Valuation Ring of P-adic Norm is Subring of P-adic Integers

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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