Recall that a *numerical semigroup* $S$ is a submonoid of the non-negative integers $\mathbb Z\_{\geq 0}$ whose relative complement $\mathbb Z\_{\geq 0} \setminus S$ is finite. Observe that the collection $S^\*$ of nonzero elements of $S$ constitutes an *ideal* of $S$ in the sense that $$S^\* \supseteq \{s + t \mid s \in S^\* \text{ and } t \in S\} = S^\* + S;$$ because of this, we refer to $S^\*$ as the *maximal ideal* of $S.$


Generally, if $n$ is a non-negative integer, then the $n$-fold sum of $S^\*$ is given by $nS^\* = \{s\_1 + \cdots + s\_n \mid s\_1, \dots, s\_n \in S^\*\}.$ One can verify that for each positive integer $n,$ the sets $nS^\* \setminus (n + 1) S^\*$ are finite; their cardinalities give rise to the *Hilbert function* of $S,$ defined by $H\_S(n) = \#(nS^\* \setminus (n + 1) S^\*).$


Given an infinite field $k,$ we may also define the *numerical semigroup ring* $k [\![S]\!]$ corresponding to $S.$ It is well-known that the Hilbert-Samuel multiplicity of $k [\![S]\!]$ is exactly $\min(S^\*)$; the latter is therefore called the *multiplicity* of $S,$ denoted by $\operatorname e(S).$ Bearing this in mind, by definition of the multiplicity of $k [\![S]\!],$ it follows that $H\_S(n) = \operatorname e(S)$ for all integers $n \gg 0.$ If I recall correctly, this holds for all $n \geq \operatorname e(S).$



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> **Question.** Is there a way to see that $\#(nS^\* \setminus (n + 1) S^\*)$ eventually stabilizes without resorting to computations involving the numerical semigroup ring of $S$? Put another way, is there a purely numerical semigroup-theoretic proof that $\#(nS^\* \setminus (n + 1) S^\*) = \operatorname e(S)$ for all integers $n \geq \operatorname e(S)$?
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I have noticed that this is tacitly acknowledged throughout the literature, but as yet, I have not seen it formally addressed. Using the `numericalsgps` package of the GAP System, I have accrued a substantial amount of data that suggests this could even be improved to show that $\#(nS^\* \setminus (n + 1) S^\*) = \operatorname e(S)$ for all integers $n \geq \operatorname e(S) - 1,$ but it is not obvious to me that this holds in general. I would appreciate any comments, suggestions, or references. Thanks in advance for your time and consideration.



On the Hilbert function of a numerical semigroup
I believe there is an easier proof.


**Proof.** By induction, it suffices to prove the case that $eS^\* = e + (e - 1) S^\*$: indeed, if we assume that $(n + 1) S^\* = e + nS^\*$ for some integer $n \geq e,$ then it holds that $$(n + 2) S^\* = (n + 1) S^\* + S^\* = e + nS^\* + S^\* = e + (n + 1) S\*.$$


Because the containment $eS^\* \supseteq e + (e - 1) S^\*$ is clear, we will prove the other containment. Equivalently, we will show that for every element $x \in eS^\*,$ we have that $x - e$ lies in $(e - 1) S^\*.$ Observe that every element $x \in e S^\*$ can be written as a sum $x = s\_1 + \cdots + s\_e$ of $e$ positive integers $s\_1, \dots, s\_e.$ By the Pigeonhole Principle, we can write $x = \sum\_{s \in A} s + \sum\_{s \notin A} s$ for some nonempty subset $A \subseteq \{s\_1, \dots, s\_e\}$ such that $\sum\_{s \in A} s$ is divisible by $e$: among the $e$ sums $s\_1, s\_1 + s\_2, \dots, s\_1 + \cdots + s\_e,$ either one of them is divisible by $e$ or some two of them must have the same remainder modulo $e,$ in which case their difference is divisible by $e.$ Ultimately, we conclude that $x - e = {\left(\sum\_{s \in A} s - e\right)} + \sum\_{s \notin A} s$ lies in $(e - 1) S^\*,$ as desired. QED.


