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Let $R=K[[x_1,\ldots,x_n]]$ the power series ring over a field $K$ of prime characteristic $p$. I have read that $R^{1/p}$ is a free $R$-module and its base is formed by the elements $\{\lambda^{1/p} x_1^{i_1/p} \cdots x_n^{i_n/p} \mid 0\leq i_j<p \text{ where } \lambda^{1/p} \text{ is a free basis of } K^{1/p} \text{ over } K\}$. For me is easy to see this if $K$ is $F$-finite, but I do not see this if $K$ is not $F$-finite.

Can you help me?

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 $s \cdot r^{1/p}=(s^pr)^{1/p}$.

Wágner Badilla
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    What does $R^{1/p}$ mean? – Thomas Andrews Mar 02 '23 at 21:17
  • What does $F$-finite mean? – Thomas Andrews Mar 02 '23 at 21:20
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    @ThomasAndrews $R^{1/p}$ is the set of $p$th roots of elements of $R$ taken in the algebraic closure of the fraction field of $R$. $F$-finite is the condition that the Frobenius endomorphism $F:R\rightarrow R$ by $F(r)=r^p$ is a finite ring homomorphism. – walkar Mar 02 '23 at 21:22
  • Is your concern about whether the set of elements of this form forms a basis (e.g. that $R^{1/p}$ is the $R$-span of these elements) or whether this span is really free? – walkar Mar 02 '23 at 21:26
  • Presumably, you mean $\lambda^{1/p}$ is in a basis for $K^{1/p}.$ Nothing says the basis has to be finite, so you just need to prove that $K^{1/p}$ is a vector space over $K$ to prove there is a basis. It is clear that $R^{1/p}$ contains the basis elements, and it is clear that $R^{1/p}$ is closed under addition and multiplication by elements of $\mathbb R.$ – Thomas Andrews Mar 02 '23 at 21:33
  • So given $a\in R$ you need to solve $b^p=a$ for some $b\in K^{1/p}[x_i^{1/p}].$ – Thomas Andrews Mar 02 '23 at 21:35
  • I do not understand you – Wágner Badilla Mar 02 '23 at 21:38
  • Is there another way to see that it is free? without thinking about the base that I gave – Wágner Badilla Mar 02 '23 at 22:29

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