A polynomial $f(x) \in \mathbb{Q}[x]$ is called a numerical polynomial if it satisfies the following equivalent conditions:

  1. If $a_{i} \in \mathbb{Q}$ are the unique coefficients satisfying $f = a_{0}\binom{x}{0} + \dotsb + a_{d}\binom{x}{d}$, then $a_{k} \in \mathbb{Z}$ for all $k$. (Here $\binom{x}{k} := \frac{(x-0)(x-1) \dotsb (x-(k-1))}{k!}$.)
  2. The polynomial $f$ is integer-valued, i.e. $f(\mathbb{Z}) \subseteq \mathbb{Z}$.

Is it true that for any subring $S \subseteq \mathbb{Q}$ we have $f(S) \subseteq S$?

It is equivalent to answer the question for $f(x) = \binom{x}{k}$. In this case it would be enough to know whether the expression $$ \alpha(m,n,k) := \frac{m(m-n) \dotsb (m-(k-1)n)}{k!} $$ is an integer for any integers $m,n$ (with $n \ne 0$), but I don't know if it has a combinatorial interpretation.

EDIT: The expression $\alpha(m,n,k)$ is not always an integer (in fact $\alpha(k!+1,k!,k)$ is not an integer for all $k \in \mathbb{N}$). But this does not yet give an answer to the main question, as $f(1+\frac{1}{k!}) \in \mathbb{Z}[\frac{1}{k!}]$ for $f(x) = \binom{x}{k}$.

(My motivation for this question is the proof of Stacks Project Tag 0CAP, which says $\binom{1/n}{k} \in \mathbb{Z}[1/n]$.)


1 Answer 1


Here is an argument. It is enough to prove that for any $m,n,k$ (with $n \ne 0$), the expression $\alpha(m,n,k)$ defined above is contained in $\mathbb{Z}[\frac{1}{n}]$, which reduces to the following claim.

Let $a,b$ be integers (with $b > 0$), and let $k$ be a positive integer. For any prime $p$ not dividing $b$, the product $(a+b)(a+2b)\dotsb(a+kb)$ is divisible by $p^{v_{p}(k!)}$, where $v_{p}(n)$ is the largest integer such that $p^{v_{p}(n)}$ divides $n$.

By Legendre's formula we have $v_{p}(k!) = \sum_{i=1}^{\infty} \lfloor \frac{k}{p^{i}} \rfloor$. It suffices to show that, if $\lfloor \frac{k}{p^{i}} \rfloor = e$, then in the arithmetic sequence $\{a+b,a+2b,\dotsb,a+kb\}$ there are at least $e$ numbers divisible by $p^{i}$. We can actually show the same for the subsequence $\{a+b,a+2b,\dotsc,a+ep^{i}b\}$, which is the disjoint union of $e$ subsequences (of length $p^{i}$) of the form $\{a+((j-1)p^{i}+1)b,\dotsc,a+(jp^{i})b\}$ for $j=1,\dotsc,e$, each of which contains a number divisible by $p^{i}$ since $b$ is relatively prime to $p^{i}$.

Bill Dubuque and KCd have answers to essentially the same question.


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