placing 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 x₁>x₂>. . .>x_n of the ideal gen- erated by h_a(n);h_a‡1(n);. . .;h_{a‡n 1}(n) contains (and hence is equal to)

(x^a₁;x^a‡1₂ ;. . .;x^a‡n_n ¹). p

The above results (and many computer calculations) suggest the fol- lowing conjecture.

CONJECTURE2.10. Let Aˆ fa;b;cg, where a;b;c are positive integers with a5b5c andgcd(A)ˆ1. Then pA(3)is a regular sequence in R3if and onlyif abc0 (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.

THEOREM2.11. Conjecture2.10is true if A either contains1and n with 2n7, or if it contains2and3.

PROOF. For simplicity, let us denote byp_k ande_k the corresponding symmetric polynomials in 3 variables. In the following considerations we will use the basic fact that e1;e2;e3 are algebraically independent gen- erators for the algebra of symmetric polynomials inx₁;x₂;x₃.

Assume first that Aˆ f1;n;mg with 2n7, m6ˆn, and 6jnm.

Formulas expressing the power sump_h in terms of the elementary sym- metric polynomials are well-known, see [M, Ex. 20, p. 33]. We will use the fact that every monomiale^b₁¹e^b₂²e^b₃³ withb₁‡2b₂‡3b₃ˆhappears in the expression ofp_hwith a non-zero coefficient. The casesnˆ2;3;4;5;7 are easy since in those cases p_n is a monomial ine₂ ande₃ mod (e₁). For in- stance, p₄ˆue²₂ mod (e₁) and p₅ˆue₂e₃ mod (e₁), where ustands for a non-zero integer. This is enough to show that



(p₁;p_n)

p ˆ



(e1;e2)

p if nˆ2;4;



(e1;e3)

p if nˆ3;



(e₁;e₂e₃)

p if nˆ5;7;

8>

><

>>

: where 

pI

denotes the radical of the idealI. In the casesnˆ2 ornˆ4, we havemˆ3v, hencep_mˆue^v₃mod (e₁;e₂) for some non-zero integeru. This implies that 

(p1;pn;pm)

p ˆ 

(e1;e2;e3)

p . One concludes in a similar man- ner in the casesnˆ3 andnˆ5;7.

The proof for nˆ6 is more complicated since p6 is not a monomial mod (e₁). Indeed, p₆ˆ 2e³₂‡3e²₃ mod (e₁). However, reducing p_m mod (e₁) andp₆, that is, replacinge₁with 0 ande²₃with (2=3)e³₂ we obtain

p_mˆ ame^h₂ mod (p1;p6) if mˆ2h;

a_me^h₂ ¹e₃ mod (p₁;p₆) if mˆ2h‡1;

(

whereamis an integer. The assertion that we have to prove is equivalent to the non-vanishing of coefficienta_m. The integera_mcan be computed using

the formula expressingpmin terms of theei's, see [M, Ex. 20, p. 33]. Ex- plicitly, we have

am ˆ

m^bh=3cX

bˆ0

( 1)^{h b} h b

h b 2b

(2=3)^b if mˆ2h;

m^bh=3cX

bˆ0

( 1)^{h b} h b

h b 2b‡1

(2=3)^b if mˆ2h‡1:

8>

>>

>>

<

>>

>>

>:

To show thata_m6ˆ0 form6ˆ1 and 6, we consider

f_m(x)ˆ

m^bh=3cX

bˆ0

( 1)^{h b} h b

h b 2b

x^b if mˆ2h;

m^bh=3cX

bˆ0

( 1)^{h b} h b

h b 2b‡1

x^b if mˆ2h‡1:

8>

>>

>>

<

>>

>>

>:

Since

2h h b

h b 2b

ˆ2 h b 2b

h b 1

2b 1

and

2h‡1 h b

h b 2b‡1

ˆ2 h b 2b‡1

h b 1

2b

;

the polynomialsf_m(x) are inZ[x]. We have to show that 2=3 is not a root of f_m(x) form6ˆ1;6. Ifm58, then f_m(x) is a non-zero constant. So we may assumem8. Ifmis odd, then the coefficient of the term of degree 0 in fm(x) is odd. Ifmis even andm60 (mod 6) andm610 (mod 18) then the leading coefficient of offm(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 m0 (mod 6) or m10 (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 Aˆ f2;3;dg for some d>3. Since p₁;p₂;p₃ generates the algebra of symmetric polynomials in x₁;x₂;x₃, we have p_dˆc_dp^d₁ mod (p₂;p₃) for a uniquely determined ra- tional numberc_d. The statement we have to prove is equivalent to the non- vanishing ofc_d. Sincee₂ˆ1

2p²₁ ande₃ˆ1

6p³₁ mod (p₂;p₃), from Newton's

equationpdˆe1pd 1 e2pd 2‡e3pd 3we obtain that c_dˆc_{d 1} 1

2c_{d 2}‡1

6c_{d 3} ford>3; withc₁ˆ1;c₂ˆ0;c₃ˆ0:

Solving the linear recurrence, see [S, 4.1.1], we get that cdˆa^d‡b^d‡b^d;

wherea2R, 05a51, andb;b2Care the roots of the polynomial x³ x²‡1

2x 1 6ˆ0:

We show that c_d>0 for d>3. To this end, it is enough to show that a^d>2jbj^dford>3, equivalently, (a=jbj)^d>2. Hence it is enough to prove the statement fordˆ4. With the help of a computer algebra program, we

find that (a=jbj)⁴ˆ2:17. . .. p

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

LEMMA2.12. Let Aˆ fa;b;cg, wheregcd(A)ˆ1and abc0 (mod 3).

Then the zero-set of the polynomial system associated to p_A(3)intersects f(z₁;z₂;z₃)2C³:jz₁j ˆ jz₂j ˆ jz₃jgonlyin(0;0;0).

PROOF. By contradiction, assume (z₁;z₂;z₃)2C³ is a solution of the polynomial system associated topA(3) andjz1j ˆ jz2j ˆ jz3j 6ˆ0. Dividing by z3, we may assume thatz3ˆ1 andjz1j ˆ jz2j ˆ1. Note that the only com- plex numbersw₁;w₂satisfyingjw₁j ˆ jw₂j ˆ1 andw₁‡w₂‡1ˆ0 are the two primitive third roots of unity. Hencez^a₁;z^b₁ andz^c₁ are primitive third roots of 1. Since gcd(a;b;c)ˆ1, there exist integers a;b;g such that aa‡bb‡cgˆ1. It follows thatz1ˆz^aa‡bb‡cg₁ ˆ(z^a₁)^a(z^b₁)^b(z^c₁)^gand hencez1

itself is a third root of 1. But one amonga;b;c, saya, is divisible by 3. Hence

z^a₁ˆ1 which is a contradiction. p

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

PROPOSITION2.13. Conjecture2.10holds if A contains a and at with t2 f2;3;4;5;7g.

PROOF. Let r be a primitive third root of 1. We claim that for t2 f2;3;4;5;7gthe zero-set ofp1(3) andpt(3) consists (up to multiples and

permutations) of at most (1;r;r²) (ift60 mod 3)) and (1; 1;0) (iftis odd).

The assertion follows from the fact that, for these values of t,pt(3) is a monomial in e₂(3) and e₃(3) mod (e₁(3)). From Lemma 2.5 it follows that every non-zero pointzˆ(z₁;z₂;z₃)2C³ in the zero set ofp_A(3) satisfies z₁z₂z₃6ˆ0. Hence, according to the claim above, (z^a₁;z^a₂;z^a₃) equals (1;r;r²) (up to multiples and permutations). Hencejz1j ˆ jz2j ˆ jz3jand this con-

tradicts Lemma 2.12. p

REMARK 2.14. For b2N, b>1, set f_b(x)ˆ1‡x^b‡( 1)^b(x‡1)^b. Note that f_b(x)ˆp_b(x;1; 1 x). Let Aˆ f1;a;bg with ab0 (mod 6).

Conjecture 2.10 forAis equivalent to gcd(fa(x);fb(x))ˆ1. We expectfb(x) to be irreducible inQ[x] up to the factorxand cyclotomic factors 1‡xor 1‡x‡x² which are present or not depending onbmod 6. In particular, we expectf_b(x) to be irreducible Q[x] ifb0 (mod 6). A simple compu- tation shows that Eisenstein's criterion applies tof_b(x‡1) with respect to pˆ3 forbof the form 3^u(3^v‡1) withu>0 andv0. These considera- tions imply that, ifbis of that form, thenf_bis irreducible (overQ). Hence, Conjecture 2.10 holds forAˆ f1;a;bgwherea5bandbˆ3^u(3^v‡1) with u>0 andv0.

Forn>3 the condition Q

i2Ai

0 (modn!) does not imply thatpA(n) is a regular sequence. For nˆ4, computer experiments suggest the fol- lowing conjecture.

CONJECTURE2.15. Let AN withjAj ˆ4, sayAˆ fa₁;a₂;a₃;a₄g, and assumegcd(A)ˆ1. Then p_A(4)is a regular sequence if and onlyif A satisfies the following conditions:

(1) At least two of the a_i's are even, at least one is a multiple of3, and at least one is a multiple of4.

(2) If E is the set of the even elements in A and dˆgcd(E)then the setfa=d:a2Egcontains an even number.

(3) A does not contain a subset of the formfd;2d;5dg.

REMARK 2.16. (a) Condition (1) is obviously stronger than 4!divides a1a2a3a4. For instance, the set f1;3;5;8g 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 f1;3;4;12gsatisfies (1) but not (2) andf1;2;5;12gsatisfies (1) and (2) but not (3).

(c) We can prove that conditions (1), (2) and (3) are necessary. Indeed, assumepA(4) is a regular sequence. Then (1) holds by Lemma 2.6. To get (2), consider the pointPˆ(x; x;y; y)2C⁴. ObviouslyPis a solution of the equationp_k(4)ˆ0 withkodd. Ifkis even, thenp_k(4) evaluated atPis 2p_k(2) evaluated at (x;y). Hence we see that the only common root ofp_k(2) withk2Eis (0;0). So, by Lemma 2.4, at least one element offa=d:a2Eg is even. To show that (3) is necessary, note that, by degree reasons,p5(4) is in the ideal (p1(4);p2(4)). Replacing x_ibyx^d_i, we see thatp_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 Aˆ fa;b;cg with a5b5c. Then hA(3) is a regular sequence if and onlyif the following conditions are satisfied:

(1) abc0 (mod 6).

(2) gcd(a‡1;b‡1;c‡1)ˆ1.

(3) For all t2Nwith t>2there exist d2A such that d‡260;1 (modt).

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).

A. Appendix: Non-vanishing of the coefficienta_m We want to prove that

X

bh=3c bˆ0

( 1)^{h b} h b

h b 2b

2 3

_b …A:1†

is non-zero except forhˆ3.

We may assume from now on thath>3. The idea is a 3-adic analysis of the summands. All of them are rational numbers. Ifh63;5 (mod 9), then we shall show that the 3-adic valuation of the last summand (the summand forbˆ bh=3cis smaller than the 3-adic valuations of all other summands.

In this situation, the sum cannot be zero. Similarly, ifh3;5 (mod 9), we shall show that the 3-adic valuations of the summands forbjh

3

k 2 are all larger than the 3-adic valuation of the sum of the two summands for

bˆjh 3

kandbˆjh 3

k 1. Here, summands withbjh 3

k 2 do indeed exist (since we assumed thath>3 which, together withh3;5 (mod 9), implies thathmust be at least 3‡9ˆ12), and therefore the sum cannot be zero.

We writev3

r s

for the 3-adic valuation of the rational numberr swhich, by definition, is 3^{a b}, where 3^a is the largest power of 3 dividing r, and where 3^bis the largest power of 3 dividings.

We shall use the following (well-known and easy to prove) fact: the 3- adic valuation of a binomial coefficient m

n ,v₃ 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.

Case1:h0 (mod 3):Lethˆ3k,k>1. In (A.1) replacebbyk bto obtain the sum

X^k

bˆ0

( 1)^b 2k‡b

2k‡b 3b

2 3

_{k b} : …A:2†

Let firstk61 (mod 3) (i.e.,h63 (mod 9)):

For the summand in (A.2) forbˆ0, we have v₃ 1

2k 2

3 _k!

k:

We claim that we have v3 1

2k‡b

2k‡b 3b

2 3

_{k b}!

> k for allb1:

…A:3†

To prove the claim we assume that v3(2k‡b)ˆe and that 3^{s 1}3b53^s:Clearlye0 ands2:Using these conventions, we have …A:4† v₃ 1

2k‡b

2k‡b 3b

2 3

_{k b}!

ˆb k e‡#(carries during addition of 3b and 2k 2b):

Sinceb1, the inequality (A.3) holds foreˆ0. From now on we assume thate>0.

Ifs>e>0, then from (A.4) we get v3 1

2k‡b

2k‡b 3b

2 3

k b!

3^{s 2} k e k‡3^{e 1} e k:

This proves (A.3) fore>1, since in this case we have actually 3^{e 1}>e, and thus> kin the last line of the inequality chain. It also proves (A.3) for eˆ1 ands>2, since in this case we have 3^{s 2}>3^{e 1}, and thus we have

> k‡3^{e 1} ein the inequality chain. The only case which is left open is eˆ1 andsˆ2.

Let nowse. Let us visualize the numbers 2k‡b and 3bin ternary notation,

(2k‡b)₃ˆ. . .0|‚‚‚‚‚{z‚‚‚‚‚}. . . .0

e

; (3b)₃ˆ |‚{z‚}. . .0

s

:

When we add 2k 2bto 3b, then we get 2k‡b. Since (2k‡b)₃has 0's ass- th, (s‡1)-st;. . .;e-th digit (from the right), we must have at leaste s‡1 carries when adding 2k 2band 3b. From (A.4) we then have

v₃ 1 2k‡b

2k‡b 3b

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 sˆ2 and e>0, the claim (A.3) is proved.

However, ifsˆ2, thenbˆ1 orbˆ2. Ifbˆ2, then we haveb>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 isbˆ1 ande>0. However, if bˆ1, then eˆv3(2k‡b)ˆv3(2k‡1)ˆ0 since we assumed that k61 (mod 3). This is absurd, and hence the claim is established com- pletely.

Now we address the (more complicated) case thatk1 (mod 3) (i.e., h3 (mod 9)). In this case we combine the summands in (A.2) forbˆ0

andbˆ1:

1 2k

2 3

k 1

2k‡1

2k‡1 3

2 3

k 1

ˆ 2^{k 1} 3^k

(k 1)(2k²‡k‡1)

k :

The 3-adic valuation of this expression is v3(k 1) k:

Let us write f ˆv₃(k 1), and, as before, v₃(2k‡b)ˆe and 3^{s 1}3b<3^s. Sincek1 (mod 3), we know thatf 1. We claim that for b2 we have

v3 1 2k‡b

2k‡b 3b

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)₃ˆ. . .0|‚‚‚‚‚‚‚‚‚‚{z‚‚‚‚‚‚‚‚‚‚}. . . .02

f

(2k‡b)₃ˆ. . . .0|‚‚‚{z‚‚‚}. . .00

e

(b)₃ˆ . . .2|‚‚‚{z‚‚‚}. . .21

minfe;fg

|‚‚‚‚‚‚{z‚‚‚‚‚‚}

s 1

(3b)₃ˆ . . .22::210|‚‚‚‚{z‚‚‚‚}

minfe;fg‡1

|‚‚‚‚‚‚‚‚{z‚‚‚‚‚‚‚‚}

s

For later use, we note that we must have b3^minfe;fg 2;

…A:6†

and ifs 1>eeven

b3^{s 2}‡3^minfe;fg 2:

…A:7†

Another general observation is that, ife>0, then the number of carries when adding 3band 2k 2bmust be at least

e‡x(f e)maxf0;f s‡x(s6ˆe‡1)g;

…A:8†

because there must be carries when adding up the second, third

;. . .;(e‡1)-st digits (from the right), and because (2k‡b)₃has 0's ass-th,

(s‡1)-st;. . .;f-th digit. Here,x(A)ˆ1 ifAis true andx(A)ˆ0 other- wise. The truth value x(s6ˆe‡1) occurs because if sˆe‡1 we cannot count the carry at the (sˆe‡1)-st digit twice.

(a)ef 1. From (A.4), (A.6), and (A.8), we get v₃ 1

2k‡b

2k‡b 3b

2 3

k b!

(3^f 2) k e‡e k‡f:

Moreover, iff >1 then the last inequality is strict, proving the claim (A.5) in this case. The only case which remains open isf ˆ1. Ifs>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 casesˆ2 andf ˆ1. In that case, we would havebˆ1, a contradiction.

(b)f sˆe‡12. From (A.4), (A.6), and (A.8), we get v₃ 1

2k‡b

2k‡b 3b

2 3

k b!

3^e 2 k e‡(e‡f s) k‡f‡3^{s 1} s 2

> k‡f;

as long ass>2. On the other hand, ifsˆ2 and, hence,eˆ1, then we would havebˆ1, a contradiction.

(c)f s>e‡12. From (A.4), (A.7), and (A.8), we get v₃ 1

2k‡b

2k‡b 3b

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; sinces>2 ande1.

(d)s>f >e1. From (A.4), (A.7), and (A.8), we get v3 1

2k‡b

2k‡b 3b

2

3

k b!

3^{s 2}‡3^e 2 k e‡e 3^{f 1}‡3^e 2 k

> k‡f; sincef >1 ande1.

(e)f >eˆ0. Our previous visualization of the involved numbers then collapses to

(2k)₃ˆ. . .0|‚‚‚‚‚‚‚‚‚‚{z‚‚‚‚‚‚‚‚‚‚}. . . .02

f

(2k‡b)₃ˆ. . . .: (b)₃ˆ . . . .|‚‚‚‚‚{z‚‚‚‚‚}

s 1

(3b)₃ˆ . . . .|‚‚‚‚‚‚‚{z‚‚‚‚‚‚‚}0

s

Instead of (A.6) or (A.7), we have nowb3^{s 2}. The number of carries when adding 3band 2k 2bmust still be at least maxf0;f s‡1g. Hence, from (A.4) we get

v₃ 1 2k‡b

2k‡b 3b

2 3

k b!

ˆ3^{s 2} k‡maxf0;f s‡1g:

…A:9†

Iff <s, then we obtain from (A.9) that v₃ 1

2k‡b

2k‡b 3b

2 3

k b!

3^{f 1} k k‡f;

sincef 1. This inequality chain is in fact strict as long as s>f ‡1 or f >1. Thus, the only case which is open isf ˆ1 andsˆ2. In that case, we must necessarily havebˆ2. So, instead ofb3^{s 2}to obtain (A.9), we could have usedb>1ˆ3^{s 2}, again leading to a strict inequality.

Iff s, then we obtain from (A.9) that v₃ 1

2k‡b

2k‡b 3b

2 3

k b!

3^{s 2} k‡f s‡1 k‡f;

sinces2. This inequality chain is in fact strict whenevers3. Ifsˆ2 then the same argument as in the previous paragraph leads also to a strict inequality.

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

Case2:h1 (mod 3). Lethˆ3k‡1,k1. In (A.1) replacebbyk b

to obtain the sum X^k

bˆ0

( 1)^b‡1 2k‡b‡1

2k‡b‡1 3b‡1

2 3

k b: …A:10†

For the summand in (A.10) forbˆ0, we have

v₃ 1

2k‡1(2k‡1) 2 3

k!

ˆ k:

We claim that we have

v3 1

2k‡b‡1

2k‡b‡1 3b‡1

2 3

k b!

> k

for allb1. To see this we argue as in Case 1. Everything works in com- plete 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 3band 2k 2bin Case 1; namely when adding the first digits), and this implies that the complications that we had in Case 1 when k1 (mod 3) do not arise here.

Case3:h2 (mod 3). Lethˆ3k‡2,k1. In (A.1) replacebbyk b to obtain the sum

X^k

bˆ0

( 1)^b 2k‡b‡2

2k‡b‡2 3b‡2

2 3

k b: …A:11†

For the summand in (A.11) forbˆ0, we have v3 1

2k‡2

(2k‡2)(2k‡1) 2

2 3

k!

ˆ k‡v3(2k‡1):

Let us writef ˆv3(2k‡1). Thus, v3 1

2k‡2

(2k‡2)(2k‡1) 2

2 3

k!

ˆ k‡f: …A:12†

Similarly to earlier, we assume that eˆv₃(2k‡b‡2) and that 3^{s 1}3b‡3<3^s. Then the analogue of (A.4) is

…A:13† v3 1 2k‡b‡2

2k‡b‡2 3b‡2

2 3

k b!

ˆb k e‡#(carries during addition of 3b‡2 and 2k 2b):

For convenience, we visualize the involved numbers, (2k‡1)₃ˆ. . .0|‚‚‚‚‚‚‚‚‚‚{z‚‚‚‚‚‚‚‚‚‚}. . . .00

f

(2k‡b‡2)₃ˆ. . . .0|‚‚‚{z‚‚‚}. . .00

e

(b‡1)₃ˆ . . .0|‚‚‚{z‚‚‚}. . .00

minfe;fg

|‚‚‚‚‚‚{z‚‚‚‚‚‚}

s 1

(3b‡2)₃ˆ . . .22::|‚‚‚‚{z‚‚‚‚}222

minfe;fg‡1

|‚‚‚‚‚‚‚‚{z‚‚‚‚‚‚‚‚}

s

The visualization of (3b‡2)₃has to be taken with a grain of salt because, if bˆ3^{s 2} 1, then (3b‡2)₃has onlys 1 digits.

For later use, we note that we must haves 1>minfe;fgand b3^{s 2} 1‡x(eˆ0):

…A:14†

Moreover, ifsˆ2, thenb‡1 has just one digit and therefore necessarily bˆ1. Another general observation is that the number of carries when adding 3band 2k 2bmust be at least

e‡x(f e)(x(e>0)‡maxf0;f s‡1g);

…A:15†

because there must be carries when adding up the first, second,. . .,e-th digits (from the right), because iff 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

v₃ 1

2k‡b‡2

2k‡b‡2 3b‡2

2 3

k b! …A:16†

3^{s 2} 1‡x(eˆ0) k e‡(e‡x(f e)(x(e>0)‡maxf0;f s‡1g)) k‡3^{s 2} 1‡x(f e)‡x(f e)maxf0;f s‡1g:

Let firstk61 (mod 3) (i.e.,h65 (mod 9)) or, equivalently,f ˆ0. If we do the according simplifications in (A.16), then we arrive at

v3 1

2k‡b‡2

2k‡b‡2 3b‡2

2 3

k b!

k‡3^{s 2} 1 k:

Ifs>3 the last inequality is in fact a strict inequality. Ifsˆ2 then nec- essarilybˆ1. In that case we could have usedb>0ˆ3^{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 k1 (mod 3) (i.e., h5 (mod 9)). Equivalently, f 1. We combine the summands in (A.11) forbˆ0 andbˆ1:

1 2k‡2

2k‡2 2

2 3

k 1

2k‡3

2k‡3 5

2 3

k 1

ˆ 2^{k 2}

53^k(2k‡1)(2k³‡k² k 10):

The 3-adic valuation of this expression is

v3(2k‡1) kˆ k‡f: We claim that we have

v₃ 1

2k‡b‡2

2k‡b‡2 3b‡2

2 3

k b!

> k‡f

for allb2. It should be noted thatb2 impliess3.

(a)f s3. Sinces 1>minfe;fg, we must havef >e. Thus, from (A.16), we get

v₃ 1

2k‡b‡2

2k‡b‡2 3b‡2

2 3

k b!

k‡3^{s 2}‡(f s‡1)

> k‡f; sinces3.

(b)s>f 0. From (A.16), we get

v₃ 1

2k‡b‡2

2k‡b‡2 3b‡2

2 3

k b!

k‡3^{s 2} 1

k‡maxf1;3^{f 1}g 1 k‡f;

as long asf 1. Iff 3, the inequality chain is in fact strict. Iff ˆ0 or f ˆ1 then, because ofs3, we could have used 3^{s 2}>1 to obtain that the 3-adic valuation in question must be at least k‡f ˆ k. If f ˆ2 and s>3, then we could have used the estimation 3^{s 2}>3^{f 1}instead. The only remaining case isf ˆ2 andsˆ3. Since we must haves 1>minfe;fg, ths

only options foreareeˆ0 oreˆ1. Ifeˆ0 then, using (A.14) and (A.15) in (A.13), we obtain

v3 1

2k‡b‡2

2k‡b‡2 3b‡2

2 3

k b!

ˆ3^{s 2} k> k‡2ˆ k‡f: Finally, ifeˆ1, then from the visualization of (3b‡2)₃we 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

v3 1 2k‡b‡2

2k‡b‡2 3b‡2

2 3

k b!

ˆ3^{s 2} 1 k 1‡2> k‡2ˆ k‡f: This completes the proof of our claim.

REFERENCES

[DZ] R. DVORNICICH- U. ZANNIER,Solution of a problem about symmetric functions, Rocky Mountain J. Math.,33, no. 4 (2003), pp. 1279-1287.

[LP] A. LASCOUX - P. PRAGACZ, Jacobians of symmetric polynomials, Ann.

Comb.6, no. 2 (2002), pp. 169-172.

[M] I. G. MACDONALD, Symmetric functions and Hall polynomials, second edition, Oxford Mathematical Monographs. Oxford Science Publications.

The Clarendon Press, Oxford University Press, New York, 1995.

[S] R. STANLEY, Enumerative combinatorics, Vol.1, Cambridge Studies in Advanced Mathematics, 49. Cambridge University Press, Cambridge, 1997.

Manoscritto pervenuto in redazione il 22 gennaio 2008.