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
- 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction ... (previous) ... (next): $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.5$