Leigh.Samphier/Sandbox/P-adic Expansion Representative of P-adic Number is Unique
Theorem
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers as a quotient of Cauchy sequences.
That is, $\Q_p$ is the quotient ring $\CC \, \big / \NN$ where:
- $\CC$ denotes the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$
- $\NN$ denotes the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$.
Let $\mathbf a$ be an equivalence class in $\Q_p$.
Let $\displaystyle \sum_{i \mathop = m}^\infty d_i p^i$ and $\displaystyle \sum_{i \mathop = k}^\infty e_i p^i$ be $p$-adic expansions that represent $\mathbf a$.
Then:
- $(1) \quad m = k$
- $(2) \quad \forall i \ge m : d_i = e_i$
That is, the $p$-adic expansions $\displaystyle \sum_{i \mathop = m}^\infty d_i p^i$ and $\displaystyle \sum_{i \mathop = k}^\infty e_i p^i$ are identical.
Proof
Let $\textbf p$ denote the equivalence class that is identified with the prime number $p$.
By Rational Numbers are Dense Subfield of P-adic Numbers:
- the constant sequence $\tuple {p, p, p, \dots}$ represents $\mathbf p \in \Q_p$.
From P-adic Number times P-adic Norm is P-adic Unit there exists $n \in \Z$:
- $\mathbf p^n \mathbf a \in \Z^\times_p$
- $\norm {\mathbf a}_p = p^n$
where $\Z^\times_p$ is the set of $p$-adic units.
Let $l = -n$.
From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient:
- $l = \min \set {i: i \ge m \land d_i \ne 0}$
and
- $l = \min \set {i: i \ge k \land e_i \ne 0}$
Now:
\(\ds m\) | \(<\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds d_m\) | \(\ne\) | \(\ds 0\) | Definition of $p$-adic Expansion | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds m\) | \(=\) | \(\ds l\) | Definition of $l$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds l\) | \(<\) | \(\ds 0\) |
Similarly:
- $k < 0 \leadsto k = l$
Proof of statement $(1)$
Case $l < 0$
Let $l < 0$.
Then:
\(\ds l\) | \(<\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds m\) | \(<\) | \(\ds 0\) | as $m \le l$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds m\) | \(=\) | \(\ds l\) | from above |
Similarly:
- $l < 0 \leadsto k = l$
So:
- $m = l = k$.
$\Box$
Case $l \ge 0$
Let $l \ge 0$.
Then:
\(\ds l\) | \(\ge\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds m\) | \(\not <\) | \(\ds 0\) | as $m < 0 \leadsto l < 0$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds m\) | \(=\) | \(\ds 0\) | Definition of $p$-adic Expansion |
Similarly:
- $l \ge 0 \leadsto k = 0$
So:
- $m = 0 = k$.
$\Box$
Proof of statement $(2)$
By definition of $l$, for all $i$ such that $m \le i < l$:
- $d_i = 0$
- $e_i = 0$
So for all $m \le i < l$:
- $d_i = e_i$.
Consider the series:
- $\displaystyle \sum_{i \mathop = l}^\infty d_i p^i$
From P-adic Expansion Less Intial Zero Terms Represents Same P-adic Number
- $\displaystyle \sum_{i \mathop = l}^\infty d_i p^i$
is a representative of $\mathbf a$
From Multiple Rule for Cauchy Sequences in Normed Division Ring:
- $\displaystyle p^{-l} \sum_{i \mathop = l}^\infty d_i p^i = \sum_{i \mathop = l}^\infty d_i p^{i - l} = \sum_{i \mathop = 0}^\infty d_{i + l} p^i$
is a representative of $\mathbf p^{-l} \mathbf a \in \Z^\times_p$.
By definition of a $p$-adic expansion:
- $\forall i \in N : 0 \le d_{i + 1} < p - 1$
Thus the series:
- $\displaystyle \sum_{i \mathop = 0}^\infty d_{i + l} p^i$
is a $p$-adic expansion that represents $\mathbf p^{-l} \mathbf a \in \Z^\times_p$.
Similarly the series:
- $\displaystyle \sum_{i \mathop = 0}^\infty e_{i + l} p^i$
is a $p$-adic expansion that represents $\mathbf p^{-l} \mathbf a \in \Z^\times_p$.
From Equivalence Class in P-adic Integers Contains Unique P-adic Expansion it follows that:
- $\forall i \in N : d_{i + 1} = e_{i + 1}$
That is:
- $\forall i \ge l : d_i = e_i$
The result follows.
$\blacksquare$