# Is any nonarchimedean field containing all roots of unity perfectoid?

Say $K$ is a complete nonarchimedean extension of $\mathbf{Q}_p$, i.e., it is the fraction field of a $p$-adically complete and $p$-torsionfree rank $1$ valuation ring. Assume that the residue field of $K$ is perfect, and that one of the following two conditions holds:

a) $\mu_{p^\infty} \in K$.
b) $p$ admits a compatible system of $p$-power roots in $K$.

Is $K$ perfectoid?

## 1 Answer

Not necessarily. For example, if you take any perfectoid field $K$, the $p$-adic completion of the field $K(T)$ won't be perfectoid.

• Thanks! Can you clarify what you mean by the $p$-adic completion of $K(T)$? In particular, what is the valuation ring? – clueless May 12 '18 at 15:13
• The $p$-adic completion of $K(T)$ will include power series such that the coefficients converge to 0 $p$-adically. For example, something like $\sum_0^\infty p^nT^n$ will be there. The valuation ring will depend on the topology you put on $K(T)$. If you put the $p$-adic topology on, it'll be the $p$-adic completion of $\mathcal O_K(T)$. In particular, things like $1/T$ will be in the valuation ring. If you put the $(p,T)$-adic topology on, it'll be the $p$-adic completion of $\mathcal O_K[T]$. – user124375 May 13 '18 at 16:22