I have read in many places that one of the most basic examples of perfectoid field is the completion of $\mathbb Q_p(p^{1/p^\infty})=\cup_{n\ge 1}\mathbb Q_p(p^{1/p^n}).$ However, I am not sure what exactly does $\mathbb Q_p(p^{1/p^n})$ mean. Does it mean adjoining all the $p^n$-th power roots of $p$, or just one of them? As $\mathbb Q_p$ does not have a primitive $p^n$ root of unity, I believe there is a difference.
Thanks a lot!
To be precise: let $\mathbb C_p$ be the completion of $\overline{\mathbb Q}_p$, the algebraic closure of $\mathbb Q_p$.
Now, inductively construct $p^n$-th roots of $p$ as follows: when $n=0$, let $x_0:=p$. For $n\ge1$, let $x_n$ be a root of the polynomial $X^p-x_{n-1}=0$, which has a solution in $\mathbb C_p$ since $\mathbb C_p$ is algebraically closed. This produces a compatible system of $p^n$-th roots of $p$: $\{x_n\}_{n\ge0}$.
Now, we can construct the field $\mathbb Q_p(\{x_n\}_{n\ge0})$, and denote the completion as $K$. Now $K\subseteq\mathbb C_p$ is perfectoid, since it is a complete topological field with topology induced by a nondiscrete valuation of rank $1$ (namely, the one restricted from $\mathbb C_p$), and $K^\circ$, the completion of $\mathbb Z_p[\{x_n\}_{n\ge0}]$, and on the quotient $$K^\circ/p\cong \mathbb Z_p[\{x_n\}_{n\ge0}]/p\cong\mathbb F_p[\{X_n\}_{n\ge0}]/(X_1^p=0,X_n^p=X_{n-1}),$$ the Frobenius is surjective.