A trailer for p-adic analysis, second half: Mahler coefficients

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

\displaystyle  \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

\displaystyle  a_n = \sum_{k=0}^n \binom nk (-1)^{n-k} f(n).  \ \ \ \ \ (1)

Then {\lim_n a_n = 0} and

\displaystyle  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”.

\displaystyle  \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)}, \dots 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

\displaystyle  \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

\displaystyle  \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

\displaystyle  \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

\displaystyle  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

\displaystyle  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

\displaystyle  \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

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