Degree of $[K:K^p]$ and Completion Suppose $K$ is a field endowed with a non-archimedian absolute value.  Assume $K$ has characteristic $p>0$ and that $[K:K^p] < \infty$.  Let $L$ be the completion of $K$ with respect to this absolute value.  Is it always true that $[L:L^p] = [K:K^p]$?
 A: In fact, BCnrd's comment says that for dvr's, only for non-excellent one's $[L:L^p]\ne [K:K^p]$ can happen. Actually, suppose $d=[K:K^p]$ is finite. Consider the canonical map $K\otimes_{K^p} L^p \to L$. The source is a $L^p$-vector space of dimension $d$, its image is therefore closed ($L^p$ is complete) and contains $K$, so the map is surjective. Therefore $[L:L^p]=d$ if and only if the above map is an isomorphism, or equivalently if $K$ and $L^p$ are linearly disjoint over $K^p$. Extracting $p$-th root in all these fields, this is also equivalent to $K^{1/p}$ and $L$ linearly disjoint over $K$. Which is also equivalent to $L$ separable (i.e. geometrically reduced) over $K$. This is the definition for the dvr of $K$ to be excellent. 
Example of non-excellent dvr with finite $[K : K^p]$: Let $s$  be an element of $\mathbb F_p((t))$   transcendent over $\mathbb F_p(t)$, let $K=\mathbb F_p(t,s)⊂\mathbb F_p((t))$ be endowed with the $t$-adic valuation. Then $[K:K^p]=p^2$  and $[L:L^p]=p$. If we take any number of algebraically independent elements $s_1,...,s_n$  (instead of just one $s$) over $\mathbb F_p(t)$, then we have $[K:K^p]=p^{n+1}$.
