## I.10: Almost mathematics (II)

I continue by introducing standard notions from commutative algebra to the context of almost mathematics. For the references to the proof of the following definition/proposition, see Proposition 4.7 in Scholze’s “Perfectoid spaces”‘. I follow closely the exposition in the Scholze’s paper in this blog post (consult it for all references).

Definition 42 Let ${A}$ be a ${K^{\circ a}}$-algebra

1. An ${A}$-module ${M}$ is flat if the functor ${X \mapsto M \otimes _A X}$ on ${A}$ is exact. If ${R}$ is a ${K^{\circ}}$-algebra and ${N}$ is an ${R}$-module, then ${R^a}$-module ${N^a}$ is flat if and only if for all ${R}$-modules ${X}$ and all ${i >0}$, the module ${\mathrm{Tor} ^R _i(N,X)}$ is almost zero.
2. An ${A}$-module ${M}$ is almost projective if the functor ${X \mapsto al\mathrm{Hom} _A(M,X)}$ on ${A}$-modules is exat. If ${R}$ is a ${K^{\circ}}$-algebra and ${N}$ is an ${R}$-module, then ${N^a}$ is almost projective over ${R^a}$ if and only if for all ${R}$-modules ${X}$ and all ${i >0}$, the module ${\mathrm{Ext} ^R _i(N,X)}$ is almost zero.
3. If ${R}$ is a ${K^{\circ}}$-algebra and ${N}$ is an ${R}$-module then we say ${M=N^a}$ is almost finitely generated (resp. almost finitely presented) ${R^a}$-module if and only if for all ${\epsilon \in \mathfrak{m}}$, there is some finitely generated (resp. finitely presented) ${R}$-module ${N_{\epsilon}}$ with a map ${f _{\epsilon}: N_{\epsilon} \rightarrow N_{\epsilon}}$ such that the kernel and cokernel of ${f_{\epsilon}}$ are annigilated by ${\epsilon}$. We say ${M}$ is uniformly almost finitely generated if there is some integer ${n}$ such that ${N_{\epsilon}}$ can be chosen to be generated by ${n}$ elements for all ${\epsilon}$.

Proposition 43 Let ${A}$ be a ${K^{\circ a}}$-algebra. Then an ${A}$-module ${M}$ is flat and almost finitely presented if and only if it is almost projective and almost finitely generated.

We shall call such ${A}$-modules ${M}$ finite projective. If moreover ${M}$ is uniformly almost finitely generated, we say ${M}$ is uniformly finite projective. For such modules we have a good notion of rank.

Theorem 44 Let ${A}$ be a ${K^{\circ a}}$-algebra and let ${M}$ be a uniformly finite projective ${A}$-module. Then there is a unique decomposition ${A = A_0 \times A_1 \times ... \times A_k}$ such that for each ${i=0,...,k}$ the ${A_i}$-module ${M_i = M \otimes _A A_i}$ has the property that ${\wedge ^i M_i}$ is invertible, and ${\wedge ^{i+1} M_i = 0}$. Here, we call ${A}$-module ${L}$ invertible if ${L \otimes _A al\mathrm{Hom} _A(L,A)= A}$.

We introduce the notion of étale morphisms

Definition 45 Let ${A}$ be a ${K^{\circ a}}$-algebra and let ${B}$ be an ${A}$-algebra. Let ${\mu : B\otimes _A B \rightarrow B}$ denote the multiplication morphism.

(1) The morphism ${A \rightarrow B}$ is unramified if there is an element ${e \in (B\otimes _A B)_{*}}$ such that ${e^2 = e}$, ${\mu(e) = 1}$ and ${xe =0}$ for all ${x\in \ker (\mu)_{*}}$.

(2) The morphism ${A \rightarrow B}$ is étale if it is unramified and ${B}$ is a flat ${A}$-module.

Definition 46 A morphism ${A \rightarrow B}$ of ${K^{\circ a}}$-algebras is finite étale if it is étale and ${B}$ is an almost finitely presented ${A}$-module. We write ${A_{fet}}$ for the category of finite étale ${A}$-algebras.

Observe that any finite étale ${A}$-algebra is also finite projective ${A}$-module. There is an equivalent characterization of finite étale morphisms in terms of trace morphisms. For a ${K^{\circ a}}$-algebra ${A}$ and a finite projective ${A}$-module ${P}$ we define ${P^{*} = al\mathrm{Hom}(P,A)}$ which is also a finite projective ${A}$-module. Moreover ${P{**} \simeq P}$ canonically and there is an isomorphism

$\displaystyle \mathrm{End}(P)^a = P \otimes _A P^{*}$

In particular, we get a trace morphism ${\mathrm{tr} _{P/A} : \mathrm{End}(P)^a \rightarrow A}$

Definition 47 Let ${A}$ be a ${K^{\circ a}}$-algebra and let ${B}$ be an ${A}$-algebra such that ${B}$ is a finite projective ${A}$-module. We define the trace form as the bilinear form

$\displaystyle t _{B/A} : B\otimes _A B \rightarrow A$

given by the composition of ${\mu : B \otimes _A B \rightarrow B}$ and the map ${B \rightarrow A}$ sending any ${b\in B}$ to the trace of the endomorphism ${b' \mapsto bb'}$ of ${B}$.

We remark that the last map is defined on ${(.)_{*}}$ as we cannot talk about elements ${b}$ of an almost object ${B}$. That is, we construct a map ${B _{*} \rightarrow \mathrm{End} _{A_{*}}(B_{*})^a}$ in the same way and then pass to the almost setting by taking ${(.)^a}$.

Theorem 48 If ${A,B}$ are as in the definition above, then ${A \rightarrow B}$ is finite étale if and only if the trace map is a perfect pairing, that is it induces an isomorphism ${B \simeq B^{*}}$.

We finish by saying that finite étale covers lift uniquely over nilpotents (recall ${\varpi}$ was a uniformizer of our base field):

Theorem 49 Let ${A}$ be a ${K^{\circ a}}$-algebra. Assume that ${A}$ is flat over ${K^{\circ a}}$ and ${\varpi}$-adically complete, that is

$\displaystyle A \simeq \varprojlim _n A/\varpi ^n$

Then the functor ${B \mapsto B \otimes _A A/\varpi}$ induces an equivalence of categories ${A _{fet} \simeq (A/\varpi)_{fet}}$. Any ${B \in A_{fet}}$ is again flat over ${K^{\circ a}}$ and ${\varpi}$-adically complete. Moreover ${B}$ is a uniformly finite projective ${A}$-module if and only if ${B \otimes _A A/\varpi}$ is a uniformly finite projective ${A/\varpi}$-module.

## I.9: Almost mathematics (I)

I describe basics of almost mathematics which is useful in many ways in applications. Often certain maps are “almost isomorphisms”‘ rather that isomorphisms. Let ${K}$ be a perfectoid field, let ${\mathfrak{m} = K^{\circ \circ} \subset K^{\circ}}$ be the subset of topologically nilpotent elements. It is also the unique maximal ideal of ${K^{\circ}}$ and can be identified with the set ${\{x\in K | |x| <1\}}$. The idea behind almost mathematics is to do things up to ${\mathfrak{m}}$-torsion. I will follow closely Scholze’s [Perfectoid spaces] in exposition, which in turn cites mostly Gabber-Ramero’s book – for precise references consult Section 4 in [Perfectoid spaces].

Definition 38 Let ${M}$ be a ${K^{\circ}}$-module. An element ${x \in M}$ is almost zero if ${\mathfrak{m} x = 0}$. The module ${M}$ is almost zero if all of its elements are almost zero or equivalently ${\mathfrak{m} M = 0}$.

One can prove that the full subcategory of almost zero objects in the category $K^{\circ}-\mod$ of ${K^{\circ}}$-modules is thick, i.e. closed under extensions (and triangulated).

Define the category of almost ${K^{\circ}}$-modules as $K^{\circ a}-\mod = K^{\circ}-\mod$ / (${\mathfrak{m}}$-torsion). We have a localization functor ${M \mapsto M^a}$ from $K^{\circ }-\mod$ to $K^{\circ a}-\mod$, with kernel being exactly the category of almost zero modules. In this way we have defined both objects and morphisms. A result of Gabber-Ramero yields:

Proposition 39 Let ${M,N}$ be two ${K^{\circ}}$-modules. Then

$\displaystyle \mathrm{Hom} _{K^{\circ a}}(M^a,N^a) = \mathrm{Hom} _{K^{\circ }}(\mathfrak{m} \otimes M,N)$

In particular ${\mathrm{Hom} _{K^{\circ a}}(X,Y)}$ has a natural structure of ${K^{\circ}}$-module for any two ${K^{\circ a}}$-modules ${X,Y}$. The module ${\mathrm{Hom} _{K^{\circ a}}(X,Y)}$ has no almost zero elements.

For two ${K^{\circ a}}$-modules ${M,N}$ we define ${al\mathrm{Hom}(X,Y) = \mathrm{Hom}(X,Y) ^a}$. We then have

Proposition 40 The category $K^{\circ a}-\mod$ is an abelian tensor category, where we define kernels, cokernels and tensor products in the unique way compatible with their definition in $K^{\circ}-\mod$, that is

$\displaystyle M^{a} \otimes N^a = (M \otimes N)^a$

for any two ${K^{\circ}}$-modules ${M,N}$. For any $L,M,N \in K^{\circ a}-\mod$ there is a functorial isomorphism

$\displaystyle \mathrm{Hom}(L, al\mathrm{Hom}(M,N))= \mathrm{Hom}(L\otimes M,N)$

This means that $K^{\circ a} -\mod$ has all properties of the category of modules over a ring and thus one can define the notion of ${K^{\circ a}}$-algebra. Any ${K^{\circ}}$-algebra ${R}$ defines a ${K^{\circ a}}$-algebra ${R^{a}}$ as the tensor products are compatible (multiplication). Localisation extents to a functor from ${R}$-modules to ${R^a}$-modules.

Proposition 41 There is a right adjoint

$\displaystyle K^{\circ a}-\mod \mapsto K^{\circ}-\mod : M\mapsto M_*$

to the localization functor ${M\mapsto M^a}$ given by the functor of almost elements

$\displaystyle M_* = \mathrm{Hom} _{K^{\circ a}}(K^{\circ a},M)$

The adjunction morphism ${(M_*)^a \mapsto M}$ is an isomorphism. If ${M}$ is a ${K^{\circ}}$-module, then ${(M^a)_* = \mathrm{Hom} (\mathfrak{m},M)}$.

If ${A}$ is a ${K^{\circ a}}$-algebra, then ${A_*}$ has a natural structure as ${K^{\circ}}$-algebra and ${A^a _* = A}$. In particular, any ${K^{\circ a}}$-algebra comes via localization from a ${K^{\circ}}$-algebra. Furthermore the functor ${M\mapsto M_*}$ induces a functor from ${A}$-modules to ${A_*}$-modules, and one can see that also all ${A}$-modules come via localization from ${A_*}$-modules. The category of ${A}$-modules is again an abelian tensor category, and all properties about the category of ${K^{\circ a}}$-modules stay true for the category of ${A}$-modules. Observe that we could equivalently define ${A}$-algebras as algebras over the category of ${A}$-modules, or as ${K^{\circ a}}$-algebras ${B}$ with an algebra morphism ${A \rightarrow B}$.

## I.8: Tilting

Tilting is the key construction for perfectoid rings which allows to naturally associate a perfectoid (perfect) ring in characteristic ${p}$ to a perfectoid ring of any characteristic (in particular, ${0}$). Thus, some of the geometrical problems in characteristic ${0}$ might be translated to characteristic ${p}$, solved there and the result can be translated again to characteristic ${0}$. This was done with the weight-monodromy conjecture by Peter Scholze. The following blog post (as was the previous) is based mostly on notes of Jared Weinstein from Scholze’s lectures at Berkeley.

Let ${R}$ be a perfectoid Tate ring. The tilt of ${R}$ is

$\displaystyle R^{\flat} = \varprojlim _{x \mapsto x^p} R$

with the inverse limit topology. A priori this is only a topological multiplicative monoid, but we can endow it with a ring structure, by defining the addition

$\displaystyle (x^{(0)},x^{(1)},...)+(y^{(0)},y^{(1)},...) = (z^{(0)}, z^{(1)},...)$

where

$\displaystyle z^{(i)} = \lim _{n \rightarrow \infty}(x^{(i+n)}+y^{(i+n)})^{p^n} \in R$

We have to check that it is well-defined.

Lemma 33 The limit ${z^{(i)}}$ exists and defines a ring structure, which makes ${R^{\flat}}$ into a topological ${\mathbb{F}_p}$-algebra that is a perfect uniform Tate ring. The subset ${R^{\flat \circ}}$ of power-bounded elements is given by the topological ring isomorphism

$\displaystyle R^{\flat \circ} = \varprojlim _{\Phi} R^{\circ} \simeq \varprojlim _{\Phi} R^{\circ} /p \simeq \varprojlim _{\Phi} R^{\circ} / \varpi$

where ${\varpi \in R}$ is a pseudo-uniformizer which divides ${p}$ in ${R^{\circ}}$. Moreover there exists a pseudo-uniformizer ${\varpi \in R}$ with ${\varpi ^p | p}$ in ${R^{\circ}}$ which admits a sequence of ${p}$-th power roots ${\varpi ^{1/p^{n}}}$, giving rise to an element

$\displaystyle \varpi ^{\flat} = (\varpi, \varpi ^{1/p},...) \in R^{\flat \circ}$

which is a pseudo-uniformizer of ${R^{\flat}}$. We have ${R^{\flat} = R^{\flat \circ}[1/\varpi ^{\flat}]}$.

Proof: By definition ${R^{\flat}}$ is perfect. Let ${\varpi _0}$ be a pseudo-uniformizer of ${R}$. Let us check that the maps

$\displaystyle \varprojlim _{\Phi} R^{\circ} \rightarrow \varprojlim _{\Phi} R^{\circ} /p \rightarrow \varprojlim _{\Phi} R^{\circ} / \varpi _0$

are isomorphisms. The key is that any sequence ${(\bar{x}_0,\bar{x}_1,...) \in \varprojlim _{\Phi} R^{\circ} / \varpi _0}$ lifts uniquely to a sequence ${(x_0,x_1,...) \in \varprojlim _{\Phi} R^{\circ} }$. This is standard: ${x_i = \lim _{n \rightarrow \infty} \tilde{x} _{n+i} ^{p^n}}$, where ${\tilde{x}_j \in R^{\circ}}$ is any lift of ${\bar{x}_j}$. Hence ${R^{\flat \circ}}$ is a well-defined ring.

Now assume ${\varpi _0 ^p |p}$ in ${R^{\circ}}$. We construct ${\varpi ^{\flat}}$. In fact the preimage of ${\varpi _0}$ under ${R^{\flat \circ} = \varprojlim _{\Phi} R^{\circ} / \varpi _0 ^p \rightarrow R^{\circ} / \varpi _0 ^p}$ is an element ${\varpi ^{\flat}}$ with the right properties. It is congruent to ${\varpi _0}$ modulo ${\varpi _0 ^p}$ and therefore it is also a pseudo-uniformizer. Then the projection ${\varpi}$ of ${\varpi ^{\flat}}$ to the ${0}$-th coordinate is the desired pseudo-uniformizer of ${R^{\circ}}$. $\Box$

As in the last line of the proof above, we have a continuous multiplicative (but not additive) map

$\displaystyle R^{\flat} = \varprojlim _{\Phi} R \rightarrow R$

which projects onto the ${0}$-th coordinate. We denote it by ${f \mapsto f^{\sharp}}$. This projection defines a ring isomorphism ${R^{\flat \circ} / \varpi ^{\flat} \simeq R^{\circ} / \varpi}$. Let us note the following fact which follows from this isomorphism, the nilpotence conditions and a fact that an open subring is integrally closed in ${R^{\circ}}$ (resp. ${R^{\circ \flat}}$) if it its image in the quotient ${R^{\circ} / \varpi}$ (resp. ${R^{\flat \circ} / \varpi ^{\flat}}$) is integrally closed.

Lemma 34 The set of rings of integral elements ${R ^+ \subset R^{\circ}}$ is in bijection with the set of rings of integral elements ${R^{\flat +} \subset R^{\flat \circ}}$ by ${R^{\flat +} = \varprojlim _{x \mapsto x^p} R^+}$. Also ${R ^{\flat +} / \varpi ^{\flat} = R^{+} / \varpi}$.

We say that a Huber pair ${(R,R^+)}$ is perfectoid if ${R}$ is perfectoid. We have the following theorem due to Scholze (see also Kedlaya-Liu’s work)

Proposition 35 Let ${(R,R^+)}$ be a perfectoid Huber pair. Then for all rational subsets ${U \subset X = \mathrm{Spa} (R,R^+)}$, the ${\mathcal{O}_X(U)}$ is again perfectoid. In particular ${(R,R^+)}$ is stably uniform, hence sheafy (by Buzzard-Verberkmoes, as explained earlier).

The following two fundamental theorems are due to Scholze. We leave the without proof. A reader should consult Scholze’s “Perfectoid spaces”‘.

Theorem 36 Let ${(R,R^+)}$ be a perfectoid Huber pair with tilt ${(R^{\flat}, R^{\flat +})}$. There is a homeomorphism

$\displaystyle \mathrm{Spa} (R,R^+) \simeq \mathrm{Spa} (R^{\flat}, R^{\flat +})$

which sends ${x}$ to ${x ^{\flat}}$ such that that ${|f(x^{\flat})| = |f^{\sharp}(x)|}$. This homeomorphism preserves rational subsets.

Theorem 37 Let ${R}$ be a perfectoid ring with tilt ${R^{\flat}}$. Then there is an equivalence of categories between perfectoid ${R}$ algebras and perfectoid ${R^{\flat}}$-algebras via ${S \mapsto S^{\flat}}$.

## I.7: Perfectoid rings

An important subclass of adic spaces are perfectoid spaces which form a basis for the pro-etale topology of locally noetherian adic spaces (Proposition 4.8 in Scholze’s “p-adic Hodge theory for rigid-analytic varieties”). Perfectoid spaces are glued from perfectoid algebras, hence I will start by explaining a notion of perfectoid rings. The word “perfectoid”‘ refers to their “almost perfect structure”‘ – the Frobenius map is surjective and there is a “tilt construction”‘ which allows for a passage between algebras in characteristic ${0}$ and characteristic ${p}$.

We recall that a ring ${R}$ is uniform if its subring of power-bounded elements ${R ^{\circ} \subset R}$ is bounded.

Definition 27 A complete Tate ring ${R}$ is perfectoid if ${R}$ is uniform and there exists a pseudo-uniformizer ${\varpi \in R}$ such that ${\varpi ^p | p}$ in ${R^{\circ}}$ and such that the ${p}$-th power Frobenius map

$\displaystyle \Phi: R ^{\circ} / \varpi \rightarrow R^{\circ} / \varpi ^{p}$

is an isomorphism.

In fact, for any complete Tate ring ${R}$ and ${\varpi}$ as above ${\Phi: R ^{\circ} / \varpi \rightarrow R^{\circ} / \varpi ^{p}}$ is necessarily injective. Indeed, if ${x \in R^{\circ}}$ satisfies ${x^p = \varpi ^p y}$ for some ${y \in R^{\circ}}$ then the element ${x/ \varpi \in R}$ lies in ${R^{\circ}}$ since its ${p}$-th power does. Hence, we really demand surjectivity in the definition of a perfectoid ring. In fact, the surjectivity is equivalent to the surjectivity of the Frobenius map

$\displaystyle R^{\circ} / (p, \varpi ^n) \rightarrow R^{\circ} / (p, \varpi ^{np})$

for any ${n \geq 1}$. The statement follows by induction from ${n=1}$.

In what follows we denote by ${\Phi}$ any map ${R /I \rightarrow R / J }$ induced by ${x \mapsto x^{p}}$ where ${R}$ is a ring and ${I,J \subset R}$ are two ideals containing ${p}$ such that ${I^p \subset J}$.

Remark 28 Discretely valued non-archimedean fields ${K}$ of residue characteristic ${p}$ are not perfectoid (hence, for example ${\mathbb{Q}_p}$). Indeed, ${K^{\circ}/\varpi}$ and ${K^{\circ} / \varpi ^{p}}$ are Artin local rings of different lengths and hence cannot be isomorphic.

Standard examples of perfectoid rings include:

1. ${\mathbb{Q}_p ^{cyc}}$ – the completion of ${\mathbb{Q}_p(\mu _{p^{\infty}})}$.
2. The ${t}$-adic completion of ${\mathbb{F}_p((t))(t^{1/p^{\infty}})}$, which we denote by ${\mathbb{F}_p((t^{1/p^{\infty}}))}$.
3. ${\mathbb{Q}_p ^{cyc}\langle T^{1/p^{\infty}} \rangle = (\mathbb{Z}_p ^{cyc}[ T^{1/p^{\infty}}])^{\wedge}[1/p]}$, where the completion is ${p}$-adic.

A natural question is whether one can give a more general definition of perfectoid Huber rings which do not have to be Tate. So that for example ${\mathbb{Z} _p ^{cyc}[[T^{1/p^{\infty}}]]}$, the ${(p,T)}$-adic completion of ${\mathbb{Z}_p ^{cyc}[T^{1/p^{\infty}}]}$, would be perfectoid.

Proposition 29

Let ${R}$ be a topological ring with ${pR =0}$. The following are equivalent:

1. ${R}$ is perfectoid.
2. ${R}$ is a perfect uniform complete Tate ring.

Here, perfect means that the Frobenius ${\Phi: R \rightarrow R}$ is an isomorphism.

Proof: If ${R}$ is perfectoid then ${\Phi : R ^{\circ} / \varpi \rightarrow R^{\circ} / \varpi ^{p}}$ is an isomorphism and hence also ${R ^{\circ} / \varpi ^n \rightarrow R^{\circ} / \varpi ^{np}}$ by induction. We conclude by taking inverse limit, using completeness and inverting ${\varpi}$.

If ${R}$ is perfect, then ${\varpi ^p |p=0}$ is satisfied automatically. If ${x \in R^{\circ}}$ then ${x ^p \in R^{\circ}}$ so that ${\Phi : R^{\circ} \rightarrow R^{\circ}}$ is an isomorphism. Thus ${\Phi : R ^{\circ} / \varpi \rightarrow R^{\circ} / \varpi ^{p}}$ is surjective. $\Box$

Definition 30 A perfectoid field is a perfectoid ring which is a non-archimedean field.

Let us note the following criterion

Proposition 31 Let ${K}$ be a non-archimedean field. Then ${K}$ is perfectoid if and only if the following conditions hold:

1. ${K}$ is not discretely valued.
2. ${|p| <1}$
3. ${\Phi: \mathcal{O} _K / p \rightarrow \mathcal{O} _K /p}$ is surjective, where ${\mathcal{O} _K}$ is the ring of integers of ${K}$.

Let us finish with another characterisation of perfectoid rings

Proposition 32 Let ${R}$ be a complete uniform Tate ring.

1. If there exists a pseudo-uniformiser ${\varpi \in R}$ such that ${\varpi ^p |p}$ and the Frobenius map ${\Phi: R ^{\circ} / p \rightarrow R^{\circ} /p}$is surjective, then ${R}$ is a perfectoid ring.
2. Conversely, if ${R}$ is a perfectoid ring then ${\Phi: R^{\circ} /p \rightarrow R^{\circ} /p}$ is surjective under the additional assumption that the ideal ${pR^{\circ} \subset R^{\circ}}$ is closed.

Proof: (1) is easy and (2) comes down to showing an isomorphism

$\displaystyle R^{\circ}/p \rightarrow \varprojlim _n R^{\circ} / (p, \varpi ^n)$

$\Box$

## I.6: Sheafy rings

In the definition of adic spaces, one demands ${\mathcal{O}_X}$ to be a sheaf. Hence it is neccessary to know for which ${X}$ the presheaf ${\mathcal{O}_X}$ is a sheaf. The first criterion is due to Huber (from “`A generalization of formal schemes and rigid-anaytic varieties”‘). Recall that a Huber ring is Tate if it contains a topologically nilpotent unit.

Proposition 23 Let ${(A, A^+)}$ be a Huber pair. Then ${\mathcal{O} _{\mathrm{Spa}(A,A^+)}}$ is a sheaf, and its higher cohomology vanishes on rational subdomains, if either of the following conditions holds:

1. ${\hat{A}}$ has a noetherian ring of definition.
2. ${A}$ is a strongly noetherian Tate ring (i.e. for all ${n}$ the ring ${A\langle t_1,...,t_n \rangle}$ are noetherian).

Huber uses this theorem to show that locally noetherian formal schemes give rise to adic spaces (point (1) ensures that) as well as rigid-analytic varieties (by point (2)). On the other hand Scholze proved that ${\mathcal{O} _{\mathrm{Spa}(A,A^+)}}$ is a sheaf when ${A}$ is perfectoid, in particular highly non-noetherian (we will discuss perfectoid spaces later on). Thus we would like to have a criterion for being sheafy which would apply to both these contexts: noetherian and non-noetherian. Before stating it, let us introduce two definitions. For what follows, we let ${(A,A^+)}$ be a Huber pair with ${A}$ Tate.

Definition 24 The pair ${(A,A^+)}$ is uniform if the open subring ${A^{\circ}}$ of power-bounded elements is bounded.

An example of non-uniform Tate ring is ${\mathbb{Q}_p[\varepsilon]/(\varepsilon ^2)}$ with the ${p}$-adic topology. We have in this case ${A^{\circ} = \mathbb{Z}_p \oplus \mathbb{Q}_p \cdot \varepsilon}$, which is not bounded as it contains a ${\mathbb{Q}_p}$-line.

Definition 25 The pair ${(A,A^+)}$ is stably uniform if for all rational subsets ${U \subset \mathrm{Spa}(A,A^+)}$, the ring ${\mathcal{O}_{\mathrm{Spa}(A,A^+)}(U)}$ is unifom.

There are examples of pairs which are stably uniform, but not uniform. The main result of this section is the theorem of Buzzard and Verberkmoes:

Theorem 26 Let ${(A,A^+)}$ be a stably uniform Huber pair with ${A}$ Tate. Then ${\mathcal{O}_{\mathrm{Spa}(A,A^+)}}$ is a sheaf and its higher sheaf cohomology vanishes on rational domains.

We have followed in this statement the seminar of Conrad (see L13). Observe that there is no finiteness assumption in the theorem which permits to apply it to perfectoid algebras. Let us now give a sketch of the proof of the sheaf property.

There are two steps. Let ${X = \mathrm{Spa} (A, A^+)}$. First, one verifies the sheaf property for the cover of the form ${U = \{ v \in X \ : \ v(t) \geq 1\}}$ and ${V = \{v \in X \ : \ v(t) \leq 1 \}}$, where ${t}$ is any element of ${A}$, only assuming that ${(A,A^+)}$ is uniform. Second, one reduces the general case to this case, adding a stably uniform hypothesis.

In the first step we want to prove that

$\displaystyle 0 \rightarrow \mathcal{O} _X(X) \rightarrow \mathcal{O} _X(U) \oplus \mathcal{O}_X(V) \rightarrow \mathcal{O}_X(U \cap V) \rightarrow 0$

is exact. But before completion, this sequence is

$\displaystyle 0 \rightarrow A \rightarrow ^f A(t/1) \oplus A(1/t) \rightarrow ^g A(t/1,1/t) \rightarrow 0$

where ${f}$ is the localisation map to both factors and ${g}$ is the substraciton of the second factor from the first. It is not hard to establish the exactness of this sequence. In order to conclude, we want to know that when we complete, we still end up with an exact sequence. This is not automatic! In order to do so, we need to prove that both ${f}$ and ${g}$ are strict, i.e. a map of topological abelian groups ${h : B \rightarrow C}$ is strict if the quotient topology on ${h(B)}$ from ${B}$ coincides with the subspace topology from ${C}$. It is equivalent to ${h}$ being continuous and the induced map ${B \rightarrow h(B)}$ being open. In our setting ${g}$ is always strict, but in order to prove that ${f}$ is strict we have to use the uniformity. We leave it to the reader.

The second step, deducing the full theorem, follows the reasoning of Tate. By induction one checks the sheaf property for Laurent covers i.e. ${U_I}$, where ${I}$ is a subset of ${\{1,...,n\}}$, ${t_1,...,t_n \in A}$ and

$\displaystyle U_I = \{v \in X \ : \ v(t_i) \leq 1 \textrm{ for } i \in I, \ v(t_i) \geq 1 \textrm{ for } i \notin I \}$

Then by using a refinement one passes to any rational domains and then to arbitrary covers by rational domains. See § 8.2 of [BGR].

## D.1: Overconvergent modular forms

Together with David Hansen and Christian Johansson we have uploaded our paper “Overconvergent modular forms and perfectoid Shimura curves” to arXiv (http://arxiv.org/abs/1507.04875).

I mention it here as we use adic spaces and perfectoid spaces in a crucial way. We basically re-work the theory of overconvergent modular forms by using the pro-etale site and the prior work of Scholze on the modular curves at the infinite level. This allows us to be explicit – we construct an analogue of the complex coordinate “z” in the p-adic setting and use it to construct sheaves of overconvergent modular forms. We introduce the overconvergent Eichler-Shimura map after Andreatta-Iovita-Stevens and analyse its properties using our description of the sheaves of overconvergent modular forms. We collect some general results about adic spaces in the appendix.

Today I’m writing about an important class of adic spaces, namely analytic adic spaces. An important property is that all specializations on maps between analytic adic spaces are vertical. Analytic adic spaces should provide the correct level of generality for arithmetic considerations. Most of this part is taken from Conrad’s seminar (week 7 and week 6).

Definition 17 Let ${A}$ be a Huber ring. A point ${v \in \mathrm{Cont} (A)}$ is called analytic if its support ${\mathrm{supp}(v) \in \mathrm{Spec}(A)}$ is not an open ideal of ${A}$.

Lemma 18 Let ${A}$ be a Huber ring with a couple of definition ${(A_0,I)}$. A point ${v \in \mathrm{Spv}(A)}$ is non-analytic if and only if ${v(I)=0}$.

Proof: The support ${\mathrm{supp}(v)}$ is open in ${A}$ if and only if it contains ${I^n}$ for sufficiently large ${n}$. But ${\mathrm{supp}(v)}$ is radical so this is equivalent to ${\mathrm{supp}(v)}$ containing ${I}$, that is exactly ${v(I)=0}$. $\Box$

Let us denote by ${\mathrm{Cont}(A)_{\mathrm{an}}}$ the subspace of analytic points in ${\mathrm{Cont}(A)}$ endowed with the subspace topology.

Corollary 19 Let ${A}$ be a Tate ring. Then ${\mathrm{Cont}(A) = \mathrm{Cont}(A)_{\mathrm{an}}}$.

Proof: Let ${u}$ be the topologically nilpotent unit in ${A}$. Then ${v \in \mathrm{Spv}(A)}$ cannot kill ${u}$, even though ${u^n \in I}$ for all sufficiently large ${n}$. This implies ${v(I)\not = 0}$ and hence the result by the previous lemma. $\Box$

Let us now describe the structure of ${\mathrm{Cont}(A)_{\mathrm{an}}}$. We still denote by ${A}$ a Huber ring. Let ${(A_0, I)}$ be its couple of definitions and let ${\{f_1,...,f_n\}}$ be a set of generators of ${I}$. Let ${v \in \mathrm{Cont}(A) _{\mathrm{an}}}$. Then

$\displaystyle \sum ^n _{i=1} A_0 f_i = I \not \subset \mathrm{supp}(v)$

Hence for some ${r}$ we must have ${v(f_{r}) \not = 0}$. We choose ${r}$ for which ${v(f_r)}$ is maximal and hence ${v(f_j) \leq v(f_r) \not = 0}$ for all ${j}$. We can then form ${A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right)}$. There is a natural map ${A \rightarrow A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right)}$ which induces the map on ${\mathrm{Cont}(-)}$:

Proposition 20 The image of the natural continuous map

$\displaystyle \mathrm{Cont} \left(A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right)\right) \rightarrow \mathrm{Cont}(A)$

is the subspace of ${\mathrm{Cont}(A)_{\mathrm{an}}}$ defined by ${\{v \in \mathrm{Cont}(A)_{\mathrm{an}} | \ \ v(f_j) \leq v(f_r) \textrm{ for all j}\}}$.

Using this result we get

Proposition 21 Let ${A}$ be a Huber ring with ${f_1,...,f_n}$ as before. We have

$\displaystyle \mathrm{Cont}(A)_{\mathrm{an}} = \bigcup ^n _{i=1} \mathrm{image} \left(\mathrm{Cont} \left(A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right) \right) \right)$

where

$\displaystyle \mathrm{image} \left(\mathrm{Cont} \left(A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right) \right) \right) = \mathrm{Spv}(A) \cap \mathrm{Spv} \left(A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right)\right)$

I finish by stating a theorem on specializations

Theorem 22 Let ${A}$ be a Huber ring. Then all specializations inside ${\mathrm{Cont}(A)_{\mathrm{an}}}$ are vertical. In particular, if ${A}$ is Tate, then all specializations in ${\mathrm{Cont}(A)}$ are vertical.

Proof: One recalls a structure theorem for specializations which says that we can write each specialization as a vertical specialization of a horizontal specialization. Then it is enough to prove that there are no proper horizontal specializations. This is done for ${v\in \mathrm{Cont}(A)_{\mathrm{an}}}$ by choosing ${H \subset \Gamma _v}$ which contains ${v(A)_{\geq 1}}$. Hence ${v_{| H}}$ is horizontal specialization of ${v}$ and it lies in ${\mathrm{Cont}(A)}$. Assume that ${H\not =\Gamma _v}$ and derive a contradiction by showing that ${\mathrm{supp}(v_{|H})}$ is open, hence ${v}$ could not be analytic. $\Box$