Unique Integer Close to Rational in Valuation Ring of P-adic Norm

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Theorem

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime number $p$.

Let $x \in \Q$ such that $\norm{x}_p \le 1$.


Then for all $i \in \N$ there exists a unique $\alpha \in \Z$ such that:

$(1): \quad \norm {x - \alpha}_p \le p^{-i}$
$(2): \quad 0 \le \alpha \le p^i - 1$


Proof

Let $i \in \N$.


From Integer Arbitrarily Close to Rational in Valuation Ring of P-adic Norm:

$\exists \mathop {\alpha'} \in \Z: \norm{x - \alpha'}_p \le p^{-i}$


By Integer is Congruent to Integer less than Modulus, then there exists $\alpha \in \Z$:

$\alpha \equiv \alpha' \pmod {p^i}$.
$0 \le \alpha \le p^i - 1$

Then $\norm {\alpha' - \alpha}_p \le p^{-i}$


Hence:

\(\ds \norm {x - \alpha}_p\) \(=\) \(\ds \norm {\paren {x - \alpha'} + \paren {\alpha' - \alpha} }_p\)
\(\ds \) \(\le\) \(\ds \max \set {\norm {x - \alpha'}_p, \norm {\alpha' - \alpha}_p }\) Non-Archimedean Norm Axiom $\text N 4$: Ultrametric Inequality
\(\ds \) \(\le\) \(\ds p^{-i}\)


Now suppose $\beta \in \Z$ satisfies:

$(\text a): \quad 0 \le \beta \le p^i - 1$
$(\text b): \quad \norm {x -\beta}_p \le p^{-i}$

Then:

\(\ds \norm {\alpha - \beta}_p\) \(=\) \(\ds \norm {\paren{\alpha - x} + \paren {x - \beta} }_p\)
\(\ds \) \(\le\) \(\ds \max \set {\norm{\alpha - x}_p, \: \norm {x - \beta}_p}\) Non-Archimedean Norm Axiom $\text N 4$: Ultrametric Inequality
\(\ds \) \(\le\) \(\ds \max \set {\norm {x - \alpha}_p, \: \norm {x - \beta}_p}\) Norm of Negative
\(\ds \) \(\le\) \(\ds p^{-i}\)

Hence $p^i \divides \alpha - \beta$, or equivalently, $\alpha \equiv \beta \pmod {p^i}$

By Initial Segment of Natural Numbers forms Complete Residue System then $\alpha = \beta$.

The result follows.

$\blacksquare$


Also see


Sources

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