# 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

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

$\Box$