Prime characteristic $p$ and Power series ring

Asked Mar 02 '23 at 23:47

Active Mar 03 '23 at 19:03

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Let $R=K[[x_1,\ldots,x_n]]$ the power series ring over a field $K$ of prime characteristic $p$. If we take $K$ if $F$-finite, I know show that $R^{1/p}$ is a free $R$-module. If $K$ not is $F$-finite, is $R^{1/p}$ a free $R$-module?

Notation. $R^{1/p}$ is the set of $p$- th roots of elements of $R$ ,i.e, $R^{1/p}=\{r^{1/p} \mid r \in R\}$. Its $R$-module structure is given by $s \cdot r^{1/p}=(s^pr)^{1/p}$.

edited Mar 03 '23 at 19:03

asked Mar 02 '23 at 23:47

Wágner Badilla

1

You should probably define $R^{1/p}$ here too. I can't guess what the notation is, and if Thomas Andrews doesn't recognize it either then we know for sure it is obscure. – rschwieb Mar 03 '23 at 02:47
Why don't you try writing out a proof in the $F$-finite case? – math54321 Mar 03 '23 at 18:15
The $F$-Finito case is simple and very common, my question is when it is not $F$-infinite. – Wágner Badilla Mar 03 '23 at 19:09

Prime characteristic $p$ and Power series ring

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