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 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.

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    $\begingroup$ math.uni-bonn.de/people/scholze/Geometrization Not an expert, but these lectures should be related. $\endgroup$ – Achim Krause Dec 7 '20 at 20:10
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    $\begingroup$ My understanding is that the FF curve parametrizes untilts. I remember finding Morrow's Bourbaki seminar on this very useful. $\endgroup$ – Geordie Williamson Dec 8 '20 at 1:20
  • $\begingroup$ @GeordieWilliamson Thanks, I was aware of the FF curve and also the seminar by Morrow but I wasn't able to find something which addresses this question. $\endgroup$ – Kostas Kartas Dec 8 '20 at 11:11

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:

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

  2. 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 totally ramified finite 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 a 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).

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    $\begingroup$ It's not true that $\mathbb Q_p$ has only countably many algebraic extensions. Indeed, it even has uncountably many unramified extensions, despite having one only one in each degree. Indeed, for any prime $\ell$ let $K_\ell$ be the union of unramified extensions of degree a power of $\ell$. Then for any subset of primes we can consider a suitable compositum and get pairwise distinct extensions this way. By throwing in some ramification we can also get uncountably many algebraic extensions with perfectoid completions. $\endgroup$ – Wojowu Apr 18 at 20:47
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    $\begingroup$ Sorry for that nonsense, I shouldn't try to do math on a sunday evening ;-). $\endgroup$ – Peter Scholze Apr 18 at 21:19
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    $\begingroup$ I'm most likely making a mistake here, but would the positive answer to the question be consistent with the existence of very large pro-p totally ramified extensions of $\mathbb{Q}_p$? Namely, by Janssen-Wingberg, the surjection from $G_{\mathbb{Q}_p}$ to the tame Galois group has a section, so the field $K$ of invariants in $\overline{\mathbb{Q}_p}$ under the image of such section is a (non-Galois) union of totally ramified p-extensions. However, by tilting correspondence, $G^{tame}_{\mathbb{Q}_p}=Gal(\overline{\mathbb{Q}_p}/K)=Gal(\overline{\widehat{K}^{\flat}}/\widehat{K}^{\flat})$. $\endgroup$ – SashaP Apr 19 at 1:40
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    $\begingroup$ So if $\widehat{K}^{\flat}$ was isomorphic to $\mathbb{F}_p((t^{1/p^{\infty}}))$ that would imply that the absolute Galois group of the latter field is two-step solvable which is not the case. $\endgroup$ – SashaP Apr 19 at 1:40

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