P-adic Numbers are Generated Ring Extension 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.

Then:
 * $Q_p = \Z_p \sqbrk {1 / p}$

where $\Z_p \sqbrk {1 / p}$ denotes the ring extension generated by $1 / p$.

Proof
Let $a \in \Q_p$.

From Leigh.Samphier/Sandbox/Every P-adic Number is Integer Power of p times P-adic Integer, there exists $n \in \N$ such that $p^n a \in \Z_p$

Since $n \in \N$ and $p^n a \in \Z_p$, let $f \paren{X} \in \Z_p \sqbrk X$ be the polynomial:
 * $\paren {p^n a} X^n$

Then:
 * $\map f {1 / p} = \paren {p^n a} \paren {1 / p}^n = a$.

Hence:
 * $a \in \Z_p \sqbrk {1 / p}$.

Since $a \in \Q_p$ was arbitrary it follows that $\Q_p \subseteq \Z_p \sqbrk {1 / p }$.

By definition of a generated ring extension, $\Z_p \sqbrk {1 / p } \subseteq \Q_p$.

The result follows.