# Does perfect fraction field imply perfect residue field?

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).

The answer is no. Let $$F$$ be any imperfect subfield of a perfect field $$F'$$. Let $$B$$ be the ring of integral 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.
• Or just the subring $F[[t]]+tF'[[t]]$ of $F'[[t]]$?
• @YCor This one doesn't have perfect fraction field ($t$ doesn't have a $p$th root) Commented May 14, 2021 at 21:13