Let $K$ be a global field and $v$ be a non-archimedean place of $K$. Given a field $k$, let $k^{\mathrm{sep}}$ be a separable closure inside an algebraic closure $\overline k$.
I would like to know why $K^{\mathrm{sep}}$ dense in $K_v^{\mathrm{sep}}$.
I think that $\overline K$ is dense in $\overline{K_v}$ as shown on p. 33 of Non-Archimedean Operator Theory, so it would settle the case where $K$ is perfect.
I've heard that Krasner's lemma can help, but I don't see how. Thanks.
Prove the following as an exercise (no Krasner's lemma needed):
Now here is an outline on how to proceed:
Let $\alpha\in K_v^\mathrm{sep}$, let $\alpha_2,\dots,\alpha_n$ be the conjugates of $\alpha$ and let $f\in K_v[x]$ be the minimum polynomial of $\alpha$ over $K_v$. Choose $g\in K[x]$ to be an approximation of $f$ so that each coefficient of $g-f$ has absolute value smaller than $\min_{i=2,\dots,n}|\alpha-\alpha_i|/|\alpha|^n$. Conclude that $|g(\alpha)|<\min_{i=2,\dots,n}|\alpha-\alpha_i|$. By writing $g$ as a product of linear polynomials deduce that $g$ has a root $\beta$ for which $|\alpha-\beta|<\min_{i=2,\dots,n}|\alpha-\alpha_i|$, and use Krasner's lemma to deduce that $\alpha\in K_v(\beta)$, which means by the lemma above you can approximate $\alpha$ by elements of $K(\beta)$ and hence by elements of $K^\mathrm{sep}$.