Note on a theorem of Eliahou

Abstract

From a result (Wilf’s conjecture and Macaulay’s theorem, 2017, Theorem 5.11) of Eliahou on the growth of the Hilbert function of a standard graded algebra we derive an inequality related to a question in Wilf (Am Math Mon 85, 1978). This enables us to construct a (to our knowledge new) class of numerical semigroups of embedding dimension 5 and arbitrarily high Cohen-Macaulay-type, for which the question of Wilf  (1978) has an affirmative answer.

1 Introduction

In [1], Eliahou deduces the following statement from Macaulay’s theorem on the growth of the Hilbert function of a graded algebra:

Theorem

[1, Theorem 5.11] Let $R=\bigoplus _{i\ge 0}R_i$ be a standard graded algebra over a field k with Hilbert function $h_n=\dim _kR_n$, $n\in \mathbb {N}$. For each $n\ge 0$,

$$\begin{aligned} nh_n\le h_1(1+h_1+\ldots +h_{n-1}) \end{aligned}$$

(1)

$\square$

Let $P=k[X_1,\ldots , X_e]$ be the polynomial ring over k in $e\ge 1$ variables $X_1,\ldots , X_e$. We are interested in rings of the form $R=P/I$ where I is a monomial ideal of P.

For $a=(a_1,\ldots , a_e)\in \mathbb {N}^e$ and $C\subseteq \mathbb {N}^e$ we denote by $\deg a=a_1+\ldots +a_e$ the degree of a and set $C_n=\{a\in C\vert \deg a=n\}$.

Let $C\subseteq \mathbb {N}^e$ be a monoid ideal, i. e. $C+\mathbb {N}^e\subseteq C$, and $I=I(C):=\left( \left\{ \left. X_1^{a_1}\cdot \ldots \cdot X_e^{a_e}\right| \left( a_1,\ldots , a_e\right) \in C\right\} \right)$ the corresponding monomial ideal of P. Then $\#\left( \mathbb {N}^e\setminus C\right) _n, n\in \mathbb {N}$ is the Hilbert function of $R=P/I$ and the above theorem implies

Proposition 1.1

Let C be an ideal in $\mathbb {N}^e$. For each $n\in \mathbb {N}$

$$\begin{aligned} n\cdot \#\left( \mathbb {N}^e\setminus C\right) _n\le e\left( 1+\left( \mathbb {N}^e\setminus C\right) _1+\ldots +\left( \mathbb {N}^e\setminus C\right) _{n-1}\right) . \end{aligned}$$

(2)

$\square$

From this we will deduce inequalities related to a question of Wilf.

2 Contributions to Wilf’s conjecture on numerical semigroups

Let $S=\mathbb {N}\cdot g_1+\ldots +\mathbb {N}\cdot g_e\ne \mathbb {N}$ be a numerical semigroup with minimal generating set

$$\begin{aligned} g_1<\ldots <g_e, \gcd \left( g_1, \ldots , g_e\right) =1. \end{aligned}$$

We set

$$\begin{aligned} c=c(S):=\min \{s\in S\vert s+\mathbb {N}\subseteq S\}, \end{aligned}$$

sometimes called the conductor of S and

$$\begin{aligned} L=L(S):=S\cap \{0,\ldots , c(S)-1\}, \end{aligned}$$

the part of S which is to the left of c. Wilf asks in [5] if the inequality

$$\begin{aligned} \frac{\#L}{c}\ge \frac{1}{e} \end{aligned}$$

(3)

holds for all numerical semigroups. Assertion (3) is also known as Wilf’s conjecture. In [1] it has been shown that (3) is asymptotically true as the genus $g(S):=\#(\mathbb {N}\setminus S)$ goes to infinity.

We endow $\mathbb {N}^e$ with the lexicographic order LEX, i. e. $a<_{\text {LEX}}b$ if the leftmost nonzero component of $a-b$ is negative. Consider the canonical surjection

$$\begin{aligned} \pi =\pi _S: \mathbb {N}^e\rightarrow S, \left( a_1, \ldots , a_e\right) \mapsto \sum _{i=1}^ea_ig_i. \end{aligned}$$

For $s\in S, M\subseteq S$ and $n\in \mathbb {N}$ we set

$$\begin{aligned} s':=\max _{{\text {LEX}}}\left( \pi ^{-1}(s)\right) , M'=\{m'\vert m\in M\}\text { and }M^n:=\pi \left( {M'}_n\right) . \end{aligned}$$

Hence $S=\mathop {\dot{\bigcup }}_{n\in \mathbb {N}}S^n$ and $\pi$ maps $M'$ bijectively to M, ${M'}_n$ to $M^n$.

   By $n_0:=\lceil \frac{c}{g_1}\rceil$ we denote the smallest integer n such that $ng_1\ge c$.

Proposition 2.1

• (a)
  $$\begin{aligned} \frac{\#S^0+\ldots +\#S^{n_0-1}}{n_0\cdot \#S^{n_0}}\ge \frac{1}{e}. \end{aligned}$$

  (4)

• (b) $n_0$ is the smallest integer n such that $L\subseteq S^0\cup \ldots \cup S^{n-1}$.

• (c) For each $n\in \mathbb {N}$,

  $$\begin{aligned} \#S^n\le \#S^{n+1}\le g_1\text { and }\min S^n=ng_1. \end{aligned}$$
• (d)
  $$\begin{aligned}\#S^{n_0}=g_1.\end{aligned}$$

• (e) $S^{n_0+n}=ng_1+S^{n_0}$, hence $\#S^{n_0+n}=g_1$ for all $n\ge 0$.

$\square$

In particular the numerator from the left side of (4) is $\ge \#L$ and the denominator is $\ge c$. We get the following weak form of Wilf’s inequality

$$\begin{aligned} \frac{\#S^0+\ldots +\#S^{n_0-1}}{c}\ge \frac{1}{e}. \end{aligned}$$

(5)

Moreover we have

Corollary 2.2

Wilf’s conjecture holds whenever

$$\begin{aligned} S^{n_0}=\{c, c+1, \ldots , c+g_1-1\}. \end{aligned}$$

For instance, this is the case if $g_1$ divides c and

$$\begin{aligned} (n_0-1)g_e<n_0g_1-1, \end{aligned}$$

(6)

particularly for the following.

Examples

• (a) S is generated by consecutive numbers

  $$\begin{aligned} g_1, g_1+1, \ldots , g_1+e-1=g_e, 2\le e\le g_1. \end{aligned}$$

  In his diploma thesis [3], Knebl mentioned, that

  $$\begin{aligned} n_0=\left\lceil \frac{g_1-1}{e-1}\right\rceil \text { and }c=n_0g_1, \end{aligned}$$

  as is easily seen. Further

  $$\begin{aligned} \#L=n_0\cdot \frac{2+\left( n_0-1\right) (e-1)}{2}. \end{aligned}$$

  Notice that $n_0=\left\lceil \frac{g_1-1}{e-1}\right\rceil$ implies $\left( n_0-1\right) (e-1)<g_1-1$, hence

  $$\begin{aligned} \left( n_0-1\right) g_e<n_0g_1-1 \end{aligned}$$

  and (6) holds.

  But since $n_0(e-1)\ge g_1-1$ and $g_1\ge e\ge 2$, elementary computation also shows

  $$\begin{aligned} \frac{\#L}{c}=\frac{2+\left( n_0-1\right) (e-1)}{2g_1}\ge \frac{2+g_1-e}{2g_1}\ge \frac{1}{e}. \end{aligned}$$

• (b) For $n\ge 2$ let

  $$\begin{aligned} S=S(n):=\left\langle n^2,n^2+1,n^2+n,n^2+n+1\right\rangle . \end{aligned}$$

  Then $e=4$, $n_0=n-1$, $c=(n-1)\cdot n^2$ and

  $$\begin{aligned} \left( n_0-1\right) g_e=n^3-n^2-n-2<n^3-n^2-1=n_0g_1-1. \end{aligned}$$

  Further

  $$\begin{aligned} \#L=\frac{(n-1)n(2n-1)}{6} \end{aligned}$$

  and again we see immediately that

  $$\begin{aligned} \frac{\#L}{c}=\frac{1}{3}-\frac{1}{6n}\ge \frac{1}{4}\text { for }n\ge 2. \end{aligned}$$

  For the following recall that for general S the set of pseudo-Frobenius numbers is defined as

  $$\begin{aligned} \left\{ \left. l\in \mathbb {N}\setminus S\right| l+S\setminus \{0\}\subseteq S\right\} . \end{aligned}$$

  Using the notion block of length $\mathbf l$ for a subset of S consisting of l consecutive natural numbers, we can describe L in the situation of example (b):

  Let E be the minimal set of generators of S. Then

  $$\begin{aligned} L=\{0\}~\dot{\cup }~E~\dot{\cup }~2E~\dot{\cup }~\ldots ~\dot{\cup }~(n_0-1)E \end{aligned}$$

  (7)

  where $lE=S^l=S\cap [lg_1,(l+1)g_1[, (l=1, \ldots , n_0-1)$. From left to right, in $S^l$ there are always

  • $l+1$ blocks of length $l+1$, where between two consecutive such blocks the numbers of gaps is $n-l-1$, followed by

  • $(n^2-1)-l(n+1)$ gaps.

  This is repeated for l running from 1 to $n-1$. (Look at the figure below.)

  From this description it is easy to see that the pseudo-Frobenius numbers of S are precisely those gaps which arise from the case $l=n-2$ in the description above; the number of those gaps is $(n-2)+(n^2-1)-(n-2)\cdot (n+1)=2n-1$; i. e. the type of S(n) is $2n-1$; in particular, for this class of examples the type gets arbitrarily high though the number of generators is always 4 (cf. Backelin’s examples in [2]).

$\square$

Remark

In [1], Eliahou shows that Wilf’s conjecture holds true for all semigroups which have $n_0\le 3$. Furthermore, he shows in theorem 7.1 with the help of his theorem 5.11 (see above) that “Wilf” is true for all semigroups having property (7). By [1, Corollary 7.2], this includes all semigroups which have $n_0\ge 4$ and $(n_0-1)g_e<c$. Therefore, our previous examples are not new.

However, by a slight modification of example (b) we will arrive at a class of semigroups which do fulfill the hypothesis of 2.2 Corollary, but not (7):

For odd $n\ge 5$, we enlarge $S=S(n)$ to

$$\begin{aligned} S_+:=S+\mathbb {N}g_5, g_5=\frac{3n^2+n}{2}. \end{aligned}$$

Since $g_4<g_5<2g_1$,

$$\begin{aligned} E_+:=\{g_1,g_2,g_3,g_4,g_5\} \end{aligned}$$

is the minimal generating set of $S_+$.

Recall that $n_0=n_0(S)=n-1$ and $c=c(S)=(n-1)n^2$.

From the block structure of S we will derive (with some effort):

Proposition 2.3

The semigroup $S_+$ satisfies

• (i) $n_0(S_+)=n-1$ and $c(S_+)=c=(n-1)n^2$

• (ii) $S_+$ does not fulfill (7).

• (iii) ${S_+}^{n-1}=\{c, c+1, \ldots , c+g_1-1\}$, the hypothesis of 2.2 Corollary for $S_+$.

• (iv) The type of $S_+$ is at least $n+1$.

Throughout this paper we use the convention ${x\atopwithdelims ()e-1}=0$ for all integers $x<e-1$.

For $b^1, \ldots , b^l\in \mathbb {N}^e$ let $\left| b^1, \ldots , b^l\right|$ be the degree of $\max \left\{ b^1, \ldots , b^l\right\} \in \mathbb {N}^e$, where the maximum is taken coordinatewise if $l\ge 1$ and is $(0, \ldots , 0)$ for the empty set.

An application of 1.1 to a $\mathbb {N}^e$-ideal $\bigcup _{i=1}^m\left( a^i+\mathbb {N}^e\right)$, $\left\{ a^1,\ldots , a^m\right\} \subseteq \mathbb {N}^e$, $m\in \mathbb {N}$ together with Sylvester’s inclusion–exclusion principle will give us

Proposition 2.4

• (a) For each $n\in \mathbb {N}$,

  $$\begin{aligned} \sum _{l=0}^m(-1)^{l+1}\sum _{i_1<\ldots <i_l}\left| a^{i_1},\ldots ,a^{i_l}\right| {e-1+n-\left| a^{i_1},\ldots ,a^{i_l}\right| \atopwithdelims ()e-1}\ge 0. \end{aligned}$$

  (8)

• (b) In the situation of 2.1, for each $n\in \mathbb {N}_+$,

  $$\begin{aligned} \frac{\#S^0+\ldots +\#S^{n-1}}{\sum _{l=0}^m(-1)^l\sum _{i_1<\ldots <i_l}\left( n-\left| a^{i_1},\ldots ,a^{i_l}\right| \right) {e-1+n-\left| a^{i_1},\ldots ,a^{i_l}\right| \atopwithdelims ()e-1}}=\frac{1}{e}.\quad \end{aligned}$$

  (9)

$\square$

Since every ideal of $\mathbb {N}^e$ has the form $\bigcup _{i=1}^m\left( a^i+\mathbb {N}^e\right)$, conversely (2) also follows from (8) together with Sylvester’s inclusion–exclusion principle. However we could not find any direct (combinatorial) proof of (8) which does not make use of Macaulay’s theorem.

Also note that (9) determines, in contrast to (4), the precise denominator for which the ratio equals $\frac{1}{e}$; this is done for each n, not just for $n_0$.

3 Proof of 2.1 Proposition

Let $A={\text {Ap}}(S,g_1):=\{s\in S\vert s-g_1\not \in S\}$ the Apéry set of S with respect to $g_1$. A consists precisely of the smallest representatives in S of the congruence classes modulo $g_1$.

It is easy to see that there exists a subset $\tilde{A}\subseteq \mathbb {N}^{e-1}$ with $A'=\{0\}\times \tilde{A}$ and $S'=\mathbb {N}\times \tilde{A}$.

Lemma 3.1

$\mathbb {N}^e\setminus S'$ is an ideal of $\mathbb {N}^e$.

Proof

Let $a\in \mathbb {N}^e\setminus S'$ and $x\in \mathbb {N}^e$ be arbitrary. There exists $b\in \mathbb {N}^e$ with $\pi (b)=\pi (a)$ and $b>_{{\text {LEX}}}a$. Hence $\pi (b+x)=\pi (a+x)$ and $b+x>_{{\text {LEX}}}a+x$, i. e. $a+x\in \mathbb {N}^e\setminus S'$. $\square$

Remark

Replacing everywhere in the proof of [6, theorem 1] “lexicographically minimal” by “lexicographically maximal” we also obtain, that $\mathbb {N}^{e-1}\setminus \tilde{A}$ is an ideal, and so is $\mathbb {N}^e\setminus \mathbb {N}\times \tilde{A}$.

Lemma 3.2

• (a) For every $n\in \mathbb {N}$,

  • (i) $\min S^n=ng_1$.

  • (ii) $g_1+S^n\subseteq S^{n+1}$, in particular, $\#S^n\le \#S^{n+1}$.

  • (iii) $\#S^n\le g_1$; more precisely, different elements of $S^n$ belong to different congruence classes modulo $g_1$.

• (b) $\#S^{n_0}=g_1$.

Proof

(a) (i) and (ii) are clear.

(iii) Different elements s and t of S with $s\equiv t\mod g_1$ are of the form $s=lg_1+a$, $t=mg_1+a$, $a\in A$; therefore, $\deg s'-\deg t'=l-m\ne 0$.

(b) By (a) (i), $n_0g_1, n_0g_1+1, \ldots , n_0g_1+g_1-1$ are not in $S^n$ for each $n\ge n_0+1$; hence these elements are in $S^0\cup \ldots \cup S^{n_0}$. Therefore, by (a) (ii) and (iii), each congruence class modulo $g_1$ shows up exactly once in $S^{n_0}$. $\square$

Proof of 2.1 Proposition

• (a) Apply 1.1 to the ideal $C=\mathbb {N}^e\setminus S'$.

• (b) By 3.2 (a) (i) $L\cap S^n=\emptyset$ for each $n\ge n_0$, and $\left( n_0-1\right) g_1\in L\cap S^{n_0-1}$. Claim (b) follows.

• (c) and (d) are parts of 3.2 Lemma.

• (e) is a consequence of [3.2 Lemma (a) (ii), (iii) and (b)].$\square$

4 Proof of 2.2 Corollary and of 2.3 Proposition

Corollary 2.2 immediately follows from formula (4) and

Lemma 4.1

$$\begin{aligned} S^{n_0}=\{c, c+1, \ldots , c+g_1-1\} \end{aligned}$$

if and only if

$$\begin{aligned} L=S^0\dot{\cup }\ldots \dot{\cup }S^{n_0-1}. \end{aligned}$$

In any case, then $c=\min S^{n_0}=n_0g_1$ and $\#L=\#S^0+\ldots +\#S^{n_0-1}$ and (4) is identified with (3).

Proof of 4.1 Lemma

Obviously $L=S^0\dot{\cup }\ldots \dot{\cup }S^{n_0-1}$ if and only if $S\setminus L=\mathop {\dot{\bigcup }}_{n\in \mathbb {N}}S^{n_0+n}$.

This implies $S^{n_0}=\{c, c+1, \ldots , c+g_1-1\}$, since $S\setminus L=c+\mathbb {N}$, $\min S^{n_0}=n_0g_1$, $\#S^{n_0}=g_1$ and $\min S^n\ge (n_0+1)g_1$ for all $n\ge n_0+1$.

Conversely, if $S^{n_0}=\{c, c+1, \ldots , c+g_1-1\}$, then $c=\min S^{n_0}=n_0g_1$ and $S\setminus L=c+\mathbb {N}=\mathop {\dot{\bigcup }}_{n\in \mathbb {N}}\left( ng_1+S^{n_0}\right) =\mathop {\dot{\bigcup }}_{n\in \mathbb {N}}S^{n_0+n}$ (by 2.1 e)). $\square$

Proof of 2.3 Proposition

• (i) We show that $c(S_+)=c$, in particular, $n_0(S_+)=n-1$: Since $2g_5=2g_1+g_3$,

  $$\begin{aligned} S_+=S\cup (g_5+S). \end{aligned}$$

  Therefore, it suffices to show that $x:=c-1-g_5\not \in S$: We have seen that S fulfills (7). Further

  $$\begin{aligned} x=(n-3)n^2+\frac{n-2}{2}(n+1), \end{aligned}$$

  therefore, because of $n>4$,

  $$\begin{aligned} (n-3)g_1=(n-3)n^2<x<(n-3)n^2+(n-3)(n+1)=(n-3)g_4. \end{aligned}$$

  If x was in S, by (7) we would have $x\in (n-3)E$, i. e.

  $$\begin{aligned} x=ag_4+bg_3+cg_2+dg_1\text { where }a,b,c,d\in \mathbb {N}\text { and }a+b+c+d=n-3. \end{aligned}$$

  This would imply

  $$\begin{aligned} x-(n-3)n^2=\frac{(n-2)(n+1)}{2}=a(n+1)+bn+c, \end{aligned}$$

  hence

  $$\begin{aligned} n(n-1-2(a+b))=2(a+c+1). \end{aligned}$$

  Since n is odd, this would imply that n divides $a+c+1$, contradicting

  $$\begin{aligned} 1\le a+c+1\le a+b+c+d+1=n-2. \end{aligned}$$

• (ii): By (i), $n_0(S_+)=n-1$ and $c(S_+)=(n-1)n^2$. Furthermore, $2g_5=2g_1+g_3\in 2E_+\cap 3E_+$ and $n_0(S_+)-1\ge 3$. Hence

  $$\begin{aligned} L(S_+)\ne \{0\}~\dot{\cup }~E_+~\dot{\cup }~2E_+~\dot{\cup }~3E_+~\dot{\cup }~\ldots ~\dot{\cup }~(n_0(S_+)-1)E_+ \end{aligned}$$

  and $S_+$ is not among the examples from [1, Theorem 7.1].

• (iii): Because of $c(S_+)=c(S)=c$ we have to show that

  $$\begin{aligned} {S_+}^{n-1}=\{c,c+1,\ldots ,c+g_1-1\}=:R. \end{aligned}$$

  $\#R=\#{S_+}^{n-1}=g_1$ by 2.1 (d), therefore it suffices to show that $R\subseteq {S_+}^{n-1}$:

For arbitrary $s\in R$ let $s'=(a_1,a_2,a_3,a_4,a_5)$ be, just like above, the LEX-maximal element with $\pi (s')=\sum _{i=1}^5a_ig_i=s$. It remains to show that $\deg s'=n-1$:

Since $s<ng_1$, we have $\deg s'=\le n-1$. We have seen that $R=S^{n-1}$.

We distinguish between the following cases:

• (a) If $a_5=0$, then $(a_1,a_2,a_3,a_4)$ is, by construction, the LEX-maximal element in $\mathbb {N}^4$ with $\sum _{i=1}^4a_ig_i=s$ and hence $\deg s'=a_1+a_2+a_3+a_4=n-1$ because of $R=S^{n-1}$.

• (b) The case $a_5\ge 2$ cannot occur, because then we would have

  $$\begin{aligned} \tilde{s}:=(a_1+2,a_2,a_3+1,a_4,a_5-2)>_{\text {LEX}}s', \pi (\tilde{s})=s\text {, contradiction.} \end{aligned}$$

• (c) $a_5=1$: Since $ng_1>s\ge (n-1)g_1$ and $g_1<g_5<2g_1$, we have

  $$\begin{aligned} t:=s-g_5>(n-3)g_1\text { and }t\in L. \end{aligned}$$

  Therefore, because of (7), $t\in (n-3)E~\dot{\cup }~(n-2)E$, and hence

  $$\begin{aligned} n-2\le \deg t'+1=\deg s'\le n-1; \end{aligned}$$

  it remains to exclude the case $\deg s'=n-2$: The latter would imply

  $$\begin{aligned} u:=s-g_5-a_1g_1=a_2g_2+a_3g_3+a_4g_4\in ((n-3)-a_1)E. \end{aligned}$$

  Because of $g_5<2g_1$, we would have

  $$\begin{aligned} u>s-(2+a_1)g_1=:v\text { and }v\in [lg_1,(l+1)g_1[, l:=n-3-a_1. \end{aligned}$$

  Both elements u and v would be in the interval $[lg_1,(l+1)g_1[$, with $u>v$, $u\in S$ and $u\equiv v{\text {mod}} n$ because of $n|g_5$, $n|g_1$.

From the block structure of S as described above we would get $s-(2+a_1)g_1=v\in S$, contradicting the LEX-maximality of $s'$.

• (iv): From the block structure of S one sees, that there are no elements in S which are less than $(n-2)n^2$ and congruent to $-1$ or $-2$ modulo n. Consequently, since n divides $g_5$, the pseudo-Frobenius numbers $(n-2)n^2+kn-1$, $k\in \{1,\ldots ,n\}$ and $(n-1)n^2-2$ of S are still gaps of $S_+$, hence ${\text {type}}(S_+)\ge n+1$.

$\square$

Note that in general the bound “${\text {type}}(S_+)\ge n+1$” is not strict: For instance, in case $n=5$, 35 is a pseudo-Frobenius number of $S_+$ but not of S.

5 Proof of 2.4 Proposition

In the following, summation over the empty set is defined to be zero.

Lemma 5.1

For every $N\in \mathbb {N}$,

$$\begin{aligned} e\left( {e-1\atopwithdelims ()e-1}+{e\atopwithdelims ()e-1}+\ldots +{e+N-2\atopwithdelims ()e-1}\right) =N\cdot {e+N-1\atopwithdelims ()e-1} \end{aligned}$$

(10)

Proof

Induction on N (for a ’combinatorial’ proof see [4, Exercise 1.3a) (”Solutions”)]). $\square$

Lemma 5.2

Let $m,n\in \mathbb {N}$ and $a^1, \ldots , a^m\in \mathbb {N}^e$. The number of elements of degree n in $\mathbb {N}^e\setminus \left( \bigcup _{i=1}^ma^i+\mathbb {N}^e\right)$ is

$$\begin{aligned} \sum _{l=0}^m(-1)^l\sum _{i_1<\ldots <i_l}{e-1+n-\left| a^{i_1},\ldots ,a^{i_l}\right| \atopwithdelims ()e-1}. \end{aligned}$$

(11)

Proof

For each $a\in \mathbb {N}^e$, $\#\left( a+\mathbb {N}^e\right) _n={e-1+n-\deg a\atopwithdelims ()e-1}$, therefore, Sylvester’s inclusion–exclusion principle gives formula (11). $\square$

Proof of 2.4

Let $C:=\left( \bigcup _{i=1}^ma^i+\mathbb {N}^e\right)$. By Lemmas 5.1 and 5.2,

$$\begin{aligned}&\ e\left( 1+\#\left( \mathbb {N}^e\setminus C\right) _1+\ldots +\#\left( \mathbb {N}^e\setminus C\right) _{n-1}\right) \\&\quad \buildrel \text {(11)}\over =e\cdot \left( \sum _{l=0}^m(-1)^l\sum _{i_1<\ldots<i_l}\left( {e-1\atopwithdelims ()e-1}+\ldots +{e+n-\left| a^{i_1},\ldots ,a^{i_l}\right| -2\atopwithdelims ()e-1}\right) \right) \\&\quad \buildrel \text {(10)}\over =\sum _{l=0}^m(-1)^l\sum _{i_1<\ldots <i_l}\left( n-\left| a^{i_1},\ldots ,a^{i_l}\right| \right) {e+n-\left| a^{i_1},\ldots ,a^{i_l}\right| -1\atopwithdelims ()e-1}. \end{aligned}$$

From this, (b) can be read off directly by using $C:=\mathbb {N}^e\setminus S'$, and (a) follows since by (2) the first term is $\ge n\cdot \#\left( \mathbb {N}^e\setminus C\right) _n$ and the last term equals ${n\cdot \#\left( \mathbb {N}^e\setminus C\right) _n+\sum _n}$ with $\sum _n$ the left hand side of (8), using (11). $\square$

Final remark In some sense (3) and (4) are completely analogous formulas: Let

$$\begin{aligned} R=R(S):=k[X_1, \ldots , X_e]/I\left( \mathbb {N}^e\setminus S'\right) \end{aligned}$$

Ad (4) Let $\left( h_n\right) _{n\in \mathbb {N}}$ be the Hilbert function of R w. r. to the standard grading $\deg X_i=1$, $i=1, \ldots , e$. Then (4) reads as

$$\begin{aligned} n_0h_{n_0}\le e\left( h_0+h_1+\ldots +h_{n_0-1}\right) , \end{aligned}$$

Eliahou’s formula.

Ad (3) With respect to the grading $\deg X_i=g_i, i=1, \ldots , e$, R again is positively graded with Hilbert function

$$\begin{aligned} h^*_n={\left\{ \begin{array}{ll}1&{}\text { if }n\in S\\ 0&{}\text { if not}\end{array}\right. } \end{aligned}$$

and Wilf’s (proposed) inequality writes as

$$\begin{aligned} ch^*_c\le e\left( h^*_0+h^*_1+\ldots +h^*_{c-1}\right) . \end{aligned}$$

$\square$