P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient

From ProofWiki
Jump to navigation Jump to search

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

Retrieved from "https://proofwiki.org/w/index.php?title=P-adic_Norm_of_P-adic_Expansion_is_determined_by_First_Nonzero_Coefficient&oldid=657904"