Let $A$ be a local integral domain of characteristic $p$. Let $K$ be the fraction field and let $k$ be the residue field of $A$. If $K$ is perfect, is $k$ necessarily perfect?

*Thoughts*:

- If $A$ is normal, then $K$ perfect implies that in fact $A$ itself is perfect, hence $k$ is perfect.
- If $A$ is essentially of finite type over a field, then $K$ perfect implies $\dim A = 0$.
- Here is a stronger question. For a field $F$ of characteristic $p$, the $p$-rank of $F$ is $p\operatorname{-rk}(F) := \log_{p} [F:F^{p}]$ (possibly infinite). So it would be enough to know: if $x_{1},x_{2}$ are points in an $\mathbb{F}_{p}$-scheme $S$ such that $x_{2} \in \overline{\{x_{1}\}}$ (i.e. "$x_{1}$ specializes to $x_{2}$"), then is $p\operatorname{-rk}(\kappa(x_{1})) \ge p\operatorname{-rk}(\kappa(x_{2}))$?
- Here is something related: For any integral domain $A$ with fraction field $K$, we can take an algebraic closure $\overline{K}$ and consider the integral closure $\overline{A}$ of $A$ in $\overline{K}$; then $\overline{A}$ is absolutely integrally closed (0DCL); in particular all residue fields of $\overline{A}$ are algebraically closed (0DCN).