My question: It is known that $\mathbb{Q}_{p}(p^{1/p^{\infty}})$ is defined to be $\bigcup_{n>0} \mathbb{Q}_{p}(p^{1/p^{n}})$, which means adjoining all $p$-power roots of $p$ to the mixed characteristic field $\mathbb{Q}_{p}$. However, I have problem understanding the symbol $\mathbb{Q}_{p}(p^{1/p^{n}})$. How can this relate to the $p$-power roots of $p$? Why in the symbol, the power of $p$ is $1/p^{n}$? I think that $\mathbb{Q}_{p}(p^{1/p^{n}})$ is a cyclotomic extension of $\mathbb Q_p$, where $p^{1/p^{n}}$ is the primitive $n$th root of unity. But it seems that this does not make sense. And I saw in other answers that $\mathbb{Q}_{p}(p^{1/p^{n}})$ is a ramified extension. Could anyone tell me where can I learn about $\mathbb{Q}_{p}(p^{1/p^{n}})$?Answer1: By definition, $\Bbb Q_p(p^{1/p^n}) \cong \Bbb Q_p[X]/\langle X^{p^n} - p \rangle$.Answer2: The notation $F(a^{1/d})$ means the field obtained by adjoining to $F$ a solution of $x^d-a$. This is just like $a^{1/2}$ being $\sqrt{a}$: very standard algebraic notation. The notation $p^{1/d}$ is a root of $x^d-p$, not $x^d-1$. A cyclotomic extension of $F$ could be written as $F(1^{1/d})$ to mean you are adjoining a $d$th root of unity, I suppose. Why do you think $p^{1/d}$ (for $d = p^n$, say) should mean a root of unity rather than a $d$th root of $p$?This question was asked on May 29, 2021, when the author was in his freshman year and was confused about how to get started with Peter Scholze's Perfectoid geometry.