foundations of perfectoid geometry, i - stanford...

foundations of perfectoid geometry, i

ALESSANDRO MARIA MASULLO1

Abstract

We introduce Tate-perfectoid rings (the rings we’ve been calling “perfectoid”, so far)and perfectoid rings, and relate the two notions. We discuss the tilting equivalence in thegenerality of both perfectoid rings and Tate-perfectoid rings, and describe the equivalenceof the finite-étale sites over a Tate-perfectoid ring and its tilt. We globalize the results andintroduce perfectoid spaces, establishing the basic foundational features. We discuss allresults at the aforementioned level of generality, explaining the motivation for the choicesmade throughout, in light of applications we have in mind.

These notes are organized as follows.In §1 we discuss a slightly nonstandard perspective on the Zariski Nagata purity the-

orem in [SGA2], explaining the meaning of Falting’s Almost Purity and how to regardit as a “purity result”. We give motivation and describe the nature of this work, whileannouncing the results contained in the second part of it.

In §2 we discuss the needed preliminaries on perfectoid rings. We’ll explain how torelate the notion of perfectoid ring and Tate-perfectoid ring bearing in mind the geometricpicture.

In §3 we discuss the tilting equivalence in the generality of perfectoid rings stressingthe fact that almost mathematics does not come into play in the proof, which, rather,relies on a core property of perfectoid rings we discuss and ensure the deformation theorymachinery, reviewed in §B, works. On the other hand, in this section we prove the tiltingequivalence for Tate-perfectoid rings deducing the result from the equivalence for perfectoidrings, pointing out where almost mathematics (though mildly) comes into the picture andwhy.

In §4 we discuss finite étale tilts, set up the necessary preliminaries for establishingthe equivalence of the almost finite étale sites on perfectoid rings and finite étale sites onTate-perfectoid rings, and on their tilts, respectively. We prove the almost purity theoremfor Tate-perfectoid rings in positive characteristic, and draw consequences from it. Wefinally adress completely the equivalence for perfectoid fields, recovering the Fontaine-Wintenberger correspondence.

In §5 we finally introduce perfectoid spaces, exploiting the full power of the generalityadopted throughout. We globalize the tilting functor and prove that the category ofperfectoid spaces (upon fixing a base perfectoid space) contains fibered products.

In §6 we define the small étale site on a perfectoid space, and prove the global equi-valence of the étale site of a perfectoid space and its tilt, achieving the geometric form ofFaltings’ almost purity.

In the first appendix §A we discuss some loose ends, recollecting some needed toolsfrom almost mathematics, and explaining a couple of different proofs to results discussedin previous sections.

In the second appendix §B, we recollect the basics of deformation theory via the co-tangent complex, for convenience of the reader. The purpose of this handout is to makethe setup and foundational results on perfectoid spaces uniform and uniformly treated atthis level of generality, with the aim of providing a unified and consisent reference.

1The purpose of these notes is to give a uniform treatment of the theory, given the various evolutionary stepsthe theory encountered so far. I hope it helps. Comments to [email protected]. Lecture notes coveringthe prerequisites, available at http://math.stanford.edu/∼conrad/Perfseminar/. Notes explaining/expandingthe proof of finiteness of p-adic étale cohomology for proper rigid-analytic varieties, coming soon under the title“Foundations of Perfectoid Geometry, II”. Proof of Poincaré duality for p-adic cohomology of proper rigid-analyticvarieties, also included in Foundations, II.

Contents

Contents 21 Introduction 32 Perfectoid rings 173 The tilting equivalence over perfectoid rings 344 Finite étale extensions and finite étale tilts 575 Perfectoid spaces 686 Small étale sites and tilting 83A Loose ends on Almost Mathematics 101B The cotangent complex 108References 155

foundations of perfectoid geometry, i 3

1 Introduction

If we let X be a locally noetherian scheme, and Y → X a closed subscheme, in case we havea natural isomorphism πet

1 (Y ) ' πet1 (X), we have an equivalence of categories between étale

covers of Y and X respectively. Grothendieck’s strategy to tackle situations along these linesand decide whether such equivalence holds or not, is to first compare the category of étalecovers of X with that of a neighbourhood U of Y in X. Then compare the category of étalecovers of U with that of the formal completion of X along Y . As a last step, compare thecategory of étale covers of the formal completion of X along Y , with that of Y .

The very first step relies on the idea that if Y is “large enough”, then passing from étale coversof U to étale covers of X should involve addition of high codimension subvarieties, whichwill inaffect the category of étale covers. We review the Zariski - Nagata purity theorem inSGA2 from a slightly non-standard perspective. We shall discuss its proof in full detail, andthen describe the nature of Faltings’ method of almost étale extensions as being, actually, aninstance of purity in a wider sense.

1.1 Aims of this work and announcement

This introduction will be used in later lecture notes for the purposes of producing significantexamples, first, and motivate the introduction of what I call the Iwasawa topology on properformal Spf(Ainf)-schemes X, Ainf being Fontaine’s ring W (O[), O being a perfectoid ring, andO[ being its “tilt”.

The site (X/Ainf)Iw will be closely related to Iwasawa theory, and has the feature of yielding atopos theoretic cohomology theory specializing via canonical integral comparison isomorphismsto:

RΓcrys(XO/p/Acrys), RΓet(XO⊗OFrac(O),Zp), RΓcrys(XO ⊗O k/W (k)), RΓdR(XO ⊗O k)

and, notably:RΓet(XO[⊗O[Frac(O[),Zp).

Bhatt, Morrow and Scholze already produce such cohomology theory in [BMS], by constructinga complex of Ainf -modules AΩ on XZar, whose hypercohomology does the job. Their approachis concrete and relies on very explicit local calculations which suggest indeed that one should beable to assign a good Ainf -version of the infinitesimal site (this latter being bound not to work atall, though) recovering AΩ site theoretically. It turns out that, via the natural topos-theoreticprojection ε : (X/Ainf)Iw → XZar, we do have AΩ ' Rε(OXIw

) canonically and naturally in X,and the specialization theorems in [BMS] can be reproved “geometrically”.

Thanks to (X/Ainf)Iw, Faltings’ almost purity will be indeed a purity theorem, with no inter-vention of the theory of almost étale extensions, which, as we shall see at times throughoutthis volume and explicitly explain in the second part, is ill suited to work integrally.

We choose to explain here the original theory of perfectoid spaces in the setup given by PeterScholze and going back to Faltings, ie. making use of almost mathematics, but we’ll limit theuse of almost mathematics to the minimum.

We shall address the integral theory of perfectoid spaces in the second part of these notes,where we shall prove an analogue of Theorem 3.7.1 and Theorem 6.1.12, in the context ofp-adic formal schemes and their “perfectoid” Iwasawa covers. There will be no ad hoc of almostmathematics, but rather we shall give a geometric interpretation to it, which will clarify what

4 alessandro maria masullo

goes wrong in its usage on the integral level, and led Faltings to introduce almost mathematicsin the first place in order to obtain rational comparison isomorphisms.

This is also why we call these notes “Foundations of perfectoid geometry”: as they aim at notbeing limited to the world of rigid geometry, in terms of applicability. The ultimate results willbe applicable to proper smooth schemes over Spec(Z[[q]]), q being a parameter to be specializedfrom time to time, and will be made available by formal geometry. To guide the reader throughthe theory, we first address the theory of perfectoid spaces, for then moving to formal andalgebraic geometry in the second part. The naive slogan these notes aim to convey is that(integral) perfectoid geometry enables one to relate geometry locally at primes p and p′, withpossibly p 6= p′.

Along the way, we shall establish a number of foundational results for the p-adic étale co-homology of (proper) rigid analytic varieties over characteristic 0 complete algebraically closednonarchimedean fields C, including Poincaré duality and the Künneth formula, both discussedin the second part together with a new proof of finiteness of p-adic étale cohomology, the onlyexisting one being given by Peter Scholze in his p-adic Hodge Theory paper ([Sch2]).

Let’s go back to the introduction, and review purity a la Grothendieck. One can safely saythat our interpretation of Faltings’ purity is modeled on what follows.

1.2 Comparison of Et(Y ) and Et(X)

We fix our setup for the whole of the introduction as being that of a locally noetherian schemeX,I an ideal of definition of a closed immersion Y → X. We shall denote by X the formalcompletion of X along Y .

Let n be a positive integer. We set Yn to denote the scheme (Y, (OX/I n+1)|Y ), where weendow the structure sheaf (OX/I n+1)|Y with the discrete topology (I is of finite type). TheYn’s form a directed system of schemes, and we denote by X its colimit formal scheme, ie.the formal completion of X along Y . By [SGA1, §1.8.3] we know that the assignement of alocally finite type X-formal scheme S, is equivalent to the assignement of an inductive system(Sn) of Yn-schemes of finite type, together with isomorphisms of schemes Sn ' Sn+1×Yn+1 Yn.Moreover, S is an étale cover of X if and only if Sn is an étale cover of Yn for all n ≥ 1. It’san easy exercise to check that the base change functor:

Et(Yn+1)→ Et(Yn)

is an equivalence of categories for all n ≥ 1, and that we have the following:Proposition 1.2.1 The natural functor:

Et(X)→ Et(Y )

is an equivalence of categories.

The proof of the above Proposition 1.2.1 is an immediate application of the theory of thecotangent complex.

Proof. For any étale cover a : Y ′ → Y , the cotangent complex LY ′/Y vanishes in D−qcoh(OY ).It is, by definition, LOY ′/a

−1OY , and its vanishing yields unique (up to unique isomorphism)unobstructed deformations of Y ′ → Y to some Y ′1 → Y1, which can be arranged to be flat andof finite type. Flatness ensures vanishing of LY ′1/Y1

⊗LOY ′1

(OY ′1 /J · OY ′1 ) in D−qcoh(OY ), whichyields vanishing of LY ′1/Y1

. By induction, we conclude.


The upshot is that, in our setup, the categories of étale covers of Y and of X are alwaysequivalent. The reason of this is to be seen in the vanishing of the full cotangent complex of anétale morphism of schemes. Such vanishing, however, occurs in more general situations whereno finiteness assumption is being given. We briefly digress on this.

1.3 Digression on the vanishing of the full cotangent complex

For the convenience of the reader, we briefly digress on the relation between étaleness of a ringmap, together with weaker notions, and vanishing of the full cotangent complex.

This digression, as the whole of the introduction, is elementary and won’t be used in the sequel,and the reader may want to skip it at a first reading. It will be nonetheless useful to familiarizewith the usage of the cotangent complex, which will be deepened in Appendix B, and will beabsolutely crucial in the second volume of these notes.

We first discuss an example of a formally étale ring map whose cotangent complex is nontrivial(not isomorphic to a trivial complex in the derived category). We define R := Z[tr, r ∈ Q>0],and consider the ideal generated by the indeterminates indexed by all positive rational numbers,say I. Here we require tr · tr′ = trr′ , so our R is not actually a polynomial algebra over Z, butrather a quotient by a great number of relations. The restriction on r being always positiveensures that such relations are “not too many” and 1 6= 0 in R, so R is nonzero. Then, we haveI2 = I and therefore R→ R/I is formally étale (check!).

We now consider A := R/(tr · tr′), where r 6= r′ are two fixed positive rationals. Again A isnonzero.

We call J := I mod (tr · tr′). A base change of a formally étale morphism is formally étale, soA→ A/J is a formally étale ring map. J is not flat as an A-module, as the multiplication mapJ ⊗A J → J sends tr ⊗ tr′ to zero. It’s easy to show we have the following exact injection of(A/J)-modules:

0 6= ker(J ⊗ J → J) → H−2(L(A/J)/A)

and we conclude A → A/J is an example of a formally étale ring map with non-vanishingcotangent complex. Such non-vanishing had to appear in degree ≤ −2, as it is an easy task toshow that a formally étale ring map has vanishing truncated cotangent complex.Proposition 1.3.1 Let u : A → B be a ring homomorphism. The following conditions areequivalent:

(1) u is formally étale.

(2) τ≥−1LB/A = 0 in D(B).

If we require u to be of finite presentation, then being étale is equivalent to the vanishing of thefull cotangent complex.

Proof. We show that if τ≥−1LB/A = 0, then u is formally étale. Indeed, we must show thatfor any A-algebra C and nilpotent ideal I of C, given an A-algebra map

v0 : B → C/I

there exists a unique A-algebra map v : B → C lifting v0, that is, we have a bijection

HomA(B,C) = HomA(B,C/I).

We may and do assume I is square-zero, by an easy induction.


Via pullback, we reduce our consideration to the case B = C/I and I is a B-module. Recall(see §B) that the obstruction o(v0) to existence of v lies in

Ext1B(LB/A, I),

which is trivial, since τ≥−1LB/A is trivial. Hence v exists. Any other v′ lifting v0 is reachedfrom v via action of Ext1

B(LB/A, I) = Ext1B(τ≥−1LB/A, I), which is again trivial, and hence

v is unique up to automorphism. The automorphism group of v is Ext0B(τ≥−1LB/A, I) =

Ext0B(LB/A, I) = HomB(Ω1

B/A, I), which is again trivial, and we conclude v is unique (up tounique isomorphism).

We show the converse. For every B-module M we have

Ext1B(τ≥−1LB/A,M) ' ExalA(B,M)

the right side denoting the group of A-algebra extensions of B by M . We claim that if B isformally étale over A, the latter group is trivial. Indeed, take µ : J/J2 → M a B-modulehomomorphism such that, under the above bijection, a given extension:

0→M → B → B → 0

is isomorphic to the pushout of the extension:

0→ J/J2 → A[B]→ B → 0

along µ. Since B is formally étale over A, it is in particular formally smooth, so this lat-ter extension is trivial in ExalA(B, J/J2). Then, B is the trivial extension of B by M inExalA(B,M), implying that Ext1

B(τ≥−1LB/A,M) = 0 for all B-modules M . The same wayformal étaleness implies that Ext0

B(LB/A,M) = 0 for all B-modules M . Hence we have that

RHomB(τ≥−1LB/A,M) = 0

for all B-modules M , implying that τ≥−1LB/A = 0 in D(B), as desired.

We now show that if, in addition, we require beforehand u to be locally of finite presentation,then we have that u is étale if and only if LB/A = 0 in D(B) and u is locally of finite presentation.

We set up the following pushout diagram:

Au //

u

B

B // B ⊗A B

We have B ⊗A B = B ⊗LA B, so by flat base change we have that:

LB/A ⊗B (B ⊗A B)→ LB⊗AB/B

is an isomorphism in D(B).

Consider the maps:

Bα−→ B ⊗A B

β−→ B


where α(b) = b ⊗ 1 and β is the multiplication. In particular βα = id. Now we consider thecorresponding fundamental distinguished triangle:

LB⊗AB/B ⊗B⊗AB B → LB/B → LB/B⊗AB → LB⊗AB/B [1]⊗B⊗AB B

We now use the quasi-isomorphism: LB/A⊗B (B⊗AB) 'qi LB⊗AB/B due to flatness of B⊗ABover B, to get:

LB/A ⊗B (B ⊗A B)⊗B⊗AB B → LB/B → LB/B⊗AB → LB/A ⊗B (B ⊗A B)[1]⊗B⊗AB B

and now we use the fact that βα = id, to get that this is actually the following:

LB/A → LB/B → LB/B⊗AB → LB/A[1]

We are left to check that LB/B⊗AB is trivial. So far we have just used formal unramifiednessof u, but now we observe that our assumption on u being étale ensures that the diagonal4 : Spec(B)→ Spec(B)×A Spec(B) is an open immersion, thus yielding the desired vanishing.

The converse is clear, as if we have LB/A = 0 then in particular τ≥−1LB/A = 0, which, togetherwith the finite presentation assumption on u, implies u is an étale ring map.

In order to single out a class of morphisms of rings, and more in general sheaves of rings, henceschemes X → Y , satisfying vanishing of the full cotangent complex, the key observation is thatformation of the cotangent complex LX/Y is stable under flat localization.

We claim that if a morphism of scheme f : X → Y is flat with flat diagonal, then it hasvanishing full cotangent complex. We call this class of morphisms weakly étale. This classof morphism plays an essential role in the introduction of the proétale topology for schemes,and we refer the interested reader to [BSch]. We will review the foundations of the pro-étaletopology for adic spaces and schemes in the second part of these notes, discussing the étale andpro-étale topology on p-adic formal schemes as well.Proposition 1.3.2 Let u : A→ B be a weakly étale ring map. Then LB/A ' 0.

Proof. We’re required to check LB/B⊗AB ' 0 in D(B), though now we have no finite present-ation assumption on A → B(!) In fact a priori we do not even know τ≥−1LB/A = 0 becausewe are not assuming A→ B is formally étale.

We claim:B ⊗L

B⊗AB B ' B.If so, we conclude by flat base change. We know that B is (B ⊗A B)-flat, so we need to check:

B ' B ⊗B⊗AB B.

Let J be the kernel of the surjective flat ring map B⊗AB B, and upon applying (·)⊗B⊗ABBto the following short exact sequence of (B ⊗A B)-modules:

0→ J → B ⊗A B → B → 0

we obtain:0→ J ⊗B⊗AB B → B → B ⊗B⊗AB B → 0

Now the multiplication map J ⊗B⊗AB B → JB is injective by flatness of B over B ⊗A B, andsince JB = 0, we conclude.


Remark 1.3.3 We remark that in [BSch], the class of weakly étale ring maps generates thesmall pro-étale site on a scheme (coverings being given by fpqc coverings). The pro-étaletopology can be defined for adic spaces possessing a well behaved étale site, as well as forperfectoid spaces, whose definition will be engaged later. We’d like to stress here that weaklyétale maps have the feature of having vanishing full cotangent complex, as shown above. Suchvanishing is a property which, as we shall discuss, is enjoyed by any map between perfectoidrings in a suitable sense, and is at the core of the whole theory.

1.4 Comparison of Et(Y ) and Et(U)

We first need a couple of definitions.Definition 1.4.1 The setup being given, we consider the pair (X,Y ).

(1) We write Lef(X,F ) and say (X,Y ) satisfies the Lefschetz condition if and only if for eveyopen U in X containing Y , and every coherent locally free sheaf F on U , the naturalmap:

Γ(U,F )→ Γ(X, F )

is an isomorphism.

(2) We write Leff(X,Y ) and say (X,Y ) satisfies the effective Lefschetz condition, if and onlyif we have Lef(X,Y ) and, in addition, for every coherent locally free sheaf F on X thereexists an open neighbourhood U of Y and a coherent locally free sheaf F0 on U , comingwith an isomorphism F0 ' F .

These two conditions are fulfilled in two important examples.Proposition 1.4.2 Let A be a noetherian ring, and let $ ∈ rad(A) be an A-regular elementin the radical of A. We assume A is a quotient of a regular local ring, and that A is $-adicallycomplete. Let X ′ := Spec(A) and Y ′ := Spec(A/$). We set x := Spec(A/rad(A)), and defineX := X ′ − x, Y := Y ′ − x. Then:

(1) If for every prime ideal p of A with dim(A/p) = 1 we have depth(Ap) ≥ 2, then we haveLef(X,Y ).

(2) If, in addition, for every prime ideal p of A containing $ and such that dim(A/p) = 1 wehave depth(Ap) ≥ 3, then we have Leff(X,Y ).

Sketch of proof. We won’t discuss the full proof, but rather mention that part (1) goes throughshowing that if we let U to be an open neighbourhood of Y in X and E a locally free OU -module, we set Z := X − Y and let j : U → X the canonical open immersion, then j∗(E ) is acoherent OX -module. This is equivalent to showing that Hi

Z(E ′) is coherent in degrees 0 and1, where E ′ is a coherent extension of E to X.

To achieve this one uses the fact that if X is a locally noetherian scheme, Y a closed subsetof X and F a coherent OX -module, and if X has locally a dualizing complex, then as soon asone has, for all x ∈ X − Y :

Hi−c(x)(Fx) = 0

for i ∈ Z and c(x) := codim(x ∩ Y, x), then one has that the cohomology sheaf HiY (F ) is

coherent. This is [SGA2, Thm. 2.1, Exp. VIII].

To see that the above consitions are met by E ′, one first shows the complement U of Y inX is a union of finitely many closed points. Since U is open in X ′, one calls Z ′ the closedcomplement, defined by an ideal I. Since Z ′ ∩ Y ′ = x, the ring A/(I + $A) is Artin, and


we’re done. Thanks to this and the fact that A is a quotient of a regular ring, as soon as p ∈ U ,given by the ideal p ⊂ A, is such that c(p) = 1, then dim(A/p) = 2. Since E is locally free,then one has that for all points p in the support of E :

depth(Ep) = depth(OU,p).

One puts everything together by observing that if p ∈ U satisfies c(p) = 1, we have:

depth(E ′p) = depth(Ep) = depth(OU,p) = depth(Ap) ≥ 3− 2 = 1.

We have that the natural map Γ(X, j∗(E ))→ Γ(U,E ) is an isomorphism. Now set F := j∗(E ),which is coherent and of depth ≥ 2. Then if we denote by f : X → X ′ the canonical immersion,Rif∗(E ) are coherent for i = 0, 1. It follows the natural map:

Γ(X,F )→ Γ(X, F )

is also an isomorphism, and since it factors through the natural isomorphism Γ(X,F )'−→

Γ(U,E ), it follows the natural map:

Γ(U,E )→ Γ(X, E )

is an isomorphism, whence (1) is proved.

To show (2), we take E to be a coherent locally free sheaf on X, and we use formal GAGAto check that it’s algebrizable, ie. it is isomorphic to the formal completion of a coherent OX -module E. Then it’s easy to see E is locally free on a neighbourhood of Y , which achieves (2).Some details omitted.

The above basic cases are the key to show that if one has Lef(X,Y ), then for any open neigh-bourhood U of Y in X the functor:

Et(U)→ Et(Y )

is fully faithful, and if Leff(X,Y ) holds, then for every étale cover T → Y , there exists an openneighbourhood U of Y and an étale cover T ′ → U such that

T ′ ×U Y ' T.

As an immediate consequence, if Lef(X,Y ) holds and the notation being the same as above, wehave that the natural map: πet

1 (Y ) → πet1 (U) is surjective for all such U , and we can recover

πet1 (Y ) as the inverse limit of the πet

1 (U)’s over all such U ’s (upon fixing base points on Y andX).

1.5 Comparison of πet1 (U) and πet

1 (X)

We begin with the following definition.Definition 1.5.1 Let X be a scheme, and Z a closed subset of X. We set U := X − Z. Wesay the pair (X,Z) is pure if and only if for every open V → X, the functor:

Et(V )→ Et(V ∩ U)

assigned by:V ′ 7→ V ′ ×V (V ∩ U)


If X = Spec(A) is the spectrum of a local noetherian ring A, and rad(A) is the radical of A,with x = rad(A) the closed point of X, we say A is pure if the couple (X, x) is.


We finally get to the statement of the purity theorem.Theorem 1.5.2 ([SGA2, Thm. 3.4, Exp. X]) Every noetherian regular local ring of dimension≥ 2 is pure. Every noetherian local ring of dimension ≥ 3 which is a complete intersection ispure.

The first part of the statement is known as the Zariski-Nagata purity theorem.

The first key input which goes in the proof is the following Lemma.Lemma 1.5.3 Let X be a locally noetherian scheme, and U an open subscheme in X, j : U →X being the canonical open immersion. The following conditions are equivalent:

(1) For all open subschemes V → X, if we set V ′ := V ∩ U , the functor F 7→ F |V ′ fromthe category of locally free coherent OV -modules to the category of locally free coherentOV ′-modules, is fully faithful.

(2) The natural map OX → j∗(OU ) is an isomorphism.

(3) For all z ∈ Z, we have depth(OX,z) ≥ 2.

The proof is easy, and left to the reader. Upon establishing the fact that purity is insensitiveof faithfully flat extensions, and by means of the above Lemma 1.5.3, we can reduce the proofof Theorem 1.5.2 to the case the noetherian regular local ring A is complete.

At this point one has to show the following.Lemma 1.5.4 Let A be a noetherian regular local ring, and let $ ∈ rad(A) be an A-regularelement. We assume A is complete with respect to the $-adic topology, and that A is a quotientof a regular local ring. We set B := A/$A.

(1) If for all prime ideals p of A such that dim(A/p) = 1, we have depth(Ap) ≥ 2, then if Bis pure so is A.

(2) If for all prime ideals p of A such that dim(A/p) = 1, we have depth(Ap) ≥ 2, then if Ap

is pure whenever $ is not contained in p, and if depth(Ap) ≥ 3 if $ ∈ p, then if A ispure so is B.

The proof is a straightforward combination of the results in §1.4, and is left to the reader. Theabove Lemma 1.5.4 is [SGA2, Lemma 3.9, Exp. X].

We’ll comment on this Lemma soon, but first let’s finally achieve Theorem 1.5.2.

Proof of Theorem 1.5.2. We show (1) by induction on the dimension. Let A be a noetherianregular local ring of dimension 2. Set X ′ := Spec(A), x := rad(A) and X := X ′ − x. Wehave depth(A) = 2. We apply Lemma 1.5.3 to the pair (X ′, x), and hence the functor

Et(X ′)→ Et(X)

is fully faithful. We let f : W → X an étale cover defined by a coherent locally free étaleOX -algebra A′ := f∗(OW ). Calling j : X → X ′ the canonical immersion, we claim B := j∗(A

′)is a coherent OX -algebra.

We now have that the depth of B at x is ≥ 2, as it’s the direct image of an OX -modulesupported on X ′ − x. Since A is a regular ring of dimension 2, we have:

pd(B) + depth(B) = dim(A) = 2.

It follows the projective dimension of B is zero, hence B is projective, hence free. As a result,B defines a finite flat cover of X ′. The non-étale locus of such cover in X ′ is closed and defined


by a principal ideal: the discriminant ideal of B/A. By design, it’s contained in x and henceit’s empty, since dim(A) = 2.

Suppose, now, A is a noetherian regular local ring of dimension n ≥ 3. We assume (1) issettled for dimension less than n. Without loss of generality, we can assume A is complete.Let $ ∈ rad(A) such that its image in rad(A)/rad(A)2 is nonzero. Then, B := A/$A is anoetherian regular local ring of dimension n− 1, and hence (1) holds for B, as n− 1 ≥ 2. Weconclude by Lemma 1.5.4, which we can apply since A is complete. We now show (2). Let’sassume there exists a noetherian regular local ring B and a B-sequence ($1, . . . , $k) such thatA ' B/($1, . . . , $k). We proceed by induction on k. If k = 0, we conclude by (1). We assumethe result is settled for all k′ < k, k ≥ 1. Let C := B/($1, . . . , $k−1) so A ' C/$kC, and$k is C-regular. By inductive hypothesis, C is pure, and it’s enough to check Lemma applies.Since dim(C) ≥ 4, for all prime ideals p of C such that dim(C/p) = 1 we have depth(Cp) ≥ 3.Moreover, Cp is a complete intersection for k′ ≥ k − 1. By inductive hypothesis, C is pure, asdesired.

As a direct consequence, we easily deduce the following:Theorem 1.5.5 Let X be a connected locally noetherian scheme, and Y closed in X. Weassume we have Leff(X,Y ), and that for every open neighbourhood U of Y and every x ∈ X−U ,the local ring OX,x is regular of dimension ≥ 2 or a complete intersection of dimension ≥ 3.Then:

πet1 (X) ' πet

1 (Y ).

In order to re-gain the above results in the context of p-adic formal schemes, a new notion offundamental group will be needed, and will be offered by the introduction of the Iwasawa site.Faltings’ almost purity theorem will take the form of an actual purity theorem along the abovelines.

We, for the moment, stick to Faltings’ original approach and make use of his theory of almostétale extensions. We therby clarify the sense in which we can regard Faltings’ almost purity asa purity result, in the following section.

1.6 Presque pureté

In this section we explain how to regard Faltings’ almost purity theorem as, indeed, a puritytheorem, in light of the preceding sections of the introduction.

We fix a nonarchimedean field k of mixed characteristic (0, p), equipped with a nontrivial rank1 valuation. We call O ⊂ k the valuation ring of k, and let $ be a nonzero topologicallynilpotent element in O such that $p | p in O. Call, for each integer n ≥ 1, kn := k($1/pn).Set k0 := k, and we call On the ring of integers of kn, for all n ≥ 0. We call:

R+n := On[T±p

−n

1 , . . . , T±p−n

r ] and Rn := R+n [1/$].

Since Rn is On-smooth, it’s regular, and the transition maps Rn → Rn+1 are all finite étale.Let S+

0 be a finite and normal R+0 -algebra, and assume that S0 := S+

0 [1/$] is finite étale asan R0-algebra.

Essentially, we’re given a directed system (Rn)n≥0 of finite étale R0-algebras, together witha direct system (R+

n )n≥0 of integral structures which are, by design, not finite étale as R+0 -

algebras, as their reductions modulo $ are ramified, but they are upon inverting $.


S0 is given as being finite étale away from the locus $ = 0, and if we call S+n the Rn-

normalization of S+0 ⊗R+

0R+n , we have that Sn := S+

n [1/$] is Rn-finite étale, and the map R+n →

S+n is, instead, ramified along the locus $ = 0. Note that this is exactly the ramification

locus of S+0 over R+

0 , and we’re considering its pullback to S+n for all n ≥ 0.

Let’s give a measure of such ramification.

We can consider the cotangent complex LS+n /R

+nfor all n ≥ 0, and observe that:

LSn/Rn ' 0 in D−(Sn)

since Rn → Sn is étale, for all n ≥ 0. On the other hand, formation of the cotangent complexis stable under flat localization, and hence:

LSn/Rn ' LS+n /R

+n⊗R+

nRn

for all n ≥ 0. Since each one of the S+n is R+

n -finite, then LS+n /R

+nis represented in D−(S+

n ) bya complex of coherent S+

n -modules (note that all the rings R+n are coherent), and in particular

has S+n -finitely generated cohomology, implying that sufficiently large powers of $ kill the

cohomology groups of LS+n /R

+nin D−(S+

n ), for each n ≥ 0.

More precisely, fix n ≥ 0. S+n → Sn is flat, and since each cohomology group of LS+

n /R+n

isS+n -finitely generated, then

AnnS+n

(Hi(LS+n /R

+n

))[1/$] = AnnSn(Hi(LS+n /R

+n

[1/$]) = Sn, i ≤ 0.

It follows that, calling mn the radical of $ in S+n , all cohomology groups of LS+

n /R+n

aresupported on V (mn).

Let’s call Xn := Spec(R+n ), jn : Un := Spec(Rn) → Xn, and likewise X ′n := Spec(S+

n ) andj′n : U ′n := Spec(Sn) → X ′n. We eventually call X∞ the affine scheme given by lim←−Xn,X ′∞ := lim←−X

′n, and likewise for U∞ and U ′∞. We call in, i′n the closed immersions of the

complements of Un in Xn and U ′n in X ′n respectively. Likewise, we denote with j∞, j′∞, i∞,i′∞ the obvious remaining morphisms.

We form the triangles in Dqcoh(X ′n):

Rjn!j∗nLX′n/Xn → LX′n/Xn → in∗Li

∗nLX′n/Xn → Rjn!j

∗nLX′n/Xn [1]

thus deducing a quasi-isomorphism:

LX′n/Xn → in∗Li∗nLX′n/Xn ' in∗i

∗nLX′n/Xn

in Dqcoh(X ′n) (note that LX/Y is, by very design using its simplicial representative in thecategory of complexes by the Dold-Kan correspondence, represented by a complex of flat OX -modules!).

Likewise, LX′∞/X∞ ' i∞∗i∗∞LX′∞/X∞ in Dqcoh(X ′∞).

Formation of the cotangent complex of a morphism of sheaves of rings commutes with directlimits (of sheaves of rings), and hence we obtain:

LX′∞/X∞ ' lim−→ in∗i∗nLX′n/Xn

in Dqcoh(X ′∞). Let’s study LX′∞/X∞ by first specializing to a simple case.


Let’s assume we were in the good situation in which for some n, the ramification at the genericpoint of the locus V (m), with m = lim−→mn = ($,$1/p, · · · , $1/pn , · · · ), is actually zero. Thenthe Zariski-Nagata purity theorem tells us that the ramification locus of X ′n over Xn has tobe pure of codimension 1. Since there is no ramification at the codimension 1 points (!) wededuce that there is no ramification, that is, Xn → X ′n is étale.

Hence LX′n/Xn ' 0 in Dqcoh(X ′n). In particular, LX′∞/X∞ ' 0 in Dqcoh(X ′∞).

The almost purity theorem says that this result extends to the limit, that is, X ′∞ → X∞ isétale in the sense of almost mathematics, hence LX′∞/X∞ has almost zero cohomology.

Since $ is S+n -regular for all n ≥ 0, then mn⊗S+

nLX′n/Xn ' 0, which implies m⊗S+

∞LX′∞/X∞ '

0, meaning exactly that LX′∞/X∞ has almost zero cohomology.

We make the above ideas more concrete on an example, which will also shed light on a proto-typical version of a tool which will be quite crucial all over the theory: a lemma of Gabber andRamero on henselian approximation.Example 1.6.1 We call Rn := kn = Qp(p

1/pn), and Sn := k′n = kn(p1/2p), for all n. We setR+n := On, and S+

n := O′n, meaning of the notation being as before. We see that

S+n = R+

n [x]/(x2 − p1/pn+1

), n ≥ 0

and since p1/pn+1 ∈ (x2 − p1/pn+1

, 2x), then p1/pn+1

annihilates

Ω1S+n /R

+n

=S+n dx

2S+n x dx

' R+n [x]

(x2 − p1/pn+1 , 2x).

We have ring maps:R+n → R+

n [x] S+n

yielding a triangle:

LR+n [x]/R+

n⊗R+

n [x] S+n → LS+

n /R+n→ LS+

n /R+n [x] → LR+

n [x]/R+n

[1]⊗R+nS+n

which is, in D(R+n ):

Ω1R+n [x]/R+

n⊗R+

n [x] S+n [0]→ LS+

n /R+n→ LS+

n /R+n [x] ' In/I

2n[−1]

where In := (x2p1/pn+1

) ⊂ R+n [x]. It follows that if we let n grow arbotrarily large, we obtain,

notation as before, a quasi-isomorphism in D(S+∞):

Ω1R+∞[x]/R+

∞⊗R+

∞[x] S+∞[0] ' LS+

∞/R+∞

which, in particular, reads:LS+∞/R

+∞' Ω1

S+∞/R

+∞

[0].

The ideal m := (p1/pN), which singles out the support of the cohomology of LS+∞/R

+∞

onSpec(S+

∞), kills Ω1S+∞/R

+∞, as predicted.

Remark 1.6.2 We remark an important fact. In this example, it turns out that S∞ is aperfectoid R∞-algebra. However, the cotangent complex LS+

∞/R+∞

does not vanish! We’relucky, and such cotangent complex is concentrated in degree 0 in this case. We observe that:

Ω1S+∞/R

+∞⊗R+

∞(R+∞/p) ' Ω1

(S+∞/p)/(R

+∞/p)


and the right side is zero. Indeed, if R → S is a map of Fp-algebras such that on both R andS the pth power map is surjective, then Ω1

S/R ' 0. Indeed, you can write every s ∈ S as s′p,and Ω1

S/R is R-linearly spanned by the derivations ds, s ∈ S, ie. by ds′p

= 0, s′ ∈ S. This is atoy case in which we observe that, calling S := S+

∞ and R := R+∞, we have:

LS/R ⊗R (R/p) ' 0.

This implies, in particular, that the (derived) p-adic completion of LS/R is trivial (ie. theanalytic cotangent complex of R→ S vanishes). In our case this just reads Ω1

S/R ' 0.Remark 1.6.3 In Example 1.6.1 we noted how the cotangent complex captured the ramific-ation at the generic point of the locus $ = 0 in X ′∞ over X∞. In fact, we can developthe following little idea: define the (generalized) notion of discriminant of a finite locally freemorphism of schemes f : X → Y of constant rank n in terms of LX/Y , guided by the followingfact:

discX/Y ' Fitt0(H0(LX/Y )) = Fitt0(Ω1X/Y ),

the left side being the discriminant of X → Y , and the right being the 0th fitting ideal ofΩ1X/Y . For a geometric construction of discX/Y , recall that under the above assumptions we

may define a trace form (which can be defined more in general for quasi-finite flat separatedmorphisms):

Trf : f∗(OX)→ OY

and using the OY -structure on f∗(OX) we get a bilinear pairing:

f∗(OX)⊗OY f∗(OX)→ OY ,

inducing a bilinear pairing:

∧nf∗(OX)⊗OY ∧nf∗(OX)→ OY ,

whose image is a locally principal ideal sheaf on Y , discX/Y . For example, if f is étale, then,by computing discX/Y fiberwise we see discX/Y ' OY . If Y is a Dedekind scheme, then werecover the classical notion of discriminant in algebraic number theory.

We claim discX/Y := det LX/Y recovers the above definition whenever f : X → Y is finitelocally free of constant rank (the reader is encouraged to check this on his own: one needs tomake sure LX/Y is a perfect complex, which is the case if f is locally a complete intersection,or if f is smooth, syntomic, étale. In all these cases LX/Y is a perfect complex of finite locallyfree OX -modules concentrated in degrees −1, 0).

This gives motivation to consider, in any event, the full cotangent complex when studyingramification, as in Example 1.6.1, and so we’ll do.

In these notes, we explain the intervention of two key tools which are at the very core ofthe “classical” theory of perfectoid spaces. One is the theory of almost étale extensions, andparticularly Faltings’ almost purity in the positive characteristic case. As the reader willprogressively realize, the almost purity in the zero characteristic case is proved by means ofan “henselian approximation” process we shall explain at the appropriate moment in §6, whichGabber and Ramero prove as a generalization of Elkik’s [Elkik, Thm. 7] to the non-noetheriansituation. Its usefulness trascends the applications to almost mathematics. One can regard suchprinciple as a vast generalization of Krasner’s Lemma, and we now discuss an easy example toexplain how, already in the following basic case, Krasner’s Lemma intervenes substantially.


Remark 1.6.4 The setup being as in Example 1.6.1, the tilt C[ of C := lim−→ kn, comes togetherwith its integral structure C[0. Recall that

C[0 ' lim←−x 7→xp

C0/p

as multiplicative monoids. We shall soon review the definition of the multiplicative map, andof the ring structure on both sides, whose underlying monoid structure is preserved by suchmap.

The natural ring map:C[0 → C0/p

given by the projection onto the first component, allows us to lift p to an element t of its kernel,so that:

C[0/t'−→ C0/p ' Zp[p

p−∞ ]/p.

Following the philosophy of replacing p with t, we have:

C[0/t ' Fp[[t]][tp−∞ ]/t.

As a consequence,C[ = Fp((t))(t

p−∞)∧.

Let us focus on the tilting operation (·)[ in this case, by attempting a hands-on definition ofthe tilt of any finite extension C ′ of C, which we denote C ′[. Again the philosophy is replacingp with t. Let us assume C ′ is given by adjoining the roots of some polynomial, say x2 − p1/p,assuming p 6= 2, as in our Example 1.6.1. Then C ′[ should be given by adjoining the roots ofx2−t1/p. Clearly, such assignment is manifestly not well defined. C ′[ is far from being uniquelydetermined by C ′, as C ′[ is given as well by adjoining the roots of x2 + (p− 1)t1/p, whereas byno means the extension of C given by adjoining the roots of x2 + (p− 1)p1/p should be C.

Instead, let’s consider the polynomials x2 − (p1/p)1/pn = x2 − p1/pn+1

, for n ≥ 0. For eachinteger n ≥ 0, we have that the splitting field C ′n of x2 − p1/pn+1

and C ′n+1 of x2 − p1/pn+2

willbe “closer and closer”, and eventually stabilize by Krasner’s Lemma. We can eventually assign:

C ′[

:=(

lim←−C′0n/p)

[1/p] ' C[(t1/2p).

Remark 1.6.5 We conclude the introduction with one last comment about purity. Let’s stepback to Lemma 1.5.4. If we consider a perfectoid ring A, to use the notation of the Lemma(the notion of perfectoid ring to be introduced in Definition 2.3.6 and the discussion in §2)then A basically satisfies none of the assumptions of the Lemma. However, we’ll assume $ isA-regular, for reasons coming from geometry ($ will play the role of the parameter q in Z[[q]]mentioned in the introduction), so let’s focus on the basic case A is the valuation ring of aperfectoid field.

A is complete with respect to the $-adic topology, $ being in rad(A) and A-regular. Uponallowing almost étale covers in Definition 1.5.1, then we’ll see that the pair (Spec(A/$), m),m being the maximal ideal of A and m the reduction modulo $, is “almost pure” (this isProposition 4.3.2). The whole of §6 explains how this implies that the pair (Spf(A), m) is“almost pure” (this is Theorem 6.1.12), which implies the pair (Spa(A[1/$], A), vm) is pure.

Roughly, perfectoidness of A is, upon allowing for the moment almost mathematics into thepicture, a good substitute to the assumptions in Lemma 1.5.4. We’ll expand on this.


1.7 Prerequisites

The first part of these notes is largely expository, and intended to be as self contained aspossible (and reasonable), and as elementary as possible. Basic notions from rigid geometrywill be reviewed when necessary. The theory of the cotangent complex is reviewed and mostlyre-developed from scratch, and its application to deformation theory explained. Familiaritywith the basic theory of derived categories is necessary.

The second part of these notes, under the name “Foundations of perfectoid geometry, II”, buildson the first but is original in nature. It requires good familiarity with classical rigid geometry,for motivational reasons and for gaining a good understanding of the materials. Particularlyideas borrowed from Raynaud’s approach via formal models.

Familiarity with Raynaud and Gruson’s paper on flattening techniques is auspicable.

A very good familiarity with étale cohomology of schemes, good familiarity with étale cohomo-logy of p-adic analytic spaces, and étale cohomology of adic spaces, will be rewarding, thougheverything will be reviewed when necessary, and/or referenced.

For the final chapter, a very good familiarity with the theory of algebraic spaces is required.Moreover, familiarity with Voevodsky’s h-topology and h-descent, together with de Jong’stheorem on alterations, its proof, and Gabber’s improvement, is highly recommended.

We’ll try to avoid the use of higher algebra as much as possible (completely successfully in thisfirst part, slightly unsuccessfully in the second due to descent theory) to make the materialsmore immediately accessible. Note, however, that a number of the main results in the secondpart of these notes will actually find their most natural form in the language of higher algebra.


2 Perfectoid rings

For later purposes, we discuss all the results in this handouts and in subsequent lectures inthe wider generality of perfectoid rings, a class of rings we’re about to introduce. To thisextent, we discuss a new notion of perfectoidness for rings following Gabber and Ramero, andmake our new notion consistent with those already present in the literature. In §2.1 we discuss“Tate-perfectoid” rings, and in the subsequent §2.3 we define perfectoid rings and relate the twonotions. A careful study of Fontaine’s functor Ainf(·) will be needed, and discussed partiallystarting from the treatment in [BMS].

2.1 Tate-perfectoid rings

In Scholze’s original paper, he introduced perfectoid fields and perfectoid algebras over a per-fectoid field. Fontaine’s Bourbaki report [Fontaine], removed the assumption of having groundfield, which turned out to be very useful for the purposes of taking into consideration interestingexamples coming from the theory of formal schemes.

We’re about to give a uniform framework to deal with Fontaine’s notion of perfectoidness,which so far takes into account Tate Huber rings only. The reader should nevertheless keep inmind the initial case of perfectoid algebras over a perfectoid field, and more to the point, ofperfectoid fields, as this basic case will turn out to be conceptually fundamental in sight of asofisticated generalization of Krasner’s Lemma, which lies at the heart of the whole theory.

To give an outline of what the field case is about, take any arithmetically profinite extension Kof Qp with residue field κ. One can attach to it its field of norms K, which, non-canonically, isisomorphic to κ((t)). Fontaine and Wintenberger show that the Galois theory of K is identifiedwith that of K. The completion E of K is a perfectoid field, and the completion of the purelyinseparable closure of K is what we shall call its tilt E[. Another way to say that the Galoistheories of these fields coincide is that the small étale sites of K, K,E,E[ are (canonically)identified (!). This phenomenon will be likewise true for Tate-perfectoid rings and their tilt,and in fact for perfectoid spaces and their tilt, as we shall discuss in later lectures. It willhold likewise true for perfectoid rings, upon allowing a relaxed notion of étaleness coming fromalmost mathematics.

Notation and conventions As usual for Huber rings, we denote the Huber rings involvedusually with letters A,B,C,D, until we set up the theory of Tate-perfectoid rings completely,in which case we shall denote Tate-perfectoid rings with the letters R,R′, S etc.

For a Huber ring A, we always write A0 for the subring of power-bounded elements of A, andA+ shall denote any open and integrally closed subring of A0. In this section, all Huber ringswill be Tate with respect to a regular pseudo-uniformizer $ in A0, a ring of definition for A.$ will induce the $-adic topology on A0, which will in turn induce the respective Huber ringtopology on A.Definition 2.1.1 A Tate-perfectoid ring is a complete Tate ring A (Banach with topologicallynilpotent unit $ ∈ A) satisfying the following properties:

(1) A0 is bounded.

(2) There exists a topologically nilpotent unit $ with $p | p in A0.

(3) The pth power map ϕ on A0/$ is surjective.


Note that A0/$ is indeed an Fp-algebra, thanks to condition (2), so the pth power map isa ring homomorphism. As a first important remark, we note that condition (3) is actuallyindependent of $. We prove, more in general, the following:Proposition 2.1.2 For any complete Tate ring A and nonzero pseudo-uniformizer $ satisfying$p | p in A0, the Frobenius map Φ : A0/$ → A0/$p is necessarily injective. The surjectivitycondition is independent of the choice of such $.

Proof. Let x ∈ A0 be satisfying xp = $pa for some a ∈ A0. Then the element x/$ ∈ A lies inA0 because its pth power does. Hence ϕ is injective as claimed. We claim the surjectivity of ϕis equivalent to surjectivity of the (necessarily injective) pth power map:

A0/(p,$n)→ A0/(p,$np)

for any n ≥ 1. For n = 1 this coincides with surjectivity of ϕ. Suppose such map is surjectivefor some n. Then it is surjective for all 1 ≤ m < n. We only need to check surjectivity for allm > n. Let x ∈ A0/p. We can write:

x = ap +$npb, a, b ∈ A0/p.

Likewise, b = cp +$pu, for some c, u ∈ A0/p. Hence:

x = (a+$nc)p +$(n+1)pu.

Surjectivity for m = n+ 1 follows, and the claim is proved. We are now ready to conclude theproof. Suppose $′ is another pseudo-uniformizer satisfying $′p | p in A0. If we take n largeenough, then $n ∈ $′A0, and surjectivity of A0/(p,$n)→ A0/(p,$np) implies surjectivity ofA0/$′ → A0/$′

p. It follows that Φ is an isomorphism for all $ satisfying $p | p in A0, assoon as it holds for one such $. The proof is complete.

Notation Given Proposition 2.1.2 and its proof, and in the context of the same Proposition,we shall always denote by ϕ the pth power map on A0/$, and likewise on A0/p, unless otherwisespecified.Example 2.1.3 Here is an example of a Tate-perfectoid ring which doesn’t arise as an algebraover a (perfectoid) field:

A = Zcycp 〈(p/x)1/p∞〉∧[1/x].

One can take $ := x1/p, as $p = T divides p in A0. It is a good exercise to figure out therational subset U of

X := Spa(Zcycp [[x1/p∞ ]]∧,Zcyc

p [[x1/p∞ ]]∧)

which realizes A as OX(U). We note here that the ring Zcycp [[x1/p∞ ]]∧, the (p, x)-adic completion

of Zcycp [[x1/p∞ ]], “wants” to be a perfectoid ring, but it’s not a Tate Huber ring, hence doesn’t

lie in the class of Tate-perfectoid rings. The notion we shall introduce in §2.3 will include thiskind of examples.

We are now ready to (re)introduce the notion of perfectoid fields. Naively, one could assignsuch notion by saying that a perfectoid field is a perfectoid ring which, in addition, is a field.However, it is not clear at all, a priori, if such a field is nonarchimedean with respect to a rank 1valuation, and this latter property is quite essential to our purposes, for example when dealingwith a crucial Henselian approximation process aiming at “spreading” finite étale algebras overperfectoid rings out of a certain perfectoid field. We, therefore, build this condition into thedefinition, for now. In fact, it turns out that a Tate-perfectoid ring which is a field, is alwaysa perfectoid field, as we shall discuss later, for completeness. The key observation is that any


Huber ring is Banach with respect to a submultiplicative real-valued norm, and we’ll provethat for Banach rings whose underlying ring is a field, the uniformity condition does the joband makes them into nonarchimedean fields.2

Definition 2.1.4 A perfectoid field is a perfectoid ring C that is a field and its topology isdefined by a rank 1 valuation

| · | : C → R×≥0.

Example 2.1.5 Some examples and a number of non-examples:

(1) An infinite family of non-examples is given by any discretely-valued nonarchimedean fieldk of residue characteristic p. Indeed, let Ok be the valuation ring and $ be any nonzeroelement of its maximal ideal. It follows the quotients Ok/$ and Ok/$p are Artin localrings of different lengths, hence they can never be isomorphic.

(2) We consider Qp(p1/p∞) (with $ = p1/p) and Qcyc

p , with $ coming from the Z/pZ piece ofthe (Z/p2Z)×-field extension Qp(ζp2)/Qp. Their $-adic completions are both perfectoidfields.

(3) We consider:Qp〈x1/p∞〉 = lim−→

n≥1

Qp〈x1/pn〉∧ = lim−→n

Zp[x1/pn ]∧[1/p].

This is not a Tate-perfectoid ring. However,

Qcycp 〈x1/p∞〉 = Qp〈x1/p∞〉⊗ZpZ

cycp

is. This is also obtained as A[1/p], A being the p-adic completion of Zcycp [x1/p∞ ]. This

class of examples lies in the class of rings known as “Tate-preperfectoid”, and we shalldiscuss their theory in general in a later version of these notes.

A few important properties of Tate-perfectoid rings.Lemma 2.1.6 Any Tate-perfectoid ring A is reduced.

Proof. Since A = A0[ 1$ ], it’s enough to check that A0 is reduced. Choose a ∈ A0 such that

aN = 0 for N 1. Then (a/$n)N = 0 for any n ≥ 1 and N 1, and so a/$n is power-bounded for all n ≥ 1. Then a ∈

⋂$nA0, which is 0, since $ is topologically nilpotent.

Some preliminary remarks, before moving on.Remark 2.1.7 Why doesn’t this prove that any affinoid algebra is reduced? We are using thatA is endowed with the (unique) Huber ring topology restricting to the $-adic topology on A0.Note also that if A was a Tate-perfectoid algebra over a perfectoid field C, reducedness of A wasa consequence of the following simple observation. Suppose ε ∈ A0 was a nilpotent element.Then Cε ⊂ A0 is a power-bounded subset which is not bounded, a contradiction. We referthe reader to the paper [BV] for a wider explanation of the importance of the boundednesscondition on A0 to hope the Huber pair (A,A0) to be sheafy. Particularly, [BV, Prop. 13], and [BV, Prop. 17], which explains requiring boundedness of A0 is not enough to ensuresheafness of (A,A0), but we should rather require such boundedness to be stable under rationallocalization.Remark 2.1.8 In Sholze’s first paper [Sch], he defined a perfectoid field C of residue char-acteristic p > 0, by considering, rather, the pth power map on C0/pC0. This notion turns

2Complete proof being included.


out to be equivalent to our Definition 2.1.4. The only verification needed is to check that ϕ issurjective. Pick x ∈ C0, so that:

x ≡ yp +$pz, z ∈ C0.

In turn, z = wp +$pt, sox ≡ (y +$w)p +$2pt mod p.

Iterating, we get that x ≡ xpn +$np(· · · ) mod p where xn is a Cauchy sequence in C0. If thecharacteristic of C is 0, so that |p| 6= 0, then $np is in pC0 for n 0 and x is a p-th powermodulo pC0, as desired. If the characteristic is p > 0, then xn → y ∈ C0 and x = yp.

Therefore, a perfectoid field in characteristic p is perfect (the converse being clearly also true).Proposition 2.1.9 If C is a perfectoid field of charateristic p, then the pth power map ϕC0 :C0 → C0 is bijective, so C is perfect and |C×| is p-divisible.

Proof. The proof of perfectness of C discussed in Remark 2.1.8, shows that ϕ is surjective,hence bijective by the same argument as in Remark 2.1.2. As a direct consequence, if we letα ∈ |C×|, then α1/p is also in |C×|, as we can write α = |x| for some nonzero x ∈ C, and writex = yp for some nonzero y ∈ C. We conclude.

In fact, the above p-divisibility result for the value group of a perfectoid field in characteristicp > 0 holds in general (!). Not just a curious phenomenon, but rather a first manifestation of adeeply conceptual connection between perfectoid spaces of arbitrary characteristic and residuecharacteristic p > 0, and perfectoid spaces of characteristic p > 0.Proposition 2.1.10 Γ = |C×| is p-divisible for any perfectoid field.

Proof. We’ve already handled the characteristic p case, so we can restrict our attention tocharacteristic 0. Γ is generated by |p| and |p| < |x| ≤ 1. Since C0/pC0 has surjective pthpower map:

|x| ≡ yp mod pC0 yields |x| = |yp| = |y|p

by the triangle inequality. Now we only have to check the same for |p|. We write p = $pa forsome a ∈ C0. Then, |p| = |$|p|a| ∈ Γp and both |$| and |a| must be between |p| and 1, so |p|is also in Γp by convexity.

The above Remark 2.1.8 and Proposition 2.1.9 can be reproved with no base field assumption,that is, suitably, for perfectoid rings. There’s a subtlety to be careful about. We show thefollowing:Proposition 2.1.11 Let A be a topological ring with pA = 0. The following are equivalent:

(1) A is perfectoid.

(2) A is a perfect (the pth power map ϕ, which we call Frobenius in this case, is an Fp-automorphism of A0) uniform complete Tate ring.

This is the exact analogue of Proposition 2.1.9. Note that in condition (2) the notion ofperfectness is assigned saying ϕ is an isomorphism on A0, where A0 is defined because we areunder the assumption that A is a topological ring.

Proof. Let A be a complete uniform Tate ring. If A is perfect, then just take $ to be anypseudo-uniformizer. The condition $p | p is trivially satisfied, as p = 0. If x ∈ A is power-bounded, then so is xp, and conversely. This means that ϕ : A0 → A0 is an automorphism. In


particular, the pth power map A0/$ → A0/$p is surjective. For injectivity, assume x ∈ A0

with xp = $pa for some a ∈ A0. Write a = bp for some b ∈ A0. Then we have:

x = $b,

as desired. Conversely. Suppose A is perfectoid. Then the pth power map A0/$ → A0/$p isan isomorphism, and hence so is A0/$n → A0/$np by induction. Taking inverse limiits andusing completeness, we find that ϕ : A0 → A0 is an automorphism, as desired.

Note how the last part of the proof is just a reformulation of the argument in Remark 2.1.8.We now remark that if we assume pA = 0, then trivially pA = pA0 is a closed ideal in A0.We’d like to characterize Tate-perfectoid rings as being a special class of complete uniform Taterings, but such characterization in the lack of the assumption pA = 0 encounters the difficultyof having, possibly, pA0 non-closed in A0.Proposition 2.1.12 Let A be a complete uniform Tate ring.

(1) If there exists a pseudo-uniformizer $ ∈ A such that $p | p and ϕ : A0/p → A0/p issurjective, then A is Tate-perfectoid.

(2) If A is Tate-perfectoid, then ϕ : A0/p → A0/p is surjective under the additional assump-tion that the ideal pA0 ⊂ A0 is closed.

Proof. For (1), assume ϕ : A0/p → A0/p is surjective. Then ϕ : A0/$ → A0/$p is surjectiveas well. Injectivity is automatic by Proposition 2.1.2. For (2), we assume A is perfectoid suchthat pA0 is a closed ideal of A0. We let $ ∈ A be a pseudo-uniformizer such that $p | p, andsuch that ϕ : A0/$ → A0/$p is an isomorphism. By the proof of Proposition 2.1.2, the pthpower map:

A0/(p,$n)→ A0/(p,$np)

is an isomorphism for all n ≥ 1. We now take inverse limits over all n ≥ 1. The proof will becomplete upon showing that the natural map:

A0/p→ lim←−A0/(p,$n)

is an isomorphism. Since A0 is $-adically complete, so is A0/p (pA0 is closed!), and hencesuch map is surjective. We claim A0/p is also $-adically separated, whence injectivity wouldfollow. We haven’t used uniformity at all yet! Suppose x lies in the kernel. Then there existsyn, zn ∈ A0 such that:

x = $nyn + pzn, n ≥ 1

which is to say: x ∈ (p,$n) for all n ≥ 1. $ is topologically nilpotent, and since A is uniform,A0 is bounded and $nyn → 0 in A0 as n → ∞. We find that x = limn→∞ pzn lies in theclosure of pA0, which is closed. We conclude.

The natural question one can ask is whether or not there exist Tate-perfectoid rings A in whichpA0 is not closed as an ideal of A0. Here is an example:Example 2.1.13 True, we can take Example in §7 of the Stacks project’s “Examples” andreadapt it.This was suggested by Peter Scholze I think, somewhere. Another example is builtby deformation theory (look into ExalAinf

((p, ζp2 − 1)Zcycp ,Zcyc

p )). 3

3Ready. Ale type it or you’ll forget it somewhere.


2.2 Tilting Tate-perfectoid rings

In this section, we define a functor:

Tate-perfectoid rings // Tate-perfectoid Fp-algebras

perfectoid fields?

OO

// perfectoid fields of char. p?

OO

denoted A 7→ A[. Note that the above diagram contains a number of nontrivial statements init. There exists such assignment A 7→ A[ sending a Tate-perfectoid ring to a Tate-perfectoidFp-algebra, and such assignment should behave well enough to ensure that such ring A is afield if and only if A[ is. More than this, such A is a perfectoid field if and only if A[ is, whichmeans we’ll have to keep track of how this functor interacts with valuations.Theorem 2.2.1 For C a perfectoid field of characeristic 0, there exists an equivalence ofcategories:

perfectoid C-algebras → perfectoid C[-algebras .

We shall refer to such equivalence with the name of “tilting equivalence”, from now on.

The inverse functor depends on the “untilt” C[# of K[, as for different C one can obtain thesame C[. Fontaine gives an exhaustive description of all the characteristic 0 fields that give aparticular C[.Remark 2.2.2 For a perfectoid field C of characteristic 0, the equivalence

finite separable C ′/C 'finite separable D′/C[

is a Theorem due to Fontaine and Wintenberger. The equivalence respects degrees in bothdirections, and in fact the Galois theories on both sides. We shall expand on this later, asthis will turn out to be essential in the sequel. For now, the reader should keep in mind, asmotivation to believe in the above statement and in its geometric generalization, that the étalesite is insensitive of radicial base change. As a baby case, base-changing a positive characteristicfield extension along a purely inseparable field extension doesn’t affect the Galois theory, andthis should be regarded, at least in positive characteristic, as being the idea underlying theabove equivalence.

A consequence of the theorem is that there is a homeomorphism

Cont(C) ' Cont(C[)

respecting rational domains in both directions.

Choose $ ∈ A a pseudo-uniformizer such that $p | p, so that $ ∈ A0 and A0 has the $-adictopology.Definition 2.2.3 We define:

A[0 = lim←−ϕ

A0/$A0 = (an)n≥0 | an+1p = an.

Note that on A[0, we have a canonical p-th root: if a = (an)n≥0 then a1/p := (an+1)n≥0.


Lemma 2.2.4 The multiplicative map of monoids:

lim←−a 7→ap

A0 → lim←−ϕ

A0/$A0 =: A[0$

sending (a(n)) 7→ (a(n) mod $)n≥0 is a homeomorphism (of topological monoids).Remark 2.2.5 We note that the left side is independent of $, so A[0$ is independent of $ as aset. What about the additive structure? If $′ | $ in A0, then we get a commutative diagram

lim←−A0 //

""

A[0$

A[0$′

It is easy to see explicitly that replacing $ by some power doesn’t affect the process at all.

We then define A = A0[1/$].

Proof. The key is to build a continuous 0-th component of the inverse. We can then apply thisconstruction to the canonical p-th root extraction on the right side.

Let x = (xn) ∈ A[0. We initially pick any sequence of lifts (xn). Then we define

x# = lim−→n→∞

xpn

n .

Let’s first check that this is well-defined, i.e. if (x′n) is another sequence of representatives thenwe claim that

(x′n)pn

≡ xpn

n mod $n+1A0.

More generally, we claim that if t′ ≡ t mod $A0 then

(t′)pi

≡ tpi

mod $i+1A0 for alli.

For i = 0, the claim is trivially true. Assume it’s true for i− 1. Then,

(t′)pi−1

≡ tpi−1

+$ia

and therefore(t′)p

i

= tpi

+ p$i(· · · ) +$ipap.

The upshot is that xpn

n mod $n+1A0 is intrinsic to x. Also, (xn+1)p is a representative ofxn+1

p = xn, so xpn+1

n+1 = (xpn+1)pn ≡ xp

n

n mod $n+1A0. This gives the Cauchy property forthe $-adic topology.

We have constructed x 7→ x# sending A[0 → A0. It is an easy exercise to check the remainingdetails.

As a first step towards the tilting equivalence, we seek to define some $[ ∈ A[0, A[-regular,satisfying the following properties:

(1) A[0[1/$[] is Tate-perfectoid using $[ with A[0 the subring of power-bounded elements.

(2) There is a natural isomorphism:

A[0/$[ ' A0/$A0

using the 0-th projection A[0 → A0/$A[.


Morally, we want $[ = ($,$1/p, $1/p2 , . . .). Unfortunately we don’t know that we have allthese pth roots in A0 (in classical p-adic Hodge theory, when one is working with OCp , one cando this very explicitly). We try to construct an element with similar behavior.

Since ϕ on A0/$p is surjective, there exists $′1 ∈ A0 such that

($′1)p ≡ $ mod $pA0.

Therefore:

($′1)p = $ +$pA0

= $(1 +$p−1A0)

Inductively, choose $′i+1 such that ($′i+1)p ≡ $i mod $pA0, so that this is an element ofA[0. Therefore, it comes from a unique element ($1, $2, . . .) ∈ lim←−x7→xp A

0 such that $i ≡$′i mod $p. We define this element to be $[.Lemma 2.2.6 $[ is A[0-regular and topologically nilpotent in A[.

Proof. Using the multiplicative and topological identification:

A[0 ' lim←−A0/$

via$[ ↔ (0, $1, $2, . . .)

we see that ($[)pn

starts with n − 1 zeros, so $[ is topologically nilpotent. To show that $[

is not a zero-divisor in A[0 ' lim←−x 7→xp A0 (this being a multiplicative identification), it suffices

to observe that $0 is not a zero-divisor in A0 and that A0 is reduced.

Example 2.2.7 If C is a perfectoid field, we are going to see that C[ := C[0[1/$[] is aperfectoid field of characteristic p with |C[×| = |C×|, and |$[| = |$|.Remark 2.2.8 In general, $# = $0 is always $ times a unit of A0. That implies that thetopology on A[0 is the $[-adic topology.

In fact, we claim that:

($[)pn

A[0 = (0, . . . , 0, . . .) ∈ lim←−A0/$.

with zeros in the first n+ 1 slots.

This corresponds exactly to:

(a(0), . . . , a(n−1), a(n), . . .) ∈ lim←−A0 | a(n−1) ≡ 0 mod $A0.

This shows that:A[0/$[ ∼= A0/$.

Indeed, we just checked that $[ generates the kernel, so this map is injective. On the otherhand, it is surjective because A0/$A0 is surjective - starting with anything in A0, we can builda corresponding element on the left hand side.Definition 2.2.9 Define the tilt of A to be:

A[ = A[0[1/$[] ⊃ A[0

with the $[-adic topology on A[0. This is a complete Tate ring with $[ as a pseudo-uniformizerand A[0 the ring of definition.


To conclude that A[ is Tate-perfectoid, we must show that A[0 is actually the ring of power-bounded elements of A[ (it is obviously contained in it, as it is visibly a ring of definition). Forconcreteness, we remark that this is the set of p-power compatible sequences in A (just as A[0was the set of p-power compatible sequences in A0). Since

($[)# = $0 ∈ $(A0)× ⊂ A×,

then A = A0[1/$] = A0[1/($[)#]. Therefore, $[ ∈ A[0 maps to (($[)#, (($[))1/p)#, . . . , ).

This being a multiplicative identification, it extends uniquely to a multiplicative map

A[ → lim←−x 7→xp

A

using (∗∗) and that is sufficies to check that the first term in the sequence is power-bounded,as the rest follow automatically from this.

We now want to show that A[ is independent of the choice of $ and $[ and that its ring ofpower-bounded elements is precisely A[0.Proposition 2.2.10 For any two $,$′ ∈ A0 and $[, ($′)[ are associated choices in A[0, thenA[0[1/$[] ' A[0[1/($′)[].

Proof. We have already seen that ($[)# differs from $ by a unit in A0. Then we can replace$ by ($[)#, so we can assume that $ = ($[)# (that is, we have compatible p-power roots of$, and the sequence of these is $[).

The key is that we can now use topological nilpotence and the multiplicativity of all theidentifications. By topological nilpotence in A0, we can find N 0 such that ($′)N lies in$A0, say ($′)N = $ · a for a ∈ A0. (Note that a is a unit in A because $,$′ are.) Using thespecified p-power roots of $,$′ giving $[, ($′)[, we get such a sequence for a as well. Callingit a[, and using the multiplicativity, we get:

(($′)[)N = $[a[.

Likewise, one also has($[)M ∈ ($′)[ ·A[0.

This concludes the proof.

It remains to show that A[0 is actually the ring of power-bounded elements in A[.Proposition 2.2.11 We have A[0 = (A[)0.

Proof. Certainly we have: A[0 ⊂ (A[)0. Recall that we constructed the map:

A[0 → A0

sending x = (an)n 7→ limn→∞ apn

n =: x#. We likewise denote A[ = lim←−x 7→xp A → A the maptaking the 0th component as x 7→ x#.

We have a map A[ → A sending x 7→ x#. The key point is that the diagram:

A[ // A

A[0

OO

// A0

OO


is cartesian, as if a(0) is power-bounded, then so are the rest of the terms in any p-powersequence.

Suppose ξ ∈ A[ is power-bounded. We then want to conclude that ξ(0) ∈ A0 (which impliesthat ξ ∈ A[0 by the cartesian property). By definition, ξ is power-bounded if and only if

ξnn≥1 ⊂ ($[)−NA[0

for some uniformN . As this is a purely multiplicative statement, we can apply #: (ξ#)nn≥1 ⊂$−NA0, and by the same argument this is equivalent to ξ# being power-bounded in A. Weconclude.

We saw thatA[0/$[A[0 ' A0/$A0.

Corollary 2.2.12 If A0 is local, with maximal ideal m, then A[0 is also local, say with uniquemaximal ideal maximal ideal m[ and canonically A[0/m[ ' A0/m.

Proof. It suffices to show that $ lies in every maximal ideal of A0 and $[ lies in every maximalideal of A[0 (the “Jacobson radical”). Recall that this is equivalent to 1 + $x being a unit of(A0)× for all x ∈ A0, and 1 + $[y ∈ (A[0)× for all y ∈ A[0. But this is true (using the usual“geometric series” expansion) since A0 is $-adically separated and complete, and similarly forA[0.

Example 2.2.13 Let C be a Tate-perfectoid ring which is a field, e.g. a perfectoid field (butrecall that the latter is, a priori, more stringent, until we prove they are equivalent notions).Then

(C[)× = ( lim←−x 7→xp

C)× = lim←−x 7→xp

C× = C[ − 0.

This is obviously perfect. Note that if char(C) = p, then C[ = C.

Now assume that C is a perfectoid field, with | · | : C → R×≥0 defining the topology (i.e. we canchoose an inclusion C×/C0× → R×>0) as ordered abelian groups. (This isn’t quite canonical;it would be better to say that the value groups are canonically isomorphic, but this is moreconcrete.) We claim that C0[ is also a rank 1 valuation ring and we have a canonical absolutevalue | · |[ : C[ → R×≥0 inducing C[0, such that |C[×| = |C×|.

Proof. We know that $[ ∈ C[0 is a topologically nilpotent element defining the topologymaking C[0 complete. Define

| · |[ : C[ → |C|

by x 7→ |x#|. This is obviously multiplicative and definite. Also, |x|[ ≤ 1 ⇐⇒ x# ∈ C0 ⇐⇒x ∈ K[0. We need to check that:

‖x+ y‖[ ≤ max(|x|[, |y|[).

If x or y is 0 then this is trivial, so without loss of generality x, y 6= 0 and |x#| ≤ |y#| (sox/y ∈ C[0). Divide by |y|[ = |y#|, so(

x+ y

y

)#

=

(1 +

x

y

)#

∈ C0

so |x+ y|[ ≤ |y|[ = max(|x|[, |y|[).


Proposition 2.2.14 If C is a perfectoid field, then Cont(C) ∼= Cont(C[) via

v 7→ v[ : x 7→ v(x#).

Proof. Cont(C) is exactly the set of open valuation subrings of C0. But this must containtopologically nilpotent elements, hence the entire maximal ideal. These are just the valuationsubrings of the residue fields, but then you can go to the other side.

To verify that v[ is a valuation (the multiplicativity and 0 axiom are trivial, as C is a field)

v[(x+ y) ≤ max(v[(x), v[(y))

one uses the argument from before.

We conclude with an important remark.Remark 2.2.15 As we shall discuss, the whole theory of perfectoid spaces lies on a crucialpassage between arbitrary characteristic and positive characteristic p > 0, which relies on thefollowing simple fact. Let A be a Tate-perfectoid ring, and $ ∈ A a pseudo uniformizer as inDefinition 2.1.1. Then, given the constructions and results presented above, we have a naturalisomorphism:

A0/$ ' A[0/$[.

We’ll see that the marvelous correspondence between perfectoid spaces in arbitrary character-istic and characteristic p, relies on a process of “almost integral” extension which then passesthrough the above isomorphism. The observation that in a highly ramified at p algebra over Qp,as Qp(p

1/p∞)∧, p is extremely close to being zero, goes back to Deligne and roughly expressesexactly the isomorphism:

Zp[p1/p∞ ]/p ' Fp[[t]](t

1/p∞)/t,

where $ = p and $[ = t.

2.3 Perfectoid rings

For the purposes of giving a clean definition of perfectoid ring, we first introduce Fontaine’sring Ainf . We fix a prime number p, and a commutative ring S which is $-adically completeand separated for some element $ ∈ S dividing p in S. (Note in particular that S is thenp-adically separated since pnS ⊂ $nS.) We also assume that S is p-adically complete.Remark 2.3.1 We are not assuming $ is S-regular! We’ll discuss the advantages of fallinginto the particular situation in which $ is regular, later.

We let ϕ : S/p→ S/p be the pth power map, and write S[ := lim←−S/pS, the inverse limit beingformed along the pth power maps. We call S[ the “tilt” of S, consistently with the notationadopted so far in the previous section §2.1. Note that S[ is a perfect Fp-algebra with Frobeniusstill denoted by ϕ.

The following definition with S the valuation ring of the completed algebraic closure of a p-adicfield recovers an important construction of Fontaine:Definition 2.3.2 We define Ainf(S) := W (S[).


Notation For all x ∈ S[, we denote by [x] the Teichmüller representative of x in W (S[).Lemma 2.3.3 Let S be p-adically separated and complete. Then the canonical maps:

lim←−x 7→xp

S → S[ → lim←−ϕ

S/$S

are isomorphism of monoids.

Proof. The same methods from the proof of Lemma 2.2.4 can be used to give the result.

Definition 2.3.4 For all x ∈ S[, write x = (x(0), x(1), . . . ) under the identification from Lemma2.3.3, and define θS : Ainf(S) = W (S[)→ S by (w0, w1, . . . ) 7→ (w0)(0).

In Fontaine’s work (with S the valuation ring of a completed algebraic closure of a p-adic field)θS plays a central role. In particular, it is essential that θS is a ring homomorphism. In [Serre,§5, Ch. II] (see especially Proposition 10 of loc. cit.) a universal property is proved for Wittrings of perfect Fp-algebras via the notions of “p-ring” and “strict p-rings” defined there.

We cannot appeal to that universal property (say making its main construction explicit in ourcontext) to prove that θS is a ring homomorphism because in the cases of most interest thetarget S is never a “p-ring” (because neither S/pS nor S/$S are perfect, and moreover we areeven allowing that p may not be regular in S). Hence, a direct argument is required to prove:Proposition 2.3.5 The map θS is a ring homomorphism for every S as above.

Proof. Let a := (a(i))i≥0 and b := (b(i))i≥0 be arbitrary elements of the set A := lim←−x 7→xp S,and define:

a · b := (a(i)b(i))i≥0, a+ b := ( limk→∞

(a(i+k) + b(i+k))pk

)i≥0.

The above assignements are easily checked to endow A with a commutative Fp-algebra structure(as p = p · 1 = 0 in A), with 0 := (0, . . . , 0, . . . ) and 1 := (1, . . . , 1, . . . ). This being said, themonoid isomorphism in Lemma 2.3.3 is an Fp-algebra isomorphism, with S[ being endowedwith the usual Fp-algebra structure, and A with the one from above. Indeed, the canonicalmap A→ S[ induced by reduction modulo p is multiplicative, and respects the sum as p divides(pr

r′

)for all r′ = 1, . . . , pr − 1, r ≥ 1. Its inverse is given by sending a element s := (s(i))i≥0 in

S[ to the element t := (t(i))i≥0, with t(i) := limk→∞(s(i+k))pk

, s(i) being any lift of s(i) fromS/p to S. This assignment is easily checked to be independent of the choice of lifts, and torespect the Fp-algebra structures.

Under the Fp-algebra isomorphism given by the inverse S[ '−→ A, we consider the Zp-algebrahomomorphism W (S[) → W (A), readily seen to be an isomorphism by easy inspection, fol-lowed by the map θS sending (w0, . . . , wk, . . . ) to w(0)

0 . We rename this composition as θS forsimplicity, and we check this is a Zp-algebra homomorphism, thus establishing the Proposition.

Given s in W (S[), we can write it as:

s =

∞∑j=0

pj [sj ]

for sj ∈ S[ for all integers j ≥ 0, the series being convergent with respect to the p-adictopology on W (S[), and being uniquely determined by s, as W (S[) is a strict p-ring withW (S[)/p ' S[ (this latter Fp-algebra isomorphism being unique, as W (S[) is the unique flatp-adically complete and separated Zp-algebra deformation of S[).


Let, therefore, s =∑∞j=0 p

j [sj ] be any element of W (S[), where we write, for each j ≥ 0:

sj = (s(i)j )i≥0,

with s(i)j in S/p for all i ≥ 0, satisfying (s

(i+1)j )p = s

(i)j . We choose lifts t(i)j of s(i)

j to S, for alli, j ≥ 0, and we first define the map of sets φS : W (S[)→ S:

φS(s) =

∞∑j=0

pj limk→∞

(t(k)j+k)p

k

.

It is an easy verification that the above definition is independent of the choice of lifts made.Moreover, φS is a ring homomorphism as S[ is a perfect Fp-algebra and W (S[) a strict p-ringwith W (S[)/p ' S[, as in [Serre, §5, Ch. II] and [?, Lemma 4.4.1].

Given that the definition of φS doesn’t depend on the choice of lifts (t(i)j )i≥0 of (s

(i)j )i≥0 for each

j ≥ 0, we aim at choosing such lifts in a particularly convenient way, so that we shall identifyφS and θS on all elements of W (S[). From this we shall conclude θS is a ring homomorphism.

We do a trick. Adopting the definition of big Witt vectors W (R) with R any commutativering, as given in [Ked, Def. 1.1], we are given a Witt vector Frobenius F : W (S) → W (S),and we can form the inverse limit lim←−F W (S). Applying F termwise, yields a bijective ringhomomorphism F : lim←−W (S)→ lim←−W (S), which, in the last instance, will enable us to selecta good choice of lifts. We write an element x ∈ lim←−F W (S) as a sequence (xk)k≥0 of elementsof W (S) satisfying the constraint F (xk+1) = F (xk), for all integers k ≥ 0.

The canonical map S → S/p and Witt vector functoriality induce a unique ring homomorphismε : lim←−F W (S) → lim←−F W (S/p). We claim it is bijective. It is clearly surjective, and as forinjectivity, fix an integer r ≥ 1, and let’s analyze the kernelW (pS/prS) of the mapW (S/pr)→W (S/p). We claim such kernel dies upon forming the inverse limit along the Witt vectorFrobenius. Indeed, there is some large enough integer N ≥ 0 such that pN is zero inW (S/prS),and we claim F r+N kills such kernel. It is generated by the elements V i[p] for i ≥ 0, andF r+NV i[p] = pi[p]r+N−i = 0, either because pi = 0 if i ≥ N , or because [p]r+N−i = 0 ifi < N . Upon forming the limit over r ≥ 1, exchanging the order of the limits and using thefact that formation of W (·) commutes with inverse limits of rings and S is p-adically separatedand complete, we establish bijectivity of ε.

We now analyze the definition of φS . We choose lifts (t(i)j )i≥0 of (s

(i)j )i≥0, for each j ≥ 0, such

that F (t(i+1)j ) = F (t

(i)j ) for all integers i, j ≥ 0. Since ε is surjective, this is possible.

In limk→∞(t(k)j+k)p

k

, the limit is inaffected upon renaming k as k − j, thus yielding:

φS(s) =

∞∑j=0

pj limk→∞

(t(k−j)k )p

k−j=

∞∑j=0

limk→∞

pj(t(k−j)k )p

k−j= limk→∞

k∑j=0

pj(t(k−j)k )p

k−j.

Now we observe that∑kj=0 p

j(t(k−j)k )p

k−jis exactly the pkth ghost component of (tk), whence

it is independent of k (easily seen upon iteratively applying F ). This is due to the particularshape of the lifts chosen, due, in turn, to surjectivity of ε. We now choose k = 0, and obtain:

φS(s) = t(0)0 .

Since ε is injective, t(0)0 is uniquely determined by s.


Let w be as in the statement of the Proposition, that is, w ∈W (A) of the form (w0, . . . , wk, . . . ),for wk ∈ S for all integers k ≥ 0. w corresponds to w := (w0, . . . , wk, . . . ) in W (S[) underthe inverse of the isomorphism W (S[)

'−→ W (A), and w satisfies (wk+1)p = wk for all integersk ≥ 0. Since w likewise satisfies wpk+1 = wk, upon choosing s = w we deduce t(0)

0 = w(0)0 again

using injectivity of ε (ie. w is the unique image of the element (wk)k≥0 ∈ lim←−F W (S) under

the composition of ε and the isomorphism W (S[)'−→ W (A)). It follows φS = θS , whence the

contention.

In what follows, we will generally write θ rather than θS (with S understood from context).Definition 2.3.6 A ring R is perfectoid if it satisfies the following conditions: it is p-adicallyseparated and complete, it is $-adically complete and separated for some element $ ∈ R suchthat $p divides p in R, the pth power map ϕ : R/pR → R/pR is surjective, and the kernel ofϑ : Ainf(R)→ R is principal.

The category of perfectoid R-algebras, denoted R-Perf , consists of perfectoid rings R′ equippedwith a continuous ring homomorphism R → R′ and has as morphisms continuous maps oftopological R-algebras R′ → R′′.

We note we are not assuming $ is R-regular. We now study the kernel of ϑ more closely, andestablish that when $ is R-regular it is easy to decide whether or not ker(ϑ) is principal.

Notation We shall denote Tate-perfectoid rings and perfectoid rings with the lettersR,R′, S, S′,unless otherwise stated. This may cause some confusion, as perfectoid rings shall always ariseas subrings of power-bounded elements (or open integrally closed subrings of these) of Tate-perfectoid rings, roughly. No confusion will arise, as we shall clarify the notation wheneverneeded.

We sum up the considerations in §2 into the following Proposition.Proposition 2.3.7 Let S be a ring that is p-adically separated and complete as well as $-adically complete and separated with respect to some element $ ∈ S such that $p divides p inS. The following are equivalent:

(1) Every element of S/$pS is a pth power.

(2) Every element of S/pS is a pth power.

(3) Every element of S/$pS is a pth power.

If the above equivalent conditions hold, then there exist units u, v ∈ S× such that u$ and vpadmit compatible systems of p-power roots in S.

Proof. Trivially, (1) implies (2) which in turn implies (3). This latter implies (1) by writingevery x ∈ S as:

x =

∞∑i=0

xpi$pi

for some xi ∈ S. This can be done by means of an easy induction as performed in §2.1. Thenx ≡ (

∑∞i=0 xi$

i)p modulo p$S, as desired. For the last statement, we use Lemma 2.3.3 to Sand S/$p, deducing that the canonical map:

lim←−x 7→xp

S → lim←−x 7→xp

S/$p


is an isomorphism. By (1) we deduce there exists π ∈ lim←−x 7→xp S such that π(0) ≡ $ modulo$pS, respectively π(0) ≡ p modulo $pS. Write π(0) = $+$px, respectivelu π(0) = p$px, forsome x ∈ S. We conclude by noting that 1 + px ∈ S×, and likewise 1 +$x ∈ S×.

The info to keep in mind from Proposition 2.3.7 is that if we require on S that the pth powermap S/$ → S/$p is surjective, for some $ with respect to which S is $-adically complete andseparated and with the property $p | p in S, then we automatically obtain in S a compatiblesystem of p-power roots of both $ and p, upon adjusting these by units.

This turns out to be captured by surjectivity of the Witt vector Frobenius, as in the following:Proposition 2.3.8 Let S be a ring which is p-adically separated and complete as well as $-adically complete and separated with respect to some element $ ∈ S such that $p dividesp in S. Then, the equivalent conditions in Proposition 2.3.7 are also equivalent to the mapϑ : Ainf(S)→ S being surjective.

Proof. Given that $p ∈ pS, (1) in Proposition 2.3.7 implies condition (xiv)′ in [Ked2, Def.3.1]. By Theorem 3.2 combined with Corollary 3.3 in loc. cit., such condition implies that ϑis surjective. Every element of S in the image of ϑ is a pth power modulo p, hence this andsurjectivity of ϑ imply (2) in Proposition 2.3.7, which concludes the proof. The details of theproofs in [Ked2] are elementary in nature. We omit them for ease of exposition.

Note that as a direct consequence of Proposition 2.3.8, the assumption on surjectivity of thepth power map S/$ → S/$p implies surjectivity of ϑ. Back to analyzing Definition 2.3.6,we’re saying that S satisfies the initial assumptions in both Propositions 2.3.7 and 2.3.8, whichin turn imply that ϑ is surjective, so Definition 2.3.6 is genuinely asking merely a condition onthe kernel of ϑ.

The following Proposition establishes a criterion to decide whether or not ker(ϑ) is principal.Proposition 2.3.9 Let S be a ring which is p-adically separated and complete as well as $-adically complete and separated with respect to some element $ ∈ S such that $p divides p inS. Suppose the pth power map S/$S → S/$pS is an isomorphism, and $ is S-regular. Thenker(ϑ) is a principal ideal.

Proof. For every integer k ≥ 1 we have that the pth power map S/$1/pkS → S/$1/pk−1

S isan isomorphism. It follows the kernel of S[ → S/$S is generated by $[, and the natural mapS[/$[S[ → S/$S is an isomorphism. Let x ∈ Ainf(S) be such that ϑ(x) = 0.

Now we use $p divides p. It follows ξ := p + u[$[]p lies in ker(ϑ), for some u ∈ Ainf(S). Weshall identify:

Ainf(S)/(p, [$[]p) ' S[/$[S[

and composing this with the map S[/$[pS[ → S/$pS induced by the canonical projectiononto the first component S[ → S/$pS, we obtain a commutative square:

Ainf(S)/ξϑ //

S

Ainf(S)/(ξ, [$[]p) // S/$pS

Since ϑ(x) = 0, and since the bottom row in the above diagram is an isomorphism, as shownabove, the image of x under the leftmost vertical row is zero, hence we can write:

x = ξy0 + [$[]px1.


We have ϑ([$[]px1) = 0, and this is $pϑ(x1), since ϑ([$[]) = $. Since $ is S-regular, wededuce ϑ(x1) = 0, and we can induct to conclude:

x = ξ(y0 + [$[]py1 + · · · )

which implies ker(ϑ) is generated by ξ, hence it’s principal. We conclude.

Remark 2.3.10 In the proof of Proposition 2.3.9 we defined ξ to be p + u[$[]p, for someu ∈ Ainf(S), the choise of u depending on S and expressing the fact that $p divides p in S.Suppose ker(ϑ) is principal. Then we claim ξ is a generator, and moreover that it is Ainf(S)-regular.

Indeed, let ξ′ be a generator for ker(ϑ), and let ξ′ = (ξ′0, ξ′1, . . . ) be its Witt vector expansion.

We can write ξ = aξ′ for some a ∈ Ainf(S), and the claim is that a is a unit. We take a lookat the Witt vector expansions of the above identities:

($[pu0, 1 +$[p2u1, . . . ) = (ξ′0a0, ξ′0pa1 + ξ′1a

p0, . . . )

where the left side expands p + u[$[]p = ξ, and the right side expands ξ′a. We deduceξ′1a

p0 = 1 +$[p2u1 − ξ′0

pa1.

This is a unit in S[ = lim←−S/$S: it’s enough to check that the reduction modulo $ is, andthat is 1. It follows ξ generates ker(ϑ) as soon as this is a principal ideal.

Note that in the Witt vector expansion for ξ, ξ1 = 1+$[u1 is a unit in S[. We call the elementsξ′ ∈ ker(ϑ) with ξ′1 ∈ S[× preferred elements, and we’ve just shown that whenever ker(ϑ) isprincipal, any preferred element is a generator. Note that ker(ϑ) always contains ξ, hence atleast a preferred element.

We are ready to show the following Proposition, which will clarify what’s the level of generalityof Definition 2.3.6.Proposition 2.3.11 Let S be a ring which is p-adically separated and complete as well as$-adically complete and separated with respect to some element $ ∈ S such that $p divides pin S. Assume the pth power map ϕ : S/$S → S/$pS is surjective. If ker(ϑ) is principal, thenϕ is an isomorphism, and any generator of ker(ϑ) is Ainf(S)-regular.

Proof. The isomorphism property is clear from the commutative diagram:

Ainf(S)/ξϑ //

S

Ainf(S)/(ξ, [$[]p) // S/$pS

In fact, it’s enough to check the pth power map: S[/$[S[ → S[/$[pS[ is an isomorphism.We know the top row in the above diagram is an isomorphism, by the assumption and Re-mark 2.3.10. If an element in Ainf(S)/(ξ, [$[]p) goes to zero in S/$pS, by commutativity itis $p-divisible in S, hence [$[]p-divisible in Ainf(S)/ξ. Then it’s zero in Ainf(S)/(ξ, [$[]p).Surjectivity is obvious.

We’re left to check that ξ is Ainf(S)-regular. For this we either write down the Witt vectorexpansion of a relation ξa = 0, and conclude ai = 0 for all i ≥ 0 using [$[]-adic completenessand separatedness of Ainf(S), or we use the following criterion, which applies to our case sinceperfect Fp-algebras are reduced, and

∑∞i=0 ξiS

[ = S[. We conclude.


We note that for any perfect Fp-algebra S, p is regular in W (S) by Zp-flatness. We prove thefollowing:Proposition 2.3.12 Let S be a reduced Fp-algebra, and let x := (xn)n≥0 ∈ W (S) be anyelement. Form the ideal J :=

∑n≥0 xnS. If J is not contained in any minimal prime ideal of

S then x is W (S)-regular.

Proof. Let Min(S) denote the set of all minimal primes of S. Since S is reduced, callingS′ :=

∏p∈Min(S) Sp we get that the natural map S → S′ is injective. We, therefore, get an

injective map W (S) → W (S′), and we’re reduced to check that the image of x in W (Sp) isregular for every minimal prime p of S. But if J is not contained in any such prime, then thisis clear.

Proposition 2.3.9, Proposition 2.3.11 and Proposition 2.3.8 imply that in the case $ is regular,then Definition 2.3.6 is equivalent to the notion of a $-adically complete and separated ring Rwith $p dividing p, and such that the pth power map R/$ → R/$p is an isomorphism. Notethat this latter condition is equivalent to asking that the pth power map R/$1/p → R/$ is anisomorphism.

This latter notion of perfectoidness is closely related with that of Tate-perfectoid rings, and weshall now explain how.Proposition 2.3.13 Let R be a $-adically complete and separated with $ a regular element.Assume $p divides p, and that R is perfectoid (so R is p-adically separated and complete).Form R[1/$] endowed with the unique Huber ring topology making R open and bounded withthe $-adic topology. Then R[1/$] is Tate-perfectoid.

Proof. Note that for any open integrally closed subring R+ ⊂ R0, for any x ∈ R0 there existsa sufficiently large integer n ≥ 0 such that $nx ∈ R+, ie. the discrepancy between R+ and R0

is almost zero. The result is proved by imitating the argument in Lemma 3.6.11.

The above Proposition 2.3.13 explains how to produce a Tate-perfectoid ring out of a perfectoidring, when $ is regular. The case in which $ is a regular element seems, therefore, to be quiteinteresting for geometric purposes, as Tate-perfectoid rings naturally come into play whenstudying the locus $ 6= 0 in Spf(R), for perfectoid R.

The following Proposition clarifies that the regularity assumption on $ is exactly the assump-tion we need to gain an exhaustive correlation between Tate-perfectoid rings and perfectoidrings.Proposition 2.3.14 Let R be a complete Huber ring which is Tate with respect to a pseudo-uniformizer $, with a ring of integral elements R+ ⊂ R. If R is Tate-perfectoid, then R+ isperfectoid. Conversely, if R+ is perfectoid and bounded in R, then R is Tate-perfectoid.

Proof. Assume R is Tate-perfectoid. As a first step, we prove R0 is perfectoid. Since it’sbounded, then it is $-adically complete and separated (separatedness follows from the factthat $ is topologically nilpotent). Since $ is R0-regular, by Proposition 2.3.9 it’s enough tocheck that the pth power map R0/$ → R0/$p is an isomorphism. It’s certainly surjective(since R is Tate-perfectoid). Suppose x ∈ R0 is such that xp = $py for some y ∈ R0. Thenz := x/$ ∈ R is such that zp is power-bounded, since it is y. It follows z ∈ R0, and hencex ∈ $R0, as desired. It follows R0 is perfectoid.

We now show that R+ is perfectoid for all open integrally closed subrings R+ ⊂ R0. Note thatthe discrepancy between R+ and R0 is almost zero, and we claim that such discrepancy doesn’taffect perfectoidness.


Since $R0 consists of topologically nilpotent elements, $R0 is contained in R+ by opennessof this latter. By Proposition 2.3.8, we know every element of R0/p$R0 is a pth power. Letx ∈ R+ be any element, and write x = yp + p$z for some y, z ∈ R0.

Since$z ∈ R+, we have that yp = x−p$z is in R+ as well, and hence so is y. It follows that thepth power map R+/p→ R+/p is surjective, hence so is the pth power map R+/$ → R+/$p,by Proposition 2.3.7. For injectivity, the argument is the same as that for R0, so we deduceR+ is perfectoid, as desired.

Conversely, we know R+ is bounded by assumption. Since $R0 ⊂ R+, it follows that so isR0. We only need to check that some topologically nilpotent unit $ ∈ R is such that $p

divides p in R0, and yields a surjective pth power map R0/$ → R0/$p. Such surjectivity isachieved by a similar argument as before to prove that the pth power map on R0/p is surjectiveand using Proposition 2.3.7. The only issue is to check that we can arrange $ to be a unit.On the other hand, any topologically nilpotent unit π ∈ R is in R+, and by design of R+

we have that the map ϑ : Ainf(R+) → R+ is surjective with kernel generated by a preferred

element ξ ∈ Ainf(R+). We claim there is an element π[ ∈ (R+)[ and a unit in (R+)× such

that ϑ([π[]) = uπ. For sufficiently large integer n ≥ 0, $ := ϑ([π[1/pn

]) will do the job. Onecan construct such π[ by inductively constructing its Witt vector expansion to ensure that itsimage under ϑ is a unit multiple of π. Some details omitted and left to the reader.

Assumption Throughout, we shall assume that in Definition 2.3.6 $ is regular.

The above is motivated by the fact that such assumption gives an immediate and exhaustiverelation between perfectoid and Tate-perfectoid rings, which the intervention of adic spaces isgoing to rely on.

As an immediate consequence, we’ll have the base change theorem for the cotangent complexavailable in many situations. We remark that, for perfectoid R, p is always regular in Ainf(R),and the whole theory still turns out to work as explained throughout these notes, the keybeing that for any map of perfectoid rings R → R′, we have L(R′/p)/(R/p) ' 0 in D(R/p), andProposition 2.3.7.

We never need to pursue the slightly more general path allowing $ to be a zerodivisor in aperfectoid R anyway.

It will be useful to have a gadget to keep track of regularity of $ in an intrinsic way. For thispurpose, we introduce the completed perfection of the ring Fp[[x]], which we call Λ. We regardΛ as being the completed direct limit of Fp[[x]] along the pth power maps, and we denote by x[the element (x, x1/p, · · · ). We call Λ0 the reduction modulo x of Λ. Note that Λ is a perfectoidring.

We note that given a perfectoid ring R, with tilt R[, $[ is R[-regular if and only if the mapΛ→ R[ assigned by x[ 7→ $[ is flat. We won’t mention x[ anymore when mentioning the mapΛ→ R[, as the assignment of such map will be clear from the context.Remark 2.3.15 Going back to the introduction (§1.1), note that Fp[[x]] is Z[[x]]/(p), where xwas called q there.


3 The tilting equivalence over perfectoid rings

We fix a perfectoid ring R, and its tilt R[. Let’s call m = ($,$1/p, $1/p2 , . . . , $1/pn , . . . ) ⊂ R.As declared in §2.1, R we’re given an R-regular element $ with the property that $p | p in R,and R is $-adically complete and separated. In addition, the pth power map R/p → R/p issurjective, and the kernel of the map ϑ : Ainf(R)→ R is principal. This is Definition 2.3.6.

3.1 Setup and synopsis

We introduce the main contents of this section, explaining its purposes. The main Theorem ofthis section is the following:Theorem 3.1.1 (Tilting equivalence) Let R-Perf be the category of perfectoid R-algebras,and likewise for R[-Perf . Then the tilting functor (·)[ induces an equivalence of categories:

R-Perf'−→ R[-Perf

whose quasi-inverse is denoted by (·)# and called “untilting”.

We shall call the above equivalence “tilting equivalence”, and we’ll discuss its global form interms of perfectoid spaces in the next lecture. We outline here a brief synopsis of the strategyto show Theorem 3.1.1.

We are going to exhibit (·)[ on perfectoid R-algebras, as a composition of functors, and sep-arately prove that the factors are equivalences. By Remark 2.2.8, we recall we have a regularelement 0 6= $[ in R[ with the following property: tilting yields a canonical isomorphism

R/$ ' R[/$[.

Such isomorphism is, a priori, all the poor man is given, and should indeed be regarded asbeing the core of the tilting equivalence. We encourage the reader to step back to §2.1 andreview the very construction of $[. Such a posteriori review should at this point reveal thatthis isomorphism together with the properties of $[, is made possible by the “high ramifiedness”of R as a Zp-algebra (as the existence and properties of $[ are due to existence of a compatiblesystem of p-power roots of $ in R).

It is clearly hopeless to extend the isomorphism to an isomorphism R ' R[ (eg. when R is ofcharacteristic 0). On the other hand, once we assign the notion of perfectoid (R/$)-algebras,such isomorphism yields tautologically an equivalence of categories

(R/$)-Perf'−→ (R[/$[)-Perf .

It makes sense to aim at lifting the above equivalence to an equivalence:

R-Perf ' R[-Perf .

Endowing R[1/$] with the unique Huber ring topology making R open and bounded, it makessense to also aim at filting the above equivalence to an equivalence:

R[1/$]-Perf'−→ R[[1/$[]-Perf

which is, for later purposes, our ultimate goal.


Such aim means extending schemes over the locus $ = 0 in Spec(R), corresponding to thelocus $[ = 0 in Spec(R[), to Spec(R)-schemes and Spec(R[)-schemes respectively, func-torially, and such functorial correspondence should be an equivalence. It turns out that per-fectoid algebras over a perfectoid ring have mod $ reduction which enjoys a very import-ant deformation-theoretic property, which we call connectiveness. Roughly, their cotangentcomplex vanishes, ensuring unobstructedness and uniqueness (up to unique isomorphism) ofdeformations along nilpotent closed immersions.

We’ll see it’s not here that almost mathematics comes into the picture substantially, althoughwe will reprove the tilting equivalence also in its almost version. We shall do so because thisserves to showing the tilting equivalence in its form for Tate-perfectoid rings, but more thanthis because of an important later purpose: dealing with étale extensions of Tate-perfectoidrings, which turn out to be equivalent to almost étale extensions of perfectoid rings.

We’ll discuss examples of finite étale maps of mod $ perfectoid rings R/$ → R′/$ which donot come as the reduction mod $ of some algebraically étale map R→ R′ of perfectoid rings.We would like to extend the tilting equivalence to the small finite étale sites of R and its tiltR[, but according to the examples we’re mentioning, this is not quite possible.

The obstruction to liftability of such maps lies in the ramification along the loci $ = 0 inSpec(R) and $[ = 0 in Spec(R[), as discussed in the introduction, and the naive idea is to“get rid of it”.

We expand on such obstruction to liftability below.

This issue is solved upon allowing almost mathematics into the picture. Once we assign the no-tion of perfectoid almost (R/$)-algebras, denoted (Ra/$)-Perf , and likewise for (R[a/$[)-Perf(Definition 3.5.8), we will be able to, first, lift the tautological equivalence

(Ra/$)-Perf ' (R[a/$[)-Perf

to the almost integral level, ie. upon assigning the notion of perfectoid Ra-algebras, and likewisefor perfectoid R[a-algebras (Definition 3.3.3), to an equivalence:

Ra-Perf'−→ R[a-Perf .

This step relies again on “almost” deformation theory, and goes exactly as before, and is dealtwith in §3.6, where we prove that the “reduction mod $” functors:

R-Perf → (R/$)-Perf

andRa-Perf → (Ra/$)-Perf

are equivalences of categories.

Once we achieve the equivalence:

R-Perf'−→ R[-Perf

we would like to work with algebras over the Tate-perfectoid rings R[1/$] and R[[1/$[] (again,endowing R[1/$] with the unique Huber ring topology making R into an open and boundedsubring, and likewise for R[[1/$[]). It turns out that we can restrict the above equivalence tothe open loci Spec(R[1/$]) → Spec(R) and Spec(R[[1/$[]) → Spec(R[), extending Theorem3.1.1 to the case of algebras over R[1/$] and R[[1/$[]. This recovers the tilting equivalencefor Tate-perfectoid rings, as known before in the literature. The reason why it is important to


have such an equivalence for algebras over R[1/$] and R[[1/$[] is in light of the equivalenceof the finite étale sites over these rings.

As mentioned above, it’s not possible to obtain an equivalence along the lines of Rfet '(R/$)fet. Nevertheless, upon allowing almost mathematics into the picture, we do obtainthe following equivalence of categories:

Rafet ' (Ra/$)fet

which naturally lifts to an equivalence of categories:

Rafet ' R[afet

via the tautological equivalence (Ra/$)fet ' (R[a/$[)fet.

“Ignoring$-torsion” is what lets the open loci Spec(R[1/$])→ Spec(R) and Spec(R[[1/$[])→Spec(R[) inherit the equivalence Rafet ' R[afet, and such operation transforms the notion of al-most finite étaleness into finite étaleness, thus giving an equivalence of categories R[1/$]fet 'R[[1/$[]fet. This last step is quite nontrivial (ie. checking that pulling back the above equi-valence along the open immersion Spec(R[1/$]) → Spec(R) still yields an equivalence of theexpected form, and transforms the notion of almost finite étaleness into finite étaleness), andrelies on the almost purity theorem.

Inverting $ is equivalent to getting rid of not just the locus $ = 0, but of the intersection ofall irreducible components in Spec(R) containing $ = 0, that is, of V (m) ⊂ Spec(R). Thisprocess is enpowered by Lemma 3.2.1.

The need to work out the tilting equivalence for almost algebras is due to the strategy we’lladopt to show the equivalence of the finite étale sites of R[1/$] and R[[1/$[], which requiresforgetful functors from categories of finite étale algebras over perfectoid almost algebras, to thecorresponding categories of perfectoid almost algebras over a fixed perfectoid ring.

To begin with, we’ll then deal with the full proof of the tilting equivalence both for actual per-fectoid algebras over a perfectoid ring and its tilt, and for almost algebras. Shortly thereafter,we shall prove the same equivalence for Tate-perfectoid algebras over a Tate-perfectoid ring.

To sum up, once we assign all the necessary notions of perfectoid almost algebras, the setup isas in the following diagrams:

R-Perf

(·)[

88'§3.6

// (R/$)−Perf ' (R[/$[)-Perf R[-Perf'(§3.6)[oo

(·)#

xx


expressing the tilting equivalence for perfectoid rings.

Ra-Perf

(·)[

88'§3.6// (Ra/$)−Perf ' (R[a/$[)-Perf R[a-Perf'

(§3.6)[oo

(·)#

ww

expressing the tilting equivalence for perfectoid almost algebras rings.

R-Perf

(·)[

44

S 7→S0a

::'§3.2

// R0a-Perf

A 7→A∗[1/$]

yy

'§3.6// (R0a/$)−Perf ' (R[0a/$[)-Perf R[0a-Perf

A 7→A∗[1/$]

::'(§3.6)[oo R[-Perf'

(§3.2)[oo

(·)#

tt

S 7→S0a

ww

expressing the tilting equivalence for Tate-perfectoid rings, and the compatibility between theabove.Remark 3.1.2 We remark that the equivalence Rafet ' (R/$)fet should be regarded as astrengthening of Faltings’ almost purity. We are going to show it in the generality of Tate-perfectoid rings. We shall call the above equivalence “almost purity”, in light of the introduction.

Fix R a Tate-perfectoid ring. Hence, R0 will be a perfectoid ring. We remark that it is inap-propriate (and in fact false) to claim that almost purity “restricts” along the Zariski localizationSpec(R)→ Spec(R0) (hence Spec(R[)→ Spec(R[0)), as this ignores continuity of the structuremap of perfectoid R-algebras and R[-algebras with respect to the Huber ring topologies. To beexact, the equivalence R0a

fet ' (R0a/$)fet is purely algebraic in nature, and upon giving sense toaffine schemes and formal schemes in the context of almost mathematics, it yields an equival-ence between finite étale covers of the “perfectoid affine scheme” Spec(R0a/$) and finite étalecovers of the “perfectoid affine formal scheme” Spf(R0a) respectively. Likewise for the tiltedobjects. Such equivalence almost lifts to an equivalence with finite étale covers of the affinoidperfectoid adic space (to be defined in §5) Spa(R0, R0) (hence Spa(R[0, R[0)), that is, it lifts tothe desired equivalence with actual finite étale covers of Spa(R,R0) (hence Spa(R[, R[0)) up toignoring the behaviour of the maps Spa(S0, S0)→ Spa(R0, R0) and its tilt, for Tate-perfectoidS0[1/$], along the locus $ = 0 and its tilt. The slogan to keep in mind is that almostpurity yields liftability of almost finite étale covers of Spec(R/$) to actual finite étale coversof Spa(R0, R0) away from $ = 0, ie. of Spa(R,R0).

3.2 Away from $ = 0

Let R be a perfectoid ring, and call m = lim−→$p−nR. We note that m is a flat R-module,and, using the terminology in [GR], (R,m) is a flat almost structure. This has the advantagethat the functors between the categories of R-modules and Ra-modules have simpler form and


preserve exactness whenever it makes sense and is possible. We don’t need to review thesefacts from the general theory of almost mathematics here, but the interested reader can referto [GR], for a very wide background on almost mathematics.

The following Lemma expresses the fact that, for R Tate-perfectoid, “ignoring $-torsion” allowssafe passage from the category of perfectoid R-algebras to the category of perfectoid R0a-algebras and conversely, the geometric meaning being, given an R0-algebra S, “crushing allpoints of Spa(S[1/$], S) which specialize to a given x ∈ Spa(S[1/$], S) to its unique rank1 generization reconstructs Spa(S[1/$], S[1/$]0)”. You can visualize this in the case of theperfectoid field C given by the completion of the fraction field of Qcyc

p 〈t1/p∞〉, with valuation

ring C0 and residue field κ. The space of continuous valuations on C is homeomorphic to P1κ,

and specifying a point in it is equivalent to specifying an open and integrally closed subring C+

of C0 by [L10, Thm. 10.3.6]. Passage from C-algebras to C0a-algebras ignores the discrepancybetween C+ and C0.

We begin with a key lemma, which essentially says that given an Ra-module M , in absenceof $-torsion in M∗ as an R-module, we can pick any other $-torsionfree R-module N , almostisomorphic toM∗, and the module of almost elementsM∗ ofM can be described as the moduleof those elements x ∈ N [1/$] which “almost lie” in N .Lemma 3.2.1 If N is a $-torsion free R-module, then we have an R-module injection:

N → x ∈ N [1/$] | ∀ε ∈ m : εx ∈ N =: Napp

with almost zero cokernel.

If M = Na, we have M∗ = Napp.

In the above Lemma, Napp stands for “approximate”.

Proof. Since N is $-torsion free, the natural R-module map N → N [1/$] is injective. Itfactors through Napp → N [1/$], thus giving that N is an R-submodule of x ∈ N [1/$] |∀ε ∈ m : εx ∈ N. The cokernel is, by design, killed by m, and hence it’s almost zero as anR-module.

This, in particular, means that if M = Na, with N a $-torsion free R-module, then Napp isalmost isomorphic to M∗.

We observe that M∗ = HomR(m, N) independently of N , as long as N and M∗ are almostisomorphic R-modules. Here we don’t even need N is $-torsion free. We note that the naturalmap M∗ → HomR[1/$](R[1/$], N [1/$]) = N [1/$] is injective. Indeed, if x : m → N is anR-module map which gets killed upon inverting $, then for all ε ∈ m, x(ε) lies in the kernel ofN → N [1/$]. Since N is $-torsion free, such kernel is trivial. Therefore, x already killed m,hence was the zero map.

On the other hand, the natural map N →M∗ is an almost isomorphism. We have an R-moduleinjection:

M∗ → N0 := x ∈ N [1/$] | mx ⊆M∗.

Let x ∈ N0. Then mx ⊆M∗, and m(mx) ⊆ mM∗ ⊆ N . On the other hand, m(mx) = m2x = mx,and therefore x is in Napp. We deduce we have an R-module injection:

M∗ → Napp

with almost zero cokernel.


We need to show that given an x ∈ N [1/$] satisfying mx ⊂ N , then x ∈ M∗. The aboveR-module map M∗ → Napp coincides with the map:

HomR(m, N)→ HomR[1/$](R[1/$], N [1/$])

which extends an R-linear map x : m→ N to an R[1/$]-linear map x⊗R[1/$].

We’re required to show that for all t ∈ R[1/$], if for all ε ∈ m we have ε · x(t) ∈ N , then x isthe scalar extension of an R-linear map x0 : m → N along R → R[1/$]. We restrict x alongthe map R→ R[1/$], thus regarding x as an R-linear map. By R-linearity of x, we know theimage of m ⊂ R[1/$] under x lies in N . Therefore, we call x0 the restriction of x to m as anR-module map. We claim x′ := x0 ⊗ R[1/$] recovers x. By design of x0, the image of x′ − xis almost zero. In particular, it’s killed by $2. On the other hand, $(x′ − x) ∈ N , and sinceN is $-torsion free, we conclude $(x′ − x) = 0. For the same reason we deduce x′ = x. Thisconcludes the proof of the Lemma.

Let x ∈ R, and M be an Ra-module. Then we define multiplication by x on M by applyingthe functor (·)a : R-Mod→ Ra-Mod to multiplication by x on M∗. We need a definition.Definition 3.2.2 Let M be an Ra-module. Given x ∈ R, we say M is x-torsion free if andonly if multiplication by x on M∗ has trivial kernel, ie. if M∗ is x-torsion free.

We note that the functor (·)∗ is left-exact, because it is right-adjoint to (·)a. If we consider themultiplication-by-x map on an Ra-module M , for some x ∈ R, then

ker(×x)∗ = ker(M∗×x−−→M∗).

It follows that if M is of the form Na for some x-torsion free R-module N , then M is x-torsionfree, and hence M∗ is.Remark 3.2.3 The way Lemma 3.2.1 should be regarded is as follows. Let M be an Ra-module such that it is $-torsion free. Then for any $-torsion free R-module N such thatNa = M , we have:

M∗ = x ∈ N [1/$] | mx ⊆ N.

This is clearly false if N is not $-torsion free! LetM := Ra/$. Then we can choose N = R/$,and form N [1/$] = 0. On the other hand (Ra/$)∗ 6= 0. This was a case where M was $-torsion. Let M := Ra ⊕Ra/$. Then:

M∗ = Ra∗ ⊕ (Ra/$)∗.

On the other hand, if we let N = R⊕R/$, we do have Na = M , but:

Napp = x ∈ N [1/$] | mx ⊆ N = x ∈ R[1/$] | mx ⊆ R = Ra∗

which is not almost isomorphic to M∗.Example 3.2.4 It is not so trivial, though true, to show R0a

∗ = R0 in our case where R isTate-perfectoid. Let R → S be a morphism of perfectoid rings. We claim S0a is a $-torsionfree R0a-module. Indeed, let A := S0a, and consider:

A∗ = HomR0(m, S0).

Then A∗ has no nontrivial almost zero elements. On the other hand, suppose z ∈ A∗ is in thekernel of multiplication by $. We have A∗ ⊂ S0, and we claim z is an almost zero element ofS0. We only need to check that $p−nz = 0 for all n ≥ 0. We have:

($p−nz)pn

= $zpn

= ($z)zpn−1 = 0,


and since S0 is reduced. It follows z ∈ A∗ is almost zero, and therefore, it is zero. At thispoint, the first part of Lemma 3.2.1 for N = S0, predicts that S0a

∗ contains S0. On the otherhand, as above, S0a

∗ is clearly made of power-bounded elements, and hence S0a∗ = S0. We are

going to reprove this in a different and more concrete way in the sequel.Remark 3.2.5 Let C a perfectoid field, and C0 the valuation ring. Then all C0-modulesare flat if and only if they are $-torsion free. This implies that our definition of perfectoidalmost algebras will be consistent with the one given in [Sch], where we worked with perfectoidC-algebras.Example 3.2.6 Consider R = A[(x/$)1/p∞ ]∧[1/x], with A = C〈x〉, C a perfectoid field. R isa Tate-perfectoid ring with R0 = A0[(x/$)1/p∞ ] with A0 = C0〈x〉. The ideal J = ($,x) is afinitely presented R0-module. R0 is a domain, and the generic fiber of J is 1-dimensional. Itfollows J is flat if and only if it is invertible. However, if p is the unique maximal ideal in R0

containing $ and x, J/pJ is 2-dimensional over the residue field, hence J can’t be invertible.This is an example of a $-torsion free R0-module which is not R0-flat. Nevertheless, Lemma3.2.1 applies to J , and gives:

(Ja)∗ = z ∈ J [1/$] | mz ⊆ J = z ∈ xR | mz ⊂ J

Lemma 3.2.7 Let M be a $-torsion free Ra-module. Then, for all x ∈ R we have:

(xM)∗ = xM∗.

Moreover, M∗/xM∗ ⊂ (M/xM)∗, and for every ε ∈ m the image of the reduction map:

(M/xεM)∗ → (M/xM)∗

equals M∗/xM∗.

Proof. Note that (xM∗)a = xM , and that xM∗ is a $-torsion free R-module, because M∗ is,

and $ is a regular element. Therefore, we can apply Lemma 3.2.1 with N = xM∗, to obtain:

(xM)∗ = HomR(m, xM∗) = y ∈ xM∗[1/$] | ∀ε ∈ m : εy ∈ xM∗.

The right side equals xM∗. Now since (·)∗ is right-adjoint to (·)a, it is left-exact. Given thesequence of Ra-modules:

0→ xM →M →M/xM → 0

we deduce the inclusionM∗/xM∗ ⊂ (M/xM)∗, using that (xM)∗ = xM∗, as just shown above.

Let m′ ∈ (M/xεM)∗ be a lift of m ∈ (M/xM)∗. We regard m′ as being an element ofHomR0(m,M∗/xεM∗), using the fact that M∗/xεM∗ and (M/xεM)∗ are almost isomorphic.We evaluate m′ on ε, and call n := m′(ε), an element of M∗/xεM∗. Let n′ be a lift of n to M∗.We claim n′ is divisible by ε. It is enough to check that ε′n′ is divisible by ε for all ε′ ∈ m. Wehave:

ε′n = ε′m′(ε) = εm′(ε′) ∈ εM∗/xεM∗and hence ε′n′ ∈ εM∗, as claimed. It follows that u := n′/ε ∈ M∗ is the desired lift ofm ∈ (M/xM)∗. Indeed, multiplication by ε induces an injection

(M/xM)∗ → (M/xεM)∗

using left exactness of (·)∗, and the images agree: n′ maps to εm = m′(ε) = n in (M/xεM)∗.

We conclude with an essential Lemma.


Lemma 3.2.8 Let M be a $-torsion free Ra-module. Then M is $-adically complete if andonly if M∗ is $-adically complete.

Proof. The functors (·)∗ and (·)a both admit left adjoints, given by (·)a and (·)! = m ⊗ (·)∗respectively. Therefore, they both commute with inverse limits. Suppose M is $-adicallycomplete. Then:

M∗ = (lim←−M/$nM)∗ = lim←−(M/$nM)∗ = lim←−M∗/$nM∗,

using Lemma 3.2.7 in the last equality. Therefore, M∗ is $-adically complete. Conversely, ifM∗ is $-adically complete, we have:

M = (M∗)a = (lim←−M∗/$

nM∗)a = lim←−(M∗/$

nM∗)a = lim←−M/$nM.

3.3 Perfectoid R and (R/$)-algebras and almost algebras

In this section we assign the notion of perfectoid algebras over Ra and Ra/$, as well as R/$.

To start off, we fix a perfectoid field C of arbitrary characteristic. Let R be a Tate-perfectoidC-algebra.

We briefly recall the notion of perfectoid C0a-algebras, and perfectoid (C0a/$)-algebras, inorder to build from there. Note that since C[ is again a perfectoid field, we are automaticallyalso given the notions of perfectoid C[0a-algebras and perfectoid (C[0a/$[)-algebras.

Notation We agree beforehand to use the letters A,B to denote almost algebras, as opposedto the letters R,S to denote actual algebras.Definition 3.3.1 C is fixed, and C0 is its valuation ring.

(1) Let A be a C0-algebra. We say A is perfectoid if and only if it is a flat $-adically completeand separated as a C0-module, and the following “perfectness” condition is satisfied: themod $ pth power map induces a surjection

A/$1/p → A/$.

Morphisms of perfectoid C0-algebras are morphisms of K0-algebras.

(2) We say a C0a-algebra A is “perfectoid” if and only if the almost version of the aboveconditions is met by A. Morphisms of perfectoid C0a-algebras are morphisms of C0a-algebras.

(3) Likewise, let A denote a (C0/$)-algebra. We say A is perfectoid if and only if it is flat asa (C0/$)-module, and the following “perfectness” condition is satisfied:

A/$1/p ' A

the map being the pth power map. Morphisms of perfectoid (C0/$)-algebras are morph-isms of (C0/$)-algebras.

(4) We say a (C0a/$)-algebra A is “perfectoid” if and only if the almost version of the aboveconditions is met by A. Morphisms of perfectoid (C0a/$)-algebras are morphisms of(C0a/$)-algebras.


Remark 3.3.2 We remark that the flatness condition built into the above definitions is mo-tivated by the fact that since R is a C-algebra, then R0 is automatically C0-flat, as C0 is avaluation ring, and hence R0a is C0a-flat, likewise R0a/$ over C0a/$.

The first difficulty which appears naturally at once, for the purposes of avoiding presence of thebase field C all over the place, is to give a suitable notion of perfectoid almost algebras. We fixa perfectoid ring R, and denote by A an Ra-algebra. If A comes from a perfectoid R-algebraS as A = Sa, then it is not true in general that A is Ra-flat.

The reader may see the example in §3.4.Definition 3.3.3 Let R be a perfectoid ring. A perfectoid Ra-algebra is a $-adically completeand separated Ra-algebra A such that A/$1/p ' A/$ under the pth power map, and suchthat $ is A-regular. Morphisms of perfectoid Ra-algebras are morphisms of Ra-algebras.

A functor:R-Perf → Ra-Perf

is now defined. When R is a Tate-perfectoid ring, hence R0 a perfectoid ring, then we’ll seeR-Perf is equivalent to R0a-Perf , and we’ll have to be careful in showing that we can recoverthe subring of power-bounded elements, when showing the functor A 7→ A∗[1/$] on perfectoidR0a-algebras, has essential image in R-Perf , and is a quasi-inverse to S 7→ S0a. The crux willalso consist in deducing the “pth power map isomorphism” from the almost level.

We remark that Definition 3.3.3 is quite non-surprising, because perfectoidness of Tate-perfectoidrings R comes from the conditions imposed on R0, rather than on R, being careful to requiringR0 to be bounded. As long as our definition of perfectoid Ra-algebras takes care of bounded-ness of R in R[1/$] (which is the case by Lemma 3.2.1, as shown in the sequel) we are goodto go. The crux will be singling out the correct definition of perfectoid (Ra/$)-algebras.

3.4 An example

We address the necessity of re-assigning the notion of perfectoid almost algebras. We firstconsider the following example.Example 3.4.1 We consider the formal perfectoid unit discX := Spa(Zcyc

p [[x1/p∞ ]],Zcycp [[x1/p∞ ]])

and its rational domains:

Un := X(pn/x)

They are cut as the “perfection” of |p|n ≤ |x|, ie.

Un = Spa(Zcycp 〈(pn/x)1/p∞〉[1/x],Zcyc

p 〈(pn/x)1/p∞〉).

Pick any point s : Spa(κ(s), κ(s)+) → Un. Since Un is an analytic adic space, s is an analyticpoint, and hence, as we shall soon see, the completion of the residue field κ(s) is a perfectoidfield. Likewise, the completion of Zcyc

p 〈(pn/x)1/p∞〉[1/x] is a Tate-perfectoid ring. The morph-ism s can be arranged to be non-flat, for example by picking s to be any higher rank point (ifs was flat, its image should be closed under generization). The morpism of perfectoid rings:

Zcycp 〈(pn/x)1/p∞〉∧ → κ(s)+∧

is not almost flat. This example suggests a rather general procedure to produce examples alongthese lines.


We give an example of a morphism of Tate-perfectoid rings R → R′ which is not flat, andsuch that R0 → R′

0 and R0a → R′0a are not flat. This shows that we must assign the notion

of perfectoid almost algebras with no flatness assumption built into it, and such that the oldnotion allowing a perfectoid base field, is recovered as a special case.

We first give a synopsis of the strategy, and then discuss the details.

Example 3.4.2 First, recall that given a morphism of adic spaces Y f−→ X, we say f is flat ata point y ∈ Y if and only if the natural map OX,f(y) → OY,y is flat, and we say f is flat if it isflat at every point.

We shall consider a nonflat closed immersion of affinoid adic spaces Z → X (from which weshall build a nonflat map between two perfectoid spaces). More precisely, we want to showthere exists a morphism between two affinoid perfectoid spaces:

Spa(R′, R′0)→ Spa(R,R0)

for Huber pairs (R,R0) and (R′, R′0), for R and R′ Tate-perfectoid, enjoying the following

property: R → R′ is not flat. We claim non-flatness of R → R′ ensures the map R0a → R′0a

is not flat, in the sense of almost mathematics. Let’s first show this, before moving on.

Indeed, since m is R0-flat the functor (·)! is exact on R0a-modules. Suppose R′0a is a flatR0a-module. Then R′0a! is flat as an R0-module, and using Lemma 3.2.1 upon inverting $, wededuce R→ R′ is flat, a contradiction.

Note that it’s here that flatness fails to be automatically enjoyed by R0a → R′0a for a random

Tate-perfectoid ring R and a perfectoid R-algebra R′. If R was a perfectoid field, then R→ R′

was automatically flat, and likewise both R0 → R′0 and R0a → R′

0a.

Let, now, A be a uniform Tate-Huber ring, with A+ ⊂ A0 an open integrally closed subring,and such that (A∧, A∧+) is a perfectoid Tate-Huber pair. Let I ⊂ A be a (non pure, toensure non-flatness of the quotient map) ideal. We let B := A/I, and B+ the integral closureof A+ in B. We claim (B∧, B∧+) is a perfectoid Tate-Huber pair. Indeed, we observe thatZ := Spa(B,B+) is the cofiltered limit of its rational neighbourhoods Ui ⊂ X := Spa(A,A+).To see this, take f1, . . . , fk ∈ I, and consider the subsets cut out by the inequalities |fi| ≤ p−n,i = 1, . . . , k. Z is their intersection. We claim we have:

B∧+ = ( lim−→Z⊂Ui⊂X

O+X(Ui))

∧.

In fact, any f ∈ I becomes infinitely p-divisible in the colimit, and is therefore killed by thecompletion. Since each one of the OX(Ui) is perfectoid, because by assumption A∧ is, then sois B∧, as desired.

We have constructed a nonflat map of adic spaces

Z := Spa(B,B+)→ Spa(A,A+) =: X

such that, calling Z := Spa(B∧, B∧+) and X := Spa(A∧, A∧+), we have a nonflat map ofperfectoid spaces:

Z → X.

We call R := A∧ and R′ := B∧, and get our example, as desired. Clearly, this situation cannotoccur when X is Spa(C,C+) for C a perfectoid field.

The new definition of almost perfectoid algebra will have to substitute the flatness conditionwith a weaker condition!


3.5 Connectiveness

We claim we aren’t keeping track of a very important property of perfectoid rings, yet. Suchproperty becomes apparent primarily in positive characteristic.

We first recall a vanishing result for the cotangent complex of a ring map, under a derived“relative perfectness” assumption.Theorem 3.5.1 Let R1 → R2 be a morphism of Fp-algebras. We denote by R′1 the ring R1

regarded as an algebra over itself under the pth power map, that is, with R1-algebra structuregiven by ϕ1 : R1 → R1. Similarly for R′2. We assume that the relative Frobenius ΦR2/R1

induces an isomorphism:R′1 ⊗L

R1R2

'−→ R′2[0]

in D(R′1). Then, we have: LR2/R1' 0 in D(R′1).

A general remark.Remark 3.5.2 Since R′1 and R2 are Tor-independent over R1, the functorial comparison mapin D(R′1):

(·)⊗LR1R′1

'−→ (·)⊗LR2R′2

is an isomorphism. It follows the conclusion in the above Theorem 3.5.1 is, equivalently, avanishing statement in D(R1). Generalities about derived change of rings are discussed inLemma B.1.1.

We briefly recall the proof from Lecture 20.

Summary of the proof. Let R•2 → R2 be a simplicial resolution by free R1-algebras. We havethe relative Frobenius:

ΦR•2/R1: R′1 ⊗R1

R•2 → R•′

2

We identifyRn2 with a polynomial algebra in possibly infinitely many indeterminates: R1[t1, . . . , tk, . . . ],and the relative Frobenius ΦRn2 /R1

is given by the R′1-algebra map sending tk to tpk. By as-sumption, ΦR•2/R1

induces a quasi-isomorphism of simplicial R′1-algebras, which implies thatΦR•2/R1

yields an isomorphism in D−(R′1):

R′1 ⊗LR1

LR2/R1' LR′2/R′1 .

However, the explicit description of ΦR•2/R1shows that the map induced by this on each module

of differentials sends dtk to dtpk = 0, and hence is the zero map. It follows LR′2/R′1 ' 0, andupon identifying this with LR2/R1

, we conclude.

We need the following:Definition 3.5.3 Let B be an A-algebra.

(1) We say B is a connective A-algebra if and only if LB/A ' 0 in D(A).

(2) In the case A is an almost algebra, then we say B is connective if and only if LB!!/A!!is

almost quasi-isomorphic to the trivial complex in D(A!!).

Suppose now A is a topological ring endowed with the I-adic topology, I ⊆ A being a finitelygenerated ideal. Let A → B be a continuous (adic) morphism of (adic) topological rings, andassume A and B are complete with respect to the respective adic topologies. We say B isanalytically connective as an A-algebra if and only if the analytic cotangent complex

LanB/A := R lim←−L(B/InB)/(A/In)

is quasi-isomorphic to the trivial complex in D(A).


We note that for any ring or almost algebra A, A is connective as an A-algebra under theidentity map. Likewise for complete adic rings.Lemma 3.5.4 Let R→ R′ be a map of perfectoid rings. Then LR′/R⊗L

R (R/p) ' 0 in D(R/p).In particular, the derived p-adic completion of LR′/R satisfies LR′/R ' 0 in D(R).

Proof. Note that, canonically:

R′[0] ' R⊗LAinf (R) Ainf(R

′)

because Ainf(R′) is ξ-torsion free as an Ainf(R)-module. More precisely, Tor

Ainf (R)i (Ainf(R

′),Ainf(R)/ξ) =

0 for i > 1 by an consideration, and the only claim to check is TorAinf (R)1 (Ainf(R

′),Ainf(R)/ξ) =0, which is true since ξ is regular. By the base change theorem for the cotangent complex it’senough to show

LAinf (R′)/Ainf (R) ⊗LR (R/p) ' 0

in D(R/p). Note here that the statement:

R′[0] ' R⊗LAinf (R) Ainf(R

′)

in D(R′) is only intended to encode the fact that R and Ainf(R′) are Tor-independent over

Ainf(R), and that the obvious diagram is cocartesian. Thus we can apply the base changetheorem, keeping in mind that the derived base change (·)⊗L

Ainf (R) R is canonically isomorphicto (·)⊗Ainf (R′) R

′, the isomorphism being in D(R). On the other hand, we have:

LAinf (R′)/Ainf (R) ⊗LR (R/p) ' LR′[/R[ ' 0

the first quasi-isomorphism being due to vanishing of TorAinf (R)i (Ainf(R

′), R/p) for i > 0 (argueagain using 0 → R → R → R/p → 0), and last vanishing being due to derived relativeperfectness of the Fp-algebras R[ and R′

[, hence following from Theorem 3.5.1. We notethat relative perfectness of R[ and R′

[ is a nontrivial fact, and it is settled in the followingProposition.

Proposition 3.5.5 Let A be a perfect Fp-algebra, and B and C arbitrary perfect A-algebras.Then we have: TorAi (B,C) = 0 for all i > 0.

Proof. We write B = A′/I where A′ is a perfectified polynomial ring over A. Since A′ is A-flat,we have

B ⊗LA C = B ⊗L

A′ (A′ ⊗A C),

and upon renaming A′ ⊗A C, which is perfect, as C, and renaming A′ as A, we can assume Bis of the form A/I where I is a perfect ideal. Writing I as the filtered colimit of perfections offinitely generated ideals and using compatibility of Tor’s with filtered colimits, we can assumeI is the perfection of a finitely generated ideal.

By inducting on the number of generators of I (prior to its perfection) we can arrange that Iis the perfection of a principal ideal xA.

In this special case, the ideal I is always A-flat. Indeed, let An = A for n ≥ 1 and define themap of A-modules An → An+1 as multiplication by x1/pn−1/pn+1

. The maps An = A→ I viamultiplication by x1/pn are compatible with the transition maps, thus yielding an A-modulemap:


lim−→An → I

that is clearly surjective. We claim it is also injective. Indeed, if x1/pna = 0 for some a, thenx1/pn−1/pn+1

a = 0 because its pth power is a multiple of x1/pn−1−1/pna = (x1/pn)p−2(x1/pna) =0.

Therefore, I is expressed as an A-linear filtered colimit of free modules of rank 1, hence flatA-modules, and therefore I is A-flat, as claimed.

Thus, from the short exact sequence of A-modules:

0→ I → A→ A/I = B → 0

it follows thatTorAi (B,C) = 0 for i > 1.

We are left to show the vanishing of TorA1 (B,C), which is the kernel of the A-module map:

I ⊗A C → C.

By the above description of I, and since tensor products commute with colimits, we identifyI ⊗A C with IC, since C is also a perfect ring. Hence, the above map is injective, and theTorA1 (B,C) vanishes as well. This concludes the proof.

For a more conceptual proof of Proposition 3.5.5, the reader may see Appendix A. Note thatLemma 3.5.4 shows that any morphism of perfectoid rings R → R′ is analytically connective,and in fact connective if the rings are of positive characteristic p.Remark 3.5.6 LetR be a perfectoid ring, R′ a perfectoidR-algebra. We claim that L(R′/$)/(R/$)

vanishes, ie. R′/$ is a connective (R/$)-algebra. As a first step, we want to compareLR′/R ⊗L

R (R/p) and L(R′/p)/(R/p) in D(R/p).Lemma 3.5.7 We let R→ R′ be a map of perfectoid rings. Then the natural map:

LR′/R ⊗LR (R/p)→ L(R′/p)/(R/p)

is a quasi-isomorphism in D(R/p).

Proof. The only interesting case being the one with R of characteristic 0, we only need to checkTorRi (R′, R/p) is zero for all i > 0. We get exact sequences:

TorRi (R′, R)→ TorRi (R′, R/p)→ TorRi−1(R′, R)

and deduce TorRi (R′, R/p) = 0 for i > 1. On the other hand, TorR1 (R′, R/p) is the kernel ofmultiplication by p on R′, which is p-torsion free.

We are ready to give the following:Definition 3.5.8 Let R be a perfectoid ring.

(1) Let A be an (R/$)-algebra. We say A is perfectoid if it is Λ0-flat and the pth power map:A/$1/p → A is an isomorphism. Morphisms of perfectoid (R/$)-algebras are morphismsof (R/$)-algebras.


(2) Let A be an (Ra/$)-algebra. We say A is perfectoid if it is Λa0-flat and the pth powermap: A/$1/p → A is an isomorphism. Morphisms of perfectoid (Ra/$)-algebras aremorphisms of (Ra/$)-algebras.

Recall that, calling S := lim←−A, the limit being formed along the pth power maps, saying Ais Λ0-flat means that $[ is S-regular. This condition expresses the fact that we’re assuming$ in Definition 2.3.6 is regular, an assumption which is due to our aim of switching betweenperfectoid and Tate-perfectoid rings when needed, for geometric purposes.

Note that given the above definition, the natural question is whether all perfectoid (R/$)-algebras come as the reduction mod $ of some perfectoid R-algebra, leaving out uniquenessaspects.Tt’s clear where we’re going. The next step will be showing that all perfectoid (R/$)-algebras are connective, and likewise for almost algebras.

We first check that the reduction modulo $ of a perfectoid R-algebra R′ is connective over(R/$), to show consistency of the above claim.Proposition 3.5.9 Let R be a perfectoid ring, and R′ a perfectoid R-algebra. Then A :=R′/$ is a connective perfectoid (R/$)-algebra. Likewise for a perfectoid Ra-algebra B and itsreduction modulo $.

Proof. In order to repeat the argument in Remark 3.5.6, we are required to check that given amap of perfectoid rings R→ R′, then LR′/R ⊗L

R (R/$) ' 0 in D(R/$).

The canonical ring isomorphism Ainf(R)/(ξ, [$[])'−→ R/$ sends the relation ξ = [$[p]+pu = 0,

for some element u ∈ Ainf(R)×, to the relation $ = 0 together with p = 0. Therefore, thenatural map Ainf(R)→ R/$ factors as a composition of ring homomorphisms:

Ainf → Ainf(R)/p ' R[ → R/$

the last map being the projection onto the first component in R[. By functoriality of formationof the cotangent complex in the map Ainf(R)→ Ainf(R

′), the natural map:

LAinf (R′)/Ainf (R) ⊗LAinf (R) (R/$)→ L(R′/$)/(R/$)

which is a quasi-isomorphism in D(R/$) because $ is Ainf(R′)-regular, is identified with the

map:

(LAinf (R′)/Ainf (R) ⊗LAinf (R) (R/p))⊗L

R/p (R/$)→ L(R′/$)/(R/$).

We must be careful here, and we use the formalism of derived changes of rings reviewed in .

The derived base change functor

(·)⊗LAinf (R) (R/$) : D(Ainf(R))→ D(R/$)

factors through the base change

(·)⊗LAinf (R) (R/p) : D(Ainf(R))→ D(R/p)

meaning that the composition of derived base changes:

D(Ainf(R))→ D(R/p)→ D(R/$)

is given by (·)⊗LAinf (R) (R/$).


Therefore, we have quasi-isomorphisms in D(R/$):

L(R′/$)/(R/$)'←− LAinf (R′)/Ainf (R)⊗L

Ainf (R)(R/$)'−→ (LAinf (R′)/Ainf (R)⊗L

Ainf (R)(R/p))⊗LR/p(R/$)

and using the quasi-isomorphism:

LAinf (R′)/Ainf (R) ⊗LAinf (R) (R/p)

'−→ LR′[/R[ ' 0

in D(R/p) (the last vanishing being Lemma 3.5.7), it follows L(R′/$)/(R/$) ' 0 in D(R/$), asdesired.

In particular, R′/$ is a connective (R/$)-algebra, and R′a/$ is a connective (Ra/$)-algebra.In both cases, we conclude.

A subtle class of examples.Example 3.5.10 The pth power map on a ring R of characteristic p > 0, is not flat in general.It is an exercise to show that the pth power map on noetherian R is flat if and only if R isregular (this is a celebrated theorem of Kunz, the easy direction being an easy consequence ofthe Cohen structure Theorem, and the hard one relying on the fact that flatness of the pthpower map passes to completions, and on results on finiteness of Tor-dimension).

Let C be a perfectoid field of characteristic p, and C0 its valuation ring. Since C is notdiscretely valued, C0 is not regular (its maximal ideal is idempotent!). However, the pth powermap C0 → C0 is flat, since it is an Fp-algebra automorphism. In general, for every perfectring the pth power map is flat. This implies the ring is regular only in the noetherian case.

Suppose R is an Fp-algebra with flat pth power map. Then we can form Rperf := lim−→R, thetransition maps being the pth power maps. We claim Rperf is R-flat. Indeed, Rperf is a directlimit of flat R-modules (each finite layer is regarded as an R-module under an appropriatepower of the pth power map, and this R-module structure is trivially compatible with thetransition maps), hence flat. If we assume that R is a complete local ring, then Rperf is alsolocal, and R→ Rperf is faithfully flat, so regularity of R is always inherited from Rperf (becauseRperf -modules have finite Tor-dimension if and only if R-modules do). Hence, one can constructnon-perfect examples of non-regular Fp-algebras with non-flat pth power map, which come witha faithfully flat map to a (non-noetherian) non-regular Fp-algebra with flat pth power map.For example, take Fp[[x]].

One can consider the subring:

R := Fp[[u]] + vFp((u))[[v]] ⊂ Fp((u))((v))

which is a valuation ring with respect to the obvious rank 2 valuation. R is a local ring, andFrobenius is not flat because otherwise R→ Rperf would be faithfully flat. On the other hand,the R-module M := R/Rp is killed upon tensoring by Rperf over R, though it’s not zero. Itfollows R is not regular, which is confirmed by the fact that it’s a higher rank valuation ring.

Note that R is non-noetherian (since it’s non discrete).

As a result, despite, for perfectoid field C, C0 is not regular, the pth power map C0 → C0 isa honest (continuous) ring map, and hence we expect C0 to be a perfectoid C0-algebra underthe pth power map, according to Definition 2.3.6. Indeed, this is the case consistently with theold definition of perfectoid C0-algebras (which included flatness among the assumptions).

Given that the pth power map on C0 is flat, and by naturality of the pth power map, it’s carriedover to the pth power map C0/$ → C0/$ under reduction modulo $, and we deduce this map


is also flat (though not an isomorphism). Therefore, again our definition of perfectoid (R/$)-algebras is consistent with the old definition, which included flatness among the assumptions,and C0/$ is perfectoid as an algebra over itself not only via the identity, but also the pthpower map.

We are left to check that perfectoid (R/$)-algebras are indeed connective, in order to havethat reduction modulo $ yields indeed equivalences of categories R-Perf → (R/$)-Perf andRa-Perf → (Ra/$)-Perf .Lemma 3.5.11 Let R be a perfectoid ring, and A be a perfectoid (R/$)-algebra. Then A isconnective.

Proof. Let S := lim←−A, the inverse limit being taken along the pth power map isomorphismsA/$1/p ' A. We have the quasi-isomorphism in D(W (S)):

W (S)⊗LAinf (R) (R/$) ' A[0]

because [$[] is W (S)-regular, as $[ is in S. Such quasi-isomorphism factors through thequasi-isomorphism in D(W (S)):

W (S)⊗LAinf (R) (R/p) ' (W (S)/p)[0] ' S[0].

Hence LA/(R/$) ' 0 in D(R/$), because LS/R[ ' 0 in D(R[).

3.6 Deformations and tilting

The main result of this section is the following.Theorem 3.6.1 Let R be a perfectoid ring.

(1) The reduction mod $ functor:




Ra-Perf → (Ra/$)-Perf


To achieve Proposition 3.6.1, we first deal with the following Lemma, which expresses the fullpower of connectiveness of mod $ perfectoid algebras and almost algebras.Lemma 3.6.2 Let R → R′ be a morphism of perfectoid rings. Then, for every n ≥ 1, thecotangent complex L(R′/$n)/(R/$n) vanishes in D(R/$n).

Proof. By the same reasoning in Lemma 3.5.7, we have a quasi-isomorphism in D(R/$n):

LR′/R ⊗LR (R/$n)

'−→ L(R′/$n)/(R/$n)

and hence it’s enough to check that LR′/R ⊗LR (R/$n) ' 0.

We consider the following short exact sequence of complexes of (R/$n)-modules:


0→ ($n−1R/$nR)[0]→ (R/$n)[0]→ (R/$n−1)[0]→ 0

Upon derived tensoring it by LR′/R ⊗LR (R/$n), the right side becomes L(R′/$n−1)/(R/$n−1),

which is quasi-isomorphic to zero by inductive hypothesis.

We regard the ring R/$ as an (R/$n)-module under reduction modulo $, and we have an(R/$n)-module isomorphism:

R/$ → $n−1R/$nR

given by sendig x ∈ R/$ to the class $n−1(x+$nR) +$nR. We have:

(LR′/R ⊗LR (R/$n))⊗L

R/$n (R/$) ' L(R′/$)/(R/$)

in D(R/$), and the latter vanishes again by Lemma 3.5.7. It follows L(R′/$n)/(R/$n) ' 0 inD(R/$n), as desired.

Remark 3.6.3 The above Lemma 3.6.2 clarifies what it means for any perfectoid R-algebraR′ to be analytically connective. We have unique (up to unique isomorphism) unobstruc-ted deformations of all the reductions modulo powers of $. This observation is the key inmany calculations in étale cohomology of rigid analytic varieties by means of perfectoid spacesSpa(R[1/$], R), R some perfectoid ring, as one can avoid dealing with continuous cohomologyby performing the needed calculations along (any of) the loci $n = 0 in Spf(R).

We still need to show Theorem 3.6.1, and Lemma 3.6.2 suggest a strategy.

Let A be a perfectoid (R/$)-algebra. Given the surjection with square-zero kernel R/$2 →R/$, we’d like to lift A =: A0 to an (R/$2)-algebra A1 uniquely up to unique isomorphisms.The obstruction to existence of such lifting lies in

Ext2A0

(LA0/(R/$), $/$2)

which vanishes, since LA0/(R/$) vanishes in D(R/$). Such vanishing also ensures uniquenessof the existing lifting A, as the (nonempty!) set of such liftings is an Ext1

A0(LA0/(R/$), $/$

2)-torsor. Any such lift has automorphism group isomorphic to Ext0

A0(LA0/(R/$), $/$

2), whichis, again, trivial. It follows A1 exists and is unique up to unique isomorphism. Let’s assume thatwe can draw an additional piece of information from the above vanishing, that is: LA1/(R/$2)

also vanishes in D(R/$2). Then, inductively, this process yields a projective system (An)n≥0

of (R/$n)n≥0-algebras, where A0 = A is a perfectoid (R/$)-algebra. Set R′ := lim←−An, whichis naturally an algebra over R ' lim←−R/$

n.

Proceeding under the assumption that LAn/(R/$n) ' 0 in D(R/$n) for each n ≥ 1, we provethe following:Proposition 3.6.4 R′ is a perfectoid R-algebra.

Proof. We need to check that R′ is $-adically complete, that the pth power map R′/p→ R′/pis surjective, and that the kernel of ϑ : Ainf(R)→ R is principal. The first condition is obviousby design of R′. We observe that ∩k≥0$

kR′ = (0). Indeed, if we call I := ∩k≥0$kR′, and

R′′ := R′/I, then R′/$k ' R′′/$k, for all k ≥ 1, and by uniqueness of R/$k-deformations ofR′/$ up to unique isomorphism, for all k ≥ 1, it follows R′′ ' R′, ie. I = (0).

Moreover, checking that the pth power map R′/p→ R′/p is surjective is equivalent to checkingthat the pth power map R′/$ → R′/$ is surjective, upon checking that $p divides p in R′.Since tensor products are right-exact on (R/p)-modules, since $p divides p in R, so does inR′, and we have:

R′/$1/p = A/$1/p ' A = R′/$,


so in particular the pth power map R′/$ → R′/$ is surjective. To show that ker(ϑ) is principal,it’s enough to check that $ is not a zerodivisor in R′, because the assertion follows from thefact that the pth power map R′/$1/p → R′/$ is an isomorphism. This is explained in Remark3.6.5 below.

Remark 3.6.5 We recall for clarity how we use the cotangent complex to deal with deform-ations along square-zero closed immersions. We’re given a surjective ring map A A0 withkernel I, and a ring map A0 → B0, together with a B0-module J and a B0-module surjectionB0 ⊗A0

I J . We assume I2 (and hence J2) is zero. We’d like to lift the map A0 → B0 to amap from A to some ring B which comes with a surjection to B0 which is A-linear, and has aB0-module isomorphism from J to the kernel of B → B0.

B completes the following diagram:

0 // J // B // B0// 0

0 // I

OO

// A //

OO

A0//

OO

0

Note that here we require the natural map I ⊗A0B0 → J to be an surjection because we want

the rightmost square to be cartesian.

Additionally, we require it to be an isomorphism, for the following reason. In the case the baseperfectoid ring R is the valuation ring of a perfectoid field C, we’ll know that any perfectoid(R/$)-algebra is flat, and the above isomorphism condition enables our deformation theory tokeep track of flatness by the following equivalence: given a surjection of rings A A0 withkernel I which is nilpotent, then any A-moduleM is A-flat if and only ifM⊗AA0 is A0-flat andTorA1 (M,A0) = 0. Note that in this equivalence there’s no noetherianity assumption, and thatthe condition TorA1 (M,A0) = 0 means exactly that the natural surjective map I ⊗AM IMis an isomorphism. In this situation, flatness of the reduction of M ensures flatness of M , ie.flatness lifts along nilimmersions.

Deformations of the datum (A → A0, I) are governed by RHomB0(LB0/A0

, J) in degrees 0,1 and 2. We remark that in our situation we have unobstructed deformations, unique up tounique isomorphism (anyway any deformation B comes with an isomorphism B ⊗A A0

'−→ B0,and isomorphisms of deformations are required to act trivially on B ⊗A A0 = B0). Therefore,in our case we always have a deformation B, with a unique isomorphism B ⊗A A0 ' B0, andan isomorphism B0 ⊗A0

I ' J , and these data are unique, up to unique isomorphism.

In our case of Proposition 3.6.4, we have a perfectoid (R/$)-algebra A, which comes with a flatmap Λ0 → A which factors through a flat map Λ0 → R/$. Note that obviously this doesn’timply at all the map R/$ → A is flat.

The maps:Λ0 → R/$ → A

yield a distinguished triangle:

LR[/Λ0⊗L

Λ0A→ LA/Λ0

→ LA/(R[/$) → LR[/Λ0[1]⊗L

Λ0A

which implies that the map Λ0 → A uniquely lifts to a flat map Λ→ R′[ which factors through

a flat map Λ→ R[, implying $[ is R′[-regular. This yields $ is R-regular.Remark 3.6.6 As a final comment, connectiveness of perfectoid (R/$)-algebras is what en-sures not only existence and uniqueness (up to unique isomorphism) of a $-adically complete


R-algebra R′ whose reduction mod $ recovers the given (R/$)-algebra, and trivially satis-fies the condition on the pth power map inducing an isomorphism R′/$1/p ' R′/$, but alsoensures that $ is not a zerodivisor in R′, hence implying the kernel of ϑ : Ainf(R

′) → R′ isprincipal, and therefore R′ is perfectoid.

We’re left to show the following:Proposition 3.6.7 Fix R be a perfectoid ring, and A1 a perfectoid (R/$)-algebra. Let A bethe unique (R/$2)-deformation of A1. Then:

LA/(R/$2) ' 0

in D(R/$2).

Proof. We have a short exact sequence of (R/$2)-modules:

0→ $R/$2R→ R/$2 → R/$ → 0

where the left side is (R/$2)-module isomorphic to (R/$), regarded as an (R/$2)-moduleunder reduction mod $. Proceeding as in Lemma 3.6.2, we’re required to check that

LA/(R/$2) ⊗LR/$2 (R/$) ' 0

in D(R/$).

We claim this latter is LA0/(R/$), so we conclude.

The pth power map induces an isomorphism

R/(p,$2/p)'−→ R/(p,$2)

and we can form the inverse limit along the pth power maps:

S := lim←−R/(p,$2)

and consider, being S perfect as an Fp-algebra, the ring of Witt vectors W (S). The pth powermap also induces an isomorphism:

A/(p,$2/p)'−→ A/p

so we can form the inverse limit along the pth power maps:

S′ := lim←−A/p,

a perfect Fp-algebra, and consider the ring of Witt vectorsW (S′). By Witt vector functoriality,we are given a natural map W (S) → W (S′). Now, by perfectness of S′, W (S′)/p ' S′, andsince p is W (S′)-regular we also have, :

W (S′)⊗LW (S) (W (S)/p) ' S′[0]

in D(W (S′)). We call [$2[] the Teichmüller lift of (0, $2/p, $2/p2 , . . . ) ∈ S′, and [$[] theTeichmüller lift of ($,$1/p, . . . ) ∈ S′. Note that [$[]2 = [$2[] in W (S′), and that [$2[] isW (S′)-regular, because $[ is regular in S′ and by Proposition 2.3.12. We have, therefore:

W (S′)/(p, [$[]2) ' A

and we deduce, in D(W (S′)):

W (S′)⊗LW (S) (W (S)/(p, [$[]2)) ' A[0]


together with:W (S′)⊗L

W (S) (W (S)/(p, [$[])) ' A1[0].

We remark that we are not endowing the left sides wth the structure of a derived ring, butrather the above quasi-isomorphism are used to concisely express Tor-independence, whichenables us to use the base change theorem for the cotangent complex.

Therefore, we conclude that in D(W (S′)):

LW (S′)/W (S) ⊗LW (S) (W (S)/(p, [$[]2)) ' LA/(R/$2)

and

(LW (S′)/W (S)⊗LW (S) (W (S)/(p, [$[]2)))⊗L

W (S)/(p,[$[]2)W (S)/(p, [$[]) ' LA/(R/$2)⊗L (R/$).

The left side is quasi-isomorphic in D(A1) to LW (S′)/W (S) ⊗LW (S) (W (S)/(p, [$[])), which is in

turn quasi-isomorphic in D(A1) to LA1/(R/$), as desired. Indeed, the desired vanishing alsooccurs in D(R/$).

Remark 3.6.8 Note that the same argument showing the above Proposition 3.6.7, provedthat if we have Ak an (R/$k)-algebra deformation of A1 for all k = 1, . . . , n− 1, n ≥ 2 beinga fixed integer, and if LAk/(R/$k) ' 0 in D(R/$k) for all such k, then not only we deduce aunique (R/$n)-algebra deformation of A0 (hence of Ak for all such k) exists and is unique upto unique isomorphism, but also LAn/(R/$n) ' 0 in D(R/$n).

To sum up, given the vanishing of LAn−1/(R/$n−1) in D(R/$n−1), there exists a unique (up tounique isomorphism) (R/$n)-deformation An of An−1. Given the vanishing of LA1/(R/$) inD(R/$), we form the short exact sequence of (R/$n)-modules:

0→ R/$ → R/$n → R/$n−1 → 0

and choose a representative of LAn/(R/$n) in the category of complexes of (R/$n)-modulesmade of flat (R/$n)-modules (this is possible). Tensoring by such representative, we’re reducedto check vanishing of the cohomology of the outer complexes, which amounts to checking that:

LAn/(R/$n) ⊗LR/$n (R/$n−1) ' 0

in D(R/$n−1) and:LAn/(R/$n) ⊗L

R/$n (R/$) ' 0

in D(R/$). The above argument shows that these are, respectively, quasi isomorphic toLAn−1/(R/$n−1) and LA1/(R/$) in the respective derived categories, and we conclude.

We note that the arguments from above achieve Theorem 3.6.1.Theorem 3.6.9 Let R be a perfectoid ring.





Ra-Perf → (Ra/$)-Perf



Proof. For (1) we only observe that vanishing of the full cotangent complex of the structuremap of any perfectoid (R/$)-algebra A ensures existence of a perfectoid (by Proposition 3.6.4)R-algebra R′ reducing to A modulo $, thus showing essential surjectivity. Full faithfullness isagain ensured by such vanishing, as deformations of maps between perfectoid A0 := (R/$)-algebras B0 → B′0 along a square-zero closed immersion A→ A0 with ideal I are governed bythe complex in D(B0):

RHomB0(LB0/A0, B′0 ⊗A0 I) ' 0.

For (2), we observe for any perfectoid (Ra/$)-algebra A there is an (R/$)-algebra A′ suchthat A′a = A. Moreover, we know the natural map

LA′/(R/$) → LA!!/(Ra/$)!!

is an almost quasi-isomorphism in D(R/$). If we show that A′ can be chosen to be perfectoid,then vanishing of the former allows us to repeat the arguments for almost algebras. We addressthis last aspect in the remaining of this section.

Remark 3.6.10 This, together with the tautological equivalence (R/$)-Perf ' (R[/$[)-Perf ,concludes the proof of Theorem 3.1.1:

R-Perf ' R[-Perf

and its almost version.

Let’s fix a perfectoid (Ra/$)-algebra A. We aim at showing that there exists a perfectoid(R/$)-algebra A′ with the property that A′a = A. Of course, A is of this form for some(R/$)-algebra A′, but we need this to be perfectoid. Since the functor (·)a from R-algebras toalmost R-algebras is exact, we reduce to investigate existence, given a perfectoid Ra-algebra A,of a perfectoid R-algebra A′ such that A′a = A as perfectoid Ra-algebras. In light of Lemma3.2.8, which says that being $-adically complete for almost modules can be checked on almostelements, a good candidate for A′ will be A∗. $ will be regular in A∗ since it is in A, as Ais perfectoid by assumption and (·)∗ is left exact (hence the kernel of multiplication by $ onA∗ is almost zero, but A∗ contains no nonzero almost zero elements). We are left to show thefollowing.Lemma 3.6.11 Let A be a perfectoid Ra-algebra. Then, the pth power map :

A∗/$1/p → A∗/$

is an isomorphism.

Proof. It is an almost isomorphism, and we claim, first, it is injective. Suppose x ∈ A∗ is suchthat xp ∈ $A∗. Since xp ∈ $A∗, then for all ε ∈ m we have εx ∈ $1/pA∗ by almost injectivity.We can apply Lemma 3.2.7 to $1/pA∗ to conclude it equals ($1/pA)∗ as an R-module. Bythe same Lemma, we deduce x ∈ ($1/pA)∗ = $1/pA∗, which establishes injectivity. Forsurjectivity, let us endow the ring R′ := A∗[1/$] with the unique Huber ring topology makingA∗ open and bounded as a subring. Suppose z ∈ R′ is such that zp ∈ A∗. Then there existssome positive integer N such that $N/px ∈ A∗, as A∗ is open. We call z′ := $N/pz, anddeduce z′p ∈ $A∗, so that, by injectivity, z′ ∈ $1/pA∗. Given that z is not $-torsion, as $is regular, we get $

N−1p z ∈ A∗, and by inducting on N we conclude. We are ready to show

surjectivity of the pth power map A∗/$1/p → A∗/$. By almost surjectivity, it is enough toshow that the composition:

A∗/$1/p → A∗/$ → A∗/m


is surjective. Let x ∈ A∗. By almost surjectivity we get some y ∈ A∗ such that:

$1/px ≡ yp mod $A∗.

We call z := y/$1/p2 , and get that zp ≡ x modulo $p−1p A∗. In particular, zp ∈ A∗. By the

above reasoning, we deduce z ∈ A∗, and since x is zp modulo $p−1p A∗, hence in particular

modulo mA∗, we achieve surjectivity, as desired.

The proof of the above Lemma has a very important consequence.Lemma 3.6.12 Let R be a Tate-perfectoid ring, and set A := R0a. Then R0 = A∗.

Proof. It is clear that A∗ consists of power-bounded elements, hence we have A∗ ⊆ R0. Thecrux is the reverse inclusion. Let x ∈ R. For all ε ∈ m, we get that εx is topologically nilpotent,and hence by openness of A∗ we deduce (εx)p

N ∈ A∗ for sufficiently large N ≥ 0. But thenεx ∈ A∗ (!) and since A∗ has trivial $-torsion because R is perfectoid, we can apply Lemma3.2.7 to A∗ and deduce that x ∈ A∗, as desired.

3.7 Tilting the open locus Spa(R,R)− $ = 0

As remarked in the previous section, we don’t strictly need almost mathematics to show thetilting equivalence over general perfectoid rings. However, for geometric purposes, the adicspaces Spa(R,R) for R a perfectoid ring is not well suited to possess a well behaved étale site(in fact to possess one at all). We’re going to discuss examples, but for now we just leavethe needed geometric understanding to the reader, and deduce we’re going to rather focus onthe open locus Spa(R[1/$], R) → Spa(R,R), where R[1/$] is given the unique Huber ringtopology making R into an open and bounded subring.

Note that R is endowed with the $-adic topology, and $ is a regular element dividing p in R.It follows R[1/$] is Tate-perfectoid, and hence we focus on this case here. From now until theend of this section, R will denote a Tate-perfectoid ring. We know its subring of power-boundedelements R0 is perfectoid, as well as any other open and integrally closed subring R+. In fact,any ring of definition R0 ⊂ R will be perfectoid.

We address the following Theorem.Theorem 3.7.1 Let R be a Tate-perfectoid ring, R[ its tilt. Then tilting induces an equivalenceof categories:

R-Perf → R[-Perf .

To show this, we’d like to claim that since we know R0-Perf ' R[0-Perf from Theorem3.1.1, then it’s enough to check that “taking power bounded subrings” yields an equivalence ofcategories R-Perf → R0-Perf . There’s an issue. Let S be any perfectoid R0-algebra, endowedwith the $-adic topology, and form S[1/$] endowing it with the unique Huber ring topologyinduced by the topology on S and making S into an open and bounded subring. Who on earthtells you that S0 = S? In fact, we may take any other open integrally closed subring S+ ⊂ S0.S+ is again a perfectoid R0-algebra, and we have S+[1/$] = S[1/$].

Given a Tate-perfectoid ring R, there’s a discrepancy between R0 and any R+ ⊂ R0 open andintegrally closed, which is controlled by almost mathematics. In fact, let x ∈ R0. Then $xis topologically nilpotent, since x is power-bounded. It follows ($x)n ∈ R+ for large enoughn ≥ 1, and since R+ is integrally closed in R0, then $x ∈ R+. Likewise for all p-power rootsof $, and therefore mx ⊂ R+. More precisely, we’re singling out the following set:

x ∈ R | mx ⊂ R+.


By Lemma 3.2.1, this set doesn’t depend on R+ now! and exactly recovers (R+a)∗ = (R0a)∗,because $ is regular in both R0 and R+, and R0[1/$] = R = R+[1/$].

This suggests the following proposition, which is the only missing step we’re left to check toachieve Theorem 3.7.1.Proposition 3.7.2 Let R be a Tate-perfectoid ring. Then, the assignment (·)0a is a functorR-Perf → R0a-Perf , and is, in fact, an equivalence of categories. Its quasi inverse is theassignment (·)∗[1/$].

Proof. Lemma 3.6.11, Lemma 3.6.12 and Lemma 3.2.1 do the job.

Remark 3.7.3 We already remarked that our definition of Tate-perfectoid rings and almostrings recover the original definitions of perfectoid algebras and almost algebras over a perfectoidfield. We note that regularity of $ was automatic in any such algebra, and inherited byregularity of $ in C0, C being the perfectoid base field, by flatness, which was built into thedefinitions. Such condition was introduced because given any perfectoid C-algebra R, then R0

was automatically C0-flat, since C0 is a valuation ring. The generality adopted here achievesthe tilting correspondence for perfectoid spaces as Spa(Zcyc

p [[x1/p∞ ]]∧,Zcycp [[x1/p∞ ]]∧), or the

open locus p 6= 0 inside it: Spa(Qcycp [[x1/p∞ ]]∧,Zcyc

p [[x1/p∞ ]]∧), or the open locus x 6= 0,etc. As well as any (other) rational subdomain inside it. None of these lives over a perfectoidpoint. We note that, strictly speaking, perfectoid spaces will be appropriately defined only inthe next chapters, and we refer the reader to §5.Remark 3.7.4 We finally note that, from the proof of the tilting equivalence for Tate-perfectoid rings, the untilting functor (·)# can be made explicit. Given a Tate-perfectoidR-algebra R′, R′[ its tilt, we have R′[# = (Ainf(R

′0) ⊗Ainf (R0) R0)[1/$]. This is exactly Fon-

taine’s functor.


4 Finite étale extensions and finite étale tilts

In this chapter we discuss Faltings’ theory of almost étale extensions in the generality of per-fectoid rings and of Tate-perfectoid rings, the ultimate purpose being the proof of Faltings’almost purity theorem, in the geometric framework provided by perfectoid spaces.

4.1 Setup and synopsis

From now on, we fix a Tate-perfectoid ring R. Consider the categories of finite étale algebrasRfet and R[fet. We want to show that the tilting functor S 7→ S[ induces an equivalence ofcategories between them. Such equivalence shall be called (affinoid) equivalence of the finite-étale sites, and will be eventually proved in the next sections, along with its global geometricform.

We begin setting up the following diagram with the aim of filling in the dotted arrows withfunctors:

R′fet

// R′0afet

// R′0afet

R′[0afet

R′[0afet

oo R′[fet

oo

R-Perf '// R0a-Perf '

// R0a/$ −Perf R[0a/$[-Perf R[0a-Perf'oo R[-Perf'

oo

where the bottom row of equivalences expresses the “tilting equivalence” R-Perf ' R[-Perf .Recall R is Tate-perfectoid, and therefore R0 is perfectoid and we do have the tilting equivalencein the form R0-Perf ' R[0-Perf , but to take care of the discrepancy between R0 and any openintegrally closed subring R+ ⊂ R0, we need to appeal to almost mathematics, as R+a-Perf 'R0a-Perf , and this latter is equivalent to R-Perf . Likewise on the tilted side.

We need to define the dotted arrows, the vertical ones aiming at being forgetful functors,and the horizontal ones aiming at being, again, equivalences. We shall deduce from this theequivalence of categories Rfet ' R[fet, with the aim of “geometrizing” it, in mind.Remark 4.1.1 We recall that tilting sends perfectoid fields to perfectoid fields. If R is Tate-perfectoid, R[ is a field if and only if R is. This was shown by considering the norm

‖x‖R = sup|t|−1 | t ∈ k×, tx ∈ R0,

and noting that ‖ · ‖R is multiplicative if and only if ‖ · ‖R[ is multiplicative. It followsthat equivalence of the finite étale sites over R and R[ restricts to the Fontaine-WintenbergerTheorem, as we shall discuss at the very end of these notes.

In §4.2 we prove Faltings’ almost purity in positive characteristic p > 0, and we draw all of itsconsequences in §4.3. Finally, we prove the full almost purity theorem in the case of perfectoidfields, in §4.5.

4.2 Almost Purity in positive characteristic

We give a proof of Faltings’ almost purity in the generality of Tate-perfectoid rings of charac-teristic p > 0.


Theorem 4.2.1 Let R be a Tate-perfectoid ring of characteristic p > 0. Then Rfet → R0afet is

an equivalence of categories.

The meaning of the above theorem is as follows. If we let S be a finite-étale R-algebra, Sinherits perfectoidness from R, and moreover the induced map R0 → S0 is finite-étale in thesense of almost mathematics.

An example from the classical computations of Tate in Galois cohomology.Example 4.2.2 LetK∞ be a totally ramified extension of Qp with Galois group Zp (a ramifiedZp-extension). Take L∞ to be a finite extension of K∞. Then let OK∞ and OL∞ be thecorresponding valuation rings. Tate showed in his seminal paper [Tate]:

Tr(OL∞) ⊃ mK∞

implying that the corresponding almost extension

OK∞ → OL∞

is finite étale in the sense of almost mathematics. In a later version of these notes we shallrephrase Tate’s paper by means of almost étale extensions and perfectoid spaces. If we have aproper smooth adic space X over a nonarchimedean field k, it comes with an étale map to ad-dimensional torus Td for some integer d ≥ 0. The pullback X of X along the proétale coverTd is a perfectoid space, Td being constructed as being perfectoid. The basic result:

Hicont(Z

dp, H

0proet(X, Zp)) ' Hi

proet(X, Zp)

for all i ≥ 0 and compatibly with cup products, following from the Cartan-Leray spectralsequence, allows an analysis of Tate’s normalized traces (it turns out that Tate’s normalizedtraces are a very special instance of trace maps on lisse Zp-sheaves on the pro-étale site of acertain X), and in general of the results in Tate’s paper.Remark 4.2.3 An important aside. Let R be any ring and t ∈ R a regular element. Thenany finitely generated R[1/t]-module M carries a unique topology which endows any finitelygenerated R-submodule with the t-adic topology, and makes it into an open submodule.

In order to show the Theorem, we discuss a slightly more general result, thus changing notation.

Let V be a perfect Fp-algebra, and let m be an ideal satisfying m2 = m. Recall that we callthe couple (V,m) a basic setup. Let t ∈ m be a regular element, and assume that m satisfies:

m = lim−→ tp−nV.

Among the good properties of such m there is flatness, and the fact that the Nth powers of thegenerators of m still generate m.

We let V a-Et be the category of uniformly almost finite projective étale V a-algebras.

More in detail, its objects are V a-algebras which are, in addition, almost projective V a-modules(recall how important it is to stress almost projectiveness, as the category of V a-modules doesnot contain enough categorical projectives) and are uniformly almost finitely generated as V a-modules. Morphisms are morphisms of V a-algebras.

We denote with V [1/t]-Et the category of finite étale V [1/t]-algebras, finite-étaleness being inthe sense of commutative algebra.

We now define the functor V a-Et→ V [1/t]-Et:

A 7→ A∗[1/t]


between these two categories.

The main theorem we show in this section is the following:Theorem 4.2.4 The above functor

(·)∗[1/t] : V a-Et→ V [1/t]-Et


We note that we are not under the assumption that V is an algebra over a field k other thanFp. We shall apply this result to the case V is a Tate-perfectoid ring R.

Proof. The goal is to construct a quasi-inverse functor.

Let R be a finite étale V [1/t]-algebra. Since V is perfect, and m ⊂ V contains all p-power rootsof t, by design, then V [1/t] is also perfect, and so is R, as perfectness is stable under flat basechange.

We choose some finite V [1/t]-subalgebra R0 ⊂ R such that R0[1/t] = R. We denote by ϕ thepth power map on R, and we now define:

S := lim−→n≥0

ϕ−n(R0).

Claim 1 The assignment of S does not depend on the choice of R0.

Proof of Claim 1. Let R′0 ⊂ R be another choice of a finite V [1/t]-subalgebra of R sharing thesame property of R0. We have:

tNR0 ⊆ R′0 ⊆ t−NR0

for a sufficiently large integer N ≥ 0. Indeed, to see this we note that the containmentR′0 ⊂ R0[1/t] means that, for sufficiently large N , we have that t−NR0 contains R′0. Uponexchanging the roles of R0 and R′0 we deduce the desired remaining containment. It follows:

tNp−nϕ−n(R0) ⊂ ϕ−n(R′0) ⊂ t−Np

−nϕ−n(R0)

for all n ≥ 0, and we conclude upon passing to direct limits.

Claim 2 Sa is an unramified V a-algebra.

Proof of Claim 2. Since R is unramified as a V [1/t]-algebra, we deduce existence of an idem-potent e ∈ R⊗V [1/t] R which annihilates the kernel of the (co)diagonal map (ie. the multiplic-ation map R⊗V [1/t] R→ R), and which is sent to 1 by such map. Geometrically, the locus ofinvertibility of e in Spec(R⊗V [1/t] R) is clopen and isomorphic to Spec(R).

For large enough N , we have:tNe ∈ R0 ⊗V R0.

Hence, for all n ≥ 0 we obtain:

tNp−ne ∈ ϕ−n(R0)⊗V ϕ−n(R0)

meaning thate ∈ ((S ⊗V S)a)∗.


It is a straightforward exercise for the reader to check that e kills

ker(S ⊗V S)a → Sa)∗

and is sent to 1Sa by the above map.

We step back for a moment, to be precise. We introduce for convenience the (flat) V -module:

m := m⊗V m,

and observe that m is V -module isomorphic to m under the multiplication map. Note that ifwe substitute tp

−nwith tNp

−n, there is no harm done, as these latter still generate m.

We claim we can attach to every ε⊗ δ ∈ m, an element eε ∈ (Sa ⊗V a Sa)∗ V -linearly, so that

ε⊗ δ 7→ δ · eε

yields an element of the set

HomV (m, (Sa ⊗V a Sa)∗) = HomV a(V a, Sa ⊗V a Sa)) = (Sa ⊗V a Sa)∗

and we claim we can arrange such element, which we shall denote e ∈ (Sa ⊗V a Sa)∗, is idem-potent, is sent to 1Sa by the multiplication map, and kills its kernel.

It’s enough to choose ε = tNp−m

and δ = tNp−n

for nonnegative integers m,n, and set:

tNp−m⊗ tNp

−n7→ tNp

−men,

where en is the map V a → Sa given by multiplication by tNp−ne, e being our idempotent in

R⊗V [1/t] R. This proves that S is V -unramified in the sense of almost mathematics.

Claim 3 Sa is a uniformly almost finite V a-algebra.

Proof of Claim 3. For large enough N ≥ 0, we have

R0 ⊂ ϕ−1(R0) ⊂ t−NR0.

Applying ϕ−1 on both sides n times, we obtain:

ϕ−n(R0) ⊂ ϕ−(n+1)(R0) ⊂ t−Np−nϕ−n(R0).

Inductively, we obtain:

ϕ−(n+k)(R0) ⊂k∏j=0

t−Np−(n+j)

ϕ−n(R0) ⊂ t−Np−(n−1)

ϕ−n(R0).

Since there is no dependency on k on the right-hand side, upon passing to the direct limit overk we obtain

S ⊂ t−Np−(n−1)

ϕ−n(R0)

for all n. The claim follows.


Claim 4 Let T be the R-normalization of V . Then Sa = T a.

Proof of Claim 4. We already have that all the elements of S are integral over V , because theyare obtained via images of elements of R0 via the inverse of ϕ, and R0 is V -finite. Then, wehave the containment S ⊂ T . We need to show T a → Sa. Endow R with the unique topologyrestricting to the t-adic topology on R0, such that R0 is open as a V -submodule. We claim Sa∗is the subring of all power-bounded elements. If this is true, then T is contained in Sa∗ , and itwould follow T a = Sa. Indeed, T is made of power-bounded elements: to see this, we may pickx ∈ T . x is integral over V , implying that there is a polynomial relation over V :

xn + vn−1xn−1 + · · ·+ v0 = 0.

We define U := R0 and U ′ := R0, and get that xk, k = 0, . . . , nU ⊂ U ′, which means x ispower-bounded.

We prove that Sa∗ is the set of power-bounded elements of R. Sa∗ can be characterized as theset of those x ∈ R such that mx ⊂ S (Lemma 3.2.1). This implies that Sa∗ consists of power-bounded elements. On the other hand, if x ∈ R is power-bounded, then δx is topologicallynilpotent for all δ ∈ m. Since R0 is open in R, it follows that for sufficiently largeN , (δx)N ∈ R0.We choose N = pk large enough, and obtain: ϕk(δx) ∈ R0, that is δx ∈ S, ie. x ∈ Sa∗ , asdesired.

Claim 5 Sa is almost projective as a V a-module.

Proof of Claim 5. As a special case of the above Claim 4, we get that ifW is the normalizationof V in V [1/t], then W a = V a. Let

TrR/V [1/t] : R→ V [1/t]

be the trace map. We claim that its image is exactly W . This is seen assuming R has finiteconstant rank n (because it is almost projective and finite) and then after a faithfully flat basechange we can assume R is V [1/t]-free. Then Tr restricts to a morphism T : Sa∗ → V a∗ . ByClaim 2, for all δ ∈ m

δe =

n∑i=1

xi ⊗ yi

for xi, yi ∈ S. We have, for all b ∈ Sa∗ ,

δb =

n∑i=1

xiT (byi).

We can define morphisms α : Sa → (V a)n and β : (V a)n → Sa by

α(b) = ((T (byi))

and β(v1, ..., vn) =∑xivi, for v1, ..., vn ∈ V a∗ , and then applying the functor (·)a, thus getting

β α = δ · 1Sa , which achieves Claim 5.


Claim 6 The functor (·)∗[1/t] : V a-Et→ V [1/t]-Et is fully faithful.

Proof. Given a finite étale V a-algebra A, since A∗ ⊂ A∗[1/t], then we deduce that the naturalmap:

(∗) HomV a(A,B)→ HomV [1/t](A∗[1/t], B∗[1/t])

is injective. In fact, we have HomV a(Aa∗, Ba∗ ) = HomV (A∗, B∗).

Since A and B are almost finite and étale over V a, the pth power map on A and B is anautomorphism. Let ψ : A∗[1/t] → B∗[1/t] be any map of V [1/t]-algebras. Since A is almostfinite, ψ(A∗) ⊂ t−NB∗ for large enough N ≥ 0. The pth power map commutes with allhomomorphisms, and therefore:

ψ(A∗) ⊂ t−Np−nϕ−nB∗ (B∗) = t−Np

−nB∗

for all n ≥ 0. Therefore, ψ induces a morphism ψa : A→ B, yelding also surjectivity of (∗), asdesired.

As a result, it follows that R 7→ Sa is a quasi-inverse of the functor (·)∗[1/t] : V a-Et →V [1/t]-Et. Theorem 4.2.4 is proved.

To fix ideas, we sum up the above proof in a simplified version, in the following Example.Example 4.2.5 Let V be any Fp-algebra, with an ideal m ⊂ V such that m2 = m, and taket ∈ m to be a regular element such that m is the perfection of the principal ideal tV .

Let A be a flat V -algebra, integrally closed in A[1/t]. Let B′ be a finite étale A[1/t]-algebra,and let B be the normalization of A in B′. The map A → B will not be étale in general. Weassume A is perfect, and we claim Aa → Ba is finite étale in the sense of almost mathematics.

Indeed, let e ∈ B′ ⊗A B′ be a diagonal idempotent. There is N > 0 such that tNe ∈ B ⊗A B,and since A is perfect, B′ is perfect, and then B is perfect. Applying the inverse of the pthpower map automorphism repeatedly, we get:

tNp−ne ∈ B ⊗A B,

and if we iterate we get e ∈ (Ba ⊗Aa Ba)∗. Then, Aa → Ba is unramified. Now take ε ∈ m.We have:

εe =

N∑i=1

xi ⊗ yi ∈ B ⊗A B

Consider the maps as before:B → A⊕N → B

Their composition is given by multiplication by ε, and since N does not depend on ε, weobtain that B is uniformly almost finitely presented almost projective. The idea to keep inmind is that perfectness ensures that such diagonal idempotent is automatically a diagonalidempotent for the integral structure, in the sense of almost mathematics. Given this key step,the rest follows fairly easily! In the last instance, the above is due to the fact that m is theperfection of a principal ideal, and hence the basic setup (V,m) makes it extremely easy tocheck almost-trivialty of modules in characteristic p.


4.3 Consequences of the positive characteristic Almost Purity

We fix a Tate-perfectoid ring R and its tilt R[. A consequence of the Almost Purity Theoremis that, given a perfectoid R[-algebra R′[, the tilt of a perfectoid R-algebra R′, the categoryR′[fet comes naturally (in R) with the forgetful functor

R′[fet → R[-Perf

sending a finite-étale R′[-algebra S to itself, regarded as a perfectoid R[-algebra. Moreover, weeventually have a functor:

R′[fet → R′

[0afet ,

given by sending such S to S0a. By Almost Purity, now we know that S0a is indeed an objectof the category R′[0afet , and since the assignment S 7→ S0a is manifestly functorial, we obtainthe desired functor.

Moreover, by design of perfectoid R[0a-algebras, since, by Almost Purity (again!), S is perfect-oid as a R[-algebra, then S0a is a perfectoid R[0a-algebra, and therefore the category R′

[0afet

comes naturally with the forgetful functor:

R′[0afet → R[0a-Perf .

Finally, Almost Purity provides us with one more piece of information. The functor R′[fet →R′[0afet is actually an equivalence of categories.

Along the same lines, we would like to obtain a functor

R′[0afet → (R′

[0a/$[)fet

together with the forgetful functor

(R′[0a/$[)fet → (R[0a/$[)-Perf .

In order to do this, and to prove that such functor is an equivalence of categories, we clearlyhave to show the following two essential results:Theorem 4.3.1 Let B be a finite étale A-algebra, where A is a perfectoid R0a-algebra. ThenB/$ is a finite étale A/$-algebra. Moreover, B 7→ B/$ = B ⊗A A/$ induces an equivalenceof categories Afet → (A/$)fet.Proposition 4.3.2 Let A be a perfectoid (R0a/$)-algebra. Take B to be a finite étale algebraover A. Then B is perfectoid as a (R0a/$)-algebra.

The reader may regard Proposition 4.3.2 as a mod $ version of the Almost Purity Theorem.

We postpone their proofs to the end (see §A).

By Theorem 4.3.1, the natural tautological equivalences:

(A[0a/$[)fet ' (A0a/$)fet and (R[0a/$[)-Perf ' (R0a/$)-Perf

yield the forgetful functor:(A0a/$)fet → (R0a/$)-Perf


It’s not over. Proposition 4.3.2 means that if I’m given a finite-étale algebra B over A = A/$(which comes as B/$ for some A-algebra B), then since A is perfectoid, by assumption, wecan conclude that so is B. Then Theorem 4.3.1 adds more information: we know the functorB 7→ B yields a natural equivalence of categories Afet

'−→ Afet.

We claim B is (isomorphic to, hence itself) perfectoid as a R0a-algebra. We need to recallhow such B is constructed in the first place, which will be done in the proof of Theorem 4.3.1.Such construction, which involves once again the vanishing of a cotangent complex and henceunobstructedness of flat deformations of almost algebras, yields a B which is automatically flatand $-adically complete. Perfectoidness of B as an R0a-algebra amounts to showing that thepth power map:

B/$1/p → B/$

is an isomorphism. But B/$ = B, which is perfectoid as a (R0a/$)-algebra, hence isomorphicto B/$1/p under the pth power map. We have:

B/$1/p = B/($,$1/p) = B/$1/p

which exactly says that the pth power map B/$1/p → B/$ is an isomorphism, as desired.

The upshot is that, given the equivalence of categories, notation being as above:

Afet → (A/$)fet

and a finite-étale (A/$)-algebra B, hence perfectoid as a (R0a/$)-algebra, then any lift (uniqueup to isomorphism) B to Afet is itself perfectoid, as a R0a-algebra. As a consequence, thecombination of Theorem 4.3.1 and Proposition 4.3.2 provides us with the forgetful functor:

Afet → R0a-Perf .

At the end of this section, we shall also prove the following:Proposition 4.3.3 Let A be a perfectoid R0a-algebra, and B a finite-étale A-algebra. SupposeA = R′

0a for some perfectoid R-algebra R′. Then B∗[1/$] is finite-étale as an R′-algebra.

Given the Proposition, we are, therefore, in the position to complete the diagram we introducedbefore, the following way:

R′fet

// R′0afet

A 7→A∗[1/$]

ee

' // R′0afet

R′[0afet

R′[0afet

A7→A∗[1/$[]

88

'oo R′[fet

'oo

(·)#

rr



oo

Remark 4.3.4 Keeping track of the definitions of each functor, and going back to the proof ofthe tilting equivalence in §3, we observe that the untilting functor (·)# : R′

[fet → R′fet coincides

with the composition of the functors R′[fet → R′[0afet , given by S[ 7→ S[0a. Of R′[0afet → R′[0afet

given by B[ 7→ B[/$[. Of the tautological equivalence R′[0afet ' R′0afet induced by thecanonical isomorphism:

R′0a/$ ' R′[0a/$[


which lives over the canonical isomorphism R0a/$ ' R[0a/$[. Of the inverse functor of theequivalence R′0afet → R′0afet, which is given by the unique flat $-adically complete lift B ofB/$ to R′0afet (provided by the vanishing of L

B!!/R′0a!!in D(R0a

!!)). And, finally, of the functor

R′0afet → R′fet, given by A 7→ A∗[1/$].

This latter functor lands, a priori, in the category of perfectoid R-algebras with a (continuous)R-algebra map from R′. It is by Proposition 4.3.3 that we can say that its essential imagelies in the category R′fet, which also shows that the untilting functor on R′[fet has its essentialimage in the category of perfectoid R-algebras with a (continuous) R-algebra map from R′,lying in the (full) subcategory of finite étale R′-algebras. The very same argument we used toshow Claim 6 in Theorem 4.2.4, applies to show that the “generic fiber” functor R′0afet → R′fet

is fully faithful, which implies that

(·)# : R′[fet → R′fet

is fully faithful! To conclude that tilting induces an equivalence of categories, we are left tocheck essential surjectivity of (·)#, which will be done in §6, exploiting the global theory ofperfectoid spaces.Example 4.3.5 We digress a little on some advantages one can draw from knowledge of thestructure of the functors (·)[ and (·)#, that is, their factorization into the functors shown inthe diagram from before.

Let C be a perfectoid field. Then R = C〈T 1/p∞〉 is a perfectoid C-algebra.

Although we did this earlier, this really comes out of the tilting equivalence. We have:

R0 = C0[T 1/p∞ ]∧,

is $-adically complete. Then

R0/$ = (C0/$)[T 1/p∞ ] = (C[0/$[)[T 1/p∞ ].

Observing that C[〈T 1/p∞〉 is a perfect C[-algebra, by going backwards through the tiltingequivalence we conclude that R is perfectoid. The key idea, which will be used quite manytimes in the sequel, is to check perfectoidness in the almost setting. That is, to check whether ornot an algebra R′ over a Tate-perfectoid ring R is perfectoid, it is sufficient to check whether ornot either R′0a is a perfectoid R0a-algebra, or R′0a/$ is a perfectoid R0a/$-algebra, taking careof how the subring of power-bounded elements of R transforms along the various equivalences.The upshot is that these intermediate equivalences simplify the situation a lot, as modding outby $ gets rid of any complication of topological nature and reduces any perfectoidness questionto algebra.

4.4 Finite étale tilts over Tate-perfectoid rings

We fix a Tate-perfectoid ring R, with tilt R[. Suppose that R′ is a perfectoid R-algebra, andwrite: A = R′

0a. We denote by R′[ its tilt, as usual. We recollect the results obtained so farin the following Theorem (see Remark 4.3.4):Theorem 4.4.1 There is a fully faithful functor

(·)# : R′[fet → R′fet.


More precisely, we have the following diagram:

R′fet

// R′0afet

A7→A∗[1/$]

ee

' // R′0afet

R′[0afet

R′[0afet

A7→A∗[1/$[]

88

'oo R′[fet

'oo

(·)#

rr



oo

where the top leftmost functor (·)∗[1/$] is fully faithful.

Note that the functor mentioned in the first part of the Theorem is the quasi-inverse of thetilting functor on the category of all perfectoid R′-algebras (hence R-algebras), but we do notknow yet whether or not the tilting functor has a quasi-inverse if restricted to the categoryR′fet. This question will be investigated in the remaining of these notes, and partially solvedin the next §4.5.

4.5 Fontaine-Wintenberger

In §6, we shall see how important is to achieve essential surjectivity of the functor in Theorem4.4.1. Via a result of Gabber and Ramero [GR, Thm. ], we can reduce the general caseof perfectoid algebras over a Tate-perfectoid ring, to the known case of perfectoid fields (theresidue field of a perfectoid space at any point being a perfectoid field, as we shall discuss).

Essentially, the principle is encode into the following equivalence of categories:

2- lim−→x∈Spa(S,S+)

O+X(Spa(S, S+))[$−1])fet

'−→ κ(x)∧fet

where X is a perfectoid space over Spa(R,R0), with R Tate-perfectoid, and Spa(S, S+) runsover all affinoid open domains in X containing x.

It is therefore essential to prove the following:Theorem 4.5.1 Let C be a perfectoid field of characteristic 0. Then

C[fet → Cfet

is essentially surjective.

Proof. We call (·)# : C[-Perf → C-Perf the inverse functor given by the tilting equivalence,which yelds a fully faithful functor C[fet → Cfet by Theorem 4.4.1. Let us make this functorexplicit via Remark 3.7.4.

Let A[ be a perfectoid C[-algebra, and B[ a finite étale A[-algebra. Then we have:

(B[)# = W (B[0)⊗W (C[0) C

where W (C[0)→ C is given by the universal property of Witt rings and W (C[0)→W (B[0) isgiven by functoriality.

Since (B[)# tilts to B[, then it is a field if and only if B[ is, and given that it is already aperfectoid C-algebra, then it is a perfectoid field if and only if B[ is. We know, furthermore,that a perfectoid field is algebraically closed if and only if its tilt is.


Let E := C[ be the completion of an algebraic closure of C[. E is complete and perfect, andthen it is perfectoid. There exists a perfectoid field E# which tilts to E, and hence E# isalgebraically closed, and it contains C as E contains C[. Any finite extension C[ ⊂ F ⊂ Egives the untilt extension C ⊂ F# ⊂ E#, where F# is finite over C. Any element of E iscontained in some finite extension of C[, since E is algebraically closed. Then

N :=⋃

[F :C[]

F# ⊂ E#

is dense. By Krasner’s Lemma, N is algebraically closed. Therefore, any finite extension Lof C is contained in N . We can take the Galois closure of L in E#, to get a finite Galoisextension F of C[ such that L is contained in F#. The functor F 7→ F# preserves degreeand automorphisms, and then F# is still Galois. L is given as the fixed field of F# by somesubgroup H of Gal(F#/C) = Gal(F/C[), which gives a finite extension L[ := FH of C[ thatuntilts to L. We have, in fact,

(L[)# ⊂ (F#)H = L

and we have equality, since these two fields have the same degree.


5 Perfectoid spaces

We fix a Tate-perfectoid ring R, and call S := Spa(R,R0). S is an example of what we’re goingto call affinoid perfectoid space. We denote S[ := Spa(R[, R[0).

In this lecture we define global perfectoid spaces, and globalize the tilting functor to establishthe equivalence of perfectoid spaces over S and perfectoid spaces over S[. We discuss extremelyimportant properties of such spaces and the relations with their tilt, together with existenceof fibered products in the category of perfectoid spaces (for which the main point is to verifyuniformity of completed tensor products, achieved by using the intermediate steps in the proofof the tilting equivalence!)

The sheaf properties for perfectoid spaces are deduced directly from the fact that Tate-perfectoidHuber pairs, yet to be defined, are stably uniform, hence sheafy in the sense of [BV].

5.1 Affinoid perfectoid spaces

We are eventually in the position to define global perfectoid spaces.Remark 5.1.1 We recall a few facts. Suppose we are given a perfectoid ring R, with theopen, bounded and integrally closed subring R0. Let R+ ⊂ R0 be, again, an open (in R) andintegrally closed subring of R0. This is equivalent to saying that

R+/m ⊂ R0/m

is an integrally closed subring, m being generated by all the p-power roots of a pseudo-uniformizer $ ∈ R0 which is a unit in R. But, canonically:

R0/m ' R[0/m[,

so the choice of R+ likewise corresponds exactly to an open, bounded and integrally closedsubring R[+ ⊂ R[0. Note that all topologocally nilpotent elements of R are contained in R+,because R+ is open and integrally closed, so

mR0 ⊂ R+ ⊂ R0

and hence the inclusion of R+ into R0 is an almost-isomorphism! Thus, we have an equivalenceof categories

R+a-Perf'−→ R0a-Perf .

Definition 5.1.2 A Tate-perfectoid Huber pair is a Huber pair (R,R+) such that R is aTate-perfectoid ring. We shall abbreviate the terminology calling such pairs perfectoid pairs.

Via the above bijective recipe at the level of choices of open bounded integrally closed subrings,the tilting equivalence between perfectoid R-algebras and perfectoid R[-algebras naturally ex-tends to an equivalence between categories of perfectoid pairs over (R,R+) and (R[, R[+). Weshall denote by:

(R,R+)-Perf

the category whose objects are perfectoid pairs (R′, R′+

) which come with a morphism of Huberpairs (R,R+)→ (R′, R′

+), and whose morphisms are morphisms of Huber pairs. Likewise for

(R[, R[+).


Proposition 5.1.3 Tilting (R′, R′+

) 7→ (R′[, R′

[+), and the continuous projection x 7→ x#

induces an isomorphismR[+/$[ ' R+/$.

MoreoverR[+ = lim←−

x7→xpR+.

Proof. Let (R′, R′+

) be a perfectoid pair over (R,R+). We claim R[+ is the unique (up toisomorphism) open and integrally closed subring of R[ corresponding to R+ in R. This is nota tautology. Indeed, being $ topologically nilpotent, open and integrally closed subrings of R0

are in bijection with the integrally closed subrings of R0/$, which is isomorphic to R[0/$[.Such isomorphism carries this bijection to the integrally closed subrings of R[0/$[, which are,in turn, in bijection with the open and integrally closed subrings of R[0.

The claim follows from the fact that R[+/$[ ' R+/$, and from the fact that R[+ =lim←−x 7→xp R

+, as shown before.

The tilting equivalence for usual perfectoid R-algebras now establishes the Proposition.

Theorem 5.1.4 Let (R,R+) be a perfectoid pair, with tilt (R[, R[+). Define:

X := Spa(R,R+) and X[ := Spa(R[, R[+),

equipped with the presheaves OX ,O+X and OX[ ,O

+X[

respectively.

(1) We have a homeomorphism |X| ' |X[| given by sendngs x ∈ X to the points x[ ∈ X[

characterized byf 7→ |f(x[)| := |f#(x)|.

This homeomorphism identifies rational subsets in both directions. Moreover, the com-pleted residue fields at x and x[ are perfectoid fields and are naturally tilts of each other(so in particular their value groups are naturally identified).

(2) Let U ⊂ X be a rational subset, with tilt U [ ⊂ X[. Then the complete Huber pair over(R,R+):

(OX(U),O+X(U))

is perfectoid, and its tilt is uniquely isomorphic to (OX[(U[),O+

X[(U [)) functorially in U .

(3) The presheaves OX and OX[ are sheaves.

(4) The cohomology group Hi(X,O+X) is almost zero for all i > 1.

We prove this result in many steps, using the tilting correspondence throughout. One of thestriking features about the proof is that certain problems internal to the characteristic-0 storyseem hopeless to establish by direct arguments; over and over again we use the precise natureof the tilting correspondence, especially its use of categorical equivalences with categories ofalmost-algebras. Difficult problems in characteristc 0 will be reduced to problems in almost-mathematics in characteristic p, which in turn will be solved by working with perfectoid algebrasin characteristic p (where we can access the additivity of the p-power map, etc.).

The natural map |X| → |X[| is continuous

First, we need to show x[ indeed induces a valuation on R′[. The only thing that is not

straightforward to show is the ultrametric inequality. A natural short-cut idea is to identify


f 7→ |f(x)| with the valuation associated to the completed residue field at x and to try to “tilt”this field to directly define x[ and thereby access the settled case of perfectoid fields. But thisidea cannot be employed because we have not yet shown that the completed residue field atevery point of X is in fact a perfectoid field (and so has a tilt!). The perfectoid property ofsuch completed residue fields will be established later on, but its proof will use the full forceof the tilting equivalence! So we instead simply adapt some of the approximation argumentsused in the earlier considerations with perfectoid fields.

We fix f, g ∈ R′[ for which we want to prove that |(f + g)#(x)| ≤ max(|f#(x)|, |g#(x)|) in

Γx ∪ 0. Up to rescaling f and g by a big power of a chosen pseudo-uniformizer $[ of K[, wecan (by continuity of points in adic spectra) assume f, g ∈ R′[+. We have

f# mod $ = f mod $[

under the identification R′+/$ ' R′

[+/$[, and likewise for g and f + g, so f# + g# ≡

(f+g)# mod $R′+. Since the perfectoid ring R′[ in characteristic p is perfect, so the integrallyclosed R′[+ is also perfect, we may apply the preceding congruence with f and g replaced withtheir unique pmth-roots in R′[+ for any m ≥ 0. Thus, for all such m we have:

|(f + g)(x[)|1/pm

= |(f1/pm + g1/pm)#(x)|≤ max|$(x)|, |((f#)1/pm + (g#)1/pm)(x)|≤ max|$(x)|, |f#(x)|1/p

m

, |g#(x)|1/pm

,

so raising to the pmth-power gives that |(f + g)(x[)| ≤ max|$(x)|pm , |f(x[)|, |g(x[)|. Sincethis holds for all m, and |$(x)| is topologically nilpotent in Γx, it follows by consideringarbitrarily large m that |(f + g)(x[)| ≤ max(|f(x[)|, |g(x[)|) in Γx ∪ 0. This completes theverification that x[ makes sense as a valuation on R′[, and by design of R′[ we see easily thatx[ is continuous on R′[ and bounded by 1 on R′[+. Hence, x[ ∈ Spa(R′

[, R′

[+) as desired.

Let us consider a rational subset of X[ := Spa(R′[, R′

[+), for (R′, R′

+) a perfectoid pair:

X[

(f1, . . . , fn

g

), 〈f1, . . . , fn〉 = R[, g ∈ R′[.

We recall the argument saying that one can add fn+1 = ($[)N , for some large N , withoutchanging the rational subspace. Indeed, by definition, there exist elements a1, . . . , an ∈ R′

[

such thatn∑i=1

aifi = 1.

Since R′[+ ⊂ R[ is open, we have that for N sufficiently large, ($[)Nai ∈ R′[+ for all i. Forany ξ ∈ X[(f1/g, . . . , fn/g) we have

|($[)N (ξ)| =

∣∣∣∣∣n∑i=1

(($[)Nai)(ξ)fi(ξ)

∣∣∣∣∣ ≤ max|(($[)Nai)(ξ)||fi(ξ)| ≤ |g(ξ)|.

This implies that we can append$[N to the list of fi’s (i.e., assume fn = $[N ) without affectingX[(f1/g, . . . , fn/g) but achieving the property that f#

n = $N , whence the f#i ’s generate the

unit ideal of R′! It is therefore now obvious that we have the following:

Lemma 5.1.5 For f1, . . . , fn, g ∈ R′[, the preimage U# ⊂ X of U := X[(f1/g, . . . , fn/g)

under the map X → X[ is equal to X(f#1 /g, . . . , f

#n /g), and if U is rational then U# is

rational.


This proves that X → X[ is continuous. We next turn to a description of structure presheaveson certain rational domains.Remark 5.1.6 In the following arguments, we make use of the equivalences of categories:

R-Perf ' R0a-Perf ' (R0a/$)-Perf

the following way. Given a $-adically complete Huber R-algebra R′, we sometimes need tocheck it is perfectoid but computing R′0, or simply computing with R′0 even if we can determineit, may be very difficult. We will often instead verify corresponding conditions for R′0a/$ andthen use the concrete descriptions of the functors involved in the above equivalences to deducethat R′0a/$ is a perfectoid R0a/$-algebra, and consequently that R′0a is a perfectoid R0a-algebra.

To conclude that R′ is a perfectoid R-algebra, we will verify in cases of interest that thetopological R-algebra obtained from applying the inverse functor R0a-Perf → R-Perf to R′0a

is naturally isomorphic to R′, whence R′ is perfectoid! The key to doing this final miraclestep is to use the formula for that inverse functor, which arose in Remark 3.7.4 after provingthe absolutely fundamental fact that for a perfectoid R0a-algebra A, if we equip the R-algebraR′ := A∗[$

−1] with the unique Huber ring structure making A∗ open and bounded thennecessarily A∗ = R′

0 and R′ is perfectoid.

Lemma 5.1.7 Let X = Spa(R′, R′+

), and choose a rational domain U ⊂ X[ with correspond-ing preimage U# ⊂ X. Write U = X[(f1/g, . . . , fn/g) with fi, g ∈ R′

[0 and fn = $[N forsome N > 0. By abuse of notation, let (f#

i /g#)1/pm denote (f

1/pm

i )#/(g1/pm)# in R′[1/g#].

(1) For the subring

R′0[(F#/g#)1/p∞ ] := R′

0[(f#

1 /g#)1/p∞ , . . . , (f#

n /g#)1/p∞ ] ⊂ R′[1/g#],

let R′0〈(F#/g#)1/p∞〉 denote its $-adic completion. Then R′0〈(F#/g#)1/p∞〉a is Tate-

perfectoid as an R′0a-algebra.

(2) The algebra OX(U#) is a perfectoid R-algebra, with associated perfectoid R0a-algebra

OX(U#)0a ' R′0〈(F#/g#)1/p∞〉a.

(3) The tilt of OX(U#) is naturally identified with OX[(U) functorially in U .

Proof. To show (1), we shall construct isomorphisms

R′0[(F#/g#)1/p∞ ]/$ ' R′[0[(F/g)1/p∞ ]/$[

andR′

0[(F#/g#)1/p∞ ]/$1/p ' R′[0[(F/g)1/p∞ ]/($[)1/p

compatibly with the pth power map between left sides and between right sides, as then thedesired “almost isomorphism” property on the left side is reduced to its analogue on the rightside; i.e., it would reduce the problem to the case over R[, which is to say the case of char-acteristic p. We begin by assuming R is of characteristic p > 0. We already know thatR′

0〈(F#/g#)1/p∞〉 is $-adically complete and separated, with $ regular, as a R0-algebra, andso is R′0〈(F#/g#)1/p∞〉a as an R0a-algebra.

Consider the presentation:

0→ I ′ → R′0[T

1/p∞

1 , . . . , T 1/p∞

n ]→ R′0[(F#/g#)1/p∞ ]→ 0


in which I ′ is just the kernel of the surjection R′0[T1/p∞

1 , . . . , T1/p∞

n ] R′0[(F#/g#)1/p∞ ]. We

do not need to describe I ′ explicitly, but rader consider the ideal I ⊂ I ′ ofR′0[T1/p∞

1 , . . . , T1/p∞

n ],generated by the elements T 1/pm

i (g1/pm)# − (f1/pm

i )# as an ideal.

We claim the natural map

R′0[T

1/p∞

1 , . . . , T 1/p∞

n ]/I → R′0[(F#/g#)1/p∞ ]

is an almost isomorphism. LetM be the kernel of the above map, as an R′0[T1/p∞

1 , . . . , T1/p∞

n ]-module. Let f ∈ M . We know that the above map is an isomorphism after inverting $,because inverting $ also inverts g#, which implies that f becomes zero in M [1/$]. There-fore, there exists n ≥ 0 such that $nf = 0 in R′

0[T

1/p∞

1 , . . . , T1/p∞

n ]/I. Call f any lift toR′

0[T

1/p∞

1 , . . . , T1/p∞

n ]. Then $nf ∈ I, and by definition of I and the fact that R′0 is perfectsince we are in characteristic p, we get that

($n/pkf)pk

∈ I

implies $n/pkf ∈ I for all k ≥ 0. Therefore, since the $n/pk , k ≥ 0 also generate m, f isalmost zero, implying that M is. Since the map:

R′0[T

1/p∞

1 , . . . , T 1/p∞

n ]/I → R′0[F#/g#]

is surjective, we conclude.

We claim that the pth power map gives an actual isomorphism:

R′0[T

1/p∞

1 , . . . , T 1/p∞

n ]/(I,$1/p)'−→ R′

0[T

1/p∞

1 , . . . , T 1/p∞

n ]/(I,$).

It clearly induces an isomorphism:

R′0[T

1/p∞

1 , . . . , T 1/p∞

n ]/$1/p = (R′0/$1/p)[T

1/p∞

1 , . . . , T 1/p∞

n ]'−→ (R′

0/$)[T

1/p∞

1 , . . . , T 1/p∞

n ]

= R′0[T

1/p∞

1 , . . . , T 1/p∞

n ]/$

because R′ is perfectoid. We denote by ϕ the pth power map, as usual. Reduction modulo Iyelds the following cocartesian diagram:

R′0[T

1/p∞

1 , . . . , T1/p∞

n ]/$1/p

ϕ// R′

0[T

1/p∞

1 , . . . , T1/p∞

n ]/$

R′0[T

1/p∞

1 , . . . , T1/p∞

n ]/(I,$1/p)ϕ// R′

0[T

1/p∞

1 , . . . , T1/p∞

n ]/(I,$)

Assume that some f ∈ R′0[T1/p∞

1 , . . . , T1/p∞

n ] is such that fp ∈ (I,$). It means that fp =

h+$h0, for some h0 ∈ R′0[T1/p∞

1 , . . . , T1/p∞

n ] and h ∈ I. However, R′0 is perfect, and hence sois R′0[T

1/p∞

1 , . . . , T1/p∞

n ]. Therefore we can write h0 as hp1 for some h1 ∈ R′0[T1/p∞

1 , . . . , T1/p∞

n ].

On the other hand, I is generated by all the elements g#1/pm

T1/pm

i − f#i

1/pm

, and R′0 is

perfect. So likewise we can write h as h′p for some h′ ∈ R′0[T1/p∞

1 , . . . , T1/p∞

n ]. Since we arein characteristic p, it follows we can write

fp = h′p

+ ($1/p)php1,


and given that Frobenius is an isomorphism on R′0[T1/p∞

1 , . . . , T1/p∞

n ] we conclude that

f = h′ +$1/ph1

that is, f ∈ (I,$1/p). Therefore, reduction modulo I yelds that the bottom horizontal arrowin the above diagram is an isomorphism. On the other hand, the vertical arrows in the diagrambelow are almost isomorphisms:

R′0[T

1/p∞

1 , . . . , T1/p∞

n ]/(I,$1/p)

ϕ// R′

0[T

1/p∞

1 , . . . , T1/p∞

n ]/(I,$)

R′0[(F#/g#)1/p∞ ]/$1/p

ϕ// R′

0[(F#/g#)1/p∞ ]/$

and we conclude that the pth power map induces an isomorphisms of R0a-algebras

R′0[(F#/g#)1/p∞ ]a/$1/p ' R′0[(F#/g#)1/p∞ ]a/$

which is what we wanted.

We now give for granted the fact that part (1) implies part (2), which will be settled in Lemma5.1.8 below.

To obtain (1) for general Tate-perfectoid R, observe that having achieved it in characteristicp means that we have it for X[, that is, we know that (OX[(U),O+

X[(U)) is a perfectoid pair.

One argues the same way using the result in the tilted situation, but the difference is that theideal I will now be generated by the elements T 1/pm

i (g1/pm)# − (f1/pm

i )#. Let (S, S+) be anyuntilt of (OX[(U),O+

X[(U)). The map

Spa(S, S+)→ X

factors through U#, thus yelding a factorization

(R′, R′+

)→ (OX(U#),O+X(U#))→ (S, S+).

The composite map

R′0〈T 1/p∞

1 , . . . , T 1/p∞

n 〉a → R′0〈(F#/g#)1/p∞〉a → OX(U#)0a → S0a

is a map of perfectoid R0a-algebras, which is the untilt of the composite map:

R′[0〈T 1/p∞

1 , . . . , T 1/p∞

n 〉a → R′[0〈T 1/p∞

1 , . . . , T 1/p∞

n 〉a/I[ → OX[(U)0a

I[ being the ideal generated by g1/pmT1/pm

i − f1/pm

i , for i = 1, . . . , n and all m ≥ 0. We knowthat:

R′[0〈T 1/p∞

1 , . . . , T 1/p∞

n 〉/(I[, $[) = R′0〈T 1/p∞

1 , . . . , T 1/p∞

n 〉/(I,$).

By the proof in the characteristic p case, we have that the map:

R′[0〈T 1/p∞

1 , . . . , T 1/p∞

n 〉a/(I[, $[)→ OX[(U)0a/$[

is an isomorphism, and this map is the tilt of the composite

R′0〈T 1/p∞

1 , . . . , T 1/p∞

n 〉a/(I,$)→ R′0〈(F#/g#)1/p∞〉a/$ → S0a/$.


Being the former composite an isomorphism, so is this latter. The first map:

R′0〈T 1/p∞

1 , . . . , T 1/p∞

n 〉a/(I,$)→ R′0〈(F#/g#)1/p∞〉a/$

is surjective, and hence it is an isomorphism together with the map:

R′0〈(F#/g#)1/p∞〉a/$ → S0a/$.

This achieves part (1) in general, and hence also part (2). This last isomorphism also achievespart (3).

Lemma 5.1.8 With same notation as in Lemma 5.1.7 we have:

$kR′0[(F#/g#)1/p∞ ] ⊂ R′0[f#

1 /g#, . . . , f#

n /g#]

for sufficiently large k > 0.

Proof. We endow R′[1/g#] with the topology generated by the base of open neighbourhoodsof 0

$kR′0[f#

1 /g#, . . . , f#

n /g#],

since R′0 is open and bounded, and hence we can choose it to be a ring of definition of R′ as aHuber ring. However, R′0[(F#/g#)1/p∞ ] is also open and bounded with the subspace topologyof R′[1/g#] because it contains R′0[f#

1 /g#, . . . , f#

n /g#], and is contained in the integral closure

S of R′0[f#1 /g

#, . . . , f#n /g

#] in R′[1/g#], which we claim is bounded. We show this at the endof the proof.

Therefore, the choice of R′0[(F#/g#)1/p∞ ] as a ring of definition for R′[1/g#] with the $-adic topology doesn’t change the Huber ring topology on R′[1/g#]. In particular, for k > 0sufficiently large we have

$kR′0[(F#/g#)1/p∞ ] ⊂ R′0[f#

1 /g#, . . . , f#

n /g#]

as desired.

By part (1) of Lemma 5.1.7, completely established, R′〈(F#/g#)1/p∞〉 = R′[1/g#] is Tate-perfectoid, and therefore the integral closure S′ of R′0〈(F#/g#)1/p∞〉 is bounded. We havecontinuous maps:

R′0[f#

1 /g#, . . . , f#

n /g#] → R′

0〈(F#/g#)1/p∞〉 → S′ → R′[1/g#].

Being S′ integrally closed, it contains S, which is, therefore, bounded with respect to the Huberring topology assigned by the ring of definition R′

0〈(F#/g#)1/p∞〉 with its $-adic topology.On the other hand S′ itself is bounded with respect to the Huber ring topology assigned onR′[1/g#] by R′0[f#

1 /g#, . . . , f#

n /g#] with its $-adic topology, implying that there exists some

sufficiently large k > 0 such that:

S′ ⊂ $−kR′0[f#1 /g

#, . . . , f#n /g

#],

and henceR′

0[(F#/g#)1/p∞ ] ⊂ S ⊂ $−kR′0[f#

1 /g#, . . . , f#

n /g#],

as desired.


We obtain the inclusions:

R′0〈f#

1 /g#, . . . , f#

n /g#〉 ⊂ R′0〈(F#/g#)1/p∞〉 ⊂ R′〈f#

1 /g#, . . . , f#

n /g#〉 = OX(U).

It follows that, from part (1) of Lemma 5.1.7,

OX(U) = R′0〈(F#/g#)1/p∞〉[1/$]

is Tate-perfectoid, with desiredR0a-algebra OX(U)0a. This achieves the proof of the implication(1) implies (2) in Lemma 5.1.7.Remark 5.1.9 This proves that given a perfectoid R-algebra R′, then R′ is stably uniform.Hence, by the results in [BV], Spa(R′, R′

+) is an adic space, in the sense that the structure

presheaf is a sheaf.

The Approximation Lemma

In general, the map f 7→ f# extensively discussed in the previous sections is not surjective.However, we would like to know when the valuation on R′ sending f ∈ R′ to |f(x)|, forx ∈ X = Spa(R′, R′

+), is close to being of the form g 7→ |g#(x)|, because this last is supposed

to induce the desired homeomorphism X ' X[ by sending x 7→ x[. It turns out that one canfind g# approximating f , so that the two maps coincide on all but those points at which |f(x)|and |g#(x)| are small.

Lemma 5.1.10 (Approximation) Let O = R〈T 1/p∞

0 , . . . , T1/p∞

n 〉. Let f ∈ O0 be a homogen-eous element of degree d ∈ Z[1/p]. Pick any rational number c ≥ 0 and any ε > 0. Then thereexists an element

gc,ε ∈ O[0 = R[0〈T 1/p∞

0 , . . . , T 1/p∞

n 〉

homogeneous and of the same degree d, such that for all points x ∈ X = Spa(O,O0), we have

|f(x)− g#c,ε(x)| ≤ |$|1−ε max(|f(x)|, |$|c).

As a first consequence we make precise the intuition discussed above. If ε < 1, it means that|f(x)| and |g#

c,ε(x)| are small, and we have, for all x ∈ X = Spa(O,O0),

max(|f(x)|, |$|c) = max(|g#c,ε(x)|, |$|c).

We have:Proposition 5.1.11 Let (R′, R′

+) be a perfectoid pair, with tilt (R′

[, R′

[+), and let X =

Spa(R′, R′+

), X[ = Spa(R′[, R′

[+).

(1) For any f ∈ R′ and c ≥ 0 rational, ε > 0, there exists gc,ε ∈ R′[ such that for all x ∈ X,we have

|f(x)− g#c,ε(x)| ≤ |$|1−ε max(|f(x)|, |$|c).

(2) For any x ∈ X, the completed residue field κ(x)∧ is a perfectoid field.

(3) The morphism |X| → |X[| induces a homeomorphism, indentifying rational subsets inboth directions.

Before proving the Proposition, we observe the following.


Remark 5.1.12 Let X be an adic space. Let us express X as a triple:

(X,OX , | · |xx∈X),

where vx is the valuation on κ(x) with valuation ring κ(x)+, attached to each point x ∈ X. Weconsider x, x′ ∈ X such that x is a specialization of x′ in X. The specialization map X → Xsending x′ to x, yields a morphism of triples:

(X,OX , | · |x)→ (X,OX , | · |x)

making the following diagram commute:

R

R

κ(x) //

vx

κ(x′)

vx′

Γx ∪ 0 ∃// Γx′ ∪ 0

meaning that there exists a (necessarily unique!) morphism of ordered abelian groups α : Γx →Γx′ making the above diagram commutative. It follows that if, for some c and ε, the followinginequality holds:

|f(x)− g#c,ε(x)|x ≤ |$|1−εx ·max(|f(x)|x, |$|cx)

then the following holds:

α(|f(x)− g#c,ε(x)|x) ≤ α(|$|1−εx ) ·max(α(|f(x)|x), α(|$|cx))

We can write α(|f(x)− g#c,ε(x)|x) = |f(x′)− gc,ε(x′)|x′ , and at the cost of enlarging ε, we can

arrange that

α(|$|1−εx ) ·max(α(|f(x)|x), α(|$|cx)) ≥ |$|1−εx′ ·max(|f(x′)|x′ , |$|cx′).

Proof of Proposition 5.1.11. The rank-one points in Spa(R′, R′+

) are all contained in Spa(R′, R′0),

and if we increase slightly ε, by Remark 5.1.12 it suffices to check the inequality at closed points,so we can assume R′+ = R′

0. Up to enlarging c as well, we can assume f is power-bounded,and that c is an integer. We can write

f = g#0 +$g#

1 + · · ·+$cg#c +$c+1fc+1

for some g0, . . . , gc ∈ R′[0 and fc+1 ∈ R′0, which we may assume to be zero. We have a map:

R〈T 1/p∞

0 , . . . , T 1/p∞

n 〉 → R′

sending T 1/pm

i to (g1/pm

i )#. f is the image of T0 +$T1 + · · ·+$cTc. We are done with (1) bythe Approximation Lemma. Roughly, we can call ϕ the map:

X → DperfR

where DperfR is the perfectoid unit disc over R. We have f ∈ Γ(X,OX), and we are applying

the Approximation Lemma to ϕ∗f ∈ Γ(DperfR ,ODperf

R).


We prove (2) in the characteristic p case first. We know already that OX(U)0a is perfectoidfor any rational subset U . It follows that the $-adic completion of O0a

X,x is perfectoid as aR0a-algebra, and then κ(x)∧ is perfectoid as a R-algebra. It is obviously a nonarchimedeanfield, so we conclude.

The meaning of (1), and hence the reason why one wants the Approximation Lemma, is thatthe preimage of any rational subset of X[ is a rational subset of X. This is important, becausewe already know that X, as an adic space, is spectral, and hence T0. Then the continuous map|X| → |X[| is injective. Indeed, suppose two points x, x′ ∈ |X| go to the same x[ ∈ |X[|. Takeany open neighbourhood U of x[, which we can assume to be rational. Its preimage in |X|must contain both x and x′, which, if distinct, cannot be separated. Hence x = x′.

To analyze completed residue fields, pick an x ∈ X. The valuation induced by x factors alwaysthe following way

R′[ → κ(x)∧ → Γx

and now we see why we were interested in proving that the completed residue field was per-fectoid.

We can untilt both R′[ and κ(x)∧ to R′ and some perfectoid field κ over R.

On the other hand, there is some R′+, unique up to isomorphism, which tilts to R′[+ ⊂ R′[0,

and respectively some κ+ ⊂ κ0 which tilts to κ(x)+,∧. This is due to Proposition 5.1.3.

It follows that the adic point:

x : Spa(κ(x), κ(x)+)→ Spa(R′[, R′

[+)

untilts to a pointx# : Spa(κ, κ+)→ Spa(R′, R′

+)

which implies that the map |X| → |X[| is surjective, as desired.

Once we have obtained that this is a homeomorphism, (2) follows in the general case the verysame way we did in characteristic p.

Remark 5.1.13 Arguing with the untilts of R′[+ and κ(x)+,∧ respectively, is equivalent tosaying that we can untilt valuations, thus yelding the following equivalent variant of the argu-ment.

We know that continuous valuations:

κ(x)∧ → Γx ∪ 0

are mapped bijectively, up to equivalence, to continuous valuations:

κ→ Γx ∪ 0

via |x| = |x#|. So we can also untilt the valuation κ(x)∧ → Γx ∪0, meaning that |X| → |X[|is not just injective, but also surjective.

We end this section with the proof of the Approximation Lemma.

Proof of Lemma 5.1.10. We fix f in O0, and ε > 0, which we can assume to be in Z[1/p] andless than 1. Inductively, we shall prove that for all c we can find some ε(c) > 0 and some

gc ∈ O[0 = R[0〈T 1/p∞

0 , . . . , T 1/p∞

n 〉


homogeneous of degree d, such that for all x ∈ X = Spa(O,O0) we have:

|f(x)− g#c (x)| ≤ |$|1−ε+ε(c) max(|f(x)|, |$|c).

We will clarify why we need ε(c) along the way! Fix a rational number a in (0, ε)∩Z[1/p], andincrease c to c′ := c+ a and argue by induction on c. If c = 0, we just choose ε(c) = ε and get:

|f(x)− g#0 (x)| ≤ |$|max(|f(x)|, 1) = |$|

for any g0 ∈ O[0. We can replace ε(c) by something smaller, so without loss of generality letus assume ε(c) ≤ ε− a, and ε(c) ∈ Z[1/p].

Consider the rational subset Uc ⊂ X[ given by x ∈ X[ | |gc(x)| ≤ |$[|c. Recall we areassuming the statement true for c, and claim it’s true for c′. Its preimage U#

c ⊂ X is given byx ∈ X | |f(x)| ≤ |$|c. The difference h := f − g#

c is such that

h ∈ $c+1−ε+ε(c)O+X(U#

c ).

We know by Proposition that:

OX(U#c )0a = O0〈(g#

c /$c)1/p∞〉a.

Now, h is homogeneous, and then it almost lies in the $-adic completion of⊕u∈Z[1/p]∩[0,1]

($[)c+1−ε+ε(c)(gc/($[)c)u · O0

deg=d−du.

It follows that there exist elements ru ∈ O0 homogeneous of degree d− du, such that ru → 0,and

h =∑

u∈Z[1/p]∩[0,1]

($[)c+1−ε+ε(c′)(gc/($[)c)uru

where we choose 0 < ε(c′) < ε(c), ε(c′) ∈ Z[1/p]. Now we choose su ∈ O[0 such that su ishomogeneous of degree d− du, and $ divides ru − s#

u . Define:

gc′ := gc +∑

u∈Z[1/p]∩[0,1]

($[)c+1−ε+ε(c′)(gc/($[)c)usu.

We claim that gc′ does the job, that is: for all x ∈ X, we have

|f(x)− g#c′ (x)| ≤ |$|1−ε+ε(c

′) max(|f(x)|, |$|c′).

We therefore see where the ε(c) comes from: we loose a constant due to almost mathematics.

Assume first that |f(x)| > |$|c. Then

|g#c (x)| = |f(x)| > |$|c.

We are reduced to show that

∣∣∣∣(($[)c+1−ε+ε(c′)(gc/($[)c)usu

)#

(x)

∣∣∣∣ ≤ |$|1−ε+ε(c′)|f(x)|.


Since |su(x)| ≤ 1, the left hand side is maximal when u = 1, in which case it equals the righthand side, so we conclude in this case.

Now assume |f(x)| ≤ |$|c. We claim:

|f(x)− g#c′ (x)| ≤ |$|c

′+1−ε+ε(c′).

By design,c+ 1 > c′ + 1− ε+ ε(c′).

Then it is enough to check that we have

f − g#c′ ∈ $

c+1OX(U#c )0.

We have:

gc′/($[)c = gc/($

[)c +∑u

($[)1−ε+ε(c′)(gc/($[)c)usu,

with all summands being in OX[(Uc)0. Modulo $, we get:

g#c′ /$

c = g#c /$

c +∑u

$1−ε+ε(c′)(g#c /$

c)uru

in OX(U#c )0, up to multiples of $. Multiply by $c, and get, up to multiples of $c+1,

f − g#c′ = f − g#

c − h,

which is zero modulo $c+1. This gives the desired estimate also in this case, and we conclude.

5.2 Global tilting equivalence

Let Y ⊂ X be a subset of X. We shall indicate by Y [ the corresponding subset of X[ underthe tilting equivalence. We prove the following:

Proposition 5.2.1 Let (R′, R′+

) be a perfectoid pair over (R,R+), with tilt (R′[, R′

[+), and

let X = Spa(R′, R′+

), X[ = Spa(R′[, R′

[+). Then for all rational U ⊂ X, the pair:

(OX(U),O+X(U))

is a perfectoid pair with tilt (OX[(U[),O+

X[(U [)).

Remark 5.2.2 This also proves that perfectoid pairs are stably uniform, and hence by [BV]they are sheafy! We deduce that OX , and hence O+

X is a sheaf for any affinoid adic space Xgiven by the adic spectrum of a perfectoid pair.

Proof. Since X → X[ is a homeomorphism identifying rational subsets, and since (V #)[ = Vfor all rational subsets V ⊂ X[, by Proposition 5.1.3, (OX(U),O+

X(U)) is Tate-Huber perfectoid,for V ⊂ X rational. We know that for every complete Tate-Huber pair (S, S+) over (R,R+),withX(F/g)) =: U ⊂ X = Spa(R′, R′

+) a rational subset, and with a map (R′, R′

+)→ (S, S+)

such that the induced mapSpa(S, S+)→ X


factors through U , there is a unique map:

(R′〈F/g〉, B∧)→ (S, S+)

making the obvious diagram commute, where B is the integral closure of R′+[F/g] in R′[F/g].Tilting preserves this universal property, which is satisfied by (OX[(U

[),O+X[

(U [)) the obiousway.

We have assigned affinoid adic spaces X = Spa(R′, R′+

) to perfectoid pairs (R′, R′+

) over(R,R+). We call such spaces affinoid perfectoid spaces, and calling S := Spa(R,R+), we’vethus assigned the notion of affinoid perfectoid S-spaces. We can eventually give the following:Definition 5.2.3 A perfectoid space over S is an adic space which is locally isomorphic toan affinoid perfectoid S-space. Morphisms of perfectoid spaces over S are morphisms of adicspaces over S. For an affinoid perfectoid space X over S, tilting yelds an affinoid perfectoidspace X[ over S[.

We give the following definition, to establish what we mean when we say a perfectoid space X[

over S[ is the tilt of a perfectoid space X over S.Definition 5.2.4 We say a perfectoid space X[ over S[ is the tilt of a perfectoid space X overS if and only if the following natural bijection holds true for all perfectoid pairs over (R,R+):

Hom(Spa(R′, R′+

), X) ' Hom(Spa(R′[, R′

[+), X[).

As a formal consequence of the results in the preceding sections, we obtain the following:Theorem 5.2.5 Any perfectoid space X over S admits a tilt X[ over S[, unique up to uniqueisomorphism. This induces an equivalence between the categories of perfectoid spaces over Sand perfectoid spaces over S[. The underlying topological spaces |X| and |X[| are naturallyhomeomorphic, the homeomorphism preserving rational subsets in both directions. A perfectoidspace X over S is affinoid perfectoid if and only if X[ is affinoid perfectoid over S[. Finally,for any affinoid perfectoid subspace U ⊂ X, the pair:

(OX(U),O+X(U))

is a perfectoid pair with tilt(OX[(U

[),O+X[

(U [)).

Moreover:U ' Spa(OX(U),O+

X(U)) and U [ ' Spa(OX[(U[),O+

X[(U [)).

Remark 5.2.6 We remark the importance of Proposition 5.1.3. The datum of R′+, togetherwith R′, behaves well with respect to tilting, and this is indeed, in the last instance, whatenables us to globalize the tilting functor which was defined on perfectoid and Tate-perfectoidrings, to perfectoid pairs, first, and then to global perfectoid spaces. The tilting correspondencefor perfectoid algebras, and subsequently the equivalence of finite-étale sites for algebras only,is already striking, but it is more striking the fact that it carries on to global perfectoid spaces,as, in addition, it yelds a correspondence between the integral structures.

5.3 Fibered products of perfectoid spaces

We do not have a good notion of fibered products in the category of adic spaces! We can definefibered products of adic spaces over S, X → Y ← Z, provided at least one between X and Zis locally of finite type.


We wish to be more precise about existence of fibered products of adic spaces.

First, a definition:Definition 5.3.1 Let A be a complete Huber ring. We say A is sheafy if for every integralstructure A+ ⊂ A, the presheaf OA is a sheaf. We say A is stably sheafy if B is sheafy for everyA-algebra B of topologically finite type.

We note every perfectoid pair over (R,R+) is obviously stably sheafy.Definition 5.3.2 Let X be an adic space. We say X is stable if there exists an open coveringby affinoid domains of the form Spa(A,A+), where A is stably sheafy.

Every perfectoid space over S, in particular, is a stable adic space. We have the followingsufficient condition for existence of fibered products of adic spaces.Theorem 5.3.3 Let f : X → Y , and Z → Y be morphisms of adic spaces. Assume Y is stable,and that f is locally of finite type. Then the fibered product X ×Y Z exists in the category ofadic spaces. Moreover, X ×Y Z is stable as an adic space.

In the proof of the above Theorem, which the reader may want to work out as an exercise, theassumption on f being locally of finite type just ensures that if we set X = Spa(A,A+), Y =Spa(B,B+), Z = Spa(C,C+), we can have the following situation. Define (D,D+) to be thefollowing: D = (A ⊗B C)∧, and D+ is the completion of the integral closure of the image ofA+ ⊗B+ C+ in D. Then Spa(D,D+) is stable. This is indeed due to the fact that A is oftopologically finite type over B. But we do not need any finiteness assumption to ensure Dis stably sheafy, if we prove its perfectoid as an R-algebra, say, from knowledge that (A,A+),(B,B+) and (C,C+) are. Indeed, this is exactly what happens, and we have the following:Proposition 5.3.4 Let X and Z be perfectoid spaces over S with morphisms of adic spaces toa perfectoid space Y over S, respectively. Then the fibered product

X ×Y Z

exists in the category of adic spaces over S, and is a perfectoid space.Remark 5.3.5 In the following proof, we make use of the equivalences of categories:

R-Perf ' R0a-Perf ' (R0a/$)-Perf

the following way. Given a $-adically complete Huber R-algebra R′, we want to check it isperfectoid, and instead of checking the required conditions are satisfied by R′, we check thatthe corresponding conditions under the composite of the above two equivalences are satisfiedby R′

0a/$, thus concluding that R′0a/$, and hence R′0a, is a perfectoid R0a/$-algebra,

respectively a perfectoid R0a-algebra. Now, to conclude that R′ is indeed a perfectoid R-algebra, we need to make sure that the inverse functor R′0a-Perf → R′-Perf takes R′0a to R′itself up to (unique) isomorphism. This is indeed the case: we proved that given a perfectoidR0a-algebra A, then if we equip R′ := A∗[$

−1] with the unique Huber ring structure makingA∗ open and bounded, we have A∗ = R′

0 and R′ is perfectoid.

The main tool was Lemma 3.2.1.

It is sufficient to check that R′0a∗ [$−1] = R′, for our R′. To show this it is sufficient to showthat R′0a∗ = R′

0. We know already that R′0a∗ and R′0 are almost isomorphic as R0-modules.By Lemma 3.2.1, they are isomorphic.

Proof. We want to construct fibered products in the affinoid case, and then glue, so the situationis the following. We have affinoid perfectoid spaces X = Spa(A,A+), Y = Spa(B,B+) and


Z = Spa(C,C+), and we want to construct P := X ×Y Z. P is given by Spa(D,D+), where Dis the completion of A⊗B C. Now take the image of A+ ⊗B+ C+ in D, and complete it. Thatdefines D+. We claim that (A0a ⊗B0a C0a)∧ is a perfectoid R0a-algebra. It is enough to checkthat A0a⊗B0a C0a/$ is perfectoid as a R0a/$-algebra, and that $ is (A0a⊗B0a C0a)∧-regular,and this means that we are reduced to the characteristic p case. Here the first condition isclear, and it is enough to check that A0a ⊗B0a C0a is $-torsion free. If we have an f such that$f = 0, then $1/pf1/p = 0 since A0a ⊗B0a C0a is perfect. But then $1/pf = 0, implyingthat f is killed by every $1/pm . But then it is zero, as its reduction modulo $1/pm is alwayszero, and hence if we lift each of these reductions to give a sequence in A0a ⊗B0a C0a, thistends to zero, which equals f , then. In particular, (D,D+) is a perfectoid pair, and by densityarguments and the universal property of tensor product, we deduce that it satisfies the desireduniversal property.


6 Small étale sites and tilting

In this section we prove Faltings’ almost purity theorem. The proof we present relies on thegeometric framework provided by perfectoid spaces. We shall discuss the proof in full detail, andsubsequently prove some needed preliminaries which we temporarily give for granted throughoutthe proof, but one. This is what we shall call the Henselian approximation Lemma, and providesus with a crucial lifting principle for Henselian pairs. This result is at the heart of many keysteps in the theory, and we devote a separate handout to explain its proof, which generalizesresults of Elkik to the non - noetherian setting.

Stepping back to the almost purity theorem, as for the tilting correspondence, the next stepwill be globalizing it to show the equivalence between the étale site of a perfectoid space andits tilt. In the last instance, this equivalence formally carrying on to the respective étale topoi,we are given canonical isomorphisms between the étale cohomology of a perfectoid space in anycharacteristic, and the étale cohomology of a perfectoid space in characteristic p: the tilt.

We will see a first application of this marvelous principle in the second part of lecture notes,when we set up the machinery needed to achieve finiteness of étale cohomology of adic spaces.More precisely, we shall prove that given any nonarchimedean field k and any proper adic spaceX over k, its p-adic étale cohomology groups Hi

et(X,Zp) are finitely generated Zp-modules (andmore generally for Zp-lisse coefficients).

The proof will actually establish more. Once we introduce the pro-étale topology for adicspaces, and if we pick a proper formal model X over Spf(Ok), Ok being the valuation ring ofk, with X as general fiber, we can consider the map of nearby cycles ν : Xproet → XZar.

The proof of the finiteness statement above actually shows that the complex Rν∗(O+X/p) has

coherent cohomology in the sense of almost mathematics.

We shall discuss these results in the generality of proper smooth adic spaces X/k, and thendeduce them in the general case either with a naive trick with (canonical) resolution of sin-gularities, or showing that surjective maps of quasi-compact quasi-separated adic spaces overSpa(k,Ok) are universally of cohomological descent, and we deduce almost coherence of thecohomology of Rν∗(O+

X/p) from the smooth case using hypercovers by smooth adic spaces, thusreducing the task to the smooth case.

At the heart of the proofs of the above results is the availability of characteristic p methods onperfectoid spaces and their étale sites and topoi.

6.1 The étale topology on perfectoid spaces

We need to define the notion of étale morphism for perfectoid spaces. This is slightly prob-lematic for a number of reasons, one of which being reducedness of perfectoid spaces. We shalldraw motivation from algebraic geometry and use a more flexible notion of étaleness, which bywork of Huber turns out to be equivalent to the classical one for locally noetherian adic spaces.

Many of the classical results in rigid geometry (as Temkin’s work on generically étale alterationsof formal models of rigid spaces, as well as Temkin’s work on relative curves) can be adapted tothis notion of étaleness, thus becoming available for applications when dealing with perfectoidcovers of rigid analytic varieties (with respect to the proétale topology, yet to be defined).

Before starting, we recall a fundamental result from algebraic geometry, which will guide us infinding a good notion of étale morphisms for adic spaces which is flexible enough to work forperfectoid spaces.


Theorem 6.1.1 (Zariski’s Main Theorem) Suppose f : X → Y is a proper morphism oflocally noetherian schemes.

(1) The set of points of X that are isolated in their fiber forms an open subset X0 ⊂ X.

(2) The morphism f |X0 : X0 → Y factors into an open immersion followed by a finite morph-ism:

X0

f

open// Y ′

finite~~

Y

The proof makes use of the Stein factorization for f , and so, in the last instance, Zariski’sMain Theorem is a consequence of the Theorem on formal functions, as discussed in [EGA III,§4.4.3].

We will draw motivation from Theorem 6.1.1, and shall get back to this in a moment, but letus first give a few definitions.Definition 6.1.2 Let (A,A+) and (B,B+) be complete Huber psirs, and let f : (A,A+) →(B,B+) be a morphism of complete Huber pairs. We say f is finite if it is of topologicallyfinite type, and the induced ring homomorphism A+ → B+ is integral, together with A→ B.Remark 6.1.3 We note that from Definition 6.1.2 it follows at once that B is finite over A,and carries the natural A-module topology, and B+ is the integral closure of A+ in B.Example 6.1.4 One can construct finite Huber rings over (A,A+) by simply taking C to bea finite A-algebra with its A-module topology, and assigning C+ as being the integral closureof A+ in C. The upshot is that (C,C+) is a complete and finite Huber ring over (A,A+).Definition 6.1.5 A morphism of adic spaces f : X → Y is called finite if, for every y ∈ Y ,there exists an open affinoid neighbourhood V of y in Y such that U := f−1(V ) is affinoid,and the morphism of Huber pairs

(OY (V ),O+Y (V ))→ (OX(U),O+

X(U))

is finite.Remark 6.1.6 In [Hub, §1.4], Huber points out that from Definition 6.1.5, it follows at oncethat if f is finite one has, for every open affinoid subspace V of Y , that U := f−1(V ) isaffinoid, and (OX(U),O+

X(U)) is finite over (OY (V ),O+Y (V )). We caution the reader that it is

not yet clear if this holds true in reasonably good generality, that is, it does not heavily relyon noetherianity assumptions. Moreover, there is no cohomological criterion for affinoidness ofadic spaces available. This is related to the fact that there are compact separated k-analyticspaces (k a nonarchimedean field) which are not affinoid, but for which every coherent sheafis acyclic. We refer the reader to Q. Liu’s counterexample in [QLiu]. In any event Huber’sapproach to the proof of the above statement makes use of quasi-coherent sheaves of finite type,whereas we know a good theory of quasi-coherent sheaves on adic spaces does not exist in fullgenerality.

We are ready to give the following first definition of étale morphisms of adic spaces:Definition 6.1.7 A morphism f : X → Y of adic spaces is called étale if f is locally of finitepresentation, and if, for any Huber pair (A,A+) and square-zero ideal I ⊂ A, given a morphism

Spa(A,A+)→ Y

the natural map

HomY (Spa(A,A+), X)→ HomY (Spa(A/I,A+/I ∩A+), X)


is a bijection.

We notice that due to Definition 6.1.5, the category of finite morphisms to Spa(A,A+) in thecategory of morphisms of adic spaces is naturally equivalent to the category of finite morphismsto Spec(A) in the category of morphisms of schemes.

Recall that a finite morphism of schemes Z → S is étale if, for any finite morphism of schemesZ ′ → S and any closed subscheme Z ′0 ⊂ Z ′ cut out by a square-zero ideal I → OZ′ , the naturalmap

HomS(Z ′, Z)→ HomS(Z ′0, Z)

is a bijection. One can prove the following:Proposition 6.1.8 Let Z → S be a finite morphism of adic spaces. It is étale if, for any finitemorphism of adic spaces Z ′ → S and any closed adic subspace Z ′0 ⊂ Z ′ defined by a square-zeroideal of Γ(Z ′,OZ′), the natural map

HomS(Z,Z ′)→ HomS(Z,Z ′0)

is a bijection.

This is just telling us that the categories of finite étale morphisms to Spa(A,A+) in the cat-egory of morphisms of adic spaces and finite étale morphisms to Spec(A) in the category ofmorphisms of schemes are equivalent. Here we find the first issue in giving a good notion offinite étale morphisms of of adic spaces which works for perfectoid spaces. Perfectoid R-algebrasare reduced! So we cannot give a definition like the one we have just discussed for adic spaces,as it turns out to rely on liftings of nilpotents.

We now exploit the following theorem of Huber.Proposition 6.1.9 Let X → Y be an étale morphism of locally noetherian affinoid adic spaces.Then every y ∈ Y has an open neighbourhood U ⊂ Y such that the restriction of f to f−1(U)has a factorization

f−1(U)

f##

i open// Z

h finite étale

U

This is proved using Zariski’s Main Theorem, and the proof can be found on [Hub, § 1.6].We agree that our definition of étale morphism of adic spaces will be, then, the following one,specializing to the case our adic spaces are always over a field k, for the moment.Definition 6.1.10 A morphism f : X → Y of locally noetherian adic spaces is called étale iffor any point x ∈ X, there exist open neighbourhoods U and V of x and f(x) respectively, anda factorization for f |U

U

f |U

i open// W

h finite étale~~

V

We make use of this definition in the case of perfectoid spaces. We fix a Tate-perfectoid ringR, and regard all perfectoid spaces in the sequel of this section as being over Spa(R,R+) withno further mention.


Notation We won’t use the letter S to denote the fixed perfectoid base Spa(R,R+), as we’llneed S to denote certain finite étale algebras over a Tate-perfectoid ring.

From Theorem 4.3.4, we know that there is a fully faithful functor R′[fet → R′fet, for a perfectoidR-algebra R′. Furthermore, its essential image consists exactly of those finite étale covers S ofR′ such that S is perfectoid, and S0a if finite étale over R′0a.

Indeed, let R′ be a perfectoid R-algebra, and fix A a perfectoid R′0a-algebra which is finite étale.Then A∗[1/$] is formally unramified by design, since there exists e ∈ (A ⊗R′0a A)∗[1/$] =A∗[1/$]⊗R′A∗[1/$] idempotent, which kills the kernel of the multiplication map and on whichthe multiplication map is 1. Moreover, A∗ is almost finitely generated projective, meaning thatthere are maps for all ε ∈ m with the property that

A∗ → R′⊕n → A∗

is multiplication by ε. Fix any such ε, and invert $. Therefore we invert also ε, and we realizeA∗[1/$] as the direct summand of a finite free R′-module. It follows A is finite étale over R′,and note also that it is perfectoid.

First, we are therefore led to assign a stronger notion of étaleness for morphisms for perfectoidspaces, which will turn out to be equivalent to the above one, and this will be exactly thecontent of the almost purity theorem.Definition 6.1.11

(1) A morphism of perfectoid pairs (R′, R′+

) → (S, S+) is called strongly finite étale if it isfinite étale in the sense of Definition 6.1.2 and, in addition, S0a is a finite étale R′0a-algebra.

(2) A morphism f : X → Y of perfectoid spaces is called strongly finite étale if there is a coverof Y by open affinoid perfectoid subspaces V ⊂ Y such that the preimage U := f−1(V ) isaffinoid perfectoid, and the associated morphism of perfectoid pairs

(OY (V ),O+Y (V ))→ (OX(U),O+

X(U))

is strongly finite étale.

(3) A morphism f : X → Y of perfectoid spaces is called strongly étale if for any point x ∈ Xthere are open neighbourhoods U and V of x and f(x) respectively, and a factorizationfor f |U :

U

f |U

i open// Z

h

V

where h is strongly finite étale.

This means that Definition 6.1.11 is quite a good shot, as we have that f : X → Y is stronglyfinite étale, resp. strongly étale, if and only if the tilt f [ : X[ → Y [ is! We recall that incharacteristic p we have already achieved the almost purity theorem, which, for the sake ofclarity, we restate here again in the general case.Theorem 6.1.12 (Almost Purity) Let R′ be a perfectoid R-algebra. For any finite étalecover S/R′, S is perfectoid and S0a is finite and étale over R′0a in the sense of almost math-ematics.


As pointed out before, Theorem 6.1.12 is equivalent to the fact that the two étaleness notionsgiven in Definition 6.1.7 and Definition 6.1.11 are actually equivalent (ie. every étale morphismof perfectoid spaces is strongly étale).

Before turning to the proof of the almost purity theorem and, beforehand, its setup, we digresson an essential ingredient of the whole theory, which is essentially due to Lorenzo Ramero andOfer Gabber, generalizing results of Elkik.

6.2 Henselian approximation.

We briefly recall the following Definition.Definition 6.2.1 Let I be an ideal of a commutative ring A. The pair (A, I) is called Henselianif and only if the following equivalent conditions are satisfied:

(1) I ⊂ J(A), where J(A) is the Jacobson radical of A, and for every two relatively primemonic polynomials g, h ∈ A[T ], where A = A/I, and monic lifting f ∈ A[T ] of g · h, thereexist unique monic liftings g, h ∈ A[T ] such that f = g ·h. Namely, Hensel’s Lemma holdstrue.

(2) If B is a finite A-algebra, then Idem(B) ' Idem(B/IB), where Idem(B) is the set ofidempotents in B, and ' means bijection.

(3) If A′ is an étale A-algebra, and σ ∈ HomAlgA(A′, A/I), then there exists a unique σ ∈HomAlgA(A′, A) which lifts σ, that is

Hom(A′, A)→ Hom(A′, A/I)

is bijective.

We also say A is Henselian along I, meaning the same.Example 6.2.2 $-adically complete R0-algebras are obviously Henselian along ($).Theorem 6.2.3 Suppose (Ai) is a direct system of R0-algebras which are Henselian along ($).Then lim−→Ai is Henselian along ($) as an R0-algebra.

Proof. We have:lim−→(Ai/$) = A/$

with A := lim−→Ai. One must just check (1) from Definition 6.2.1. First of all, given that$ ∈ J(Ai) for all i, we certainly have $ ∈ J(A). Write any two relatively prime, monicpolynomials f, g ∈ (A/$)[T ] as lim−→ fi and lim−→ gi, with fi and gi in (Ai/$)[T ], for each i, monicand relatively prime. Since each Ai is Henselian along ($), we can lift uniquely each fi and gito some fi, gi ∈ Ai[T ], and calling f := lim−→ fi and g := lim−→ gi, f and g do the job.

We need the following result from the book of Gabber and Ramero ([GR, Prop. 5.4.54]).Lemma 6.2.4 Let A be an R0-algebra which is Henselian along ($). Then the categoriesof finite étale A[1/$]-algebras and A∧[1/$]-algebras are equivalent. A∧ indicates the $-adiccompletion of A.

This result is both striking and crucial all over the place in the topics we are going to deal withhere onwards. The complete proof is discussed in a separate handout.

The reader may regard it as a vast generalization of Krasner’s Lemma.


6.3 Geometric Faltings’ almost purity

Let us state the main Theorem of this section, which we eventually prove in full detail.

Theorem 6.3.1 Let (R′, R′+

) be a perfectoid pair over (R,R+) with tilt (R′[, R′

[+) over

(R,R+). For any finite étale cover S/R′, S is perfectoid and S0a is finite and étale overR′

0a. Moreover, S0a is a uniformly almost finitely generated R′0a-module.

We first outilne the strategy of the proof.

Synopsis of the proof

We fix a perfectoid R-algebra R′, for Tate-perfectoid R. Recall that R0 is a perfectoid ring,likewise R′0. Our focus on Tate-perfectoid rings is due to the fact that our statements aregoing to be geometric in nature.

We only need to check that the inverse functor

(·)# : R′[fet → R′fet

provided by the tilting correspondence and almost purity in characteristic p > 0 (see §4.3),already achieved, is essentially surjective.

We aim at proving that a finite étale R′-algebra S is, therefore, strongly finite étale. To do sowe shall construct a covering Ui → X of X by open affinoid perfectoids such that for every i

S ⊗R′ OX(Ui) =: Si

is strongly finite étale. At this point, the datum of the Si’s will lead us to define affinoidperfectoid spaces Vi → X strongly finite étale over X, which glue to a global perfectoid spaceY → X which is strongly finite étale over X by design.

A crucial result saying that if X is affinoid (as in our case) and Y → X is a strongly finiteétale map of perfectoid spaces, then Y is affinoid, applies, and gives Y = Spa(A,A+), for(R′, R′

+) → (A,A+) strongly finite étale. By flatness of S over R′ one realizes at once that

A = S, and we conclude!

Bearing the above outline in mind, we shall discuss all the proof recalling step by step someneeded results, which will be proved separately in the last sections.

Proof. Perfectoid spaces over U which are strongly finite étale, are the same datum as finiteétale OX(U)0a-algebras, which are, in turn, a full subcategory of OX(U)-algebras. Who tellsus that they are a full subcategory of the category of finite étale OX(U)-algebras?

Indeed, this is the case.

Here we are also using the fact that if Y → X is a strongly finite étale morphism of perfectoidspaces, for any affinoid perfectoid subspace U ⊂ X, its preimage V ⊂ Y is affinoid perfectoid,and the morphism of Huber pairs over (R,R+):

(OX(U),O+X(U))→ (OY (V ),O+

Y (V ))

is strongly finite étale! Therefore, in particular, OX(V )0a∗ [1/$] = OX(V ) is finite étale over

OX(U), as desired.

We can assume U = X.


Step 1. We fix a finite étale R′-algebra S. First we check that for any x ∈ X we can find anaffinoid perfectoid neighbourhood U ⊂ X containing x, and a strongly finite étale cover V → Usuch that S ⊗R′ OX(U) is the untilt of OX[(V

[).

We use the following:Lemma 6.3.2 Let (Ai) be a filtered direct system of complete R0-algebras, and let A be thecompletion of the direct limit, which is again a complete R0-algebra. Then we have an equival-ence of categories:

A[1/$]fet ' 2- lim−→Ai[1/$]fet.

In particular, if (Ri) is a filtered direct system of perfectoid R-algebras, and R′ is the completionof the direct limit, then:

R′fet ' 2- lim−→(Ri)fet.

The above Lemma is telling us the following.

Let I = U ⊂ X | x ∈ U, where all U ’s are affinoid perfectoid, and call i = U , andAi := OX(U). We know Ai is a perfectoid R-algebra, and so we obtain an equivalence ofcategories between 2- lim−→(Ai)fet and (lim−→Ai)

∧fet. But:

lim−→Ai = lim−→x∈U

OX(U) = OX,x.

We need the following refinement, whose proof is postponed to the last section:Lemma 6.3.3 Fix x ∈ X. Then we have the following equivalence of categories:

2- lim−→x∈U

OX(U)fet ' κ(x)∧fet.

κ(x)∧ is a perfectoid field by Proposition 5.1.11.

We also recall the following Lemma, which we prove in a moment.

Roughly, it says that we know essential surjectivity of the untilting functor in the case R is aperfectoid field, and this base case is at the heart of the whole argument, in fact.

Lemma 6.3.4 The fully faithful functor κ(x[)∧fet → κ(x)

∧fet is an equivalence of categories.

Consider the finite étale κ(x)∧-algebra

S ⊗R′ κ(x)∧.

It is an object of the category κ(x)∧fet, and by the case of perfectoid fields, that is, Lemma 6.3.3,we know S ⊗R′ κ(x)∧ is a perfectoid κ(x)∧-algebra (!) and therefore we can tilt it to

(S ⊗R′ κ(x)∧

)[.

We set up the following diagram:

2- lim−→x∈U (OX(U))fet' // κ(x)

∧fet

2- lim−→x[∈U ′ (OX[(U′))fet '

// κ(x[)∧fet


where we use the fact κ(x)∧,[ ' κ(x[)∧, such isomorphism being unique.

(S⊗R′κ(x)∧

)[ spreads out to some open rational U ′ ofX[, by essential surjectivity of the bottomarrow, yelding a finite étale OX[(U

′)-algebra S′(U ′). By the equivalence κ(x)∧fet ' κ(x[)

∧fet,

we deduce that the untilt S′(U ′)# (unique up to unique isomorphism) which is an object ofOX(U ′

#)fet, has the property that its κ(x)∧-fiber is canonically isomorphic to the κ(x)∧-fiber

of S. By full faithfulness of the top arrow, we deduce existence of some open rational subsetUx → X such that:

S ⊗R′ OX(Ux)

is the untilt of a finite étale algebra over OX[(U′) for some U ′ rational in X[. By the almost

purity theorem in characteristic p > 0, already achieved, we deduce that such S ⊗R′ OX(Ux),which we call S(Ux), is strongly finite étale over OX(Ux).

This construction has been performed after fixing a point x ∈ X at the beginning, and thereforeyelds a cover of X by rational domains Ui → X, which, since X is affinoid, we may assumeis finite, such that

Si := S(Ui)

is strongly finite étale as an OX(Ui)-algebra, for all i. For each i, we let S+i to be the integral

closure of O+X(Ui) in Si, and then we set:

Vi = Spa(Si, S+i ).

Step 2. We find an appropriate affinoid perfectoid space Y strongly finite étale over X.

We state the following result, whose proof just amounts to saying that base-change preservesfinite étale morphisms, essentially using, also, that finite projective modules over perfectoidR-algebras are complete.Lemma 6.3.5 Let X = Spa(A,A+), Y = Spa(B,B+) and Z = Spa(C,C+) be affinoid per-fectoid spaces, with morphisms X → Y ← Z, and (A,A+) strongly finite étale over (B,B+).Then X ×Y Z = Spa(D,D+), where D = A⊗B C and D+ is the integral closure of C+ in D,and (D,D+) is strongly finite étale over (C,C+).

We’ll prove the Lemma at the end of the section.

Consider the pullback:V ′i := U ′ ×Ui Vi //

Vi

U ′

// Ui

for some open rational U ′ ⊂ Ui. V ′i is again strongly finite étale over U ′ by the Lemma, and itis affinoid perfectoid, of the form V ′i = Spa(S′i, S

′+i ). We have

S′i = S ⊗R′ OX(U ′).

This means that the Vi satisfy the gluing condition, and then glue in the category of perfectoidspaces over Spa(R,R+) to a perfectoid space Y → X which is strongly finite étale over X bydesign. But sinceX is affinoid perfectoid, then Y is affinoid perfectoid as well: Y = Spa(A,A+),with (A,A+) a perfectoid pair.


Step 3. R′[fet → R′fet is an equivalence.

To prove this, we must just show that A = S, because this means that the fully faithful functorR′[fet → R′fet is essentially surjective. In fact, recall that its image is exactly given by the

strongly finite étale R′-algebras, and A is strongly finite étale over R′, so if we show A = S, weobtain that any finite étale perfectoid R′-algebra S is actually strongly finite étale.

Since Y is perfectoid, OY is a sheaf, and the following sequence is exact:

0→ A→∏i

OY (Vi)→∏i,j

OY (Vi ∩ Vj)→ · · ·

On the other hand, the sheaf property of OX gives the exact sequence

0→ R′ →∏i

OX(Ui)→∏i,j

OX(Ui ∩ Uj)→ · · ·

and since S is flat over R′, then tensoring by S over R′ this latter sequence, we obtain theformer, since OY (Vi) = OX(Ui)⊗R′S, as discussed in Step 2, concluding that A = R′⊗R′S = S,as desired.

6.4 Proofs of the needed Lemmata

For the sake of clarity, we restate each Lemma and give the proof below.Lemma 6.4.1 Let (Ai) be a filtered direct system of complete R0-algebras, and let A be thecompletion of the direct limit, which is again a complete R0-algebra. Then we have an equival-ence of categories:

A[1/$]fet ' 2- lim−→Ai[1/$]fet.

In particular, if (Ri) is a filtered direct system of perfectoid R-algebras, and R′ is the completionof the direct limit, then:

R′fet ' 2- lim−→(Ri)fet.

Proof. Finite étale covers and morphisms between them (ie. squares) are finitely presentedobjects. Then

(lim−→Ai[1/$])fet ' 2- lim−→(Ai[1/$]fet).

Since lim−→Ai is Henselian along ($), the left-hand side agrees with A[1/$]fet. The result isproved.

Lemma 6.4.2 Fix x ∈ X. Then we have the following equivalence of categories:

2- lim−→x∈U

OX(U)fet ' κ(x)∧fet.

Proof. We compute:2- lim−→x∈U

OX(U)fet ' (lim−→OX(U))fet.

lim−→x∈U O+X(U) is Henselian along ($), as O+

X(U) is $-adically complete. Then we have:

(OX,x)fet ' (O+∧X,x[1/$])fet.


We have the following exact sequence of R0-modules:

0→ I → O+X,x → κ(x)+

where the last map has dense image.

We notice that if we pick f ∈ I, then |f(x)| ≤ |$(x)|, so that |$−1f(x′)| is ≤ 1 for x′ in someopen neighbourhood of x: U := x′ ∈ X | |f(x′)| ≤ |$(x′)| is such open. Then

f/$ ∈ ker(O+X,x → κ(x)+) = I

so that if we take $-adic completion I vanishes, because it is $-divisible. We claim thatO+,∧X,x → κ(x)+,∧ is an isomorphism. Reducing modulo $n, for any n ≥ 0, we get:

O+X,x/$

n → κ(x)+/$n

is an isomorphism, because I/$n = 0 and O+X,x → κ(x)+ is dense and the claim follows.

Therefore:(O+∧

X,x[1/$])fet = κ(x)+,∧[1/$]fet = κ(x)∧fet,

as desired.

The next Lemma is a direct consequence of the equivalence of étale sites in the case of perfectoidfields, which enables us to spread-out and find the desired finite étale cover V [ → U [ and untiltit to a strongly finite étale cover V → U . More in general, it makes Gabber and Ramero’sLemma fruitful.Lemma 6.4.3 The fully faithful functor κ(x[)

∧fet → κ(x)

∧fet is an equivalence of categories.

Proof. This is a direct consequence of the equivalence Cfet ' C[fet for any perfectoid field C,as discussed in §4.5.

6.5 Complements on strongly étale morphisms

We recall that the definition of a strongly étale morphism (Definition 6.1.11) of perfectoidspaces behaves very well under tilting, that is, f : X → Y is a strongly étale morphism ofperfectoid spaces if and only if f [ : X[ → Y [ is. Finiteness is also preserved by tilting, so thesame statement holds true with strongly finite étale morphisms. By the almost purity theoremin characteristic p > 0, any (finite) étale morphism is also strongly (finite) étale, and the upshotof the general version of Almost Purity is that étale morphisms of perfectoid spaces behavevery well under tilting.

We expand on the notion of strongly étale and strongly finite étale morphisms of perfectoidspaces, explaining in full detail some results which are needed in the proof of Theorem 5.1.4.The main one is the preservation of strong étaleness under rational localization.

We begin with the following:Proposition 6.5.1 Let X = Spa(A,A+), Y = Spa(B,B+) and Z = Spa(C,C+) be affinoidperfectoid spaces, with (A,A+) strongly finite étale over (B,B+). Then, we have

X ×Y Z = Spa(D,D+)

where D = A⊗B C and D+ is the integral closure of C+ in D. Moreover, (D,D+) is stronglyfinite étale over (C,C+).


Remark 6.5.2 Note that fiber products of affinoid perfectoid spaces exist by Proposition ,where they were constructed the following way. Notation being as in Proposition 6.5.1, X×Y Zwas Spa(F, F+), where

D = A⊗BC

and D+ was the integral closure of A+ ⊗B+ C+ in D. Therefore, the Proposition has alreadysome content in claiming that actually D+ can be chosen to be the integral closure of C+ in D.This has a nice consequence. Let U → Y be a rational embedding, U of the form Spa(B′, B′

+).

Then V := U ×Y Z → X is still an open embedding, as fibered products obviously preserveopen embeddings. Moreover, V is still affinoid. The upshot is that for any strongly finite étalemorphism of affinoid perfectoid spaces X → Y , there exist rational domains V ⊂ X and U ⊂ Ysuch that V → U is strongly finite étale.

Proof. A ⊗B C is finite projective over C, and therefore it is complete with its C-moduletopology. The universal property is a consequence of the universal property of tensor products.More precisely, suppose there exist maps (C,C+) → (F, F+) and (A,A+) → (F, F+), with(F, F+) a perfectoid pair. Then there’s a unique map D := C ⊗B A → F , and we claim thisinduces a unique map D+ → F+. We certainly have a unique map (A+ ⊗B+ C+)∼ → F+,(−)∼ meaning “integral closure in D”. The map 1⊗ id : C+ → (A+⊗B+C+)∼ certainly inducesa map D+ → (A+ ⊗B+ C+)∼, and we claim this is unique up to unique isomorphism. This istrue if and only if the induced map

D+a → (A+ ⊗B+ C+)∼,a

is unique up tp unique isomorphism, and now we know that (A+⊗B+ C+)∼,a = A0a⊗B0a C0a,which is, again, complete because it is almost finite projective over C0a. Since D+a → A0a⊗B0a

C0a is integral and D+ is open in D, we obtain that reducing modulo $ we get a canonicalisomorphism:

D+a/$ = A0a ⊗B0a C0a/$.

BothD+a and A0a⊗B0aC0a are finite étale over C0a, and thereforeD+a is a solution to the samedeformation problem as D0a, and hence we get a unique canonical isomorphism D+a = D0a,as desired.

The upshot is that the diagram:

(D,D+) (A,A+)oo

(B,B+)

OO

(C,C+)

OO

oo

is cocartesian in the category of complete Tate-Huber pairs.

Proposition 6.5.3 Let f : X → Y be a strongly finite étale, respectively strongly étale morph-ism of perfectoid spaces, and let g : Z → Y be an arbitrary morphism of perfectoid spaces.Then

X ×Y Z → Z

is a strongly finite étale, respectively strongly étale morphism of perfectoid spaces. Moreover,the underlying map of topological spaces

|X ×Y Z| → |X| ×|Y | |Z|

is surjective.


Remark 6.5.4 We first remark that the surjectivity statement is quite important when weintroduce the pro-étale topology for adic spaces, as we will point out in the next lecture, at theappropriate moment.

Proof. Suppose first that all spaces are affinoid perfectoid. By definition of a strongly finiteétale map, we can reduce to this case at least for the statement concerning strongly finite étalemorphisms, so Proposition 6.5.1 does the job. Since fibered products preserve open embeddings,the statement on strongly étale morphisms also follows. We address the surjectivity statement.

Let x and z be points of X and Z respectively, with same image y in Y under the maps X → Yand Z → Y . There exists a valuation v on D lying over x and z. Let 〈Γx〉 and 〈Γy〉 be theconvex hulls of the value groups of x and y in the value group Γv of v. Set H := 〈Γx〉 ∪ 〈Γy〉,which means H is either 〈Γx〉 or 〈Γy〉. In particular, H contains the characteristic subgroupcΓv of v (see ). We consider the vertical generization w := v|H of v. It satisfies w(a) ≤ 1 forall a ∈ D+, and it is continuous. Indeed, if we fix rings of definitions for A,B and C, and Ian ideal of definition for B (which can be chosen to be I = $B0 because all Huber pairs areTate and perfectoid) then w(I) is cofinal in Γw, because x(I) is cofinal in Γx and y(I) is cofinalin Γy. Therefore, w ∈ |X ×Y Z|, and still lies over (x, y) ∈ |X| ×|Y | |Z|, which concludes theproof.

We discuss the analogous results for (finite) étale morphisms of adic spaces over a perfectoidbase Spa(R,R+) of characteristic p > 0.Proposition 6.5.5 Let X = Spa(A,A+), Y = Spa(B,B+) and Z = Spa(C,C+) be affinoidadic spaces over Spa(R,R+), with (A,A+) finite étale over (B,B+), with Z perfectoid. Then,we have

X ×Y Z = Spa(D,D+)

where D = A⊗B C and D+ is the integral closure of C+ in D. Moreover, (D,D+) is stronglyfinite étale over (C,C+), and X ×Y Z is a perfectoid space.

Proof. Note that by the almost purity theorem in characteristic p > 0, D = A ⊗B C is aperfectoid R-algebra. It follows that X ×Y Z = Spa(D,D+) is a perfectoid space, and the restof the proof goes as in Proposition 6.5.1. We conclude.

Proposition 6.5.6 Let X → Y be a finite étale, respectively étale morphism of adic spaces,and Z → Y a map from a perfectoid space Z to Y . Then the fibered product X ×Y Z exists inthe category of adic spaces, is a perfectoid space, and the projection X×Y Z → Z is finite étale,respectively étale. Moreover, the map of underlying topological spaces |X ×Y Z| → |X| ×|Y | |Z|is surjective.

Proof. The finite étale case reduces to the case all spaces are affinoid, and hence to Proposition6.5.5. Open embeddings are trivially handled, and the proof of the surjectivity statement isexactly the same as in Proposition 6.5.1. We conclude.

Remark 6.5.7 Note that to conclude that a finite étale algebra over a perfectoid algebra isstill perfectoid, we need Almost Purity, and to avoid circular reasoning we must assume R is ofcharacteristic p > 0 to make use of the almost purity theorem, which, for now, is only availablein characteristic p > 0, as its general version relies on the results we are now discussing.

Let us state the main Theorem of this section.


Theorem 6.5.8 Let f : X → Y be a strongly finite étale morphism of perfectoid spaces. Then,for any open affinoid perfectoid V ⊂ Y , its preimage f−1(V ) =: U is still affinoid perfectoid,and the map OY (V )→ OX(U) is strongly finite étale.

To achieve the proof of the above Theorem, we need to realize perfectoid spaces in characteristicp as cofiltered limits of locally noetherian adic spaces. This results in noetherian approximationprinciple in positive characteristic, which will enable us to reduce many statements to theclassical rigid - analytic setup.

noetherian approximation of perfectoid spaces in characteristic p

For any Tate-perfectoid ring R, we may and do regard R as a perfectoid C-algebra for C theperfectoid field Fp(!($))($1/p∞)∧. Therefore, the results in this section are dealt with in thegenerality of (Tate-)perfectoid rings which are algebras over a perfectoid field C of characteristicp. We denote by ϕ the pth power map, as usual.Definition 6.5.9 A perfectoid pair (R,R+) over C is called p-finite if and only if there existsa reduced Tate-Huber pair (S, S+) over C of topologically finite type (ie. roughly S is a C-affinoid algebra) such that (R,R+) is the completed perfection of (S, S+), ie. R+ is the $-adiccompletion of lim−→ϕ

S+, and R = R+[1/$].

Remark 6.5.10 We say (S, S+) is a reduced Tate-Huber pair over C of topologically finitetype, instead of saying that S is C-affinoid, just because, to be precise, we are not consideringSp(S) but rather rKSp(S) = Spa(S, S+), as in [H2, Prop. 4.3]. We shall call such algebrasC-affinoid in the sequel.

Furthermore, note that we are saying that a Tate-Huber pair (R,R+) which is known to beperfectoid, is p-finite if it falls in the above situation, so we would not be required to check thatthe completed perfection of a C-affinoid algebra is perfectoid, for now. However, this is indeedthe case, as we show in the following Proposition.Proposition 6.5.11 Let S be a C-affinoid algebra, S+ = S0, and consider (R,R+) to be(S, S+)perf,∧, the completed perfection of (S, S+). Then (R,R+) is a perfectoid pair.

Proof. We only need to check that R+a is a perfectoid C0a-algebra, as R+ is certainly C0-flat, and therefore we have (R+a)∗ = R+ by the usual argument. R+a is C0a - flat, and itis $-adically complete by design. We are left to show that the pth power mao isomorphismcondition is satisfied. We have

R+a/$1/p = (lim−→ϕ

S+)/$1/p ' (lim−→ϕ

S+)/$ = R+a/$,

since lim−→S+ is perfect by design. We conclude.

We recall some essential facts about C-affinoid algebras.Theorem 6.5.12 Let (S, S+) be a C-affinoid algebra, and call X := Spa(S, S+). Then:

(1) The subset S+ = S0 is open and bounded.

(2) For any rational subset U ⊂ X, the Tate-Huber pair (OX(U),O+X(U)) over C is reduced

and of topologically finite type.

(3) For any finite covering Ui → X by rational domains, each cohomology group of thecomplex:

0→ OX(X)0 →∏i

OX(Ui)0 →

∏i,j

OX(Ui ∩ Uj)0 → · · ·


is killed by some power of $.

Proof. By Remark , part (1) is exactly [BGR, Thm. 1, 6.2.4], and part (2) is [BGR, Cor. 10,7.3.2]. By Tate’s aciclicity ([BGR, Thm. 1, 8.2.1]) we have that the complex:

0→ OX(X)d0−→∏i

OX(Ui)d1−→∏i,j

OX(Ui ∩ Uj)d2−→ · · ·

is exact. Therefore, ker(di) is a closed subspace of a Banach C-space, hence it is itelf Banach.Moreover, di−1 surjects onto ker(di), and by Banach Open Mapping Theorem, di−1 is an openmapping to ker(di). But therefore, the subspace topology on ker(di) coincides with the quotienttopology on im(di−1) = ker(di). Now let us consider the complex:

(∗) 0→ OX(X)0 d00−→∏i

OX(Ui)0 d01−→

∏i,j

OX(Ui ∩ Uj)0 d02−→ · · ·

Since S0 is open and bounded, and by part (2) the same is true for each one of the OX(Ui)0 and

OX(Ui ∩ Uj)0, the quotient topology on im(di−1) is generated by the basis of open neighbour-hoods of zero: $n · im(d0

i−1), n ∈ Z. Likewise, the subspace topology on ker(di) is generatedby the basis of open neighbourhoods of zero: $n · ker(d0

i ), n ∈ Z. The fact that they agreeimplies that for each i, there exists some integer Ni such that

$Ni · ker(d0i ) ⊆ im(d0

i−1).

Note that such Ni can be chosen to be uniform because X is finite dimensional, and therefore(∗) is killed by some power of $, as desired.

We need one more preliminary result:Proposition 6.5.13 Let (R,R+) be a perfectoid pair which is p-finite, given as the completedperfection of a Tate - Huber pair (S, S+) over (C,C+), for C-affinoid S. Call X := Spa(R,R+)and X0 := Spa(S, S+). Then:

(1) The map X → X0 is a homeomorphism identifying rational subsets in both directions.

(2) For any U ⊂ X rational, corresponding to U0 ⊂ X0, the perfectoid pair:

(OX(U),O+X(U))

is the completed perfection of (OX0(U0),O+

X0(U0)).

Proof. Taking the perfection and then the completion does not change the adic space X0 andrational domains, which establishes part (1). For part (2), we simply observe that the completedperfection of (OX0

(U0),O+X0

(U0)) has the universal property defining (OX(U),O+X(U)) among

all perfectoid pairs.

At this point we are ready to discuss the main result, which enables us to make use of thep-finite perfectoid pairs as a bridge between perfectoid spaces and classical rigid-geometry, toreduce many problems to this latter setup.Theorem 6.5.14

(1) Any perfectoid pair (R,R+) can be written as the completion of a filtered colimit of p-finiteones (Ri, R

+i ).


(2) This induces a homeomorphism Spa(R,R+) ' lim←− Spa(Ri, R+i ), and each rational U ⊂

X = Spa(R,R+) comes as the preimage of some rational Ui ⊂ Xi = Spa(Ri, R+i ).

(3) In this case, (OX(U),O+X(U)) is the completion of the filtered colimit of the (OXj (Uj),O

+Xj

(Uj)),where Uj is the preimage of Ui in Xj, for j ≥ i.

(4) Let Ui ⊂ Xi be some quasi-compact open subset containing the image of X. Then there issome index j such that the image of Xj is contained in Ui.

Proof. For part (1), pick any finite subset I ⊂ R+. We define the C-subalgebra SI ⊂ R to bethe image of

K〈ti, i ∈ I〉 → R

given by ti 7→ i. Then SI is a reduced quotient of K〈ti, i ∈ I〉, and we endow SI with thequotient topology. We let S+

I ⊂ SI be the set of power-bounded elements in SI . By [Tate,Theorem 5.2], this is also the set of elements of SI integral over K〈ti, i ∈ I〉. Now, we define

(RI , R+I ) := (SI , S

+I )perf,∧,

that is:R+I := (lim−→

Frob

S+I )∧ and RI := R+

I [1/$].

Note that (RI , R+I ) is a p-finite perfectoid pair over C, by Proposition 6.5.11. It comes, for

each I, with a natural map to (R,R+). We claim that

lim−→I

R+I /$

n = R+/$n

for all n ≥ 0. Indeed, this map is surjective, as we may take x in R+, and we let SI be theimage of K〈ti, i ∈ I〉 in R, with I = x. Write R+ as lim−→Frob

R+, since R+ is perfect. Itfollows that

lim−→Frob

S+I → R+

is still surjective by right-exactness of filtered colimits, and being R+ also $-adically complete,passing to $-adic completion preserves surjectivity for the same reason. It follows that thereexists I such that R+

I → R+ is surjective. We can write R+ as lim−→IR+, the transition maps

being the identity, and again right-exactness of filtered colimits yields surjectivity of lim−→IR+I →

R+, hence of lim−→IR+I /$

n = R+/$n, as desired. This map is also injective. Indeed, if x, y ∈ R+I

satisfyx− y = $nz, z ∈ R+,

then for some larger J ⊃ I containing z, we get x − y ∈ $nR+J . We deduce that R+ is the

completed filtered colimit of the (RI , R+I ).

For part (2), call (A,A+) the filtered colimit of the (Ri, R+i ) equipped with the $-adic topology.

Then we get the natural homeomorphism

Spa(A,A+) ' lim←− Spa(Ri, R+i ),

compatible with rational subspaces. But then the same thing holds true for the completedfiltered colimit.

For part (3), we simply observe that the completion of the filtered colimit of the (OXj (Uj),O+Xj

(Uj))

is a perfectoid pair which shares the same universal property describing (OX(U),O+X(U)). Fi-

nally, we prove part (4).


We know Ui contains the image of X → lim←−Xi =: X∞, and is quasi-compact. Call Uj thepreimage of Ui in Xj for all j ≥ i, and U∞ the preimage of Ui in X∞. We endow each one of theXj −Uj with the constructible topology, and likewise for their limit, to get a homeomorphism:

X∞ − U∞ = lim←−(Xi − Ui).

Both sides are compact topological spaces, with spectral transition maps (observe that since Uiis quasi-compact, Xj −Uj are constructible in Xj for each j ≥ i). Hence Xj −Uj are spectral,and if X∞ = U∞, we deduce that for some i we had Xj = Uj , as desired.

Now let X be an affinoid perfectoid space. From Theorem 6.5.14 we can write X as the limitlim−→Xi of p-finite perfectoid spaces, and if we take a (finite) étale cover Y → X, we wonder if itdescends to some finite stage. To address this question we need once again Gabber’s Lemma,which we recall.Lemma 6.5.15 Let (Ai) be a filtered direct system of complete flat K0-algebras, and let A bethe completion of the direct limit, which is again a complete flat K0-algebra. Then we have anequivalence of categories

A[1/$]fet ' 2- lim−→Ai[1/$]fet.

In particular, if (Ri) is a filtered direct system of perfectoid C-algebras, and R is the completionof the direct limit, then

Rfet ' 2- lim−→(Ri)fet.

We prove the following:Proposition 6.5.16 Let X be an affinoid perfectoid space. Write X as the limit lim−→Xi ofp-finite perfectoid spaces, and consider a finite étale cover Y → X. Then Y is already definedas a finite étale cover of some Xi.

Proof. There exist finitely many rational domains in X over which we have a finite étale cover.By Gabber’s Lemma, these rational domains and covers are already defined at some finite stage,and since X is quasi-separated, the gluing data over intersections and the cocycle condition arealso defined at some finite stage, and we conclude.

We are ready to show the fundamental Proposition 6.5.8.

Proof of Proposition 6.5.8. We can assume Y = V , and claim X is affinoid, given that f :X → Y is strongly finite étale. Tilting preserves affinoidness in both directions, so this can bechecked after tilting, that is, we reduce to the caseX is a perfectoid space in characteristic p > 0.Writing Y = lim←−Yi, where Yi = Spa(Ri, R

+i ) are p-finite perfectoid spaces, by Proposition 6.5.16

f is already defined as a finite étale cover of some Yi, and we use Proposition 6.5.1 to reduceto the p-finite case. Now write Y as Spa(R,R+), where (R,R+) is the completed perfectionof some reduced Tate-Huber pair of topologically finite type. Using Gabber’s Lemma, again fis defined at some finite stage of the completed perfection limit process, and we are reducedto the case f is a finite étale morphism of locally noetherian adic spaces over C, with Yaffinoid. This is proved in Huber’s book [Hub, Example 1.6.6(ii), Lemma 7.3(iv)]. By thetilting correspondence, OY (V ) is a perfectoid C-algebra, and we conclude.

Again by using Theorem 6.5.14 and Proposition 6.5.16, we can establish the following:Proposition 6.5.17 Let f : X → Y be an étale map of perfectoid spaces. Then, for anyx ∈ X there exist affinoid perfectoid neighbourhoods U ⊂ X of x and V ⊂ Y of f(U), and anétale morphism of affinoid noetherian adic spaces U0 → V0 over K, sich that U = U0 ×V0

V .Moreover, strongly finite étale morphisms of perfectoid spaces are open, and compositions ofstrongly finite étale morphisms are strongly finite étale.


6.6 The small étale site and étale topos on a perfectoid space X

We fix an affinoid perfectoid space S := Spa(R,R+) and its tilt S[.

We give the following:Definition 6.6.1 Let X be a perfectoid space over S. The étale site of X is the category Xet

of perfectoid spaces which are étale over X, and coverings are given by topological coverings.The associated topos is denoted by X∼et .

For a neat treatment of the étale topology and cohomology of adic spaces, the reader may referto [Hub, §2.1, 2.2, 2.3].

The results discussed in these notes imply that Xet is indeed a site, and hence X∼et is a topos.

As soon as we have a morphism of perfectoid spaces X → Y , we obtain an induced morphismof sites Xet → Yet, as well as of topoi.

Due to all the results we discussed so far, we obtain that the tilting operation not only inducesa homeomorphism of topological spaces |X| ' |X[| which is functorial in X, but also anequivalence of sites:

Xet ' X[et

which carries on to the étale topoi.

This is a key fact which turns out to be the full strength of the theory of perfectoid spaces,which will be made fruitful for the purpose of understanding p-adic étale cohomology of properrigid-analytic varieties by the introduction of the pro-étale topology.

Very roughly speaking, we deduce that X and X[ have the same étale cohomology, an assertionwe make precise in a moment. We remark that X may live in mixed characteristic, while X[

lives in equal characteristic p.

To make use of this intuition, we need comparison results between the étale sites of perfectoidspaces and adic spaces. This motivates the introduction of the pro-étale topology on adicspaces, and the fact that a wide class of adic spaces is pro-étale locally perfectoid makes thetheory fruitful and extremely powerful to compute their étale cohomology.

6.7 Cohomological vanishing and almost vanishing

We conclude with a few vanishing results.Proposition 6.7.1 Let X be a perfectoid space over S. Then, for all i > 0, Hi(X,O+

X) isalmost zero. Moreover, the assignement

U 7→ OU (U)

is a sheaf on Xet, and Hiet(X,O

0aX ) = 0 for all i > 0 if X is affinoid perfectoid.

Proof. We prove the last statement, being the proof of the first identical. This is checked justproving exactness of the complex

(∗) 0→ OX(X)0a →∏i

OVi(Vi)0a →

∏i,j

OVi×XVj (Vi ×X Vj)0a → · · ·

for any finite covering Vi of X, where each Vi is étale over X. The Vi’s are rational subsetsof some finite étale V ′i → Ui ⊂ X, and Ui is a rational subspace. We are reduced to the caseX is affinoid perfectoid and by tilting we reduce to the characteristic p > 0 case.


We first assume we know already OX is an étale sheaf on X, and therefore the complex

0→ OX(X)→∏i

OVi(Vi)→∏i,j

OVi×XVj (Vi ×X Vj)→ · · ·

is exact. We use the almost purity theorem and Banach open mapping theorem to deduce thata suitable power of $ kills the cohomology of (∗). By applying the inverse of the pth powermap, we deduce such cohomology is almost zero, and we conclude.

We are left to show that OX is a sheaf on Xet. We reduce to the case X is affinoid perfectoid.For any étale cover V → X, by the almost purity theorem we know V is again a perfectoidspace, and then it makes sense to tilt it. By tilting, we reduce to checking exactness of thecomplex:

0→ OX(X)→∏i

OVi(Vi)→∏i,j

OVi×XVj (Vi ×X Vj)→ · · ·

in characteristic p > 0, and by means of Theorem 6.5.14 and Proposition 6.5.16, we reduce tothe case X is a locally noetherian adic space over Fp(($))($1/p∞)∧, for which OX is indeed anétale sheaf.

When we define the pro-étale topology on locally noetherian adic spaces, it will turn out thatperfectoid spaces form a basis for such topology (in characteristic 0) and their well behavednessis due to the fact that they are contractible in the sense of almost mathematics, according tothe above Proposition.

It turns out that for X affinoid perfectoid, Hiet(X,O

+X/p) is almost zero for all i > 0, the proof

being similar to the above, and left to the reader as an exercise.


A Loose ends on Almost Mathematics

We prove some results which are yet to be discussed, though freely used in §3 and §6, in ordernot to make the presentation heavier.

A.1 Some tools

We recollect here a few facts we are going to need straight away from Almost Mathematics,recollecting motivation and proofs.

We first recall the following:Proposition A.1.1 Let A be a commutative ring, and I a finitely generated ideal such thatI = I2. Then V (I) → Spec(A) is a clopen immersion, and there exists an idempotent e ∈ Asuch that A/I ' Ae.

Proof. By Nakayama’s Lemma, there exists a ∈ I such that e := 1 + a kills I. We check a fewthings. First, V (I) = D(e): simply, since eI = 0, (A/I)e ' Ae, that is, V (I) ∩D(e) = D(e),and D(e) ⊂ V (I). On the other hand, if f ∈ I, then if f ∈

√eA we have fn = ai = eb =

e(eb) = e(ai) = 0, that is,√eA ∩ I = (0), and it follows V (I) ⊂ D(e).

Of course, the reader is encouraged to recall the geometric picture behind the above Proposition!

The typical situation to which we apply the above proposition is when A→ B is an unramifiedring map with B finite type over A. We’ll have that the multiplication map µB/A : B⊗AB → Bis surjective, and its kernel J is finitely generated by bi⊗1−1⊗bi, for bi finitely many A-algebragenerators for B. By unramifiedness, J2 = J , so that the proposition applies, and we havethat:

Spec(B)→ Spec(B ⊗A B)

is a clopen immersion, and we are given a unique idempotent e ∈ B ⊗A B for free.Remark A.1.2 The idempotent given by Proposition A.1.1 is unique: suppose there is anotheridempotent e′ such that D(e) = D(e′). Then if we call a := 1− e and b := 1− e′, we have

0 6= e− e′ = e(e′ + b)− e′(e+ a) = eb− e′a.

Then at least one of eb and e′a is nonzero and idempotent, and hence there is a prime ideal pwhich does not contain it. This implies that either e ∈ p and e′ /∈ p, or conversely. Contradic-tion.Example A.1.3 As a typical non-example, consider A = Zp, and B = Zp[ζp].

AnnB(Ω1B/A) = DB/A = (ζp − 1)p−2Zp[ζp] 6= B,

and this means J/J2 6= 0, implying that J 6= J2. As an example, do the same with A = Zpand B = Zp[ζ`] for ` 6= p. Then B is unramified over A, and if we consider an element x ofJ , which is 1 ⊗ ζ` − ζ` ⊗ 1, for example, then µx =: T makes J into a (B ⊗A B)[T ]-module,and its minimal polynomial pT is such that pT (x)J = 0. pT (x) = 1 + a, where a ∈ J . One canexplicitly compute pT for our x, and e will be exactly pT (x).

In light of Proposition A.1.1, we recollect here the properties that such idempotent e ∈ B⊗ABsatisfies in the case A→ B is unramified.

(1) e2 = e.


(2) e · ker(µB/A) = 0.

(3) µB/A(e) = 1B .

Proof. x := µB/A(e) must be an idempotent of B, and obviously it cannot be 0. Recallthat e = 1 + a for a ∈ J . µ(1 + a) = 1, because a goes to zero under µ.

Let us make another example:Example A.1.4 Let A → B be such that Spec(B) → Spec(A) is an étale G-torsor, for somefinite group G. Then: ∏

g∈GB ' B ⊗A B

with b⊗ b′ 7→ (b · g(b′))g∈G. e corresponds to the tuple (1, 0, · · · , 0).

A few more remarks:

(1) Write e =∑xi⊗ yi. If we consider multiplication by e, µe, this has a trace which we can

compute this way:Tr(µB/A(e)) =

∑Tr(xiyi) = Tr(1)

and if B/A is finite étale, then this is exactly the degree of B/A. Think about B/A afinite Galois extension of fields.

(2) If we takeB → A⊕N → B

the maps b 7→ (Tr(xib)) and (ai) 7→∑aiyi, the composition is the identity.

A.2 Almost unramified and almost étale maps

Recall that given a V a-algebra A, we can let B be an A-algebra (that is, a commutative unitarymonoid in the abelian tensor category (A-Mod,⊗, A), with some additional structure. For abroader discussion, the reader is referred to Lecture 18). We can also let:

B∗ := HomA(A,B) = HomA(A,Ca) = HomA∗(A!, C) = HomA∗(m⊗VA∗, C) = HomA∗(A∗,Hom(m, C)).

Note that if A = V a, we recover B∗ = HomV (m, C).

On almost elements, given an A-algebra B, we have a multiplication map:

B∗ ⊗A∗ B∗ → B∗

given by sending b⊗ b′ to bb′. Applying the functor (·)a, we obtain a map µB/A : B⊗AB → Bwhich we shall always confuse with the lower-starred version. We can view B as a B ⊗A B-algebra. We are ready to recall our first:Definition A.2.1 Let f : A→ B be a morphism of almost V -algebras.

(1) We say f is flat (resp. faithfully flat) if B is flat as an A-module (resp. if (·) ⊗A B isfaithful).

(2) We say that f is (uniformly) almost finite if B is a (uniformly) almost finitely generatedA-module.

(3) We say that f is weakly unramified (resp. unramified) if B is a flat (resp. almost projective)B ⊗A B-module.


(4) f is weakly étale (resp. étale) if it is flat and weakly unramified (resp. unramified).

We prove the following characterization of unramified maps of almost algebras, which was notdiscussed in Lecture 19.Proposition A.2.2 A morphism f : A→ B is unramified if and only if there exists an almostelement eB/A ∈ (B ⊗A B)∗ such that

(1) e2B/A = eB/A.

(2) µB/A(eB/A) = 1B.

(3) eB/A · J = 0, with J = ker(µB/A).

Proof. Let us assume f is unramified. This means that B is almost projective as a B ⊗A B-module via µB/A. Consider the short exact sequence:

0→ J → B ⊗A B → B → 0

Since B is almost projective, it (almost) splits. More precisely, for all ε ∈ m, there is an A-linearmap uε : B → B ⊗A B, with the property that µB/A uε = ×ε, that is, “multiplication by ε”,on B. We must show that there exists eε ∈ (B ⊗A B)∗ such that e2

ε = ε · eε, µB/A(eε) = ×ε,eε · J∗ = 0. We define eε := uε(1). uε is actually B⊗AB-linear. This means that the followingdiagram commutes:

B

λµ(b⊗b′)

uε // B ⊗A B

λb⊗b′

Buε// B ⊗A B

where λ denotes “left multiplication”.

In particular:e2ε = eε · eε = eε · uε(1) = uε(µ(eε) · 1).

This latter is εeε because µB/A(eε) = ε by design. We take x ∈ J∗, and compute:

x · eε = x · uε(1) = uε(µ(x)) = 0.

Our eB/A we are seeking, is an almost element. We regard it, then, as an A-homomorphism

HomA(A,B ⊗A B)

However, A is a V a-algebra, which, by design, comes with a canonical isomorphism V a⊗V aA 'A encoding the V a-multiplication on A. We obtain:

HomA(A⊗V a V a, B ⊗A B) ' HomV a(V a,HomA(A,B ⊗A B)) ' HomV (m, (B ⊗A B)∗),

where m = m⊗V m is actually isomorphic to m, which we assume to be flat as a V -module. Itis introduced only to make the argument cleaner.

m→ (B ⊗A B)∗ is everything we are required to assign, then. We set, for ε, δ ∈ m,

ε⊗ δ 7→ δ · eε : A→ B ⊗A B.

A few little verifications, given that δ − eδ, ε− eε ∈ J , give

δeε = εeδ.

The assignement is clearly V -linear, and we conclude. Conversely, if we are given eB/A, thenwe can define ub := eB/A (1⊗ b) : A→ B ⊗A B, and b 7→ ub is a map

B∗ → HomA(A,B ⊗A B) = (B ⊗A B)∗.

Applying (−)a we conclude.


A.3 Almost traces

We have, given A a V a-algebra, and P an A-module, a natural map:

ωP/A : P ⊗A P∨ → EndA(P )a.

If P is almost finitely generated projective, then ωP/A is an isomorphism. We briefly recollectthe main ideas in the proof. We have, by assumption on P , that the composition:

P → AI → P

is multiplication by ε for all ε ∈ m and all I := I(ε). Apply the natural transformation

P ⊗A (−)∨ → HomA(−, P )a

and with a diagram chase we get that its kernel and cokernel are killed by ε. Since ε ∈ m isarbitrary, they are almost zero, as required.

We define the trace morphism of P to be:

tP/A := evP/A ω−1P/A : EndA(P )a → P ⊗A P∨ → A,

the map evP/A : P ⊗A P∨ → A being the evaluation map.

We also define:TrB/A := tB/A µ,

where µ is multiplication. We introduce:

TB/A := TrB/A µB/A : B ⊗A B → A

which induces a bilinear pairing τB/A : B → B∨ by τB/A(b) = T (b⊗ (−)). We say T is a perfectpairing if and only if τB/A is an isomorphism.

We finally recall the following useful fact, whose proof was quite involved and used the notionof almost differents and transitivity of Tr with respect to successive extensions.Proposition A.3.1 Let A → B be a map of V a-algebras, such that B is uniformly almostfinitely generated almost projective as an A-module. Then A → B is étale if and only if thetrace pairing is perfect.

A.4 Finite étale lifts and purely inseparable morphisms

We address the following two results, which are respectively Theorem 4.3.1 and Proposition4.3.2. We fix a perfectoid ring R throughout.Theorem A.4.1 Let B be a finite étale A-algebra, where A is a perfectoid Ra-algebra. ThenB/$ is a finite étale A/$-algebra. Moreover, B 7→ B/$ = B ⊗A A/$ induces an equivalenceof categories Afet → (A/$)fet.Proposition A.4.2 Let A be a perfectoid (Ra/$)-algebra. Take B to be a finite étale algebraover A. Then B is perfectoid as an (Ra/$)-algebra.

We begin with the proof of Theorem A.4.1.

Proof of Theorem A.4.1. Since B is finite étale as an A-algebra, its reduction modulo$ is finiteétale as an (A/$)-algebra, because étaleness is stable under base change and likewise finiteness


(in the sense of almost mathematics). Let B be a finite étale (A/$)-algebra. By étaleness,LB/(A/$) ' 0 in D(A/$). It’s easy to check that, for each (unique) deformation of B at everyfinite layer, the corresponding cotangent complex vanishes, and hence there exists a unique (upto unique isomorphism) flat $-adically complete A-algebra B such that B/$ ' B as (A/$)-algebras. B is unramified over A. We claim B/$n is unramified as an (A/$n)-algebra, in thesense of almost mathematics. We call An := A/$n, Bn := B/$n, for all n ≥ 1. Grantingthis for a moment, we obtain almost idempotents en ∈ (Bn ⊗An Bn)∗ for all n ≥ 1, such that,calling µn : Bn ⊗An Bn → Bn the multiplication map and In its kernel, µn(en) = 1Bn anden(In)∗ = 0. The en’s are compatible, since each one of them is unique and the multiplicationmaps are clearly compatible. Therefore, we obtain an almost idempotent

e ∈ (B ⊗A B)∗ = (lim←−Bn ⊗An Bn)∗ = lim←−(Bn ⊗An Bn)∗.

Its properties are easily deduced from the fact that the multiplication map µ : B ⊗A B → B isthe inverse limit of the projective system of maps (µn)n≥1, and likewise I = ker(µ) is lim←− In.We deduce B is unramified as an A-algebra in the sense of almost mathematics.

We know B is unramified as an (A/$)-algebra. To let the induction start, we show thefollowing:

Claim Let f : V → V ′ be a morphism of Ra-algebras, and I, J ⊂ Ra any two ideals. Iff ⊗ idV/I and f ⊗ idA/J are unramified, then so is f ⊗ idA/IJ .

The Claim does the job with V = A and V ′ = B, and with the choice I = $A and J = $n−1B,to show B/$n is unramified as an (A/$n)-algebra, given that both B/$ and B/$n−1 are,as an (A/$)-algebra and (A/$n−1)-algebra respectively. We show the Claim in the followingLemma A.4.3.

We claim B is an almost finitely generated A-module. It’s easy to check inductively that eachone of the (B/$n) is almost finitely generated as an (A/$n)-module. In particular, B/$2 is asan (A/$2)-module. It means there exist almost elements b1, . . . , bk in B whose images in B/$2

generate an (A/$2)-submodule M of B/$2 with the property that m(B/$2) ⊂ M ⊂ B/$2.Call M the A-submodule of B almost generated by the bi’s. We have (B/$) → B/M isalmost surjective, hence, since B is an M -extension of B/M as an A-module, is almost finitelygenerated. Since B/$ is almost finitely presented as an (A/$)-module, by the same argumentapplied to the kernel of an almost surjection (A/$)⊕N → B/$ for some integer N ≥ 0 wededuce B is almost finitely presented as an A-module. Hence B is finite étale over A, and inparticular it is almost finite projective, by the usual trace argument.

It follows that the assignment B 7→ B, which is functorial due to vanishing of the cotangentcomplex, is a quasi-inverse to the reduction modulo $ functor, thus achieving the Theorem.

Lemma A.4.3 Let f : V → V ′ be a morphism of Ra-algebras, and I, J ⊂ Ra any two ideals.If f ⊗ idV/I and f ⊗ idV/J are unramified, then so is f ⊗ idV/IJ .

Proof. Recall from Definition A.2.1 that we are required to check that V ′/IJ is an almostprojective (V ′/IJ)⊗V/IJ (V ′/IJ)-module. First we claim that if f ⊗ idV/I and f ⊗ idV/J areweakly unramified, then so is f ⊗ idV/IJ .

Let A be any V -algebra, and set A′ := A⊗V V ′. For every A′ ⊗A A′-module M , we claim wehave a natural isomorphism:

TorV′⊗V V ′

1 (V ′,M) ' TorA′⊗AA′

1 (A′,M).


Indeed, the short exact sequence of V ′ ⊗V V ′-modules:

0→ IV ′/V → V ′ ⊗V V ′ → V ′ → 0

is split exact as a sequence of V ′-modules (hence as a sequence of V -modules). Tensoring byany V -module N we still get an exact sequence, then. It follows:

TorV′⊗V V ′

1 (V ′, N ⊗V V ′ ⊗V V ′) = 0

for every V -module N . The morphism V ′ ⊗V V ′ → A′ ⊗A A′ yields a base change spectralsequence:

E2ij := TorA

′⊗AA′i (TorV

′⊗V V ′j (V ′, A′ ⊗A A′),M)⇒ TorV

′⊗V V ′i+j (V ′,M)

for every A′ ⊗A A′-module M . By the above reasoning we have E2i1 = 0 for all i ≥ 0, and the

claim follows.

To show the statement on weak unramifiedness, it’s enough to check the case when IJ = 0 (ie.rename V/IJ as V ). We only need to check that for every V ′ ⊗V V ′-module N :

TorV′⊗V V ′

1 (V ′, N) = 0.

We can reduce to the case I kills N or J kills N . Say IN = 0. Calling A := V/I andA′ := V ′/IV ′, we deduce:

TorV′⊗V V ′

1 (V ′, N) ' TorA′⊗AA′

1 (A′, N)

and the latter module is zero by design of f ⊗ idA′ .

It follows V ′ is flat as a V ′ ⊗V V ′-module in the sense of almost mathematics. We need tocheck it’s also almost projective. Certainly, we know that, calling I ′ := I(V ′ ⊗V V ′) andJ ′ := J(V ′⊗V V ′), V ′/(IV ′∩JV ′) is almost projective as a (V ′⊗V V ′/(I ′∩J ′))-module. Since(I ′ ∩ J ′)2 ⊂ I ′J ′ = 0, we conclude because of the following Lemma.

Lemma A.4.4 Let A → B a surjective map of Ra-algebras with nilpotent kernel, and M aflat A-module. If M ⊗A B is almost projective as a B-module, then so is M as an A-module.

Proof. We are required to check that Ext1A(M,N) is almost zero for all A-modules N . Let I

be the kernel of A → B. We can assume IN = 0. Let ξ ∈ Ext1A(M,N) be represented by an

extension:0→ N → Q→M → 0

of A-modules. Upon tensoring by B over A and using flatness of M as an A-module, we obtainthe exact sequence of B-modules:

0→ N ⊗A B → Q⊗A B →M ⊗A B → 0

which implies ξ comes from an element of Ext1B(M ⊗A B,N ⊗A B). This latter is almost zero

by assumption. We conclude.

We finally address the proof of Proposition A.4.2.

We need the following:


Definition A.4.5 Let V be an Fp-algebra and ϕ : V → V the pth power map, B a V -algebra. Let B(m) := (ϕm)∗B, i.e. B(m) is B regarded as a V -algebra under the structure map

Vϕm−−→ V → B.

Let f : A → B be a morphism of V -algebras. We say f is invertible up to ϕm if and only ifthere exists f ′ : B → A with the property that f ′ f = ϕmA and f f ′ = ϕmB , ϕA and ϕB beingthe Fp-linear pth power maps on A and B respectively (ie. the absolute Frobenii).

The Proposition is settled by the following:Proposition A.4.6 Let f : A→ B be an étale morphism of Ra-algebras.

(1) If f is invertible up to some power of ϕ, then it is an isomorphism.

(2) For every integer m ≥ 0, the following diagram is cocartesian:

A

ϕmA

f// B

ϕmB

A(m)f(m)

// B(m)

The motivation for (1) is that a radicial morhpism of schemes which is étale (and surjective) isan isomorphism. (For field extensions, this says that a purely inseparable field extension whichis separable, is trivial).

Proof. To check (1), we first show that f is faithfully flat. As f is flat, we are only required tocheck that M ⊗A B = 0 implies M = 0. We can check this for M = A/I, I ⊂ A an arbitraryideal. Upon base changing along the map A → A/I, we reduce to show that if B = 0, thenA = 0.

By assumption, A∗ → B∗ is invertible up to ϕm, for some integer m ≥ 0. Then ϕmA∗ = 0,implying A∗ = 0. It follows A = 0, because A ' Aa∗. We conclude f is faithfully flat.

Now we need to show that f is an isomorphism. Note that by the foregoing, the proof thatf is an isomorphism is complete in the case f is an epimorphism, as we’ve just shown it’s amonomorphism.

We have the maps:B

h−→ B ⊗A BµB/A−−−→ B

where the first map is the base change h of f along the A-algebra structure on B given byf itself. It’s an easy exercise left to the reader to check that h is invertible up to ϕm, andthat, given this, then µB/A is invertible up to ϕm. On the other hand, µB/A is flat with flatdiagonal (check this latter property using a change of rings spectral sequence!), and these werethe only assumptions we used above. Therefore, we conclude µB/A is an isomorphism, sinceit’s an epimorphism. It follows h is an isomorphism, and by faithful flatness so is f , as desired.

To show (2), it’s enough to check that the morphism g : B⊗AA(m) → B(m) induced by ϕmB andf(m) is flat with flat diagonal. If so, upon showing it’s invertible by some power of ϕ, then by(1) it’s an isomorphism, which achieves the Proposition. We know ϕmA and ϕmB are invertibleup to ϕm, and that B⊗A ϕmA : B → B⊗AA(m) is invertible up to ϕm. It’s an easy verificationleft to the reader to check that, therefore, g is invertible up to ϕ2m, and we conclude.


B The cotangent complex

In this appendix we recollect what we need from deformation theory by means of Illusie’scotangent complex. We shall not review the construction and the main theorems, but ratherrefer the reader to [Cot] for a wider background. The explanation of the needed results fromobstruction theory is completely self contained and the theorems directly proved.

We begin with some needed preliminaries on derived base change.

B.1 Preliminaries on derived change of rings

Let R→ R′ be a ring map. Let N• be a complex of R′-modules. We use Spaltenstein’s K-flatresolutions (we refer the reader to [Sp] for a wider background, which won’t be needed for ourpurposes).

We recall that a complex of R-modules K• is called K-flat if, for every acyclic complex of R-modules L•, the complex Tot(K•⊗RL•) is acyclic. It’s an easy verification to check that everycomplex of R-modules N• comes with a quasi-isomorphism K• → N• from a K-flat complex.One checks that the total complex of the tensor product of K-flat complexes is still K-flat, andthat filtered colimits of K-flat complexes are K-flat. Complexes of projective R-modules areK-flat.

We define a functor:(·)⊗L

R N• : D(R)→ D(R′)

as the left derived functor of the functor: Comp(R-Mod)→ Comp(R′-Mod),M• 7→ Tot(M•⊗RN•). In particular, taking N• = R′[0] we obtain a derived base change functor:

(·)⊗LR R

′ : D(R)→ D(R′).

To sum up, for every complex of R-modules M• we can choose a K-flat resolution K• → M•

and setM• ⊗L

R N• := Tot(K• ⊗R N•).

Lemma B.1.1 The construction above is independent of choices and defines an exact functorof triangulated categories: (·)⊗L

R N• : D(R)→ D(R′). There is a functorial isomorphism:

E• ⊗LR N

• = (E• ⊗LR R

′)⊗LR′ N

•

for E• in D(R).

Proof. To prove the existence of the derived functor (·) ⊗LR N• we use the general theory.

Set D = Comp(R-Mod) and D′ = D(R′). We write F : D → D′ the exact functor oftriangulated categories defined by the rule F (M•) = Tot(M• ⊗R N•). We let S be the setof quasi-isomorphisms in D. Check that LF is everywhere defined. Thus we obtain a derivedfunctor:

LF : D(R) = S−1D→ D′ = D(R′)

Finally, check that LF (K•) = F (K•) = Tot(K• ⊗R N•) when K• is K-flat, ie. LF is indeedcomputed in the way described above. The complex K• ⊗R R′ is a K-flat complex of R′-modules. Hence:

(K• ⊗LR R

′)⊗LR′ N

• = Tot((K• ⊗R R′)⊗R′ N•) = Tot(K• ⊗R′ N•) = K• ⊗LR′ N

•

which proves the final statement of the lemma.


Lemma B.1.2 Let R→ R′ be a ring map. Let f : N• → N ′• be a morphism of complexes of

R′-modules (ie. in the category Comp(R′)). Then f induces a natural tranformation:

1⊗ f : (·)⊗LR′ N

• → −⊗LR′ N

′•

If f is a quasi-isomorphism, then 1⊗ f is an isomorphism of functors.

Proof. The functors are computed by evaluating on K-flat complexes K•, and therefore we canuse the functoriality:

Tot(K• ⊗R N•)→ Tot(K• ⊗R N ′•)

to define the transformation. The last statement an easy verification, and is left to the reader.

Lemma B.1.3 Let A→ B → C be ring maps. Let N• be a complex of B-modules and N ′• acomplex of C-modules. The compositions of the functors:

D(A)(·)⊗L

AN•

−−−−−−→ D(B)(·)⊗L

AN′•

−−−−−−→ D(C)

is the functor: (·)⊗LA (N• ⊗L

B N′•) : D(A) → D(C). If N , N ′, N ′′ are modules over A, B, C

respectively, then we have:

(N ⊗LA N

′)⊗LB N

′′ = N ⊗LA (N ′ ⊗L

B N′′) = (N ⊗L

A C)⊗LC (N ′ ⊗L

B N′′)

in D(C). We also have a canonical isomorphism:

(N ⊗LA N

′)⊗LB N

′′ → (N ⊗LA N

′′)⊗LC (N ′ ⊗L

B C)

up to a universal sign.

Consider now a commutative diagram:

A // A′

R //

OO

R′

OO

of rings. Given an object K of D(A) we can consider its restriction to an object of D(R). Wecan then take the derived change of rings of K to an object of D(A′) and D(R′). We claimthere is a functorial comparison map:

(∗) K ⊗LR R

′ → K ⊗LA A

′

in D(R′). To construct this comparison map choose a K-flat complex K• of A-modules rep-resenting K. Next, choose a quasi-isomorphism E• → K• where E• is a K-flat complex ofR-modules. The map above is the map:

K ⊗LR R

′ = E• ⊗R R′ → K• ⊗A A′ = K ⊗LA A

′

In general there is no chance that this map is an isomorphism.

However, we often encounter the situation where the diagram above is a cocartesian diagramof rings. In this situation, for any A-module M we have M ⊗A A′ = M ⊗R R′. Thus (·)⊗R R′is equal to (·)⊗A A′ as a functor A-Mod→ A′-Mod. In general this equality does not extendto derived tensor products. In other words, the comparison map is not an isomorphism. Asimple example is given by: R = Z, A = R′ = A′ = Z/p and K• = A[0]. Clearly, a necessarycondition is that TorRi (A,R′) = 0 for all i > 0, ie. that A and R′ are Tor-independent over R.


Lemma B.1.4 The comparison map (∗) is an isomorphism if A′ = A ⊗R R′ and A and R′are Tor independent over R.

Proof. We choose a free resolution F • → R′ of R′ as an R-module. Because A and R′ are Torindependent over R we see that F • ⊗R A is a free A-module resolution of A′ over A. By ourgeneral construction of the derived tensor product above we see that:

K• ⊗A A′ ' Tot(K• ⊗A (F • ⊗R A)) = Tot(K• ⊗R F •) ' Tot(E• ⊗R F •) ' E• ⊗R R′

as desired.

The following Lemma will be useful.Lemma B.1.5 Consider a commutative diagram of rings

A′ R′ //oo B′

A

OO

Roo

OO

// B

OO

Assume that R′ is flat over R and A′ is flat over A⊗R R′ and B′ is flat over R′ ⊗R B. Then

TorRi (A,B)⊗(A⊗RB) (A′ ⊗R′ B′) = TorR′

i (A′, B′)

Proof. There are canonical maps by functoriality:

TorRi (A,B)→ TorR′

i (A⊗R R′, B ⊗R R′)→ TorR′

i (A′, B′)

These induce a map from left to right in the formula of the lemma.

Take a free resolution F• → A of A as an R-module. Then we see that F•⊗RR′ is a resolutionof A⊗RR′. Hence TorR

′

i (A⊗RR′, B⊗RR′) is computed by F•⊗RB⊗RR′. By our assumptionthat R′ is flat over R, this computes TorRi (A,B) ⊗R R′. Thus TorR

′

i (A ⊗R R′, B ⊗R R′) =TorRi (A,B)⊗R R′ (using only flatness of R′ over R).

By Lazard’s theorem we can write A′, resp. B′ as a filtered colimit of finite free A⊗R R′, resp.B ⊗R R′-modules. Say A′ = lim−→Mi and B′ = lim−→Nj . The result above gives

TorR′

i (Mi, Nj) = TorRi (A,B)⊗A⊗RB (Mi ⊗R′ Nj)

as one can see by writing everything out in terms of bases. Taking the colimit we get the resultof the lemma.

Let us therefore discuss in detail a few facts about the functor Ω1(−)/(−) from a purely functorial

perspective, though they are very well known.

B.2 Relative differentials

For the moment we stick to the case of ring maps, and define the category RngMor to bethe category of morphisms of rings, whose morphisms are commutative squares. Let us alsodefine the category RngMod whose objects are couples (B,M), where B is a ring and M is aB-module, and morphisms are couples of maps

(α, µ) : (B,M)→ (B′,M ′)


with α : B → B′ a ring map, and µ a map which satisfies the property:

α(b) · µ(m) = µ(b ·m).

Equivalently, one can view β as a map M ⊗B B′ → M ′ which is B′-linear. We consider thefunctor

Ω1(−)/(−) : RngMor→ RngMod

taking a ring map A→ B to the pair (B,ΩB/A), where a morphism of ring maps

A

// A′

Bα// B′

is sent to the morphisms of ring-modules (α,dB/Aα), with dB/Aα as usual. Ω(−)/(−) is indeeda functor, which we may construct explicitly the following way. We let A→ B be a ring map,and form the kernel of the multiplication map

0→ IB/A → B ⊗A B → B → 0

We let Ω1B/A := IB/A/I

2B/A, which is a B-module via the usual isomorphism

IB/A/I2B/A ⊗B⊗AB B 'B IB/A/I

2B/A.

We define dB/A(b) := b⊗ 1− 1⊗ b modulo I2B/A, for all b ∈ B. The construction is manifestly

functorial in A → B, and easily sheafifies, to give a functor Ω1(−)/(−) from the category of

morphisms of sheaves of rings to the category of couples of sheaves of rings and sheaves ofmodules. Since such construction involves a tensor product, which commutes with all colimits,and formation of a kernel, which commutes with filtered colimits, we conclude at once thatΩ1

(−)/(−) commutes with filtered colimits of morphisms of rings (resp. sheaves of rings). Onemay deduce therefore that formation of the relative differentials commutes with localizations(which are filtered colimits)! In particular, given a morphism of sheaves of rings A → B on aspace X, the stalk at x ∈ X of the sheaf Ω1

B/A is just Ω1Bx/Ax

.

Actually, formation of the relative differentials commutes with all colimits! We now discussthe proof of this basic fact, and comment on it. We remark that as soon as we construct thecotangent complex, we shall see how arbitrary base change fails for it in the category of rings,sheaves of rings on a site, ...

This is quite annoying at first, given that, instead, we’ll soon prove that the relative differen-tials commute with arbitrary base change, as a consequence of the fact that their formationcommutes with all colimits. However, such failure has a conceptual (and in fact “geometric”)meaning, which will be explained in Appendix .Proposition B.2.1 The functor Ω1

(−)/(−) : RngMor→ RngMod is left adjoint to the func-tor D(−)(−) : RngMod→ RngMor sending a couple (B,M) to the ring map B → DB(M),where DB(M) is the trivial square-zero extension of B by M .

We shall soon review algebra extensions, but let us first recall how DB(M) is defined. We aregiven a ring B and a B-module M . We form the B-module B ⊕M , and endow it with thefollowing amalgamated product:

(b,m) · (b′,m′) := (bb′, bm′ + b′m).


This is easily seen to define a ring structure on DB(M), and in fact a B-algebra structure viathe map sending any b ∈ B to (b, 0). Obviously the sequence of B-modules

0→M → DB(M)→ B → 0

is split (the above is a section to the last map), where the B-module M is now a square-zeroideal of DB(M). Indeed, (0,m) · (0,m′) = (0, 0) for all m,m′ ∈M .

The construction of DB(M) is manifestly natural in the couple (B,M), and defines a functorD(−)(−) : RngMod→ RngMor.

We are ready to prove Proposition B.2.1.

Proof. Let (B′,M) be any object of RngMod, and A→ B any object of RngMor. We claimthat we have a bijection:

HomRngMod((B,Ω1B/A), (B′,M)) ' HomRngMor(A→ B,B′ → DB′(M)).

Call f the ring map A → B. Pick a map (α, µ) : (B,Ω1B/A) → (B′,M). α is a map B → B′,

and on the other hand we have already a map B′ → DB′(M) given by b′ 7→ (b′, 0). If we takethe composition A f−→ B

α−→ B′, we therefore get a map A αf−−→ B′, and we can also form a map

B → DB′(M)

by letting any b ∈ B go to (α(b), µ(db)) (it is immediate to check that this assignment is a ringhomomorphism, due to the fact that M is a square-zero ideal of DB′(M)). The upshot is thatfrom a map (α, µ) in HomRngMod((B,Ω1

B/A), (B′,M)) we produced a commutative square:

A

αf

f// B

α×(µd)

B′b′ 7→(b′,0)

// DB′(M)

Commutativity is readily checked: (α× (µ d)) f(a) = (α(f(a)), µ(df(a))), and on the otherhand αf(a) 7→ (α(f(a)), 0) in DB′(M). These two are equal for all a ∈ A because df(a) = 0.Notice that it seemed, according to our introduction of the functor Ω1

(−)/(−) and the categoriesRngMor and RngMod, that we were forgetting the A-linearity of the natural universalderivation d : B → Ω1

B/A, but in fact this is not the case: we really need such A-linearity, orthe diagram above would not be commutative (and hence Ω1

(−)/(−) would not be a functor, sosuch linearity is actually built into functoriality of Ω1

(−)/(−)). Conversely, let us consider a map(A→ B)→ (B′ → DB′(M)) given by the commutative square:

A

αf

f// B

α×d

B′b′ 7→(b′,0)

// DB′(M)

for some d : B → M making the diagram commutative. The reader may check that in orderto make the square commute d must be an A-derivation B →M . Therefore we may just sendsuch square to the morphism in RngMod given by:

α× (db 7→ db) : (B,Ω1B/A)→ (B′,M).

These two assignments are inverse to each other, thus proving the claim.


As a consequence of Proposition B.2.1, by general nonsense we deduce that formation ofΩ1

(−)/(−) commutes with all colimits in the category RngMor, that is, given any small categoryI and a functor I → RngMor assigned by

i 7→ (Ai → Bi)

we have, calling A := lim−→Ai → B := lim−→Bi the corresponding unique map between colimits,

lim−→Ω1Bi/Ai

' Ω1B/A.

As anticipated, as a consequence of commutativity with all colimits we now show the followingproperty of Ω1

(−)/(−):Proposition B.2.2 Let the following cocartesian diagram be given:

A

// A′

B // B′

Then,Ω1B′/A′ = Ω1

B/A ⊗A A′

andΩ1B′/A = (Ω1

B/A ⊗B B′)⊕ (Ω1

A′/A ⊗A′ B′).

Proof. Let us consider the following (discrete) small category I = 0 → 1, 0 → 2, and thefunctor I → RngMor given by (0 → 1) 7→ (A → B) =: (A0 → B1) and (0 → 2) 7→ (A →A′) =: (A0 → A2). Then

lim−→I

(Ai → Bi) = (A→ B ⊗A A′)

and thereforeΩ1

lim−→I(Ai→Bi) = Ω1

B′/A.

On the other hand, we have commutativity of Ω1(−)/(−) with all colimits, and so the above is

equal tolim−→I

Ω1Bi/Ai

.

To compute this, let us see what happens when we apply the functor Ω1(−)/(−) to our diagram

I → RngMor, which therefore yelds a diagram I → RngMod, and in this latter category wecompute the colimit.

We have the diagram (0 → 1) 7→ Ω1B/A, (0 → 2) 7→ Ω1

A′/A, so that now the diagram is madeof two isolated points. Now we should really point out that the category RngMod containscoproducts. Given two objects (B,M) and (B′,M ′), their coproduct is just

(B′′ := B ⊗Z B′, (M ⊗B B′′)⊕ (M ′ ⊗B′ B′′))

as one checks by means of the expected universal property, using the universal propertiesof tensor product of rings, tensor product and coproduct of modules. We step back to ourcomputation, and observe that the colimit of Ω1

Bi/Aiis a coproduct, which is

(B,Ω1B/A)

∐(A′,Ω1

A′/A) = (B′, (Ω1B/A ⊗B B

′)⊕ (Ω1A′/A ⊗A′ B

′)),


as desired.

Now consider instead the small category I := (0 → 1) → (2 → 3), and the functor I →RngMor assigned by (0→ 1) 7→ (A→ B) and (2→ 3) 7→ (A′ → B′). We have

Ω1lim−→I

(Ai→Bi) = Ω1B′/A′ ,

and lim−→IΩ1Bi/Ai

is just lim−→((B,ΩB/A)→ (B′,ΩB′/A′)), which is lim−→(B → B′,Ω1B/A → Ω1

B′/A′),where the map Ω1

B/A → Ω1B′/A′ is required to be (B → B′)-linear by definition of RngMod.

By the universal property of tensor product of B-modules, such colimit is (B′,Ω1B/A ⊗B B

′),whence

Ω1B′/A′ = Ω1

B/A ⊗B B′,

which is in turn Ω1B/A ⊗A A

′ since our initial diagram is cocartesian.

The above proof is actually equivalent to the “usual” proof using the universal property ofdifferentials. However, it stresses a crucial fact: formation of the sheaf of relative differentialscommute with all colimits because, roughly, it represents a functor in the category RngMod.Instead, as we shall see, the relative cotangent complex of, say, a morphism of sheaves of ringsA→ B, does not enjoy the same property in the derived category D(B). Roughly, such derivedcategory is not its most natural habitat. We will see instead that formation of the cotangentcomplex commutes with all homotopy colimits.

B.3 Simplicial methods

As we shall better explain later, it turns out that simplicial rings and sheaves of rings are mostsuitable to approximate smoothness of a map of ordinary rings or sheaves of rings. This sectionis devoted to explaining the needed facts from simplicial methods, and prove the most relevant.

We fix an abelian category A, and consider the category of simplicial objects in A, that is,functors 4op → A, whose morphisms are natural transformations of functors. We denote suchcategory by sA.

Let A be an object of sA, and define the following object of the category C(A) of complexesof A:

Tot(A)−n := An,

with boundary map:dn : Tot(A)n → Tot(A)n−1

given by

dn :=

n∑i=0

(−1)i∂in : An → An−1.

One can check immediately that dn−1dn = 0. One just fixes two indexes 0 ≤ i < j ≤ n andconsiders ∂ : [n− 1] → [n] to be the unique morphism in 4 missing i and j in [n]. ∂ has twofactorizations: ∂ = ∂jn∂

in−1, and ∂ = ∂in∂

j−1n−1, and in the sum defining dn−1dn they both appear

with opposite sign.

Tot(A) is usually called the total complex of A, or the un-normalized complex of A, for reasonswe now explain. Let’s in fact define the normalized complex of A:

Ner(A)−n :=

n⋂i=1

ker(∂in : An → An−1)


with boundary map given by dn|Ner(A)n , which is immediately seen to coincide with ∂0n. Ner(A)

is a subcomplex of Tot(A), and formation of both complexes is manifestly functorial in A.

Let us call C≥0(A) the category of cochain complexes of A in nonnegative degree. We thereforehave functors:

Tot : sA→ C≤0(A)

andNer : sA→ C≤0(A)

withH−n : C≥0(A)→ A.

We now study the relation between Tot(A) and Ner(A).

We have the following:Proposition B.3.1 Let A be an object in sA. Then Tot(A) and Ner(A) are quasi-isomorphic.

To understand better the relation between Tot(A) and Ner(A), we define the so called degen-erate complex of A by setting

D(A)−n :=

n−1∑i=0

im(sin−1 : An−1 → An)

where the sum is taken over the degeneracies. D(A) is a subcomplex of Tot(A). This is seenby showing that

dn(im(sjn)) ⊆ D(A)−n+1,

an immediate consequence of the simplicial relations.

We have the following:Proposition B.3.2 The total complex Tot(A) of A ∈ sA splits naturally as a direct sum ofNer(A) and D(A).

The degenerate subcomplex is acyclic, and therefore the projection Tot(A) → Ner(A) is aquasi-isomorphism.

B.4 The Dold-Kan equivalence

The Dold-Kan equivalence will be our main tool to stick to reality. I will explain better what Imean by this later in Appendix A. For the moment, recall we have constructed functors Tot(−)and Ner(−) from the category of simplicial objects of A to the category C≤0(A). We shall nowdescribe in good detail the Dold-Kan equivalence and the reason why it holds true:Theorem B.4.1 (Dold-Kan-Puppe) The functor Ner(−) : sA→ C≤0(A) is an equivalenceof categories.

Sketch of proof. We describe the quasi-inverse functor, which we call K. Let P be an object ofC≤0(A). We define

K(P )n :=⊕

σ:[n]→[m]

P−m

where the sum runs over all surjections σ : [n] → [m] for n fixed, in 4. Call A := K(P ) forsimplicity. We need to construct faces and degeneracies, and then show the simplicial identities.


Given σ : [n] → [k], let σ′ denote the insertion map P−k → An. We define, for τ : [n] → [m],A(τ) by requiring the following diagram:

AmA(τ)

// An

P−k

σ′

OO

σ′τ ′

==

to commute for all σ : [m] → [k] in 4. Now pick a morphism ∂in : [n − 1] → [n] in 4. Wewant to define the face map din : An → An−1 out of it. We distinguish a few cases. Pick anysurjective morphism in 4

σ : [n]→ [k].

In the case the composition σ∂n?j is also surjective, we ask dinσ′ = (σ∂in)′. If not, we writeσ∂in = ∂jkτ for some unique τ : [n− 1]→ [k − 1] surjective, and ∂jk : [k − 1]→ [k] injective. Inthis case and j = 0, we require dinσ′ to be the composition

P−k → P−k+1τ ′−→ Ak−1

the first map being the boundary map. We let dinσ′ to be zero by definition, if j > 0.

An easy verification shows that Ner(−) and K(−) are quasi-inverses,

B.5 Algebra extensions

We now introduce a crucial tool in deformation theory, which is the group of algebra extensionsof an algebra by a module. We shall soon discuss how such extensions and a careful analysis ofthem are at the heart of some basic deformation situations. We do not intend to repeat thateverything can be done essentially the same way for sheaves of rings over a ringed topos, stack,algebraic space, all over the place. We will rather focus on the case of sheaves of rings/algebrason schemes, and point out modifications when needed to deal with the other cases of interest.Until further notice, the reader may read sheaves of rings whenever he reads rings.

Let A→ B be a ring map, and M a B-module. An A-algebra extension of B by M is a shortexact sequence of A-modules:

0→M → B′ → B → 0

where M carries a B-module structure, and B′ is not just an A-module, but actually an A-algebra, where M is an ideal of B′. We denote the set of all such extensions by ExalA(B,M).This is not just a set. One can define the notion of a morphism of extensions, once we fixA, B and M : given two extensions B′ and B′′ of B by M , a morphism between them is anisomorphism of A-algebras B′ → B′′ inducing the identity on M and B, that is:

0 // M // B′

∼

// B // 0

0 // M // B′′ // B // 0

Every morphism in ExalA(B,M) is an isomorphism, thus making ExalA(B,M) into a group-oid. If A and B are sheaves of rings, then ExalA(B,M) is actually a category fibered ingroupoids over the Zariski site of Spec(A). Indeed, it is a Picard stack. We shall not deal with


this perspective in this section: instead, we focus on the isomorphism classes of ExalA(B,M),which we shall denote ExalA(B,M). We shall deal with Picard stacks/groupoids in the sequel.

As it is universally known, ExalA(B,M) is a group under Baer sums: the sum of two extensionsis just the A-extension of B by M given by (B′⊕B′′)/(m,−m) | m ∈M, the pullback of B′and B′′ over B, with the usual product law. In fact it is a B-module, where given any b ∈ B,multiplication by b is defined by pushing out an extension

0 // M // B′ // B // 0

along multiplication by bµb : M →M.

It is readily checked that such assignments indeed defines a map

B ⊗Z ExalA(B,M)→ ExalA(B,M)

assigning a B-module structure on the commutative group ExalA(B,M).

It will be useful to know explicitly the group of automorphisms of the trivial extension of B byM :Proposition B.5.1 Let A→ B be a ring map, and M a B-module. Let d ∈ DerA(B,M), andsend it to the morphism:

0 // M // DB(M)

id×d

// B // 0

0 // M // DB(M) // B // 0

yelds a group isomorphism between the automorphism group of the trivial extension of B by Mover A, and DerA(B,M).

Proof. Pick two such automorphisms, and take their difference: given f, g : DB(M)→ DB(M),form f − g : DB(M) → DB(M), and call it d. Certainly, d(a) = 0 for all a ∈ A since f and gboth restrict to the identity on A. The same way, d kills M as an ideal of DB(M), so it factorsuniquely through B. d is of the form(

α βγ δ

): B ⊕M → B ⊕M

as a map of A-modules, where β = 0, and α = δ = id. So actually the datum of d is the datumof an A-linear map d : B →M . We check the Leibniz rule. We have, for all b, b′ ∈ B:

d(bb′) = f(bb′)− g(bb′)

= f(b)f(b′)− g(b)g(b′)

= f(b)(f(b′)− g(b′)) + g(b′)(f(b)− g(b))

= bd(b′) + b′d(b)

using the fact that f, g restrict to the identity on B. It is therefore clear what the map mustbe from DerA(B,M) to the automorphism group of the trivial extension in question. It sendsa derivation d to id+d. Manifestly, such map is injective, and surjectivity is proved above! Weneed to check that it’s a group homomorphism, and observe we haven’t used yet the fact thatM is square-zero as an ideal of DB(M). We have:

bb′ + d(bb′) = bb′ + bd(b′) + b′d(b)

= (b+ d(b)) · (b′ + d(b′))

using the fact that d(b)d(b′) = 0.


We are going to need the following fundamental exact sequence, which we are going to reinter-pret after we introduce the relative cotangent complex:

Theorem B.5.2 Let A α−→ Bβ−→ C be ring maps. Then for any C-module M we have the

following exact sequence, natural in M :

0→ DerB(C,M)→ DerA(C,M)→ DerA(B,M)∂−→

→ ExalB(C,M)→ ExalA(C,M)→ ExalA(B,M)

The map ∂ sends an A-linear derivation d to the isomorphism class of the trivial B-algebraextension of C by M , where DC(M) is made into a B-algebra via the map β × d.

Proof. We check exactness only at ExalB(C,M). Let C ′ be a B-extension of C by M which issent to the trivial A-extension DC(M) of C by M . This means that there exists an A-algebrasection s : C → C ′ which makes the extension

0→M → C ′ → C → 0

split as an A-extension, that is, isomorphic to the trivial extension (hence the same elementof the group ExalA(C,M)). Let us be more precise. We call i and i′ the B-algebra structuremorphisms of C and C ′ respectively. We are saying that in the following diagram:

A // Bsi //

i′// C

i′ and si coincide if precomposed with the structure map A→ B. Define

d(b) := i′(b)− si(b), b ∈ B.

Such assignement is manifestly a map d : B →M which is A-linear, given the above diagram!It satisfies the Liebniz rule:

d(bb′) = i′(bb′)− si(bb′)= i′(b)i′(b′)− si(b)si(b′)= i′(b)(i′(b′)− si(b′)) + si(b′)(i′(b)− si(b))= b · d(b′) + b′ · d(b)

(note that i′(b′) is a lift of i(b) to C ′, and likewise for si(b′) and i(b′). All we need to show is thatthe B-extension C ′ of C by M , given that it goes to zero in ExalA(C,M), actually comes froman element in DerA(B,M), in fact, from d! Recall how ∂ is made. It suffices to check that themap DC(M)→ C ′ given by the assignment (c,m) 7→ s(c) +m is a B-algebra homomorphism,where DC(M) is a B-algebra via i × d. This is an easy verification (note that one needs tocheck that DC(M)→ C ′ is a homomorphism, which needs a little verification).

Grothendieck’s questionLet A → B → C be two given ring maps. Then we have the following exact sequence ofC-modules:

Ω1B/A ⊗B C → Ω1

C/A → Ω1C/B → 0

Grothendieck knew already in [EGA IV] that such sequence was part of a long exact sequenceyeld by taking cohomology of some suitable (negatively graded) complex. He was able to con-struct a two-term complex whose cohomology yelded a six-term long exact sequence terminating


with the above three terms. Such complex already captured a lot of information for the pur-poses of deformation theory (see Remark ), but some deformation problems indeed neededthe introduction of a full complex, whose truncation in degree −1 was the one constructed byGrothendieck.

The upshot is that the relative cotangent complex of the map A → B is the complex we arelooking for, and it has good functoriality properties: for example given two ring maps as above,we have a fundamental distinguished triangle in D(C), and taking H0 we get the desired fulllong-exact sequence. The so called “truncated cotangent complex” will be the one constructedby Grothendieck’s initially.

Before exploring the above problem, we are going to need a few facts concerning when theabove exact sequence is also left-exact. In particular, we are going to need such situationsrephrased in terms of algebra extensions, so this section is devoted to working out the neededfacts.

B.6 The truncated cotangent complex

We now explain the first attempt to give an answer to Grothendieck’s question, which was infact made by Grothendieck himself. We shall construct a two-term complex

τ≥−1LB/A = (L−1 → L0)

attached to any ring map A → B, which we shall call the truncated cotangent complex, asthe notation suggests. As soon as we construct the actual full cotangent complex, we shall atonce recover the truncated version by... truncating. The following construction is naive buteffective: roughly, it will achieve the expected goal as explain in the previous paragraph, buton the other hand it will still be unsatisfactory for various reasons.

Let us consider the ring B as a set (or, the sheaf of rings B as a sheaf of sets). I should denoteit differently to avoid confusion. Will fix this. Let us form the polynomial algebra A[B] overA, that is,

A[xb, b ∈ B].

We have a natural surjective map A[B]→ B, with kernel I.Definition B.6.1 Let A → B be a ring map. We define the truncated cotangent complex ofB over A to be the two-term complex of B-modules:

τ≥−1LB/A := (I/I2 → Ω1A[B]/A ⊗A[B] B).

Note that I/I2 is clearly a B-module given that I ⊗A[B] B ' I/I2. The complex τ≥−1LB/Acomes with a natural augmentation τ≥−1LB/A → Ω1

B/A, a property which will be enjoyed alsoby the full cotangent complex.

We are going to do two things to get a first handle on Definition B.6.1: we shall explore thefunctoriality properties of A[B], and we shall see that if we substitute A[B] with any formallysmooth A-algebra C with a surjective A-algebra map C → B, then

τ≥−1LB/A 'qi (J/J2 → Ω1C/A ⊗C B)

with J := ker(C B).

This latter nice property of τ≥−1LB/A will shed some light on the intimate meaning of theconstruction of the full complex!, so the reader should really care about it.

We begin by pursuing the first task.


Free modules and algebras

Given any set S (or sheaf of sets S) and ring A, we can form the free A-module over S bysimply considering AS =

∐S A. The sheaf-theoretic analogue is clearly given by the sheaf

associated to the presheaf U 7→ A(U)S(U) (clarify what U is. Case of locally ringed topoi. Caseof algebraic spaces).

The assignment sending a couple (A,S), with A a ring and S a sheaf of sets, to AS is functorial,in the sense that it yelds a functor

Rng × Set→ RngMod

which manifestly preserves filtered colimits.

As an immediate consequence, preservation of filtered colimits yelds that, for S a sheaf of setsand A a sheaf of rings on a locally ringed space X, say,

(AS)x = ASxx ,

for all points x ∈ X. To get a handle on the surjection A[B] → B of the previous paragraph,and on its functoriality properties, we start by dealing with the case of free modules, andobserve that if we fix an A-module M , the forgetful functor ModA → Set has a left-adjointgiven exactly by the free A-module functor, that is, we have the following bijection

HomSet(M,M) ' HomModA(AM ,M).

The natural A-module map AM →M corresponding to the set-theoretic identity M →M is asurjection.

A further step. We have introduced the category RngMod, and we have a forgetful functor

RngMor→ RngMod

given by sending a ring map A→ B to the couple (A,B). Such functor has a left-adjoint, givenby the functor sending a couple (A,M), for a ring A and an A-module M , to the symmetricalgebra of M over A: SymA(M).

In the case of sheaves of rings, the symmetric algebra functor is the sheaf associated to thepresheaf

U 7→ SymA(U)(M(U)).

Sheafification is an exact functor, and by construction of the symmetric algebra (tensor productsand kernels), such functor commutes with filtered colimits (tensor products commute with allcolimits, but kernels commute just with filtered colimits), so once again, if A is a sheaf of ringson a locally ringed space X andM is a sheaf of A-modules on X, we have, for all points x ∈ X,

(SymAM)x = SymAx(Mx).

Left-adjointness of the symmetric algebra functor also ensures that such functor carries colimitsin the category ModA to colimits in the category of A-algebras! In particular, since coproductsof A-modules are colimits, we get

SymA(M ⊕M ′) = SymA(M)⊗A SymA(M ′).


We form the composition of the functors

(A,S) 7→ AS 7→ SymA(AS) =: A[S].

As remarked before, since such composition is the composition of two left adjoint functors, itis itself a left adjoint. Indeed, one can consider the forgetful functor from the category AlgAto the category Set, and the free A-algebra functor is its left-adjoint. Given the properties ofthe free A-module functor and of the symmetric A-algebra functor, the free A-algebra functorcommutes with filtered colimits, so that we get, for a sheaf of rings A on a locally ringed spaceX and a sheaf of sets S on X,

(A[S])x = Ax[Sx]

for all points x ∈ X. Another consequence of commutativity with filtered colimits is that, givena morphism f : X → Y of locally ringed spaces, we have

f−1(A[S]) = (f−1A)(f−1S).

Differentials of polynomial algebras

Let A be a fixed ring, and M any A-module. Call B := SymA(M). We definitely have a mapwhich is natural in M and A→ B:

M ⊗A B → Ω1B/A

given on elementary tensors by m ⊗ b 7→ bdm. The result we now prove is that this map isactually an isomorphism of B-modules.Proposition B.6.2 The natural map M ⊗A B → Ω1

B/A described above is an isomorphism ofB-modules.

Proof. We let M ′ be a B-module and pick an A-linear derivation d : B → M ′. Since B isgraded and its degree 1 piece is isomorphic to M as an A-module, by restriction we get a mapd′ : M →M ′ of A-modules. Since M generates B as an A-algebra (essential!) d is completelydetermined by d′, and by A-linearity of this latter and the Leibniz rule, it is also determineduniquely. Conversely, given an A-module map ϕ : M → M ′, we can cook up an A-linearderivation:

d(m1 ⊗ · · · ⊗mt) :=

t∑i=1

ϕ(mi)m1 ⊗ · · · ⊗mi−1 ⊗mi+1 ⊗ · · · ⊗mt

(the only difficulty is the Leibniz rule, but it is easy to check it). Clearly d′ = ϕ, which achievesthe following natural bijection:

DerA(B,M ′) = HomA(M,M ′).

Using the extension-restriction adjunction we get

HomB(Ω1B/A,M

′) = DerA(B,M ′) = HomB(M ⊗A B,M ′)

which says that Ω1B/A and M ⊗A B represent the same functor, ie. satisfy the same universal

property. Hence they are isomorphic as B-modules, as desired.

As a typical example, we can choose A = k a field, M := kx1,...,xn the free module overa finite set, and then B = k[x1, . . . , xn]. We have Ω1

B/A has dx1, . . . ,dxn as a basis as aB-module, as known.


B.7 Main facts about the truncated cotangent complex

We are now going to state the main facts about the truncated version of the cotangent complex,intruduced in Definition . Such facts will be only stated and not proved, because we shall provethem all for the full cotangent complex, and rededuce them at once for the truncated cotangentcomplex. Nonetheless, we are going to use them in advance to discuss algebra extensions: thearguments won’t be circular at all because algebra extensions do not need the intruduction ofthe full complex in order to be discussed.

We start from functoriality:Proposition B.7.1 Let A → B be a ring map, and τ≥−1LB/A be its truncated cotangentcomplex. Formation of such complex is functorial in A→ B.

This is one of the only two facts we are going to show, for the following reason: in order torededuce all the facts we are now stating for the truncated cotangent complex, from thosefor the full cotangent complex, we need an isomorphism of complexes in D(B) between thetruncation at −1 of the full cotangent complex and the truncated cotangent complex. Suchisomorphism must be natural, and naturality is achieved immediately, provided formation ofthe truncated cotangent complex is functorial.

Proof. Immediate from the definition! (ah ah!)

B.8 Formally smooth replacement trick

I was trying to show myself that if A→ B is a surjective ring map with kernel I, then τ≥−1LB/Ais quasi-isomorphic to I/I2[1], and that if A→ B is formally smooth, then τ≥−1LB/A → Ω1

B/A

(ie. not just for polynomial algebras). This proof is indeed given in general once we showvery nice properties of the full cotangent complex which, roughly, says that formation of thecomplex is local in nature (with respect to what topologies is to be specified).

We present here a trick which gives a very concrete proof of both facts, and at the same timegives more insight into the construction of the full cotangent complex itself, as discussed inSection . The trick is divided into two lemmas.Lemma B.8.1 (formally smooth replacement - I) Let A → B be a ring map withtwo factorizations through ring maps C → B and D → B respectively, which come with acomparison map C → D making the obvious diagram commute. Suppose C → B is surjectivewith kernel I and D → B is surjective with kernel J . There is a natural base-change map

I/I2 //

Ω1C/A ⊗C B

J/J2 // Ω1D/A ⊗A B

which is a quasi-isomorphism as soon as either C → D is onto and D is formally smooth asan A-algebra, or both C and D are formally smooth.

The proof of this is not conceptually hard, but long.


Proof. We first proceed under the assumption that, in the following diagram:

C // D

A

OO

// B

the map C → D is onto, and D is formally smooth. Call K the kernel of C → D. Given thatD is formally smooth, the following sequence is short exact:

0→ K/K2 → Ω1C/A ⊗C D → Ω1

D/A → 0

where Ω1D/A is flat. Such flatness implies that if we apply the functor (−)⊗D B we get a short

exact sequence0→ (K/K2)⊗D B → Ω1

C/A ⊗C B → Ω1D/A ⊗D B → 0

Recall that C → B and D → B are surjective. We have, therefore, that the above short exactsequence is:

0→ (K/IK)→ Ω1C/A/IΩ1

C/A → Ω1D/A/JΩ1

D/A → 0

where we have:(K/K2)⊗D B = K ⊗C B = K/(IK).

We claim that IK = I2 ∩ K. We certainly have IK ⊆ I2 ∩ K. Suppose we have i, i′ ∈ Isuch that ii′ ∈ K − IK. We observe that d(ii′) = idi′ + i′di, which is zero in Ω1

C/A/IΩ1C/A,

is nonzero in K/(IK), contradiction, since the map K/(IK) → Ω1C/A/IΩ1

C/A is injective. Wetherefore have the following commutative diagram:

0 // K/(I2 ∩K) // I/I2

// J/J2

// 0

0 // K/(IK) // Ω1C/A ⊗C B // Ω1

D/A ⊗D B // 0

To see that the first row in the above diagram is exact, form the following diagram of maps ofC-modules:

0

0

0

0 // I2 ∩K //

I2

// J2

// 0

0 // K

// I

// J

// 0

0 // K/(I2 ∩K)

// I/I2

// J/J2

// 0

0 0 0

where by the Snake Lemma we conclude. We know that taking H0 of both complexes we getΩ1B/A. We need to see that the above map of complexes induces an isomorphism on H1, and

this is again a consequence of the Snake Lemma.


Now we instead deal with the case C and D are formally smooth. First of all we can alwaysfactor the map C → D into C → C[D] → D, where the map C[D] → B is surjective withkernel L. We have maps of complexes

I/I2 //

Ω1C/A ⊗C B

L/L2 //

Ω1C[D]/A ⊗C[D] B

J/J2 // Ω1D/A ⊗D B

where we know from the previous case that the bottom map is a quasi-isomorphism. We willsee a gigantic generalization of this in a few sections! We are reduced to show that the topmap is a quasi-isomorphism.

Now recall the situation:C[D]

C //

""

==

D

B

where all the downwards maps are surjective. For all d ∈ D, we can choose cd ∈ C which isa lift of the image of xd ∈ C[D] under the map C[D] → D. We can send xd to cd and get aC-algebra map

C[D]→ C

which is surjective, and yelding a factorization of the identity map C → C. We obtain mapsof complexes:

I/I2 //

Ω1C/A ⊗C B

L/L2 //

Ω1C[D]/A ⊗C[D] B

I/I2 // Ω1C/A ⊗C B

where the composition is the identity, hence a quasi-isomorphism, and the bottom one is aquasi-isomorphism by the previous case. We conclude that the top map is a quasi-isomorphism,which is exactly what we wanted to prove.

One can modify the proof of this to include the case of sheaves of rings, by arguing on stalks,since tensor products and kernels commute with taking stalks (which are filtered colimits).

Lemma B.8.2 (formally smooth replacement - II) Let A→ B be a ring map which canbe factored through an A-algebra map C → B, with C formally smooth and C → B surjective


with kernel J . We have a quasi-isomorphism:

τ≥−1LB/A 'qi (J/J2 → Ω1C/A ⊗C B).

I present a proof which I currently don’t know how to streamline for the case of sheaves ofrings (ie. I can prove it for sheaves of rings in the only possible way I could modify the prooffor rings! I’m sure there’s a way easier proof).

Proof. Call α : C → B and β : A[B]→ B. Let I be the kernel of β. We call S := C∐A[B] as

a set, and consider D := A[S]. D is still formally smooth over A, and now we claim that wecan complete the following diagram with solid arrows instead of dotted ones:

Dγ//

δ

C

α

A[B]β// B

For all f ∈ A[B], choose some cf ∈ C such that α(cf ) = β(c), which is possible by surjectivityof α. Same thing using surjectivity of β, to choose, for all c ∈ C, an element fc ∈ A[B] suchthat β(fc) = α(c). It is uniquely determined an A-algebra map γ : D → A[B] by settingγ(f) := cf ∈ C for all f ∈ A[B] ⊂ S, and γ(fc) = c for all c ∈ C ⊂ S. Analogously for δ.The composition δα = γβ : D → B has kernel K. We apply Lemma B.8.2 to get the desiredquasi-isomorphism of complexes of B-modules.

Two extremely important corollaries:Corollary B.8.3 Let A→ B be a ring map which is formally smooth. Then LB/A → Ω1

B/A[0]is a quasi-isomorphism of complexes of B-modules.

Proof. We simply let C = B in Lemma B.8.2.

Corollary B.8.4 If A → B is a ring map which is surjective with kernel I, then LB/A 'qi

I/I2[1].

Proof. Just take C = A in Lemma B.8.2.

A pleasant consequence which will be proved in full generality later:Corollary B.8.5 Let A→ B be of finite type. Then H0(τ≥−1LB/A) = Ω1

B/A is a B-module offinite type, and if A is Noetherian, so is H−1(τ≥−1LB/A). If A→ B is of finite presentation, theresult holds true substituting the finite presentation condition to the finite type one everywhere.

Proof. The first statement is known already, while the second is not. Since A → B is finitetype, there is a polynomial A-algebra C in finitely many indeterminates, and a factorization

A→ C → B

where C → B is onto. Since A is Noetherian, so is B and C and the kernel J of C → B is afinitely generated C-module. It follows J/J2 is a finitely generated B-module, and given thatC is formally smooth we can apply Lemma B.8.2 and get a quasi-isomorphism of τ≥−1LB/Aand

(J/J2 → Ω1C/A ⊗C B).

Given that the boundary map is a map of modules of finite type over a Noetherian ring, thecohomology of this complex is definitely made of finite type B-modules.


We now state the following:Theorem B.8.6 Let the following diagram of ring maps be given:

A

// A′

B // B′

and assume such diagram is cocartesian. Then, the natural base-change map

τ≥−1LB/A ⊗B B′ → τ≥−1LB′/A′

yelds an isomorphism on degree zero cohomology, and yelds a surjection between the respectivedegree −1 cohomology modules. If A→ A′ is flat, then it is a quasi-isomorphism.

The above Theorem is an immediate consequence of a very general statement saying that assoon as the maps A → A′ and A → B are tor-independent, then the base-change map is aquasi-isomorphism between LB/A ⊗B B′ and LB′/A′ . Note that no derived tensor product isneeded, because LB/A will be a complex of flat B-modules. In fact, projective B-modules.

We state a last fact on the truncated cotangent complex:Proposition B.8.7 Let A → B → C be two ring maps. Then we have the following naturalexact sequence of C-modules:

H−1(τ≥−1LB/A ⊗B C)→ H−1(τ≥−1LC/A)→ H−1(τ≥−1LC/B)→→ H0(τ≥−1LB/A ⊗B C)→ H0(τ≥−1LC/A)→ H0(τ≥−1LC/B)→ 0

and for any C-module M we have the natural exact sequence of C-modules:

HomC(τ≥−1LC/B ,M)→ HomC(τ≥−1LC/A,M)→ HomC(τ≥−1LB/A ⊗B C)→→ Ext1

C(τ≥−1LC/B ,M)→ Ext1C(τ≥−1LC/A,M)→ Ext1

C(τ≥−1LB/A ⊗B C)→ 0

This result will be a direct consequence of the existence of the fundamental distinguishedtriangle for the full cotangent complex in D(C). Notice that Proposition B.8.7 is exactly theresult in [?] due to the preliminary work of Grothendieck.

B.9 The main Theorem on algebra extensions

We are eventually ready to deal with algebra extensions, and let us state immediately thefollowing basic:Theorem B.9.1 Let A → B be a ring map (or a map of sheaves of rings) and let M be aB-module. Then there is an isomorphism of B-modules which is natural in M :

ExalA(B,M) = Ext1B(τ≥−1LB/A,M).

We are going to prove Theorem B.9.1 in two different ways. The former is new to the author (ie.dated two days ago :) ) and the latter is just the exact same proof of Appendix B, rephrasedwith no mention of Picard groupoids.

Let us first of all show that the Theorem holds true in two special cases.


Proposition B.9.2 Theorem B.9.1 holds true if either A → B is surjective map of rings orsheaves of rings, or formally smooth.

Proof. We first focus on ring maps.

Step 1: ring maps. By Corollary B.8.4 we have that if A → B is surjective with kernel I,then τ≥−1LB/A = I/I2[1]. Therefore,

Ext1B(τ≥−1LB/A,M) = HomB(I/I2,M)

and we also know already that

ExalA(B,M) ' HomB(I/I2,M).

Now we supposeA→ B is formally smooth. Then τ≥−1LB/A → Ω1B/A[0] is a quasi-isomorphism,

and we getExt1

B(τ≥−1LB/A,M) = Ext1B(Ω1

B/A,M).

However, since B is a formally smooth A-algebra, Ω1B/A is projective (famous long footnote in

EGA IV). It follows Ext1B(Ω1

B/A,M) = 0, and since ExalA(B,M) is also trivial, we conclude.

Step 2: sheaves of rings. The argument given for a surjective ring maps can be repeatedfor sheaves of rings on a locally ringed space X. Let us now consider a formally smooth mapof sheaves of rings A→ B. We have, again, that τ≥−1LB/A → Ω1

B/A[0] is a quasi-isomorphism,and Ω1

B/A is projective only locally. This implies that the higher sheaves Ext vanish, so thatthe local-to-global spectral sequence for Ext degenerates and yelds:

Ext1B(Ω1

B/A,M) = H1(X,HomB(Ω1B/A,M)).

Now we claim we have the following isomorphism:

ExalA(B,M) = H1(X,HomB(Ω1B/A,M)).

We compute this latter using Cech cohomology. Choose an open cover Ui of X and denoteby Uij the double intersection Ui ∩ Uj , as usual. An element of H1(X,HomB(Ω1

B/A,M)) is acollection of maps

αij : Ω1B/A|Uij →M |Uij

satisfying the cocycle condition on triple intersections. These data are representatives of equi-valence classes modulo the equivalence relation given naturally by refining the cover. The mapsαij determine uniquely an A-linear derivation

dij : B|Uij →M |Uij

and the idea is to glue the A-algebra extension we can cook up with each one of these. We canconsider the trivial extension DB(M) on each one of the Ui, and glue them together by usingthe maps dij . Given that the automorphism group of the trivial extension is DerA(B,M). Thisgives a natural map

H1(X,HomB(Ω1B/A,M))→ ExalA(B,M),

which is an isomorphism because every A-algebra extension, is locally trvial! due to formallsmoothness.


We now prove the general case.

Proof of Theorem B.9.1. Let A→ B be any ring map. We can always factor it as A→ A[B]→B, and A[B] is formally smooth. We use Proposition B.8.7 and the known cases to obtain:

DerA(A[B],M)∂ // ExalA[B](B,M) // ExalA(B,M) // ExalA(A[B],M)

HomB(LA[B]/A ⊗A[B] B,M)

'

OO

// Ext1B(LB/A[B],M)

'

OO

// Ext1B(LB/A,M) //

OO

Ext1A[B](LA[B]/A,M)

'

OO

Here the left vertical arrow is the composition of

HomA[B](τ≥−1LA[B]/A ⊗A[B] B,M) = HomB(H0(τ≥−1LA[B]/A ⊗A[B] B),M)

= HomB(H0(τ≥−1LA[B]/A)⊗A[B] B,M)

= HomB(Ω1A[B]/A ⊗A[B] B,M)

= HomA[B](Ω1A[B]/A,M)

= DerA(A[B],M).

I’m going to add the check of commutativity of the left square using the explicit description of∂ we know, and of the map

HomB(τ≥−1LA[B]/A ⊗A[B] B,M)→ Ext1B(τ≥−1LB/A[B],M).

But we know that both groups at the right are trivial! which implies the dotted arrow is anisomorphism, as desired. The case of sheaves of rings is a little tricky. Roughly, one choosesa factorization of any map of sheaves of rings A → B of the form A → A[B] → B. This way,all the stalks of A and B are formally smooth. The dotted arrow now has not been introducedyet actually.

B.10 Construction of the relative cotangent complex

We discuss the construction of the complex. Motivation for the reason why this constructionwe are about to perform indeed is the right thing to do will be given in Appendix A.

Functorial trivial cofibration

To start off, let us consider two categories C and D, and two adjoint functors (F,G):

CF−→ D

G−→ C

such adjunction is uniquely determined by either the unit

ε : 1→ FG

or the counitη : GF → 1

both natural transformations of functors (ie. not necessarily isomorphisms). We have alreadydiscussed in Section 1 a number of adjunctions, and we are mainly interested in the following:


let us consider C the category of A-algebras, where A is either a ring or a sheaf of rings, andlet D be the category of sets. We let F be the forgetful functor, and G be the free A-algebrafunctor sending a set S to A[S]. We have seen already that (F,G) is an adjoint pair, so itcomes with a unit and a counit.

Resolution via adjunction

Given an object B of C, we are able to produce an object PA(B)• of sC using the adjunction(F,G). It is sufficient to set:

PA(B)n := (GF )n+1(B)

and for all 0 ≤ i ≤ n− 1 we define

∂i : PA(B)n−1 → PA(B)n

to be the B-section of the following natural transformation:

G(FG)i 1 (FG)n−iF1G1(FG)iε1(FG)n−i1F−−−−−−−−−−−−−−−−−→ G(FG)i (FG) (FG)n−iF

and we define for 0 ≤ i ≤ n+ 1:

σi : PA(B)n+1 → PA(B)n

to be the B-section of the following natural transformation

(GF )i (GF ) (GF )n−i+11(GF )iη1(FG)n−i+1

−−−−−−−−−−−−−−→ (GF )i (GF )n−i+1

One can check that the datum(PA(B)•, ∂i, σi)

is a simplicial A-algebra, as desired.

Augmentation

We have a natural transformation

ηn+1 : (GF )n+1 → 1

which yelds, for all n, a morphism

αn : PA(B)n → B

which satisfy the required relations defining a morphism of simplicial rings

α : PA(B)• → B

(which is a natural transformation of functors 4op → C. Here we regard B as a constantsimplicial A-algebra, where all faces and degeneracies are the identity. Therefore, the simplicialA-algebra PA(B)• comes with an augmentation to B. We shall prove that such augmentedsimplicial A-algebra has acyclic normalized complex, that is, PA(B)• is a resolution of B overA. In the terminology we shall discuss in Appendix A, PA(B)• → B is a trivial cofibration,one of the ingredients to define a model structure on the category of simplicial A-algebras.


We are eventually ready to define the relative cotangent complex of a morphism of rings orsheaves of rings on a space T .

We let A → B be a morphism of rings or sheaves of rings on T . We consider the simplicialB-module

Ω1• := Ω1

PA(B)•/A⊗PA(B)• B

whose simplicial structure maps are induced by those of A → PA(B)• by functoriality, A andB being regarded as constant simplicial rings.Definition B.10.1 We define the relative cotangent complex of A→ B to be:

LB/A := Ner(Ω1•).

We shall explain the meaning of this assignment in Appendix A. For now let us observe that theDold-Kan equivalence is to be regarded only as a dictionary, and therefore the definition of therelative cotangent complex of A→ B could already be Ω1

•. Indeed, it will turn out that this is“the right way” to deal with cotangent complexes, as one can put a model category structure onthe category of simplicial rings and find a generalization of schemes such that every morphismof schemes appears to be “smooth” in the category of such generalized schemes. The correctanalog of Ω1

(−)/(−) in such category will measure how far was the initial morphism of schemesfrom being smooth, roughly, and will be its cotangent complex. It is not a complex yet, but bymeans of the Dold-Kan equivalence we can read the information contained in it by producinga complex in a suitable derived category, rather than a simplicial sheaf of rings...Remark B.10.2 We remark that by definition of LB/A, functoriality follows at once, that is,given a commutative diagram of ring maps (or morphisms of sheaves of rings):

A

// B

A′ // B′

we are given a base change map

LB/A ⊗LB B

′ → LB′/A′ .

We also note that since LB/A is made, by definition, of flat B-modules, we have a naturalidentification in D(B):

LB/A ⊗LB B

′ ' LB/A ⊗B B′

so we shall always omit derived tensor products and pullbacks when dealing with base-changefor the cotangent complex.

As a final remark, we observe that formation of the free A-algebra over a set or sheaf ofsets commutes with all filtered colimits (since formation of the symmetric algebra does), andsince Ω1

(−)/(−) on the category of ring maps commutes with all colimits, and formation of thenormalized complex involves formation of kernels, which commutes with all filtered colimits,we conclude that formation of the cotangent complex of a ring map or a morphism of sheavesof rings commutes with all filtered colimits! That is, given a filtered small category I, and adiagram I → RngMor, we get:

lim−→I

LBi/Ai ' LB/A,

where (A→ B) = lim−→(Ai → Bi).


A pleasant consequence of this is that taking stalks and forming the cotangent complex arecommuting operations.Remark B.10.3 We remark, furthermore, that it is not at all true that formation of thecotangent complex of a ring map commutes with all colimits. Cexample: Z → Z → Z/pZand use fact that cokernel is a colimit. This is due to the fact that the functor sending a ringmap to its cotangent complex is not a left-adjoint, or (equivalently!) LB/A does not representa functor on RngMor. Instead, Ω1

(−)/(−) will do such thing on the category of morphisms ofsimplicial rings with a model structure, so formation of the cotangent complex will commutewith all colimits in such category, ie. with all homotopy colimits!

B.11 Properties of the relative cotangent complex

Let T be a ringed topos, and A a simplicial ring of T . A simplicial A-moduleM can be regardedas a bi-simplicial object in the category of sheaves of rings of T by simply observing that foreach integer p ≥ 0, Mp,• is an Ap-module functorially in [p] (that is, a simplicial Ap-module),and M•,q is an A-module.

How to regard an A-module M as a “trivial” simplicial A-module? By simply defining M•,• :=M . One can also take the diagonal of a simplicial A-module viewed as a bi-simplicial ring, thusyelding functors

ModA → simplModA4−→ModA

We stress again that A is a simplicial ring itself!

Now we observe that the category of A-modules is an abelian category. Let D≤0(A) be the fullsubcategory of its derived category D(A) of complexes whose cohomology is trivial in positivedegree. We recall that if we have E,F ∈ ob(D≤0(A)) with another object L ∈ ob(D≤0(A))which is quasi-isomorphic to E, if L is A-flat then

E ⊗LA F → L⊗A F

is a quasi-isomorphism.

We discuss in [Ill, Lemma 3.3.2.1]. We use Illusie’s terminology, and say that an arrow ofsimplicial objects in T is a quasi-isomorphism if it yelds a quasi-isomorphism of the respectivenerves (or, equivalently, of the respective total complexes).Lemma B.11.1 ( [Ill, 3.3.2.1]) Let L be an A-module, where A is a simplicial ring in T . Let,also, f : E → F be a quasi-isomorphism of A-modules. If L is A-flat, or if E and F are A-flat,then

L⊗ f : L⊗A E → L⊗A F


Proof. We tensor by f the standard free A-module simplicial resolution

FA(L)• → L

and we get a commutative diagram of maps of simplicial modules:

FA(L)⊗ E //

FA(L)⊗ F

L⊗ E // L⊗ F


where L⊗E ad L⊗ F are regarded as “trivial” simplicial A-modules, as explained above. Letus explain, for example, how the left vertical arrow is made. It is indeed a family of arrowsindexed by integers n ≥ 0:

FAp(Lp)n ⊗Ap Ep → Lp ⊗Ap Epfor a fixed integer p ≥ 0. We can apply the Kunneth formula to each one of these, and byflatness of either Lp over Ap, or of Ep and Fp over Ap, these are quasi-isomorphisms, implyingthat the left vertical arrow is a quasi-isomorphism. Likewise for the right-vertical arrow. Againby the case A is constant, we deduce the top horizontal arrow is a quasi-isomorphism, andsetting up the spectral sequence for double complexes we get that the bottom horizontal arrowis a quasi-isomorphism as well!

Now, given again A a simplicial ring in T , we define the bifunctor (−) ⊗À (−) to be thecomposition

(−)⊗À (−) : D(A)×D(A)(−)⊗L

A(−)−−−−−−→ D(A)4−→ D(A)

We deduce at once that if L→ E is a quasi-isomorphism, with L A-flat, then

L⊗A F → E ⊗À F


We observe that if A is constant we recover the usual derived tensor product!

B.12 The fundamental distinguished triangle

Existence of the so called “fundamental distinguished triangle” is truly at the heart of the wholetheory. Let T be a ringed topos, and A → B → C be morphisms of sheaves of rings. In thissection we explain how to construct the following distinguished triangle in D≤0(T ):

LB/A ⊗LB C → LC/A → LC/B → LB/A ⊗L

B C

Simplicial resolutionsTo start off, we set up the following diagram:

PA(B)• //

Q(C)4•,•

A

<<

// B //

%%

C

PB(C)•

OO

PA(B)• denotes as usual the standard simplicial resolution of B over A, and likewise for PB(C)•.We observe that for all integers n ≥ 0, C is a PA(B)n-algebra. Hence we may take the simplicialresolution of C as a PA(B)n-algebra for all such n’s, thus yelding

Q(C)n,• → C

which is functorial in n, that is, assigns a functor

Q(C)•,• : 4op ×4op → T .


We take its diagonal, and get Q• := Q(C)4•,•. We write

P• := PA(B)• and R• := PB(C)•

for simplicity.Remark B.12.1 Fix an integer n ≥ 0. We have that Qn is a polynomial algebra over Pn,and on the other hand Rn is a polynomial algebra over B. The morphism of sheaves of A-algebras Pn → B induces, by functoriality of the standard simplicial resolution construction, amorphism of sheaves of rings

Qn = Pn[· · · [C] · · · ]→ B[· · · [C] · · · ] = Rn.

More precisely, the morphism Pn → B induces directly the above morphism, and functorialityof the standard simplicial resolution construction ensures compatibility of such morphism withthe simplicial relations, thus yelding a morphism of simplicial objects Q• → R•.

We prove the following lemma:Lemma B.12.2 Let A → B be a morphism of sheaves of rings in T , and let PA(B)• be thestandard simplicial resolution of B over A. Then the nerve of the simplicial sheaf of ringsPA(B)• is quasi-isomorphic to B, seen as the constant complex.

Proof. Call F the forgetful functor from the category of sheaves of A-algebras of T to Set, andG its left adjoint: the “free A-algebra” functor A[−]. We first assume B is free, plugging A[B]in place of B, regarding B as a sheaf of sets. Let ε : 1 → FG be the unit of the adjunction.Recall that PA(W )n is nothing but (GF )n+1(W ) = (GF )n+1G(B). We let

Hn := 1(GF )n+1 1G ε : (GF )n+1G→ (GF )n+2G.

The compositions∂i Hn and Hn−1 ∂i

are equal for all i = 0, . . . , n, and moreover the composition ∂n+1 Hn is the identity on(GF )n+1G. It follows

(

n+1∑i=0

∂i) Hn −Hn−1 (

n∑i=0

∂i) : (GF )n+1G→ (GF )n+1G

yelds a homotopy between the identity on Tot(PA(A[B])•) and the zero map, thus yelding theresult. Let now B be a sheaf of A-algebras. B is the coequalizer of the maps ∂0 and ∂1:

A[A[B]]∂0 //

∂1

// A[B] // B

Equivalently, B is the coequalizer of the maps ∂0 − ∂1 and 0, that is:

A[A[B]]0 //

∂0−∂1// A[B] // B

and now ∂0 − ∂1 augments the standard simplicial resolution of A[B]. We are reduced to thefree case, and by the above discussion we conclude.

Remark B.12.3 Recall that given a simplicial object X• in an abelian category A, its nerveNer(A•) was quasi-isomorphic to the total complex Tot(A•). We deduce from Lemma B.12.2that Ner(R•) is quasi-isomorphic to the constant complex C, and likewise for Ner(P•) and B.


We need the following:Lemma B.12.4 The notation being that from above, we have a quasi-isomorphism betweenthe constant complex C and Ner(Q•).

Proof. Given Q•,•, form the double complex c(Q•,•) whose (−p,−q)-th entry is simply Qp,q,and whose horizontal and vertical differentials are, respectively,

dhorp :=

p∑i=0

(−1)i∂hori : Qp,q → Qp−1,q

and

dverq :=

q∑i=0

(−1)i∂veri : Qp,q → Qp,q−1

The cohomology of total complex Tot(c(Q•,•)), the differential of this latter being

dn :=∑

−p−q=n(dhorp + (−1)pdver

q ),

s computed by a spectral sequence degenerating at the first page to the bi-complex

0 C0 0

On the other hand, the Eilenberg-Zilber Theorem ensures precisely:

Tot(Q•) 'qi Tot(c(Q•,•))

where 'qi stands for “quasi-isomorphism”. On the other hand, we also know

Tot(Q•) 'qi Ner(Q•)

which concludes the proof.

The case of polynomial algebrasWe now assume B is of the form: B = A[S], for some sheaf of sets S of T .Theorem B.12.5 We have a quasi-isomorphism of complexes of sheaves of A[S]-modules:

LA[S]/A 'qi Ω1A[S]/A.

Proof. The proof follows the proof of Lemma B.12.2, applying the functor

Ω1(−)/A ⊗(−) A[S]

to the standard simplicial resolution of A[S] over A, and obtaining as in Lemma B.12.2 maps:

Hn : Ω1PA(A[S])n/A

⊗PA(A[S])n A[S]→ Ω1PA(A[S])n+1/A

⊗PA(A[S])n+1A[S]

yelding a homotopy between the identity and the zero map on LA[S]/A. Thus this latter ismapped quasi-isomorphically to its cohomology in degree zero, which is exactly Ω1

A[S]/A.


A fundamental lemmaWe describe a construction which turns out to be handy, as explained in [Ill, § III-1.2]. IfA → B is a morphism of sheaves of rings of T , we know how to construct the standardsimplicial resolution PA(B)• of B over A. Note that the word “resolution” should better beintroduced after proving Lemma B.12.2. I’ll fix this.

Now let, instead, A• → B• be a morphism of simplicial objects in the category of sheaves ofrings of T . Fix an integer p ≥ 0. For any such p we have a functorial assignment

[q] 7→ PAp(Bp)q,

which assigns a functor4op ×4op → T

I should better specify the correct target category (will do so when polishing the notes, duringChristmas!). Composition with the functor Ω1

(−)/Ap⊗(−) Bp, which is in turn functorial in [p],

yelds a bisimplicial object in the category of sheaves of rings of T :

([p], [q]) 7→ Lp,q := Ω1PAp (Bp)q/Ap

⊗PAp (Bp)q Bp.

We define L4B•/A• to be

L4B•/A• := Ner(L4•,•).

Note that L4B•/A• is just notation, to remind that this is the nerve of the simplicial objectconstructed taking the diagonal of the bisimplicial sheaf of rings L•,•.Remark B.12.6 If we let A• and B• be, respectively, A and B viewed as constant simplicialobjects, we recover

L4B•/A• = LB/A.

We prove the following essential lemma:Lemma B.12.7 We consider a commutative diagram of arrows of simplicial sheaves of ringsof T :

A•

// B•

A′• // B′•

Assume the two vertical arrows induce quasi-isomorphisms

Ner(A•) 'qi Ner(A′•)

andNer(B•) 'qi Ner(B′•).

Then the following natural maps of simplicial sheaves of rings of T induce quasi-isomorphismsas well:

1.PA•(B•)

4• → PA′•(B

′•)4•

2.Ω1PA• (B•)

4• /A•

→ Ω1PA′•

(B′•)4• /A′•

Moreover, the following natural map of complexes of sheaves of rings of T is a quasi-isomorphism:

L4B•/A• → L4B′•/A′•.

The proof is actually elementary, and left to the reader.


Construction of the fundamental distinguished triangle

Recall once again the situation:

P• //

Q•

A

>>

// B //

!!

C

R•

OO

whereP• := PA(B)•, Q• := Q(C)4•,•, R• := PB(C)•.

We focus on the arrowsA→ P• → Q•

where A is to be seen as a constant simplicial sheaf of rings of T . These yeld a short exactsequence of simplicial sheaves of modules:

0→ Ω1P•/A

⊗P• Q• → Ω1Q•/A

→ Ω1Q•/P•

→ 0

where for all p ≥ 0, Ω1Qp/Pp

is flat as a Qp-module, being Qp a polynomial algebra sheaf ofrings over Pp.

We recall that by construction Qp is free over Pp for all integers p ≥ 0. This implies, byTheorem B.12.5, a quasi-isomorphism

L4Q•/P• → Ner(Ω1Q•/P•

).

For the same reason, we obtain a quasi-isomorphism:

L4P•/A → Ner(Ω1P•/A

).

Finally, since Qp is a polynomial algebra sheaf over Pp for all integers p ≥ 0, and Pp is apolynomial algebra sheaf over A for all such p, then so is Qp. Hence, as above, we have aquasi-isomorphism:

L4Q•/A → Ner(Ω1Q•/A

).

We tensor the above sequence along the augmentation Q• → C, where C is to be seen as aconstant simplicial sheaf of A-algebras. Using flatness we deduce the following sequence ofsimplicial sheaves of modules is still exact:

0→ Ω1P•/A

⊗P• Q• ⊗Q• C → Ω1Q•/A

⊗Q• C → Ω1Q•/P•

⊗Q• C → 0

By the above remarks and Lemma B.12.7 we obtain:

L4P•/A ⊗`P• Q• ⊗

`Q• C → L4Q•/A → L4Q•/P• → L4P•/A ⊗

`P• Q• ⊗

`Q• C[1]

the notation being that of [Ill, § 3.3.1], where Illusie explains derived tensor products of simpli-cial sheaves. By the Kunneth formula, Lemma B.12.7 and Lemma B.12.2, the above triangleis termwise quasi-isomorphic to:

L4B/A ⊗`B C ⊗`C C → L4C/A → L4C/B → L4B/A ⊗

`B C ⊗`C C[1]


and since A, B and C are constant simplicial objects, we have that the above triangle is equalto the following:

LB/A ⊗LB C → LC/A → LC/B → LB/A ⊗L

B C[1]

as desired.

B.13 Base change

Theorem B.13.1 Let T be a ringed topos, and consider the following commutative diagramof arrows of simplicial rings of T :

B // B′

A

OO

// A′

OO

We assume B⊗`AA′ → B′ in D(AlgA) is an isomorphism. Then the following canonical naturalmaps are quasi-isomorphisms:

1.L4B/A ⊗B B

′ → L4B′/A′ .

2.(L4B/A ⊗B B

′)⊕ (B′ ⊗A′ L4A′/A)→ L4B′/A.

Proof. To simplify notation, we set

P := PA(B)4, P ′ := PA′(B′)4, P 1 := P ⊗A A′A′.

We set up the following commutative diagram:

P //

P 1

// P ′

~~

B // B′

A

OO

FF

// A′

FF

OO

FF

Now the assumption B ⊗`A A′ ' B′ implies that P 1 → B′ is a quasi-isomorphism. Since theabove diagram commutes, then P 1 → P ′ is a quasi-isomorphism as well. We have the followingnatural maps:

Ω1P/A ⊗P P

1 → Ω1P 1/A′ → Ω1

P ′/A′

where the first one is an isomorphism, and since P 1 → P ′ is a quasi-isomorphism, we deducethe second one is a quasi-isomorphism by Lemma B.12.7. Now we extend scalars to B′, andsince

Ner(Ω1P/A) = L4B/A

andNer(Ω1

P ′/A′ ⊗B′) = L4B′/A′

we obtainL4B/A ⊗B B

′ → L4B′/A′


is a quasi-isomorphism, as desired.

Now let us show (2). Set Q := PA(A′)4, R := P ⊗A Q, and R′ := PA(B′)4.

P //

R

~~

// R′

wwB // B′

A

OO ??

BB

// A′

OO

Q

OO

oo

The assumption B⊗`AA′ ' B′ implies that R→ B′ is a quasi-isomorphism. By commutativityof the above diagram, R → R′ is a quasi-isomorphism. On the other hand, P and Q arepolynomial algebras over A, and then so is R. We have natural maps

(Ω1P/A ⊗P R)⊕ (R⊗Q Ω1

Q/A)→ Ω1R/A → Ω1

R′/A

where the former is an isomorphism, and the latter is a quasi-isomorphism due to LemmaB.12.7. Extend scalars to B′, and use Kunneth to conclude.

We now observe that if A, B and B′ are constant simplicial rings, then since B ⊗`A A′ =B ⊗L

A A′ ' B′, the above Theorem becomes the following:

Theorem B.13.2 We consider the following commutative diagram of arrows of ringed topoi:

X

f

X ′uoo

f ′

Y Y ′oo

This yelds a commutative diagram of rings of X ′:

B // B′

A

OO

// A′

OO

We assume TorAi (B,A′) = 0 for all i > 0, and LB′/(B⊗AA′) = 0. Then the canonical naturalmaps:

1. u∗LX/Y → LX′/Y ′

2. u∗LX/Y ⊕ (f ′)∗LY ′/Y → LX′/Y

are quasi-isomorphisms.

Proof. DefineB1 := B ⊗A A′.

The fundamental triangle forA′ → B1 → B′

is:LB1/A′ ⊗B1

B′ → LB′/A′ → LB′/B1= 0


thus implying the first map is a quasi-isomorphism. The fundamental triangle for A→ B1 → B′

isLB1/A ⊗B1

B′ → LB′/A → LB′/B1= 0

and likewise the first map is a quasi-isomorphism. Now by Theorem B.13.1 we get quasi-isomorphisms:

LB/A ⊗B B1 'qi LB1/A′ and (LB/A ⊗B B1)⊕ (B1 ⊗A′ LA′/A) 'qi LB1/A.

If we extend scalars to B′, these stay quasi-isomorphisms by the Kunneth formula, given thatLB/A is B-flat as well as LB1/A′ is B1-flat. Now by Tor-independence of A′ and B over A, wehave

B ⊗LA A

′ 'qi B′

and Theorem B.13.1 applies. It follows that (1) and (2) are quasi-isomorphisms.

B.14 Quasi-coherence and coherence

In this section we prove the following:Theorem B.14.1 Let X → Y be a morphism of schemes. Then not only LX/Y has quasi-coherent cohomology sheaves, but it has quasi-coherent terms itself.

Proof. We begin calling B := OX and A := f−1OY , so that LX/Y = LB/A. We can assume Xand Y are affine, and consider the natural map

LB(X)/A(X) ⊗OX(X) OX → LB/A.

Such map is a localization. Indeed, pick x ∈ X and compute the stalk at x of both sides: wehave

LB(X)/A(X) ⊗OX(X) OX,x → LBx/Ax .

But since localizations are filtered colimits, and formation of the cotangent complex ommuteswith filtered colimits, we conclude that the above map is stalkwise a quasi-isomorphism, im-plying that

LB/A = ˜LB/A(X),

as desired. This obviously implies that LX/Y has always quasi-coherent cohomology.

We now introduce some finiteness conditions. Roughly, we expect that in the case Y is locallyNoetherian and X → Y is locally of finite type, then LX/Y has coherent cohomology, in analogyto what happens for Ω1

X/Y .Theorem B.14.2 Suppose X → Y is a morphism of schemes locally of finite type, with Ylocally Noetherian. Then the cohomology sheaves of LX/Y are coherent.

Proof. Again the question is local in both X and Y , so let’s assume they are both affine.André proves in [?, Prop. 17.2] that for any ring map A→ B of finite type, with A Noetherian,there exists a simplicial resolution ΦA(B)• → B over A with ΦA(B)n of finite type over Afor all n. The idea is to consider the free A-algebra over A-generators of B, then inductivelyconstruct ΦA(B)n by taking A-generators for the intersection of all the kernels of the faces andconsider the free algebra over them, and suitably define faces and degeneracies in such a waythat they satisfy the simplicial identities. We shall very explicitly describe the construction inAppendix B, by means of an algorithm together with some explicit computations. It followsthat Ω1

ΦA(B)n/A⊗ΦA(B)n B is of finite presentation as a B-module for all n, and hence its


normalized complex has finite presentation cohomoloy modules. Note that coherence of thecohomology sheaves of LX/Y is meant the following way: LX/Y , as an object of D(OX), has arepresentative in D(OX) with coherent cohomology.

B.15 Special cases

We specialize the base-change theorem from the previous section. Let us first assume that inthe following commutative diagram of morphisms of ringed topoi:

B // B′

A

OO

// A′

OO

u is the identity, thus yelding a diagram of rings of X ′:

B B

A

OO

// A′

OO

We have the following:Lemma B.15.1 Suppose B ⊗L

A A′ 'qi B. Then we have

LB/A 'qi LB/A′

in D(B). Likewise, if A′ = B we have that under the same assumption LB/A = 0 in D(B).

Proof. The proof is immediate by Theorem B.13.1.

As a consequence of the lemma above, we obtain the following:Theorem B.15.2 Suppose A→ B and B ⊗A B → B are flat maps of rings. Then LB/A = 0.

Proof. We set up the following diagram:

B // B′ = B ⊗A B

A

OO

// A′ = B

OO

where the assumptions of Theorem B.13.1 are satisfied!

As a corollary, we get:Corollary B.15.3 Let A→ B be a morphism of rings. If A→ B is étale, then the cotangentcomplex LB/A is zero in D(B). If B = S−1A for a multiplicative part S ⊂ A, then the cotangentcomplex LB/A is again zero in D(B).

We also obtain the following important:Corollary B.15.4 Let A → B be a morphism of rings. Supose S ⊂ A and S′ ⊂ B aremultiplicative parts such that S is mapped into S′. Then

LS′−1B/S−1A 'qi LB/A ⊗B S′−1B.

Proof. This is just Theorem B.13.1 with B′ = S′−1B and A′ = S−1A.


The case of smooth ring maps

Theorem B.15.5 Suppose A→ B is a smooth ring map. Then LB/A 'qi Ω1B/A[0].

Proof. We know already that the two complexes agree in degree 0. We need to check that thecohomology of LB/A in nonzero degree is trivial. By Corollary B.15.4, since LBg/A = (LB/A)g,for all g ∈ B, it is sufficient to show this locally on Spec(B). We can therefore assume A→ Bis standard smooth: we can factor A→ B as A→ A[x1, . . . , xn]→ B, where the latter map isétale. By base-change we get

LB/A = LA[x1,...,xn]/A ⊗A[x1,...,xn] B,

and the conclusion follows from knowledge of the cotangent complex of a polynomial algebramap.

Locally complete intersections

Lemma B.15.6 Let A = Z[x]→ B = Z be the ring map which sends x to 0. Let I = (x) ⊂ A.Then LB/A is quasi-isomorphic to I/I2[1].

Proof. Consider the distinguished triangle

LZ[x]/Z ⊗Z[x] Z→ LZ/Z → LZ/Z[x] → LZ[x]/Z ⊗Z[x] Z[1]

The complex LZ[x]/Z is quasi-isomorphic to Ω1Z[x]/Z by Theorem B.15.5. The complex LZ/Z

is zero in D(Z). Thus we see that LB/A has only one nonzero cohomology group which is[(x)/(x)2][1].

We finally prove the following:Theorem B.15.7 Let A→ B be a surjective ring map whose kernel I is generated by a regularsequence. Then LB/A is quasi-isomorphic to I/I2[1].

Proof. This is true if I = (0). If I = (f) is generated by a single nonzerodivisor, then considerthe ring map Z[x]→ A which sends x to f . By assumption we have B = A⊗L

Z[x] Z. Thus weobtain LB/A = I/I2[1] from Lemmas B.13.1 and B.15.6.

We prove the general case by induction. Suppose that we have I = (f1, . . . , fr) where f1, . . . , fris a regular sequence. Set C = A/(f1, . . . , fr−1). By induction the result is true for A → Cand C → B. We have a distinguished triangle

LC/A ⊗LC B → LB/A → LB/C → LC/A ⊗L

C B[1]

which shows that LB/A has only one nonzero cohomology group which is I/I2.

The reader can work out the proof of the relevant statement in the general case, by flat descent.


B.16 Absolute deformations

We are given a commutative diagram of morphisms of schemes:

X0

f0

i // X

f

S0j// S

where i is a square-zero thickening of ideal I and j is a square-zero thickening of ideal J .

We notice that the maps f0 and f are the same map on topological spaces, and by abuse ofnotation let us denote again with f0: it will be clear when there will be a distinction. Thenatural map f−1

0 OS0→ OX0

yelds a natural map f−10 J → I, and hence a natural map

f∗0 J = f−10 J ⊗f−1

0 OSOX → I,

which is surjective if and only if the diagram is cartesian in the category of locally ringed spaces.Note that since both I and J are square-zero, so is f∗0 J , and hence we can regard f∗0 J → I asa map of OX0-modules.

Instead of assuming such diagram is given, let us assume we are only given the following data:solid arrows as below:

X0

f0

i // X

f

S0j// S

where j is a given square-zero thickening of ideal J , and we are also given a fixed OX0-module Iand a map of OX0-modules f∗0 J → I. The most naive question one can ask himself is whethersuch data are equivalent to the datum of a diagram as at the very beginning, or not.

First of all let us observe that the word “equivalent” is a little subtle, as manifestly an extensionof the solid arrows to a complete square, if it exists, is defined up to automorphisms. Let uscall such morphism f in the completed square a solution to the problem of lifting f0 to an S-morphism making such square commute (abbreviated AD). An automorphism of such solutionwill be an automorphism of the schemeX which commutes with both f and i, as in the followingdiagram:

X

a∼

~~

yy

X0

f0

i // X

f

S0j// S

Such automorphisms form a group. We can ask a refined question: what such group is, andwhether or not a solution to the problem exists, and under what circumstances. The answer isvery precise, and is the content of the following:


Theorem B.16.1 There exists an element

o(f0) ∈ Ext2OX0

(LX0/S0, I)

whose vanishing is necessary and sufficient for the existence of a solution to the problem (AD).If o(f0) vanishes, the set of isomorphism classes of the solutions to (AD) is a torsor underExt1

OX0(LX0/S0

, I), and the automorphism group of any such solution is canonically isomorphicto HomOX0

(Ω1X0/S0

, I).

We now prove the theorem, and then discuss a crucial situation in which properties of f0 areinherited by any solution f to (AD).

Proof. This is the typical way to use the cotangent complex. We recognise that the problemis actually a problem of algebra extensions. Indeed, the morphism i is an isomorphism on thelevel of topological spaces: everything happens on the level of nilpotents. We let A0 := f−1

0 OS0 ,A := f−1OS and M := f−1J . We also call B0 := OX0 and B := OX (whose existence mustbe discussed). We have LX0/S0

= LB0/A0, and a solution to the problem (SCP) is exactly an

A-algebra extension B of B0 by I, as below:

0 // M

// A

h

// A0

h0

// 0

0 // I // B // B0// 0

where h0 := f#0 and h := f#.

We know such extensions are classified by the Picard groupoid ExalA(B0, I), which is equivalentto the Picard groupoid

τ≤1RHomB0(LB0/A, I)[1]

so we need to get a handle on this latter. Given that the map A→ B0 factors through A0 bydesign, we can get a handle on RHomB0

(LB0/A, I)[1] by using the distingushed triangle

LA0/A ⊗A0B0 → LB0/A → LB0/A0

→ LA0/A ⊗A0B0[1].

Applying the functor RHomB0(−, I)[1], we get maps in D(B0):

RHomB0(LA0/A⊗A0B0, I)→ RHomB0

(LB0/A0, I)[1]→ RHomB0

(LB0/A, I)[1]→ RHomB0(LA0/A⊗A0B0, I)[1].

Taking H0 we obtain the sequence:

HomB0(LB0/A0

, I)→ HomB0(LB0/A, I)→ HomB0

(LA0/A ⊗A0B0, I)→

→ Ext1B0

(LB0/A0, I)→ Ext1

B0(LB0/A, I)→ Ext1

B0(LA0/A ⊗A0

B0, I)→∂−→ Ext2

B0(LB0/A0

, I).

Now we observe that

HomB0(LA0/A ⊗A0 B0, I) = HomB0

(Ω1A0/A

⊗A0 B0, I) = 0

because A→ A0 is surjective, which implies Ω1A0/A

= 0. We know that the set of isomorphismclasses of the Picard groupoid ExalA0

(B0, I) is isomorphic (as a group under Baer sums) toExt1

B0(LB0/A0

, I), and likewise

ExalA(B0, I) ' Ext1B0

(LB0/A, I).


Now since A → A0 is surjective with kernel M , we have H−1(LA0/A) = M/M2 = M , sinceM2 = 0. More precisely,

τ≥−1LA0/A = M [1].

Therefore we have the canonical isomorphism:

H0RHomB0(LA0/A ⊗A0

B0, I)[−1] = H0RHomB0(LA0/A[1]⊗A0

B0, I)

= H0RHomB0(M [0]⊗A0

B0, I)

= HomB0(M ⊗A0

B0, I)

= HomA0(M, I)

where the last canonical isomorphism is the usual extension-restriction adjunction. The upshotis that our sequence becomes the following:

HomB0(Ω1

B0/A0, I)→ HomB0

(Ω1B0/A

, I)→ 0→

→ ExalA0(B0, I)→ ExalA(B0, I)→ HomA0

(M, I)∂−→ Ext2

B0(LB0/A0

, I).

where we now describe the maps involved.

Description of the map Ext1B0

(LB0/A0, I)→ Ext1

B0(LB0/A, I)

The map A→ A0 yelds a natural map LB0/A → LB0/A0, and hence a map

τ≥−1LB0/A → τ≥−1LB0/A0

and, finally, a map

τ≤1RHomB0(LB0/A0

, I)[1]→ τ≤1RHomB0(LB0/A, I)[1]

which we now describe explicitly. The complex on the left is

HomB0(Ω1

A0[B0]/A0⊗A0[B0] B0, I)→ HomB0

(K0/K20 , I)

in degrees −1 and 0 respectively, with K0 := ker(A[B0] B0).

The same way, the complex on the right is

HomB0(Ω1

A[B0]/A ⊗A[B0] B0, I)→ HomB0(K/K2, I)

with K := ker(A[B0] B0).

Therefore, the map τ≤1RHomB0(LB0/A0

, I)[1]→ τ≤1RHomB0(LB0/A, I)[1] is:

RHomB0(LB0/A0

, I)[1]

HomB0(Ω1

A0[B0]/A0⊗A0[B0] B0, I)

ϕ

// HomB0(K0/K

20 , I)

ψ

RHomB0(LB0/A, I)[1] HomB0

(Ω1A[B0]/A ⊗A[B0] B0, I) // HomB0

(K/K2, I)

Since we have Ω1A0[B0]/A0

' Ω1A[B0]/A ⊗A A0, ϕ is nothing but the map DerA0

(A0[B0], I) →DerA(A[B0], I) given by precomposing an A0-linear derivation D0 : A0[B0] → I with the


reduction map A[B0] → A0[B0]. Likewise, ψ is given by precomposition with A[B0]/K2 →A0[B0]/K2

0 , and ψ is the map which induces

Ext1B0

(LB0/A0, I)→ Ext1

B0(LB0/A, I)

so given , we are good to go. The map

ExalA0(B0, I)→ ExalA(B0, I)

just sends an A0-algebra extension of B0 by I (constructed by pushing out the extension

0→ K0/K20 → A0[B0]/K2

0 → B0 → 0

along µ0 : K0/K20 → I) to the same extension regarded as an A-algebra extension, which is

identified with the pushout of

0→ K/K2 → A[B0]/K2 → B0 → 0

along the composition K/K2 → K0/K20µ0−→ I.

Description of the map Ext1B0

(LB0/A, I)→ Ext1B0

(LA0/A ⊗A0B0, I)

Finally, the map ExalA(B0, I)→ HomA0(M, I) is just the composition of the natural map

ExalA(B0, I)→ ExalA(A0, I)

and the isomorphism ExalA(A0, I) = HomB0(M, I). The first map is given by pullback along

A0 → B0 of an A-algebra extension of B0 by I to an A-algebra extension of A0 by I. The lastis given as follows. We consider an A-algebra extension of A0 by I:

0 // I // A′ // A0// 0

A

v

OO >>

Given m ∈ M , v(m) must be contained in I by commutativity of the above diagram. Thenthe map ExalA(A0, I)

'−→ HomA0(M, I) is given by sending the extension A′ to the restriction

v|M . In the end it follows that the map

ExalA(B0, I)→ HomA0(M, I)

sends and A-algebra extension B of B0 by I to the A0-linear homomorphism v|M : M → Iwhere v is given as in the diagram below:

0 // I // B // B0// 0

A

v

OO >>

We are ready to drow our conclusions. A solution to (AD) is an element of ExalA(B0, I), so wewant to first establish when this set is nonempty. It is if and only if, given the above discussion,


there exists some u ∈ HomA0(M, I) which goes to zero in Ext2

B0(LB0/A0

, I) under ∂. We aregiven u by our very problem, so a solution to our problem exists if and only if

o(f0) := ∂(u) ∈ Ext2B0

(LB0/A0, I)

vanishes, and if such obstruction vanishes then the set of isomorphism classes of such solutionsis manifestly a torsor under Ext1

B0(LB0/A0

, I). The action is given explicitly by sending anisomorphism class of A0-algebra extensions of B0 by I to an isomorphism class of A-algebraextension of B0 by I as explained above, and then acting with such extension via the grouplaw on ExalA(B0, I).

Finally, an automorphism of a solution to the problem (AD) is an automorphism of ringsα : B → B making the usual diagram of extensions from above commutative. Given such α,we can define an A0-linear derivation D : B0 → I by letting D(b) := α(b′) − b′, for any liftb′ of b ∈ B0 to B. D corresponds to an element of HomB0

(Ω1B0/A0

, I). Conversely, given anA0-linear derivation D, we can define α : B → B by α(b′) := b′ +D(b), with b the image of b′under B′ → B. The Theorem is proved.

B.17 Flat absolute deformations

Suppose now that in (AD) f0 is flat, and call the new problem (FAD). We wonder whether ornot we are able to deform it to a flat f completing the square.

More precisely, given the solid arrows as below:

X0

f0

i // X

f

S0j// S

with f0 flat, we wonder if there exists a flat deformation f of f0 making the above diagramcommutative. Note we are not requiring the diagram to be cartesian! However, recall the localcriterion of flatness:Proposition B.17.1 Let A → A0 be a surjective ring map with nilpotent kernel I. For anyA-module M , the following are equivalent:

(1) M is a flat A-module.

(2) TorA1 (M,A0) = 0 and M ⊗A A0 is a flat A-module.

Note that the above result can be restated for sheaves of rings, rather than just rings, assumingthe kernel I has nilpotent stalks, which is all we need.

Proof. (1) implies (2) because flatness is stable under base change. To see the converse, notethat the ring case also implies the same results for sheaves of rings, since we can check it onstalks. Since I is nilpotent, by the usual devissage argument we reduce to check TorA1 (M,N) = 0for all A-modules N killed by I.

In this case, the functor (·) ⊗A N factors as the composition of the functors (·) ⊗A A0 andthe functor (·) ⊗A0 N . We claim that the first left derived functor of (·) ⊗A N vanishes. Byassumption, the first left derived functor of (·)AA0 vanishes on M , and since M ⊗A A0 is A0-flat, then the first left derived functor of (·) ⊗A0

N vanishes on M ⊗A A0. We can use theGrothendieck spectral sequence and conclude.


Given a diagram:

X0

f0

i // X

f

S0j// S

where j is a nilpotent closed immersion, flatness of f is equivalent to flatness of its reductiong0 : X ×S S0 → S0 to S0 together with the condition Tor

g−10 OS

1 (OX , g−10 OS0) = 0. There’s no

mention of f0.

Our problem becomes: among all the solutions to (AD), is there a flat f if f0 is flat? GivenProposition B.17.1, we are reduced to the case where the commutative diagram in question iscartesian in the category of locally ringed spaces.

Let us consider the following exact sequence of OS-modules:

(∗) 0→ J → OS → j∗OS0→ 0

Since |X0| is the topological space of any solution to (FAD), we keep calling f−10 (−) the

topological pullback functor. (FAD) has a solution making the square cartesian if and only ifthere exists an X f−→ S with the property that OX is a flat f−1

0 OS-module, and

i∗OX0= OX ⊗f−1

0 OSf−1

0 OS0.

The flatness property is in turn satisfied if and only if Torf−10 OS

1 (OX , f−10 OS0) = 0 and OX⊗f−1

0 OS

f−10 OS0 = OX0 is flat as an f−1

0 OS0 -module.

If we apply (−)⊗f−10 OS

OX to f−10 (∗), flatness yelds the following short exact sequence:

0→ f−10 J ⊗f−1

0 OSOX → OX → j∗OS0

⊗f−10 OS

OX → 0

where f−10 J ⊗f−1

0 OSOX = f∗J and j∗OS0

⊗f−10 OS

OX = i∗OX0.

The natural map of OX -modules f∗J → I is therefore an isomorphism, and since both I andJ are square-zero, it can be regarded as the map of OX0

-modules f∗0 J → I.

Conversely, if the natural map f∗0 J → I is an isomorphism, not only any solution to (AD) forthe diagram of solid arrows:

X0

f0

i // X

f

S0j// S

will make the diagram cartesian in the category of locally ringed spaces, but if f0 is flat thenso will be f , thus yelding a solution to (FAD).

Let us therefore rephrase our deformation problem: given a diagram of solid arrows:

X0

f0

i // X

f

S0j// S


where j is a square-zero closed immersion of ideal J , and I is a fixed OX0-module isomorphic

to f∗0 J as an OX0-module, we seek deformations of f0 to a flat f making the diagram cartesian.

We finally observe that up to automorphisms of any solution, we may assume f∗0 J and I areequal. Now if we look back to the proof of Theorem B.16.1, we notice that the only modificationto make is that the obstruction to existence of a solution to (FAD) is

o(f0) := ∂(id) ∈ Ext2OX0

(LX0/S0, f∗0 J)

with id ∈ HomOX0(f∗0 J, f

∗0 J).

Let us restate the theorem, for completeness:Theorem B.17.2 There exists an element

o(f0) ∈ Ext2OX0

(LX0/S0, f∗0 J)

whose vanishing is necessary and sufficient for the existence of a solution to the problem (FAD)as stated above, that is, a flat morphism f of S-schemes which is a deformation of the flatmorphism f0 of S0-schemes. If o(f0) vanishes, the set of isomorphism classes of the solutionsto (FAD) is a torsor under Ext1

OX0(LX0/S0

, f∗0 J), and the automorphism group of any suchsolution is canonically isomorphic to HomOX0

(Ω1X0/S0

, f∗0 J).Example B.17.3 We observe that Theorem B.17.2 can be rephrased into the following:Theorem B.17.4 Let A → A0 a surjection of rings with square-zero kernel I, and considerthe problem of lifting a flat A0-algebra B0 to a flat A-algebra B such that B⊗AA0 ' B0. Thereexists an element

o ∈ Ext2B0

(LB0/A0, I ⊗A0 B0)

whose vanishing is necessary and sufficient for the existence of a solution to the problem asstated above. If o vanishes, the set of isomorphism classes of such solutions is a torsor underExt1

B0(LB0/A0

, I ⊗A0B0), and the automorphism group of any such solution is canonically

isomorphic to HomB0(Ω1

B0/A0, I ⊗A0

B0).

Using the vanishing criterion of , we know that given any perfect Fp-algebra R, then LR/Fp ' 0in D(R) (perfectness implies the vanishing of the complex). We can choose A0 := Fp, A1 =Z/p2Z, the map A1 → A0 being reduction modulo p, and we let B0 := R. Certainly R isFp-flat, and the Theorem tells us that there exists a unique lift of R/Fp to a flat A1-algebraB1. Note that LR/Fp is trivial termwise, and not just in D(R)!

By flat base change we have that aciclicity of LBi/Ai implies that of LBi+1/Ai+1.

We therefore have, in particular, LBi/Ai ' 0 in D(Bi) for all i, as we can run an induction, andwe get a projective system (Ai → Bi) of flat algebras, where Ai := Z/pi+1Z and Bi⊗AiAi−1 'Bi−1, thus yelding in particular Bi ⊗A1

A0 ' B0. Letting B := lim←−Bi and A := lim←−Ai = Zp,we get that B is p-adically complete, unique (due to uniqueness of liftings at each finite step)and flat.

For flatness, we can use the local criterion: we have B/p ' R, which is A/p = Fp-flat, andmoreover TorA1 (B,A/p) = 0 because the map A→ A given by multiplication by p stays injectiveon B.

The limit B is nothing but the Witt vectors W (R) of R. The known properties of W (R) nowcome for free. We shall expand on this example later.

As remarked before, Theorem B.17.2 applies to the special case of Fp-algebras. More precisely,any Fp-algebra A is clearly Fp-flat, and if A satisfies LA/Fp ' 0 in D(A), then there existsa unique (up to unique isomorphism) Z/p2Z-deformation A′ of A which is Z/p2Z-flat. If it


happens that the cotangent complex of A′ over Z/p2Z is also trivial, and iteratively so is thecotangent complex of each finite layer deformation, then we can recognize W (A) to be theunique flat p-adically complete deformation of A, as being the inverse limit of all finite layerdeformations.

If A is perfect, then Proposition 3.5.5 ensures the successive vanishing property is met. Let’sfix ideas on an example.Example B.17.5 LetA := Fp[x, y]/(xy), andA1 := (Z/p2)[x, y]/(p−xy), A2 := (Z/p2Z)[x, y]/(xy).Both A1 and A2 reduce to A modulo p, and the reason is that LA/Fp is nontrivial. In fact, thequotient map Fp[x, y] A is a complete intersection, and its cotangent complex is perfect andconcentrated in degree −1. The distinguished triangle for the maps Fp → Fp[x, y]→ A reads:

Ω1Fp[x,y]/Fp

[0]→ LA/Fp → (xy)/(xy)2[−1]

whence LA/Fp is nontrivial and concentrated in degrees−1, 0. It follows that Ext2A(LA/Fp ,M) =

0 for all A-modulesM , and hence indeed we have unobstructed deformations, whence existenceof A1 and A2. Nevertheless, we have no uniqueness, though we can control the discrepancybetween any two solutions to the deformation problem at hand via the Ext1

A(LB/A, (p)/(p)2)-

action, and the automorphism group HomA(LA/Fp , (p)/(p)2) of any such solution.

B.18 Embedded deformations

We now consider the following diagram:

X0

h0

f0

i // X

f

h

S0j//

g0

S

g

T0k // T

where i, j, k are square-zero closed embeddings with ideals I, J and K respectively.

The situation we have in mind is rather different now. We fix X0i−→ X, and wonder if f exists,

making the above diagram commute. Roughly, we do not have the freedom of choosing i, thatis, of choosing an isomorphism of f−1

0 OS0-modules I ' ker(OX OX0

). We fix it in advance,

and wonder if there exists an f which realizes (OX , Ii#−→ ker(OX OX0

)) as an object of thegroupoid ExalS(X0, I). It suffices to check this on isomorphism classes of ExalS(X0, I).

We call such an f an embedded deformation of f0, and the problem encoded into the followingdiagram the (ED) problem, for simplicity:

X0

h0

f0

i // X

f

h

S0j//

g0

S

g

T0k // T


Theorem B.18.1 There exists an obstruction class

o(i, j, u, f0) ∈ Ext1OX0

(f∗0 LS0/T0, I)

whose vanishing is necessary and sufficient for the problem (ED) to have a solution f . Whensuch obstruction vanishes, the set of all solutions to (ED) is a torsor under the group HomOX0

(f∗0 LS0/T0, I).

Note that the solutions to (ED) have no automorphisms! This is due to the fact that theforgetful functors Sch/S0 → Sch/T0 and Sch/S → Sch/T are faithful.

Proof. We rephrase the problem more conveniently, to make clear the fact that it is actuallya problem of algebra extensions. We give names: let us call A0 := OT0

, A := OT , B0 := OS0,

B := OS , C0 := OX0 and C := OX . We can view them all as sheaves of rings on the Zariskitopos on Spec(Z), say. We form the diagram of morphisms of algebra extensions:

ξC 0 // I // C // C0// 0

ξB

OO

0 // J

v

OO

// B

f

OO

// B0

f0

OO

// 0

ξA

OO

0 // K

EE

u

OO

// A

EE

g

OO

// A0

EE

g0

OO

// 0

where we use again f0, f, g0, g with abuse of notation. Let us call, likewise, h the map A→ Cand h0 the map A0 → C0, where h0 = f0 g0 by assumption. Also, call w : K → I thecomposition v u. As before we observe we have a natural map

K ⊗A0 B0 → J.

Our problem becomes finding an arrow f : B → C completing the above diagram, that is,finding a morphism of extensions ξB → ξC . We denote by

εB ∈ Ext1B0

(LB0/A, J), εC ∈ Ext1C0

(LC0/A, I)

the elements corresponding to the isomorphism classes of extensions ξB ∈ ExalA(B0, J) andξC ∈ ExalA(C0, I) via the group isomorphisms

Ext1B0

(LB0/A, J) ' ExalA(B0, J), Ext1C0

(LC0/A, I) ' ExalA(C0, I).

We also call v∗ the natural map

Ext1B0

(LB0/A, J)→ Ext1B0

(LB0/A, I)∼−→ Ext1

C0(LB0/A ⊗B0 C0, I)

where the last isomorphism is given by the usual extension-restriction adjunction, and f∗0 thenatural map

Ext1C0

(LC0/A, I)→ Ext1C0

(LB0/A ⊗B0 C0, I).

Let us explain these two maps a bit more. We can apply the morphism of Picard groupoids

v∗ : ExalA(B0, J)→ ExalA(B0, I)

which is given by pushing out an A-extension of B0 by J along v : J → I, and we can alsoapply the morphism of Picard groupoids

f∗0 : ExalA(C0, I)→ ExalA(B0, I)


which is simply given by pullback along f0 of A-extensions of C0 by I. Bearing in mindour initial diagram, in order to find a solution to (ED) it is necessary and sufficient that theextensions

ξB ∈ ExalA(B0, J) and ξC ∈ ExalA(C0, I)

land in the same isomorphism class in ExalA(B0, I), that is:

v∗(ξB) ' f∗0 (ξC).

Equivalently, their difference in ExalA(B0, I) is isomorphic to the trivial extension of B0 by I,that is:

v∗(εB)− f∗0 (εC) = 0

in ExalA(B0, I) ' Ext1B0

(LB0/A, I). We want to cook up a morphism ξB → ξC from triviality ofthe isomorphism class v∗(εB)− f∗0 (εC) = 0 in Ext1

B0(LB0/A, I), and this is done by considering

the fundamental distinguished triangle for the maps of sheaves of rings A → A0 → B0. Wehave:

0→ Ext1B0

(LB0/A0, I)→ Ext1

B0(LB0/A, I)→ Ext1

B0(LA0/A ⊗A0

B0, I)

or equivalently:

0→ ExalA0(B0, I)→ ExalA(B0, I)→ HomB0(K ⊗A0

B0, I)

and now we must use our assumptions from (ED): we have that the morphism K → I ofA0-modules induced respectively by the A-extensions of B0 by I given by v∗(ξB) and f∗0 (ξC)is exactly w. Indeed, consider first v∗(ξB). This extension is just the pushout of ξB alongv : J → I, which is

0→ I → I ⊕v B → B0 → 0

Certainly the morphism ξA → v∗(ξB) in ExalA, the category of A-extensions, yelds a morphismof A0-modules K → I which by design is just the composition of u and v, which is w. Likewise,f∗0 (ξC) is just the pullback of ξC along f0, which is

0→ I → C ×C0,f0 B0 → B0 → 0

where I remains untouched. Therefore, once again, the morphism ξA → f∗0 (ξC) in ExalA yeldsa map of A0-modules K → I which is the composition of u and v, that is, w. Given that

HomB0(K ⊗A0

B0, I) ' ExalA(A0, I)

the isomorphism being given by the composition of the canonical extension-restriction adjunc-tion isomorphism HomB0

(K⊗A0B0, I) ' HomA0

(K, I) and the map sending µ ∈ HomA0(K, I)

to µ∗(ξA), we see that the last map

ExalA(B0, I)→ HomB0(K ⊗A0 B0, I)

is nothing but pullback along g0. We observe that g∗0v∗(ξB) is

0→ I → (I ⊕v B)×B0,g0 A0 → A0 → 0

and that g∗0f∗0 (ξC) = h∗0(ξC) is

0→ I → C ×C0,h0 A0 → A0 → 0

A diagram chase shows that (I ⊕v B) ×B0,g0 A0 ' C ×C0,h0A0 as A-algebras, and hence

h∗0(ξC) ' g∗0v∗(ξB) in ExalA(A0, I), since the maps induced on A0 and I are the identity. This


means that the image of v∗(ξB)− f∗0 (ξC) under the map ExalA(B0, I)→ HomB0(K ⊗A0

B0, I)is zero, meaning that

o := v∗(εB)− f∗0 (εC) ∈ Ext1B0

(LB0/A0, I) ' Ext1

C0(LB0/A0

⊗B0 C0, I).

As we already observed, a solution to (ED) exists if and only if o = 0, and in this case it isclear that the set of such solutions is a DerA0

(B0, I)-torsor. Given that

DerA0(B0, I) ' HomB0

(LB0/A0, I) ' HomC0

(LB0/A0⊗B0

C0, I),

we conclude.

B.19 Deformations of almost algebras

We fix a commutative ring V which contains an ideal m such that m2 = m.

We begin with some preliminaries on almost mathematics. First of all recall that the functor:

(·)∗ : V a-Mod→ V -Mod

is right adjoint to the localization functor (·)a, while the functor:

(·)! : V a-Mod→ V -Mod

assigned by (·)! := (·)∗ ⊗V m, is left adjoint to (·)a.

We fix A be a V a-algebra. Let B be any A-algebra. Multiplication on B∗ is inherited by B!,which is therefore made into a non-unital ring. We first cook up a unital ring out of B!.

Let’s endow the V -module V ⊕B! with the ring structure determined by:

(v, b) · (v′, b′) := (vv′, vb′ + v′b+ bb′), v, v′ ∈ V and b, b′ ∈ B!.

It’s easy to check that V ⊕B! is a unital ring, the 1 being given by (1, 0). We call m := m⊗V m,and we now assume m is a flat V -module. This implies, in particular, that m ' m as V -modules,though it will be convenient to make use of m.

The V -submodule of V ⊕B! generated by the elements of the form (xy,−x⊗y⊗1) for arbitraryx, y ∈ m, forms an ideal of V ⊕B! which is isomorphic to m.

We define B!! to be the quotient (V ⊕B!)/m, which fits into the following short exact sequenceof V -modules:

0→ m→ V ⊕B! → B!! → 0

We briefly discuss the following:Proposition B.19.1 The functor (·)!! : A-Alg → A!!-Alg is left adjoint to the localizationfunctor (·)a : A!!-Alg→ A-Alg.

Proof. Let B be an A-algebra, C an A!!-algebra, and f : B → Ca be a morphism of A-algebras.Since (·)∗ is left adjoint to localization, we obtain a natural A∗-linear map B! → C. Togetherwith the structure map V → C, it yields a map f : V ⊕ B! → C which is easily seen torespect the ring structures. f kills the ideal m in V ⊕B!, and hence it factors through a map ofA!!-algebras B!! → C. Conversely, such a map induces a morphism of A-algebras by applying(·)a. It is easily checked that these two operations are inverse to each other, which achieves theProposition.


We begin with the following Lemma on algebra extensions:Lemma B.19.2 Let B be an A-algebra, and M a B-module. The functor:

ExalA!!(B!!,M!)→ ExalA(B,M)

assigned by applying (·)a, is an equivalence of categories. Such equivalence induces a naturalgroup isomorphism ExalA!!

(B!!,M!)'−→ ExalA(B,M).

Proof. The last assertion is an immediate consequence of the first. To show this, we set:

ξ := (0→M → E → B → 0)

for any object of ExalA(B,M). Since m is V -flat, the sequence:

ξ! := (0→M! → E!! → B!! → 0)

is exact (check!). This assignement is a quasi-inverse to the functor (·)a : ExalA!!(B!!,M!) →

ExalA(B,M).

The above Lemma is at the heart of our study of deformations of almost algebras. As a firstpiece of info we can drow out of it, the solution to any of the deformation problems discussed inthe previous sections and rephrased for almost algebras will be found in some goup of extensionsof almost algebras, and the above Lemma B.19.2 relates such groups to groups of extensions ofclassical algebras by classical modules.

The main Theorem of this section is the following:Theorem B.19.3 Let I ⊂ A a square-zero ideal. Let A0 be A/I, and B0 an A0-algebra. Thereexists an obstruction class:

o ∈ Ext2B0!!

(LB0!!/A0!!, B0!! ⊗A0!!

I!)

which vanishes precisely when there exists an A-algebra B such that B ⊗A A0 ' B0. If thereexists such a deformation, then the set of all isomorphism classes of such deformations formsa torsor under Ext1

B0!!(LB0!!/A0!!

, B0!!⊗A0!!I!), and every deformation has automorphism group

isomorphic to HomB0!!(LB0!!/A0!!

, B0!! ⊗A0!!I!).

Proof. Since I is square-zero, so is I! as an ideal of A!!. Moreover, A0!! ' A!!/I!. By TheoremB.16.1, there exists an obstruction class:

o ∈ Ext2B0!!

(LB0!!/A0!!, B0!! ⊗A0!!

I!)

to existence of an A!!-algebra deformation D of B0!!. If there exists such a deformation, thenthe set of all isomorphism classes of all such deformations forms a torsor under

Ext1B0!!

(LB0!!/A0!!, B0!! ⊗A0!!

I!),

and every deformation has automorphism group isomorphic to

HomB0!!(LB0!!/A0!!

, B0!! ⊗A0!!I!).

Recall that the key in showing the above lies in the following exact sequence, and the naturalidentification of its terms as the hypercohomology groups of the complex

RHomB0!!(LB0!!/A0!!

, B0!! ⊗A0!!I!) :


HomB0!!(Ω1

B0!!/A0!!, B0!! ⊗A0!!

I!)→ HomB0!!(Ω1

B0!!/A!!, B0!! ⊗A0!!

I!)→ 0→

→ ExalA0!!(B0!!, I!)→ ExalA!!

(B0!!, I!)→ HomA0!!(I!, I!)

∂−→ Ext2B0!!

(LB0!!/A0!!, B0!! ⊗A0!!

I!).

Da solves the deformation problem of finding a square-zero A-deformation of B0 as an A0-algebra. Indeed, D yields an element ξD ∈ ExalA!!

(B0!!, I!), and Lemma B.19.2 does the job.The equivalence of algebra extensions groupoids of Lemma B.19.2 reduces the task of showingthat the isomorphism classes of deformations of A0-algebras are an Ext1

B0!!(LB0!!/A0!!

, B0!!⊗A0!!

I!)-torsor (provided any such cass exists) to the task of showing that the isomorphism classesof deformations of A0!!-algebras are an Ext1

B0!!(LB0!!/A0!!

, B0!! ⊗A0!!I!)-torsor, which is true.

Likewise for the statement concerning the automorphism group of any such deformation. Weconclude.

For applications, we need the following:Lemma B.19.4 Let A→ B be a V a-algebra homomorphism. Then the natural map:

m⊗V LB!!/A!!→ LB!!/A!!

is a quasi-isomorphism in D(B!!).

Proof. The V -algebra maps V → A!! → B!! yields a distinguished triangle for the cotangentcomplex, which readuces at once the task of checking the Lemma to the case A!! = V . Wedenote P• := PV (B!!) the free simplicial V -algebra which comes with an augmentation map toB!! which yields a quasi-isomorphism between the respective normalized complexes. We havean exact sequence of simplicial V -modules:

0→ s.m→ s.V ⊕ (P a• )! → (P a• )!! → 0

where s.(·) denotes the constant simplicial module. The augmentation map (P a• )! → (Ba!!)! ' B!

is a quasi-isomorphism and upon tensoring by m we deduce that (P a• )!! → B!! is a quasiisomorphism, whence (P a• )!! → P• is also a quasi-isomorphism. We have:

Pn ' Sym(Fn), n ≥ 0

where Fn is a free V -module. The map (P an )!! → Pn is identified with Sym(m ⊗V Fn) →Sym(Fn), whence:

Ω1Pan!!/V

⊗Pan!!Pn → Ω1

Pn/V

is identified with m ⊗V Ω1Pn/V

→ Ω1Pn/V

. It is an easy consequence that Ω1(Pa• )!!/V

→ Ω1P•/V

yields a quasi-isomorphism between normalized complexes. Since m is V -flat and Ω1P•/V

→Ω1P•/V

⊗P• B!! = LB!!/V is a quasi-isomorphism, we get the desired conclusion.

As a consequence of Lemma B.19.4 and Lemma 3.6.11, if R is a perfectoid ring and A isa perfectoid (R/$)-algebra, then LaA/(R/$)!

is almost quasi-isomorphic to LAa!!/(Ra/$)!! . Itfollows that vanishing of LA/(R/$) implies unobstructedness and uniqueness (up to uniqueisomorphism) of deformations of perfectoid almost algebras as well.Remark B.19.5 Note that Lemma B.19.4 says that LB!!/A!!

actually depends only on A→ B.Upon clarifying what D(B) is, LB!!/A!!

literally is an object of D(B), and one can develop thetheory entirely in the setting of almost mathematics.


References

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Department of Mathematics, Stanford University, Building 380, Sloan Hall, Stanford,CA 94305 USA

E-mail address: [email protected]

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