Recall that an integral domain $R$ with quotient field $K$ is an almost valuation domain if for every $0 \not= x \in K$, there is a positive integer $n$ (depending on $x$) such that $x^n \in R$ or $x^{−n} \in R$. Now for a prime number $p$ let $R := \mathbb{Z}p +XF[[X]]$ and $F := \overline{\mathbb{Z}p}$ be the algebraic closure of $\mathbb{Z}p$. Why is $R$ an almost valuation domain and for every positive integer $n$, there is $a \in F$ such that $a^n\not\in\mathbb{ Z}p$ and $a^{-n}\not\in\mathbb{ Z}p$?
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I assume that $\mathbb Zp$ means the field with $p$ elements, and denote it by $\mathbb F_p$.
Since $\lambda= (\lambda X)/X$, the fraction field of $R$ is equal to the fraction field of $F[[X]]$, which is the ring of Laurent power series $F((X))$.
Let $f=\sum_{n=m}^{\infty} a_n X^n \in F((X))$ with $a_m\neq 0$. After possibly replacing $f$ by its inverse we may assume that $m\ge 0$. Let $q$ be the number of elements in $\mathbb{F}_p(a_m)\subset F$. Then $f^{q-1}= X^{m(q-1)} (1 + X\sum_{n=0}^{\infty} b_n X^n)$, since $a_m^{q-1}=1$. Thus $f^{q-1}= X^{m(q-1)} u$, where $u$ is a unit in $R$.
For the last question you can just take $f=1+X$: clearly $f^n \not\in \mathbb{F}_p$ for any $n>0$, and if $f^{-n}\in \mathbb F_p$ then so is its inverse.
