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.
Let $a$ be a $p$-adic number, that is left coset, in $\Q_p$.
Let $\ds \sum_{i \mathop = m}^\infty d_i p^i$ and $\ds \sum_{i \mathop = k}^\infty e_i p^i$ be $p$-adic expansions that represent $a$.
Then:
- $(1) \quad m = k$
- $(2) \quad \forall i \ge m : d_i = e_i$
That is, the $p$-adic expansions $\ds \sum_{i \mathop = m}^\infty d_i p^i$ and $\ds \sum_{i \mathop = k}^\infty e_i p^i$ are identical.
Proof
From P-adic Number times P-adic Norm is P-adic Unit there exists $n \in \Z$:
- $p^n a \in \Z^\times_p$
- $\norm a_p = p^n$
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:
- $\ds \sum_{i \mathop = l}^\infty d_i p^i$
From P-adic Expansion Less Intial Zero Terms Represents Same P-adic Number
- $\ds \sum_{i \mathop = l}^\infty d_i p^i$
is a representative of $a$
From Multiple Rule for Cauchy Sequences in Normed Division Ring:
- $\ds p^{-l} \sum_{i \mathop = l}^\infty d_i p^i = \sum_{i \mathop = l}^\infty d_i p^{i - l} = \sum_{i \mathop = 0}^\infty d_{i + l} p^i$
is a representative of $p^{-l} a \in \Z^\times_p$.
By definition of a $p$-adic expansion:
- $\forall i \in N : 0 \le d_{i + 1} < p - 1$
Thus the series:
- $\ds \sum_{i \mathop = 0}^\infty d_{i + l} p^i$
is a $p$-adic expansion that represents $p^{-l} a \in \Z^\times_p$.
Similarly the series:
- $\ds \sum_{i \mathop = 0}^\infty e_{i + l} p^i$
is a $p$-adic expansion that represents $p^{-l} a \in \Z^\times_p$.
From P-adic Integer has Unique P-adic Expansion Representative 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.
$\blacksquare$