# The Nonarchimedean Scottish Book

The original Scottish Book was a compilation of problems, primarily concerning functional analysis, assembled at the Scottish Cafe in Lvov during the 1930s and 1940s. This web site is an analogous compilation of problems concerning nonarchimedean functional analysis and related topics, including analytic geometry (especially adic spaces) and perfectoid rings, fields, and spaces.

An ideal entry in this list includes the name of the proposer, the date of the initial post, a detailed statement of the problem, and attributed comments (including references as appropriate; these can be listed at the bottom of the page).

For security reasons, editing this wiki requires logging into the server. I have created a communal login credential for editing this page; please contact me for the details. Alternatively, I am wiling to accept entries by email and post them manually. (If someone with more free time than me wants to reimplementing this using better technology, I would support this. One option is CoCalc; see this link for a preview.)

## Problem 1

Proposed by Kiran S. Kedlaya, 17 December 2015.

Statement: Let K be a nonarchimedean commutative Banach ring whose underlying ring is a field. Suppose in addition that K is uniform, i.e., its norm is equivalent to a power-multiplicative norm. Is K necessarily a nonarchimedean field, that is, is the topology on K defined by some multiplicative norm?

Proposer's comment (updated 28 Apr 16): This question acquires a negative answer if the uniform condition is omitted. See arXiv:1602.09004v1.

Comment (Kedlaya, 26 Dec 15, updated 28 Apr 16): Without loss of generality, one may assume K is a Banach algebra over a (nontrivially normed) nonarchimedean field F. The question acquires a positive answer if one adds the assumption that F is not discretely valued; from this, the positive answer follows for perfectoid rings. That is, any perfectoid ring which is a field is a perfectoid field. See arXiv:1602.09004v1.

## Problem 2

Proposed by Kiran S. Kedlaya, 17 December 2015.

Statement: Let (A,A^+) be a Huber pair which is Tate (i.e., A contains a topologically nilpotent unit). Suppose that Spa(A,A^+) is a perfectoid space (for some prime p). Is A necessarily a perfectoid algebra?

Comment (Kedlaya, 18 Dec 2015): This is [Sch2, Conjecture 2.16], but a counterexample is known [BV, Proposition 13]. However, if one further assumes that A is uniform (or even stably uniform), the problem is open in general; see [KL2, Remark 3.6.27]. If A is of characteristic p and sheafy, then the statement holds; see [BV, Corollary 10], [KL1, Proposition 3.1.16].

Comment (Kedlaya, 18 Feb 2016): If (A,A^+) is sheafy, Spa(A,A^+) is a perfectoid space, and the tilt of Spa(A,A^+) is an affinoid perfectoid space, then A is a perfectoid algebra. For A over a perfectoid field, this follows from [Sch1, Proposition 6.17]; the general case is handled by [KL2, Proposition 3.4.10].

Comment (KSK, 10 Sep 2020): If Spa(A,A^+) is an affinoid subdomain of a affinoid perfectoid space, then A is a perfectoid ring. Sketch of proof: say the ambient affinoid perfectoid space is X = Spa(B,B^+). For each rank-1 point of X not in Spa(A,A^+), it is easy to construct a rational subspace of X missing that point but containing Spa(A,A^+). The completed direct limit of these gives an affinoid perfectoid subdomain of X containing Spa(A,A^+) with exactly the same rank-1 points; the ring of global sections on this space must then equal A.

## Problem 3

Proposed by David Hansen, 17 December 2015.

Statement: Let A be a perfectoid Tate ring with an action of a finite group G. Is the fixed subring A^G perfectoid?

Proposer's comment: This is true if A is of characteristic p, by an easy argument. This is also true if A is a Q_p-algebra, by a more subtle argument of Kedlaya.

Comment (Kedlaya, 21 Feb 16): The case where A is over a perfectoid field is [Shen, Proposition 2.5]. The general case is [KL2, Theorem 3.3.24].

## Problem 4

Proposed by David Hansen, 18 December 2015.

Statement: Let A be a sheafy Tate ring, and suppose some Zariski-open and dense subset of some Spa(A,A^+) is a perfectoid space. Is A perfectoid?

Proposer's comment: This is a stronger version of Problem 2.

Comment (Kedlaya, 27 Apr 16): I believe a counterexample against this question can be made by taking A = K{T^{1/p^infty}}[T^{1/2}] (for p>2).

Comment (KSK, 2 Dec 16): The previous example is also an example of a stably uniform Tate ring which is not perfectoid, but for which every point of Spa(A,A^+) has perfectoid residue field.

## Problem 5

Proposed by David Hansen, 18 December 2015.

Statement: Let A be a stably uniform Tate ring over Q_p, and let R^+ \subset R be a perfectoid Tate ring in characteristic p. Consider the Tate ring (W(R^+) \widehat{\otimes} A^\circ)[1/p], where the completion is taken for the p-adic topology. Is this ring stably uniform?

Proposer's comment: The answer is yes when A is finite etale over some Q_p<X_1,...,X_n>.

Comment (KSK, 7 Jul 20): see Problem 32 for a related question.

## Problem 6

Proposed by David Hansen, 18 December 2015.

Statement: Call a Tate ring A "sousperfectoid" if there exists a perfectoid Tate ring B and a continuous A-algebra map A-->B which admits a continuous A-Banach module splitting. This class includes perfectoid Tate rings, and any sousperfectoid ring is stably uniform and hence sheafy. If R is sousperfectoid, then any rational localization of R is sousperfectoid, and so are R<X>, R<X^1/p^infty> and R' for any finite etale R'/R.

Is there an example of a stably uniform Tate ring which is not sousperfectoid?

Comment (Kedlaya, 18 Dec 15): For A to be sousperfectoid, a necessary condition is that A be seminormal in the sense of Greco-Traverso and Swan: for any y, z in A for which y^3 = z^2, there is a unique x in A with x^2 = y, x^3 = z. (This will be shown in [KL2].)

Comment (Kedlaya, 18 Dec 15): Let X = Spa(A,A^+) be a classical affinoid algebra over a p-adic field. Let f: Y -> X be a resolution of singularities. If A is seminormal and R^1 f_* O_Y = 0, then I believe I can prove that A is sousperfectoid. For example, this holds if X has rational singularities because then all of the R^i f_* O_Y vanish. (Updated 27 Dec 16: this now works assuming only that A is seminormal.)

## Problem 7

Proposed by Kiran S. Kedlaya, 18 December 2015.

Statement: Let (A,A^+) be a sheafy uniform (Tate) Huber pair. Is (A,A^+) necessarily stably uniform?

Proposer's comment: This question is taken from [KL1, Remark 2.8.11].

## Problem 8

Proposed by Kiran S. Kedlaya, 18 December 2015.

Statement: Let (A,A^+) be a Huber pair. Is the property of (A,A^+) being sheafy, or stably uniform, independent of the choice of A^+?

Comment (Hansen, 20 December 2015): Stable uniformity is independent of the choice of A^+. Let A^+ \subseteq A'^+ be two choices of rings of integral elements, so we have an associated inclusion X'=Spa(A,A'^+) \subseteq X=Spa(A,A^+) of affinoid pre-adic spaces. Then the map {rational subsets of X} --> {rational subsets of X'} given by U --> U' = U \cap X' is surjective, and O_X(U)=O_X'(U') for any U.

Comment (Gabber, 13 Sep 2016): Sheafiness is independent of the choice of A^+. By Lemma 2.6 of Huber, "A generalization of formal schemes...", every open covering can be refined to a standard rational covering, and the collection of such coverings does not depend on A^+.

## Problem 9

Proposed by Kiran S. Kedlaya, 18 December 2015.

Statement: Let (A,A^+) be a stably uniform Huber pair. Is it true that for any finite etale morphism (A,A^+) -> (B,B^+), the pair (B,B^+) is again stably uniform?

Proposer's comment: This is true if A is sousperfectoid (see problem 6), as then B is as well.

## Problem 10

Proposed by Kiran S. Kedlaya, 18 December 2015.

Statement: Let (A,A^+) be a Huber pair such that Spa(A,A^+) is analytic (i.e., none of its points carries the trivial valuation). Does it follow that A is Tate?

Proposer's comment: It can be shown that Spa(A,A^+) is analytic if and only if the topologically nilpotent elements of A generate the unit ideal; see [KL1, Remark 2.3.9]. The problem is to then deduce (or refute) that there exists a single topologically nilpotent element generating the unit ideal.

Comment (KSK, 19 Dec 16): A negative answer is given by taking the quotient of the weighted Tate algebra Q<a/t,b/t,x/t^{-1},y/t^{-1}> (for some t>1) by the ideal (ax+by-1). (Essentially the same counterexample was suggested by Gabber.)

## Problem 11

Proposed by Brian Conrad, 18 December 2015.

Statement: Let k'/k be an extension of fields complete with respect to nontrivial nonarchimedean absolute values. Let i: U -> X be an open immersion of separated rigid analytic spaces over k. Is the base extension i': U' -> X' of i from k to k' also an open immersion?

Proposer's comment: This is Problem 20 from this list. The analogous question for Berkovich or Huber spaces trivially admits an affirmative answer; in the rigid context, one only sees directly that i is an injective local isomorphism, and that if it quasicompact and/or a Zariski open embedding, then i' is again an open immersion.

## Problem 12

Proposed by David Hansen, 20 December 2015.

Statement: Let A be a Tate ring with an action of a finite group G. Is the natural map Cont(A)/G-->Cont(A^G) always a homeomorphism? (Here we give Cont(A)/G the quotient topology.) Likewise for pairs (A,A^+) and the map Spa(A,A^+)/G-->Spa(A^G,A^{+G}).

Proposer's comment: This map is always surjective (use the integrality of A over A^G). It seems likely one can answer this affirmatively when Cont(A) has finite Krull dimension, by adapting the argument in Section 6.4 of [CHJ].

Comment (David Hansen, 9 February 2016, modified 25 October 2016): These maps are homeomorphisms in all generality. Details recorded in [H].

## Problem 13

Proposed by David Hansen, 20 December 2015.

Statement: Let A be a perfectoid Tate ring over Q_p. Does A contain a perfectoid field?

Comment (KSK, 25 Oct 16): Gabber exhibits a counterexample by taking Z_p[t], adjoining t^{1/p^n} and p(1+t)^{1/p^n} for all n, p-adically completing, then inverting p. By specializing at t=0 and t=p/(1-p), one sees that A contains no complete subfield larger than Q_p.

Here is a sketch of a second approach to constructing a counterexample. Let R be the completed perfect closure of F_p((t))<T>. In the relative Robba ring over R, one can find an element t satisfying phi(t) = p(1 + [T])t. This element gives rise to an untilt of R which cannot contain a perfectoid field: if it did, then there would exist a unit u in W(R) such that phi(u)*t/u belongs to the Witt ring of a subfield of R, but this can be seen to be impossible by direct computation mod p^2.

## Problem 14

Proposed by Kiran S. Kedlaya, 26 December 2015.

Statement: Let K be an infinite algebraic extension of Q_p. If K is arithmetically profinite, then the completion of K is a perfectoid field. Does the converse hold?

Proposer's comment: This question arises from discussions with Brian Lawrence.

Comment (Ofer Gabber, 13 Sep 2016): The completion of K is perfectoid iff K is a deeply ramified extension in the sense of Coates-Greenberg, which is equivalent (for separable algebraic extensions of a henselian DVR with perfect residue field of characteristic p>0) to being deeply ramified in the sense of Gabber-Ramero, "Almost Ring Theory". For the relationship with arithmetically profinite extensions, see I.B. Fesenko, "On deeply ramified extensions".

## Problem 15

Proposed by Kiran S. Kedlaya, 29 December 2015.

Statement: Let R be a perfect valuation ring of characteristic p which is not a field. Is the ring W(R) coherent?

Proposer's comment: Almost surely the answer is negative in all cases. This is known whenever the value group of R is not equal to the real numbers; the proof is similar to the proof that R[ [t] ] is not coherent (by theorems of Jondrup-Small, Vasconcelos, Anderson-Watkins; see [AW]).

## Problem 16

Proposed by David Hansen, 5 January 2016.

Statement: Let X be a diamond, with an action of a finite group G. Under what conditions is the quotient sheaf X/G a diamond? Under what conditions does a categorical quotient of X by G exist?

Proposer's comment: In the analogous setting of *separated* algebraic spaces, a categorical quotient of a space X by a finite group G always exists, with no conditions on the group action (Deligne).

Comment (David Hansen, 15 January 2016): Forgot to mention - the sheaf quotient X/G is a diamond if G acts freely on geometric points of X; this is proved in Proposition 4.3.2 of [W].

## Problem 17

Proposed by Kiran S. Kedlaya, 11 January 2016 (based on a suggestion of Peter Wear).

Statement: Let (A,A^+) -> (B,B^+) be a rational localization of adic Banach rings. Is it always the case that the contraction of any maximal ideal of B is a maximal ideal of A? If not, what about the case where A is strongly noetherian? or perfectoid?

Comment (Ofer Gabber, 13 Sep 2016): for a counterexample where A is strongly noetherian, take A = Z_p[ [t] ][1/p], A^+ = Z_p[ [t] ] and consider the localization |t| >= 1. To modify this into a counterexample where A is perfectoid, adjoin p-power roots of p and t and take p-adic completions.

## Problem 18

Proposed by Kiran S. Kedlaya, 22 January 2016 (based on a suggestion of Ruochuan Liu).

Statement: Let (A,A^+) be an adic affinoid algebra over Q_p. Let (B,B^+) be the total space of a Z_p-local system L over Spa(A,A^+). Under what conditions is B perfectoid?

Proposer's comment: In the case where A is a p-adic field (i.e., a complete discretely valued field with perfect residue field), Sen's theorem on ramification in p-adic Lie extensions implies that B is perfectoid if and only if L does not trivialize over the maximal unramified extension of A. However, in the general case a more complicated criterion is needed; for instance, one must rule out the possibility of a local system obtained by pullback from a lower-dimensional base.

## Problem 19

Proposed by Kiran S. Kedlaya, 3 February 2016.

Statement: Let (A,A^+) be a Huber pair which is sheafy but not necessarily Tate. Does it follow that the structure sheaf is acyclic? Does one have glueing for finite projective modules?

Proposer's comment: for both statements, the Tate case is resolved affirmatively in [KL1]. A related result is shown in [Ked1, Theorem 7.14, Theorem 7.15]: if one assumes sheafiness for the reified adic spectrum, then acyclicity and glueing hold without requiring a topologically nilpotent unit.

Comment (KSK, 25 Oct 16): Gabber's notes indicate affirmative answers to both questions. Details will be posted later.

## Problem 20

Proposed by KSK, 27 April 2016.

Statement: Let A be a perfectoid ring. Let B be the seminormalization (in the sense of Swan) of a finite A-algebra. Is the completion (or failing that, the uniform completion) of B a perfectoid algebra?

Proposer's comment: This question arises from discussions with Xu Shen.

## Problem 21

Proposed by KSK, 28 April 2016.

Statement: Let (A,A^+) be a Huber pair such that every point of X = Spa(A,A^+) has perfectoid residue field. Under what conditions can one conclude that X is a perfectoid space, or that A is a perfectoid ring (compare problem 2)?

Proposer's comment: An important special case is that of a completed direct limit of a finite etale tower of affinoid algebras over a perfectoid field.

## Problem 22

Proposed by KSK, 28 April 2016.

Statement: Let f: A -> B be a morphism of perfectoid rings corresponding to the morphism g: R -> S of perfect rings. Is it the case that f is finite if and only if g is finite?

Proposer's comment: The corresponding statement holds if "finite" is replaced by "finite etale" (by almost purity) or "surjective" (see [KL1, KL2]).

## Problem 23

Proposed by KSK, 28 April 2016.

Statement: Let K be a completed algebraic closure of F_p((t)). Let f: K -> K be a continuous endomorphism of K such that f(t) is not integral over the completion of the maximal tamely ramified extension of F_p((t)) in K. Does it follows that f is not surjective?

Proposer's comment: Some examples where f is not surjective are given in arXiv:1506.00742v6. The corresponding question for the completed perfect closure of F_p((t)) has an affirmative answer; see arXiv:1602.09051v1.

## Problem 24

Proposed by KSK, 16 Sept 2016.

Statement: Let (A, A^+) -> (B, B^+) be a rational localization of Huber-Tate pairs. Is the morphism A -> B of rings necessarily flat?

Proposer's comment: Huber showed this when A is strongly noetherian. If A is stably uniform, one can show that A -> B satisfies a weaker condition called "pseudoflat"; see [KL2].

Comment (KSK, 30 Dec 16): the pseudoflatness argument can be extended to the case where A is (Tate) and sheafy. This will appear both in [Ked2] and in a subsequent version of [KL2].

Comment (Alex Mathers via KSK, 23 Oct 19): In [Sta, Tag 0AL8] one finds an example of a ring R and an element f such that the map R[ [x] ] -> R_f[ [x] ] is not flat. In this example, the map R((x)) -> R_f((x)) is also not flat; for the x-adic topology on R((x)), this can be interpreted as the rational localization corresponding to the space |f| >= 1.

## Problem 25

Proposed by KSK, 22 Sept 2016.

Statement: Let X be a rigid analytic space over a perfectoid field K. Let f: X -> X be a finite flat morphism. Under what conditions is the inverse limit along f (in the "similar to" sense) a perfectoid space?

Proposer's comment: This is related to (but probably simpler than) Problem 18. Typical examples where the inverse limit is perfectoid include the p-th power map on a toric variety or the multiplication-by-p map on an abelian variety. Even the case where X is one-dimensional would be of some interest.

## Problem 26

Proposed by KSK, 2 Jan 2017.

Statement: Thanks to examples of Qing Liu, it is known that the analogue of Serre's cohomological criterion for affinity of noetherian schemes (namely the acyclicity of coherent sheaves) fails to pick out affinoid subspaces among all quasicompact rigid analytic spaces. Is there a similar failure for affinoid perfectoid spaces among perfectoid spaces?

## Problem 27

Proposed by KSK, 11 Jun 2017.

Statement: Let (A,A^+) be a Huber pair with A Tate. Is H^1(Spa(A,A^+), O^+) always annihilated by some topologically nilpotent unit of A?

Proposer's comment: I expect this to be false, but it does hold for affinoid algebras over a mixed-characteristic nonarchimedean field. This can be deduced by equating this statement with the completeness of the pro-etale H^1 of the completed structure sheaf, for which see [KL2, Section 8].

Comment (Gabber, 11 Sep 17): in the statement and the previous comment, it should be assumed that the reduced quotient of A is seminormal. For a reduced but non-seminormal affinoid algebra over a nonarchimedean field, one can compare with the seminormalization to see that the torsion in H^1(Spa(A,A^+), O^+) is unbounded.

## Problem 28

Proposed by KSK (based on a question of Peter Scholze), 11 Oct 2017.

Statement: Do there exist a Huber pair (A,A^+) with A Tate (or even perfectoid) and an element f of A such that multiplication by f is a strict inclusion on A (that is, f is not a zero-divisor and fA is a closed ideal), but f restricts to zero on some rational subspace of Spa(A,A^+)?

Proposer's comment: this is motivated by a related construction of Scholze. Using the Hodge-Tate period map, one can exhibit a perfectoid ring A and an element f of such that multiplication by f is a strict inclusion on A, but not on every rational localization of A. This will appear in the final version of [Ked2].

## Problem 29

Proposed by Bogdan Zavyalov (email to KSK), 6 Jun 2019.

Statement: Let A -> B be a flat continuous morphism of complete, strongly noetherian Tate-Huber rings. Does it follow that A<T> -> B<T> is flat?

Proposer's comment: This is easily shown to be true in the case of topologically finite type algebras over a complete rank-1 valuation field.

Comment (Gabber, 26 Mar 21): This can fail even when A and B have noetherian rings of definition. For example, let k be a field; take f,g in t k[ [t] ] with zero constant term such that t,f,g are algebraically independent over k; form the continuous k-algebra map A_0 = k[ [x,y,z] ] -> B_0 = k(u)[ [t] ] taking x,y,z to ut,uf,ug; equip A_0 with the x-adic topology and B_0 with the t-adic topology; and take A = A_0[1/x], B = B_0[1/t]. The map A<T> -> B<T> is not flat because the height-1 prime (1-Tu) contracts to the height-2 prime (y-f(Tx)/T, z-g(Tx)/T).

## Problem 30

Proposed by Bogdan Zavyalov (email to KSK), 6 Jun 2019.

Statement: Let X = Spa(A, A^+) be the adic spectrum of a complete, strongly noetherian Tate-Huber pair. For any point x in X, is the stalk O_{X,x} of the structure sheaf at x noetherian?

Proposer's comment: By an argument of Temkin, this is known in the case of a topologically finitely generated Tate-Huber pair over a complete rank-1 valuation field.

Comment (Ofer Gabber, 27 Mar 2021): [FK, Theorem II.8.3.6] covers the cases when A has a ring of definition which is either noetherian, or topologically of finite type over a microbial complete valuation ring. (More precisely, the given reference requires taking A^+ to be the integral closure of the ring of definition; but using [FK, Theorem II.3.2.17(3)] and the strongly noetherian condition, we see that the specialization maps are faithfully flat. So we may reduce to rank-1 points and thus eliminate the dependence on the choice of A^+.)

## Problem 31

Proposed by KSK (via David Hansen), 29 Nov 2019.

Statement: Let A be an adic ring with finitely generated ideal of definition I. Suppose that for all n >= 0, the complement of the zero locus of I in Spec A<T_1,...,T_n> is noetherian (Hansen says that A is "strongly noetherian outside I"). Does it follow that A is sheafy?

Proposer's comment: See the above link for further discussion.

Comment (Bogdan Zavyalov, 10 Sep 2020, updated 27 Mar 2021): The correct statement is that if A is a Huber ring admitting a ring of definition A_0 with an ideal of definition I such that A_0<T_1,...,T_n> is noetherian outside I for all n, then A is sheafy. See [Z].

## Problem 32

Proposed by KSK, 7 Jul 2020.

Statement: Let A be a perfectoid ring over a perfectoid field K. Let B be an affinoid algebra over K. Is the completed tensor product of A with B over K sheafy?

Proposer's comment: if B is seminormal, it is weakly sousperfectoid in the sense of [HK], and this property is preserved by the completed tensor product.

## Problem 33

Proposed by KSK, 7 Jul 2020.

Statement: Let (R,R^+) be a perfectoid Huber pair in characteristic p. Equip W(R^+) with the (p, [x])-topology for some topologically nilpotent unit x in R. Is W(R^+) sheafy as a Huber ring?

Proposer's comment: by [Ked2, Lemma 3.1.3], this is true away from the non-analytic locus (i.e., the zero locus of (p, [x])).

## Problem 34

Proposed by KSK, 10 Sep 2020.

Statement: Let Y -> X be a morphism of adic spaces such that X is a perfectoid space, Y is uniform, and the fiber over each point of X is a perfectoid space. Is Y a perfectoid space?

Proposer's comment: if X is affinoid perfectoid, Y is affinoid and uniform, and each fiber is an affinoid perfectoid space, then Y is an affinoid perfectoid space. Hence in any given instance, it would suffice to solve Problem 2 for subspaces of a single fiber.

## Problem 35

Proposed by KSK, 29 Oct 2020.

Statement: Let (A, A^+) be a Tate Huber pair. Does there always exist a stably uniform Huber pair (B, B^+) such that (A, A^+) -> (B, B^+) induces a homeomorphism of adic spectra and isomorphisms of complete residue fields?

Proposer's comment: One can construct a specific Huber pair out of (A, A^+) which must be isomorphic to (B, B^+) if the latter exists. This might be useful for constructing a counterexample.

## Problem 36

Proposed by KSK, 5 Feb 2021.

Statement: Let K be a completed algebraic closure of F_p((t)). Let f: K -> K be an endomorphism. Under what conditions is f guaranteed to be an isomorphism? In particular, is it necessary and/or sufficient that f(t) is algebraic over the completion of the maximal tame extension of F_p((t))?

Proposer's comment: It is known that there exist examples where f is not surjective, based on a method of Matignon-Reversat; see [KT]. This is related to the question of when an untilt of the tilt of C_p (a completed algebraic closure of Q_p) is isomorphic to C_p; the examples of [KT] show that some of these untilts are strictly larger. A related result is that if K is a completed perfect closure of F_p((t)), an endomorphism of K is an automorphism iff it is a composition of a power of Frobenius with an automorphism of F_p((t)); see [Ked3].

## Problem 37

Proposed by KSK, 24 Feb 2021.

Statement: Let A be a complete, Tate, strongly noetherian Huber ring. Does A being excellent (as an underlying ring) imply that A<T> is excellent?

Proposer's comment: I would guess the answer is "no". Gabber suggests that a negative answer may occur even if we also assume that A has a noetherian ring of definition, as this ring itself may not be excellent (or even quasi-excellent).

## Problem 38

Proposed by Ehsan Shahoseini, 24 Feb 2021.

Statement: Let K be a perfectoid field of characteristic 0. Let E be a perfectoid subfield of the tilt K^b of K. Is E necessarily the tilt of a perfectoid subfield of K?

Comment (Ofer Gabber, 27 Mar 2021): To obtain a negative answer with [K^b:E] finite, let K be the completion of Q_p(mu_{p^infty}). The tilt of K is the completed perfect closure of F_p((t)). For any integer m>1 coprime to p, the completed perfect closure of F_p((t^m)) is a perfectoid subfield E of K of index m. If E is the tilt of a subfield F of K, then by Ax-Sen-Tate, F is itself the completion of a subfield of Q_p(mu_{p^infty}); we can rule this out by requiring m to be coprime to p(p-1).

Comment (KSK and Ehsan Shahoseini, 8 Apr 2021): One can also obtain negative answers using the existence of nonsurjective endomorphisms of C_p^b (see Problem 36). If K = C_p, then we can find a subfield E of K^b which is again the completed algebraic closure of a power series field, but which is strictly smaller than K^b. Such a field cannot be the tilt of a perfectoid subfield of K: such a field would itself have to be algebraically closed (and complete), but there are no complete algebraically closed fields between Q_p and C_p.