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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).
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1 Answer 1

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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.

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  • $\begingroup$ Or just the subring $F[[t]]+tF'[[t]]$ of $F'[[t]]$? $\endgroup$
    – YCor
    Commented May 14, 2021 at 20:53
  • $\begingroup$ @YCor This one doesn't have perfect fraction field ($t$ doesn't have a $p$th root) $\endgroup$
    – Wojowu
    Commented May 14, 2021 at 21:13
  • $\begingroup$ Oh, indeed, thanks. $\endgroup$
    – YCor
    Commented May 14, 2021 at 21:14

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