Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Lemma 1

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 $n, m \in Z$, such that $n < m$.


Then:

$\forall y \in \Q_p: \norm{y}_p \le p^{-n}$ if and only if there exists $i \in \Z$:
$(1)\quad0 \le i \le p^\paren{m-n}-1$
$(2)\quad\norm{y - i p^n}_p \le p^{-m}$


Proof

Necessary Condition

Let $y \in \Q_p$.

Let $\norm{y}_p \le p^{-n}$.


Now:

\(\displaystyle \norm{y}_p\) \(\le\) \(\displaystyle p^{-n}\)
\(\, \displaystyle \leadsto \, \) \(\displaystyle p^n \norm{y}_p\) \(\le\) \(\displaystyle 1\) Multiply both sides by $p^n$
\(\, \displaystyle \leadsto \, \) \(\displaystyle \norm{p^{-n} }_p \norm{y}_p\) \(\le\) \(\displaystyle 1\) Definition of $p$-adic norm
\(\, \displaystyle \leadsto \, \) \(\displaystyle \norm{p^{-n} y}_p\) \(\le\) \(\displaystyle 1\) Norm axiom (N2) : (Mulitplicativity)
\(\, \displaystyle \leadsto \, \) \(\displaystyle \map {B_1^-} {p^{-n}y}\) \(=\) \(\displaystyle \map {B_1^-} 0\) Characterization of Closed Ball in P-adic Numbers


From Integers are Dense in Unit Ball of P-adic Numbers:

$\exists \mathop k \in \Z : \norm{p^{-n} y - k}_p \le p^\paren{n-m}$


From Residue Classes form Partition of Integers:

$\exists \mathop 0 \le i \le p^\paren{m-n}-1 : p^\paren{m-n} \divides k - i$

By definition of the $p$-adic norm: $\norm{k - i}_p \le p^\paren{n-m}$


It follows that:

\(\displaystyle \norm{p^{-n} y - i}_p\) \(\le\) \(\displaystyle \max \set{\norm{p^{-n} y - k}_p, \norm{i - k}_p}\) Corollary to P-adic Metric on P-adic Numbers is Non-Archimedean Metric
\(\displaystyle \) \(\le\) \(\displaystyle p^\paren{n-m}\)
\(\, \displaystyle \leadsto \, \) \(\displaystyle \norm{p^{-n} }_p \norm{y - i p^n}_p\) \(\le\) \(\displaystyle p^\paren{n-m}\) Norm axiom (N2) : (Mulitplicativity)
\(\, \displaystyle \leadsto \, \) \(\displaystyle p^n \norm{y - i p^n}_p\) \(\le\) \(\displaystyle p^\paren{n-m}\) Definition of $p$-adic norm
\(\, \displaystyle \leadsto \, \) \(\displaystyle \norm{y - i p^n}_p\) \(\le\) \(\displaystyle p^{-m}\) Divide both sides by $p^{-n}$


$\Box$

Sufficient Condition

Let $y\in \Q_p$.

Let there exist $i \in \Z$:

$(1)\quad0 \le i \le p^\paren{m-n}-1$
$(2)\quad\norm{y - i p^n}_p \le p^{-m}$


Now:

\(\displaystyle \norm{ y }_p\) \(=\) \(\displaystyle \norm{ y - i p^n + i p^n}_p\)
\(\displaystyle \) \(\le\) \(\displaystyle \max \set {\norm{ y - i p^n}_p, \norm{i p^n}_p}\) Norm axiom $(N4)$ : (Ultrametric Inequality)

By assumption:

$\norm{ y - i p^n} \le p^{-m} \le p^{-n}$

and:

\(\displaystyle \norm{i p^n}_p\) \(=\) \(\displaystyle \norm{i}_p \norm{p^n}_p\) Norm axiom $(N2)$ : (Multiplicativity)
\(\displaystyle \) \(\le\) \(\displaystyle 1 \cdot p^{-n}\) As $i \in \Z \subseteq \Z_p$
\(\displaystyle \) \(=\) \(\displaystyle p^{-n}\)

Hence:

$\max \set {\norm{ y - i p^n}_p, \norm{i p^n}_p} \le p^{-n}$.

So:

$\norm{ y }_p \le p^{-n}$.


$\blacksquare$