# The Hilbert-Kunz Function for Binomial Hypersurfaces

Authors: Shyamashree Upadhyay
Journal: Algebra
 (2014)

---

## Abstract

I give an iterative closed form formula for the Hilbert-Kunz function for any binomial hypersurface in general, over any field of arbitrary positive characteristic. I prove that the Hilbert-Kunz multiplicity associated with any binomial hypersurface over any field of arbitrary positive characteristic is rational. As an example, I also prove the well known fact that for 1-dimensional binomial hypersurfaces the Hilbert-Kunz multiplicity is a positive integer and give a precise account of the integer.


---

## Body

## 1. Introduction

Let(
A
,
n
) be a Noetherian local ring of dimension d and of prime characteristic p
>
0. Let I be an n-primary ideal and q
=
p
n where n
>
0 is an integer. The “Hilbert-Kunz function” of A with respect to I is defined as
(1)
H
K
I
,
A
q
=
l
A
I
q
,
where I
[
q
]
=
nth Frobenious power of I, that is, the ideal generated by x
q, x
∈
I.The associated Hilbert-Kunz multiplicity is defined to be(2)
e
H
K
(
I
)
=
lim
⁡
q
→
∞
H
K
I
,
A
(
q
)
q
d
.Monsky showed in his paper [1] that the limit exists and is a real constant. Further he showed that
(3)
H
K
I
,
A
q
=
e
H
K
I
q
d
+
O
q
d
-
1
.
Several authors have investigated e
H
K
(
I
). They showed that e
H
K
(
I
) is a rational number for certain special kind of binomial hypersurfaces [2], cubic curves and surfaces [3] and full flag varieties, and elliptic curves [4].A binomial hypersurface is defined in the beginning of Section4. In this paper, we are interested in giving an iterative closed form formula for the Hilbert-Kunz function for any binomial hypersurface in general. Our methods work for any positive characteristic. Our work generalizes the work of Conca [2]. In [2], Conca computes the Hilbert-Kunz function of monomial ideals and also of those Binomial hypersurfaces whose terms defining the hypersurface are relatively prime. In [2], Conca also proves that the Hilbert-Kunz multiplicity associated with these special Binomial Hypersurfaces is always rational. In this paper, we prove that the Hilbert-Kunz multiplicity associated with any binomial hypersurface over any field of arbitrary positive characteristic is rational. The rationality result appears in Eto’s work (Theorem 2.2, [5]), where the multiplicity is interpreted as a volume of a polytope. Here we give a different proof for the rationality result. Our work also generalizes the work [6] of Watanabe, where he deals only with normal toric varieties.The organization of this work is more or less clear from the table of contents. But to be precise, until Section4 begins, the matter of this work holds true for any hypersurface over any field of positive characteristic and need not have to be a binomial hypersurface! Also, the filtration introduced in Section 2.1 is effective for any general ideal J of a polynomial ring S
=
k
[
x
1
,
…
,
x
m
] where k is a field of arbitrary prime characteristic p
>
0, not just when the ideal J is generated by a single polynomial. The iterative closed form formula for the Hilbert-Kunz function for binomial hypersurfaces appears in Theorem 26 in Section 4.2.5. After that, in Section 4.4, we discuss the example of the 1-dimensional case. In the example of the 1-dimensional case, I prove the well known fact that for any 1-dimensional binomial hypersurface, the associated Hilbert-Kunz multiplicity is a positive integer and I give a precise account of the integer. In Section 4.5, I give an expression for the Hilbert-Kunz multiplicity for a general binomial hypersurface (see (67)).


## 2. Stating the Problem

For the rest of this paper we will assume thatS
=
k
[
x
1
,
…
,
x
m
] where k is a field of arbitrary prime characteristic p
>
0 and J is an arbitrary ideal in S. Let R
=
S
/
J. Put m
^
:
=
m
+
J where m
=
(
x
1
,
…
,
x
m
). We will assume that J
⊆
m; otherwise, m
^
=
R. Consider
(4)
H
K
m
^
R
m
^
,
R
m
^
p
n
=
l
R
m
^
m
^
R
m
^
p
n
.Note that the ringsR
m
^
/
(
m
^
R
m
^
)
[
p
n
] and R
/
(
m
[
p
n
]
+
J
) are isomorphic, where m
[
p
n
]
=
(
x
1
p
n
,
…
,
x
m
p
n
). Hence the Hilbert-Kunz function H
K
m
^
R
m
^
,
R
m
^
(
p
n
) is given by
(5)
H
K
m
^
R
m
^
,
R
m
^
p
n
=
l
R
m
p
n
+
J
.
We are presently interested in the case when R is a hypersurface, that is, when the ideal J is generated by a single polynomial.


### 2.1. A Filtration for Computing the Lengthl
(
R
/
(
m
p
n
+
J
)
)

LetS denote the set of all m-tuples (
u
1
,
…
,
u
m
) such that for each i
∈
{
1
,
…
,
m
}, u
i
∈
{
0,1
,
…
,
p
n
-
1
}. Let us arrange the elements of the set S in the lexicographic order; that is,we say that(
u
1
,
…
,
u
m
)
<
lex
(
u
1
′
,
…
,
u
m
′
) if in the vector difference (
u
1
′
,
…
,
u
m
′
)
-
(
u
1
,
…
,
u
m
), the leftmost nonzero entry is >
0. It is easy to check that the set S equipped with the lexicographic order <
lex is in fact a totally ordered set, and this set contains p
m
n many elements. Let us denote these p
m
n many elements of the set S arranged in the <
lex order by U
0
<
lex
U
1
<
lex
U
2
<
lex
⋯
<
lex
U
p
m
n
-
1.LetM
:
=
{
x
1
j
1
⋯
x
m
j
m
∣
0
≤
j
i
<
p
n
∀
1
≤
i
≤
m
}. There are p
m
n many elements in the set M. For any U
=
(
u
1
,
u
2
,
…
,
u
m
)
∈
S, let us denote by X
p
n
-
1
-
U the element of M given by x
1
p
n
-
1
-
u
1
x
2
p
n
-
1
-
u
2
⋯
x
m
p
n
-
1
-
u
m. The set S is in one-to-one correspondence with the set M via the map U
↦
X
p
n
-
1
-
U. Let us denote by X
p
n
-
1
-
U
j the image of the element U
j under this correspondence.For anyi
∈
{
1
,
…
,
m
} and any U
j
=
(
u
j
,
1
,
…
,
u
j
,
m
)
∈
S, let us denote by X
i
p
n
-
1
-
U
j the product x
1
p
n
-
1
-
u
j
,
1
⋯
x
i
p
n
-
1
-
u
j
,
i, that is, the product in X
p
n
-
1
-
U
j up to the ith variable.Given anyj
∈
{
0
,
…
,
p
m
n
-
1
} and any i
∈
{
1
,
…
,
m
}, define
(6)
B
j
,
i
≔
X
i
p
n
-
1
-
U
j
x
i
if
u
j
,
i
>
0
,
0
otherwise.Definition 1.
The idealsI
j: for any j
∈
{
0
,
…
,
p
m
n
-
1
}, let I
j
≔
〈
B
j
,
i
∣
1
≤
i
≤
m
〉
+
L where L
=
〈
x
1
p
n
,
x
2
p
n
,
…
,
x
m
p
n
〉.
DeclareI
p
m
n to be the polynomial ring S itself.Remark 2.
From the definition of the lexicographic order<
lex on S, it follows easily that for any j
∈
{
0
,
…
,
p
m
n
-
2
}, either X
p
n
-
1
-
U
j
=
X
p
n
-
1
-
U
j
+
1
x
m or there exists k
∈
{
1
,
…
,
m
-
1
} such that X
k
p
n
-
1
-
U
j
=
X
k
p
n
-
1
-
U
j
+
1
x
k, X
p
n
-
1
-
U
j
=
X
k
p
n
-
1
-
U
j, and X
p
n
-
1
-
U
j
+
1
=
X
k
p
n
-
1
-
U
j
+
1
(
x
k
+
1
⋯
x
m
)
p
n
-
1.Notation 1.
For anyj
∈
{
0
,
…
,
p
m
n
-
1
}, let i
(
j
)
≔
max
⁡
{
i
∣
1
≤
i
≤
m
,
B
j
,
i
≠
0
}.Lemma 3.
(i) For anyj
∈
{
0
,
…
,
p
m
n
-
1
}, X
p
n
-
1
-
U
j
∈
I
j
+
1;
(ii)B
j
,
k
=
B
j
+
1
,
k for all j
∈
{
0
,
…
,
p
m
n
-
2
} and for all k
<
i
(
j
+
1
).Proof.
(i) It follows from Remark2 that X
p
n
-
1
-
U
j
=
B
j
+
1
,
i
(
j
+
1
). Hence X
p
n
-
1
-
U
j
∈
I
j
+
1;
(ii) the proof of this follows from Remark2.Lemma 4.
For anyj
∈
{
0
,
…
,
p
m
n
-
1
}, I
j
⊆
I
j
+
1.Proof.
AsI
j
≔
〈
B
j
,
i
∣
1
≤
i
≤
m
〉
+
L (where L
=
〈
x
1
p
n
,
x
2
p
n
,
…
,
x
m
p
n
〉), it is enough to show that B
j
,
i
∈
I
j
+
1 for all j
∈
{
0
,
…
,
p
m
n
-
1
} and for all i
∈
{
1
,
…
,
m
}.
Case  1 (when
X
p
n
-
1
-
U
j
=
X
p
n
-
1
-
U
j
+
1
x
m
). In this case, clearly B
j
,
i
=
B
j
+
1
,
i for all 1
≤
i
≤
m
-
1. And B
j
,
m is either equal to 0 or equal to B
j
+
1
,
m
x
m. Hence in either case, B
j
,
m
∈
I
j
+
1. Therefore, I
j
⊆
I
j
+
1.
Case  2 (when
X
p
n
-
1
-
U
j
≠
X
p
n
-
1
-
U
j
+
1
x
m
). In this case, there exists k
∈
{
1
,
…
,
m
-
1
} such that X
k
p
n
-
1
-
U
j
=
X
k
p
n
-
1
-
U
j
+
1
x
k, X
p
n
-
1
-
U
j
=
X
k
p
n
-
1
-
U
j, and X
p
n
-
1
-
U
j
+
1
=
X
k
p
n
-
1
-
U
j
+
1
(
x
k
+
1
⋯
x
m
)
p
n
-
1.
Clearly then,B
j
,
i
=
B
j
+
1
,
i for all i
<
k. Also, B
j
,
k is either equal to 0 or equal to B
j
+
1
,
k
x
k. Hence in either case, B
j
,
k
∈
I
j
+
1. For i
>
k, B
j
,
i
=
B
j
+
1
,
k
x
i. Hence B
j
,
i
∈
I
j
+
1 for all i
>
k. Therefore, I
j
⊆
I
j
+
1.Notation 2.
LetM be any S-module. Let λ
(
M
) denote thelength of M with respect to S.Lemma 5.
Letj
∈
{
0
,
…
,
p
m
n
-
1
}. Put M
j
+
1
≔
I
j
+
1
/
I
j. Then λ
(
M
j
+
1
)
=
1.Proof.
Fixj. We will show that for each i
∈
{
1
,
…
,
m
}, x
i
I
j
+
1
⊆
I
j. We claim that for any i
,
k
∈
{
1
,
…
,
m
}, x
i
B
j
+
1
,
k
∈
I
j.
Case  1 (when
k
>
i
(
j
+
1
)
). In this case, we have B
j
+
1
,
k
=
0. Hence x
i
B
j
+
1
,
k
∈
I
j for any k
>
i
(
j
+
1
) and for any i
∈
{
1
,
…
,
m
}.
Case  2 (when
k
≤
i
(
j
+
1
)
). It follows from Lemma 3 that B
j
,
k
=
B
j
+
1
,
k for all k
<
i
(
j
+
1
). Therefore B
j
+
1
,
k
∈
I
j for all k
<
i
(
j
+
1
). This implies that x
i
B
j
+
1
,
k
∈
I
j for all k
<
i
(
j
+
1
) and for all i
∈
{
1
,
…
,
m
}.
It remains to show thatx
i
B
j
+
1
,
i
(
j
+
1
)
∈
I
j for all i
∈
{
1
,
…
,
m
}. It follows from the proof of Lemma 3(i) that X
p
n
-
1
-
U
j
=
B
j
+
1
,
i
(
j
+
1
). Hence for any i
∈
{
1
,
…
,
m
}, x
i
B
j
+
1
,
i
(
j
+
1
)
=
x
i
X
p
n
-
1
-
U
j is either a multiple of B
j
,
i or a multiple of x
i
p
n. Therefore x
i
B
j
+
1
,
i
(
j
+
1
)
∈
I
j for all i
∈
{
1
,
…
,
m
}.Lemma 6.
For allj
∈
{
0
,
…
,
p
m
n
-
1
}, I
j
+
1
=
〈
X
p
n
-
1
-
U
j
〉
+
I
j.Proof.
It follows from Lemma3 that X
p
n
-
1
-
U
j
∈
I
j
+
1 for all j
∈
{
0
,
…
,
p
m
n
-
1
}. It also follows from the same lemma that B
j
,
k
=
B
j
+
1
,
k for all j
∈
{
0
,
…
,
p
m
n
-
2
} and for all k
<
i
(
j
+
1
). Therefore the generators B
j
+
1
,
k (where k
≠
i
(
j
+
1
)) of the ideal I
j
+
1 belong to I
j. Due to degree reasons, X
p
n
-
1
-
U
j
=
B
j
+
1
,
i
(
j
+
1
) does not belong to I
j. Therefore X
p
n
-
1
-
U
j is the only generator of the ideal I
j
+
1 which does not belong to I
j. Hence I
j
+
1
=
〈
X
p
n
-
1
-
U
j
〉
+
I
j.It follows from Lemmas4, 5, and 6 above that the following chain
(7)
I
0
⊊
I
1
⊊
⋯
⊊
I
p
m
n
-
1
⊊
I
p
m
n
is a filtration of ideals such that λ
(
I
j
+
1
/
I
j
)
=
1.Therefore reducing modulo the idealJ, it follows that the following chain
(8)
0
=
I
0
+
J
I
0
+
J
⊆
I
1
+
J
I
0
+
J
⊆
I
2
+
J
I
0
+
J
⊆
⋯
⊆
I
p
m
n
+
J
I
0
+
J
=
R
m
[
p
n
]
+
J
is afiltration of ideals, which is having the property thateach successive quotient either iszero or is generated by a single element. The total number of successive quotients of the above mentionedfiltration which are generated by a single element equals the length l
(
R
/
(
m
[
p
n
]
+
J
)
). The following corollary to Lemma 6 above gives a precise account of which successive quotient of the above mentioned filtration is generated by a single element and which successive quotient is zero.Corollary 7.
For anyj
∈
{
0
,
…
,
p
m
n
-
1
}, λ
(
(
I
j
+
1
+
J
)
/
(
I
j
+
J
)
)
≤
1, and λ
(
(
I
j
+
1
+
J
)
/
(
I
j
+
J
)
)
=
0 if and only if X
p
n
-
1
-
U
j
∈
I
j
+
J.Proof.
Fix anyj
∈
{
0
,
…
,
p
m
n
-
1
}, it follows from Lemma 5 that λ
(
I
j
+
1
/
I
j
)
=
1. Therefore, reducing modulo the ideal J, we have λ
(
(
I
j
+
1
+
J
)
/
(
I
j
+
J
)
)
≤
1. Let M
j
+
1 be as in Lemma 5. It follows from Lemma 6 that X
p
n
-
1
-
U
j generates M
j
+
1. Therefore reducing modulo the ideal J, it follows that the quotient (
I
j
+
1
+
J
)
/
(
I
j
+
J
) is zero if and only if X
p
n
-
1
-
U
j
∈
I
j
+
J and is generated by the single element X
p
n
-
1
-
U
j otherwise.Hence for knowing that which successive quotient of the above mentioned filtration is generated by single element and which one is zero, the only condition that needs to be verified is the following:X
p
n
-
1
-
U
j
∈
I
j
+
J or not for every j
∈
{
0
,
…
,
p
m
n
-
1
}?This is thekey checking condition for computing the length l
(
R
/
(
m
[
p
n
]
+
J
)
).


## 2.1. A Filtration for Computing the Lengthl
(
R
/
(
m
p
n
+
J
)
)

LetS denote the set of all m-tuples (
u
1
,
…
,
u
m
) such that for each i
∈
{
1
,
…
,
m
}, u
i
∈
{
0,1
,
…
,
p
n
-
1
}. Let us arrange the elements of the set S in the lexicographic order; that is,we say that(
u
1
,
…
,
u
m
)
<
lex
(
u
1
′
,
…
,
u
m
′
) if in the vector difference (
u
1
′
,
…
,
u
m
′
)
-
(
u
1
,
…
,
u
m
), the leftmost nonzero entry is >
0. It is easy to check that the set S equipped with the lexicographic order <
lex is in fact a totally ordered set, and this set contains p
m
n many elements. Let us denote these p
m
n many elements of the set S arranged in the <
lex order by U
0
<
lex
U
1
<
lex
U
2
<
lex
⋯
<
lex
U
p
m
n
-
1.LetM
:
=
{
x
1
j
1
⋯
x
m
j
m
∣
0
≤
j
i
<
p
n
∀
1
≤
i
≤
m
}. There are p
m
n many elements in the set M. For any U
=
(
u
1
,
u
2
,
…
,
u
m
)
∈
S, let us denote by X
p
n
-
1
-
U the element of M given by x
1
p
n
-
1
-
u
1
x
2
p
n
-
1
-
u
2
⋯
x
m
p
n
-
1
-
u
m. The set S is in one-to-one correspondence with the set M via the map U
↦
X
p
n
-
1
-
U. Let us denote by X
p
n
-
1
-
U
j the image of the element U
j under this correspondence.For anyi
∈
{
1
,
…
,
m
} and any U
j
=
(
u
j
,
1
,
…
,
u
j
,
m
)
∈
S, let us denote by X
i
p
n
-
1
-
U
j the product x
1
p
n
-
1
-
u
j
,
1
⋯
x
i
p
n
-
1
-
u
j
,
i, that is, the product in X
p
n
-
1
-
U
j up to the ith variable.Given anyj
∈
{
0
,
…
,
p
m
n
-
1
} and any i
∈
{
1
,
…
,
m
}, define
(6)
B
j
,
i
≔
X
i
p
n
-
1
-
U
j
x
i
if
u
j
,
i
>
0
,
0
otherwise.Definition 1.
The idealsI
j: for any j
∈
{
0
,
…
,
p
m
n
-
1
}, let I
j
≔
〈
B
j
,
i
∣
1
≤
i
≤
m
〉
+
L where L
=
〈
x
1
p
n
,
x
2
p
n
,
…
,
x
m
p
n
〉.
DeclareI
p
m
n to be the polynomial ring S itself.Remark 2.
From the definition of the lexicographic order<
lex on S, it follows easily that for any j
∈
{
0
,
…
,
p
m
n
-
2
}, either X
p
n
-
1
-
U
j
=
X
p
n
-
1
-
U
j
+
1
x
m or there exists k
∈
{
1
,
…
,
m
-
1
} such that X
k
p
n
-
1
-
U
j
=
X
k
p
n
-
1
-
U
j
+
1
x
k, X
p
n
-
1
-
U
j
=
X
k
p
n
-
1
-
U
j, and X
p
n
-
1
-
U
j
+
1
=
X
k
p
n
-
1
-
U
j
+
1
(
x
k
+
1
⋯
x
m
)
p
n
-
1.Notation 1.
For anyj
∈
{
0
,
…
,
p
m
n
-
1
}, let i
(
j
)
≔
max
⁡
{
i
∣
1
≤
i
≤
m
,
B
j
,
i
≠
0
}.Lemma 3.
(i) For anyj
∈
{
0
,
…
,
p
m
n
-
1
}, X
p
n
-
1
-
U
j
∈
I
j
+
1;
(ii)B
j
,
k
=
B
j
+
1
,
k for all j
∈
{
0
,
…
,
p
m
n
-
2
} and for all k
<
i
(
j
+
1
).Proof.
(i) It follows from Remark2 that X
p
n
-
1
-
U
j
=
B
j
+
1
,
i
(
j
+
1
). Hence X
p
n
-
1
-
U
j
∈
I
j
+
1;
(ii) the proof of this follows from Remark2.Lemma 4.
For anyj
∈
{
0
,
…
,
p
m
n
-
1
}, I
j
⊆
I
j
+
1.Proof.
AsI
j
≔
〈
B
j
,
i
∣
1
≤
i
≤
m
〉
+
L (where L
=
〈
x
1
p
n
,
x
2
p
n
,
…
,
x
m
p
n
〉), it is enough to show that B
j
,
i
∈
I
j
+
1 for all j
∈
{
0
,
…
,
p
m
n
-
1
} and for all i
∈
{
1
,
…
,
m
}.
Case  1 (when
X
p
n
-
1
-
U
j
=
X
p
n
-
1
-
U
j
+
1
x
m
). In this case, clearly B
j
,
i
=
B
j
+
1
,
i for all 1
≤
i
≤
m
-
1. And B
j
,
m is either equal to 0 or equal to B
j
+
1
,
m
x
m. Hence in either case, B
j
,
m
∈
I
j
+
1. Therefore, I
j
⊆
I
j
+
1.
Case  2 (when
X
p
n
-
1
-
U
j
≠
X
p
n
-
1
-
U
j
+
1
x
m
). In this case, there exists k
∈
{
1
,
…
,
m
-
1
} such that X
k
p
n
-
1
-
U
j
=
X
k
p
n
-
1
-
U
j
+
1
x
k, X
p
n
-
1
-
U
j
=
X
k
p
n
-
1
-
U
j, and X
p
n
-
1
-
U
j
+
1
=
X
k
p
n
-
1
-
U
j
+
1
(
x
k
+
1
⋯
x
m
)
p
n
-
1.
Clearly then,B
j
,
i
=
B
j
+
1
,
i for all i
<
k. Also, B
j
,
k is either equal to 0 or equal to B
j
+
1
,
k
x
k. Hence in either case, B
j
,
k
∈
I
j
+
1. For i
>
k, B
j
,
i
=
B
j
+
1
,
k
x
i. Hence B
j
,
i
∈
I
j
+
1 for all i
>
k. Therefore, I
j
⊆
I
j
+
1.Notation 2.
LetM be any S-module. Let λ
(
M
) denote thelength of M with respect to S.Lemma 5.
Letj
∈
{
0
,
…
,
p
m
n
-
1
}. Put M
j
+
1
≔
I
j
+
1
/
I
j. Then λ
(
M
j
+
1
)
=
1.Proof.
Fixj. We will show that for each i
∈
{
1
,
…
,
m
}, x
i
I
j
+
1
⊆
I
j. We claim that for any i
,
k
∈
{
1
,
…
,
m
}, x
i
B
j
+
1
,
k
∈
I
j.
Case  1 (when
k
>
i
(
j
+
1
)
). In this case, we have B
j
+
1
,
k
=
0. Hence x
i
B
j
+
1
,
k
∈
I
j for any k
>
i
(
j
+
1
) and for any i
∈
{
1
,
…
,
m
}.
Case  2 (when
k
≤
i
(
j
+
1
)
). It follows from Lemma 3 that B
j
,
k
=
B
j
+
1
,
k for all k
<
i
(
j
+
1
). Therefore B
j
+
1
,
k
∈
I
j for all k
<
i
(
j
+
1
). This implies that x
i
B
j
+
1
,
k
∈
I
j for all k
<
i
(
j
+
1
) and for all i
∈
{
1
,
…
,
m
}.
It remains to show thatx
i
B
j
+
1
,
i
(
j
+
1
)
∈
I
j for all i
∈
{
1
,
…
,
m
}. It follows from the proof of Lemma 3(i) that X
p
n
-
1
-
U
j
=
B
j
+
1
,
i
(
j
+
1
). Hence for any i
∈
{
1
,
…
,
m
}, x
i
B
j
+
1
,
i
(
j
+
1
)
=
x
i
X
p
n
-
1
-
U
j is either a multiple of B
j
,
i or a multiple of x
i
p
n. Therefore x
i
B
j
+
1
,
i
(
j
+
1
)
∈
I
j for all i
∈
{
1
,
…
,
m
}.Lemma 6.
For allj
∈
{
0
,
…
,
p
m
n
-
1
}, I
j
+
1
=
〈
X
p
n
-
1
-
U
j
〉
+
I
j.Proof.
It follows from Lemma3 that X
p
n
-
1
-
U
j
∈
I
j
+
1 for all j
∈
{
0
,
…
,
p
m
n
-
1
}. It also follows from the same lemma that B
j
,
k
=
B
j
+
1
,
k for all j
∈
{
0
,
…
,
p
m
n
-
2
} and for all k
<
i
(
j
+
1
). Therefore the generators B
j
+
1
,
k (where k
≠
i
(
j
+
1
)) of the ideal I
j
+
1 belong to I
j. Due to degree reasons, X
p
n
-
1
-
U
j
=
B
j
+
1
,
i
(
j
+
1
) does not belong to I
j. Therefore X
p
n
-
1
-
U
j is the only generator of the ideal I
j
+
1 which does not belong to I
j. Hence I
j
+
1
=
〈
X
p
n
-
1
-
U
j
〉
+
I
j.It follows from Lemmas4, 5, and 6 above that the following chain
(7)
I
0
⊊
I
1
⊊
⋯
⊊
I
p
m
n
-
1
⊊
I
p
m
n
is a filtration of ideals such that λ
(
I
j
+
1
/
I
j
)
=
1.Therefore reducing modulo the idealJ, it follows that the following chain
(8)
0
=
I
0
+
J
I
0
+
J
⊆
I
1
+
J
I
0
+
J
⊆
I
2
+
J
I
0
+
J
⊆
⋯
⊆
I
p
m
n
+
J
I
0
+
J
=
R
m
[
p
n
]
+
J
is afiltration of ideals, which is having the property thateach successive quotient either iszero or is generated by a single element. The total number of successive quotients of the above mentionedfiltration which are generated by a single element equals the length l
(
R
/
(
m
[
p
n
]
+
J
)
). The following corollary to Lemma 6 above gives a precise account of which successive quotient of the above mentioned filtration is generated by a single element and which successive quotient is zero.Corollary 7.
For anyj
∈
{
0
,
…
,
p
m
n
-
1
}, λ
(
(
I
j
+
1
+
J
)
/
(
I
j
+
J
)
)
≤
1, and λ
(
(
I
j
+
1
+
J
)
/
(
I
j
+
J
)
)
=
0 if and only if X
p
n
-
1
-
U
j
∈
I
j
+
J.Proof.
Fix anyj
∈
{
0
,
…
,
p
m
n
-
1
}, it follows from Lemma 5 that λ
(
I
j
+
1
/
I
j
)
=
1. Therefore, reducing modulo the ideal J, we have λ
(
(
I
j
+
1
+
J
)
/
(
I
j
+
J
)
)
≤
1. Let M
j
+
1 be as in Lemma 5. It follows from Lemma 6 that X
p
n
-
1
-
U
j generates M
j
+
1. Therefore reducing modulo the ideal J, it follows that the quotient (
I
j
+
1
+
J
)
/
(
I
j
+
J
) is zero if and only if X
p
n
-
1
-
U
j
∈
I
j
+
J and is generated by the single element X
p
n
-
1
-
U
j otherwise.Hence for knowing that which successive quotient of the above mentioned filtration is generated by single element and which one is zero, the only condition that needs to be verified is the following:X
p
n
-
1
-
U
j
∈
I
j
+
J or not for every j
∈
{
0
,
…
,
p
m
n
-
1
}?This is thekey checking condition for computing the length l
(
R
/
(
m
[
p
n
]
+
J
)
).


## 3. A Procedure for Doing the Key Check

We are interested in the case when the idealJ is generated by a single polynomial, say J
=
(
f
). In this section, we will first define a term order ⊳ on the set of all monomials in the variables x
1
,
…
,
x
m, and then with respect to ⊳, we will arrange the terms of the polynomial f, and with the help of all this notation, we will describe a way of doing the key checking.


### 3.1. The Term Order⊳

To order the terms of the polynomialf, let us put an order (denote it by ⊳) on the set of all monomials in the variables x
1
,
…
,
x
m as follows.(i)
Setx
1
⊳
⋯
⊳
x
m.
(ii)
On the set of all monomials in the variablesx
1
,
…
,
x
m, ⊳ is the degree lexicographic order with respect to the order ⊳ defined on the variables x
1
,
…
,
x
m. Say, the polynomial f has F many terms. Let us denote the most initial (with respect to ⊳) term of f by [
F
], the next most initial term of f by [
F
-
1
], …, and so on till the least initial term [
1
]. Hence we have
(9)
J
=
f
=
F
+
F
-
1
+
⋯
+
1
.
For example, when m
=
5 and f
=
x
1
3
+
x
2
3
+
x
1
x
2
x
3
+
x
4
x
5, then F
=
4, 4
=
x
1
3, [
3
]
=
x
1
x
2
x
3, [
2
]
=
x
2
3, and [
1
]
=
x
4
x
5.Remark 8.
Note here that the termsF
]
,
[
F
-
1
,
…
,
[
1
], of f are assumed to be containing scalar coefficients.


### 3.2. The Main Theorem

Given any term[
τ
] of the polynomial f, define [
τ
]
-
1
≔
1
/
[
τ
]. Recall the set {
M
:
=
X
p
n
-
1
-
U
j
∣
j
∈
{
0
,
…
,
p
m
n
-
1
}
}. Let A be an arbitrary element of the set M. Say the monomial A chosen above from the set M equals X
p
n
-
1
-
U
j for some j
∈
{
0
,
…
,
p
m
n
-
1
}. The key checking condition for the monomial A says that “A
∈
I
j
+
J or not.” Let us denote by A
c the ideal I
j. We call A
c theideal of convergence corresponding to
A. Given any A
∈
M, we need to check whether or not A
∈
A
c
+
J. Theorem 10 below provides a way for doing this checking.Definition 9.
A monomial in the variablesx
1
,
…
,
x
m of the type
(10)
A
τ
1
-
1
⋯
τ
r
-
1
α
1
⋯
α
r
,
where [
τ
i
] and [
α
i
] are terms of the polynomial f and r is any nonnegative integer, is called acombination monomial in A and f if A
[
τ
1
]
-
1
⋯
[
τ
r
]
-
1
[
α
1
]
⋯
[
α
r
] contains no negative powers of any of the underlying variables x
1
,
…
,
x
m. In particular, the monomial A itself is a combination monomial in A and f.Theorem 10.
Letf
=
[
F
]
+
[
F
-
1
]
+
⋯
+
[
1
]. A
∈
A
c
+
J if and only if one of the following mutually exclusive conditions holds.(i)
The term[
1
] of f divides the monomial A.
(ii)
The term[
1
] of f does not divide the monomial A, but there exists a polynomial h whose terms are of the type c
B
[
τ
]
-
1 where c is a nonzero scalar, B is a combination monomial in A and f such that B
∉
A
c, [
τ
] is a term of f which divides B, such that the product f
·
h equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c where a is a nonzero scalar.Proof.
If condition (i) of the theorem holds, the proof follows easily from the construction of the idealA
c. If condition (ii) of the theorem holds, then it is easy to see that A
∈
A
c
+
J.
Conversely, supposeA
∈
A
c
+
J. Note that due to degree reasons, the monomial A can never belong to the ideal A
c (the proof of this follows easily from the construction of the ideal A
c). Therefore the fact that A
∈
A
c
+
J implies that there exists a polynomial (say g) such that the product f
·
g equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c where a is a nonzero scalar in the ground field k.
Since the productf
·
g contains a term of the type a
A (for some nonzero scalar a), therefore the polynomial g contains finitely many terms of the form c
A
[
τ
]
-
1 where c is a nonzero scalar and [
τ
] is a term of f such that [
τ
] divides A. Say c
1
A
[
τ
1
]
-
1
+
⋯
+
c
l
A
[
τ
l
]
-
1 is the part of the polynomial g which has all its terms of the form c
A
[
τ
]
-
1 where c is some nonzero scalar and [
τ
] is some term of f which divides A. If at least one of the [
τ
i
]’s appearing in the expression c
1
A
[
τ
1
]
-
1
+
⋯
+
c
l
A
[
τ
l
]
-
1 is equal to [
1
], then we can conclude that the term [
1
] divides A and we are done.
If not, then look at the portion (say,g
1) of the polynomial g which consists of sum of all terms of the type c
B
[
τ
]
-
1 where c is a nonzero scalar, B is a combination monomial in A and f such that B
∉
A
c, and [
τ
] is a term of f which divides B. Clearly then g
1 contains the sum c
1
A
[
τ
1
]
-
1
+
⋯
+
c
l
A
[
τ
l
]
-
1. If f
·
g
1 equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c for some nonzero scalar a, then condition (ii) of the theorem holds and with h
=
g
1 and we are done.
So assume that the productf
·
g
1 contains at least one term (say, D
1) which is a combination monomial in A and f not belonging to A
c. Since the product f
·
g equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c for some nonzero scalar a, therefore there exists a nonzero term E in g such that E is not a term of the polynomial g
1 and E
[
τ
]
=
b
D
1 for some nonzero scalar b and some term [
τ
] of f. Hence E
=
b
D
1
[
τ
]
-
1 for some nonzero scalar b and some term [
τ
] of f which divides D
1. But since D
1 is a combination monomial in A and f not belonging to A
c, it follows from the equation E
=
b
D
1
[
τ
]
-
1 that E is a term of the polynomial g
1, hence a contradiction.


## 3.1. The Term Order⊳

To order the terms of the polynomialf, let us put an order (denote it by ⊳) on the set of all monomials in the variables x
1
,
…
,
x
m as follows.(i)
Setx
1
⊳
⋯
⊳
x
m.
(ii)
On the set of all monomials in the variablesx
1
,
…
,
x
m, ⊳ is the degree lexicographic order with respect to the order ⊳ defined on the variables x
1
,
…
,
x
m. Say, the polynomial f has F many terms. Let us denote the most initial (with respect to ⊳) term of f by [
F
], the next most initial term of f by [
F
-
1
], …, and so on till the least initial term [
1
]. Hence we have
(9)
J
=
f
=
F
+
F
-
1
+
⋯
+
1
.
For example, when m
=
5 and f
=
x
1
3
+
x
2
3
+
x
1
x
2
x
3
+
x
4
x
5, then F
=
4, 4
=
x
1
3, [
3
]
=
x
1
x
2
x
3, [
2
]
=
x
2
3, and [
1
]
=
x
4
x
5.Remark 8.
Note here that the termsF
]
,
[
F
-
1
,
…
,
[
1
], of f are assumed to be containing scalar coefficients.


## 3.2. The Main Theorem

Given any term[
τ
] of the polynomial f, define [
τ
]
-
1
≔
1
/
[
τ
]. Recall the set {
M
:
=
X
p
n
-
1
-
U
j
∣
j
∈
{
0
,
…
,
p
m
n
-
1
}
}. Let A be an arbitrary element of the set M. Say the monomial A chosen above from the set M equals X
p
n
-
1
-
U
j for some j
∈
{
0
,
…
,
p
m
n
-
1
}. The key checking condition for the monomial A says that “A
∈
I
j
+
J or not.” Let us denote by A
c the ideal I
j. We call A
c theideal of convergence corresponding to
A. Given any A
∈
M, we need to check whether or not A
∈
A
c
+
J. Theorem 10 below provides a way for doing this checking.Definition 9.
A monomial in the variablesx
1
,
…
,
x
m of the type
(10)
A
τ
1
-
1
⋯
τ
r
-
1
α
1
⋯
α
r
,
where [
τ
i
] and [
α
i
] are terms of the polynomial f and r is any nonnegative integer, is called acombination monomial in A and f if A
[
τ
1
]
-
1
⋯
[
τ
r
]
-
1
[
α
1
]
⋯
[
α
r
] contains no negative powers of any of the underlying variables x
1
,
…
,
x
m. In particular, the monomial A itself is a combination monomial in A and f.Theorem 10.
Letf
=
[
F
]
+
[
F
-
1
]
+
⋯
+
[
1
]. A
∈
A
c
+
J if and only if one of the following mutually exclusive conditions holds.(i)
The term[
1
] of f divides the monomial A.
(ii)
The term[
1
] of f does not divide the monomial A, but there exists a polynomial h whose terms are of the type c
B
[
τ
]
-
1 where c is a nonzero scalar, B is a combination monomial in A and f such that B
∉
A
c, [
τ
] is a term of f which divides B, such that the product f
·
h equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c where a is a nonzero scalar.Proof.
If condition (i) of the theorem holds, the proof follows easily from the construction of the idealA
c. If condition (ii) of the theorem holds, then it is easy to see that A
∈
A
c
+
J.
Conversely, supposeA
∈
A
c
+
J. Note that due to degree reasons, the monomial A can never belong to the ideal A
c (the proof of this follows easily from the construction of the ideal A
c). Therefore the fact that A
∈
A
c
+
J implies that there exists a polynomial (say g) such that the product f
·
g equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c where a is a nonzero scalar in the ground field k.
Since the productf
·
g contains a term of the type a
A (for some nonzero scalar a), therefore the polynomial g contains finitely many terms of the form c
A
[
τ
]
-
1 where c is a nonzero scalar and [
τ
] is a term of f such that [
τ
] divides A. Say c
1
A
[
τ
1
]
-
1
+
⋯
+
c
l
A
[
τ
l
]
-
1 is the part of the polynomial g which has all its terms of the form c
A
[
τ
]
-
1 where c is some nonzero scalar and [
τ
] is some term of f which divides A. If at least one of the [
τ
i
]’s appearing in the expression c
1
A
[
τ
1
]
-
1
+
⋯
+
c
l
A
[
τ
l
]
-
1 is equal to [
1
], then we can conclude that the term [
1
] divides A and we are done.
If not, then look at the portion (say,g
1) of the polynomial g which consists of sum of all terms of the type c
B
[
τ
]
-
1 where c is a nonzero scalar, B is a combination monomial in A and f such that B
∉
A
c, and [
τ
] is a term of f which divides B. Clearly then g
1 contains the sum c
1
A
[
τ
1
]
-
1
+
⋯
+
c
l
A
[
τ
l
]
-
1. If f
·
g
1 equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c for some nonzero scalar a, then condition (ii) of the theorem holds and with h
=
g
1 and we are done.
So assume that the productf
·
g
1 contains at least one term (say, D
1) which is a combination monomial in A and f not belonging to A
c. Since the product f
·
g equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c for some nonzero scalar a, therefore there exists a nonzero term E in g such that E is not a term of the polynomial g
1 and E
[
τ
]
=
b
D
1 for some nonzero scalar b and some term [
τ
] of f. Hence E
=
b
D
1
[
τ
]
-
1 for some nonzero scalar b and some term [
τ
] of f which divides D
1. But since D
1 is a combination monomial in A and f not belonging to A
c, it follows from the equation E
=
b
D
1
[
τ
]
-
1 that E is a term of the polynomial g
1, hence a contradiction.


## 4. The Case of Binomial Hypersurfaces

In this section, we will study the case where the polynomialf contains only 2 terms; that is, f
=
[
2
]
+
[
1
]. The affine variety defined by the ideal J where J
=
(
f
)
=
(
[
2
]
+
[
1
]
) is called a binomial hypersurface.


### 4.1. The Main Theorem for Binomial Hypersurfaces

In this subsection, we will prove a theorem which will reduce the “key checking condition” mentioned above to checking of a combinatorial condition. The theorem which does this job is the following.Theorem 11.
A
∈
A
c
+
J if and only if either [
1
] divides A or there exists a positive integer M for which A
[
2
]
-
M
[
1
]
M
-
1 contains no negative powers of any of the underlying variables and A
[
2
]
-
M
[
1
]
M belongs to A
c. (Note here that by the notation [
2
]
-
M, we mean [
2
]
-
1 multiplied M times and similarly for the notation [
1
]
M.)Proof.
If[
1
] divides A, then it follows from Theorem 10 above that A
∈
A
c
+
J.
Suppose there exists a positive integerM for which A
[
2
]
-
M
[
1
]
M
-
1 has no negative powers and A
[
2
]
-
M
[
1
]
M belongs to A
c. Let M
0 denote the least positive integer M for which this happens. Then the polynomial g given by
(11)
a
1
A
2
-
1
+
a
2
A
2
-
2
1
1
+
⋯
+
a
M
0
A
2
-
M
0
1
M
0
-
1
,
where a
1 is a nonzero scalar in the ground field k and a
i
+
a
i
+
1
=
0 (mod
⁡
p) for each i
∈
{
1
,
…
,
M
0
-
1
}, is having the property that the product f
·
g equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c where a is a nonzero scalar. Hence A
∈
A
c
+
J. We now need to prove the other way round.
SupposeA
∈
A
c
+
J and [
1
] does not divide A. Then by Theorem 10, there exists a polynomial h whose terms are of the type c
B
[
τ
]
-
1 where c is a nonzero scalar, B is a combination monomial in A and f not belonging to A
c, and [
τ
] is a term of f which divides B, such that the product f
·
h equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c where a is a nonzero scalar.
It is easy to see that for the case of a binomial hypersurface, any combination monomial inA and f is either equal to A itself or it is of the type A
[
2
]
-
N
[
1
]
N for some positive integer N or it is of the type A
[
1
]
-
N
[
2
]
N for some positive integer N. It is also elementary to observe that any combination monomial in A and f of the type A
[
1
]
-
N
[
2
]
N always belongs to A
c. So any combination monomial not belonging to A
c is either equal to A itself or of the type A
[
2
]
-
N
[
1
]
N for some positive integer N. It hence follows that the terms of the polynomial h are either of the type A
[
2
]
-
N
[
1
]
N
-
1 for some positive integer N or of the type A
[
1
]
-
1. But we have assumed that the term [
1
] does not divide A, so A
[
1
]
-
1 cannot be a term of the polynomial h. Hence the terms of the polynomial h are only of the type A
[
2
]
-
N
[
1
]
N
-
1 for some positive integer N.
LetN
0 denote the largest positive integer N for which A
[
2
]
-
N
[
1
]
N
-
1 is a term of the polynomial h. Clearly then A
[
2
]
-
N
0
[
1
]
N
0
-
1 has no negative powers of any of the underlying variables x
1
,
…
,
x
m. Moreover, since the product f
·
h equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c (where a is a nonzero scalar), it follows that A
[
2
]
-
N
0
[
1
]
N
0 belongs to A
c.


### 4.2. A Formula for the Hilbert-Kunz Function for Binomial Hypersurfaces

In this subsection, we will give a closed form iterative formula for computing the Hilbert-Kunz function for any binomial hypersurface in general, over any field of positive characteristic. The notation and terminology remain the same as in the previous part of this paper.


#### 4.2.1. Notation

Recall from Section3.1 the variables x
1
,
…
,
x
m and the term order ⊳ on the set of all monomials in these variables. Let f
=
[
2
]
+
[
1
] be the polynomial defining the binomial hypersurface where [
2
] and [
1
] are defined as in Section 3.1.For anyi
∈
{
1
,
…
,
m
}, let
(12)
Δ
i
≔
the
power
of
x
i
in
the
term
1
-
the
power
of
x
i
in
the
term
[
2
]
.
Note that some of the Δ
i’s can be negative, some can be positive, and some can be 0. Without loss of generality, we can assume that Δ
1
≤
⋯
≤
Δ
m.Letr be the integer such that 0
≤
r
≤
m and, the ordered set {
x
1
⊳
⋯
⊳
x
r
} equals the set {
x
i
∣
1
≤
i
≤
m
,
Δ
i
<
0
}. For i
∈
{
1
,
…
,
r
}, let N
i
≔
x
r
+
1
-
i and Δ
N
i
≔
Δ
r
+
1
-
i. Similarly, let s denote the number of x
i’s for which Δ
i
=
0 (note that s can also be 0). Let x
r
+
1
⊳
⋯
⊳
x
r
+
s be the elements of the set {
x
i
∣
1
≤
i
≤
m
,
Δ
i
=
0
}. For i
=
{
1
,
…
,
s
}, let Z
i
≔
x
r
+
s
+
1
-
i and Δ
Z
i
≔
Δ
r
+
s
+
1
-
i. Finally, let t
≔
m
-
(
r
+
s
) and let x
r
+
s
+
1
⊳
⋯
⊳
x
m be the elements of the set {
x
i
∣
1
≤
i
≤
m
,
Δ
i
>
0
}. Let us denote by P
1
⊳
⋯
⊳
P
t the ordered set x
r
+
s
+
1
⊳
⋯
⊳
x
m and for i
=
{
1
,
…
,
t
}, let Δ
P
i
≔
Δ
r
+
s
+
i. In other words, we can say that the ordered set x
1
⊳
⋯
⊳
x
m is the same as the ordered set N
r
⊳
⋯
⊳
N
1
⊳
Z
s
⊳
⋯
⊳
Z
1
⊳
P
1
⊳
⋯
⊳
P
t. We call N
i’s thenegative difference variables, Z
i’s thezero difference variables, and P
i’s thepositive difference variables. For any i
∈
{
1
,
…
,
r
}, let N
i
,
min
⁡ and N
i
,
max
⁡ denote the minimum and the maximum powers, respectively, of the variable N
i that appears in the expression of the polynomial f. We can similarly define P
q
,
min
⁡, P
q
,
max
⁡, Z
j
,
min
⁡, and Z
j
,
max
⁡ for all q
∈
{
1
,
…
,
t
} and for all j
∈
{
1
,
…
,
s
}. Observe that since each Z
j is a zero difference variable, we have Z
j
,
min
⁡
=
Z
j
,
max
⁡ for all j
∈
{
1
,
…
,
s
}.For anyi
∈
{
1
,
…
,
r
}, let a
i
≔
-
(
Δ
N
i
) and b
i
≔
N
i
,
min
⁡. For any i
∈
{
1
,
…
,
r
} and any positive integer n, let M
a
i
,
n and r
a
i
,
n be defined by the equation p
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n.


#### 4.2.2. The NumberM
max,
A

Definition 12 (ofM
max
⁡
,
A).
LetA be an arbitrary element of the set M (this set has been defined earlier) such that [7] divides A. It follows from Theorem 11 that for those monomials A which are divisible by [7], we need to keep an account of the largest positive integer M for which the monomial A
[
2
]
-
M
[
1
]
M
-
1 has no negative power for any of the variables x
1
,
…
,
x
m. Let M
max
⁡
,
A denote this largest positive integer.


#### 4.2.3. An Explicit Account ofM
max,
A

Corollary15 below gives an explicit account of the number M
max
⁡
,
A, for any monomial A which is divisible by [7]. The number M
max
⁡
,
A is important to us because of Theorem 11 which provides us with a way to count the required length l
(
R
/
(
m
[
p
n
]
+
J
)
).Remark 13.
LetA be any monomial in M which [
2
] divides; then it is of the form
(13)
N
r
p
n
-
k
N
r
⋯
N
1
p
n
-
k
N
1
Z
s
p
n
-
k
Z
s
⋯
Z
1
p
n
-
k
Z
1
P
1
j
1
⋯
P
t
j
t
,
where p
n
-
1
≥
j
q
≥
P
q
,
min
⁡ for any q
∈
{
1
,
…
,
t
}, p
n
-
1
≥
p
n
-
k
N
i
≥
N
i
,
max
⁡ for every 1
≤
i
≤
r, and p
n
-
1
≥
p
n
-
k
Z
i
≥
Z
i
,
max
⁡ for every 1
≤
i
≤
s.
Hence for any positive integerM, A
[
2
]
-
M
[
1
]
M
-
1 is of the form
(14)
∏
i
=
1
r
N
i
p
n
-
k
N
i
-
a
i
M
-
b
i
∏
i
=
1
s
Z
i
p
n
-
k
Z
i
-
Z
i
,
min
⁡
∏
q
=
1
t
P
q
j
q
+
M
Δ
P
q
-
P
q
,
max
⁡
,
where p
n
-
1
≥
j
q
≥
P
q
,
min
⁡ for any q
∈
{
1
,
…
,
t
}, p
n
-
1
≥
p
n
-
k
N
i
≥
N
i
,
max
⁡ for every 1
≤
i
≤
r, and p
n
-
1
≥
p
n
-
k
Z
i
≥
Z
i
,
max
⁡ for every 1
≤
i
≤
s.
It is now easy to see that for anyM
≥
1, the power of any zero difference variable or any positive difference variable in A
[
2
]
-
M
[
1
]
M
-
1 is ≥
0. But the power of any negative difference variable in A
[
2
]
-
M
[
1
]
M
-
1 may be negative for some values of M.The following lemma and its corollary give an explicit account ofM
max
⁡
,
A for any monomial A in M which [
2
] divides. But for stating the lemma, we need some notation. Let i
∈
{
1
,
…
,
r
} and let n be any positive integer. If b
i
≥
a
i, then let y
a
i denote the least positive integer ≥
(
b
i
-
r
a
i
,
n
) which is divisible by a
i. Let q
a
i be defined by the equation y
a
i
=
a
i
q
a
i. Let
(15)
p
a
i
≔
0
if
b
i
<
a
i
and
r
a
i
,
n
≥
b
i
,
1
if
b
i
<
a
i
and
r
a
i
,
n
<
b
i
,
q
a
i
if
b
i
≥
a
i
clearly
here
r
a
i
,
n
<
b
i
.
And let
(16)
E
a
i
,
b
i
≔
r
a
i
,
n
-
b
i
if
p
a
i
=
0
,
(
r
a
i
,
n
-
b
i
)
+
a
i
if
p
a
i
=
1
,
y
a
i
-
(
b
i
-
r
a
i
,
n
)
if
p
a
i
=
q
a
i
.
For any real number β, let β denote the smallest positive integer ≥β. For any 1
≤
i
≤
r, let
(17)
k
~
N
i
,
last
≔
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
a
i
.Lemma 14.
For any1
≤
i
≤
r, if k
N
i is such that 1
≤
k
N
i
≤
p
n
-
N
i
,
max
⁡, then the maximum value of the positive integer M for which p
n
-
k
N
i
-
a
i
M
-
b
i is ≥0 equals(18)
M
a
i
,
n
-
p
a
i
i
f
1
≤
k
N
i
≤
1
+
E
a
i
,
b
i
M
a
i
,
n
-
p
a
i
-
1
i
f
1
+
E
a
i
,
b
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
M
a
i
,
n
-
p
a
i
-
2
i
f
1
+
E
a
i
,
b
i
+
a
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
2
a
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
i
f
1
+
E
a
i
,
b
i
+
a
i
(
k
~
N
i
-
1
)
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
l
a
s
t
i
f
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
l
a
s
t
-
1
<
k
N
i
≤
p
n
-
N
i
,
max
⁡
.Proof.
Observe that sincep
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n, we have p
n
-
k
N
i
-
a
i
M
-
b
i
≥
0 if and only if M
≤
M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
). So we need to compute the greatest positive integer M, that is, ≤M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
).
Case  1 (when  b
i
≥
a
i
). In this case, we have 0
≤
r
a
i
,
n
<
a
i
≤
b
i. The least positive integer ≥
b
i
-
r
a
i
,
n which is divisible by a
i is y
a
i. Therefore the least positive integer ≥
(
b
i
-
r
a
i
,
n
)
/
a
i is q
a
i
=
p
a
i. Therefore the greatest negative integer ≤
-
(
b
i
-
r
a
i
,
n
)
/
a
i is -
p
a
i. And hence the greatest positive integer ≤
M
a
i
,
n
-
(
b
i
-
r
a
i
,
n
)
/
a
i is M
a
i
,
n
-
p
a
i.
Sincek
N
i
≤
p
n
-
N
i
,
max
⁡, it follows that M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
)
≥
M
a
i
,
n
-
p
a
i
-
(
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i
). The smallest positive integer ≥
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i is k
~
N
i
,
last. Therefore the greatest positive integer ≤
M
a
i
,
n
-
p
a
i
-
(
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i
) is M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last. As k
N
i increases, M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
) decreases, the lower bound being equal to M
a
i
,
n
-
p
a
i
-
(
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i
). Therefore the greatest positive integer ≤
M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
) equals M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last.
Hence the greatest positive integer≤
C
i
,
n
≔
M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
) is equal to(19)
M
a
i
,
n
-
p
a
i
if
M
a
i
,
n
-
p
a
i
≤
C
i
,
n
M
a
i
,
n
-
p
a
i
-
1
if
M
a
i
,
n
-
p
a
i
-
1
≤
C
i
,
n
<
M
a
i
,
n
-
p
a
i
M
a
i
,
n
-
p
a
i
-
2
if
M
a
i
,
n
-
p
a
i
-
2
≤
C
i
,
n
<
M
a
i
,
n
-
p
a
i
-
1
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
if
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
≤
C
i
,
n
<
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
+
1
.
This is equivalent to saying that the maximum value of the positive integerM for which p
n
-
k
N
i
-
a
i
M
-
b
i is ≥0 equals(20)
M
a
i
,
n
-
p
a
i
if
1
≤
k
N
i
≤
1
+
E
a
i
,
b
i
M
a
i
,
n
-
p
a
i
-
1
if
1
+
E
a
i
,
b
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
M
a
i
,
n
-
p
a
i
-
2
if
1
+
E
a
i
,
b
i
+
a
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
2
a
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
if
1
+
E
a
i
,
b
i
+
a
i
(
k
~
N
i
-
1
)
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
if
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
last
-
1
<
k
N
i
≤
p
n
-
N
i
,
max
⁡
.
Case  2 (when
b
i
<
a
i
). In this case, we have 2 subcases.
Subcase  2.1 (when
b
i
≤
r
a
i
,
n
). Here we have p
a
i
=
0 and 0
≤
b
i
≤
r
a
i
,
n
<
a
i. Therefore 0
≤
(
r
a
i
,
n
-
b
i
)
/
a
i
<
1 and hence -
1
<
(
b
i
-
r
a
i
,
n
)
/
a
i
≤
0. Therefore the least nonnegative integer ≥
(
b
i
-
r
a
i
,
n
)
/
a
i is p
a
i
=
0. The rest of the proof is similar to the proof of Case  1.
Subcase  2.2 (when
b
i
>
r
a
i
,
n
). Here we have, p
a
i
=
1 and 0
≤
r
a
i
,
n
<
b
i
<
a
i. Therefore 0
<
(
b
i
-
r
a
i
,
n
)
/
a
i
<
1. Hence the least positive integer ≥
(
b
i
-
r
a
i
,
n
)
/
a
i is p
a
i
=
1. The rest of the proof is similar to the proof of Case  1.The following corollary is immediate.Corollary 15.
LetA be a monomial in M which [
2
] divides and let M be a positive integer. Let A and A
[
2
]
-
M
[
1
]
M
-
1 be of the form as given in Remark 13. For 1
≤
i
≤
r, let M
i
,
A be the maximum value of the positive integer M for which p
n
-
k
N
i
-
a
i
M
-
b
i is ≥
0.
ThenM
max
⁡
,
A equals the minimum of all M
i
,
As as i ranges from 1 to r.LetM
[
2
]
≔ the set of all monomials in M which are divisible by [
2
]. Let M
max
⁡
[
2
]
≔
{
M
max
⁡
,
A
∣
A
∈
M
[
2
]
}.Remark 16.
It follows from Corollary15 above that M
max
⁡
[
2
]
=
{
M
a
r
,
n
-
p
a
r
-
j
∣
0
≤
j
≤
k
~
N
r
,
last
}.


#### 4.2.4. The Length Count for a FixedM
max,
A

Remark 17.
It is easy to verify that ifM
1 is a positive integer such that A
[
2
]
-
M
1
[
1
]
M
1
-
1 contains no negative powers and M
2 is a positive integer such that M
2
≤
M
1, then A
[
2
]
-
M
2
[
1
]
M
2
-
1 contains no negative powers. It is also easy to verify that if N
1 is a positive integer such that A
[
2
]
-
N
1
[
1
]
N
1 belongs to A
c and N
2 is a positive integer such that N
1
≤
N
2, then A
[
2
]
-
N
2
[
1
]
N
2 belongs to A
c. Hence for any A
∈
M
[
2
], A
∈
A
c
+
J if and only if either [
1
] divides A or A
[
2
]
-
M
max
⁡
,
A
[
1
]
M
max
⁡
,
A belongs to A
c.Remark 18.
Recall from Remark13 as well as from Corollary 15 that given any monomial A in M
[
2
], the value of M
max
⁡
,
A depends only on the powers of the negative difference variables appearing in A. It now follows that for each i
∈
{
1
,
…
,
r
}, if we fix numbers k
N
i
0 such that p
n
-
1
≥
p
n
-
k
N
i
0
≥
N
i
,
max
⁡, then for any A in the set
(21)
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
≔
A
∣
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
where
p
n
-
1
≥
l
j
≥
0
for
every
j
∈
1
,
…
,
s
and
,
N
r
p
n
-
k
N
r
0
p
n
-
1
≥
j
q
≥
0
for
every
q
∈
1
,
…
,
t
,
the value of M
max
⁡
,
A is the same (call it M
max
⁡
0).The main result of this subsubsection is Lemma24 below which provides an account of the length count for a fixed value of M
max
⁡
,
A. But for stating the lemma, we need to introduce some notation first.For an arbitrary elementM
max
⁡ in M
max
⁡
[
2
] and for every q
∈
{
1
,
…
,
t
}, let
(22)
Mi
n
P
q
,
M
max
⁡
c
≔
Min
⁡
p
n
-
P
q
,
min
⁡
,
M
max
⁡
Δ
P
q
.
The symbol Mi
n
c in the above definition of Mi
n
P
q
,
M
max
⁡
c stands for “minimum convergent.” The reason behind this notation is explained in Remark 19 below.Remark 19.
Observe that if the power of the variableP
q in the monomial A is j
q, then the power of P
q in the monomial A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ is j
q
+
M
max
⁡
Δ
P
q. If A is divisible by [
2
], then p
n
-
1
≥
j
q
≥
P
q
,
min
⁡. Now if j
q
+
M
max
⁡
Δ
P
q
≥
p
n, then A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c. That is, if p
n
-
j
q
≤
Mi
n
P
q
,
M
max
⁡
c, then A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c. That is, if p
n
-
Mi
n
P
q
,
M
max
⁡
c
≤
j
q
≤
p
n
-
1, then A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c. So p
n
-
Mi
n
P
q
,
M
max
⁡
c is the minimum possible power of the variable P
q in the monomial A for which A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c.Recall the ordered setP
1
⊳
⋯
⊳
P
t of positive difference variables. Let A
t
+
1
≔
0 and C
t
≔
1. For any q
∈
{
1
,
…
,
t
-
1
}, let
(23)
C
q
≔
p
(
t
-
q
)
n
.
And for any q
∈
{
2
,
…
,
t
}, let
(24)
A
q
≔
p
n
-
P
q
,
min
⁡
A
q
+
1
+
P
q
,
min
⁡
C
q
.
For any M
max
⁡ in M
max
⁡
[
2
], let D
t
+
1
,
M
max
⁡
:
=
1 and D
~
t
+
1
,
M
max
⁡
≔
0. For any M
max
⁡ in M
max
⁡
[
2
] and any q
∈
{
1
,
…
,
t
}, let
(25)
D
q
,
M
max
⁡
≔
Mi
n
P
q
,
M
max
⁡
c
A
q
+
1
+
p
n
-
P
q
,
min
⁡
-
Mi
n
P
q
,
M
max
⁡
c
D
q
+
1
,
M
max
⁡
+
P
q
,
min
⁡
C
q
,
(26)
D
~
q
,
M
max
⁡
≔
Mi
n
P
q
,
M
max
⁡
c
A
q
+
1
+
(
p
n
-
Mi
n
P
q
,
M
max
⁡
c
-
P
q
,
max
⁡
)
D
~
q
+
1
,
M
max
⁡
+
P
q
,
max
⁡
-
P
q
,
min
⁡
D
q
+
1
,
M
max
⁡
+
(
P
q
,
min
⁡
)
C
q
if
Mi
n
P
q
,
M
max
⁡
c
<
p
n
-
P
q
,
min
⁡
and
P
q
,
max
⁡
<
p
n
-
Mi
n
P
q
,
M
max
⁡
c
D
q
,
M
max
⁡
if
Mi
n
P
q
,
M
max
⁡
c
<
p
n
-
P
q
,
min
⁡
and
P
q
,
max
⁡
≥
p
n
-
Mi
n
P
q
,
M
max
⁡
c
(
Mi
n
P
q
,
M
max
⁡
c
)
A
q
+
1
+
(
P
q
,
min
⁡
)
C
q
if
Mi
n
P
q
,
M
max
⁡
c
=
p
n
-
P
q
,
min
⁡
.
Recall the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) from Remark 18. For any A in this set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
), the value of M
max
⁡
,
A is the same, namely, M
max
⁡
0. Lemma 24 below gives an account of the length count for the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
). But we will break up the proof of Lemma 24 in a few short steps as follows.Lemma 20.
For everyq
∈
{
2
,
…
,
t
}, let
(27)
F
q
,
A
,
J
1
,
…
,
J
q
-
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
)
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
a
n
d
j
1
=
J
1
,
…
,
j
q
-
1
=
J
q
-
1
,
where for each k
∈
{
1
,
…
,
q
-
1
}, J
k is a fixed integer such that P
k
,
min
⁡
≤
J
k
≤
p
n
-
1 and at least one J
k is ≥
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c.
Then the total number of monomialsA in the set F
q
,
A
,
J
1
,
…
,
J
q
-
1 for which A
∉
A
c
+
J equals A
q.Proof.
The proof is by reverse induction onq, the base case of induction being q
=
t. Observe that if A is any monomial in F
t
,
A
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
A
,
J
1
,
…
,
J
t
-
1 is p
n
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
A
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
A
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
A
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(28)
p
n
-
P
t
,
min
⁡
0
+
P
t
,
min
⁡
1
=
p
n
-
P
t
,
min
⁡
A
t
+
1
+
P
t
,
min
⁡
C
t
=
A
t
.
This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.Lemma 21.
For everyq
∈
{
2
,
…
,
t
}, let
(29)
F
q
,
C
,
J
1
,
…
,
J
q
-
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
)
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
,
j
1
=
J
1
,
…
,
j
q
-
1
=
J
q
-
1
,
where for each k
∈
{
1
,
…
,
q
-
1
}, J
k is a fixed integer such that 0
≤
J
k
≤
p
n
-
1 and at least one J
k is <
P
k
,
min
⁡.
Then the total number of monomialsA in the set F
q
,
C
,
J
1
,
…
,
J
q
-
1 for which A
∉
A
c
+
J equals C
q
-
1.Proof.
The proof is by reverse induction onq, the base case of induction being q
=
t. Observe that if A is any monomial in F
t
,
C
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
≤
p
n
-
1, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
C
,
J
1
,
…
,
J
t
-
1 is p
n. Hence the total number of monomials A in the set F
t
,
C
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals p
n
=
p
(
t
-
(
t
-
1
)
)
n
=
C
t
-
1. This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.Lemma 22.
For everyq
∈
{
2
,
…
,
t
}, let
(30)
F
q
,
D
,
J
1
,
…
,
J
q
-
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
)
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
a
n
d
j
1
=
J
1
,
…
,
j
q
-
1
=
J
q
-
1
,
where for each k
∈
{
1
,
…
,
q
-
1
}, J
k is a fixed integer such that P
k
,
min
⁡
≤
J
k
<
p
n
-
Min
P
k
,
M
max
⁡
0
c.
Then the total number of monomialsA in the set F
q
,
D
,
J
1
,
…
,
J
q
-
1 for which A
∉
A
c
+
J equals (i)
D
q
,
M
max
⁡
0 if there exists at least one k
∈
{
1
,
…
,
q
-
1
} such that P
k
,
max
⁡
≥
p
n
-
Min
P
k
,
M
max
⁡
0
c;
(ii)
D
~
q
,
M
max
⁡
0 if for every k
∈
{
1
,
…
,
q
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Min
P
k
,
M
max
⁡
0
c and P
k
,
max
⁡
≤
J
k
<
p
n
-
Min
P
k
,
M
max
⁡
0
c;
(iii)
D
q
,
M
max
⁡
0 if for every k
∈
{
1
,
…
,
q
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Min
P
k
,
M
max
⁡
0
c and there exists at least one k
∈
{
1
,
…
,
q
-
1
} such that P
k
,
min
⁡
≤
J
k
<
P
k
,
max
⁡.Proof.
The proof is by reverse induction onq, the base case of induction being q
=
t. Let us first prove the base case of induction.
Case  1. When there exists at least one k
∈
{
1
,
…
,
t
-
1
} such that P
k
,
max
⁡
≥
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c.
Then since we have that for eachk
∈
{
1
,
…
,
t
-
1
}, J
k is a fixed integer such that P
k
,
min
⁡
≤
J
k
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c, it follows that monomials in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 are not divisible by [
1
].
Observe now that ifA is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then [
2
] divides A but [
1
] does not divide A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∉
A
c. Hence such a monomial A satisfies A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(31)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
1
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
D
t
+
1
,
M
max
⁡
0
+
P
t
,
min
⁡
C
t
=
D
t
,
M
max
⁡
0
.
This proves the base case of induction in Case  1.
Case  2. When for every k
∈
{
1
,
…
,
t
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c and P
k
,
max
⁡
≤
J
k
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c. In this case, there are 3 subcases.
Subcase  2.1 (
Mi
n
P
t
,
M
max
⁡
0
c
=
p
n
-
P
t
,
min
⁡
). Observe that if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡
=
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(32)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
P
t
,
min
⁡
C
t
=
D
~
t
,
M
max
⁡
0
.
This proves the base case of induction in this subcase.
Subcase  2.2 (
Mi
n
P
t
,
M
max
⁡
0
c
<
p
n
-
P
t
,
min
⁡
and
P
t
,
max
⁡
≥
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
). Observe that if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then [
2
] divides A but [
1
] does not divide A (since P
t
,
max
⁡
≥
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c) and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∉
A
c. Hence such a monomial A satisfies A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(33)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
1
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
D
t
+
1
,
M
max
⁡
0
+
P
t
,
min
⁡
C
t
=
D
t
,
M
max
⁡
0
=
D
~
t
,
M
max
⁡
0
.
This proves the base case of induction in this subcase.
Subcase  2.3 (
Mi
n
P
t
,
M
max
⁡
0
c
<
p
n
-
P
t
,
min
⁡
and
P
t
,
max
⁡
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
). Observe that if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
max
⁡
≤
j
t
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then [
1
] divides A (since P
t
,
max
⁡
≤
j
t and P
k
,
max
⁡
≤
J
k
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c for every k
∈
{
1
,
…
,
t
-
1
}). Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
max
⁡. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
<
P
t
,
max
⁡, then [
2
] divides A but [
1
] does not divide A (since j
t
<
P
t
,
max
⁡) and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∉
A
c (since j
t
<
P
t
,
max
⁡
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c). Hence such a monomial A satisfies A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
max
⁡
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(34)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
max
⁡
0
+
P
t
,
max
⁡
-
P
t
,
min
⁡
1
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
max
⁡
×
D
~
t
+
1
,
M
max
⁡
0
+
P
t
,
max
⁡
-
P
t
,
min
⁡
D
t
+
1
,
M
max
⁡
0
+
(
P
t
,
min
⁡
)
C
t
=
D
~
t
,
M
max
⁡
0
.
This proves the base case of induction in this subcase. Hence the base case of induction is proved in Case  2.
Case  3. When for every k
∈
{
1
,
…
,
t
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c and there exists at least one k
∈
{
1
,
…
,
t
-
1
} such that P
k
,
min
⁡
≤
J
k
<
P
k
,
max
⁡.
Since there exists at least onek
∈
{
1
,
…
,
t
-
1
} such that P
k
,
min
⁡
≤
J
k
<
P
k
,
max
⁡, it follows that monomials in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 are not divisible by [
1
]. The rest of the proof of this case is similar to the proof of Case  1. This proves the base case of induction in Case  3.
This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.Let(35)
F
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
.Lemma 23.
The total number of monomialsA in the set F
1 for which A
∉
A
c
+
J equals D
~
1
,
M
max
⁡
0.Proof.
LetJ
1 be an arbitrarily fixed integer such that p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
≤
J
1
≤
p
n
-
1. It then follows from Lemma 20 that for any such fixed J
1, the total number of monomials A in F
2
,
A
,
J
1 such that A
∉
A
c
+
J equals A
2. Also if J
1 is an arbitrarily fixed integer such that 0
≤
J
1
<
P
1
,
min
⁡, then it follows from Lemma 21 that for any such fixed J
1, the total number of monomials A in F
2
,
C
,
J
1 such that A
∉
A
c
+
J equals C
1. Now there are 3 cases.
Case  1 (if
Mi
n
P
1
,
M
max
⁡
0
c
=
p
n
-
P
1
,
min
⁡
). In this case, the total number of monomials A in the set F
1 for which A
∉
A
c
+
J equals
(36)
p
n
-
1
-
p
n
+
Mi
n
P
1
,
M
max
⁡
0
c
+
1
A
2
+
P
1
,
min
⁡
C
1
=
Mi
n
P
1
,
M
max
⁡
0
c
A
2
+
P
1
,
min
⁡
C
1
=
D
~
1
,
M
max
⁡
0
.
Case  2 (if
Mi
n
P
1
,
M
max
⁡
0
c
<
p
n
-
P
1
,
min
⁡
and
P
1
,
max
⁡
≥
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
). It follows from Lemma 22 that for any fixed integer J
1 such that P
1
,
min
⁡
≤
J
1
<
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c, the total number of monomials A in F
2
,
D
,
J
1 for which A
∉
A
c
+
J equals D
2
,
M
max
⁡
0. Hence in this case, the total number of monomials A in the set F
1 for which A
∉
A
c
+
J equals
(37)
p
n
-
1
-
p
n
+
Mi
n
P
1
,
M
max
⁡
0
c
+
1
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
Mi
n
P
1
,
M
max
⁡
0
c
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
D
1
,
M
max
⁡
0
=
D
~
1
,
M
max
⁡
0
.
Case  3 (if
Mi
n
P
1
,
M
max
⁡
0
c
<
p
n
-
P
1
,
min
⁡
and
P
1
,
max
⁡
<
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
). It follows from Lemma 22 that for any fixed integer J
1 such that P
1
,
max
⁡
≤
J
1
<
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c, the total number of monomials A in F
2
,
D
,
J
1 for which A
∉
A
c
+
J equals D
~
2
,
M
max
⁡
0. It again follows from Lemma 22 that for any fixed integer J
1 such that P
1
,
min
⁡
≤
J
1
<
P
1
,
max
⁡, the total number of monomials A in F
2
,
D
,
J
1 for which A
∉
A
c
+
J equals D
2
,
M
max
⁡
0. Hence in this case, the total number of monomials A in the set F
1 for which A
∉
A
c
+
J equals
(38)
p
n
-
1
-
p
n
+
Mi
n
P
1
,
M
max
⁡
0
c
+
1
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
max
⁡
D
~
2
,
M
max
⁡
0
+
P
1
,
max
⁡
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
Mi
n
P
1
,
M
max
⁡
0
c
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
max
⁡
D
~
2
,
M
max
⁡
0
+
P
1
,
max
⁡
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
D
~
1
,
M
max
⁡
0
.Lemma 24.
The total number of monomialsA in the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
s where S
M
max
⁡
0
,
s is defined inductively as follows:
(39)
S
M
max
⁡
0
,
0
≔
D
~
1
,
M
max
⁡
0
,
f
o
r
e
v
e
r
y
j
∈
1
,
…
,
s
,
S
M
max
⁡
0
,
j
≔
S
M
max
⁡
0
,
j
-
1
p
n
-
Z
j
,
min
⁡
+
p
t
+
j
-
1
n
Z
j
,
min
⁡
.Proof.
Fix an arbitrary integerL
1 such that p
n
-
1
≥
L
1
≥
Z
1
,
min
⁡. Let
(40)
P
L
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
2
=
p
n
-
1
,
l
1
=
L
1
.
It is now easy to see that the total number of monomials A in P
L
1 for which A
∉
A
c
+
J equals D
~
1
,
M
max
⁡
0
=
S
M
max
⁡
0
,
0 (the proof of this is similar to the proof of Lemma 23).
Fix an arbitrary integerM
1 such that 0
≤
M
1
<
Z
1
,
min
⁡. Let
(41)
P
M
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
2
=
p
n
-
1
,
l
1
=
M
1
.
Since 0
≤
M
1
<
Z
1
,
min
⁡, therefore any monomial A
∈
P
M
1 is divisible neither by [
1
] nor by [
2
]. Hence the total number of monomials A in P
M
1 for which A
∉
A
c
+
J equals p
t
n.
So if we defineP
1 as the subset of P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) consisting of all those monomials in P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) for which l
s
=
l
s
-
1
=
⋯
=
l
2
=
p
n
-
1, then the total number of monomials A in the set P
1 for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
1. Similarly, if we define P
2 as the subset of P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) consisting of all those monomials in it for which l
s
=
l
s
-
1
=
⋯
=
l
3
=
p
n
-
1, then the total number of monomials A in the set P
2 for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
2. Proceeding inductively, it is now easy to see that the total number of monomials A in the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
s.Definition 25.
Given anyM
max
⁡ in M
max
⁡
[
2
], let Φ
M
max
⁡
≔
S
M
max
⁡
,
s where S
M
max
⁡
,
s is defined in the same way as S
M
max
⁡
0
,
s is defined in Lemma 24 (just replace M
max
⁡
0 in Lemma 24 by M
max
⁡).


#### 4.2.5. The Formula for the Hilbert-Kunz Function

We will now state the formula for the Hilbert-Kunz function for the general binomial hypersurface that we have considered above. But for stating the formula, we need some more notation, which we provide first.SetG
P
(
t
-
1
)
≔
P
t
,
max
⁡ and for every q
∈
{
1
,
…
,
(
t
-
2
)
}, let G
P
q
≔
(
p
n
-
P
q
+
1
,
max
⁡
)
G
P
q
+
1
+
(
P
q
+
1
,
max
⁡
)
p
n
(
t
-
q
-
1
);
SetG
Z
1
≔
(
p
n
-
P
1
,
max
⁡
)
G
P
1
+
(
P
1
,
max
⁡
)
p
n
(
t
-
1
), and for every l
∈
{
2
,
…
,
s
}, let G
Z
l
≔
(
p
n
-
Z
l
-
1
,
max
⁡
)
G
Z
l
-
1
+
(
Z
l
-
1
,
max
⁡
)
p
n
(
t
+
l
-
2
);
SetG
N
1
≔
(
p
n
-
Z
s
,
max
⁡
)
G
Z
s
+
(
Z
s
,
max
⁡
)
p
n
(
t
+
s
-
1
), and for every i
∈
{
2
,
…
,
r
}, let G
N
i
≔
(
p
n
-
N
i
-
1
,
min
⁡
)
G
N
i
-
1
+
(
N
i
-
1
,
min
⁡
)
p
n
(
t
+
s
+
i
-
2
). Recall that for any i
∈
{
1
,
…
,
r
}, let a
i
≔
-
(
Δ
N
i
) and for any i
∈
{
1
,
…
,
r
} and any positive integer n, M
a
i
,
n and r
a
i
,
n are defined by the equation p
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n. Also recall the definitions of E
a
i
,
b
i and p
a
i from Section 4.2.3.For everyi
∈
{
1
,
…
,
r
}, let
(42)
D
a
i
,
a
r
≔
1
+
E
a
i
,
b
i
+
a
i
M
a
i
,
n
-
p
a
i
-
M
a
r
,
n
+
p
a
r
a
i
′
≔
p
n
-
N
i
,
max
⁡
-
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
last
-
1
,
X
~
N
i
,
last
≔
M
a
r
,
n
-
p
a
r
-
M
a
i
,
n
+
p
a
i
+
k
~
N
i
,
last
.
Clearly a
i
′
≤
a
i for every i
∈
{
1
,
…
,
r
}.For any nonnegative integerj such that M
a
r
,
n
-
p
a
r
-
j belongs to the set M
max
⁡
[
2
], let
(43)
∫
j
(
N
1
)
‍
≔
Φ
M
a
r
,
n
-
p
a
r
-
j
,
for
any
i
∈
2
,
…
,
r
,
and let
(44)
∫
j
(
N
i
)
‍
≔
D
a
i
-
1
,
a
r
+
j
a
i
-
1
∫
j
(
N
i
-
1
)
‍
+
a
i
-
1
∫
j
+
1
(
N
i
-
1
)
‍
+
⋯
+
a
i
-
1
∫
(
X
~
N
i
-
1
,
last
)
-
1
(
N
i
-
1
)
‍
+
a
i
-
1
′
∫
X
~
N
i
-
1
,
last
(
N
i
-
1
)
‍
+
N
i
-
1
,
max
⁡
-
N
i
-
1
,
min
⁡
G
N
i
-
1
+
N
i
-
1
,
min
⁡
p
n
(
s
+
t
+
i
-
2
)
if
j
<
X
~
N
i
-
1
,
last
p
n
-
N
i
-
1
,
max
⁡
∫
j
(
N
i
-
1
)
‍
+
N
i
-
1
,
max
⁡
-
N
i
-
1
,
min
⁡
G
N
i
-
1
+
N
i
-
1
,
min
⁡
p
n
(
s
+
t
+
i
-
2
)
otherwise
.Theorem 26.
The formula for the Hilbert-Kunz function of the general binomial hypersurface under consideration is(45)
D
a
r
,
a
r
∫
0
(
N
r
)
‍
+
a
r
∫
1
(
N
r
)
‍
+
⋯
+
a
r
∫
X
~
N
r
,
l
a
s
t
-
1
N
r
‍
+
a
r
′
∫
X
~
N
r
,
l
a
s
t
(
N
r
)
‍
+
N
r
,
max
⁡
-
N
r
,
min
⁡
G
N
r
+
N
r
,
min
⁡
p
n
(
s
+
t
+
r
-
1
)
.Proof.
Observe that for any fixedi
∈
{
2
,
…
,
r
} and any j
∈
Z such that 0
≤
j
≤
k
~
N
r
,
last, the number ∫
j
(
N
i
)
‍ provides an account of the number of monomials A in M which satisfy the following 3 properties simultaneously.(i)
A
∉
A
c
+
J.
(ii)
The power of each of the variablesN
i
+
1
,
…
,
N
r is p
n
-
1.
(iii)
The power of the variableN
i is p
n
-
k
N
i where
(46)
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
-
1
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
and k
~
N
i
=
M
a
i
,
n
-
p
a
i
-
M
a
r
,
n
+
p
a
r
+
j.
Therefore for anyj
∈
Z such that 0
≤
j
≤
X
~
N
r
,
last, ∫
j
(
N
r
)
‍ denotes the number of monomials A in M for which A
∉
A
c
+
J and the power of the variable N
r is p
n
-
k
N
r where 1
+
E
a
r
,
b
r
+
a
r
(
j
-
1
)
<
k
N
r
≤
1
+
E
a
r
,
b
r
+
a
r
j.
Clearly then the number of monomialsA in M for which A
∉
A
c
+
J and the power of the variable N
r is p
n
-
k
N
r where 1
≤
k
N
r
≤
1
+
E
a
r
,
b
r equals D
a
r
,
a
r
∫
0
(
N
r
)
‍. Similarly, the number of monomials A in M for which A
∉
A
c
+
J and the power of the variable N
r is p
n
-
k
N
r where 1
+
E
a
r
,
b
r
+
a
r
(
j
-
1
)
<
k
N
r
≤
1
+
E
a
r
,
b
r
+
a
r
j and 1
≤
j
≤
(
X
~
N
r
,
last
)
-
1 equals a
r
∫
j
(
N
r
)
‍. We can similarly account for the terms a
r
′
∫
X
~
N
r
,
last
(
N
r
)
‍, (
N
r
,
max
⁡
-
N
r
,
min
⁡
)
G
N
r, and (
N
r
,
min
⁡
)
p
n
(
s
+
t
+
r
-
1
) in the formula (45) of the Hilbert-Kunz function.


### 4.3. An Illustrative Example

For this example as well as for the example given in Section4.4 below, we will need the following notation.Given any real numberβ, let β denote the smallest integer ≥
β and let [
β
] denote the largest integer ≤
β. We call β and [
β
]
the upper and the lower integral parts of β.LetS
=
k
[
N
2
,
N
1
,
Z
2
,
Z
1
,
P
1
,
P
2
] where k is a field of arbitrary prime characteristic p
>
0, and let J be the ideal in S defined by J
=
(
f
) where f
=
N
2
10
N
1
6
Z
2
5
Z
1
2
P
1
3
P
2
4
+
N
2
4
N
1
2
Z
2
5
Z
1
2
P
1
6
P
2
9. The differences Δ
N
2
,
Δ
N
1
,
Δ
Z
2
,
Δ
Z
1
,
Δ
P
1, and Δ
P
2 are therefore given by -
6
,
-
4,0
,
0,3
,
5, respectively. Hence the N
i’s are the negative difference variables, the Z
i’s are the zero difference variables, and the P
i’s are the positive difference variables. Here a
1
=
-
Δ
N
1
=
4, a
2
=
-
Δ
N
2
=
6, b
1
=
N
1
,
min
⁡
=
2, and b
2
=
N
2
,
min
⁡
=
4. Let E
a
1
,
b
1
,
E
a
2
,
b
2 be as defined in Section 4.2.3 above. Here, E
a
1
,
b
1
=
E
4,2 and E
a
2
,
b
2
=
E
6,4. For any i
∈
{
1,2
}, let M
a
i
,
n and r
a
i
,
n be defined by the equation(s) p
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n. That is,
(47)
M
a
1
,
n
=
M
4
,
n
=
p
n
-
1
-
r
4
,
n
4
,
M
a
2
,
n
=
M
6
,
n
=
p
n
-
1
-
r
6
,
n
6
,
Also
k
~
N
1
,
last
=
p
n
-
6
-
1
-
E
a
1
,
b
1
4
,
k
~
N
2
,
last
=
p
n
-
10
-
1
-
E
a
2
,
b
2
6
,
the
set
M
max
⁡
2
=
M
6
,
n
-
p
6
-
j
∣
j
∈
Z
,
0
≤
j
≤
k
~
N
2
,
last
.
For every j
∈
Z such that 0
≤
j
≤
k
~
N
2
,
last, we have
(48)
Mi
n
P
1
,
M
6
,
n
-
p
6
-
j
c
≔
Min
⁡
p
n
-
3
,
M
6
,
n
-
p
6
-
j
Δ
P
1
=
3
M
6
,
n
-
p
6
-
j
,
Mi
n
P
2
,
M
6
,
n
-
p
6
-
j
c
≔
Min
⁡
p
n
-
4
,
M
6
,
n
-
p
6
-
j
Δ
P
2
=
5
M
6
,
n
-
p
6
-
j
.
Using the formulae given in Lemma 24, it follows that
(49)
S
M
6
,
n
-
p
6
-
j
,
0
=
15
p
n
-
30
M
6
,
n
-
p
6
-
j
-
54
,
S
M
6
,
n
-
p
6
-
j
,
1
=
15
p
n
-
30
M
6
,
n
-
p
6
-
j
-
54
×
p
n
-
2
+
2
p
2
n
,
S
M
6
,
n
-
p
6
-
j
,
2
=
15
p
n
-
30
M
6
,
n
-
p
6
-
j
-
54
×
p
n
-
2
p
n
-
5
+
7
p
3
n
-
10
p
2
n
.
Since for any j
∈
Z such that 0
≤
j
≤
k
~
N
2
,
last, we know that ∫
j
(
N
1
)
‍
≔
S
M
6
,
n
-
p
6
-
j
,
s and s
= the total number of zero difference variables =
2 here, therefore we have
(50)
∫
j
N
1
‍
=
15
p
n
-
30
M
6
,
n
-
p
6
-
j
-
54
×
p
n
-
2
p
n
-
5
+
7
p
3
n
-
10
p
2
n
.
For getting the formula for the Hilbert-Kunz function in this case, we need to compute ∫
j
(
N
2
)
‍ for the various values of j under consideration. A straight forward computation tells us that for any j
∈
Z such that 0
≤
j
<
X
~
N
1
,
last, we have
(51)
∫
j
N
2
‍
=
D
a
1
,
a
2
+
4
j
∫
j
N
1
‍
+
4
∑
k
=
j
+
1
X
~
N
1
,
last
-
1
∫
k
N
1
‍
+
a
1
′
∫
X
~
N
1
,
last
N
1
‍
+
6
-
2
G
N
1
+
2
p
n
2
+
2
+
2
-
2
,
where
(52)
D
a
1
,
a
2
=
1
+
E
a
1
,
b
1
+
4
M
4
,
n
-
p
4
-
M
6
,
n
+
p
6
=
1
+
E
4,2
+
4
M
4
,
n
-
p
4
-
M
6
,
n
+
p
6
a
1
′
=
p
n
-
6
-
1
-
E
a
1
,
b
1
-
4
k
~
N
1
,
last
-
1
,
X
~
N
1
,
last
=
M
6
,
n
-
p
6
-
M
4
,
n
+
p
4
+
k
~
N
1
,
last
,
G
N
1
=
19
p
3
n
-
130
p
2
n
+
372
p
n
-
360
.
Similarly, for any j
∈
Z such that j
≥
X
~
N
1
,
last, we have
(53)
∫
j
N
2
‍
=
p
n
-
6
∫
j
N
1
‍
+
6
-
2
G
N
1
+
2
p
n
2
+
2
+
2
-
2
=
p
n
-
6
p
n
-
2
p
n
-
5
+
7
p
3
n
-
10
p
2
n
15
p
n
-
30
M
6
,
n
-
p
6
-
j
-
54
×
p
n
-
2
p
n
-
5
+
7
p
3
n
-
10
p
2
n
+
4
19
p
3
n
-
130
p
2
n
+
372
p
n
-
360
+
2
p
4
n
.
Clearly then ∫
j
(
N
2
)
‍ is a rational function in p and n for all values of j. Therefore, the formula for the corresponding Hilbert-Kunz function is
(54)
D
a
2
,
a
2
∫
0
N
2
‍
+
6
∑
j
=
1
X
~
N
2
,
last
-
1
∫
j
N
2
‍
+
a
2
′
∫
X
~
N
2
,
last
N
2
‍
+
(
10
-
4
)
G
N
2
+
4
p
n
2
+
2
+
2
-
1
,
where
(55)
D
a
2
,
a
2
=
1
+
E
a
2
,
b
2
=
1
+
E
6,4
,
a
2
′
=
p
n
-
10
-
1
-
E
a
2
,
b
2
-
6
k
~
N
2
,
last
-
1
,
X
~
N
2
,
last
=
M
6
,
n
-
p
6
-
M
4
,
n
+
p
4
+
k
~
N
2
,
last
,
G
N
2
=
21
p
4
n
-
168
p
3
n
+
632
p
2
n
-
1104
p
n
+
720
,
the
formulae
of
∫
j
(
N
2
)
‍
are
given
above
.
Since all the quantities involved in the above formula for the Hilbert-Kunz function are rational functions in the variables p and n, therefore the Hilbert-Kunz function is a rational function in p and n.


### 4.4. Example of the One-Dimensional Case

In this subsection, we will discuss the case of1-dimensional binomial hypersurfaces and will observe that using the above formula for the Hilbert-Kunz function, we do get that in this case of 1-dimension, the associated Hilbert-Kunz multiplicity is an integer (e.g., see Chapter 6, Corollary 6.2 of [7]). In this case, this integer happens to be equal to the ordinary multiplicity because the Hilbert-Kunz multiplicity is equal to the ordinary multiplicity for 1-dimensional rings.In the case of1-dimensional binomial hypersurfaces, we have S
=
k
[
x
1
,
x
2
], J
=
(
f
) where f
=
[
2
]
+
[
1
], R
=
S
/
J and the terms [
2
] and [
1
] of f are monomials in the 2 variables x
1 and x
2.There are3 possible cases:(1)
x
1
⊳
x
2 where x
1 is a negative difference variable and x
2 is a positive difference variable;
(2)
x
1
⊳
x
2 where x
1 is a negative difference variable and x
2 is a zero difference variable;
(3)
x
1
⊳
x
2 where x
1 is a zero difference variable and x
2 is a positive difference variable. We will first discuss Cases  (2) and (3).In Cases  (2) and (3), it is easy to check (without using the formula above) following.The Hilbert-Kunz function evaluated atp
n
=
(
x
1
,
min
⁡
+
x
2
,
min
⁡
)
p
n
-
x
1
,
min
⁡
x
2
,
min
⁡ and The associated Hilbert-Kunz multiplicity =
x
1
,
min
⁡
+
x
2
,
min
⁡ which is an integer.We will now discuss Case  (1).Sincex
1 is a negative difference variable and x
2 is a positive difference variable, then using our notation in Section 4.2.1, we can denote x
1 by N
1 and x
2 by P
1. Then writing x
1
⊳
x
2 is the same as writing N
1
⊳
P
1. In this case, there are 3 subcases.Subcase  1.1 (when
a
1
<
Δ
P
1
). It is an exercise to check that using the above formula for the Hilbert-Kunz function, we get that for n large enough.The Hilbert-Kunz function evaluated atp
n equals
(56)
1
+
E
a
1
,
b
1
P
1
,
min
⁡
+
a
1
P
1
,
min
⁡
×
p
n
Δ
P
1
-
a
1
a
1
Δ
P
1
-
1
a
1
-
r
a
1
,
n
a
1
-
p
a
1
+
P
1
,
min
⁡
Δ
P
1
-
a
1
P
1
,
max
⁡
+
a
1
P
1
,
max
⁡
p
n
-
N
1
,
max
⁡
-
1
-
E
a
1
,
b
1
a
1
-
a
1
P
1
,
max
⁡
p
n
Δ
P
1
-
a
1
a
1
Δ
P
1
-
1
a
1
-
r
a
1
,
n
a
1
-
p
a
1
+
P
1
,
min
⁡
Δ
P
1
+
a
1
′
P
1
,
max
⁡
+
N
1
,
max
⁡
-
N
1
,
min
⁡
P
1
,
max
⁡
+
N
1
,
min
⁡
p
n
,
where for any real number β, β and [
β
] denote the upper and lower integral parts of β, respectively.Hence the associated Hilbert-Kunz multiplicity equals(57)
a
1
P
1
,
min
⁡
Δ
P
1
-
a
1
a
1
Δ
P
1
+
a
1
P
1
,
max
⁡
Δ
P
1
+
N
1
,
min
⁡
which is nothing but P
1
,
min
⁡
+
a
1
+
N
1
,
min
⁡, which is an integer.Subcase  1.2 (when
a
1
>
Δ
P
1
). It is an exercise to check that using the above formula for the Hilbert-Kunz function, we get that for n large enoughthe Hilbert-Kunz function evaluated atp
n equals
(58)
p
n
-
N
1
,
max
⁡
P
1
,
max
⁡
+
N
1
,
max
⁡
-
N
1
,
min
⁡
P
1
,
max
⁡
+
N
1
,
min
⁡
p
n
.Hence the associated Hilbert-Kunz multiplicity equals P
1
,
max
⁡
+
N
1
,
min
⁡, which is an integer.Subcase  1.3 (when
a
1
=
Δ
P
1
). Let N
~
0 be the smallest positive integer such that if k
~
>
N
~
0, then P
1
,
min
⁡
-
1
-
r
a
1
,
n
-
a
1
p
a
1
<
a
1
k
~. Similarly, let N
~
0
′ be the smallest positive integer such that if k
~
>
N
~
0
′, then P
1
,
max
⁡
-
1
-
r
a
1
,
n
-
a
1
p
a
1
<
a
1
k
~. Clearly then N
~
0
′
≥
N
~
0.It is an exercise to check that using the above formula for the Hilbert-Kunz function, we get that forn large enough,the Hilbert-Kunz function evaluated atp
n equals
(59)
1
+
E
a
1
,
b
1
P
1
,
min
⁡
+
a
1
P
1
,
min
⁡
N
~
0
+
a
1
∑
k
~
=
N
~
0
+
1
N
~
0
′
1
+
r
a
1
,
n
+
a
1
p
a
1
+
a
1
k
~
+
a
1
P
1
,
max
⁡
p
n
-
N
1
,
max
⁡
-
1
-
E
a
1
,
b
1
a
1
-
1
-
N
~
0
′
+
a
1
′
P
1
,
max
⁡
+
N
1
,
max
⁡
-
N
1
,
min
⁡
P
1
,
max
⁡
+
N
1
,
min
⁡
p
n
,where for any real number β, β denotes the upper integral part of β.Hence the associated Hilbert-Kunz multiplicity equalsP
1
,
max
⁡
+
N
1
,
min
⁡, which is an integer.


### 4.5. Rationality of the Hilbert-Kunz Multiplicity

The Hilbert-Kunz multiplicity for the general Binomial hypersurface (whose Hilbert-Kunz function is given by (45) above) is by definition equal to the coefficient of p
n
(
s
+
t
+
r
-
1
) in the formula given by (45). We can easily see from the above discussion that for any j such that 0
≤
j
≤
X
~
N
r
,
last, ∫
j
(
N
r
)
‍ is obtained iteratively by first computing ∫
j
(
N
1
)
‍ for all j, then ∫
j
N
2
‍
(
for
all
j
)
,
…, and so on till ∫
j
(
N
r
-
1
)
‍ for all j. Moreover, for any nonnegative integer j such that 0
≤
j
≤
k
~
N
r
,
last, we know that M
a
r
,
n
-
p
a
r
-
j belongs to the set M
max
⁡
[
2
] and
(60)
∫
j
(
N
1
)
‍
≔
Φ
M
a
r
,
n
-
p
a
r
-
j
=
S
M
a
r
,
n
-
p
a
r
-
j
,
s
.So let us first give a formula forS
M
a
r
,
n
-
p
a
r
-
j
,
s. I will rather give a formula for S
M
max
⁡
,
s for any M
max
⁡ in M
max
⁡
[
2
] in general and then one can put M
max
⁡
=
M
a
r
,
n
-
p
a
r
-
j in it to get the required formula for S
M
a
r
,
n
-
p
a
r
-
j
,
s.LetM
max
⁡
∈
M
max
⁡
[
2
] be arbitrary. To compute S
M
max
⁡
,
s, we first need to compute S
M
max
⁡
,
0. I will compute S
M
max
⁡
,
0 only inthe most general case, that is, when
(61)
Mi
n
P
q
,
M
max
⁡
c
<
p
n
-
P
q
,
max
⁡
<
p
n
-
P
q
,
min
⁡
∀
q
∈
1
,
…
,
t
,
∀
M
max
⁡
∈
M
max
⁡
2
.
In this case, we have
(62)
Mi
n
P
q
,
M
max
⁡
c
=
M
max
⁡
Δ
P
q
.
In all other cases, it is a tedious job to compute S
M
max
⁡
,
0 because we will have to take care of various complicated possibilities likefor someq
∈
{
1
,
…
,
t
}, we may have Mi
n
P
q
,
M
max
⁡
c
<
p
n
-
P
q
,
min
⁡ and for some other q
′
∈
{
1
,
…
,
t
}, we may have Mi
n
P
q
′
,
M
max
⁡
c
=
p
n
-
P
q
′
,
min
⁡. This makes the computation of S
M
max
⁡
,
0 highly tedious and complicated. Let us now compute S
M
max
⁡
,
0 for the most general case.For2
≤
l
≤
t, let E
l
,
min
⁡ denote the sum of all possible terms which are products of l-many items either of the type M
max
⁡
Δ
P
q or of the type P
q
,
min
⁡ where q
∈
{
1
,
…
,
t
} such that the following two properties hold.(i)
At least one of thel items in each such product is of the type M
max
⁡
Δ
P
q;
(ii)
not all of thel items in any such product are of the type M
max
⁡
Δ
P
q. Similarly, for any l
∈
{
2
,
…
,
t
}, let E
l
,
max
⁡ denote the sum of all possible terms which are products of l-many items either of the type M
max
⁡
Δ
P
q or of the type P
q
,
max
⁡ where q
∈
{
1
,
…
,
t
} such that the following two properties hold.(i)
At least one of thel items in each such product is of the type M
max
⁡
Δ
P
q;
(ii)
not all of thel items in any such product are of the type M
max
⁡
Δ
P
q. For 2
≤
l
≤
t, let T
l denote the sum of all possible terms which are products of the type ∏
i
1
,
i
2
,
…
,
i
l
∈
1
,
…
,
t
i
1
≠
i
2
≠
⋯
≠
i
l
P
i
1
,
max
⁡
P
i
2
,
max
⁡
⋯
P
i
l
,
max
⁡.Then(63)
S
M
max
⁡
,
0
=
p
t
-
1
n
∑
i
=
1
t
P
i
,
max
⁡
+
∑
l
=
2
t
-
1
l
p
t
-
l
n
E
l
,
min
⁡
-
E
l
,
max
⁡
-
T
l
.Hence the formula forΦ
M
max
⁡ (≔
S
M
max
⁡
,
s) is given by
(64)
p
t
-
1
n
∑
i
=
1
t
P
i
,
max
⁡
+
∑
l
=
2
t
-
1
l
p
t
-
l
n
×
p
t
-
1
n
∑
i
=
1
t
P
i
,
max
⁡
+
∑
l
=
2
t
-
1
l
p
t
-
l
n
E
l
,
min
⁡
-
E
l
,
max
⁡
-
T
l
∏
i
=
1
s
p
n
-
Z
i
,
min
⁡
+
∑
k
=
1
s
∑
i
1
,
i
2
,
…
,
i
k
∈
1
,
…
,
s
i
1
≠
i
2
≠
⋯
≠
i
k
Z
i
1
,
min
⁡
Z
i
2
,
min
⁡
⋯
Z
i
k
,
min
⁡
-
1
k
+
1
p
t
+
s
-
k
n
×
∑
i
1
,
i
2
,
…
,
i
k
∈
1
,
…
,
s
i
1
≠
i
2
≠
⋯
≠
i
k
Z
i
1
,
min
⁡
Z
i
2
,
min
⁡
⋯
Z
i
k
,
min
⁡
.
Since ∫
j
(
N
1
)
‍
=
Φ
M
a
r
,
n
-
p
a
r
-
j, we get the formula for ∫
j
(
N
1
)
‍ from (64) above by putting M
max
⁡
=
M
a
r
,
n
-
p
a
r
-
j in it.Let us now write down the formula (45) for the Hilbert-Kunz function, in a different way which will help us see how the formula for the Hilbert-Kunz multiplicity will look like. But for that, we first need some notation.For eachj
∈
N, let F
j
(
1
)
≔
1
,
F
j
(
2
)
≔
2
j
-
1 and for any positive integer k
≥
3, let F
j
(
k
)
≔
∑
l
=
1
j
-
1
F
l
(
k
-
1
)
+
j
F
j
(
k
-
1
).
For each0
≤
j
≤
X
~
N
r
,
last, let C
j
(
N
r
) denote the coefficient of ∫
j
(
N
r
)
‍ in the formula (45) for the Hilbert-Kunz function.
For anym
∈
{
1
,
…
,
r
}, let V
(
m
)
≔
{
D
a
r
,
a
r
,
D
a
r
-
1
,
a
r
,
…
,
D
a
m
,
a
r
}, U
(
m
)
≔
{
a
r
,
a
r
-
1
,
…
,
a
m
}, and C
0
(
N
m
)
≔
∏
k
=
m
r
D
a
k
,
a
r. For any integer j such that 1
≤
j
<
X
~
N
m
,
last and any integer l such that 1
≤
l
≤
r
-
m
+
1, let H
l
(
m
)
≔ sum of all possible terms which are products of r
-
l
-
m
+
1 many D
a
k
,
a
rs and l many a
ks where the D
a
k
,
a
rs belong to the set V
(
m
) and the a
ks belong to the set U
(
m
).Letm
∈
{
1
,
…
,
r
-
1
} be arbitrary. Then for any integer j such that 1
≤
j
<
X
~
N
m
,
last, let
(65)
C
j
(
N
m
)
≔
∑
l
=
1
r
-
m
+
1
F
j
(
l
)
H
l
(
m
)
.
And for any integer j such that X
~
N
m
,
last
≤
j
≤
X
~
N
r
,
last, let us define C
j
(
N
m
) inductively as follows.LetC
X
~
N
m
,
last
(
N
m
)
:
=
a
m
′
[
Σ
j
=
0
X
~
N
m
,
last
-
1
C
j
(
N
m
+
1
)
]
+
(
p
n
-
N
m
,
max
⁡
)
C
X
~
N
m
,
last
(
N
m
+
1
)  and for any X
~
N
m
,
last
<
j
≤
X
~
N
r
,
last, let C
j
(
N
m
)
:
=
(
p
n
-
N
m
,
max
⁡
)
C
j
(
N
m
+
1
).Now if we expand the formula ((45) above) of the Hilbert-Kunz function, assuming that X
~
N
1
,
last
≤
X
~
N
2
,
last
≤
⋯
≤
X
~
N
r
,
last, we get that the expression in (45) equals
(66)
∑
j
=
0
X
~
N
1
,
last
-
1
C
j
N
1
∫
j
N
1
‍
+
a
1
′
∑
j
=
0
X
~
N
1
,
last
-
1
C
j
N
2
+
p
n
-
N
1
,
max
⁡
C
X
~
N
1
,
last
N
2
∫
X
~
N
1
,
last
(
N
1
)
‍
+
p
n
-
N
1
,
max
⁡
∑
j
=
X
~
N
1
,
last
+
1
X
~
N
r
,
last
C
j
N
2
∫
j
N
1
‍
+
N
r
,
max
⁡
-
N
r
,
min
⁡
G
N
r
+
N
r
,
min
⁡
p
n
(
s
+
t
+
r
-
1
)
+
∑
k
=
1
r
-
1
(
N
r
-
k
,
max
⁡
-
N
r
-
k
,
min
⁡
)
G
N
r
-
k
+
(
N
r
-
k
,
max
⁡
-
N
r
-
k
,
min
⁡
)
N
r
-
k
,
min
⁡
p
n
(
s
+
t
+
r
-
k
-
1
)
×
C
0
(
N
r
-
k
+
1
)
+
∑
l
=
1
k
H
l
r
-
k
+
1
∑
j
=
1
X
~
N
r
-
k
+
1
,
last
-
1
F
j
l
+
H
l
r
-
k
+
1
∑
j
=
1
X
~
N
r
-
k
+
1
,
last
-
1
F
j
l
∑
j
=
X
~
N
r
-
k
+
1
,
last
X
~
N
r
,
last
C
j
N
r
-
k
+
1
,
where the formulae for ∫
j
(
N
1
)
‍ can be obtained from (64). The Hilbert-Kunz multiplicity is by definition the coefficient of p
n
(
s
+
t
+
r
-
1
) in the expression (66) above. Computing the coefficient of p
n
(
s
+
t
+
r
-
1
) in the expression (66) above, we get that the Hilbert-Kunz multiplicity equals
(67)
N
r
,
min
⁡
+
∑
k
=
1
r
-
1
N
r
-
k
,
min
⁡
∑
j
=
1
X
~
N
r
-
k
+
1
,
last
-
1
F
j
l
Coefficient
of
p
k
n
in
∑
l
=
1
k
H
l
r
-
k
+
1
∑
j
=
1
X
~
N
r
-
k
+
1
,
last
-
1
F
j
l
+
∑
q
=
0
t
-
1
∑
j
=
1
X
~
N
1
,
last
-
1
j
q
F
j
(
r
)
coefficient
of
j
q
p
n
t
+
s
-
1
-
q
in
∫
j
N
1
‍
×
∑
j
=
1
X
~
N
1
,
last
-
1
j
q
F
j
(
r
)
[
terms
of
H
1
1
of
degree
p
n
(
r
-
1
)
]
×
∑
j
=
1
X
~
N
1
,
last
-
1
j
q
F
j
1
+
terms
of
H
2
1
of
degree
p
n
r
-
2
×
∑
j
=
1
X
~
N
1
,
last
-
1
j
q
F
j
(
2
)
+
⋯
+
terms
of
H
r
1
of
degree
p
n
r
-
r
×
∑
j
=
1
X
~
N
1
,
last
-
1
j
q
F
j
r
.If we expand the above formula (expression (67)) of the Hilbert-Kunz multiplicity, then inside the expansion, we will get terms which are of the form:(i)
finite (by “finite”, I mean depending uponn) sums of the form ∑
j
=
ϵ
n
θ
n
j
v where v is some positive integer and ϵ
n and θ
n are upper or lower integral parts of some real numbers which are of the form “polynomials in p
n with rational coefficients”;
(ii)
polynomials inp
n with rational coefficients (including constant terms which are rational numbers);
(iii)
products of the above mentioned2 types of terms.It hence follows from the above discussion that the Hilbert-Kunz multiplicity associated with any general binomial hypersurface is always rational.


## 4.1. The Main Theorem for Binomial Hypersurfaces

In this subsection, we will prove a theorem which will reduce the “key checking condition” mentioned above to checking of a combinatorial condition. The theorem which does this job is the following.Theorem 11.
A
∈
A
c
+
J if and only if either [
1
] divides A or there exists a positive integer M for which A
[
2
]
-
M
[
1
]
M
-
1 contains no negative powers of any of the underlying variables and A
[
2
]
-
M
[
1
]
M belongs to A
c. (Note here that by the notation [
2
]
-
M, we mean [
2
]
-
1 multiplied M times and similarly for the notation [
1
]
M.)Proof.
If[
1
] divides A, then it follows from Theorem 10 above that A
∈
A
c
+
J.
Suppose there exists a positive integerM for which A
[
2
]
-
M
[
1
]
M
-
1 has no negative powers and A
[
2
]
-
M
[
1
]
M belongs to A
c. Let M
0 denote the least positive integer M for which this happens. Then the polynomial g given by
(11)
a
1
A
2
-
1
+
a
2
A
2
-
2
1
1
+
⋯
+
a
M
0
A
2
-
M
0
1
M
0
-
1
,
where a
1 is a nonzero scalar in the ground field k and a
i
+
a
i
+
1
=
0 (mod
⁡
p) for each i
∈
{
1
,
…
,
M
0
-
1
}, is having the property that the product f
·
g equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c where a is a nonzero scalar. Hence A
∈
A
c
+
J. We now need to prove the other way round.
SupposeA
∈
A
c
+
J and [
1
] does not divide A. Then by Theorem 10, there exists a polynomial h whose terms are of the type c
B
[
τ
]
-
1 where c is a nonzero scalar, B is a combination monomial in A and f not belonging to A
c, and [
τ
] is a term of f which divides B, such that the product f
·
h equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c where a is a nonzero scalar.
It is easy to see that for the case of a binomial hypersurface, any combination monomial inA and f is either equal to A itself or it is of the type A
[
2
]
-
N
[
1
]
N for some positive integer N or it is of the type A
[
1
]
-
N
[
2
]
N for some positive integer N. It is also elementary to observe that any combination monomial in A and f of the type A
[
1
]
-
N
[
2
]
N always belongs to A
c. So any combination monomial not belonging to A
c is either equal to A itself or of the type A
[
2
]
-
N
[
1
]
N for some positive integer N. It hence follows that the terms of the polynomial h are either of the type A
[
2
]
-
N
[
1
]
N
-
1 for some positive integer N or of the type A
[
1
]
-
1. But we have assumed that the term [
1
] does not divide A, so A
[
1
]
-
1 cannot be a term of the polynomial h. Hence the terms of the polynomial h are only of the type A
[
2
]
-
N
[
1
]
N
-
1 for some positive integer N.
LetN
0 denote the largest positive integer N for which A
[
2
]
-
N
[
1
]
N
-
1 is a term of the polynomial h. Clearly then A
[
2
]
-
N
0
[
1
]
N
0
-
1 has no negative powers of any of the underlying variables x
1
,
…
,
x
m. Moreover, since the product f
·
h equals a
A
+
f
i
n
i
t
e
l
y
m
a
n
y
t
e
r
m
s
a
l
l
b
e
l
o
n
g
i
n
g
t
o
A
c (where a is a nonzero scalar), it follows that A
[
2
]
-
N
0
[
1
]
N
0 belongs to A
c.


## 4.2. A Formula for the Hilbert-Kunz Function for Binomial Hypersurfaces

In this subsection, we will give a closed form iterative formula for computing the Hilbert-Kunz function for any binomial hypersurface in general, over any field of positive characteristic. The notation and terminology remain the same as in the previous part of this paper.


### 4.2.1. Notation

Recall from Section3.1 the variables x
1
,
…
,
x
m and the term order ⊳ on the set of all monomials in these variables. Let f
=
[
2
]
+
[
1
] be the polynomial defining the binomial hypersurface where [
2
] and [
1
] are defined as in Section 3.1.For anyi
∈
{
1
,
…
,
m
}, let
(12)
Δ
i
≔
the
power
of
x
i
in
the
term
1
-
the
power
of
x
i
in
the
term
[
2
]
.
Note that some of the Δ
i’s can be negative, some can be positive, and some can be 0. Without loss of generality, we can assume that Δ
1
≤
⋯
≤
Δ
m.Letr be the integer such that 0
≤
r
≤
m and, the ordered set {
x
1
⊳
⋯
⊳
x
r
} equals the set {
x
i
∣
1
≤
i
≤
m
,
Δ
i
<
0
}. For i
∈
{
1
,
…
,
r
}, let N
i
≔
x
r
+
1
-
i and Δ
N
i
≔
Δ
r
+
1
-
i. Similarly, let s denote the number of x
i’s for which Δ
i
=
0 (note that s can also be 0). Let x
r
+
1
⊳
⋯
⊳
x
r
+
s be the elements of the set {
x
i
∣
1
≤
i
≤
m
,
Δ
i
=
0
}. For i
=
{
1
,
…
,
s
}, let Z
i
≔
x
r
+
s
+
1
-
i and Δ
Z
i
≔
Δ
r
+
s
+
1
-
i. Finally, let t
≔
m
-
(
r
+
s
) and let x
r
+
s
+
1
⊳
⋯
⊳
x
m be the elements of the set {
x
i
∣
1
≤
i
≤
m
,
Δ
i
>
0
}. Let us denote by P
1
⊳
⋯
⊳
P
t the ordered set x
r
+
s
+
1
⊳
⋯
⊳
x
m and for i
=
{
1
,
…
,
t
}, let Δ
P
i
≔
Δ
r
+
s
+
i. In other words, we can say that the ordered set x
1
⊳
⋯
⊳
x
m is the same as the ordered set N
r
⊳
⋯
⊳
N
1
⊳
Z
s
⊳
⋯
⊳
Z
1
⊳
P
1
⊳
⋯
⊳
P
t. We call N
i’s thenegative difference variables, Z
i’s thezero difference variables, and P
i’s thepositive difference variables. For any i
∈
{
1
,
…
,
r
}, let N
i
,
min
⁡ and N
i
,
max
⁡ denote the minimum and the maximum powers, respectively, of the variable N
i that appears in the expression of the polynomial f. We can similarly define P
q
,
min
⁡, P
q
,
max
⁡, Z
j
,
min
⁡, and Z
j
,
max
⁡ for all q
∈
{
1
,
…
,
t
} and for all j
∈
{
1
,
…
,
s
}. Observe that since each Z
j is a zero difference variable, we have Z
j
,
min
⁡
=
Z
j
,
max
⁡ for all j
∈
{
1
,
…
,
s
}.For anyi
∈
{
1
,
…
,
r
}, let a
i
≔
-
(
Δ
N
i
) and b
i
≔
N
i
,
min
⁡. For any i
∈
{
1
,
…
,
r
} and any positive integer n, let M
a
i
,
n and r
a
i
,
n be defined by the equation p
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n.


### 4.2.2. The NumberM
max,
A

Definition 12 (ofM
max
⁡
,
A).
LetA be an arbitrary element of the set M (this set has been defined earlier) such that [7] divides A. It follows from Theorem 11 that for those monomials A which are divisible by [7], we need to keep an account of the largest positive integer M for which the monomial A
[
2
]
-
M
[
1
]
M
-
1 has no negative power for any of the variables x
1
,
…
,
x
m. Let M
max
⁡
,
A denote this largest positive integer.


### 4.2.3. An Explicit Account ofM
max,
A

Corollary15 below gives an explicit account of the number M
max
⁡
,
A, for any monomial A which is divisible by [7]. The number M
max
⁡
,
A is important to us because of Theorem 11 which provides us with a way to count the required length l
(
R
/
(
m
[
p
n
]
+
J
)
).Remark 13.
LetA be any monomial in M which [
2
] divides; then it is of the form
(13)
N
r
p
n
-
k
N
r
⋯
N
1
p
n
-
k
N
1
Z
s
p
n
-
k
Z
s
⋯
Z
1
p
n
-
k
Z
1
P
1
j
1
⋯
P
t
j
t
,
where p
n
-
1
≥
j
q
≥
P
q
,
min
⁡ for any q
∈
{
1
,
…
,
t
}, p
n
-
1
≥
p
n
-
k
N
i
≥
N
i
,
max
⁡ for every 1
≤
i
≤
r, and p
n
-
1
≥
p
n
-
k
Z
i
≥
Z
i
,
max
⁡ for every 1
≤
i
≤
s.
Hence for any positive integerM, A
[
2
]
-
M
[
1
]
M
-
1 is of the form
(14)
∏
i
=
1
r
N
i
p
n
-
k
N
i
-
a
i
M
-
b
i
∏
i
=
1
s
Z
i
p
n
-
k
Z
i
-
Z
i
,
min
⁡
∏
q
=
1
t
P
q
j
q
+
M
Δ
P
q
-
P
q
,
max
⁡
,
where p
n
-
1
≥
j
q
≥
P
q
,
min
⁡ for any q
∈
{
1
,
…
,
t
}, p
n
-
1
≥
p
n
-
k
N
i
≥
N
i
,
max
⁡ for every 1
≤
i
≤
r, and p
n
-
1
≥
p
n
-
k
Z
i
≥
Z
i
,
max
⁡ for every 1
≤
i
≤
s.
It is now easy to see that for anyM
≥
1, the power of any zero difference variable or any positive difference variable in A
[
2
]
-
M
[
1
]
M
-
1 is ≥
0. But the power of any negative difference variable in A
[
2
]
-
M
[
1
]
M
-
1 may be negative for some values of M.The following lemma and its corollary give an explicit account ofM
max
⁡
,
A for any monomial A in M which [
2
] divides. But for stating the lemma, we need some notation. Let i
∈
{
1
,
…
,
r
} and let n be any positive integer. If b
i
≥
a
i, then let y
a
i denote the least positive integer ≥
(
b
i
-
r
a
i
,
n
) which is divisible by a
i. Let q
a
i be defined by the equation y
a
i
=
a
i
q
a
i. Let
(15)
p
a
i
≔
0
if
b
i
<
a
i
and
r
a
i
,
n
≥
b
i
,
1
if
b
i
<
a
i
and
r
a
i
,
n
<
b
i
,
q
a
i
if
b
i
≥
a
i
clearly
here
r
a
i
,
n
<
b
i
.
And let
(16)
E
a
i
,
b
i
≔
r
a
i
,
n
-
b
i
if
p
a
i
=
0
,
(
r
a
i
,
n
-
b
i
)
+
a
i
if
p
a
i
=
1
,
y
a
i
-
(
b
i
-
r
a
i
,
n
)
if
p
a
i
=
q
a
i
.
For any real number β, let β denote the smallest positive integer ≥β. For any 1
≤
i
≤
r, let
(17)
k
~
N
i
,
last
≔
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
a
i
.Lemma 14.
For any1
≤
i
≤
r, if k
N
i is such that 1
≤
k
N
i
≤
p
n
-
N
i
,
max
⁡, then the maximum value of the positive integer M for which p
n
-
k
N
i
-
a
i
M
-
b
i is ≥0 equals(18)
M
a
i
,
n
-
p
a
i
i
f
1
≤
k
N
i
≤
1
+
E
a
i
,
b
i
M
a
i
,
n
-
p
a
i
-
1
i
f
1
+
E
a
i
,
b
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
M
a
i
,
n
-
p
a
i
-
2
i
f
1
+
E
a
i
,
b
i
+
a
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
2
a
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
i
f
1
+
E
a
i
,
b
i
+
a
i
(
k
~
N
i
-
1
)
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
l
a
s
t
i
f
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
l
a
s
t
-
1
<
k
N
i
≤
p
n
-
N
i
,
max
⁡
.Proof.
Observe that sincep
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n, we have p
n
-
k
N
i
-
a
i
M
-
b
i
≥
0 if and only if M
≤
M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
). So we need to compute the greatest positive integer M, that is, ≤M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
).
Case  1 (when  b
i
≥
a
i
). In this case, we have 0
≤
r
a
i
,
n
<
a
i
≤
b
i. The least positive integer ≥
b
i
-
r
a
i
,
n which is divisible by a
i is y
a
i. Therefore the least positive integer ≥
(
b
i
-
r
a
i
,
n
)
/
a
i is q
a
i
=
p
a
i. Therefore the greatest negative integer ≤
-
(
b
i
-
r
a
i
,
n
)
/
a
i is -
p
a
i. And hence the greatest positive integer ≤
M
a
i
,
n
-
(
b
i
-
r
a
i
,
n
)
/
a
i is M
a
i
,
n
-
p
a
i.
Sincek
N
i
≤
p
n
-
N
i
,
max
⁡, it follows that M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
)
≥
M
a
i
,
n
-
p
a
i
-
(
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i
). The smallest positive integer ≥
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i is k
~
N
i
,
last. Therefore the greatest positive integer ≤
M
a
i
,
n
-
p
a
i
-
(
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i
) is M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last. As k
N
i increases, M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
) decreases, the lower bound being equal to M
a
i
,
n
-
p
a
i
-
(
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i
). Therefore the greatest positive integer ≤
M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
) equals M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last.
Hence the greatest positive integer≤
C
i
,
n
≔
M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
) is equal to(19)
M
a
i
,
n
-
p
a
i
if
M
a
i
,
n
-
p
a
i
≤
C
i
,
n
M
a
i
,
n
-
p
a
i
-
1
if
M
a
i
,
n
-
p
a
i
-
1
≤
C
i
,
n
<
M
a
i
,
n
-
p
a
i
M
a
i
,
n
-
p
a
i
-
2
if
M
a
i
,
n
-
p
a
i
-
2
≤
C
i
,
n
<
M
a
i
,
n
-
p
a
i
-
1
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
if
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
≤
C
i
,
n
<
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
+
1
.
This is equivalent to saying that the maximum value of the positive integerM for which p
n
-
k
N
i
-
a
i
M
-
b
i is ≥0 equals(20)
M
a
i
,
n
-
p
a
i
if
1
≤
k
N
i
≤
1
+
E
a
i
,
b
i
M
a
i
,
n
-
p
a
i
-
1
if
1
+
E
a
i
,
b
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
M
a
i
,
n
-
p
a
i
-
2
if
1
+
E
a
i
,
b
i
+
a
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
2
a
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
if
1
+
E
a
i
,
b
i
+
a
i
(
k
~
N
i
-
1
)
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
if
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
last
-
1
<
k
N
i
≤
p
n
-
N
i
,
max
⁡
.
Case  2 (when
b
i
<
a
i
). In this case, we have 2 subcases.
Subcase  2.1 (when
b
i
≤
r
a
i
,
n
). Here we have p
a
i
=
0 and 0
≤
b
i
≤
r
a
i
,
n
<
a
i. Therefore 0
≤
(
r
a
i
,
n
-
b
i
)
/
a
i
<
1 and hence -
1
<
(
b
i
-
r
a
i
,
n
)
/
a
i
≤
0. Therefore the least nonnegative integer ≥
(
b
i
-
r
a
i
,
n
)
/
a
i is p
a
i
=
0. The rest of the proof is similar to the proof of Case  1.
Subcase  2.2 (when
b
i
>
r
a
i
,
n
). Here we have, p
a
i
=
1 and 0
≤
r
a
i
,
n
<
b
i
<
a
i. Therefore 0
<
(
b
i
-
r
a
i
,
n
)
/
a
i
<
1. Hence the least positive integer ≥
(
b
i
-
r
a
i
,
n
)
/
a
i is p
a
i
=
1. The rest of the proof is similar to the proof of Case  1.The following corollary is immediate.Corollary 15.
LetA be a monomial in M which [
2
] divides and let M be a positive integer. Let A and A
[
2
]
-
M
[
1
]
M
-
1 be of the form as given in Remark 13. For 1
≤
i
≤
r, let M
i
,
A be the maximum value of the positive integer M for which p
n
-
k
N
i
-
a
i
M
-
b
i is ≥
0.
ThenM
max
⁡
,
A equals the minimum of all M
i
,
As as i ranges from 1 to r.LetM
[
2
]
≔ the set of all monomials in M which are divisible by [
2
]. Let M
max
⁡
[
2
]
≔
{
M
max
⁡
,
A
∣
A
∈
M
[
2
]
}.Remark 16.
It follows from Corollary15 above that M
max
⁡
[
2
]
=
{
M
a
r
,
n
-
p
a
r
-
j
∣
0
≤
j
≤
k
~
N
r
,
last
}.


### 4.2.4. The Length Count for a FixedM
max,
A

Remark 17.
It is easy to verify that ifM
1 is a positive integer such that A
[
2
]
-
M
1
[
1
]
M
1
-
1 contains no negative powers and M
2 is a positive integer such that M
2
≤
M
1, then A
[
2
]
-
M
2
[
1
]
M
2
-
1 contains no negative powers. It is also easy to verify that if N
1 is a positive integer such that A
[
2
]
-
N
1
[
1
]
N
1 belongs to A
c and N
2 is a positive integer such that N
1
≤
N
2, then A
[
2
]
-
N
2
[
1
]
N
2 belongs to A
c. Hence for any A
∈
M
[
2
], A
∈
A
c
+
J if and only if either [
1
] divides A or A
[
2
]
-
M
max
⁡
,
A
[
1
]
M
max
⁡
,
A belongs to A
c.Remark 18.
Recall from Remark13 as well as from Corollary 15 that given any monomial A in M
[
2
], the value of M
max
⁡
,
A depends only on the powers of the negative difference variables appearing in A. It now follows that for each i
∈
{
1
,
…
,
r
}, if we fix numbers k
N
i
0 such that p
n
-
1
≥
p
n
-
k
N
i
0
≥
N
i
,
max
⁡, then for any A in the set
(21)
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
≔
A
∣
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
where
p
n
-
1
≥
l
j
≥
0
for
every
j
∈
1
,
…
,
s
and
,
N
r
p
n
-
k
N
r
0
p
n
-
1
≥
j
q
≥
0
for
every
q
∈
1
,
…
,
t
,
the value of M
max
⁡
,
A is the same (call it M
max
⁡
0).The main result of this subsubsection is Lemma24 below which provides an account of the length count for a fixed value of M
max
⁡
,
A. But for stating the lemma, we need to introduce some notation first.For an arbitrary elementM
max
⁡ in M
max
⁡
[
2
] and for every q
∈
{
1
,
…
,
t
}, let
(22)
Mi
n
P
q
,
M
max
⁡
c
≔
Min
⁡
p
n
-
P
q
,
min
⁡
,
M
max
⁡
Δ
P
q
.
The symbol Mi
n
c in the above definition of Mi
n
P
q
,
M
max
⁡
c stands for “minimum convergent.” The reason behind this notation is explained in Remark 19 below.Remark 19.
Observe that if the power of the variableP
q in the monomial A is j
q, then the power of P
q in the monomial A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ is j
q
+
M
max
⁡
Δ
P
q. If A is divisible by [
2
], then p
n
-
1
≥
j
q
≥
P
q
,
min
⁡. Now if j
q
+
M
max
⁡
Δ
P
q
≥
p
n, then A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c. That is, if p
n
-
j
q
≤
Mi
n
P
q
,
M
max
⁡
c, then A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c. That is, if p
n
-
Mi
n
P
q
,
M
max
⁡
c
≤
j
q
≤
p
n
-
1, then A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c. So p
n
-
Mi
n
P
q
,
M
max
⁡
c is the minimum possible power of the variable P
q in the monomial A for which A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c.Recall the ordered setP
1
⊳
⋯
⊳
P
t of positive difference variables. Let A
t
+
1
≔
0 and C
t
≔
1. For any q
∈
{
1
,
…
,
t
-
1
}, let
(23)
C
q
≔
p
(
t
-
q
)
n
.
And for any q
∈
{
2
,
…
,
t
}, let
(24)
A
q
≔
p
n
-
P
q
,
min
⁡
A
q
+
1
+
P
q
,
min
⁡
C
q
.
For any M
max
⁡ in M
max
⁡
[
2
], let D
t
+
1
,
M
max
⁡
:
=
1 and D
~
t
+
1
,
M
max
⁡
≔
0. For any M
max
⁡ in M
max
⁡
[
2
] and any q
∈
{
1
,
…
,
t
}, let
(25)
D
q
,
M
max
⁡
≔
Mi
n
P
q
,
M
max
⁡
c
A
q
+
1
+
p
n
-
P
q
,
min
⁡
-
Mi
n
P
q
,
M
max
⁡
c
D
q
+
1
,
M
max
⁡
+
P
q
,
min
⁡
C
q
,
(26)
D
~
q
,
M
max
⁡
≔
Mi
n
P
q
,
M
max
⁡
c
A
q
+
1
+
(
p
n
-
Mi
n
P
q
,
M
max
⁡
c
-
P
q
,
max
⁡
)
D
~
q
+
1
,
M
max
⁡
+
P
q
,
max
⁡
-
P
q
,
min
⁡
D
q
+
1
,
M
max
⁡
+
(
P
q
,
min
⁡
)
C
q
if
Mi
n
P
q
,
M
max
⁡
c
<
p
n
-
P
q
,
min
⁡
and
P
q
,
max
⁡
<
p
n
-
Mi
n
P
q
,
M
max
⁡
c
D
q
,
M
max
⁡
if
Mi
n
P
q
,
M
max
⁡
c
<
p
n
-
P
q
,
min
⁡
and
P
q
,
max
⁡
≥
p
n
-
Mi
n
P
q
,
M
max
⁡
c
(
Mi
n
P
q
,
M
max
⁡
c
)
A
q
+
1
+
(
P
q
,
min
⁡
)
C
q
if
Mi
n
P
q
,
M
max
⁡
c
=
p
n
-
P
q
,
min
⁡
.
Recall the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) from Remark 18. For any A in this set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
), the value of M
max
⁡
,
A is the same, namely, M
max
⁡
0. Lemma 24 below gives an account of the length count for the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
). But we will break up the proof of Lemma 24 in a few short steps as follows.Lemma 20.
For everyq
∈
{
2
,
…
,
t
}, let
(27)
F
q
,
A
,
J
1
,
…
,
J
q
-
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
)
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
a
n
d
j
1
=
J
1
,
…
,
j
q
-
1
=
J
q
-
1
,
where for each k
∈
{
1
,
…
,
q
-
1
}, J
k is a fixed integer such that P
k
,
min
⁡
≤
J
k
≤
p
n
-
1 and at least one J
k is ≥
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c.
Then the total number of monomialsA in the set F
q
,
A
,
J
1
,
…
,
J
q
-
1 for which A
∉
A
c
+
J equals A
q.Proof.
The proof is by reverse induction onq, the base case of induction being q
=
t. Observe that if A is any monomial in F
t
,
A
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
A
,
J
1
,
…
,
J
t
-
1 is p
n
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
A
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
A
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
A
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(28)
p
n
-
P
t
,
min
⁡
0
+
P
t
,
min
⁡
1
=
p
n
-
P
t
,
min
⁡
A
t
+
1
+
P
t
,
min
⁡
C
t
=
A
t
.
This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.Lemma 21.
For everyq
∈
{
2
,
…
,
t
}, let
(29)
F
q
,
C
,
J
1
,
…
,
J
q
-
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
)
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
,
j
1
=
J
1
,
…
,
j
q
-
1
=
J
q
-
1
,
where for each k
∈
{
1
,
…
,
q
-
1
}, J
k is a fixed integer such that 0
≤
J
k
≤
p
n
-
1 and at least one J
k is <
P
k
,
min
⁡.
Then the total number of monomialsA in the set F
q
,
C
,
J
1
,
…
,
J
q
-
1 for which A
∉
A
c
+
J equals C
q
-
1.Proof.
The proof is by reverse induction onq, the base case of induction being q
=
t. Observe that if A is any monomial in F
t
,
C
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
≤
p
n
-
1, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
C
,
J
1
,
…
,
J
t
-
1 is p
n. Hence the total number of monomials A in the set F
t
,
C
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals p
n
=
p
(
t
-
(
t
-
1
)
)
n
=
C
t
-
1. This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.Lemma 22.
For everyq
∈
{
2
,
…
,
t
}, let
(30)
F
q
,
D
,
J
1
,
…
,
J
q
-
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
)
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
a
n
d
j
1
=
J
1
,
…
,
j
q
-
1
=
J
q
-
1
,
where for each k
∈
{
1
,
…
,
q
-
1
}, J
k is a fixed integer such that P
k
,
min
⁡
≤
J
k
<
p
n
-
Min
P
k
,
M
max
⁡
0
c.
Then the total number of monomialsA in the set F
q
,
D
,
J
1
,
…
,
J
q
-
1 for which A
∉
A
c
+
J equals (i)
D
q
,
M
max
⁡
0 if there exists at least one k
∈
{
1
,
…
,
q
-
1
} such that P
k
,
max
⁡
≥
p
n
-
Min
P
k
,
M
max
⁡
0
c;
(ii)
D
~
q
,
M
max
⁡
0 if for every k
∈
{
1
,
…
,
q
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Min
P
k
,
M
max
⁡
0
c and P
k
,
max
⁡
≤
J
k
<
p
n
-
Min
P
k
,
M
max
⁡
0
c;
(iii)
D
q
,
M
max
⁡
0 if for every k
∈
{
1
,
…
,
q
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Min
P
k
,
M
max
⁡
0
c and there exists at least one k
∈
{
1
,
…
,
q
-
1
} such that P
k
,
min
⁡
≤
J
k
<
P
k
,
max
⁡.Proof.
The proof is by reverse induction onq, the base case of induction being q
=
t. Let us first prove the base case of induction.
Case  1. When there exists at least one k
∈
{
1
,
…
,
t
-
1
} such that P
k
,
max
⁡
≥
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c.
Then since we have that for eachk
∈
{
1
,
…
,
t
-
1
}, J
k is a fixed integer such that P
k
,
min
⁡
≤
J
k
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c, it follows that monomials in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 are not divisible by [
1
].
Observe now that ifA is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then [
2
] divides A but [
1
] does not divide A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∉
A
c. Hence such a monomial A satisfies A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(31)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
1
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
D
t
+
1
,
M
max
⁡
0
+
P
t
,
min
⁡
C
t
=
D
t
,
M
max
⁡
0
.
This proves the base case of induction in Case  1.
Case  2. When for every k
∈
{
1
,
…
,
t
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c and P
k
,
max
⁡
≤
J
k
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c. In this case, there are 3 subcases.
Subcase  2.1 (
Mi
n
P
t
,
M
max
⁡
0
c
=
p
n
-
P
t
,
min
⁡
). Observe that if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡
=
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(32)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
P
t
,
min
⁡
C
t
=
D
~
t
,
M
max
⁡
0
.
This proves the base case of induction in this subcase.
Subcase  2.2 (
Mi
n
P
t
,
M
max
⁡
0
c
<
p
n
-
P
t
,
min
⁡
and
P
t
,
max
⁡
≥
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
). Observe that if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then [
2
] divides A but [
1
] does not divide A (since P
t
,
max
⁡
≥
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c) and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∉
A
c. Hence such a monomial A satisfies A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(33)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
1
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
D
t
+
1
,
M
max
⁡
0
+
P
t
,
min
⁡
C
t
=
D
t
,
M
max
⁡
0
=
D
~
t
,
M
max
⁡
0
.
This proves the base case of induction in this subcase.
Subcase  2.3 (
Mi
n
P
t
,
M
max
⁡
0
c
<
p
n
-
P
t
,
min
⁡
and
P
t
,
max
⁡
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
). Observe that if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
max
⁡
≤
j
t
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then [
1
] divides A (since P
t
,
max
⁡
≤
j
t and P
k
,
max
⁡
≤
J
k
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c for every k
∈
{
1
,
…
,
t
-
1
}). Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
max
⁡. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
<
P
t
,
max
⁡, then [
2
] divides A but [
1
] does not divide A (since j
t
<
P
t
,
max
⁡) and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∉
A
c (since j
t
<
P
t
,
max
⁡
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c). Hence such a monomial A satisfies A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
max
⁡
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(34)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
max
⁡
0
+
P
t
,
max
⁡
-
P
t
,
min
⁡
1
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
max
⁡
×
D
~
t
+
1
,
M
max
⁡
0
+
P
t
,
max
⁡
-
P
t
,
min
⁡
D
t
+
1
,
M
max
⁡
0
+
(
P
t
,
min
⁡
)
C
t
=
D
~
t
,
M
max
⁡
0
.
This proves the base case of induction in this subcase. Hence the base case of induction is proved in Case  2.
Case  3. When for every k
∈
{
1
,
…
,
t
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c and there exists at least one k
∈
{
1
,
…
,
t
-
1
} such that P
k
,
min
⁡
≤
J
k
<
P
k
,
max
⁡.
Since there exists at least onek
∈
{
1
,
…
,
t
-
1
} such that P
k
,
min
⁡
≤
J
k
<
P
k
,
max
⁡, it follows that monomials in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 are not divisible by [
1
]. The rest of the proof of this case is similar to the proof of Case  1. This proves the base case of induction in Case  3.
This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.Let(35)
F
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
.Lemma 23.
The total number of monomialsA in the set F
1 for which A
∉
A
c
+
J equals D
~
1
,
M
max
⁡
0.Proof.
LetJ
1 be an arbitrarily fixed integer such that p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
≤
J
1
≤
p
n
-
1. It then follows from Lemma 20 that for any such fixed J
1, the total number of monomials A in F
2
,
A
,
J
1 such that A
∉
A
c
+
J equals A
2. Also if J
1 is an arbitrarily fixed integer such that 0
≤
J
1
<
P
1
,
min
⁡, then it follows from Lemma 21 that for any such fixed J
1, the total number of monomials A in F
2
,
C
,
J
1 such that A
∉
A
c
+
J equals C
1. Now there are 3 cases.
Case  1 (if
Mi
n
P
1
,
M
max
⁡
0
c
=
p
n
-
P
1
,
min
⁡
). In this case, the total number of monomials A in the set F
1 for which A
∉
A
c
+
J equals
(36)
p
n
-
1
-
p
n
+
Mi
n
P
1
,
M
max
⁡
0
c
+
1
A
2
+
P
1
,
min
⁡
C
1
=
Mi
n
P
1
,
M
max
⁡
0
c
A
2
+
P
1
,
min
⁡
C
1
=
D
~
1
,
M
max
⁡
0
.
Case  2 (if
Mi
n
P
1
,
M
max
⁡
0
c
<
p
n
-
P
1
,
min
⁡
and
P
1
,
max
⁡
≥
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
). It follows from Lemma 22 that for any fixed integer J
1 such that P
1
,
min
⁡
≤
J
1
<
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c, the total number of monomials A in F
2
,
D
,
J
1 for which A
∉
A
c
+
J equals D
2
,
M
max
⁡
0. Hence in this case, the total number of monomials A in the set F
1 for which A
∉
A
c
+
J equals
(37)
p
n
-
1
-
p
n
+
Mi
n
P
1
,
M
max
⁡
0
c
+
1
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
Mi
n
P
1
,
M
max
⁡
0
c
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
D
1
,
M
max
⁡
0
=
D
~
1
,
M
max
⁡
0
.
Case  3 (if
Mi
n
P
1
,
M
max
⁡
0
c
<
p
n
-
P
1
,
min
⁡
and
P
1
,
max
⁡
<
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
). It follows from Lemma 22 that for any fixed integer J
1 such that P
1
,
max
⁡
≤
J
1
<
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c, the total number of monomials A in F
2
,
D
,
J
1 for which A
∉
A
c
+
J equals D
~
2
,
M
max
⁡
0. It again follows from Lemma 22 that for any fixed integer J
1 such that P
1
,
min
⁡
≤
J
1
<
P
1
,
max
⁡, the total number of monomials A in F
2
,
D
,
J
1 for which A
∉
A
c
+
J equals D
2
,
M
max
⁡
0. Hence in this case, the total number of monomials A in the set F
1 for which A
∉
A
c
+
J equals
(38)
p
n
-
1
-
p
n
+
Mi
n
P
1
,
M
max
⁡
0
c
+
1
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
max
⁡
D
~
2
,
M
max
⁡
0
+
P
1
,
max
⁡
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
Mi
n
P
1
,
M
max
⁡
0
c
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
max
⁡
D
~
2
,
M
max
⁡
0
+
P
1
,
max
⁡
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
D
~
1
,
M
max
⁡
0
.Lemma 24.
The total number of monomialsA in the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
s where S
M
max
⁡
0
,
s is defined inductively as follows:
(39)
S
M
max
⁡
0
,
0
≔
D
~
1
,
M
max
⁡
0
,
f
o
r
e
v
e
r
y
j
∈
1
,
…
,
s
,
S
M
max
⁡
0
,
j
≔
S
M
max
⁡
0
,
j
-
1
p
n
-
Z
j
,
min
⁡
+
p
t
+
j
-
1
n
Z
j
,
min
⁡
.Proof.
Fix an arbitrary integerL
1 such that p
n
-
1
≥
L
1
≥
Z
1
,
min
⁡. Let
(40)
P
L
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
2
=
p
n
-
1
,
l
1
=
L
1
.
It is now easy to see that the total number of monomials A in P
L
1 for which A
∉
A
c
+
J equals D
~
1
,
M
max
⁡
0
=
S
M
max
⁡
0
,
0 (the proof of this is similar to the proof of Lemma 23).
Fix an arbitrary integerM
1 such that 0
≤
M
1
<
Z
1
,
min
⁡. Let
(41)
P
M
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
2
=
p
n
-
1
,
l
1
=
M
1
.
Since 0
≤
M
1
<
Z
1
,
min
⁡, therefore any monomial A
∈
P
M
1 is divisible neither by [
1
] nor by [
2
]. Hence the total number of monomials A in P
M
1 for which A
∉
A
c
+
J equals p
t
n.
So if we defineP
1 as the subset of P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) consisting of all those monomials in P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) for which l
s
=
l
s
-
1
=
⋯
=
l
2
=
p
n
-
1, then the total number of monomials A in the set P
1 for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
1. Similarly, if we define P
2 as the subset of P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) consisting of all those monomials in it for which l
s
=
l
s
-
1
=
⋯
=
l
3
=
p
n
-
1, then the total number of monomials A in the set P
2 for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
2. Proceeding inductively, it is now easy to see that the total number of monomials A in the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
s.Definition 25.
Given anyM
max
⁡ in M
max
⁡
[
2
], let Φ
M
max
⁡
≔
S
M
max
⁡
,
s where S
M
max
⁡
,
s is defined in the same way as S
M
max
⁡
0
,
s is defined in Lemma 24 (just replace M
max
⁡
0 in Lemma 24 by M
max
⁡).


### 4.2.5. The Formula for the Hilbert-Kunz Function

We will now state the formula for the Hilbert-Kunz function for the general binomial hypersurface that we have considered above. But for stating the formula, we need some more notation, which we provide first.SetG
P
(
t
-
1
)
≔
P
t
,
max
⁡ and for every q
∈
{
1
,
…
,
(
t
-
2
)
}, let G
P
q
≔
(
p
n
-
P
q
+
1
,
max
⁡
)
G
P
q
+
1
+
(
P
q
+
1
,
max
⁡
)
p
n
(
t
-
q
-
1
);
SetG
Z
1
≔
(
p
n
-
P
1
,
max
⁡
)
G
P
1
+
(
P
1
,
max
⁡
)
p
n
(
t
-
1
), and for every l
∈
{
2
,
…
,
s
}, let G
Z
l
≔
(
p
n
-
Z
l
-
1
,
max
⁡
)
G
Z
l
-
1
+
(
Z
l
-
1
,
max
⁡
)
p
n
(
t
+
l
-
2
);
SetG
N
1
≔
(
p
n
-
Z
s
,
max
⁡
)
G
Z
s
+
(
Z
s
,
max
⁡
)
p
n
(
t
+
s
-
1
), and for every i
∈
{
2
,
…
,
r
}, let G
N
i
≔
(
p
n
-
N
i
-
1
,
min
⁡
)
G
N
i
-
1
+
(
N
i
-
1
,
min
⁡
)
p
n
(
t
+
s
+
i
-
2
). Recall that for any i
∈
{
1
,
…
,
r
}, let a
i
≔
-
(
Δ
N
i
) and for any i
∈
{
1
,
…
,
r
} and any positive integer n, M
a
i
,
n and r
a
i
,
n are defined by the equation p
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n. Also recall the definitions of E
a
i
,
b
i and p
a
i from Section 4.2.3.For everyi
∈
{
1
,
…
,
r
}, let
(42)
D
a
i
,
a
r
≔
1
+
E
a
i
,
b
i
+
a
i
M
a
i
,
n
-
p
a
i
-
M
a
r
,
n
+
p
a
r
a
i
′
≔
p
n
-
N
i
,
max
⁡
-
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
last
-
1
,
X
~
N
i
,
last
≔
M
a
r
,
n
-
p
a
r
-
M
a
i
,
n
+
p
a
i
+
k
~
N
i
,
last
.
Clearly a
i
′
≤
a
i for every i
∈
{
1
,
…
,
r
}.For any nonnegative integerj such that M
a
r
,
n
-
p
a
r
-
j belongs to the set M
max
⁡
[
2
], let
(43)
∫
j
(
N
1
)
‍
≔
Φ
M
a
r
,
n
-
p
a
r
-
j
,
for
any
i
∈
2
,
…
,
r
,
and let
(44)
∫
j
(
N
i
)
‍
≔
D
a
i
-
1
,
a
r
+
j
a
i
-
1
∫
j
(
N
i
-
1
)
‍
+
a
i
-
1
∫
j
+
1
(
N
i
-
1
)
‍
+
⋯
+
a
i
-
1
∫
(
X
~
N
i
-
1
,
last
)
-
1
(
N
i
-
1
)
‍
+
a
i
-
1
′
∫
X
~
N
i
-
1
,
last
(
N
i
-
1
)
‍
+
N
i
-
1
,
max
⁡
-
N
i
-
1
,
min
⁡
G
N
i
-
1
+
N
i
-
1
,
min
⁡
p
n
(
s
+
t
+
i
-
2
)
if
j
<
X
~
N
i
-
1
,
last
p
n
-
N
i
-
1
,
max
⁡
∫
j
(
N
i
-
1
)
‍
+
N
i
-
1
,
max
⁡
-
N
i
-
1
,
min
⁡
G
N
i
-
1
+
N
i
-
1
,
min
⁡
p
n
(
s
+
t
+
i
-
2
)
otherwise
.Theorem 26.
The formula for the Hilbert-Kunz function of the general binomial hypersurface under consideration is(45)
D
a
r
,
a
r
∫
0
(
N
r
)
‍
+
a
r
∫
1
(
N
r
)
‍
+
⋯
+
a
r
∫
X
~
N
r
,
l
a
s
t
-
1
N
r
‍
+
a
r
′
∫
X
~
N
r
,
l
a
s
t
(
N
r
)
‍
+
N
r
,
max
⁡
-
N
r
,
min
⁡
G
N
r
+
N
r
,
min
⁡
p
n
(
s
+
t
+
r
-
1
)
.Proof.
Observe that for any fixedi
∈
{
2
,
…
,
r
} and any j
∈
Z such that 0
≤
j
≤
k
~
N
r
,
last, the number ∫
j
(
N
i
)
‍ provides an account of the number of monomials A in M which satisfy the following 3 properties simultaneously.(i)
A
∉
A
c
+
J.
(ii)
The power of each of the variablesN
i
+
1
,
…
,
N
r is p
n
-
1.
(iii)
The power of the variableN
i is p
n
-
k
N
i where
(46)
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
-
1
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
and k
~
N
i
=
M
a
i
,
n
-
p
a
i
-
M
a
r
,
n
+
p
a
r
+
j.
Therefore for anyj
∈
Z such that 0
≤
j
≤
X
~
N
r
,
last, ∫
j
(
N
r
)
‍ denotes the number of monomials A in M for which A
∉
A
c
+
J and the power of the variable N
r is p
n
-
k
N
r where 1
+
E
a
r
,
b
r
+
a
r
(
j
-
1
)
<
k
N
r
≤
1
+
E
a
r
,
b
r
+
a
r
j.
Clearly then the number of monomialsA in M for which A
∉
A
c
+
J and the power of the variable N
r is p
n
-
k
N
r where 1
≤
k
N
r
≤
1
+
E
a
r
,
b
r equals D
a
r
,
a
r
∫
0
(
N
r
)
‍. Similarly, the number of monomials A in M for which A
∉
A
c
+
J and the power of the variable N
r is p
n
-
k
N
r where 1
+
E
a
r
,
b
r
+
a
r
(
j
-
1
)
<
k
N
r
≤
1
+
E
a
r
,
b
r
+
a
r
j and 1
≤
j
≤
(
X
~
N
r
,
last
)
-
1 equals a
r
∫
j
(
N
r
)
‍. We can similarly account for the terms a
r
′
∫
X
~
N
r
,
last
(
N
r
)
‍, (
N
r
,
max
⁡
-
N
r
,
min
⁡
)
G
N
r, and (
N
r
,
min
⁡
)
p
n
(
s
+
t
+
r
-
1
) in the formula (45) of the Hilbert-Kunz function.


## 4.2.1. Notation

Recall from Section3.1 the variables x
1
,
…
,
x
m and the term order ⊳ on the set of all monomials in these variables. Let f
=
[
2
]
+
[
1
] be the polynomial defining the binomial hypersurface where [
2
] and [
1
] are defined as in Section 3.1.For anyi
∈
{
1
,
…
,
m
}, let
(12)
Δ
i
≔
the
power
of
x
i
in
the
term
1
-
the
power
of
x
i
in
the
term
[
2
]
.
Note that some of the Δ
i’s can be negative, some can be positive, and some can be 0. Without loss of generality, we can assume that Δ
1
≤
⋯
≤
Δ
m.Letr be the integer such that 0
≤
r
≤
m and, the ordered set {
x
1
⊳
⋯
⊳
x
r
} equals the set {
x
i
∣
1
≤
i
≤
m
,
Δ
i
<
0
}. For i
∈
{
1
,
…
,
r
}, let N
i
≔
x
r
+
1
-
i and Δ
N
i
≔
Δ
r
+
1
-
i. Similarly, let s denote the number of x
i’s for which Δ
i
=
0 (note that s can also be 0). Let x
r
+
1
⊳
⋯
⊳
x
r
+
s be the elements of the set {
x
i
∣
1
≤
i
≤
m
,
Δ
i
=
0
}. For i
=
{
1
,
…
,
s
}, let Z
i
≔
x
r
+
s
+
1
-
i and Δ
Z
i
≔
Δ
r
+
s
+
1
-
i. Finally, let t
≔
m
-
(
r
+
s
) and let x
r
+
s
+
1
⊳
⋯
⊳
x
m be the elements of the set {
x
i
∣
1
≤
i
≤
m
,
Δ
i
>
0
}. Let us denote by P
1
⊳
⋯
⊳
P
t the ordered set x
r
+
s
+
1
⊳
⋯
⊳
x
m and for i
=
{
1
,
…
,
t
}, let Δ
P
i
≔
Δ
r
+
s
+
i. In other words, we can say that the ordered set x
1
⊳
⋯
⊳
x
m is the same as the ordered set N
r
⊳
⋯
⊳
N
1
⊳
Z
s
⊳
⋯
⊳
Z
1
⊳
P
1
⊳
⋯
⊳
P
t. We call N
i’s thenegative difference variables, Z
i’s thezero difference variables, and P
i’s thepositive difference variables. For any i
∈
{
1
,
…
,
r
}, let N
i
,
min
⁡ and N
i
,
max
⁡ denote the minimum and the maximum powers, respectively, of the variable N
i that appears in the expression of the polynomial f. We can similarly define P
q
,
min
⁡, P
q
,
max
⁡, Z
j
,
min
⁡, and Z
j
,
max
⁡ for all q
∈
{
1
,
…
,
t
} and for all j
∈
{
1
,
…
,
s
}. Observe that since each Z
j is a zero difference variable, we have Z
j
,
min
⁡
=
Z
j
,
max
⁡ for all j
∈
{
1
,
…
,
s
}.For anyi
∈
{
1
,
…
,
r
}, let a
i
≔
-
(
Δ
N
i
) and b
i
≔
N
i
,
min
⁡. For any i
∈
{
1
,
…
,
r
} and any positive integer n, let M
a
i
,
n and r
a
i
,
n be defined by the equation p
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n.


## 4.2.2. The NumberM
max,
A

Definition 12 (ofM
max
⁡
,
A).
LetA be an arbitrary element of the set M (this set has been defined earlier) such that [7] divides A. It follows from Theorem 11 that for those monomials A which are divisible by [7], we need to keep an account of the largest positive integer M for which the monomial A
[
2
]
-
M
[
1
]
M
-
1 has no negative power for any of the variables x
1
,
…
,
x
m. Let M
max
⁡
,
A denote this largest positive integer.


## 4.2.3. An Explicit Account ofM
max,
A

Corollary15 below gives an explicit account of the number M
max
⁡
,
A, for any monomial A which is divisible by [7]. The number M
max
⁡
,
A is important to us because of Theorem 11 which provides us with a way to count the required length l
(
R
/
(
m
[
p
n
]
+
J
)
).Remark 13.
LetA be any monomial in M which [
2
] divides; then it is of the form
(13)
N
r
p
n
-
k
N
r
⋯
N
1
p
n
-
k
N
1
Z
s
p
n
-
k
Z
s
⋯
Z
1
p
n
-
k
Z
1
P
1
j
1
⋯
P
t
j
t
,
where p
n
-
1
≥
j
q
≥
P
q
,
min
⁡ for any q
∈
{
1
,
…
,
t
}, p
n
-
1
≥
p
n
-
k
N
i
≥
N
i
,
max
⁡ for every 1
≤
i
≤
r, and p
n
-
1
≥
p
n
-
k
Z
i
≥
Z
i
,
max
⁡ for every 1
≤
i
≤
s.
Hence for any positive integerM, A
[
2
]
-
M
[
1
]
M
-
1 is of the form
(14)
∏
i
=
1
r
N
i
p
n
-
k
N
i
-
a
i
M
-
b
i
∏
i
=
1
s
Z
i
p
n
-
k
Z
i
-
Z
i
,
min
⁡
∏
q
=
1
t
P
q
j
q
+
M
Δ
P
q
-
P
q
,
max
⁡
,
where p
n
-
1
≥
j
q
≥
P
q
,
min
⁡ for any q
∈
{
1
,
…
,
t
}, p
n
-
1
≥
p
n
-
k
N
i
≥
N
i
,
max
⁡ for every 1
≤
i
≤
r, and p
n
-
1
≥
p
n
-
k
Z
i
≥
Z
i
,
max
⁡ for every 1
≤
i
≤
s.
It is now easy to see that for anyM
≥
1, the power of any zero difference variable or any positive difference variable in A
[
2
]
-
M
[
1
]
M
-
1 is ≥
0. But the power of any negative difference variable in A
[
2
]
-
M
[
1
]
M
-
1 may be negative for some values of M.The following lemma and its corollary give an explicit account ofM
max
⁡
,
A for any monomial A in M which [
2
] divides. But for stating the lemma, we need some notation. Let i
∈
{
1
,
…
,
r
} and let n be any positive integer. If b
i
≥
a
i, then let y
a
i denote the least positive integer ≥
(
b
i
-
r
a
i
,
n
) which is divisible by a
i. Let q
a
i be defined by the equation y
a
i
=
a
i
q
a
i. Let
(15)
p
a
i
≔
0
if
b
i
<
a
i
and
r
a
i
,
n
≥
b
i
,
1
if
b
i
<
a
i
and
r
a
i
,
n
<
b
i
,
q
a
i
if
b
i
≥
a
i
clearly
here
r
a
i
,
n
<
b
i
.
And let
(16)
E
a
i
,
b
i
≔
r
a
i
,
n
-
b
i
if
p
a
i
=
0
,
(
r
a
i
,
n
-
b
i
)
+
a
i
if
p
a
i
=
1
,
y
a
i
-
(
b
i
-
r
a
i
,
n
)
if
p
a
i
=
q
a
i
.
For any real number β, let β denote the smallest positive integer ≥β. For any 1
≤
i
≤
r, let
(17)
k
~
N
i
,
last
≔
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
a
i
.Lemma 14.
For any1
≤
i
≤
r, if k
N
i is such that 1
≤
k
N
i
≤
p
n
-
N
i
,
max
⁡, then the maximum value of the positive integer M for which p
n
-
k
N
i
-
a
i
M
-
b
i is ≥0 equals(18)
M
a
i
,
n
-
p
a
i
i
f
1
≤
k
N
i
≤
1
+
E
a
i
,
b
i
M
a
i
,
n
-
p
a
i
-
1
i
f
1
+
E
a
i
,
b
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
M
a
i
,
n
-
p
a
i
-
2
i
f
1
+
E
a
i
,
b
i
+
a
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
2
a
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
i
f
1
+
E
a
i
,
b
i
+
a
i
(
k
~
N
i
-
1
)
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
l
a
s
t
i
f
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
l
a
s
t
-
1
<
k
N
i
≤
p
n
-
N
i
,
max
⁡
.Proof.
Observe that sincep
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n, we have p
n
-
k
N
i
-
a
i
M
-
b
i
≥
0 if and only if M
≤
M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
). So we need to compute the greatest positive integer M, that is, ≤M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
).
Case  1 (when  b
i
≥
a
i
). In this case, we have 0
≤
r
a
i
,
n
<
a
i
≤
b
i. The least positive integer ≥
b
i
-
r
a
i
,
n which is divisible by a
i is y
a
i. Therefore the least positive integer ≥
(
b
i
-
r
a
i
,
n
)
/
a
i is q
a
i
=
p
a
i. Therefore the greatest negative integer ≤
-
(
b
i
-
r
a
i
,
n
)
/
a
i is -
p
a
i. And hence the greatest positive integer ≤
M
a
i
,
n
-
(
b
i
-
r
a
i
,
n
)
/
a
i is M
a
i
,
n
-
p
a
i.
Sincek
N
i
≤
p
n
-
N
i
,
max
⁡, it follows that M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
)
≥
M
a
i
,
n
-
p
a
i
-
(
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i
). The smallest positive integer ≥
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i is k
~
N
i
,
last. Therefore the greatest positive integer ≤
M
a
i
,
n
-
p
a
i
-
(
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i
) is M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last. As k
N
i increases, M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
) decreases, the lower bound being equal to M
a
i
,
n
-
p
a
i
-
(
(
p
n
-
N
i
,
max
⁡
-
1
-
E
a
i
,
b
i
)
/
a
i
). Therefore the greatest positive integer ≤
M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
) equals M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last.
Hence the greatest positive integer≤
C
i
,
n
≔
M
a
i
,
n
-
(
(
b
i
-
r
a
i
,
n
)
/
a
i
)
-
(
(
k
N
i
-
1
)
/
a
i
) is equal to(19)
M
a
i
,
n
-
p
a
i
if
M
a
i
,
n
-
p
a
i
≤
C
i
,
n
M
a
i
,
n
-
p
a
i
-
1
if
M
a
i
,
n
-
p
a
i
-
1
≤
C
i
,
n
<
M
a
i
,
n
-
p
a
i
M
a
i
,
n
-
p
a
i
-
2
if
M
a
i
,
n
-
p
a
i
-
2
≤
C
i
,
n
<
M
a
i
,
n
-
p
a
i
-
1
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
if
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
≤
C
i
,
n
<
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
+
1
.
This is equivalent to saying that the maximum value of the positive integerM for which p
n
-
k
N
i
-
a
i
M
-
b
i is ≥0 equals(20)
M
a
i
,
n
-
p
a
i
if
1
≤
k
N
i
≤
1
+
E
a
i
,
b
i
M
a
i
,
n
-
p
a
i
-
1
if
1
+
E
a
i
,
b
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
M
a
i
,
n
-
p
a
i
-
2
if
1
+
E
a
i
,
b
i
+
a
i
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
2
a
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
if
1
+
E
a
i
,
b
i
+
a
i
(
k
~
N
i
-
1
)
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
⋮
⋮
⋮
M
a
i
,
n
-
p
a
i
-
k
~
N
i
,
last
if
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
last
-
1
<
k
N
i
≤
p
n
-
N
i
,
max
⁡
.
Case  2 (when
b
i
<
a
i
). In this case, we have 2 subcases.
Subcase  2.1 (when
b
i
≤
r
a
i
,
n
). Here we have p
a
i
=
0 and 0
≤
b
i
≤
r
a
i
,
n
<
a
i. Therefore 0
≤
(
r
a
i
,
n
-
b
i
)
/
a
i
<
1 and hence -
1
<
(
b
i
-
r
a
i
,
n
)
/
a
i
≤
0. Therefore the least nonnegative integer ≥
(
b
i
-
r
a
i
,
n
)
/
a
i is p
a
i
=
0. The rest of the proof is similar to the proof of Case  1.
Subcase  2.2 (when
b
i
>
r
a
i
,
n
). Here we have, p
a
i
=
1 and 0
≤
r
a
i
,
n
<
b
i
<
a
i. Therefore 0
<
(
b
i
-
r
a
i
,
n
)
/
a
i
<
1. Hence the least positive integer ≥
(
b
i
-
r
a
i
,
n
)
/
a
i is p
a
i
=
1. The rest of the proof is similar to the proof of Case  1.The following corollary is immediate.Corollary 15.
LetA be a monomial in M which [
2
] divides and let M be a positive integer. Let A and A
[
2
]
-
M
[
1
]
M
-
1 be of the form as given in Remark 13. For 1
≤
i
≤
r, let M
i
,
A be the maximum value of the positive integer M for which p
n
-
k
N
i
-
a
i
M
-
b
i is ≥
0.
ThenM
max
⁡
,
A equals the minimum of all M
i
,
As as i ranges from 1 to r.LetM
[
2
]
≔ the set of all monomials in M which are divisible by [
2
]. Let M
max
⁡
[
2
]
≔
{
M
max
⁡
,
A
∣
A
∈
M
[
2
]
}.Remark 16.
It follows from Corollary15 above that M
max
⁡
[
2
]
=
{
M
a
r
,
n
-
p
a
r
-
j
∣
0
≤
j
≤
k
~
N
r
,
last
}.


## 4.2.4. The Length Count for a FixedM
max,
A

Remark 17.
It is easy to verify that ifM
1 is a positive integer such that A
[
2
]
-
M
1
[
1
]
M
1
-
1 contains no negative powers and M
2 is a positive integer such that M
2
≤
M
1, then A
[
2
]
-
M
2
[
1
]
M
2
-
1 contains no negative powers. It is also easy to verify that if N
1 is a positive integer such that A
[
2
]
-
N
1
[
1
]
N
1 belongs to A
c and N
2 is a positive integer such that N
1
≤
N
2, then A
[
2
]
-
N
2
[
1
]
N
2 belongs to A
c. Hence for any A
∈
M
[
2
], A
∈
A
c
+
J if and only if either [
1
] divides A or A
[
2
]
-
M
max
⁡
,
A
[
1
]
M
max
⁡
,
A belongs to A
c.Remark 18.
Recall from Remark13 as well as from Corollary 15 that given any monomial A in M
[
2
], the value of M
max
⁡
,
A depends only on the powers of the negative difference variables appearing in A. It now follows that for each i
∈
{
1
,
…
,
r
}, if we fix numbers k
N
i
0 such that p
n
-
1
≥
p
n
-
k
N
i
0
≥
N
i
,
max
⁡, then for any A in the set
(21)
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
≔
A
∣
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
where
p
n
-
1
≥
l
j
≥
0
for
every
j
∈
1
,
…
,
s
and
,
N
r
p
n
-
k
N
r
0
p
n
-
1
≥
j
q
≥
0
for
every
q
∈
1
,
…
,
t
,
the value of M
max
⁡
,
A is the same (call it M
max
⁡
0).The main result of this subsubsection is Lemma24 below which provides an account of the length count for a fixed value of M
max
⁡
,
A. But for stating the lemma, we need to introduce some notation first.For an arbitrary elementM
max
⁡ in M
max
⁡
[
2
] and for every q
∈
{
1
,
…
,
t
}, let
(22)
Mi
n
P
q
,
M
max
⁡
c
≔
Min
⁡
p
n
-
P
q
,
min
⁡
,
M
max
⁡
Δ
P
q
.
The symbol Mi
n
c in the above definition of Mi
n
P
q
,
M
max
⁡
c stands for “minimum convergent.” The reason behind this notation is explained in Remark 19 below.Remark 19.
Observe that if the power of the variableP
q in the monomial A is j
q, then the power of P
q in the monomial A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ is j
q
+
M
max
⁡
Δ
P
q. If A is divisible by [
2
], then p
n
-
1
≥
j
q
≥
P
q
,
min
⁡. Now if j
q
+
M
max
⁡
Δ
P
q
≥
p
n, then A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c. That is, if p
n
-
j
q
≤
Mi
n
P
q
,
M
max
⁡
c, then A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c. That is, if p
n
-
Mi
n
P
q
,
M
max
⁡
c
≤
j
q
≤
p
n
-
1, then A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c. So p
n
-
Mi
n
P
q
,
M
max
⁡
c is the minimum possible power of the variable P
q in the monomial A for which A
[
2
]
-
M
max
⁡
[
1
]
M
max
⁡ belongs to A
c.Recall the ordered setP
1
⊳
⋯
⊳
P
t of positive difference variables. Let A
t
+
1
≔
0 and C
t
≔
1. For any q
∈
{
1
,
…
,
t
-
1
}, let
(23)
C
q
≔
p
(
t
-
q
)
n
.
And for any q
∈
{
2
,
…
,
t
}, let
(24)
A
q
≔
p
n
-
P
q
,
min
⁡
A
q
+
1
+
P
q
,
min
⁡
C
q
.
For any M
max
⁡ in M
max
⁡
[
2
], let D
t
+
1
,
M
max
⁡
:
=
1 and D
~
t
+
1
,
M
max
⁡
≔
0. For any M
max
⁡ in M
max
⁡
[
2
] and any q
∈
{
1
,
…
,
t
}, let
(25)
D
q
,
M
max
⁡
≔
Mi
n
P
q
,
M
max
⁡
c
A
q
+
1
+
p
n
-
P
q
,
min
⁡
-
Mi
n
P
q
,
M
max
⁡
c
D
q
+
1
,
M
max
⁡
+
P
q
,
min
⁡
C
q
,
(26)
D
~
q
,
M
max
⁡
≔
Mi
n
P
q
,
M
max
⁡
c
A
q
+
1
+
(
p
n
-
Mi
n
P
q
,
M
max
⁡
c
-
P
q
,
max
⁡
)
D
~
q
+
1
,
M
max
⁡
+
P
q
,
max
⁡
-
P
q
,
min
⁡
D
q
+
1
,
M
max
⁡
+
(
P
q
,
min
⁡
)
C
q
if
Mi
n
P
q
,
M
max
⁡
c
<
p
n
-
P
q
,
min
⁡
and
P
q
,
max
⁡
<
p
n
-
Mi
n
P
q
,
M
max
⁡
c
D
q
,
M
max
⁡
if
Mi
n
P
q
,
M
max
⁡
c
<
p
n
-
P
q
,
min
⁡
and
P
q
,
max
⁡
≥
p
n
-
Mi
n
P
q
,
M
max
⁡
c
(
Mi
n
P
q
,
M
max
⁡
c
)
A
q
+
1
+
(
P
q
,
min
⁡
)
C
q
if
Mi
n
P
q
,
M
max
⁡
c
=
p
n
-
P
q
,
min
⁡
.
Recall the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) from Remark 18. For any A in this set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
), the value of M
max
⁡
,
A is the same, namely, M
max
⁡
0. Lemma 24 below gives an account of the length count for the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
). But we will break up the proof of Lemma 24 in a few short steps as follows.Lemma 20.
For everyq
∈
{
2
,
…
,
t
}, let
(27)
F
q
,
A
,
J
1
,
…
,
J
q
-
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
)
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
a
n
d
j
1
=
J
1
,
…
,
j
q
-
1
=
J
q
-
1
,
where for each k
∈
{
1
,
…
,
q
-
1
}, J
k is a fixed integer such that P
k
,
min
⁡
≤
J
k
≤
p
n
-
1 and at least one J
k is ≥
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c.
Then the total number of monomialsA in the set F
q
,
A
,
J
1
,
…
,
J
q
-
1 for which A
∉
A
c
+
J equals A
q.Proof.
The proof is by reverse induction onq, the base case of induction being q
=
t. Observe that if A is any monomial in F
t
,
A
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
A
,
J
1
,
…
,
J
t
-
1 is p
n
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
A
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
A
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
A
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(28)
p
n
-
P
t
,
min
⁡
0
+
P
t
,
min
⁡
1
=
p
n
-
P
t
,
min
⁡
A
t
+
1
+
P
t
,
min
⁡
C
t
=
A
t
.
This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.Lemma 21.
For everyq
∈
{
2
,
…
,
t
}, let
(29)
F
q
,
C
,
J
1
,
…
,
J
q
-
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
)
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
,
j
1
=
J
1
,
…
,
j
q
-
1
=
J
q
-
1
,
where for each k
∈
{
1
,
…
,
q
-
1
}, J
k is a fixed integer such that 0
≤
J
k
≤
p
n
-
1 and at least one J
k is <
P
k
,
min
⁡.
Then the total number of monomialsA in the set F
q
,
C
,
J
1
,
…
,
J
q
-
1 for which A
∉
A
c
+
J equals C
q
-
1.Proof.
The proof is by reverse induction onq, the base case of induction being q
=
t. Observe that if A is any monomial in F
t
,
C
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
≤
p
n
-
1, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
C
,
J
1
,
…
,
J
t
-
1 is p
n. Hence the total number of monomials A in the set F
t
,
C
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals p
n
=
p
(
t
-
(
t
-
1
)
)
n
=
C
t
-
1. This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.Lemma 22.
For everyq
∈
{
2
,
…
,
t
}, let
(30)
F
q
,
D
,
J
1
,
…
,
J
q
-
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
)
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
a
n
d
j
1
=
J
1
,
…
,
j
q
-
1
=
J
q
-
1
,
where for each k
∈
{
1
,
…
,
q
-
1
}, J
k is a fixed integer such that P
k
,
min
⁡
≤
J
k
<
p
n
-
Min
P
k
,
M
max
⁡
0
c.
Then the total number of monomialsA in the set F
q
,
D
,
J
1
,
…
,
J
q
-
1 for which A
∉
A
c
+
J equals (i)
D
q
,
M
max
⁡
0 if there exists at least one k
∈
{
1
,
…
,
q
-
1
} such that P
k
,
max
⁡
≥
p
n
-
Min
P
k
,
M
max
⁡
0
c;
(ii)
D
~
q
,
M
max
⁡
0 if for every k
∈
{
1
,
…
,
q
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Min
P
k
,
M
max
⁡
0
c and P
k
,
max
⁡
≤
J
k
<
p
n
-
Min
P
k
,
M
max
⁡
0
c;
(iii)
D
q
,
M
max
⁡
0 if for every k
∈
{
1
,
…
,
q
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Min
P
k
,
M
max
⁡
0
c and there exists at least one k
∈
{
1
,
…
,
q
-
1
} such that P
k
,
min
⁡
≤
J
k
<
P
k
,
max
⁡.Proof.
The proof is by reverse induction onq, the base case of induction being q
=
t. Let us first prove the base case of induction.
Case  1. When there exists at least one k
∈
{
1
,
…
,
t
-
1
} such that P
k
,
max
⁡
≥
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c.
Then since we have that for eachk
∈
{
1
,
…
,
t
-
1
}, J
k is a fixed integer such that P
k
,
min
⁡
≤
J
k
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c, it follows that monomials in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 are not divisible by [
1
].
Observe now that ifA is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then [
2
] divides A but [
1
] does not divide A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∉
A
c. Hence such a monomial A satisfies A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(31)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
1
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
D
t
+
1
,
M
max
⁡
0
+
P
t
,
min
⁡
C
t
=
D
t
,
M
max
⁡
0
.
This proves the base case of induction in Case  1.
Case  2. When for every k
∈
{
1
,
…
,
t
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c and P
k
,
max
⁡
≤
J
k
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c. In this case, there are 3 subcases.
Subcase  2.1 (
Mi
n
P
t
,
M
max
⁡
0
c
=
p
n
-
P
t
,
min
⁡
). Observe that if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡
=
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(32)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
P
t
,
min
⁡
C
t
=
D
~
t
,
M
max
⁡
0
.
This proves the base case of induction in this subcase.
Subcase  2.2 (
Mi
n
P
t
,
M
max
⁡
0
c
<
p
n
-
P
t
,
min
⁡
and
P
t
,
max
⁡
≥
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
). Observe that if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then [
2
] divides A but [
1
] does not divide A (since P
t
,
max
⁡
≥
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c) and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∉
A
c. Hence such a monomial A satisfies A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(33)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
1
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
min
⁡
D
t
+
1
,
M
max
⁡
0
+
P
t
,
min
⁡
C
t
=
D
t
,
M
max
⁡
0
=
D
~
t
,
M
max
⁡
0
.
This proves the base case of induction in this subcase.
Subcase  2.3 (
Mi
n
P
t
,
M
max
⁡
0
c
<
p
n
-
P
t
,
min
⁡
and
P
t
,
max
⁡
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
). Observe that if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
≤
j
t
≤
p
n
-
1, then [
2
] divides A and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∈
A
c. Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is Mi
n
P
t
,
M
max
⁡
0
c. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
max
⁡
≤
j
t
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c, then [
1
] divides A (since P
t
,
max
⁡
≤
j
t and P
k
,
max
⁡
≤
J
k
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c for every k
∈
{
1
,
…
,
t
-
1
}). Hence such a monomial A satisfies A
∈
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
max
⁡. If A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality P
t
,
min
⁡
≤
j
t
<
P
t
,
max
⁡, then [
2
] divides A but [
1
] does not divide A (since j
t
<
P
t
,
max
⁡) and A
[
2
]
-
M
max
⁡
0
[
1
]
M
max
⁡
0
∉
A
c (since j
t
<
P
t
,
max
⁡
<
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c). Hence such a monomial A satisfies A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
max
⁡
-
P
t
,
min
⁡. And if A is any monomial in F
t
,
D
,
J
1
,
…
,
J
t
-
1 such that the power j
t of the variable P
t in it satisfies the inequality 0
≤
j
t
<
P
t
,
min
⁡, then neither [
2
] nor [
1
] divides A and hence A
∉
A
c
+
J. The total number of such monomials in F
t
,
D
,
J
1
,
…
,
J
t
-
1 is P
t
,
min
⁡. Hence the total number of monomials A in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 for which A
∉
A
c
+
J equals
(34)
Mi
n
P
t
,
M
max
⁡
0
c
0
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
max
⁡
0
+
P
t
,
max
⁡
-
P
t
,
min
⁡
1
+
P
t
,
min
⁡
1
=
Mi
n
P
t
,
M
max
⁡
0
c
A
t
+
1
+
p
n
-
Mi
n
P
t
,
M
max
⁡
0
c
-
P
t
,
max
⁡
×
D
~
t
+
1
,
M
max
⁡
0
+
P
t
,
max
⁡
-
P
t
,
min
⁡
D
t
+
1
,
M
max
⁡
0
+
(
P
t
,
min
⁡
)
C
t
=
D
~
t
,
M
max
⁡
0
.
This proves the base case of induction in this subcase. Hence the base case of induction is proved in Case  2.
Case  3. When for every k
∈
{
1
,
…
,
t
-
1
}, we have P
k
,
max
⁡
<
p
n
-
Mi
n
P
k
,
M
max
⁡
0
c and there exists at least one k
∈
{
1
,
…
,
t
-
1
} such that P
k
,
min
⁡
≤
J
k
<
P
k
,
max
⁡.
Since there exists at least onek
∈
{
1
,
…
,
t
-
1
} such that P
k
,
min
⁡
≤
J
k
<
P
k
,
max
⁡, it follows that monomials in the set F
t
,
D
,
J
1
,
…
,
J
t
-
1 are not divisible by [
1
]. The rest of the proof of this case is similar to the proof of Case  1. This proves the base case of induction in Case  3.
This proves the base case of induction. The rest of the proof follows by reverse induction and is left to the reader.Let(35)
F
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
1
=
p
n
-
1
.Lemma 23.
The total number of monomialsA in the set F
1 for which A
∉
A
c
+
J equals D
~
1
,
M
max
⁡
0.Proof.
LetJ
1 be an arbitrarily fixed integer such that p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
≤
J
1
≤
p
n
-
1. It then follows from Lemma 20 that for any such fixed J
1, the total number of monomials A in F
2
,
A
,
J
1 such that A
∉
A
c
+
J equals A
2. Also if J
1 is an arbitrarily fixed integer such that 0
≤
J
1
<
P
1
,
min
⁡, then it follows from Lemma 21 that for any such fixed J
1, the total number of monomials A in F
2
,
C
,
J
1 such that A
∉
A
c
+
J equals C
1. Now there are 3 cases.
Case  1 (if
Mi
n
P
1
,
M
max
⁡
0
c
=
p
n
-
P
1
,
min
⁡
). In this case, the total number of monomials A in the set F
1 for which A
∉
A
c
+
J equals
(36)
p
n
-
1
-
p
n
+
Mi
n
P
1
,
M
max
⁡
0
c
+
1
A
2
+
P
1
,
min
⁡
C
1
=
Mi
n
P
1
,
M
max
⁡
0
c
A
2
+
P
1
,
min
⁡
C
1
=
D
~
1
,
M
max
⁡
0
.
Case  2 (if
Mi
n
P
1
,
M
max
⁡
0
c
<
p
n
-
P
1
,
min
⁡
and
P
1
,
max
⁡
≥
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
). It follows from Lemma 22 that for any fixed integer J
1 such that P
1
,
min
⁡
≤
J
1
<
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c, the total number of monomials A in F
2
,
D
,
J
1 for which A
∉
A
c
+
J equals D
2
,
M
max
⁡
0. Hence in this case, the total number of monomials A in the set F
1 for which A
∉
A
c
+
J equals
(37)
p
n
-
1
-
p
n
+
Mi
n
P
1
,
M
max
⁡
0
c
+
1
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
Mi
n
P
1
,
M
max
⁡
0
c
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
D
1
,
M
max
⁡
0
=
D
~
1
,
M
max
⁡
0
.
Case  3 (if
Mi
n
P
1
,
M
max
⁡
0
c
<
p
n
-
P
1
,
min
⁡
and
P
1
,
max
⁡
<
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
). It follows from Lemma 22 that for any fixed integer J
1 such that P
1
,
max
⁡
≤
J
1
<
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c, the total number of monomials A in F
2
,
D
,
J
1 for which A
∉
A
c
+
J equals D
~
2
,
M
max
⁡
0. It again follows from Lemma 22 that for any fixed integer J
1 such that P
1
,
min
⁡
≤
J
1
<
P
1
,
max
⁡, the total number of monomials A in F
2
,
D
,
J
1 for which A
∉
A
c
+
J equals D
2
,
M
max
⁡
0. Hence in this case, the total number of monomials A in the set F
1 for which A
∉
A
c
+
J equals
(38)
p
n
-
1
-
p
n
+
Mi
n
P
1
,
M
max
⁡
0
c
+
1
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
max
⁡
D
~
2
,
M
max
⁡
0
+
P
1
,
max
⁡
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
Mi
n
P
1
,
M
max
⁡
0
c
A
2
+
p
n
-
Mi
n
P
1
,
M
max
⁡
0
c
-
P
1
,
max
⁡
D
~
2
,
M
max
⁡
0
+
P
1
,
max
⁡
-
P
1
,
min
⁡
D
2
,
M
max
⁡
0
+
P
1
,
min
⁡
C
1
=
D
~
1
,
M
max
⁡
0
.Lemma 24.
The total number of monomialsA in the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
s where S
M
max
⁡
0
,
s is defined inductively as follows:
(39)
S
M
max
⁡
0
,
0
≔
D
~
1
,
M
max
⁡
0
,
f
o
r
e
v
e
r
y
j
∈
1
,
…
,
s
,
S
M
max
⁡
0
,
j
≔
S
M
max
⁡
0
,
j
-
1
p
n
-
Z
j
,
min
⁡
+
p
t
+
j
-
1
n
Z
j
,
min
⁡
.Proof.
Fix an arbitrary integerL
1 such that p
n
-
1
≥
L
1
≥
Z
1
,
min
⁡. Let
(40)
P
L
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
2
=
p
n
-
1
,
l
1
=
L
1
.
It is now easy to see that the total number of monomials A in P
L
1 for which A
∉
A
c
+
J equals D
~
1
,
M
max
⁡
0
=
S
M
max
⁡
0
,
0 (the proof of this is similar to the proof of Lemma 23).
Fix an arbitrary integerM
1 such that 0
≤
M
1
<
Z
1
,
min
⁡. Let
(41)
P
M
1
≔
A
=
N
r
p
n
-
k
N
r
0
⋯
N
1
p
n
-
k
N
1
0
Z
s
l
s
⋯
Z
1
l
1
P
1
j
1
⋯
P
t
j
t
∈
P
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
∣
l
s
=
l
s
-
1
=
⋯
N
r
p
n
-
k
N
r
0
=
l
2
=
p
n
-
1
,
l
1
=
M
1
.
Since 0
≤
M
1
<
Z
1
,
min
⁡, therefore any monomial A
∈
P
M
1 is divisible neither by [
1
] nor by [
2
]. Hence the total number of monomials A in P
M
1 for which A
∉
A
c
+
J equals p
t
n.
So if we defineP
1 as the subset of P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) consisting of all those monomials in P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) for which l
s
=
l
s
-
1
=
⋯
=
l
2
=
p
n
-
1, then the total number of monomials A in the set P
1 for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
1. Similarly, if we define P
2 as the subset of P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) consisting of all those monomials in it for which l
s
=
l
s
-
1
=
⋯
=
l
3
=
p
n
-
1, then the total number of monomials A in the set P
2 for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
2. Proceeding inductively, it is now easy to see that the total number of monomials A in the set P
(
k
N
1
0
,
k
N
2
0
,
…
,
k
N
r
0
) for which A
∉
A
c
+
J equals S
M
max
⁡
0
,
s.Definition 25.
Given anyM
max
⁡ in M
max
⁡
[
2
], let Φ
M
max
⁡
≔
S
M
max
⁡
,
s where S
M
max
⁡
,
s is defined in the same way as S
M
max
⁡
0
,
s is defined in Lemma 24 (just replace M
max
⁡
0 in Lemma 24 by M
max
⁡).


## 4.2.5. The Formula for the Hilbert-Kunz Function

We will now state the formula for the Hilbert-Kunz function for the general binomial hypersurface that we have considered above. But for stating the formula, we need some more notation, which we provide first.SetG
P
(
t
-
1
)
≔
P
t
,
max
⁡ and for every q
∈
{
1
,
…
,
(
t
-
2
)
}, let G
P
q
≔
(
p
n
-
P
q
+
1
,
max
⁡
)
G
P
q
+
1
+
(
P
q
+
1
,
max
⁡
)
p
n
(
t
-
q
-
1
);
SetG
Z
1
≔
(
p
n
-
P
1
,
max
⁡
)
G
P
1
+
(
P
1
,
max
⁡
)
p
n
(
t
-
1
), and for every l
∈
{
2
,
…
,
s
}, let G
Z
l
≔
(
p
n
-
Z
l
-
1
,
max
⁡
)
G
Z
l
-
1
+
(
Z
l
-
1
,
max
⁡
)
p
n
(
t
+
l
-
2
);
SetG
N
1
≔
(
p
n
-
Z
s
,
max
⁡
)
G
Z
s
+
(
Z
s
,
max
⁡
)
p
n
(
t
+
s
-
1
), and for every i
∈
{
2
,
…
,
r
}, let G
N
i
≔
(
p
n
-
N
i
-
1
,
min
⁡
)
G
N
i
-
1
+
(
N
i
-
1
,
min
⁡
)
p
n
(
t
+
s
+
i
-
2
). Recall that for any i
∈
{
1
,
…
,
r
}, let a
i
≔
-
(
Δ
N
i
) and for any i
∈
{
1
,
…
,
r
} and any positive integer n, M
a
i
,
n and r
a
i
,
n are defined by the equation p
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n. Also recall the definitions of E
a
i
,
b
i and p
a
i from Section 4.2.3.For everyi
∈
{
1
,
…
,
r
}, let
(42)
D
a
i
,
a
r
≔
1
+
E
a
i
,
b
i
+
a
i
M
a
i
,
n
-
p
a
i
-
M
a
r
,
n
+
p
a
r
a
i
′
≔
p
n
-
N
i
,
max
⁡
-
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
last
-
1
,
X
~
N
i
,
last
≔
M
a
r
,
n
-
p
a
r
-
M
a
i
,
n
+
p
a
i
+
k
~
N
i
,
last
.
Clearly a
i
′
≤
a
i for every i
∈
{
1
,
…
,
r
}.For any nonnegative integerj such that M
a
r
,
n
-
p
a
r
-
j belongs to the set M
max
⁡
[
2
], let
(43)
∫
j
(
N
1
)
‍
≔
Φ
M
a
r
,
n
-
p
a
r
-
j
,
for
any
i
∈
2
,
…
,
r
,
and let
(44)
∫
j
(
N
i
)
‍
≔
D
a
i
-
1
,
a
r
+
j
a
i
-
1
∫
j
(
N
i
-
1
)
‍
+
a
i
-
1
∫
j
+
1
(
N
i
-
1
)
‍
+
⋯
+
a
i
-
1
∫
(
X
~
N
i
-
1
,
last
)
-
1
(
N
i
-
1
)
‍
+
a
i
-
1
′
∫
X
~
N
i
-
1
,
last
(
N
i
-
1
)
‍
+
N
i
-
1
,
max
⁡
-
N
i
-
1
,
min
⁡
G
N
i
-
1
+
N
i
-
1
,
min
⁡
p
n
(
s
+
t
+
i
-
2
)
if
j
<
X
~
N
i
-
1
,
last
p
n
-
N
i
-
1
,
max
⁡
∫
j
(
N
i
-
1
)
‍
+
N
i
-
1
,
max
⁡
-
N
i
-
1
,
min
⁡
G
N
i
-
1
+
N
i
-
1
,
min
⁡
p
n
(
s
+
t
+
i
-
2
)
otherwise
.Theorem 26.
The formula for the Hilbert-Kunz function of the general binomial hypersurface under consideration is(45)
D
a
r
,
a
r
∫
0
(
N
r
)
‍
+
a
r
∫
1
(
N
r
)
‍
+
⋯
+
a
r
∫
X
~
N
r
,
l
a
s
t
-
1
N
r
‍
+
a
r
′
∫
X
~
N
r
,
l
a
s
t
(
N
r
)
‍
+
N
r
,
max
⁡
-
N
r
,
min
⁡
G
N
r
+
N
r
,
min
⁡
p
n
(
s
+
t
+
r
-
1
)
.Proof.
Observe that for any fixedi
∈
{
2
,
…
,
r
} and any j
∈
Z such that 0
≤
j
≤
k
~
N
r
,
last, the number ∫
j
(
N
i
)
‍ provides an account of the number of monomials A in M which satisfy the following 3 properties simultaneously.(i)
A
∉
A
c
+
J.
(ii)
The power of each of the variablesN
i
+
1
,
…
,
N
r is p
n
-
1.
(iii)
The power of the variableN
i is p
n
-
k
N
i where
(46)
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
-
1
<
k
N
i
≤
1
+
E
a
i
,
b
i
+
a
i
k
~
N
i
,
and k
~
N
i
=
M
a
i
,
n
-
p
a
i
-
M
a
r
,
n
+
p
a
r
+
j.
Therefore for anyj
∈
Z such that 0
≤
j
≤
X
~
N
r
,
last, ∫
j
(
N
r
)
‍ denotes the number of monomials A in M for which A
∉
A
c
+
J and the power of the variable N
r is p
n
-
k
N
r where 1
+
E
a
r
,
b
r
+
a
r
(
j
-
1
)
<
k
N
r
≤
1
+
E
a
r
,
b
r
+
a
r
j.
Clearly then the number of monomialsA in M for which A
∉
A
c
+
J and the power of the variable N
r is p
n
-
k
N
r where 1
≤
k
N
r
≤
1
+
E
a
r
,
b
r equals D
a
r
,
a
r
∫
0
(
N
r
)
‍. Similarly, the number of monomials A in M for which A
∉
A
c
+
J and the power of the variable N
r is p
n
-
k
N
r where 1
+
E
a
r
,
b
r
+
a
r
(
j
-
1
)
<
k
N
r
≤
1
+
E
a
r
,
b
r
+
a
r
j and 1
≤
j
≤
(
X
~
N
r
,
last
)
-
1 equals a
r
∫
j
(
N
r
)
‍. We can similarly account for the terms a
r
′
∫
X
~
N
r
,
last
(
N
r
)
‍, (
N
r
,
max
⁡
-
N
r
,
min
⁡
)
G
N
r, and (
N
r
,
min
⁡
)
p
n
(
s
+
t
+
r
-
1
) in the formula (45) of the Hilbert-Kunz function.


## 4.3. An Illustrative Example

For this example as well as for the example given in Section4.4 below, we will need the following notation.Given any real numberβ, let β denote the smallest integer ≥
β and let [
β
] denote the largest integer ≤
β. We call β and [
β
]
the upper and the lower integral parts of β.LetS
=
k
[
N
2
,
N
1
,
Z
2
,
Z
1
,
P
1
,
P
2
] where k is a field of arbitrary prime characteristic p
>
0, and let J be the ideal in S defined by J
=
(
f
) where f
=
N
2
10
N
1
6
Z
2
5
Z
1
2
P
1
3
P
2
4
+
N
2
4
N
1
2
Z
2
5
Z
1
2
P
1
6
P
2
9. The differences Δ
N
2
,
Δ
N
1
,
Δ
Z
2
,
Δ
Z
1
,
Δ
P
1, and Δ
P
2 are therefore given by -
6
,
-
4,0
,
0,3
,
5, respectively. Hence the N
i’s are the negative difference variables, the Z
i’s are the zero difference variables, and the P
i’s are the positive difference variables. Here a
1
=
-
Δ
N
1
=
4, a
2
=
-
Δ
N
2
=
6, b
1
=
N
1
,
min
⁡
=
2, and b
2
=
N
2
,
min
⁡
=
4. Let E
a
1
,
b
1
,
E
a
2
,
b
2 be as defined in Section 4.2.3 above. Here, E
a
1
,
b
1
=
E
4,2 and E
a
2
,
b
2
=
E
6,4. For any i
∈
{
1,2
}, let M
a
i
,
n and r
a
i
,
n be defined by the equation(s) p
n
-
1
=
a
i
M
a
i
,
n
+
r
a
i
,
n. That is,
(47)
M
a
1
,
n
=
M
4
,
n
=
p
n
-
1
-
r
4
,
n
4
,
M
a
2
,
n
=
M
6
,
n
=
p
n
-
1
-
r
6
,
n
6
,
Also
k
~
N
1
,
last
=
p
n
-
6
-
1
-
E
a
1
,
b
1
4
,
k
~
N
2
,
last
=
p
n
-
10
-
1
-
E
a
2
,
b
2
6
,
the
set
M
max
⁡
2
=
M
6
,
n
-
p
6
-
j
∣
j
∈
Z
,
0
≤
j
≤
k
~
N
2
,
last
.
For every j
∈
Z such that 0
≤
j
≤
k
~
N
2
,
last, we have
(48)
Mi
n
P
1
,
M
6
,
n
-
p
6
-
j
c
≔
Min
⁡
p
n
-
3
,
M
6
,
n
-
p
6
-
j
Δ
P
1
=
3
M
6
,
n
-
p
6
-
j
,
Mi
n
P
2
,
M
6
,
n
-
p
6
-
j
c
≔
Min
⁡
p
n
-
4
,
M
6
,
n
-
p
6
-
j
Δ
P
2
=
5
M
6
,
n
-
p
6
-
j
.
Using the formulae given in Lemma 24, it follows that
(49)
S
M
6
,
n
-
p
6
-
j
,
0
=
15
p
n
-
30
M
6
,
n
-
p
6
-
j
-
54
,
S
M
6
,
n
-
p
6
-
j
,
1
=
15
p
n
-
30
M
6
,
n
-
p
6
-
j
-
54
×
p
n
-
2
+
2
p
2
n
,
S
M
6
,
n
-
p
6
-
j
,
2
=
15
p
n
-
30
M
6
,
n
-
p
6
-
j
-
54
×
p
n
-
2
p
n
-
5
+
7
p
3
n
-
10
p
2
n
.
Since for any j
∈
Z such that 0
≤
j
≤
k
~
N
2
,
last, we know that ∫
j
(
N
1
)
‍
≔
S
M
6
,
n
-
p
6
-
j
,
s and s
= the total number of zero difference variables =
2 here, therefore we have
(50)
∫
j
N
1
‍
=
15
p
n
-
30
M
6
,
n
-
p
6
-
j
-
54
×
p
n
-
2
p
n
-
5
+
7
p
3
n
-
10
p
2
n
.
For getting the formula for the Hilbert-Kunz function in this case, we need to compute ∫
j
(
N
2
)
‍ for the various values of j under consideration. A straight forward computation tells us that for any j
∈
Z such that 0
≤
j
<
X
~
N
1
,
last, we have
(51)
∫
j
N
2
‍
=
D
a
1
,
a
2
+
4
j
∫
j
N
1
‍
+
4
∑
k
=
j
+
1
X
~
N
1
,
last
-
1
∫
k
N
1
‍
+
a
1
′
∫
X
~
N
1
,
last
N
1
‍
+
6
-
2
G
N
1
+
2
p
n
2
+
2
+
2
-
2
,
where
(52)
D
a
1
,
a
2
=
1
+
E
a
1
,
b
1
+
4
M
4
,
n
-
p
4
-
M
6
,
n
+
p
6
=
1
+
E
4,2
+
4
M
4
,
n
-
p
4
-
M
6
,
n
+
p
6
a
1
′
=
p
n
-
6
-
1
-
E
a
1
,
b
1
-
4
k
~
N
1
,
last
-
1
,
X
~
N
1
,
last
=
M
6
,
n
-
p
6
-
M
4
,
n
+
p
4
+
k
~
N
1
,
last
,
G
N
1
=
19
p
3
n
-
130
p
2
n
+
372
p
n
-
360
.
Similarly, for any j
∈
Z such that j
≥
X
~
N
1
,
last, we have
(53)
∫
j
N
2
‍
=
p
n
-
6
∫
j
N
1
‍
+
6
-
2
G
N
1
+
2
p
n
2
+
2
+
2
-
2
=
p
n
-
6
p
n
-
2
p
n
-
5
+
7
p
3
n
-
10
p
2
n
15
p
n
-
30
M
6
,
n
-
p
6
-
j
-
54
×
p
n
-
2
p
n
-
5
+
7
p
3
n
-
10
p
2
n
+
4
19
p
3
n
-
130
p
2
n
+
372
p
n
-
360
+
2
p
4
n
.
Clearly then ∫
j
(
N
2
)
‍ is a rational function in p and n for all values of j. Therefore, the formula for the corresponding Hilbert-Kunz function is
(54)
D
a
2
,
a
2
∫
0
N
2
‍
+
6
∑
j
=
1
X
~
N
2
,
last
-
1
∫
j
N
2
‍
+
a
2
′
∫
X
~
N
2
,
last
N
2
‍
+
(
10
-
4
)
G
N
2
+
4
p
n
2
+
2
+
2
-
1
,
where
(55)
D
a
2
,
a
2
=
1
+
E
a
2
,
b
2
=
1
+
E
6,4
,
a
2
′
=
p
n
-
10
-
1
-
E
a
2
,
b
2
-
6
k
~
N
2
,
last
-
1
,
X
~
N
2
,
last
=
M
6
,
n
-
p
6
-
M
4
,
n
+
p
4
+
k
~
N
2
,
last
,
G
N
2
=
21
p
4
n
-
168
p
3
n
+
632
p
2
n
-
1104
p
n
+
720
,
the
formulae
of
∫
j
(
N
2
)
‍
are
given
above
.
Since all the quantities involved in the above formula for the Hilbert-Kunz function are rational functions in the variables p and n, therefore the Hilbert-Kunz function is a rational function in p and n.


## 4.4. Example of the One-Dimensional Case

In this subsection, we will discuss the case of1-dimensional binomial hypersurfaces and will observe that using the above formula for the Hilbert-Kunz function, we do get that in this case of 1-dimension, the associated Hilbert-Kunz multiplicity is an integer (e.g., see Chapter 6, Corollary 6.2 of [7]). In this case, this integer happens to be equal to the ordinary multiplicity because the Hilbert-Kunz multiplicity is equal to the ordinary multiplicity for 1-dimensional rings.In the case of1-dimensional binomial hypersurfaces, we have S
=
k
[
x
1
,
x
2
], J
=
(
f
) where f
=
[
2
]
+
[
1
], R
=
S
/
J and the terms [
2
] and [
1
] of f are monomials in the 2 variables x
1 and x
2.There are3 possible cases:(1)
x
1
⊳
x
2 where x
1 is a negative difference variable and x
2 is a positive difference variable;
(2)
x
1
⊳
x
2 where x
1 is a negative difference variable and x
2 is a zero difference variable;
(3)
x
1
⊳
x
2 where x
1 is a zero difference variable and x
2 is a positive difference variable. We will first discuss Cases  (2) and (3).In Cases  (2) and (3), it is easy to check (without using the formula above) following.The Hilbert-Kunz function evaluated atp
n
=
(
x
1
,
min
⁡
+
x
2
,
min
⁡
)
p
n
-
x
1
,
min
⁡
x
2
,
min
⁡ and The associated Hilbert-Kunz multiplicity =
x
1
,
min
⁡
+
x
2
,
min
⁡ which is an integer.We will now discuss Case  (1).Sincex
1 is a negative difference variable and x
2 is a positive difference variable, then using our notation in Section 4.2.1, we can denote x
1 by N
1 and x
2 by P
1. Then writing x
1
⊳
x
2 is the same as writing N
1
⊳
P
1. In this case, there are 3 subcases.Subcase  1.1 (when
a
1
<
Δ
P
1
). It is an exercise to check that using the above formula for the Hilbert-Kunz function, we get that for n large enough.The Hilbert-Kunz function evaluated atp
n equals
(56)
1
+
E
a
1
,
b
1
P
1
,
min
⁡
+
a
1
P
1
,
min
⁡
×
p
n
Δ
P
1
-
a
1
a
1
Δ
P
1
-
1
a
1
-
r
a
1
,
n
a
1
-
p
a
1
+
P
1
,
min
⁡
Δ
P
1
-
a
1
P
1
,
max
⁡
+
a
1
P
1
,
max
⁡
p
n
-
N
1
,
max
⁡
-
1
-
E
a
1
,
b
1
a
1
-
a
1
P
1
,
max
⁡
p
n
Δ
P
1
-
a
1
a
1
Δ
P
1
-
1
a
1
-
r
a
1
,
n
a
1
-
p
a
1
+
P
1
,
min
⁡
Δ
P
1
+
a
1
′
P
1
,
max
⁡
+
N
1
,
max
⁡
-
N
1
,
min
⁡
P
1
,
max
⁡
+
N
1
,
min
⁡
p
n
,
where for any real number β, β and [
β
] denote the upper and lower integral parts of β, respectively.Hence the associated Hilbert-Kunz multiplicity equals(57)
a
1
P
1
,
min
⁡
Δ
P
1
-
a
1
a
1
Δ
P
1
+
a
1
P
1
,
max
⁡
Δ
P
1
+
N
1
,
min
⁡
which is nothing but P
1
,
min
⁡
+
a
1
+
N
1
,
min
⁡, which is an integer.Subcase  1.2 (when
a
1
>
Δ
P
1
). It is an exercise to check that using the above formula for the Hilbert-Kunz function, we get that for n large enoughthe Hilbert-Kunz function evaluated atp
n equals
(58)
p
n
-
N
1
,
max
⁡
P
1
,
max
⁡
+
N
1
,
max
⁡
-
N
1
,
min
⁡
P
1
,
max
⁡
+
N
1
,
min
⁡
p
n
.Hence the associated Hilbert-Kunz multiplicity equals P
1
,
max
⁡
+
N
1
,
min
⁡, which is an integer.Subcase  1.3 (when
a
1
=
Δ
P
1
). Let N
~
0 be the smallest positive integer such that if k
~
>
N
~
0, then P
1
,
min
⁡
-
1
-
r
a
1
,
n
-
a
1
p
a
1
<
a
1
k
~. Similarly, let N
~
0
′ be the smallest positive integer such that if k
~
>
N
~
0
′, then P
1
,
max
⁡
-
1
-
r
a
1
,
n
-
a
1
p
a
1
<
a
1
k
~. Clearly then N
~
0
′
≥
N
~
0.It is an exercise to check that using the above formula for the Hilbert-Kunz function, we get that forn large enough,the Hilbert-Kunz function evaluated atp
n equals
(59)
1
+
E
a
1
,
b
1
P
1
,
min
⁡
+
a
1
P
1
,
min
⁡
N
~
0
+
a
1
∑
k
~
=
N
~
0
+
1
N
~
0
′
1
+
r
a
1
,
n
+
a
1
p
a
1
+
a
1
k
~
+
a
1
P
1
,
max
⁡
p
n
-
N
1
,
max
⁡
-
1
-
E
a
1
,
b
1
a
1
-
1
-
N
~
0
′
+
a
1
′
P
1
,
max
⁡
+
N
1
,
max
⁡
-
N
1
,
min
⁡
P
1
,
max
⁡
+
N
1
,
min
⁡
p
n
,where for any real number β, β denotes the upper integral part of β.Hence the associated Hilbert-Kunz multiplicity equalsP
1
,
max
⁡
+
N
1
,
min
⁡, which is an integer.


## 4.5. Rationality of the Hilbert-Kunz Multiplicity

The Hilbert-Kunz multiplicity for the general Binomial hypersurface (whose Hilbert-Kunz function is given by (45) above) is by definition equal to the coefficient of p
n
(
s
+
t
+
r
-
1
) in the formula given by (45). We can easily see from the above discussion that for any j such that 0
≤
j
≤
X
~
N
r
,
last, ∫
j
(
N
r
)
‍ is obtained iteratively by first computing ∫
j
(
N
1
)
‍ for all j, then ∫
j
N
2
‍
(
for
all
j
)
,
…, and so on till ∫
j
(
N
r
-
1
)
‍ for all j. Moreover, for any nonnegative integer j such that 0
≤
j
≤
k
~
N
r
,
last, we know that M
a
r
,
n
-
p
a
r
-
j belongs to the set M
max
⁡
[
2
] and
(60)
∫
j
(
N
1
)
‍
≔
Φ
M
a
r
,
n
-
p
a
r
-
j
=
S
M
a
r
,
n
-
p
a
r
-
j
,
s
.So let us first give a formula forS
M
a
r
,
n
-
p
a
r
-
j
,
s. I will rather give a formula for S
M
max
⁡
,
s for any M
max
⁡ in M
max
⁡
[
2
] in general and then one can put M
max
⁡
=
M
a
r
,
n
-
p
a
r
-
j in it to get the required formula for S
M
a
r
,
n
-
p
a
r
-
j
,
s.LetM
max
⁡
∈
M
max
⁡
[
2
] be arbitrary. To compute S
M
max
⁡
,
s, we first need to compute S
M
max
⁡
,
0. I will compute S
M
max
⁡
,
0 only inthe most general case, that is, when
(61)
Mi
n
P
q
,
M
max
⁡
c
<
p
n
-
P
q
,
max
⁡
<
p
n
-
P
q
,
min
⁡
∀
q
∈
1
,
…
,
t
,
∀
M
max
⁡
∈
M
max
⁡
2
.
In this case, we have
(62)
Mi
n
P
q
,
M
max
⁡
c
=
M
max
⁡
Δ
P
q
.
In all other cases, it is a tedious job to compute S
M
max
⁡
,
0 because we will have to take care of various complicated possibilities likefor someq
∈
{
1
,
…
,
t
}, we may have Mi
n
P
q
,
M
max
⁡
c
<
p
n
-
P
q
,
min
⁡ and for some other q
′
∈
{
1
,
…
,
t
}, we may have Mi
n
P
q
′
,
M
max
⁡
c
=
p
n
-
P
q
′
,
min
⁡. This makes the computation of S
M
max
⁡
,
0 highly tedious and complicated. Let us now compute S
M
max
⁡
,
0 for the most general case.For2
≤
l
≤
t, let E
l
,
min
⁡ denote the sum of all possible terms which are products of l-many items either of the type M
max
⁡
Δ
P
q or of the type P
q
,
min
⁡ where q
∈
{
1
,
…
,
t
} such that the following two properties hold.(i)
At least one of thel items in each such product is of the type M
max
⁡
Δ
P
q;
(ii)
not all of thel items in any such product are of the type M
max
⁡
Δ
P
q. Similarly, for any l
∈
{
2
,
…
,
t
}, let E
l
,
max
⁡ denote the sum of all possible terms which are products of l-many items either of the type M
max
⁡
Δ
P
q or of the type P
q
,
max
⁡ where q
∈
{
1
,
…
,
t
} such that the following two properties hold.(i)
At least one of thel items in each such product is of the type M
max
⁡
Δ
P
q;
(ii)
not all of thel items in any such product are of the type M
max
⁡
Δ
P
q. For 2
≤
l
≤
t, let T
l denote the sum of all possible terms which are products of the type ∏
i
1
,
i
2
,
…
,
i
l
∈
1
,
…
,
t
i
1
≠
i
2
≠
⋯
≠
i
l
P
i
1
,
max
⁡
P
i
2
,
max
⁡
⋯
P
i
l
,
max
⁡.Then(63)
S
M
max
⁡
,
0
=
p
t
-
1
n
∑
i
=
1
t
P
i
,
max
⁡
+
∑
l
=
2
t
-
1
l
p
t
-
l
n
E
l
,
min
⁡
-
E
l
,
max
⁡
-
T
l
.Hence the formula forΦ
M
max
⁡ (≔
S
M
max
⁡
,
s) is given by
(64)
p
t
-
1
n
∑
i
=
1
t
P
i
,
max
⁡
+
∑
l
=
2
t
-
1
l
p
t
-
l
n
×
p
t
-
1
n
∑
i
=
1
t
P
i
,
max
⁡
+
∑
l
=
2
t
-
1
l
p
t
-
l
n
E
l
,
min
⁡
-
E
l
,
max
⁡
-
T
l
∏
i
=
1
s
p
n
-
Z
i
,
min
⁡
+
∑
k
=
1
s
∑
i
1
,
i
2
,
…
,
i
k
∈
1
,
…
,
s
i
1
≠
i
2
≠
⋯
≠
i
k
Z
i
1
,
min
⁡
Z
i
2
,
min
⁡
⋯
Z
i
k
,
min
⁡
-
1
k
+
1
p
t
+
s
-
k
n
×
∑
i
1
,
i
2
,
…
,
i
k
∈
1
,
…
,
s
i
1
≠
i
2
≠
⋯
≠
i
k
Z
i
1
,
min
⁡
Z
i
2
,
min
⁡
⋯
Z
i
k
,
min
⁡
.
Since ∫
j
(
N
1
)
‍
=
Φ
M
a
r
,
n
-
p
a
r
-
j, we get the formula for ∫
j
(
N
1
)
‍ from (64) above by putting M
max
⁡
=
M
a
r
,
n
-
p
a
r
-
j in it.Let us now write down the formula (45) for the Hilbert-Kunz function, in a different way which will help us see how the formula for the Hilbert-Kunz multiplicity will look like. But for that, we first need some notation.For eachj
∈
N, let F
j
(
1
)
≔
1
,
F
j
(
2
)
≔
2
j
-
1 and for any positive integer k
≥
3, let F
j
(
k
)
≔
∑
l
=
1
j
-
1
F
l
(
k
-
1
)
+
j
F
j
(
k
-
1
).
For each0
≤
j
≤
X
~
N
r
,
last, let C
j
(
N
r
) denote the coefficient of ∫
j
(
N
r
)
‍ in the formula (45) for the Hilbert-Kunz function.
For anym
∈
{
1
,
…
,
r
}, let V
(
m
)
≔
{
D
a
r
,
a
r
,
D
a
r
-
1
,
a
r
,
…
,
D
a
m
,
a
r
}, U
(
m
)
≔
{
a
r
,
a
r
-
1
,
…
,
a
m
}, and C
0
(
N
m
)
≔
∏
k
=
m
r
D
a
k
,
a
r. For any integer j such that 1
≤
j
<
X
~
N
m
,
last and any integer l such that 1
≤
l
≤
r
-
m
+
1, let H
l
(
m
)
≔ sum of all possible terms which are products of r
-
l
-
m
+
1 many D
a
k
,
a
rs and l many a
ks where the D
a
k
,
a
rs belong to the set V
(
m
) and the a
ks belong to the set U
(
m
).Letm
∈
{
1
,
…
,
r
-
1
} be arbitrary. Then for any integer j such that 1
≤
j
<
X
~
N
m
,
last, let
(65)
C
j
(
N
m
)
≔
∑
l
=
1
r
-
m
+
1
F
j
(
l
)
H
l
(
m
)
.
And for any integer j such that X
~
N
m
,
last
≤
j
≤
X
~
N
r
,
last, let us define C
j
(
N
m
) inductively as follows.LetC
X
~
N
m
,
last
(
N
m
)
:
=
a
m
′
[
Σ
j
=
0
X
~
N
m
,
last
-
1
C
j
(
N
m
+
1
)
]
+
(
p
n
-
N
m
,
max
⁡
)
C
X
~
N
m
,
last
(
N
m
+
1
)  and for any X
~
N
m
,
last
<
j
≤
X
~
N
r
,
last, let C
j
(
N
m
)
:
=
(
p
n
-
N
m
,
max
⁡
)
C
j
(
N
m
+
1
).Now if we expand the formula ((45) above) of the Hilbert-Kunz function, assuming that X
~
N
1
,
last
≤
X
~
N
2
,
last
≤
⋯
≤
X
~
N
r
,
last, we get that the expression in (45) equals
(66)
∑
j
=
0
X
~
N
1
,
last
-
1
C
j
N
1
∫
j
N
1
‍
+
a
1
′
∑
j
=
0
X
~
N
1
,
last
-
1
C
j
N
2
+
p
n
-
N
1
,
max
⁡
C
X
~
N
1
,
last
N
2
∫
X
~
N
1
,
last
(
N
1
)
‍
+
p
n
-
N
1
,
max
⁡
∑
j
=
X
~
N
1
,
last
+
1
X
~
N
r
,
last
C
j
N
2
∫
j
N
1
‍
+
N
r
,
max
⁡
-
N
r
,
min
⁡
G
N
r
+
N
r
,
min
⁡
p
n
(
s
+
t
+
r
-
1
)
+
∑
k
=
1
r
-
1
(
N
r
-
k
,
max
⁡
-
N
r
-
k
,
min
⁡
)
G
N
r
-
k
+
(
N
r
-
k
,
max
⁡
-
N
r
-
k
,
min
⁡
)
N
r
-
k
,
min
⁡
p
n
(
s
+
t
+
r
-
k
-
1
)
×
C
0
(
N
r
-
k
+
1
)
+
∑
l
=
1
k
H
l
r
-
k
+
1
∑
j
=
1
X
~
N
r
-
k
+
1
,
last
-
1
F
j
l
+
H
l
r
-
k
+
1
∑
j
=
1
X
~
N
r
-
k
+
1
,
last
-
1
F
j
l
∑
j
=
X
~
N
r
-
k
+
1
,
last
X
~
N
r
,
last
C
j
N
r
-
k
+
1
,
where the formulae for ∫
j
(
N
1
)
‍ can be obtained from (64). The Hilbert-Kunz multiplicity is by definition the coefficient of p
n
(
s
+
t
+
r
-
1
) in the expression (66) above. Computing the coefficient of p
n
(
s
+
t
+
r
-
1
) in the expression (66) above, we get that the Hilbert-Kunz multiplicity equals
(67)
N
r
,
min
⁡
+
∑
k
=
1
r
-
1
N
r
-
k
,
min
⁡
∑
j
=
1
X
~
N
r
-
k
+
1
,
last
-
1
F
j
l
Coefficient
of
p
k
n
in
∑
l
=
1
k
H
l
r
-
k
+
1
∑
j
=
1
X
~
N
r
-
k
+
1
,
last
-
1
F
j
l
+
∑
q
=
0
t
-
1
∑
j
=
1
X
~
N
1
,
last
-
1
j
q
F
j
(
r
)
coefficient
of
j
q
p
n
t
+
s
-
1
-
q
in
∫
j
N
1
‍
×
∑
j
=
1
X
~
N
1
,
last
-
1
j
q
F
j
(
r
)
[
terms
of
H
1
1
of
degree
p
n
(
r
-
1
)
]
×
∑
j
=
1
X
~
N
1
,
last
-
1
j
q
F
j
1
+
terms
of
H
2
1
of
degree
p
n
r
-
2
×
∑
j
=
1
X
~
N
1
,
last
-
1
j
q
F
j
(
2
)
+
⋯
+
terms
of
H
r
1
of
degree
p
n
r
-
r
×
∑
j
=
1
X
~
N
1
,
last
-
1
j
q
F
j
r
.If we expand the above formula (expression (67)) of the Hilbert-Kunz multiplicity, then inside the expansion, we will get terms which are of the form:(i)
finite (by “finite”, I mean depending uponn) sums of the form ∑
j
=
ϵ
n
θ
n
j
v where v is some positive integer and ϵ
n and θ
n are upper or lower integral parts of some real numbers which are of the form “polynomials in p
n with rational coefficients”;
(ii)
polynomials inp
n with rational coefficients (including constant terms which are rational numbers);
(iii)
products of the above mentioned2 types of terms.It hence follows from the above discussion that the Hilbert-Kunz multiplicity associated with any general binomial hypersurface is always rational.



---

*Source: 525467-2014-11-27.xml*