Valuation Ideal of P-adic Norm on Rationals

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Theorem

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


The induced valuation ideal on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the set:

$\PP = p \Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$

where $\Z_{\ideal p}$ is the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$


Proof

Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on $\Q$.

Then:

\(\ds \PP\) \(=\) \(\ds \set {\dfrac a b \in \Q : \norm{\dfrac a b}_p < 1}\) Definition of Valuation Ideal Induced by Non-Archimedean Norm
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \set {\dfrac a b \in \Q : \map {\nu_p} {\dfrac a b} > 0}\) Definition of $p$-adic Norm
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \set {\dfrac a b \in \Q : \map {\nu_p} a - \map {\nu_p} b > 0}\) Definition of $p$-adic Valuation on Rationals
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \set {\dfrac a b \in \Q : \map {\nu_p} a > \map {\nu_p} b}\)


Let $\dfrac a b \in \Q$ be in canonical form.

Then $a \perp b$


Suppose $p \divides a$.

Then $p \nmid b$.

Hence:

$\map {\nu_p} a > 0 = \map {\nu_p} b$


Suppose $p \nmid a$.

Then:

$\map {\nu_p} b \ge 0 = \map {\nu_p} a$


So:

$\map {\nu_p} a > \map {\nu_p} b$ if and only if $p \nmid b$ and $p \divides a$

Hence:

$\PP = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$


So:

\(\ds \dfrac a b \in \PP\) \(\leadstoandfrom\) \(\ds p \nmid b, p \divides a\)
\(\ds \) \(\leadstoandfrom\) \(\ds p \nmid b, \exists a' \in \Z: a = p a'\)
\(\ds \) \(\leadstoandfrom\) \(\ds \exists a' \in \Z: a = p a', \dfrac {a'} b \in \Z_{\ideal p}\) Valuation Ring of P-adic Norm on Rationals
\(\ds \) \(\leadstoandfrom\) \(\ds \dfrac a b \in p \Z_{\ideal p}\)

Hence:

$\PP = p \Z_{\ideal p}$

$\blacksquare$


Sources

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