How to find the ring of integers of $\mathbb{Q}_{p}(p^{1/p^{\infty}})$ and $\mathbb{F}_{p}((t))(t^{1/p^{\infty}})$?

Question

Maybe this is a silly question. Consider the field $\mathbb{Q}_{p}(p^{1/p^{\infty}})$, why the ring of integers of it is $\mathbb{Z}_{p}[p^{1/p^{\infty}}]$? And why the ring of integers of the field $\mathbb{F}_{p}((t))(t^{1/p^{\infty}})$ is $\mathbb{F}_{p}[t^{1/p^{\infty}}]$? Is there a way to calculate these? Note that in this case, ring of integers is the same as the valuation ring.

Answer 1

Note that we have $\Bbb Q_p(p^{1/p^{\infty}})=\bigcup_{n=1}^\infty \Bbb Q_p(p^{1/p^n})$ and $\Bbb Z_p[p^{1/p^\infty}]=\bigcup_{n=1}^\infty \Bbb Z_p[p^{1/p^n}]$. Using this, it is easy to see that it suffices to show that $\Bbb Z_p[p^{1/p^n}]$ is the ring of integers in $\Bbb Q_p(p^{1/p^n})$. It is a standard fact about local fields that if $L/K$ is a totally ramified extension of local fields and $\pi \in L$ is a uniformiser, then $\mathcal O_L=\mathcal O_K[\pi]$. Clearly $\Bbb Q_p(p^{1/p^n}) / \Bbb Q_p$ is totally ramified with uniformiser $p^{1/p^n}$, so the result follows.

For $\Bbb F_p((t))(t^{1/p^\infty})$, the same argument works.