Why $\mathbb{Z}_{p}[p^{1/p^{\infty}}]/p$ $\cong$ $\mathbb{F}_{p}[t^{1/p^{\infty}}]/t$? I'm reading Peter Scholze's Perfectoid Space, and I'm confused with the isomorphism $\mathbb{Z}_{p}[p^{1/p^{\infty}}]/p$ $\cong$ $\mathbb{F}_{p}[t^{1/p^{\infty}}]/t$. What is the meaning of $\mathbb{Z}_{p}[p^{1/p^{\infty}}]/p$ and $\mathbb{F}_{p}[t^{1/p^{\infty}}]/t$? What's more, how to define the completions of $\mathbb{Q}_{p}(p^{1/p^{\infty}})$ and $\mathbb{F}_{p}((t))(t^{1/p^{\infty}})$?
 A: I assume you mean the completions of these rings.
Have you thougt about it from a functor of points perspective? What is a continuous ring map from $\mathbb{Z}_p[p^{1/p^{\infty}}]/(p)$ to a non-zero complete ring $R$? Well a continuous ring map $\mathbb{Z}_p[p^{\frac{1}{p^\infty}}]\to R$ is easily seen to be equivalent to a sequence of elements $x_0,\ldots,x_n,\ldots\in R$ with $x_0=p$, and $x_{i+1}^p=x_i$. The condition that this map factorize through $\mathbb{Z}_p[p^{1/p^{\infty}}]/(p)$ is then just the extra condition that $p=0$. What is a continuous ring map $\mathbb{F}_p[t^{\frac{1}{p^\infty}}]/(t)\to R$? Well, a continuous ring map $\mathbb{F}_p[t^{\frac{1}{p^\infty}}]\to R$ can only exist if $p=0$ in $R$. In this case such a ring map is equivalent to a sequence $y_0,\ldots,y_n,\ldots\in R$ such that $t\mapsto y_0$ and $y_{i+1}^p=y_i$. To factorize through $\mathbb{F}_p[t^{\frac{1}{p^\infty}}]/(t)$ is then equivalent to the claim that $y_0=0$.
So, in other words we have formed functorial bijections
$$\mathrm{Hom}_\text{cont}(\mathbb{Z}_p[p^{\frac{1}{p^\infty}}]/(p),R)=\left\{(x_i)\in\prod_{i=0}^\infty R: x_0=p,\, x_{i+1}^p=x_i\text{ for all }i,\text{ and } p=0\right\}$$
and
$$\text{Hom}_\text{cont}(\mathbb{F}_p[t^{\frac{1}{p^\infty}}]/(t),R)=\left\{(y_i)\in\prod_{i=0}^\infty R: p=0,\, y_0=0,\, \text{ and } y_{i+1}^p=y_i\text{ for all }i\right\}$$
So they have the same moduli descriptions. So, you know from Yoneda that they are abstractly isomorphic as complete topological rings. Can you see how to use this to construct a topological ring isomorphism (i.e. trace through Yoneda)?
EDIT: Presumably you know what $\mathbb{Z}_p[p^{\frac{1}{p^n}}]$ is. It's literally the smallest subring of $\overline{\mathbb{Q}}_p$ containing $\mathbb{Z}_p$ and $p^{\frac{1}{p^n}}$. It's a complete topological ring with the $p$-adic topology. One can then define $\mathbb{Z}_p[p^{\frac{1}{p^\infty}}]$ to be the directed limit of the $\mathbb{Z}_p[p^{\frac{1}{p^n}}]$ as $n$ varies, or what amounts to the same thing, the 'union' of $\mathbb{Z}_p[p^{\frac{1}{p^n}}]$, as $n$ varies, as subrings of $\overline{\mathbb{Q}}_p$. This is not a complete topological ring, and so one usually defines $\mathbb{Z}_p[\![p^{\frac{1}{p^\infty}}]\!]$ to be its $p$-adic completion. Note that naturally $\mathbb{Z}_p[p^{\frac{1}{p^\infty}}]/(p)\cong \mathbb{Z}_p[\![p^{\frac{1}{p^\infty}}]\!]/(p)$ and so I was pretending that you meant the latter in the above--it doesn't really matter. The definition of $\mathbb{F}_p[t^{\frac{1}{p^\infty}}]$ and $\mathbb{F}_p[\![t^{\frac{1}{p^\infty}}]\!]$ is analogous (with $p$ replaced by $t$).
A: While I was writing this Alex Youcis produced an answer above describing the completions and it's quite nice, so I'll omit that part. Instead I'll focus on explaining to you what the rings $\mathbb{Z}_p[p^{1/p^{\infty}}]$ and $\mathbb{F}_p[t^{1/p^{\infty}}]$ are by giving them explicit constructions and descriptions.
Begin by noting that if $A$ is a commutative ring with identity, the ring
$$
A[x^{p^{-\infty}}] = A[x^{1/p^{\infty}}] = \lim_{\substack{\longrightarrow \\ n \in \mathbb{N}}}A[x^{p^{-n}}]
$$
is determined by formally adding all $p^n$-th roots of the indeterminate $x$ for all $n \in \mathbb{N}$. In this way you get that
$$
\mathbb{F}_p[t^{p^{-\infty}}] = \mathbb{F}_p[t^{1/p^{\infty}}] = \lim_{\substack{\longrightarrow \\ n \in \mathbb{N}}}\mathbb{F}_p[t^{p^{-n}}]
$$
is the ring induced by taking $\mathbb{F}_p[t]$ and adding $p^n$-th roots of $t$ for all $n \in \mathbb{N}$. The arithmetic is induced by saying that
$$
t^{p^{-k}}t^{p^{-n}} = t^{1/p^k}t^{1/p^n} = t^{1/p^k + 1/p^n} = t^{(p^k + p^n)/p^{n+k}}
$$
as in the usual cases of exponent laws.
In the cases you care about you get similarly that
$$
\mathbb{Z}_p[p^{p^{-\infty}}] = \lim_{\substack{\longrightarrow \\ n \in \mathbb{N}}}\mathbb{Z}_p[p^{p^{-n}}]
$$
is the extension ring of $\mathbb{Z}_p$ induced by adding a $p^n$-th root of $p$ to $\mathbb{Z}_p$ for each $n \in \mathbb{N}$. The multiplication is of course induced by the usual rules of exponential arithmetic, i.e., by taking
$$
p^{p^{-k}}p^{p^{-n}} = p^{1/p^k}p^{1/p^n} = p^{1/p^k + 1/p^n} = p^{(p^k + p^n)/p^{k+n}}.
$$
This rings are described classically (at least in principle or in theory) in Serre's book Local Fields on pages 37 -- 39 under the terms $p$-rings and strict $p$-rings.
