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](https://stacks.math.columbia.edu/tag/0DCL)); in particular all residue fields of $\overline{A}$ are algebraically closed ([0DCN](https://stacks.math.columbia.edu/tag/0DCN)).



Does perfect fraction field imply perfect residue field?
The answer is no. Let $F$ be any imperfect subfield of a perfect field $F'$. Let $B$ be the ring of integral [Puiseux series](https://en.wikipedia.org/wiki/Puiseux_series) over $F'$ (integral meaning ones involving only nonnegative powers of $T$) and let $A$ be the subring consisting of those series whose coefficient of $T^0$ belongs to $F$. $A$ and $B$ have the same fraction field, which is perfect since $B$ is perfect. However the residue field of $A$ is isomorphic to $F$, and so isn't perfect.


