P-adic Numbers are Generated Ring Extension 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.


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 P-adic Number times Integer Power of p is 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 $\map f X \in \Z_p \sqbrk X$ 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.

$\blacksquare$


Sources