Given examples of complete non-Archimedean field of characteristic $p$ s.t. $K$ is not $F$-finite

Question

Let $K$ be a complete non-Archimedean field of characteristic $p$ with valuation ring $A$. We say a ring $R$ is $F$-finite if it is of characteristic $p$ and the absolute Frobenius map $F_R$ is finite.

We know if $A$ is $F$-finite then $K$ is $F$-finite as a localization of $F$-finite map. So the case $A$ is $F$-finite and $K$ is not $F$-finite is not possible. But I can't deduce about other situations. So I would like to know examples for the other three cases:

1. $A$ is not $F$-finite, $K$ is $F$-finite.
2. $A$ is not $F$-finite, $K$ is not $F$-finite.
3. $A$ is $F$-finite, $K$ is $F$-finite.

The case 3 is probably the most simple one, $K=\mathbb{F}_p((t)),A=\mathbb{F}_p[[t]]$ should do the job. For case 1 and 2, $A$ is better to be non-Noetherian because $F$-finite ring is excellent (by Kunz) so Noetherian. Thanks in advance!

Answer 1

We want to know if $K/K^p$ is a finite extension and if $A$ is a finitely generated $A^p$-module.

• With $A= \Bbb{F}_p[[t]]$ both are.

• With $A= \Bbb{F}_p(x_1,x_2,x_3,\ldots)[[t]]$ neither are.

• Let $K$ be the completion of $\Bbb{F}_p[x,y,x^{-1},y^{-1}]$ for the valuation $v(x^ny^m) = n+\sqrt2 m$ ie. $$K = \{ \sum_{j\ge 1} c_j t^{b_j},c_j\in \Bbb{F}_p, b_j\in \Bbb{Z+\sqrt2\ Z}, \lim_{j\to \infty} b_j=\infty\}$$ $v(\sum_{j\ge 1} c_j t^{b_j})=\inf \{ b_j,c_j\ne 0\}$.

  Then $[K:K^p]=p^2$, while $A$ is not a finitely generated $A^p$-module:

  If $A=A^p+\sum_{m=1}^M a_m A^p$ then wlog $v(a_m)>0$ so that for $v(\alpha)<\inf v(a_m))$ we'd have $v(\alpha)\in \Bbb{pZ+\sqrt2\ p Z}$, contradicting that the elements in $(\Bbb{Z+\sqrt2\ Z})_{\ge 0}, \not \in (\Bbb{pZ+\sqrt2\ p Z})_{\ge 0}$ can be arbitrary small.