P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient
Theorem
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\ds \sum_{i \mathop = m}^\infty d_i p^i$ be a $p$-adic expansion.
Let $a$ be the $p$-adic number, that is left coset, in $\Q_p$ containing $\ds \sum_{i \mathop = m}^\infty d_i p^i$.
Let $l$ be the index of the first non-zero coefficient in the $p$-adic expansion:
- $l = \min \set {i: i \ge m \land d_i \ne 0}$
Then:
- $\norm a_p = p^{-l}$
Proof
For all $n \ge m$, let:
- $\alpha_n = \ds \sum_{i \mathop = m}^n d_i p^i$
By assumption:
- $\sequence{\alpha_n}$ is a representative of $a$
By definition of the induced norm:
- $\norm a_p = \ds \lim_{n \mathop \to \infty} \norm {\alpha_n}_p$
From Eventually Constant Sequence Converges to Constant it is sufficient to show:
- $\forall n \ge l + 1 : \norm {\alpha_n}_p = p^{-l}$
Let $n \ge l + 1$.
Then:
\(\ds \norm {\sum_{i \mathop = m}^n d_i p^i}_p\) | \(=\) | \(\ds \norm {\paren {\sum_{i \mathop = m}^{l - 1} d_i p^i} + d_l p^l + \paren {\sum_{i \mathop = l + 1}^n d_i p^i} }_p\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\paren {\sum_{i \mathop = m}^{l - 1} 0 \cdot p^i } + d_l p^l + \paren {\sum_{i \mathop = l + 1}^n d_i p^i} }_p\) | by choice of $l$, for all $i : m \le i < l \implies d_i = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {d_l p^l + \paren {\sum_{i \mathop = l + 1}^n d_i p^i} }_p\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \max \set {\norm {d_l p^l }_p, \norm {\sum_{i \mathop = l + 1}^n d_i p^i}_p}\) | Non-Archimedean Norm Axiom $\text N 4$: Ultrametric Inequality |
The sum $\ds \sum_{i \mathop = l + 1}^n d_i p^i$ can be rewritten:
- $\ds \sum_{i \mathop = l + 1}^n d_i p^i = p^{l + 1} \sum_{i \mathop = l + 1}^n d_i p^{i - \paren {l + 1} }$
By definition of divisor:
- $\ds p^{l + 1} \divides \sum_{i \mathop = l + 1}^n d_i p^i$
Then:
\(\ds \norm {\sum_{i \mathop = l + 1}^n d_i p^i}_p\) | \(\le\) | \(\ds p^{-\paren {l + 1} }\) | Definition of $p$-adic norm | |||||||||||
\(\ds \) | \(<\) | \(\ds p^{-l}\) | Power Function on Base between Zero and One is Strictly Decreasing |
By definition of a $p$-adic expansion:
- $0 < d_l < p$
Then:
- $p^l \divides d_l p^l$
- $p^{l + 1} \nmid d_l p^l$
By definition of $p$-adic norm:
- $\norm {d_l p^l}_p = p^{-l}$
Thus:
- $\ds \norm {\sum_{i \mathop = l + 1}^n d_i p^i}_p < p^{-l} = \norm {d_l p^l}_p$
Finally:
\(\ds \norm {\sum_{i \mathop = m}^n d_i p^i}_p\) | \(=\) | \(\ds \max \set {\norm {d_l p^l}_p, \norm {\sum_{i \mathop = l + 1}^n d_i p^i}_p}\) | Three Points in Ultrametric Space have Two Equal Distances: Corollary $2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {d_l p^l}_p\) | as $\ds \norm {\sum_{i \mathop = l + 1}^n d_i p^i}_p < \norm {d_l p^l}_p$ | |||||||||||
\(\ds \) | \(=\) | \(\ds p^{-l}\) | derived earlier |
The result follows.
$\blacksquare$
Sources
- 2007: Svetlana Katok: p-adic Analysis Compared with Real ... (previous) ... (next): $\S 1.4$ The field of $p$-adic numbers $\Q_p$: Lemma $1.31$