= Solution
One form of <Krasner's lemma> is: if $K$ is a complete non-Archimedean field, $\alpha$ is separable over $K$, and $\beta$ is algebraic with
$$
|\beta-\alpha|<\min_{\alpha'\ne\alpha}|\alpha-\alpha'|,
$$
where $\alpha'$ ranges over the other $K$-conjugates, then $K(\alpha)\subseteq K(\beta)$. If $\alpha\in K$, the conclusion is immediate.
To prove the nontrivial case, let $\sigma$ be an automorphism of an algebraic closure fixing $K(\beta)$. The <unique extension of an absolute value to a finite extension> implies that $\sigma$ preserves the extended absolute value. Therefore
$$
|\sigma(\alpha)-\alpha|\leq\max\bigl(|\sigma(\alpha)-\beta|,|\beta-\alpha|\bigr)=|\beta-\alpha|.
$$
This is strictly less than the distance to every other conjugate, so $\sigma(\alpha)=\alpha$. As $\alpha$ is separable over $K(\beta)$, being fixed by all these automorphisms puts it in $K(\beta)$. \b[This proves $\boxed{K(\alpha)\subseteq K(\beta)}$.] The strict inequality is what prevents switching to another conjugate.
Back to article page
