# 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 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.

• math.uni-bonn.de/people/scholze/Geometrization Not an expert, but these lectures should be related. – Achim Krause Dec 7 '20 at 20:10
• My understanding is that the FF curve parametrizes untilts. I remember finding Morrow's Bourbaki seminar on this very useful. – Geordie Williamson Dec 8 '20 at 1:20
• @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. – 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).

• 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. – Wojowu Apr 18 at 20:47
• Sorry for that nonsense, I shouldn't try to do math on a sunday evening ;-). – Peter Scholze Apr 18 at 21:19
• 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, $G^{tame}_{\mathbb{Q}_p}=Gal(\overline{\mathbb{Q}_p}/K)=Gal(\overline{\widehat{K}^{\flat}}/\widehat{K}^{\flat})$. – SashaP Apr 19 at 1:40
• 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. – SashaP Apr 19 at 1:40