How many untilts? I read the following passage in Endomorphisms of power series fields and residue fields of Fargues-Fontaine curves by Kedlaya-Temkin:
"One can construct many algebraic extensions of $\mathbb{Q}_p$ whose completions $K$ tilt to the completed perfect closure of a power series field over $\mathbb{F}_p$."
They then give the two classical examples of the $p$-adic completions of $\mathbb{Q}_p(p^{1/p^{\infty}})$ and $\mathbb{Q}_p(\zeta_{p^{\infty}})$.
This made me wonder:
Questions: (a) What other algebraic extensions of $\mathbb{Q}_p$ tilt to $\mathbb{F}_p((t^{1/p^{\infty}}))$?
(b) How many are there (up to isomorphism)?
To be precise: Question (b) asks for the cardinality of the set of untilts of $\mathbb{F}_p((t^{1/p^{\infty}}))$ which are completions of algebraic extensions of $\mathbb{Q}_p$.
I suspect that this question already has an answer in the works of Fargues–Fontaine but my scientific French is too poor to understand if this is the case.
 A: This specific question is probably not addressed in the literature; let's try to figure it out!
Let $K$ be an algebraic extension of $\mathbb Q_p$ such that the tilt of $\widehat{K}$ is isomorphic to $\mathbb F_p((t^{1/p^\infty}))$. We can observe the following:

*

*Tilting preserves residue fields, so necessarily $K$ has residue field $\mathbb F_p$, i.e. is totally ramified over $\mathbb Q_p$.


*Tilting preserves value groups, so the value group of $K$ is isomorphic to $\mathbb Z[\tfrac 1p]$. In particular, there can only be a finite amount of ramification of degree prime to $p$.
In particular, $K$ is pro-$p$ and totally ramified of infinite degree over a finite totally tamely ramified extension $K_0$ of $\mathbb Q_p$. Conversely, all such $K$ have perfectoid completion $\widehat{K}$.
As SashaP comments below, these conditions are however not yet sufficient for $\widehat{K}^\flat$ to be isomorphic to $\mathbb F_p((t^{1/p^\infty}))$. For example, one can find such an extension $K/\mathbb Q_p$ for which $\mathrm{Gal}_K$ maps isomorphically to the tame quotient of $\mathrm{Gal}_{\mathbb Q_p}$; but the tame quotient of $\mathrm{Gal}_{\mathbb Q_p}$ cannot be isomorphic to $\mathrm{Gal}_{\mathbb F_p((t))}$. It seems to be an interesting question to isolate those $K$ which have tilt isomorphic to $\mathbb F_p((t^{1/p^\infty}))$!
One definitely gets examples by taking any tower $K_0=\mathbb Q_p$, $K_1$, $K_2$, ... such that each $K_{i+1}/K_i$ is a degree $p$ extension given by extracting a $p$-th root of a uniformizer. At each step, this gives finitely many distinct choices, so there are at least $2^{\aleph_0}$ such extensions. This is also an evident upper bound, so this gives an answer to b).
