Let $v:K\to \Gamma$ be a valuation on a field $K$. I know that if the valuation ring $\mathcal O_v$ is henselian and $k$ has characteristic zero, then $k$ has a lift to $K$. I.e. there is a section $s:k\to \mathcal O_v$ of the residue map.
My question is, if $K$ is algebraically closed (then $k$ is algebraically closed) and $\mathrm{Char}(k)= \mathrm{Char}(K)$, is there a lifting of $k$? (Note that I don't want to assume that $\mathrm{Char}(k)=0$ ).
Yes, there is: let $P$ be the common prime field of $K$ and $k$. Choose a transcendence basis $\overline{T}$ of the field extension $k/P$. For every $\overline{t}\in \overline{T}$ choose an element $t\in O_v$ such that $t+M_v=\overline{t}$. The resulting subset $T\subset O_v$ then is algebraically independent over $P$. Consequently the restriction $v|_{P(T)}$ to the rational function field $P(T)$ is trivial. Let $K_0$ be the algebraic closure of $P(T)$ in $K$; it is algebraically closed since $K$ is so. The restriction $v|_{K_0}$ is trivial, since it is a prolongation of the trivial valuation $v|_{P(T)}$ to an algebraic extension of $P(T)$. Hence the residue field $\overline{K_0}$ is an algebraically closed, algebraic extension of $P(\overline{T})$, thus the algebraic closure of $P(\overline{T})$ in $k$, which equals $k$ by the choice of $\overline{T}$.