In the previous post we defined $p$-adic numbers. This post will state (mostly without proof) some more surprising results about continuous functions $f \colon \mathbb Z_p \rightarrow \mathbb Q_p$. Then we give the famous proof of the Skolem-Mahler-Lech theorem using $p$-adic analysis.

1. Digression on $\mathbb C_p$

Before I go on, I want to mention that $\mathbb Q_p$ is not algebraically closed. So, we can take its algebraic closure $\overline{\mathbb Q_p}$ — but this field is now no longer complete (in the topological sense). However, we can then take the completion of this space to obtain $\mathbb C_p$. In general, completing an algebraically closed field remains algebraically closed, and so there is a larger space $\mathbb C_p$ which is algebraically closed and complete. This space is called the $p$-adic complex numbers.

We won’t need $\mathbb C_p$ at all in what follows, so you can forget everything you just read.

2. Mahler coefficients: a description of continuous functions on $\mathbb Z_p$

One of the big surprises of $p$-adic analysis is that we can concretely describe all continuous functions $\mathbb Z_p \rightarrow \mathbb Q_p$. They are given by a basis of functions $$\binom xn \overset{\mathrm{def}}{=} \frac{x(x-1) \dots (x-(n-1))}{n!}$$ in the following way.

Theorem 1 (Mahler; see Schikhof Theorem 51.1 and Exercise 51.B)

Let $f \colon \mathbb Z_p \rightarrow \mathbb Q_p$ be continuous, and define $$a_n = \sum_{k=0}^n \binom nk (-1)^{n-k} f(n) \quad (1).$$ Then $\lim_n a_n = 0$ and $$f(x) = \sum_{n \ge 0} a_n \binom xn.$$ Conversely, if $a_n$ is any sequence converging to zero, then $f(x) = \sum_{n \ge 0} a_n \binom xn$ defines a continuous function satisfying (1).

The $a_i$ are called the Mahler coefficients of $f$.

Exercise 2. Last post we proved that if $f \colon \mathbb Z_p \rightarrow \mathbb Q_p$ is continuous and $f(n) = (-1)^n$ for every $n \in \mathbb Z_{\ge 0}$ then $p = 2$. Re-prove this using Mahler’s theorem, and this time show conversely that a unique such $f$ exists when $p=2$.

You’ll note that these are the same finite differences that one uses on polynomials in high school math contests, which is why they are also called “Mahler differences”.

$$\begin{aligned} a_0 &= f(0) \\ a_1 &= f(1) - f(0) \\ a_2 &= f(2) - 2f(1) - f(0) \\ a_3 &= f(3) - 3f(2) + 3f(1) - f(0). \end{aligned}$$

Thus, one can think of $a_n \rightarrow 0$ as saying that the values of $f(0)$, $f(1)$, … behave like a polynomial modulo $p^e$ for every $e \ge 0$. Amusingly, this fact was used on a USA TST in 2011:

Exercise 3 (USA TST 2011/3)

Let $p$ be a prime. We say that a sequence of integers $(z_n)_{n=0}^\infty$ is a $p$-pod if for each $e \geq 0$, there is an $N \geq 0$ such that whenever $m \geq N$, $p^e$ divides the sum $$\sum_{k=0}^m (-1)^k \binom mk z_k.$$ Prove that if both sequences $(x_n)_{n=0}^\infty$ and $(y_n)_{n=0}^\infty$ are $p$-pods, then the sequence $(x_n y_n)_{n=0}^\infty$ is a $p$-pod.

3. Analytic functions

We say that a function $f \colon \mathbb Z_p \rightarrow \mathbb Q_p$ is analytic if it has a power series expansion $$\sum_{n \ge 0} c_n x^n \quad c_n \in \mathbb Q_p \qquad\text{ converging for } x \in \mathbb Z_p.$$ As before there is a characterization in terms of the Mahler coefficients:

Theorem 4 (Schikhof Theorem 54.4)

The function $f(x) = \sum_{n \ge 0} a_n \binom xn$ is analytic if and only if $$\lim_{n \rightarrow \infty} \frac{a_n}{n!} = 0.$$

Just as holomorphic functions have finitely many zeros, we have the following result on analytic functions on $\mathbb Z_p$.

Theorem 5 (Strassmann’s theorem)

Let $f \colon \mathbb Z_p \rightarrow \mathbb Q_p$ be analytic. Then $f$ has finitely many zeros.

4. Skolem-Mahler-Lech

We close off with an application of the analyticity results above.

Theorem 6 (Skolem-Mahler-Lech)

Let $(x_i)_{i \ge 0}$ be an integral linear recurrence. Then the zero set of $x_i$ is eventually periodic.

Proof: According to the theory of linear recurrences, there exists a matrix $A$ such that we can write $x_i$ as a dot product $$x_i = \left< A^i u, v \right>.$$ Let $p$ be a prime not dividing $\det A$. Let $T$ be an integer such that $A^T \equiv \mathbf{1} \pmod p$.

Fix any $0 \le r < N$. We will prove that either all the terms $$f(n) = x_{nT+r} \qquad n = 0, 1, \dots$$ are zero, or at most finitely many of them are. This will conclude the proof.

Let $A^T = \mathbf{1} + pB$ for some integer matrix $B$. We have

$$\begin{aligned} f(n) &= \left< A^{nT+r} u, v \right> = \left< (\mathbf1 + pB)^n A^r u, v \right> \\ &= \sum_{k \ge 0} \binom nk \cdot p^n \left< B^n A^r u, v \right> \\ &= \sum_{k \ge 0} a_n \binom nk \qquad \text{ where } a_n = p^n \left< B^n A^r u, v \right> \in p^n \mathbb Z. \end{aligned}$$

Thus we have written $f$ in Mahler form. Initially, we define $f \colon \mathbb Z_{\ge 0} \rightarrow \mathbb Z$, but by Mahler’s theorem (since $\lim_n a_n = 0$) it follows that $f$ extends to a function $f \colon \mathbb Z_p \rightarrow \mathbb Q_p$. Also, we can check that $\lim_n \frac{a_n}{n!} = 0$ hence $f$ is even analytic.

Thus by Strassman’s theorem, $f$ is either identically zero, or else it has finitely many zeros, as desired. $\Box$