The perfectoid open unit disk

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1 The perfectoid open unit disk Peter Schneider Neckarbischofsheim, May 2015 These are expanded notes of a talk given at the workshop The Galois Group of Q p as Geometric Fundamental Group. Its purpose was to explain 3.2 and some initial steps in 4 of the paper [Wei]. 1 Perfectoid generic fibers Let K be a perfectoid field with ring of integers o K of residue characteristic p. We fix an element 0 ϖ o K such that p ϖ < 1. We consider an o K -algebra S which is complete and Hausdorff with respect to the I-adic topology for some ideal I S. For simplicity we assume that ϖ I (cf. Remark 1.6). We equip the factor ring S/ϖS with the I/ϖS-adic topology. The ojective limit of o K /ϖo K -algebras S := lim (.) p S/ϖS = lim (... (.) p S/ϖS (.)p... (.) p S/ϖS) equipped with the ojective limit topology is a topological o K -algebra, which is perfect. We denote by Φ S the p-frobenius automorphism of S. Remark 1.1. The rings S/ϖS and S are complete. Proof. The vertical maps in the commutative diagrams S/I n+1 ϖ S/I n+1 (S/ϖS)/(I/ϖS) n+1 0 S/I n ϖ S/I n (S/ϖS)/(I/ϖS) n 0 are surjective. Hence the lower horizontal map in the commutative diagram S S/ϖS = lim n S/In lim n (S/ϖS)/(I/ϖS) n is surjective. Since the left vertical map is bijective by assumption the right vertical map at least is surjective. This amounts to the completeness of S/ϖS. The completeness of S then is a formal consequence. 1

2 A fundamental system of open neighbourhoods of zero in S is given by the ideals One checks that J n,i := {(..., α 1, α 0 ) S : α i (I/ϖS) n } for n 1 and i 0. Φ i (J S n,0 ) = J n,i, Φ i (J S 1,0 ) J pi 1,0 J p i,0, and hence Φ i+j (J 1,0 ) J p j,i. This shows the following. Remark 1.2. The descending sequence of ideals Φ n S (J 1,0 ), for n 0, forms a fundamental systems of open neighbourhoods of zero in S. In particular, Φ S is a topological automorphism. Let α = (..., α i,..., α 0 ) S be any element. We choose elements a i, a i S such that a i mod ϖs = α i = a i mod ϖs. The usual standard computation shows that the sequences (a pi i ) i and (a p i i ) i are ϖ-adic Cauchy sequences and that their difference (a pi i a p i i ) i is a ϖ-adic zero sequence. Since ϖ I the same then holds true I-adically. But S is I-adically complete and Hausdorff. It follows that S S α α := lim a pi i i is a well defined multiplicative map such that α mod ϖs = α 0. In fact, we also could have used the following observation (which I learned from L. Ramero). Remark 1.3. The ring S is ϖ-adically complete and Hausdorff. Proof. The Hausdorff operty is clear since n ϖn S n In = 0. Let (a i ) i in S be a Cauchy sequence for the ϖ-adic and hence for the I-adic topology. It has an I-adic limit a. We have to show that a also is the ϖ-adic limit of the sequence. For this we may replace the original sequence by any convenient subsequence. Hence we may assume that a i+1 a i ϖ i S for any i 1. Let a i+1 a i = ϖ i z i with z i S. For j 1 we then have a i+j a i = (a i+j a i+j 1 ) (a i+1 a i ) = ϖ i (ϖ j 1 z i+j ϖz i+1 + z i ). We put y i,j := ϖ j 1 z i+j ϖz i+1 + z i and obtain y i,j y i,m = ϖ j 1 z i+j ϖ m z i+m ϖ m S for j > m. It follows that the limit y i := lim j y i,j with respect to the I-adic topology exists. We finally compute a a i = lim j (a i+j a i ) = ϖ i lim j y i,j = ϖ i y i ϖ i S. This shows that the sequence (a i ) i converges to a also in the ϖ-adic topology. Lemma 1.4. The map : S S is continuous. Proof. We show that (α + J n+1,n ) α + I n+1. If β = (..., β 0 ) α + J n+1,n then β n α n = c mod ϖs for some c I n+1. For i n we choose a i, b i S such that a i mod ϖs = α i, b i mod ϖs = β i, and b n a n I n+1. We have a pi i a pn n mod ϖ n+1 S and b pi i b pn n mod ϖ n+1 S. It follows that b pi i a pi i I n+1 for any i n and hence that β α I n+1. 2 S

3 Lemma 1.5. The map lim S (.) p S (..., a i,..., a 0 ) (..., a i mod ϖs,..., a 0 mod ϖs) is a multiplicative homeomorphism with inverse α (..., (α 1/pi ),..., α ). In particular, S is Hausdorff. 1 Proof. Of course, the asserted ojection map is multiplicative and continuous. Let α = (..., α i,..., α 0 ) S and choose a i S such that a i mod ϖs = α i. Then α 1/pi = (..., α i ) and hence (α 1/pi ) = lim j a pj i+j, and we compute ((α 1/pi+1 ) ) p = lim j a pj+1 i+1+j = lim j apj i+j = (α1/pi ). Since (α 1/pi ) mod ϖs = α i, we see that the asserted candidate for the inverse is, at least, a left inverse of the ojection map. It is continuous by Remark 1.2 and Lemma 1.4. It remains to check that the ojection map is injective. Let (..., a i,..., a 0 ) and (..., b i,..., b 0 ) elements in lim S such that a i b i mod ϖs for any i 0. We deduce that a i = a pj b i mod ϖ j+1 S for any j 0. It follows that a i b i j Ij = 0. i+j bpj i+j = Remark 1.6. If the topology of the topological o K -algebra S is the I -adic one for some ideal I S then it also is the I-adic one for I := ϖs + I. Proof. Being a topological o K -algebra we must have ϖ m I for some m 1. It follows that I m I. Lemma 1.5 and Remark 1.6 show that the topological o K -algebra S does not depend on the choice of the element ϖ. It will be technically convenient to assume in the following that p ϖ 2 < ϖ < 1 and ϖ I. Then ps ϖ 2 S I 2. Following [GR] 13.1 we now will make additional assumptions on S. The first one is (1) (S/pS) p = S/pS. Remark 1.7. Suppose that K has characteristic zero; we then have (S/pS) p = S/pS if and only if (S/ϖS) p = S/ϖS. Proof. The other direction being trivial we assume that (S/ϖS) p = S/ϖS. Because of the density of the value group we find an element ϖ 1 K such that ϖ 1/p ϖ 1 < 1. It follows that ϖs ϖ p 1 S and hence that (S/ϖp 1 S)p = S/ϖ p 1S. Now let a S be any element. Inductively we find elements (b n ) n 0 and (a n ) n 1 in S such that 1 It seems that S/ϖS need not to be Hausdorff. a = b p 0 + ϖp 1 a 1 a 1 = b p 1 + ϖp 1 a 2. a n = b p n + ϖ p 1 a n+1. 3

4 It follows that there exist elements (c n ) n 0 in S such that a c p n + ϖ p(n+1) 1 a n+1 mod ps for any n 0. But ϖ p(n+1) 1 p and hence ϖ p(n+1) 1 S ps for sufficiently large n (this is where the characteristic zero assumption is needed). This shows that (S/pS) p = S/pS. Our further assumptions are: (2) (3) I is finitely generated. ker(s/i (.)p S/I) is a finitely generated ideal in S/I. We denote by I I 1 S the finitely generated ideal such that I 1 /I = ker(s/i (.)p S/I). Then, of course, the map (4) S/I 1 (.) p = S/I is a well defined isomorphism. Lemma 1.8. Assuming (1), (2), and (3) we have: i. There are generators I 1 = (s 1,..., s r ) such that I = (s p 1,..., sp r). ii. The I 1 -adic topology on S coincides with the I-adic one. Proof. We begin by picking generators I = (a 1,..., a m ). As a consequence of assumption (1) we have (S/I 2 ) p = S/I 2. Hence we find elements s j S such that a j s p j I2. It follows that s p 1,..., sp m generate the S-module I/I 2. But, since S is I-adically complete and Hausdorff, the ideal I is contained in the Jacobson radical of S (for any a I and b S the series i 0 (ab)i converges to an inverse of 1 ab). Therefore the Nakayama lemma applies and shows that I = (s p 1,..., sp m). By construction we have {s 1,..., s m } I 1. Since I 1 is finitely generated we may enlarge this set to a finite set of generators s 1 = b 1,..., s m = b m, s m+1,..., s r of I 1 which satisfies the assertion i. Since J I I 1 the I 1 -adic topology coincides with the I-adic one. in S. We now consider the ideal I 1 := {(..., α 1, α 0 ) S : α 0 I 1 /ϖs} Proposition 1.9. Assuming (1), (2), and (3) we have: i. The ideal I 1 is finitely generated. ii. The topology of S is the I1 -adic one. 4

5 Proof. We fix generators I 1 = (s 1,..., s r ) as in Lemma 1.8.i. By the assumption (1) we find elements σ j = (..., σ j,1, σ j,0 ) S, for 1 j r, such that σ j,0 = s j mod ϖs. Obviously, the ideal J := (σ 1,..., σ r ) is contained in I 1. Hence Φ i S (J) Φ i S (I 1) = {(..., α 1, α 0 ) S : α i I 1 /ϖs} for i 0. Claim: The natural map J I1 /Φ S (I 1 ) is surjective. Let α = (..., α 0 ) I1. Since α 0 I 1 /ϖs there are γ j,0 S/ϖS such that α 0 = γ 1,0 σ 1, γ r,0 σ r,0. Again by (1) we have elements γ j S of the form γ = (..., γ j,0 ). Then α r γ j σ j = (..., β 1, 0). j=1 Using the bijectivity of (4) the vanishing β p 1 = 0 in S/ϖS implies that β 1 I 1 /ϖs. We conclude that r α γ j σ j + Φ S (I1) J + Φ S (I1), j=1 which establishes the claim. We see that all the natural maps Φ i (J)/Φ i+1 (J) Φ i (I S S S 1 )/Φi+1 (I S 1 ), for i 0, are surjective. Remark 1.2 and Lemma 1.5 together imply that i Φi (J) = S i Φi (I S 1 ) = 0. On the other hand it follows from Lemma 1.8.ii and Remark 1.2 that the topology of S is defined by the system of open zero neighbourhoods Φ i (I S 1 ). Furthermore, since J is finitely generated, say with r generators, we have (5) Φ pi r S (J) J pir Φ i S (J) Φ i S (I 1) for any i 0. First of all this shows that the J-adic topology on S is finer than the given topology. From Remark 1.1 and Lemma 1.5 we know that S is complete and Hausdorff. In this situation it follows, by a generalization of the argument in the oof of Remark 1.3 (cf. [GR] Lemma ), that S also is complete and Hausdorff for the J-adic topology. Secondly (5) shows that the J-adic topology on S coincides with the topology for which the Φ i (J) S form a fundamental system of open zero neighbourhoods. Hence S is complete and Hausdorff with respect to this latter filtration as well. At this point we have verified all assumptions of [B-CA] III 2.8 Cor. 2 for the inclusion J I1 viewed as a map of filtered abelian groups with respect to the filtrations (Φ i (J)) S i and (Φ i (I S 1 )) i, respectively. The above surjectivity therefore implies that J = I1. In particular, I 1 is finitely generated. Going back once more to (5) we also deduce that the topology of S is the I1 -adic one. We recall ([Sch] Remark 3.5) that in a perfectoid field we can choose the element ϖ, without changing its absolute value, in such a way that there is an element ϖ o K such that (((ϖ ) 1/pi ) ) pi = ϖ for any i 0. In the following we work with such a pair (ϖ, ϖ ). 5

6 The generic fiber Spf(S) ad of Spf(S) (only assuming (2)) as a (not necessarily honest) adic space over K is constructed as follows. Choose the finitely generated ideal I as well as generators I = (ϖ, s 1,..., s r ). For any n 1 let Ŝ n := ϖ-adic completion of S[Y 1,..., Y r ]/(s n 1 ϖy 1,..., s n r ϖy r ) (cf. Remark 1.3), Ŝ + n := integral closure of Ŝn in K ok Ŝ n. There are natural open immersions of adic spaces Spa(K ok Ŝ n, Ŝ+ n ) Spa(K ok Ŝ n+1, Ŝ+ n+1 ) over K. Gluing them along these immersions defines the adic space Spf(S) ad. One checks that the resulting space is independent of the choices I = (ϖ, s 1,..., s r ). If we assume (1), (2), and (3) then, because of Prop. 1.9, we similarly have the adic generic fiber Spf(S ) ad of Spf(S ) over K. Proposition Suppose that S satisfies (1), (2), and (3). Then Spf(S) ad and Spf(S ) ad are perfectoid spaces over K and K, respectively, and Spf(S ) ad = (Spf(S) ad ). Proof. (Sketch) The oof basically consists in checking that the arguments in the oof of [Sch] Lemma 6.4 generalize to the esent situation. We first consider the space Spf(S ) ad. Let ϖ, σ 1,..., σ r be generators of a finitely generated ideal of definition of S. Since S is perfect it makes sense to introduce (S ) n,j := ϖ -adic completion of S [Y 1,..., Y r ]/((σ n 1 ) 1/pj (ϖ ) 1/pj Y 1,..., (σ n r ) 1/pj (ϖ ) 1/pj Y r ) for any n 1 and j 0. We then have the sequence of homomorphisms (S ) n = (S ) n,0... (S ) n,j S ) n,j+1... with respect to the transition maps sending Y i to Y p i. One checks that after inverting ϖ these transition maps become isomorphisms. Hence inside K ok (S ) n we have the o K -subalgebra (S ) n := ϖ -adic closure of the union of the images of the (S ) n,j. By construction (S ) n is ϖ -adically complete and Hausdorff and is a flat (since ϖ-torsion free) o K -algebra. It satisfies K ok (S ) n = K ok (S ) n. At this point the arguments in the oof of [Sch] Lemma 6.4, part (i)(k of characteristic p) show that (S ) n a is a perfectoid o a -algebra. It then follows from [Sch] Thm. 5.2 that K K ok (S ) n is a perfectoid K - algebra. Therefore Spa(K ok (S ) n, (S ) n) is an affinoid perfectoid (and hence honest by [Sch] Thm. 6.3) adic space over K. By a gluing argument we finally conclude that Spf(S ) ad is a perfectoid (and hence honest) adic space over K. In order to treat Spf(S) ad we first of all make, in the discussion above, a more specific choice for the defining ideal of S. Let I be a defining ideal of S satisfying (1), (2), and (3) and pick generators I 1 = (s 1,..., s r ) = (ϖ, s 1,..., s r ) as in Lemma 1.8. Choose elements σ j = (..., s j mod ϖs) S. In Prop. 1.9 we have seen that I 1 = (ϖ, σ 1,..., σ r ) is a defining ideal of S. Since σ j s j mod ϖs we have I 1 = (ϖ, σ 1,..., σ r), which is a defining ideal of S 6

7 by Lemma 1.8.ii. Now one shows as in the oof of [Sch] Lemma 6.4, part (i),(ii)(general K) that the adic spaces Spa(K ok Ŝ n, Ŝ+ n ) are affinoid perfectoid over K with tilts Spa(K ok (S ) n, (S ) n) (this uses the result in the characteristic p case). Again by gluing Spf(S) ad is perfectoid. Since the tilting construction commutes with gluing, by [Sch] Prop. 6.17, the isomorphism in the assertion follows as well. 2 The perfectoid open unit disk Let L/Q p be a finite extension with ring of integers o and residue field k of cardinality q = p f. We also fix, once and for all, a ime element π of o. Moreover, we choose a Frobenius power series φ for π, i.e., a formal power series φ(x) o[[x]] which satisfies φ(x) = πx + higher terms and φ(x) X q mod πo[[x]]. The Lubin-Tate formal group law F φ is the unique commutative one dimensional formal group law F φ o[[x, Y ]] such that φ End o (F φ ) is a formal endomorphism of φ. There is a unique homomorphism of rings o End o (F φ ) a [a] φ (X) = ax + higher terms such that [π] φ = φ. In this way the formal o-scheme H L := Spf(o[[X]]) (for the maximal ideal as an ideal of definition) becomes a formal o-module. Its generic fiber over L is the open unit disk D L viewed a a rigid or adic space. We form the ojective limit H L := lim H φ L := lim(... Spf(φ) H L Spf(φ) H L Spf(φ)... Spf(φ) H L ) in the category of formal o-schemes. Equivalently H L = Spf( ˆR L ) where ˆR L is the (π, X)-adic completion of the o-algebra This is a formal L-vector space. Remark 2.1. ˆRL is flat over o, and R L := lim o[[x]] = lim(o[[x]] φ... φ o[[x]] φ...). φ k o ˆRL = ˆR k := X-adic completion of lim k[[x]]. Proof. In the ring o[[x]] we have the obvious implication: If Xf 1 = πf 2, then f 1 πo[[x]] (and f 2 Xo[[X]]). Using the defining operties of a Frobenius power series one checks that this still holds true in R L : If f 1, f 2 R L such that Xf 1 = πf 2, then f 1 πr L (and f 2 XR L ). 7

8 We now consider the commutative diagram 0 K n+1 R L /(π n+1, X n+1 ) π R L /(π n+1, X n+1 ) R L /(π, X n+1 ) K n R L /(π n, X n ) π R L /(π n, X n ) R L /(π, X n ) 0, where the terms in the left column are defined to be the kernels of multiplication by π. The above observation implies that the vertical map in the first column, indeed, is the zero map. Since countable ojective systems with zero transition maps have zero ojective limit as well as zero lim 1 -term, passing to the ojective limit with respect to n in the above exact sequence gives the short exact sequence 0 lim R L /(π n, X n ) π lim R L /(π n, X n ) lim R L /(π, X n ) 0 On the one hand, since the ideal sequences (π n, X n ) and (π, X) n are cofinal, we have ˆR L = lim R L/(π n, X n ). On the other hand, using that φ(x) X q mod πo[[x]] we compute lim R L/(π, X n ) = lim(r L /πr L )/(X n ) = lim[(lim k[[x]])/(x n )]. Hence we obtain the short exact sequence 0 ˆR π L ˆR L lim[(lim k[[x]])/(x n )] 0. In particular, ˆR L is π-torsion free and therefore flat over o. H ad L We let D L := denote the generic fiber of HL as an adic space over L. We briefly recall the construction: For any n 1 let ˆR L,n denote the π-adic completion of ˆR L [Y ]/(X n πy ). There are obvious open immersions of adic spaces Spa(L o ˆRL,n, ˆR L,n ) Spa(L o ˆR L,n+1, ˆR L,n+1 ) 2. Gluing the Spa(L o ˆRL,n, ˆR L,n ) along these immersions gives the space D L. Let K be a perfectoid field K which contains L. In the following a subscript K (instead of L) indicates the base extension from L to K (or from o to the ring of integers o K ) of any of the above objects. We choose the element ϖ o K such that π ϖ 2 < 1. Then I := (ϖ, X) is a defining ideal of ˆR K. We will verify that I S := ˆR K also satisfies the conditions (1) and (3) in section 1. As in the oof of Remark 2.1 we obtain (6) ˆR K /ϖ ˆR K = X-adic completion of lim o K /ϖo K [[X]] = X-adic completion of o K /ϖo K k ˆRk = o K /ϖo K ˆ k ˆRk. To describe this ring completely explicitly let N 0 [q 1 ] denote the set of all rational numbers of the form l q with l, m 0. Then m ˆR K /ϖ ˆR K = ring of all series ā i X i with coefficients ā i o K /ϖo K such that i N 0 [q 1 ] for any n 0 there are only finitely many 0 i < n with ā i 0. 2 Is ˆR L,n integrally closed in L o ˆRL,n? 8

9 This description immediately shows that the map is surjective on ˆR K /ϖ ˆR K. Because of Remark 1.7 this settles the condition (1). We also see that the cosets in ˆR K /I are reesented by the elements in the set { ā l X l/qm : m 1}. 0 l<q m Hence the kernel of on ˆR K /I is reesented by the cosets in { ā l X l/qm : m 1, ā q l = 0} = {X1/q q m 1 l<q m 0 l<q m q m 1 bl X l/qm : m 1, b q l = 0}. If ϖ 1 o K is such that ϖ 1 q = ϖ then it follows that I 1 = (ϖ 1, X 1/q ), which settles the condition (3). Lemma 2.2. DK is a perfectoid space over K (called the perfectoid open unit disk), and D K = Spf(o K ˆ k ˆRk ) ad. Proof. By the above discussion Prop applies and gives that D K is perfectoid with D K = Spf( ˆR K )ad. By (6) and since ˆR k is perfect there is an obvious homomorphism o K k ˆRk = lim(o K /ϖo K ) k ˆRk lim(o K /ϖo K ˆ k ˆRk ) = ˆR K. This extends to an isomorphism o K ˆ k ˆRk = ˆR K where the left hand side is completed with respect to the (ϖ, X)-adic topology. The reason is that one has (ϖ ) qm o K = {(..., α 1, α 0 ) lim o K /ϖo K : α m =... = α 0 = 0} for any m 0. Of course, the tilt D K is a perfectoid space over the tilt K. The final purpose of these notes is to explain a second structure as a perfectoid space over the field ˆL once we remove the origin from D K. We first recall the definition of the field L. Let L be an algebraic closure of L with maximal ideal M in its ring of integers. The formal o-module structure of H L induces an actual o-module structure on M. For any n 1 let F n := {z M : [π n ] φ (z) = 0} be the π n -torsion submodule. By adjoining these subsets to L we obtain the tower of algebraic extensions L L 1 := L(F 1 )... L n := L(F n )... L := n L n L. From Lubin-Tate theory (cf. [LT]) we know: a. The extension L n /L does not depend on the choice of φ. b. F n is a free o/π n o-module of rank one. c. L n /L is a totally ramified Galois extension of degree (q 1)q n 1. 9

10 d. The map is an isomorphism. χ L,n : Gal(L n /L) = (o/π n o) σ a such that σ(z) = [a] φ (z) for any z F n e. Any generator z n of F n as an o/π n o-module generates the ring of integers o Ln in L n as an o-algebra and is a ime element of o Ln. Lemma 2.3. The completion ˆL of the field L is perfectoid. Proof. The operty c. above implies that the value group ˆL is dense in R >0. Furthermore the generators in e. can be chosen in such a way that [π] φ (z n+1 ) = z n for any n 1. Since φ(x) X q mod πo[[x]], it follows that z q n+1 [π] φ(z n+1 ) = z n mod πo L. Again by e. the z n generate the o-algebra o L. Hence their cosets generate the k-algebra o L /πo L. We conclude that (o L /πo L ) q = o L /πo L. We therefore may form the tilt ˆL of ˆL. As a consequence of b. above the o-module T := lim(... [π] φ(.) [π] φ (.) [π] φ (.) F n+1 F n... [π] φ(.) F 1 ) is free of rank one. Since φ(x) X q mod πo[[x]] we have Therefore y q m+1 y m mod πo L for any m 1 and any (y n ) n 1 T. ι : T oˆl (y n ) n 1 (..., y n mod πo L,..., y 1 mod πo L, 0) is a well defined map (but not a homomorphism). We fix a generator t = (z n ) n of the o-module T. It follows from e. above that ι(t) is neither zero nor a unit in oˆl. Hence we have the well defined homomorphism of k-algebras which extends to an embedding of fields k[[z]] oˆl f(z) f(ι(t)), k((z)) ˆL. The image E L of the latter is a complete nonarchimedean discretely valued field of characteristic p with residue field k. It is called the field of norms of L. The image of the former map is its ring of integers o EL. One can show: E L is independent of the choice of the generator t. ˆL is the completion of the perfect hull of E L. This implies that sending X to ι(t) gives an isomorphism (7) ˆRk = oˆl. 10

11 On the other hand sending ι(t) to Z gives a homomorphism o EL o K [[Z]]. Using Lemma 2.2 and (7) we obtain isomorphisms (8) D K = Spf(oK ˆ k ˆRk ) ad = Spf(o K ˆ k o EL ˆ oel oˆl ) ad = Spf(oK [[Z]] ˆ oel oˆl ) ad. Note that Spf(o K [[Z]]) is the formal open unit disk over o K. To go further we have to remove the origin from the open unit disk, i.e., we have to start with the adic space D K := D K \ {0}. It has a perfectoid version D K. To compute its tilt we observe that now we may view the punctured open unit disk D (over K ) as an adic space over E K L via the map E L O(D K ) Z ι(t) D K. In this setting the formula (8) becomes the following isomorphism. Technically the additional complication comes from the fact that D is no longer the generic fiber of an affine formal K scheme. So one has to exhaust D by affinoid annuli and has to redo everything above for K them. Proposition 2.4. ( D K ) = D K ˆ EL ˆL. The formal o-module structure of H L induces an action of the multiplicative monoid o\{0} on the rings ˆR L, ˆRk, and ˆR K and consequently on the perfectoid spaces D K and D K and their tilts. Under the isomorphism in Lemma 2.2 this action on the left hand side corresponds to the action on the right hand side through the factor ˆR k. The action of π on ˆR k is the q-frobenius. On the other hand the Galois group Gal(L /L) naturally acts on ˆL. The ojective limit of the isomorphisms χ L,n in d. above is an isomorphism χ L : Gal(L /L) = o. Hence we have an o -action on ˆL. Letting π act as the q-frobenius this extends to an action of the monoid o \ {0} on ˆL. One checks that the isomorphism (7) is equivariant for the o \ {0}-actions on the two sides. The latter implies that in the isomorphism in Prop. 2.4 the action of π on the left hand side corresponds on the right hand side to the map (φ 1 id) φ q, where φ q is the (absolute) q-frobenius and φ on D is given by the q-frobenius on the coefficients K and φ(z) = Z. K References [B-CA] Bourbaki N.: Commutative Algebra. Hermann 1972 [GR] Gabber O., Ramero L.: Foundations for almost ring theory - Release 6.8. Preint 2014 [LT] Lubin J., Tate J.: Formal complex multiplication in local fields. Ann. Math. 8, (1965) [Sch] Scholze P.: Perfectoid Spaces. Publ. Math. IHES 116, (2012) [Wei] Weinstein J.: Gal(Q p /Q p ) as geometric fundamental group. arxiv: v1 11