TITLE: Why is $K_{\upsilon}|K$ separable for a global field $K$?
QUESTION [6 upvotes]: I asked this question on math.stackexchange but maybe it fits here better. If not, I apologize in advance and will remove the question.
Let $K$ be a global field and $\upsilon$ a prime of $K$. Then in a paper, which I read, it is used that $K_{\upsilon}|K$ is separable, where $K_{\upsilon}|K$ is the completion. This is obviously true if $charK=0$.
Why is $K_{\upsilon}|K$ separable if $charK=p>0$?
Things I tried but could not get to work (maybe I missed something and they would work):

Use that $\hat A \cap Frac(A)^{alg} \subset A^h$ for an excellent ring $A$ (this is mentioned here henselization and completion)
Use Exercise 1 of section II.4 of Serre's Local Fields: If $A$ is a DVR for which every finite purely inseparable extension $Frac(A)$ has integral closure finite over $A$ then the completion of $Frac(A)$ is separable over $Frac(A)$
Try to use that $K_{\upsilon}=\mathbb F_{p^n}((T))$ 

The statement seems to me like it should be in a book somewhere but I just can't find it.
Thanks in advance.

REPLY [5 votes]: Since there is already an answer, I want to give my slightly different answer in a special case: Say $K = k(t)$ where $k$ is a finite field and $K_v = k((t))$. We only need to check that
$$
K_v \otimes_K K^{1/p}
$$
is reduced, see Tag 030W. Since $K^{1/p} = k(t^{1/p})$ this is true because the tensor product is visibly equal to $k((t^{1/p}))$. I claim this argument works in general too, but since we have a fully written out proof above, I won't insist. Enjoy!