Polynomial ring with $p$-th root of unity I am trying to understand Example 1.3 in the paper of Kedlaya New methods for (phi, Gamma)-modules, available in his site.
Firstly, Kedlaya defines two objects
$$ A:= \Bbb Z [x^{p^{-\infty}}] :=\bigcup_{n \ge 1} \Bbb Z[x^{p^{-n}}], \quad \hat{A}=\text{ $p$-adic completion of } A$$
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Q1: How does one actually define $\Bbb Z[x^{1/k}]?$
My attempt: It should be the universal object mapping into a ring which has an element satisfyig
$$ x^k=x$$
A model of such an object should be the the polynomial ring with one variable
$$\Bbb Z[x]/(x^k-x) $$

Q2: Kedlaya claims $\hat{A}/(p) \cong \Bbb F_p [x^{p^{-\infty}}] $. How is this so  ?
My problem: I do not understand the description of $\hat{A}$ - or does this folllow formally.
 A: $\mathbb{Z}[x^{1/k}]$ is simply (up to isomorphism) the ring of polynomials in the variable $x^{1/k}$. So $A$ is the ring of the finite sums $\sum_{k \geq 0}{a_kx^{k/p^n}}$, for all $n \geq 0$.
The product, for instance, is defined by $\sum_{k \geq 0}{a_kx^{k/p^n}} \cdot \sum_{l \geq 0}{b_lx^{l/p^m}}=\sum_{s \geq 0}{\left(\sum_{p^mk+p^nl=s}{a_kb_l}\right)x^{s/p^{n+m}}}$.
Thus, $\hat{A}$ is the ring of all the series $\sum_{\alpha}{a_{\alpha}x^{\alpha}}$, where $\alpha$ is a nonnegative rational with only a power of $p$ as its denominator, $a_{\alpha} \in \mathbb{Z}_p$, and $\{\alpha,\, |a_{\alpha}|_p > \epsilon\}$ finite for each $\epsilon > 0$.
When you quotient out by $(p)$, by the above description, you only get a finite sum $\sum_{\alpha}{a_{\alpha}x^{\alpha}}$, where $a_{\alpha} \in \mathbb{F}_p$ and $\alpha \in p^{-\mathbb{N}}\mathbb{N}$. These sums are exactly the elements of $\mathbb{F}_p[x^{p^{-\infty}}]$.
A: Each ${\mathbb Z}[x^{p^{-k}}]$ is an ordinary polynomial ring in one variable $x_k := x^{p^{-k}}$, and the union in question is formally the direct limit over the maps $\varphi_k: {\mathbb Z}[x_k]\to {\mathbb Z}[x_{k+1}]$ given by $x_{k}\mapsto x_{k+1}^p$. Alternatively, you might view it as the polynomial ring over the monoid ${\mathbb N}[\frac{1}{p}]\subset{\mathbb Q}$.
