GENERATORS OF TRUNCATED SYMMETRIC POLYNOMIALS

arXiv:1011.6068v5 [] 24 May 2016

FEDERICO GALETTO

Abstract. Adem and Reichstein introduced the ideal of truncated symmetric
polynomials to present the permutation invariant subring in the cohomology
of a finite product of projective spaces. Building upon their work, I describe
a generating set of the ideal of truncated symmetric polynomials in arbitrary
positive characteristic, and offer a conjecture for minimal generators.

Contents
1. Introduction
2. Symmetric polynomials and partitions
3. Truncated symmetric polynomials
4. Generators in positive characteristic
5. A conjecture for minimal generators
6. Acknowledgements
References

1
2
6
7
10
11
11

1. Introduction
Let F be a field, and let R = F[x1 , . . . , xn ] be the polynomial ring in n variables
with coefficients in F. The symmetric group Sn acts on R by permuting the variables. Denote by RSn the invariant subring, i.e., the ring of symmetric polynomials.
The ideal of truncated symmetric polynomials in RSn is defined by
Sn
In,d = (xd+1
, . . . , xd+1
.
n )R ∩ R
1

The ideal of truncated symmetric polynomials was introduced by A. Adem and
Z. Reichstein [1] in the following geometric context. Let CP d be the complex
projective d-space, and let BU (n) be the classifying space of the unitary group.
The symmetric group Sn acts on the n-fold product (CP d )n by permuting the
factors. Consider the induced action on the cohomology ring H ∗ ((CP d )n , F) and
let H ∗ ((CP d )n , F)Sn be the invariant subring. The two authors show there is a
Date: May 25, 2016.
2010 Mathematics Subject Classification. 13A15, 55R35.
Key words and phrases. symmetric polynomial, positive characteristic, cohomology, classifying
space, unitary group.
1

2

FEDERICO GALETTO

map (CP d )n → BU (n) such that the induced map on cohomology restricts to a
ring epimorphism
H ∗ (BU (n), F) −→ H ∗ ((CP d )n , F)Sn .
The cohomology ring of BU (n) can be identified with RSn , and the kernel of
this map can be identified with In,d . Therefore RSn /In,d and H ∗ ((CP d )n , F)Sn
are isomorphic as ungraded rings (they become isomorphic as graded rings if the
grading on RSn /In,d is stretched out by a factor of 2). This result was used in
the same paper to compute the cohomology of the homotopy fiber for a natural
fibration over BU (n).
Adem and Reichstein determined a set of generators of In,d when the characteristic of F is zero and when it is strictly bigger than (n + 1)/2. Independently,
A. Conca, C. Krattenthaler and J. Watanabe [2] identified generators of In,d working over the complex numbers (although their methods hold over any field of characteristic strictly bigger than n). In addition, the ideal of truncated symmetric
polynomials has generated interest in connection to other topics, such as Lefschetz
properties [5] and invariants of Poincaré duality algebras [8].
In this paper, I describe a generating set of In,d in arbitrary characteristic. Given
a partition λ, let mλ ∈ RSn denote the monomial symmetric polynomial associated
with λ. The shorthand notation (. . . , ab , . . .) denotes a partition with the part a
repeated b times. The main result reads as follows.
Theorem. Let t = max{i ∈ N | pi 6 n} and let q0 , . . . , qt ∈ N>0 be such that,
∀i ∈ {0, . . . , t}, n = qi pi + ri with 0 6 ri < pi . For every i ∈ {0, . . . , t}, define an
ideal of RSn
Jn,d,i = (m((d+1)pi ) , . . . , m((d+qi )pi ) ).
Then In,d = Jn,d,0 + Jn,d,1 + . . . + Jn,d,t .
The next section contains a brief review of symmetric polynomials and some
computational results. In section 3, I review the algebraic results of Adem and
Reichstein. The proof of the main theorem is in section 4. Finally, in section
5, I present a conjecture for a minimal generating set of the ideal of truncated
symmetric polynomials.
2. Symmetric polynomials and partitions
Let F be a field of characteristic p. For n ∈ N>0 , set R = F[x1 , . . . , xn ], the
polynomial ring in n indeterminates over F. The symmetric group Sn acts on R
by permuting the variables. Let RSn ⊆ R be the invariant subring, i.e., the ring of
symmetric polynomials.
Recall that a partition is a sequence
λ = (λ1 , λ2 , . . . , λr , . . .)

GENERATORS OF TRUNCATED SYMMETRIC POLYNOMIALS

3

of non-negative integers in non-increasing order:
λ1 > λ2 > . . . > λr > . . .
and containing only finitely many nonzero terms. We identify two such sequences
which differ only by a string of zeroes at the end. The nonzero numbers λi are
called the parts of λ; the number of parts of λ is called the length of λ and denoted
l(λ).
For a partition λ with l(λ) 6 n, define the monomial symmetric polynomial on
λ to be the polynomial,
X
αn
1
mλ =
xα
1 . . . xn
summed over all distinct permutations α = (α1 , . . . , αn ) of λ = (λ1 , . . . , λn ). The
polynomials mλ are clearly symmetric. Moreover, if L6n denotes the set of all
partitions with length smaller than or equal to n, then {mλ | λ ∈ L6n } is a basis
of RSn as an F-vector space [6, §I.2, pp. 18-19]. If l(λ) > n, then mλ = 0.
Example 2.1. Suppose n = 3. Then
m(1,1) = x1 x2 + x1 x3 + x2 x3 ,
m(2,1) = x21 x2 + x21 x3 + x1 x22 + x1 x23 + x22 x3 + x2 x23 .
Example 2.2 (Power sums).
pi = xi1 + . . . + xin = m(i)
Example 2.3 (Elementary symmetric polynomials).
X
ei =
xj1 . . . xji = m(1,...,1 )
|{z}
16j <...<j 6n
1

i

i times

The following statement contains a multiplication formula for the polynomials
mλ . The proof follows directly from the definitions.
Proposition 2.4. For any λ, µ ∈ L6n ,
mλ mµ =

X

cν mν ,

ν∈L6n

where cν is the number of different ways to write
(ν1 , . . . , νn ) = (α1 , . . . , αn ) + (β1 , . . . , βn )
with (α1 , . . . , αn ) a permutation of λ = (λ1 , . . . , λn ) and (β1 , . . . , βn ) a permutation
of µ = (µ1 , . . . , µn ).
Example 2.5. Suppose n = 3. Then
m(1,1) m(2,1) = m(3,2) + 2m(3,1,1) + 2m(2,2,1) ,

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FEDERICO GALETTO

since
(3, 2, 0) = (1, 1, 0) + (2, 1, 0),
(3, 1, 1) = (1, 1, 0) + (2, 0, 1) = (1, 0, 1) + (2, 1, 0),
(2, 2, 1) = (0, 1, 1) + (2, 1, 0) = (1, 0, 1) + (1, 2, 0).
Let λ1 , . . . , λk be integers such that
λ1 > λ2 > . . . > λk > 0
wk
1
and let w1 , . . . , wk be positive integers. Denote by (λw
1 , . . . , λk ) the partition λ
having wi parts equal to λi for every i ∈ {1, . . . , k}. The number wi will be referred
to as the multiplicity of λi in λ. In particular, λ1 will be called the leading part of
λ, denoted by lp(λ), and w1 will be called the leading multiplicity of λ, denoted by
lm(λ). Notice l(λ) = w1 + . . . + wk .
We provide here a list of formulas that will be employed later on. The first one
is a classical result by Newton (see [7]).

Proposition 2.6 (Newton’s identities). For any s > n,
m(s) =

n
X
(−1)j−1 m(1j ) m(s−j) .
j=1

Example 2.7. Let n = 3 and s = 4. Then
m(4) = m(1) m(3) − m(1,1) m(2) + m(1,1,1) m(1) .
wk
1
Lemma 2.8. Let λ = (λw
1 , . . . , λk ) ∈ L6n , with k > 1. Then
X
X
w
w
m(λw
aµ mµ +
bν m ν ,
1 m
(λ 2 ,...,λ k ) = mλ +
1 )
2

k

lp(µ)>λ1
lm(µ)=w1
l(µ)<l(λ)

lp(ν)>λ1
lm(ν)<w1

for some aµ , bν ∈ F.
Proof. Expanding the product on the left hand side in the basis of monomial symmetric polynomials, we obtain
X
wk =
w2
m(λw
cθ mθ ,
1 )m
(λ
,...,λ
)
1
2

k

θ∈L6n

for some cθ ∈ F. We proceed to identify the partitions θ for which the coefficient
cθ may be non-zero.
Consider first the case when θ = λ. The only way to obtain the partition λ as a
wk
w2
1
sum of permutations of (λw
1 ) and (λ2 , . . . , λk ) is
wk
2
(λw1 , 0, . . . , 0) + (0w1 , λw
2 , . . . , λk , 0, . . . , 0),

where, abusing notation, 0w1 means the entry 0 is repeated w1 times. Thus cθ = 1
by Proposition 2.4.

GENERATORS OF TRUNCATED SYMMETRIC POLYNOMIALS

5

Now suppose θ 6= λ and cθ 6= 0. Since λ1 > . . . > λk > 0, lp(θ) = λ1 + λj for
some j ∈ {2, . . . , k}. Note that λ1 + λj > λ1 , hence lp(θ) > λ1 .
If there is an index j ∈ {2, . . . , k} such that wj > w1 , then we have a partition
w −w1

j
2
µ = ((λ1 + λj )w1 , λw
2 , . . . , λj

k
, . . . , λw
k )=

w −w1

w2
w1
j
1
= (λw
1 , 0, . . . , 0) + (λj , λ2 , . . . , λj

k
, . . . , λw
k )

Therefore, by Proposition 2.4, the element mµ may appear with non-zero coefficient
in our expansion. Observe that lm(µ) = w1 and µ has length
w1 + w2 + . . . + (wj − w1 ) + . . . + wk =

k
X

wj < l(λ).

j=2

Similarly, we may obtain other partitions µ with the same properties. We collect
them all in a single summation as in the statement of the lemma.
The remaining terms are of the form bν mν where, as observed, lp(ν) > λ1 and,
in addition, lm(ν) < w1 . We collect them in a single summation to obtain the
desired formula.

Example 2.9. Let n = 6 and λ = (3, 3, 2, 2, 1). Then
m(3,3) m(2,2,2,1) = m(3,3,2,2,2,1) +
+ m(5,5,2,1) +
+ m(5,4,2,2) + m(5,3,2,2,1) + m(4,3,2,2,2) .
wk
1
Lemma 2.10. Let λ = (λw
1 , . . . , λk ) ∈ L6n . If s is a positive integer such that
s + l(λ) 6 n, then


X
X
s + w1
s
m(λ1 ) mλ =
m(λs+w1 ,λw2 ,...,λwk ) +
aµ mµ +
bν mν ,
1
2
k
s
lp(µ)>λ1
lm(µ)=s
l(µ)=l(λ)

lp(ν)>λ1
lm(ν)<s

for some aµ , bν ∈ F.
Proof. The proof follows from the same reasoning used to prove Lemma 2.8. The
only aspect that requires further clarification is the binomial coefficient that appears
in front of
m(λs+w1 ,λw2 ,...,λwk ) .
1

2

k

This polynomial arises whenever a permutation of (λs1 , 0, . . . , 0) is chosen that has
s of its first s + w1 entries occupied by λ1 , and the remaining ones by 0, while
wk
1
simultaneously a permutation of (λw
1 , . . . , λk ) is chosen that has the same s among
its first s + w1 entries occupied by 0, and the remaining ones by λ1 . The number
of times this occurs is the number of ways to choose s entries among the first
s + w1 .


6

FEDERICO GALETTO

Example 2.11. Let n = 6, λ = (32 , 2, 1) and s = 2. Then
m(32 ) m(32 ,2,1) = 6m(34 ,2,1) +
+ m(62 ,2,1) +
+ (m(6,5,3,1) + m(6,4,3,2) + 2m(6,32 ,2,1) +
+ m(5,4,32 ) + 3m(5,33 ,1) + 3m(4,33 ,2) ).
We are interested in the case when the binomial coefficient appearing in Lemma
2.10 is invertible in F. This is always the case when F has characteristic zero. The
following result is helpful when F has characteristic p > 0.
Theorem 2.12 (Lucas). Let u and v be non-negative integers, p a prime, and
u = ut pt + ut−1 pt−1 + . . . + u1 p + u0 ,
v = vt pt + vt−1 pt−1 + . . . + v1 p + v0 ,
the base p expansions of u and v. Then
  Y
t  
uj
u
≡
(mod p).
vj
v
j=0
A proof can be found in [3, §6].
3. Truncated symmetric polynomials
For a given non-negative integer d, consider the ideal (xd+1
, . . . , xd+1
n ) ⊆ R.
1
Denote by In,d the intersection
Sn
(xd+1
, . . . , xd+1
.
n )∩R
1

Since RSn is a subring of R, In,d is an ideal of RSn . Call In,d the ideal of truncated
symmetric polynomials.
In this section, I will recall ideas and results previously presented in [1]. For the
convenience of the reader, all statements and proofs will be phrased in the language
and notation of this paper.
The following proposition was recorded as a simple observation without proof in
the paper of Adem and Reichstein. I fill in the details of the proof below.
Proposition 3.1.
(a) Let λ ∈ L6n . Then mλ ∈ In,d if and only if lp(λ) > d + 1.
(b) The set
{mλ | λ ∈ L6n , lp(λ) > d + 1}
spans In,d over F.

GENERATORS OF TRUNCATED SYMMETRIC POLYNOMIALS

7

Proof.

(a) Let λ = (λ1 , . . . , λn ). If lp(λ) > d + 1, then λ1 > d + 1. Hence,
for any permutation (α1 , . . . , αn ) of (λ1 , . . . , λn ), there is some index i such
d+1
αn
1 α2
that αi > d+1 and xα
, . . . , xd+1
n ). Therefore mλ ∈ In,d .
1 x2 . . . xn ∈ (x1
Vice versa, assume mλ ∈ In,d . Then mλ ∈ (xd+1
, . . . , xd+1
n ). Note that
1
d+1
d+1
{x1 , . . . , xn } is the reduced Gröbner basis of the ideal (xd+1
, . . . , xd+1
n )
1
in the lexicographic term ordering on R. Therefore the leading term of mλ is
divisible by xd+1
for some i. Since the leading term of mλ is xλ1 1 xλ2 2 . . . xλnn ,
i
it follows that lp(λ) = λ1 > λi > d + 1.
(b) Let 0 6= g ∈ In,d . Since the monomial symmetric polynomials form a basis
of RSn over F, g can be written uniquely as a linear combination
X
g=
cθ mθ ,
θ∈L6n

for some cθ ∈ F. The set {θ ∈ L6n | cθ 6= 0} is nonempty, finite, and totally
ordered lexicographically. Let λ be its maximum. Then xλ1 1 xλ2 2 . . . xλnn is
the leading term of g. Reasoning as in part (a), deduce that lp(λ) > d + 1.
Now repeat this process with g − cλ mλ to show that the remaining mθ in
the expression of g also have leading part at least d + 1. Therefore g is in
the span of the desired set.

The next result corresponds to [1, Lemma 5.2].
Proposition 3.2. Let Jn,d = (m(d+1) , . . . , m(d+n) ) ⊆ RSn . Then Jn,d contains
every mλ with lp(λ) > d + 1 and lm(λ) ∈ F× .
The last result of this section is an immediate consequence of Proposition 3.2.
It is part of Adem and Reichstein’s main theorem [1, Thm. 5.1(a)].
Corollary 3.3. If p = 0 or n < p, then In,d = (m(d+1) , . . . , m(d+n) ).
4. Generators in positive characteristic
As Adem and Reichstein already noticed, the polynomials m(d+1) , . . . , m(d+n)
alone do not generate In,d when n > p. For n 6 2p − 1, the two authors showed
that taking m((d+1)p ) in addition to m(d+1) , . . . , m(d+n) is enough to generate In,d
[1, Thm. 5.1(b)]. The goal of this section is to describe a set of generators of In,d in
arbitrary positive characteristic. Throughout this section, p > 0 will be assumed.
The following is a generalization of Newton’s identities.
Lemma 4.1. Consider h 6 n and let q ∈ N>0 be such that n = qh + r with
0 6 r < h. For every s ∈ N>0 ,
q
X
(−1)j−1 m(1jh ) m((s+q+1−j)h ) = m((s+q+1)h ) +
j=1

X
lp(µ)>s+1
lm(µ)<h

aµ mµ ,

8

FEDERICO GALETTO

for some aµ ∈ F.
Proof. When j ∈ {1, . . . , q − 1}, the product m(1jh ) m((s+q+1−j)h ) expands to
X
m((s+q+2−j)h ,1(j−1)h ) + m((s+q+1−j)h ,1jh ) +
aµ,j mµ ,
lp(µ)=s+q+2−j
lm(µ)<h

for some aµ,j ∈ F. When j = q, the same product becomes m(1qh ) m((s+1)h ) which
expands to
X
m((s+2)h ,1(q−1)h ) +
aµ,q mµ ,
lp(µ)=s+2
lm(µ)<h

for some aµ,q ∈ F. Summing over j with alternating signs, all monomial symmetric
polynomials with leading multiplicity h cancel out except for m((s+q+1)h ) which
occurs when j = 1. The remaining terms can be combined in a single summation
to give the formula above.

Proposition 4.2. Let i ∈ N be such that pi 6 n and let q ∈ N>0 be such that
n = qpi + r with 0 6 r < pi . Let K be an ideal of RSn such that
K ⊇ {mλ | λ ∈ L6n , lp(λ) > d + 1, lm(λ) < pi }.
If J = (m((d+1)pi ) , . . . , m((d+q)pi ) ), then J +K contains every mλ with lp(λ) > d+1
and lm(λ) 6 pi .
Proof. If λ ∈ L6n has lp(λ) > d + 1 and lm(λ) < pi , then, by the hypothesis,
mλ ∈ K ⊆ J + K.
Suppose lp(λ) > d + 1 and lm(λ) = pi . The proof that mλ ∈ J + K is by
induction on l(λ). For the base case, assume l(λ) = pi , i.e., mλ = m(λpi ) for some
1
λ1 > d + 1. By definition, m(λpi ) ∈ J ⊆ J + K, when d + 1 6 λ1 6 d + q. If
1
λ1 > d + q, then we may write λ1 = s + q + 1, for some s > d. Applying Lemma
4.1 with h = pi ,
q
X

(−1)j−1 m(1jpi ) m((s+q+1−j)pi ) = m((s+q+1)pi ) +

j=1

X

aµ mµ ,

lp(µ)>s+1
lm(µ)<pi

for some aµ ∈ F. The summation on the right is in K ⊆ J + K by the hypothesis. Proceeding by induction on s, the left hand side belongs to J + K. Hence
m((s+q+1)pi ) ∈ J + K.
i

wk
i
2
For the inductive step, assume λ = (λp1 , λw
2 , . . . , λk ) with l(λ) > p and λ1 >
d + 1. By Lemma 2.8,
X
X
m(λpi ) m(λw2 ,...,λwk ) = mλ +
aµ mµ +
bν mν ,
1

2

k

lp(µ)>λ1
lm(µ)=pi
l(µ)<l(λ)

lp(ν)>λ1
lm(ν)<pi

GENERATORS OF TRUNCATED SYMMETRIC POLYNOMIALS

9

for some aµ , bν ∈ F. By the base case, m(λpi ) ∈ J + K. On the right hand side, the
1
first summation belongs to J + K by the inductive hypothesis and the second one
belongs to K ⊆ J + K by the hypothesis. Thus mλ ∈ J + K.

We are now ready to exhibit a set of generators for the ideal In,d .
Theorem 4.3. Let t = max{i ∈ N | pi 6 n} and let q0 , . . . , qt ∈ N>0 be such that,
∀i ∈ {0, . . . , t}, n = qi pi + ri with 0 6 ri < pi . For every i ∈ {0, . . . , t}, define an
ideal of RSn
Jn,d,i = (m((d+1)pi ) , . . . , m((d+qi )pi ) ).
Then In,d = Jn,d,0 + Jn,d,1 + . . . + Jn,d,t .
Example 4.4. Let p = 2, n = 7, and d = 1. Then
J7,1,0 = (m(2) , m(3) , m(4) , m(5) , m(6) , m(7) , m(8) ),
J7,1,1 = (m(22 ) , m(32 ) , m(42 ) ),
J7,1,2 = (m(24 ) ),
and I7,1 = J7,1,0 + J7,1,1 + J7,1,2 .
Proof. Define auxiliary ideals by setting
Ki := Jn,d,0 + Jn,d,1 + . . . + Jn,d,i ,
for i ∈ {0, . . . , t}. With this notation, the thesis becomes In,d = Kt . It is clear
that K0 ⊆ K1 ⊆ . . . ⊆ Kt . Moreover, Kt ⊆ In,d by Proposition 3.1(a). In light
of Proposition 3.1(b), the inclusion In,d ⊆ Kt will follow if one can show that Kt
contains every polynomial mλ with lp(λ) > d + 1.
Claim: ∀i ∈ {0, . . . , t}, the inclusion
Ki ⊇ {mλ | λ ∈ L6n , lp(λ) > d + 1, lm(λ) < pi+1 }
holds.
The claim will be proven by induction on i. For i = 0, note that
K0 = Jn,d,0 = (m(d+1) , . . . , m(d+n) ).
Hence the claim is an immediate consequence of Proposition 3.2.
Assume i > 0. By the inductive hypothesis,
Ki−1 ⊇ {mλ | λ ∈ L6n , lp(λ) > d + 1, lm(λ) < pi }.
By definition, Ki = Jn,d,i + Ki−1 . Hence, by Proposition 4.2,
(1)

Ki ⊇ {mλ | λ ∈ L6n , lp(λ) > d + 1, lm(λ) 6 pi }.

It remains to show Ki contains all monomial symmetric polynomials mλ with
wk
i
1
p < lm(λ) < pi+1 . Let λ = (λw
1 , . . . , λk ) ∈ L6n with λ1 > d + 1 and p < w1 <
i+1
p . The base p expansion of w1 is
i

w1 = hi pi + hi−1 pi−1 + . . . + h1 p + h0 ,

10

FEDERICO GALETTO

where 0 6 hj < p, ∀j ∈ {0, . . . , i}, and hi 6= 0. Set z = hi−1 pi−1 + . . . + h0 , so that
w1 = hi pi + z. By Lemma 2.10,
 
X
X
w1
m(λpi ) m (hi −1)pi +z w2
mλ +
aµ mµ +
bν mν ,
wk =
i
(λ
,λ
,...,λ
)
p
1
1
2
k
lp(µ)>λ1
lm(µ)=pi
l(µ)=l(λ)

lp(ν)>λ1
lm(ν)<pi

for some aµ , bν ∈ F. By (1), m(λpi ) and the two summations on the right hand side
1
all belong to Ki . Moreover, by Lucas’ theorem,
    i−1
 
w1
hi Y hj
≡
≡ hi
pi
1 j=0 0

(mod p).


Since 0 < hi < p, wpi1 ∈ F× ; therefore mλ ∈ Ki . This concludes the proof of the
claim.
When i = t, the claim gives
Kt ⊇ {mλ | λ ∈ L6n , lp(λ) > d + 1, lm(λ) < pt+1 }.
If λ ∈ L6n and lp(λ) > d + 1, the inequality lm(λ) 6 l(λ) 6 n < pt+1 implies
mλ ∈ Kt . Therefore In,d = Kt .

5. A conjecture for minimal generators
In this section, I consider the problem of describing minimal generators of the
ideal In,d . When p = 0 or p > n!, a solution was already given by Adem and
Reichstein (see [1, Thm. 6.1(a)] and [1, Lemma 6.2(c)]); namely, the generators of
Corollary 3.3 are minimal.
Theorem 4.3 describes a set of generators of In,d for arbitrary p > 0. In general,
this set is not minimal.
Example 5.1. Let p = 2, n = 7 and d = 1. Refer to Example 4.4 for a generating
set of I7,1 . Notice that
2

(m(2) ) =

7
X
i=1

!2
x2i

=

7
X

x4i = m(4) ,

i=1

making m(4) redundant as a generator. Similarly, m(6) and m(8) can be discarded
because m(6) = (m(3) )2 and m(8) = (m(4) )2 .
A similar computation yields

2
X
X
x4i x4j = m(42 ) .
(m(22 ) )2 = 
x2i x2j  =
16i<j67

Therefore m(42 ) is also redundant.

16i<j67

GENERATORS OF TRUNCATED SYMMETRIC POLYNOMIALS

11

In general, when p > 0, we have
(2)

(m(sk ) )p = m((ps)k ) .

This observation allows to trim the generating set of Theorem 4.3, leading to the
following.
Conjecture. Let t = max{i ∈ N | pi 6 n} and let q0 , . . . , qt ∈ N>0 be such that,
∀i ∈ {0, . . . , t}, n = qi pi + ri with 0 6 ri < pi . For every i ∈ {0, . . . , t}, define the
set

o
n

Jn,d,i = m((d+h)pi ) h ∈ {1, . . . , qi } and ∀j ∈ {1, . . . , h} d + h 6= p(d + j) .
Then Jn,d,0 ∪ Jn,d,1 ∪ . . . ∪ Jn,d,t is a minimal set of generators of In,d .
Essentially, this conjecture claims that the only possible relations among the
generators of Theorem 4.3 are of the kind given in equation (2). For small values
of p, n and d, the conjecture has been verified using the software Macaulay2 [4].
6. Acknowledgements
I am indebted to the anonymous referee, whose useful comments and suggestions
helped improve the quality of this manuscript. While working on this project, I
was partially supported by an NSERC grant.
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Federico Galetto, McMaster University, 1280 Main St W, 407 Hamilton Hall, Hamilton, ON, L8S 4K1, Canada
E-mail address: galettof@math.mcmaster.ca
URL: http://math.galetto.org

