Binomial coefficients and $p$-adic limits

Question

I was reading Schoof's Catalan's Conjecture, and on page 37, while proving Cassels' Theorem it is mentioned that by Exercise 5.7, the denominators of the binomial coefficients $\binom{\frac{p}{q}}{k}$ for $0 \leq k \leq \lfloor \frac{p}{q} \rfloor + 1$, where $p$ and $q$ are odd prime numbers, are powers of $q$.

I found a proof by Keith Conrad, but it uses $p$-adic limits, so I decided to read through Gouvea's $p$-adic Numbers: An Introduction.

I still don't know what is going on, any insight please?

Answer 1

The proof you cited does not really use $p$-adic numbers but just the $p$-adic topology on $\mathbf Q$ and continuity of polynomial functions in the $p$-adic topology.

Suppose $F$ is a field and $|\cdot|$ is an absolute value on $F$. Let $f \colon F \to F$ be a continuous function and $S \subset F$ be a subset such that $|f(s)| \leq 1$ for all $s \in S$. If $x \in F$ is a limit of points in $A$, say $x = \lim_{n \to \infty} s_n$. By continuity, $f(x) = \lim_{n \to \infty} f(s_n)$. The absolute value on $F$ is continuous, so $|f(x)| = \lim_{n \to \infty} |f(s_n)|$. Since $|f(s_n)| \leq 1$ for all $n$, we get $|f(x)| \leq 1$.

Now apply this to $F = \mathbf Q$, $|\cdot| = |\cdot|_p$ ($p$-adic absolute value), and $f(x) = \binom{x}{k}$, and $S = \mathbf Z$. Polynomials are continuous functions since addition and multiplication are continuous. A rational number whose denominator is not divisible by $p$ is a $p$-adic limit of integers: if $a/b$ is a fraction where $p \nmid b$, then for each positive integer $n$ we can solve $bs_n \equiv a \bmod p^n$ for some integer $s_n$, which means $|bs_n - a|_p \leq 1/p^n$, so $|s_n - a/b|_p = |(bs_n-a)|_p/|b|_p = |bs_n-a|_p \leq 1/p^n$. Thus $s_n \to a/b$ as $n \to \infty$. Therefore $|\binom{a/b}{k}|_p \leq 1$ since $|\binom{s_n}{k}|_k \leq 1$ for all $n$. That shows the fraction $\binom{a/b}{k}$ has no $p$ in its denominator when $p \nmid b$.

Answer 2

This was really a bit too long to put into a comment to KCd's answer, but I thought it might help to see a specific concrete example of such a limit.

We already know that $\binom{m}{n}$ is an integer when $m$ is an integer, so we could write down a specific limit of integers that approaches $\frac{a}{b}$ in the p-adic topology when $b$ and $p$ are relatively prime, because

$$p^{\varphi(b)k}(b-a) + a \equiv 0 \mod b$$ $$1*(b-a) + a \equiv 0 \mod b$$ $$b \equiv 0 \mod b$$

This means $\frac{p^{\varphi(b)k}(b-a) + a}{b}$ is an integer. Then we know that because $p^{\varphi(b)k} \to 0$ as $k \to \infty$ this means we have our sequence of integers approaching $\frac{a}{b}$,

$$\frac{p^{\varphi(b)k}(b-a) + a}{b} \to \frac{a}{b}$$.

This means our limit of integers is

$$\binom{\frac{a}{b}}{n} = \lim_{k \to \infty} \binom{\frac{p^{\varphi(b)k}(b-a) + a}{b}}{n}$$

And we have that

$$\left|\binom{\frac{a}{b}}{n}\right|_p = \lim_{k \to \infty} \left|\binom{\frac{p^{\varphi(b)k}(b-a) + a}{b}}{n}\right|_p \le 1$$