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Is the following true?

Let $p$ be a prime and let $w$ be a $(p-1)$st root of unity (not necessarily primitive). Then $$\binom{w}{n}=\frac{w(w-1)\cdots(w-n+1)}{n!}$$ is $p$-integral; i.e., it can be expressed as a polynomial in $w$ with $p$-integral coefficients. In other words, if $\Phi_k(x)$ is the $k$th cyclotomic polynomial and $k$ divides $p-1$ then the remainder of $\binom{x}{n}$ modulo $\Phi_k(x)$ is $p$-integral.

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    $\begingroup$ Yes, it is true. A slick way to see this is to work in the $p$-adic numbers. The function $f(x) = \binom{x}{n}$ is a polynomial in $x$ and thus is continuous as a map from $\mathbf Q_p$ to itself. It sends $\mathbf N$ to itself, so by passing to the $p$-adic closure of $\mathbf N$, $f$ maps $\mathbf Z_p$ to itself by $p$-adic continuity. Thus if $a$ is $p$-adically integral, so is $f(a) = \binom{a}{n}$. Now apply this to the $(p-1)$th roots of unity, which are in $\mathbf Z_p$ by Hensel’s Lemma. $\endgroup$
    – KConrad
    Jun 17, 2021 at 19:01
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    $\begingroup$ Let $w$ be root of unity of order $p-1$, $K = \mathbf Q(w)$, and $R = \mathbf Z[w]$, so $R = \mathcal O_K$. The ring you ask about is $\mathbf Z_{(p)}[w]$, where $\mathbf Z_{(p)}$ is the localization of $\mathbf Z$ at $p$ (the $p$-integral rationals). Why is $\binom{w}{n}$ in this ring? Localizing and forming integral closures commute, so the int. closure of $\mathbf Z_{(p)}$ in $K$ is $\mathbf Z_{(p)}[w]$. That int. closure is also $\bigcap_{\mathfrak p} R_{\mathfrak p}$ where $\mathfrak p$ runs over primes in $R$ lying over $p$ and $R_{\mathfrak p}$ is localization of $R$ at $\mathfrak p$. $\endgroup$
    – KConrad
    Jun 17, 2021 at 22:20
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    $\begingroup$ Let's show $\binom{w}{n} \in R_{\mathfrak p}$ for all $\mathfrak p \mid p$ in $R$. Different $\mathfrak p$-adic valuations on $K$ are the $p$-adic valuations on different embeddings $\sigma \colon K \to \overline{\mathbf Q_p}$. Each $\sigma(K)$ is in $\mathbf Q_p$ since $x^{p-1}-1$ splits over $\mathbf Q_p$. For each $(p-1)$-th root of unity $z$ in $\mathbf Q_p$, $\binom{z}{n} \in \mathbf Z_p$, so ${\rm ord}_p\binom{\sigma(w)}{n} \geq 0$ for each $\sigma$. So for each $\mathfrak p$ over $p$ in $R$, ${\rm ord}_{\mathfrak p} \binom{w}{n} \geq 0$, which means $\binom{w}{n} \in R_{\mathfrak p}$. $\endgroup$
    – KConrad
    Jun 17, 2021 at 22:21
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    $\begingroup$ Here's an example that can be done more simply: $p = 5$, so $w$ has order $4$. Set $w = i$. Why is $\mathbf Z_{(5)}[i]$ the int. closure of $\mathbf Z_{(5)}$ in $\mathbf Q(i)$? For rational $a$ and $b$, if $a+bi$ is integral over $\mathbf Z_{(5)}$ then its trace and norm for the extension $\mathbf Q(i)/\mathbf Q$ are integral over $\mathbf Z_{(5)}$, so they are in $\mathbf Z_{(5)}$ (it's integrally closed). The trace and norm are $2a$ and $a^2 + b^2$, so they are in $\mathbf Z_{(5)}$. Since $2$ is a unit, $a \in \mathbf Z_{(5)}$. Thus $b^2 \in \mathbf Z_{(5)}$, so $b \in \mathbf Z_{(5)}$. $\endgroup$
    – KConrad
    Jun 17, 2021 at 22:29
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    $\begingroup$ @KConrad: I suggest that you turn your comments into an answer, so that this question can be closed. $\endgroup$
    – GH from MO
    Jun 23, 2021 at 6:43

1 Answer 1

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Here is an elementary and explicit way to see this:

Suppose we have a set of $p$ integers $A=\{a_1, a_2,\dots, a_p\}$ which forms a complete set of residues modulo p. Then we have $$\prod_{a\in A}(x-a)=x^p-x \pmod{p}$$ which means that we can write $\prod_{a\in A}(x-a)=x^p-x+pF(x)$ for some polynomial $F(x)\in \mathbb Z[x]$. Since we have $w^p-w=0$, we can conclude that $$p^{-1}\prod_{a\in A}(w-a)=F(w)$$ so it is equal to a polynomial in $w$ with integer coefficients.

Next, let's look at what happens when we have $A=\{a_1,a_2,\dots,a_{p^k}\}$ which forms a complete set of residues modulo $p^k$. The polynomial $x^p-x$ has $p$ distinct roots modulo $p^k$, let's call them $B_0=\{b_1,b_2,\dots,b_p\}\subset A$. Let's define $B_r$ as the subset of $A$ which coincides with $\{b_1+r, b_2+r, \dots, b_p+r\}\pmod{p^k}$. Then we can write $A$ as a disjoint union: $$A=B_0\cup B_p \cup B_{2p}\cup\cdots \cup B_{p^k-p}.$$ Suppose that $\ell_r$ is defined as the exponent of $p$ in the prime factorization of $r$, i.e. $p^{\ell_r}|| r$. Then we have $$\prod_{a\in B_r} (x-a)=(x-r)^p-(x-r)+p^k G_r(x)=x^p-x+p^{\ell_r} H_r(x)$$ for some polynomials $G_r,H_r$ with integer coefficients. Therefore we can calculate by grouping $$(p^k!)^{-1}\prod_{a\in A}(w-a)=\frac{1}{p^k!}\prod_{i=0}^{p^{k-1}-1}\prod_{a\in B_{pi}}(w-a)=\frac{1}{p^k!}(p^kG_0(w))\prod_{i=1}^{p^{k-1}-1}p^{\ell_{pi}}H_{pi}(w)$$ and this is $p$-integral because the exponent of $p$ in $p^k\cdot p^{\ell_p+\ell_{2p}+\cdots+\ell_{p^k-p}}$ matches that of $p$ in $p^k!$.

In order to finish the general case in the problem, let $n$ be written in base $p$ as $n_kp^k+n_{k-1}p^{k-1}+\cdots+n_0$ and then split the set $\{0,1,\dots,n-1\}$ into $n_k$ complete sets of residues modulo $p^k$, $n_{k-1}$ complete sets of residues modulo $p^{k-1}$ and so on, and then apply the argument from the previous paragraph to each of them.

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