# Can we have a perfect field of characteristic $p$ with a non-archimedean valuation?

Motivation for the question: in Neukirch - Algebraic Number Theory (Chapter II - Section 6 - Henselian Fields) the author discusses the context of a nonarchimedean valued field $K$ and its completion $\widehat{K}$, and considers the separable closure $K_v$ of $K$ in $\widehat{K}$, so that $K \subseteq K_v \subseteq \widehat{K}$.

Then he writes the following:

When $K_v$ is algebraically closed in $\widehat K$ - hence in particular $\text{char}(K) = 0$ - this is immediately obvious [...]

(emphasis mine).

My question is about the implicit claim that if $K_v$ is algebraically closed, then $\text{char}(K) = 0$.

Now we know that $K_v$ is algebraically closed if and only if $K$ is perfect (since every $K_v$ will be the algebraic closure of $K$ and so every algebraic extension of $K$ is separable).

So we're asking whether it is true that if $K$ is perfect, then it must have characteristic $0$.

I know that all fields of characteristic $0$ are perfect. I also know that we have the example of finite fields which are also perfect fields but have prime characteristic. This does not contradict the claim above because the finite fields have only the trivial valuation.

So is there something about having a nonarchimedean valuation on a perfect field that forces its characteristic to be zero?

$$\mathbb F_p(T, T^{1/p}, T^{1/p^2}, \ldots)$$
is perfect, has characteristic $p$, and admits the $T$-adic valuation, so the answer to the question in the title is "yes".