7
$\begingroup$

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).
$\endgroup$

1 Answer 1

12
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Or just the subring $F[[t]]+tF'[[t]]$ of $F'[[t]]$? $\endgroup$
    – YCor
    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
    May 14, 2021 at 21:13
  • $\begingroup$ Oh, indeed, thanks. $\endgroup$
    – YCor
    May 14, 2021 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.