Valuation Ring of P-adic Norm is Subring of P-adic Integers
Theorem
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $\Z_{\ideal p}$ be the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$.
Then:
- $(1): \quad \Z_{\ideal p} = \Q \cap \Z_p$.
- $(2): \quad \Z_{\ideal p}$ is a subring of $\Z_p$.
Corollary
The set of integers $\Z$ is a subring of $\Z_p$.
Proof
The $p$-adic integers is defined as:
- $\Z_p = \set {x \in \Q_p: \norm x_p \le 1}$
The induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$ is defined as:
- $\Z_{\ideal p} = \set {x \in \Q: \norm x_p \le 1}$
From Rational Numbers are Dense Subfield of P-adic Numbers:
- the $p$-adic norm $\norm {\,\cdot\,}_p$ on $\Q_p$ is an extension of the $p$-adic norm $\norm {\,\cdot\,}_p$ on $\Q$.
It follows that $\Z_{\ideal p} = \Q \cap \Z_p$.
This proves $(1)$ above.
By Valuation Ring of Non-Archimedean Division Ring is Subring then $\Z_p$ is a subring of $\Q_p$.
By definition of $p$-adic integers then $\Q$ is a subring of $\Q_p$.
By Intersection of Subrings is Largest Subring Contained in all Subrings then $\Z_{\paren p}$ is a subring of $\Z_p$.
This proves $(2)$ above.
$\blacksquare$
Sources
- 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction: $\S 3.3$ Exploring $\Q_p$, Proposition $3.3.4$