Completion of a Valued Field
Theorem
Let $(k,|\cdot|)$ be a valued field.
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Then there exists a unique completion $k'$ of $k$ as a metric space and an absolute value $|\cdot|'$ such that $(k', |\cdot|')$ is a valued field.
Proof
By the completion theorem for metric spaces $k$ has a unique completion $k'$ equipped with a metric $d'$ that restricts to $d$ on $k$.
Therefore we need only show that $|\cdot|' = d'(x,0)$ is an absolute value on $k'$.
But $|\cdot|'$ is an absolute value on $k$ and a metric on $k'$.
Therefore, if $|\cdot|'$ were not an absolute value on $k'$, this would imply discontinuity of the metric, contradicting Metric is Continuous Function.
$\blacksquare$
Examples
- The completion of $\Q$ with respect to the usual absolute value is $\R$
- The completion of $\Q$ with respect to the $p$-adic absolute value is known as the field of $p$-adic numbers and denoted $\Q_p$
- By Ostrowski's Theorem there are no other completions of $\Q$ as a valued field.
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