Tilting and untilting for ideals in perfectoid rings

Abstract

For a perfectoid ring R of characteristic 0 with tilt $R^{\flat }$, we introduce and study a tilting map $(-)^{\flat }$ from the set of p-adically closed ideals of R to the set of ideals of $R^{\flat }$ and an untilting map $(-)^{\sharp }$ from the set of radical ideals of $R^{\flat }$ to the set of ideals of R. The untilting map $(-)^{\sharp }$ is defined purely algebraically and generalizes the analytically defined untilting map on closed radical ideals of a perfectoid Tate ring of characteristic p introduced in the first author’s previous work. We prove that the two maps

$$\begin{aligned} J\mapsto J^{\flat }~\text {and}~I\mapsto I^{\sharp } \end{aligned}$$

define an inclusion-preserving bijection between the set of ideals J of R such that the quotient R/J is perfectoid and the set of $p^{\flat }$-adically closed radical ideals of $R^{\flat }$, where $p^{\flat }\in R^{\flat }$ corresponds to a compatible system of p-power roots of a unit multiple of p in R. Finally, we prove that the maps $(-)^{\flat }$ and $(-)^{\sharp }$ send (closed) prime ideals to prime ideals and thus define a homeomorphism between the subspace of ${{\,\textrm{Spec}\,}}(R)$ consisting of prime ideals $\mathfrak {p}$ of R such that $R/\mathfrak {p}$ is perfectoid and the subspace of ${{\,\textrm{Spec}\,}}(R^{\flat })$ consisting of $p^{\flat }$-adically closed prime ideals of $R^{\flat }$. In particular, we obtain a generalization and a new proof of the main result of the first author’s previous work which concerned prime ideals in perfectoid Tate rings.

1 Introduction

Throughout this paper, we fix a prime number p. Since their first appearance in Scholze’s thesis [17], perfectoid algebras and spaces have revolutionized both p-adic geometry and mixed characteristic commutative algebra, leading, among many other applications, to André’s proof of the direct summand conjecture [1, 2]. In the realm of commutative algebra, there is also the notion of (integral) perfectoid rings, introduced in [5, Definition 3.5]. The class of (integral) perfectoid rings includes not only rings of power-bounded elements of perfectoid algebras in the sense of [17] (and, more generally, of perfectoid Tate rings in the sense of [11]) but also contains some rings which need not even be p-torsion-free.

In [9], one of the authors proved that, for a perfectoid Tate ring A with tilt $A^{\flat }$, there is a homeomorphism between the space of so-called spectrally reduced prime ideals of the Tate ring A (by a theorem of Bhatt and Scholze, these coincide with the prime ideals $\mathfrak {p}$ of A such that the quotient $A/\mathfrak {p}$ is again a perfectoid Tate ring, see [9, Theorem 4.4]) and the space of closed radical ideals of $A^{\flat }$, the two spaces being endowed with their respective Zariski topologies ([9, Theorem 4.16]). In particular, a perfectoid Tate ring is an integral domain if and only if its tilt is an integral domain. In this paper, we use algebraic techniques (in particular, a purely algebraic definition of the untilting operation, see Definition 3.2) to generalize the main notions and results of [9] from the case of perfectoid Tate rings to perfectoid rings in the sense of [5].

In this paper, we use perfectoid rings in the sense of [5]. For any perfectoid ring R of characteristic 0, we fix some unit multiple vp of p which admits a compatible system of p-power roots of vp in R (by [5, Lemma 3.9], such a unit multiple of p always exists) and we fix an element $p^{\flat }\in R^{\flat }$ of the tilt $R^{\flat } \cong \varprojlim \nolimits _{F} R$ of R corresponding to a compatible system of p-power roots of vp (it is adapted to the notation used in [7]. Note that a compatible system of $p$-power roots of $p$ does not necessarily exist).

Then our main result can be stated as follows. The tilting operation $(-)^\flat$ for rings and ideals is defined in Definition 3.1 and Definition 3.2.

Main Theorem A

(Theorem 3.5 and Theorem 5.5) Let R be a perfectoid ring of characteristic 0. Then the map

$$\begin{aligned} J\mapsto J^{\flat } :=\{f = (f^{(n)})_n\mid f^{(n)} \in J ~\text {for all}~ n\} \end{aligned}$$

is an inclusion-preserving bijection between the set of ideals J of R such that the quotient R/J is a perfectoid ring and the set of $p^{\flat }$-adically closed radical ideals of $R^{\flat }$.

Furthermore, the restriction of this bijection yields a homeomorphism between the subspace of ${{\,\textrm{Spec}\,}}(R)$ consisting of prime ideals $\mathfrak {p}$ of R such that the quotient $R/\mathfrak {p}$ is perfectoid and the subspace of ${{\,\textrm{Spec}\,}}(R^{\flat })$ consisting of $p^{\flat }$-adically closed prime ideals of $R^{\flat }$.

As a consequence of the above theorem, we obtain a new and purely algebraic proof of the main theorem of [9] (namely, [9, Theorem 4.16]). This proof does not make use of the homeomorphism between the Berkovich spectrum of a perfectoid Tate ring and the Berkovich spectrum of its tilt.

Notation 1.1

For any element $t\in R$ in a ring R and any R-module M we denote by $M[t^{\infty }]$ the t-power-torsion submodule of M; that is,

$$\begin{aligned} M[t^{\infty }]:=\{x \in M\mid \exists n \in \mathbb {Z}_{\ge 0}, t^n x = 0\}. \end{aligned}$$

Note that, for any M and t, the quotient $M/M[t^{\infty }]$ is equal to the image of the canonical map $M\rightarrow M[1/t]$. For a ring R of characteristic 0, we denote the image $R/R[p^{\infty }]$ of the canonical map $R\rightarrow R[1/p]$ by ${\overline{R}}$.

Let $R$ be a ring and let $a$ be an element of $R$. An ideal $I$ of $R$ is $a$-adically closed if it is a closed subset of the $a$-adically topological ring $R$ with respect to the subspace topology. An ideal $I$ of $R$ is derived $a$-complete if it is derived $a$-complete as an $R$-module.

Let $A_{0}$ be a ring and $t\in A_{0}$ be any element. The Tate ring $A_{0}[1/t]$ is the Tate ring defined by the pair of definition $(A_{0}/A_{0}[t^{\infty }], (t))$. For a Tate ring $A=A_{0}[1/t]$, we denote by $\widehat{A^{u}}$ the uniform completion of A as defined in [13, Definition 5.4]. For any topological ring A, we denote by $A^{\circ }$ the set of power-bounded elements of A.

For a perfectoid ring $R$ (in the sense of [5, Definition 3.5]), we call an element $\pi \in R$ a perfectoid element if $R$ is $\pi$-adically complete and $\pi ^p$ divides $p$ in $R$. Such an element always exists in any perfectoid ring, by definition, but it is not unique.

If $\pi \in R$ is a perfectoid element in a perfectoid ring R, then $R/R[\pi ^{\infty }]$ is again a perfectoid ring [7, 2.1.3] and the Tate ring $R[1/\pi ]=(R/R[\pi ^{\infty }])[1/\pi ]$ is perfectoid in the sense of Fontaine [11] (recall that every complete Tate ring $A$ can be endowed with a norm $\Vert \cdot \Vert _A$ defining its topology, and if $t\in A$ is a topologically nilpotent unit of A, one can choose the norm $\Vert \cdot \Vert _A$ such that t is multiplicative with respect to $\Vert \cdot \Vert _A$, see, e.g., [16, Definition 2.26 and Lemma 2.27]; in particular, every complete Tate ring is a Banach ring in the sense of [11]).

Assume that R is a perfectoid ring of characteristic 0. Let us write $(W(R^{\flat }), (\xi ))$ for the perfect prism corresponding to the perfectoid ring R via [6, Theorem 3.10], where $\xi$ is a distinguished element of $W(R^{\flat })$ with Witt vector coordinates $(\xi _{0}, \xi _{1}, \dots )$. By [5, Lemma 3.9], there exists some unit multiple $vp \in R$ of p such that vp has a compatible system of p-power roots $\{(vp)^{1/p^{n}}\}_{n\ge 0}$ in R. Furthermore, one can adjust the choice of $\xi$ so that $p^{\flat }=\xi _{0}$ where $p^\flat :=((vp)^{1/p^n})_n \in R^\flat \cong \varprojlim \nolimits _{x\mapsto x^{p}}R$, see [7, 2.1.2].

For every perfectoid ring R of characteristic 0, we fix an element $p^{\flat }\in R^{\flat }$ and a distinguished element $\xi \in W(R^{\flat })$ as above throughout this paper (in particular, we always assume that $\xi$ is chosen in such a way that $\xi _0=p^{\flat }$) and we consider R (respectively, $R^{\flat }$) as topological rings equipped with the p-adic (respectively, the $p^{\flat }$-adic) topology.

2 Perfectoid ideals

In this section, we work with the following notion of perfectoid ideals.

Definition 2.1

For any perfectoid ring $R$, an ideal $J$ of $R$ is called a perfectoid ideal if the quotient $R/J$ is a perfectoid ring. Any perfectoid ideal is p-adically closed and derived $p$-complete since any perfectoid ring is $p$-adically complete.

Recall from [6] that a semiperfectoid ring S is a derived p-complete ring that can be written as a quotient of some perfectoid ring R. Recall also from loc. cit., Corollary 7.3, that, for every semiperfectoid ring S, there exists an initial object $S \rightarrow S_{{{\,\textrm{perfd}\,}}}$ in the category of perfectoid S-algebras; the ring $S_{{{\,\textrm{perfd}\,}}}$ is called the perfectoidization of S.

Lemma 2.2

Assume that $R$ is a perfectoid ring and $J$ is a derived $p$-complete ideal of $R$. Then $J$ is a perfectoid ideal if and only if $J = J_{{{\,\textrm{perfd}\,}}}$ where $J_{{{\,\textrm{perfd}\,}}} :=\ker (R \rightarrow (R/J)_{{{\,\textrm{perfd}\,}}})$.

Proof

Since $J$ is derived $p$-complete, $R/J$ is a semiperfectoid ring and thus the perfectoidization $(R/J)_{{{\,\textrm{perfd}\,}}}$ is a universal perfectoid ring over $R/J$. By [6, Theorem 7.4], the canonical map $R \rightarrow (R/J)_{{{\,\textrm{perfd}\,}}}$ is surjective. $\square$

To check the perfectoid property of rings and ideals, we can sometimes use the following lemma.

Lemma 2.3

Let R be a semiperfectoid ring and suppose that there exists a perfectoid ring $R'$ and an injective map $R\hookrightarrow R'$. Then R is a perfectoid ring.

In particular, if a derived p-complete ideal J in a perfectoid ring R can be written as an intersection of perfectoid ideals, then J is also a perfectoid ideal.

Proof

We first deduce the second assertion of the lemma from the first. So, let J be a derived p-complete ideal in a perfectoid ring R which can be written as an intersection of a family $(J_{i})_{i}$ of perfectoid ideals. Any product of perfectoid rings is again perfectoid ([4, Example 3.8(8)]). In particular, $\prod _{i}R/J_{i}$ is a perfectoid ring, so the canonical map $R/J\hookrightarrow \prod _{i}R/J_{i}$ shows that the second assertion of the lemma is a consequence of the first.

Now let R be a semiperfectoid ring admitting an injection $R\hookrightarrow R'$ into a perfectoid ring $R'$. By the universal property of perfectoidization, the injective map $R\hookrightarrow R'$ factors through the canonical map $R\rightarrow R_{{{\,\textrm{perfd}\,}}}$. It follows that the canonical map $R\rightarrow R_{{{\,\textrm{perfd}\,}}}$ is injective. But $R\rightarrow R_{{{\,\textrm{perfd}\,}}}$ is also surjective by [6, Theorem 7.4]. $\square$

In Lemma 2.7 we will clarify the relationship between the above notion of perfectoid ideals and the notion of spectrally reduced ideals in a seminormed ring as introduced in [9]. To do this, we define the notion of spectrally reduced ideals in the setting of Tate rings and explain how this definition relates to the previous one for seminormed rings.

Definition 2.4

((cf.) [9, Definition 2.16]) For a Tate ring A, we call an ideal $\mathcal {J}\subsetneq A$ spectrally reduced if

$$\begin{aligned} \mathcal {J}=\bigcap _{\ker (\phi )\supseteq \mathcal {J}}\ker (\phi ) \end{aligned}$$

where $\phi$ ranges over the continuous multiplicative seminorms on A with $\ker (\phi )\supseteq \mathcal {J}$.

Remark 2.5

By [9, Remark 2.19], an ideal $\mathcal {J}$ of a Tate ring A is spectrally reduced in the above sense if and only if $\mathcal {J}$ is a spectrally reduced ideal of the seminormed ring $(A, \Vert \cdot \Vert )$ in the sense of [9, Definition 2.16], for some (or, equivalently, any) seminorm $\Vert \cdot \Vert$ defining the topology on A such that A admits a topologically nilpotent unit which is multiplicative with respect to the seminorm $\Vert \cdot \Vert$ (for the latter condition, see also [9, Remark 2.13]).

For example, if $(A_{0}, (t))$ is a pair of definition of a Tate ring A, then, by definition, t is a topologically nilpotent unit of A which is multiplicative with respect to the canonical extension $\Vert \cdot \Vert _A$ to A of the t-adic seminorm on $A_{0}$ (see [16, Lemma 2.27]).

In the proof of Lemma 2.7 and in Sect. 4, we use the $t$-saturation of ideals:

Definition 2.6

Let $R$ be a ring, $t\in R$ and let J be an ideal of $R$. The $t$-saturation ${\widetilde{J}}^{t}$ of $J$ is the ideal defined by

$$\begin{aligned} {\widetilde{J}}^{t} :=\{f \in R\mid t^n f \in J \text{ for } \text{ some } ~n \ge 0\}, \end{aligned}$$

that is, ${\widetilde{J}}^{t}$ is the inverse image of $J[1/t]$ via the canonical map $R \rightarrow R[1/t]$

and thus $R/{\widetilde{J}}^{t}$ is $t$-torsion-free (more precisely, $R/{\widetilde{J}}^{t}$=$(R/J)/(R/J)[t^{\infty }]$). If $t=p$, then we just write ${\widetilde{J}}$ instead of ${\widetilde{J}}^{p}$ for the p-saturation of J.

Lemma 2.7

Let R be a perfectoid ring of characteristic 0, let $\pi \in R$ be a perfectoid element in R and let J be a derived p-complete ideal of R such that $\pi \notin J$. If J is a perfectoid ideal, then the ideal $J[1/\pi ]$ of the perfectoid Tate ring $R[1/\pi ]$ is spectrally reduced, and, if we assume that R/J is $\pi$-torsion-free, the converse is also true.

In general, J is a perfectoid ideal if and only if J is a radical ideal and the ideal $J[1/\pi ]$ of the perfectoid Tate ring $R[1/\pi ]$ is spectrally reduced.

Proof

First suppose that J is a perfectoid ideal of R. By definition, the quotient $R/J$ is a perfectoid ring and thus the Tate ring $(R/J)[1/\pi ]$ is a perfectoid Tate ring. Note that $(R/J)[1/\pi ]$ is isomorphic (as a Tate ring) to the quotient of $R[1/\pi ]$ by the ideal $J[1/\pi ]$. Since every perfectoid Tate ring is uniform, we deduce that $J[1/\pi ]$ is a spectrally reduced ideal of $R[1/\pi ]$.

Conversely, first we assume that $R/J$ is $\pi$-torsion-free and $J[1/\pi ]$ is spectrally reduced. Then the quotient $R[1/\pi ]/J[1/\pi ] \cong (R/J)[1/\pi ]$ is a perfectoid Tate ring, by a theorem of Bhatt and Scholze (see [9, Theorem 4.4]). By the proof of loc. cit., $(R/J)_{{{\,\textrm{perfd}\,}}}[1/\pi ]$ is equal to the uniform completion of $(R/J)[1/\pi ]$. Since perfectoid Tate rings are uniform, we conclude that the map $(R/J)[1/\pi ]\rightarrow (R/J)_{{{\,\textrm{perfd}\,}}}[1/\pi ]$ is an isomorphism. The $\pi$-torsion-freeness of $R/J$ implies that the canonical map $R/J \rightarrow (R/J)[1/\pi ] \cong (R/J)_{{{\,\textrm{perfd}\,}}}[1/\pi ]$ is injective and, in particular, $R/J \rightarrow (R/J)_{{{\,\textrm{perfd}\,}}}$ is injective. Since $R/J \rightarrow (R/J)_{{{\,\textrm{perfd}\,}}}$ is surjective by [6, Theorem 7.4], $R/J$ is isomorphic to $(R/J)_{{{\,\textrm{perfd}\,}}}$ and thus $J$ is a perfectoid ideal of $R$.

In the general case, suppose that J is a derived p-complete radical ideal of R such that the ideal $J[1/\pi ]$ of $R[1/\pi ]$ is spectrally reduced. Note that ${\widetilde{J}}^{\pi }[1/\pi ]=J[1/\pi ]$ where ${\widetilde{J}}^{\pi }$ is the $\pi$-saturation of $J$ as defined above in Definition 2.6. Since $J[1/\pi ]$ is spectrally reduced, it is in particular (topologically) closed in $R[1/\pi ]$ and thus the isomorphism of Tate rings

$$\begin{aligned} (R/{\widetilde{J}}^{\pi })[1/\pi ]=R[1/\pi ]/{\widetilde{J}}^{\pi }[1/\pi ]=R[1/\pi ]/J[1/\pi ] \end{aligned}$$

shows that $(R/{\widetilde{J}}^{\pi })[1/\pi ]$ is complete (in fact, we know that it is a perfectoid Tate ring, by [9, Theorem 4.4]). Consequently, since the complete Tate ring $(R/{\widetilde{J}}^{\pi })[1/\pi ]$ has the pair of definition $(R/{\widetilde{J}}^{\pi }, (\pi ))$, the quotient ring $R/{\widetilde{J}}^{\pi }$ is classically $\pi$-adically complete. Since, by the definition of a perfectoid element, we have $pR \subseteq \pi ^{p}R\subseteq \pi R$, this entails that $R/{\widetilde{J}}^{\pi }$ is classically p-adically complete by [18, 090T]; a fortiori, $R/{\widetilde{J}}^{\pi }$ is derived p-complete. It follows that the ideal ${\widetilde{J}}^{\pi }$ is also derived $p$-complete. Hence we can apply the assertion of the lemma in the $\pi$-torsion-free case to see that the ideal ${\widetilde{J}}^{\pi }$ is perfectoid.

On the other hand, since J is a radical ideal, it can be written as an intersection

$$\begin{aligned} J={\widetilde{J}}^{\pi }\cap \bigcap _{\mathfrak {q}}\mathfrak {q}, \end{aligned}$$

where $\mathfrak {q}$ ranges over the prime ideals of R which contain J but do not contain ${\widetilde{J}}^{\pi }$. For such a prime ideal $\mathfrak {q}$, let $x\in {\widetilde{J}}^{\pi }$ be an element which does not belong to $\mathfrak {q}$. Then we can choose an integer $n>0$ such that $\pi ^{n}x\in J\subseteq \mathfrak {q}$. Since $\mathfrak {q}$ is prime, we conclude that $\pi \in \mathfrak {q}$ and therefore $p\in \mathfrak {q}$. But this implies that every prime ideal $\mathfrak {q}$ as above is perfectoid by Proposition 2.8 below. It follows that J is perfectoid, by Lemma 2.3 above. $\square$

In the positive characteristic case, perfectoid ideals are more simple objects.

Proposition 2.8

(cf. [9, Remark 2.21]) Let $R$ be a perfectoid ring and $J$ be an ideal of $R$. Assume that $p \in J$. Then $J$ is a perfectoid ideal if and only if it is a radical ideal.

Assume that $R$ is a perfect(oid) ring of positive characteristic $p$ and $t\in R$ is an element such that R is t-adically complete. Assume further that $t \notin J$. Then the ideal $J[1/t]$ of the complete Tate ring R[1/t] is spectrally reduced if and only if $J[1/t]$ is a closed radical ideal of $R[1/t]$.

Proof

Since $R/J$ is of positive characteristic $p$, the kernel of $R/J \rightarrow (R/J)_{{{\,\textrm{perf}\,}}} = (R/J)_{{{\,\textrm{perfd}\,}}}$ is equal to $\sqrt{J}/J$.

The last statement follows from [9, Remark 2.21]. $\square$

3 Tilting and untilting of ideals

In this section, we define the tilting and untilting of ideals, which produces a correspondence between the set of perfectoid ideals of $R$ and the set of $p^{\flat }$-adically closed radical ideals of $R^\flat$ for any perfectoid ring $R$ of characteristic 0. To do this, we start with some constructions related to perfectoid rings.

Definition 3.1

For any ring $R$, we can define the tilt of this ring,

$$\begin{aligned} R^\flat :=\varprojlim _F R/pR, \end{aligned}$$

where F denotes the Frobenius on R/pR. If $R$ is $p$-adically complete, the underlying multiplicative monoid of $R^\flat$ is isomorphic to $\varprojlim \nolimits _{x \mapsto x^p} R$ as monoids, and then $\varprojlim \nolimits _{x \mapsto x^p} R$ has a natural ring structure induced from $R^\flat$ by [5, Lemma 3.2]. The isomorphism is defined by

$$\begin{aligned} R^\flat = \varprojlim \limits _F R/pR&\longrightarrow \varprojlim \limits _{x \mapsto x^p} R \end{aligned}$$

(1)

$$\begin{aligned} x :=(\overline{x^{(n)}})_n&\longmapsto (\lim _{k \rightarrow \infty } (x^{(n+k)})^{p^k})_n. \end{aligned}$$

(2)

We define $x^\sharp :=\lim _{k \rightarrow \infty } (x^{(k)})^{p^k} \in R$ for any $x = (\overline{x^{(n)}})_n \in R^\flat$.

For any ideal $J$ of $R$, we have a theta map

$$\begin{aligned} \theta _{R/J} :W((R/J)^\flat )&\longrightarrow R/J, \\ \sum _{n \ge 0} [x_n] p^n&\longmapsto \sum _{n \ge 0} x_n^\sharp p^n. \end{aligned}$$

If the Frobenius map on $(R/J)/p(R/J)$ is surjective, then $\theta _{R/J}$ is surjective.

Next, we define the tilting and untilting of ideals as follows.

Definition 3.2

Let R be a classically p-adically complete ring. We identify the underlying multiplicative monoid of the tilt $R^{\flat }$ of R with $\varprojlim \nolimits _{x\mapsto x^{p}}R$ via (1), so an element $x\in R^{\flat }$ is a sequence $(x^{(n)})_{n}$ of elements of R with

$$\begin{aligned} (x^{(n+1)})^p=x^{(n)} \in R \end{aligned}$$

for all $n\in \mathbb {Z}_{\ge 0}$. Note that under this identification we have $x^{\sharp }=x^{(0)}$ for any $x\in R^{\flat }$. For any subset J of R, we define a subset $J^{\flat }$ of $R^{\flat }$ by

$$\begin{aligned} J^\flat :=\{x = (x^{(n)})_n \in R^\flat \mid x^{(n)} \in J~\text {for any}~n \ge 0\}. \end{aligned}$$

(3)

If J is a p-adically closed ideal of R, then $J^{\flat }$ is an ideal of $R^{\flat }$. If moreover J is a $p$-adically closed radical ideal, we have $J^{\flat }=\{x \in R^\flat \mid x^\sharp \in J\}$. We call $J^{\flat }$ the tilt of the subset (respectively, of the p-adically closed ideal) J of R. Remark that this operation $(-)^\flat$ maps prime ideals of $R$ to prime ideals of $R^\flat$ by simple observations.

Conversely, if $R$ is a perfectoid ring and $I$ is an ideal of $R^\flat$, we define a subset of $W(R^\flat )$

$$\begin{aligned} W(I) :=\left\{ {\sum _{n \ge 0} [x_n] p^n \in W(R^\flat )}\mid {x_n \in I}\right\} \end{aligned}$$

(4)

which is well-defined by the unique representation. In particular, if I is a radical ideal of $R^{\flat }$ (so that the quotient $R^{\flat }/I$ is perfect), this $W(I)$ is equal to $\ker (W(R^\flat ) \rightarrow W(R^\flat /I))$ and thus $W(I)$ is an ideal of $W(R^\flat )$. Then we can define an ideal of $R$

$$\begin{aligned} I^\sharp :=\theta _R(W(I)) \end{aligned}$$

(5)

because of the surjectivity of $\theta _R$. We call it the untilt of $I$.

In the next section (Proposition 4.7), we will show that $(-)^{\sharp }$ maps $p^{\flat }$-adically closed prime ideals of $R^{\flat }$ to prime ideals of R for any perfectoid ring R of characteristic 0, but we do not use this fact in this section.

One of the important properties of these operations is the following. Roughly speaking, the tilting and untilting operations preserve the perfectoid property of ideals under some topologically closed assumptions.

Lemma 3.3

For any p-adically complete ring R and any $p$-adically closed ideal J of R, the tilt $J^{\flat }$ of J is equal to the kernel of the homomorphism

$$\begin{aligned} \varphi ^\flat :R^\flat \rightarrow (R/J)^\flat \end{aligned}$$

corresponding to the canonical quotient map $R\rightarrow R/J$.

If R is a perfectoid ring of characteristic 0 and J is a perfectoid ideal of R, then $\varphi ^{\flat }$ induces an isomorphism $R^{\flat }/J^{\flat } \xrightarrow {\cong } (R/J)^{\flat }$. In particular, the tilt $J^{\flat }$ of any perfectoid ideal J of a perfectoid ring R is a $p^{\flat }$-adically closed radical ideal of $R^{\flat }$.

Conversely, for any perfectoid ring R of characteristic $0$ and any $p^{\flat }$-adically closed radical (and thus perfectoid) ideal I of $R^{\flat }$, the untilt $I^{\sharp }$ is also a perfectoid ideal of R.

Proof

The equality $J^{\flat }=\ker (\varphi ^{\flat })$ follows by the same argument as in the proof of [9, Lemma 4.8]. If R and R/J are perfectoid, their tilts are $p^{\flat }$-adically complete (see, for example, [7, 2.1.2]). Then the kernel $J^\flat$ is $p^\flat$-adically closed.

By topological Nakayama’s lemma, $R^\flat \rightarrow (R/J)^\flat$ is surjective and this proves that $\varphi ^\flat$ induces an isomorphism $R^\flat /J^\flat \cong (R/J)^\flat$ and $J^\flat$ is a $p^{\flat }$-adically closed radical ideal of $R^\flat$.

For a $p^\flat$-adically closed radical ideal $I \subseteq R^\flat$, the canonical map $W(R^\flat ) \rightarrow W(R^\flat /I)$ is surjective by topological Nakayama’s lemma and the kernel is given by $W(I)$ by definition. Then we have

$$\begin{aligned} R/I^\sharp \cong W(R^\flat )/((\xi ) + W(I)) \cong W(R^\flat /I)/(\xi ). \end{aligned}$$

Since $I$ is a $p^\flat$-adically closed radical ideal of $R^\flat$, the quotient $R^\flat /I$ is a $p^\flat$-adically complete¹ perfect(oid) $R^\flat$-algebra. Then $R/I^\sharp$ is a perfectoid $R$-algebra by [7, Proposition 2.1.9] or [10, Thèoréme 2.10]. $\square$

Remark 3.4

The assumption that the ideal I in Lemma 3.3 be $p^{\flat }$-adically closed is not automatic: Indeed, we show in Remark 5.8 that there exists a perfectoid ring R of characteristic 0 such that not all prime ideals of $R^{\flat }$ are $p^{\flat }$-adically closed. Furthermore, the assumption that I be $p^{\flat }$-adically closed is also necessary for the assertion of Lemma 3.3 to hold true: If I is a radical ideal of $R^{\flat }$ which is not $p^{\flat }$-adically closed in $R^{\flat }$, then $I^{\sharp }$ is not a perfectoid ideal of R. Indeed, if $I^{\sharp }$ is a perfectoid ideal, i.e., $R/I^{\sharp }\cong W(R^{\flat }/I)/(\xi )$ is a perfectoid ring, then, by [6, Theorem 3.10], the pair $(W(R^{\flat }/I), (\xi ))$ is a perfect prism. By loc. cit., Lemma 3.8(2), this entails that $W(R^\flat /I)$ is classically $(p, \xi )$-adically complete. But then $R^{\flat }/I=W(R^{\flat }/I)/(p)$ is classically $\xi _0=p^{\flat }$-adically complete and thus I must be $p^{\flat }$-adically closed.

Using the above observations, we can show the following one-to-one correspondence between the sets of perfectoid ideals and $p^\flat$-adically closed radical ideals.

Theorem 3.5

Let R be a perfectoid ring of characteristic 0. Then the tilting and untilting operations $(-)^{\flat }$ and $(-)^\sharp$ define a one-to-one correspondence between the set of perfectoid ideals of R and the set of $p^{\flat }$-adically closed radical ideals of $R^{\flat }$.

Proof

By Lemma 3.3, it suffices to show that the maps $(-)^\flat$ and $(-)^\sharp$ are inverse to each other. Take any perfectoid ideal $J$ of $R$. The perfectoid $R$-algebra $R/J$ is of the form $W((R/J)^\flat )/\xi W((R/J)^\flat )$ by [7, Proposition 2.1.9] or [10, Théorème 2.10]. Then the commutative diagram

and the fact that $\ker (\theta _R)$ and $\ker (\theta _{R/J})$ are generated by the same $\xi$ show that $J^{\flat \sharp } = J$.

Conversely, if I is a $p^{\flat }$-adically closed radical ideal of $R^{\flat }$, we have

$$\begin{aligned} R^\flat /I^{\sharp \flat } \cong (R/I^\sharp )^\flat \cong (W(R^\flat /I)/(\xi ))^\flat \cong R^\flat /I \end{aligned}$$

canonically and thus $I=I^{\sharp \flat }$. $\square$

We deduce from the above theorem a corollary which ensures that the theorem also stays true when the role of p is taken on by another perfectoid element of R. Recall that for any perfectoid element $\pi$ in a perfectoid ring R of characteristic 0, [5, Lemma 3.9] entails that there exists a unit $u\in R^{\times }$ such that $u\pi$ admits a compatible system of p-power roots in R, i.e., there exists an element $\pi ^{\flat }\in R^{\flat }$ such that $\pi ^{\flat \sharp }$ is a unit multiple of $\pi$ in R.

Corollary 3.6

Let R be a perfectoid ring of characteristic 0, let $\pi \in R$ be a perfectoid element and let $\pi ^{\flat }\in R^{\flat }$ be an element such that $\pi ^{\flat \sharp }$ is a unit multiple of $\pi$ in R. Then a radical ideal I of $R^{\flat }$ is $p^{\flat }=\xi _0$-adically closed if and only if it is $\pi ^{\flat }$-adically closed.

Proof

If I is a $p^{\flat }$-adically closed radical ideal of $R^{\flat }$, then, by Theorem 3.5, $I=J^{\flat }$ for some perfectoid ideal J of R. By Lemma 3.3, $R^{\flat }/I=(R/J)^{\flat }$. By [7, 2.1.2], $(R/J)^{\flat }$ is $\pi ^{\flat }$-adically complete. Consequently, I is $\pi ^{\flat }$-adically closed.

Conversely, once again by the discussion in [7, 2.1.2], $(\pi ^{\flat })^{p}$ divides $p^{\flat }$ in $R^{\flat }$ and thus $p^{\flat } R^{\flat } \subseteq (\pi ^{\flat })^p R^{\flat } \subseteq \pi ^{\flat } R^{\flat }$. This entails that every $\pi ^{\flat }$-adically complete $R^{\flat }$-algebra is also $p^{\flat }$-adically complete, by [18, 090T]. Hence every $\pi ^{\flat }$-adically closed ideal I of $R^{\flat }$ is also $p^{\flat }$-adically closed, as claimed. $\square$

4 Correspondence of prime ideals

For any perfectoid ring R, the tilt of a p-adically closed prime ideal of R is a prime ideal of $R^{\flat }$. This can be shown easily. However, the stability of (closed) prime ideals under the untilting operation $(-)^\sharp$ is not quite obvious. We show this fact by using an idea that appeared in the proof of [9, Theorem 4.16].

Definition 4.1

(cf. [9, Definition 3.13]) Let A be a Tate ring and let $\mathcal {J}$ be an ideal of A which is not dense in A. The spectral radical $\mathcal {J}_{{{\,\textrm{sp}\,}}}$ of $\mathcal {J}$ is the smallest spectrally reduced ideal of A containing $\mathcal {J}$. That is,

$$\begin{aligned} \mathcal {J}_{{{\,\textrm{sp}\,}}}=\bigcap _{\ker (\phi )\supseteq \mathcal {J}}\ker (\phi ), \end{aligned}$$

where $\phi$ ranges over the continuous multiplicative seminorms $\phi :A\rightarrow \mathbb {R}_{\ge 0}$ with the property that $\ker (\phi )\supseteq \mathcal {J}$.

Remark 4.2

Since any Tate ring can be equipped with a seminorm defining its topology (see Notation 1.1 or [16, Lemma 2.27]), every non-dense ideal is contained in some ideal of the form $\ker (\phi )$ for a continuous multiplicative seminorm: This is a consequence of the fact that the Berkovich spectrum of any seminormed ring is non-empty (see [9, Lemma 3.12]). This shows that the above notion of the spectral radical is well-defined. Furthermore, the spectral radical $\mathcal {J}_{{{\,\textrm{sp}\,}}}$ in the sense of the above definition agrees with the spectral radical of the seminormed ring $(A, \Vert \cdot \Vert )$ if $\Vert \cdot \Vert$ is some (or, equivalently, any) seminorm defining the topology of the Tate ring A and admitting a multiplicative topologically nilpotent unit t (see [9, Remark 2.19] or Remark 2.5). For example, we could let $\Vert \cdot \Vert$ be the seminorm defined by some pair of definition $(A_{0}, (t))$ of A.

The following lemma shows the relationship between $(-)_{{{\,\textrm{perfd}\,}}}$ and $(-)_{{{\,\textrm{sp}\,}}}$.

Lemma 4.3

Let $R$ be a perfectoid ring of characteristic $0$ and $J$ be an ideal of $R$. Assume that $J$ is derived $p$-complete (as an $R$-module).

Then the $p$-saturation $\widetilde{J_{{{\,\textrm{perfd}\,}}}}$ of $J_{{{\,\textrm{perfd}\,}}}$ in $R$ is equal to the inverse image of the spectral radical $(J[1/p])_{{{\,\textrm{sp}\,}}}$ of $J[1/p]$ in ${\overline{R}}[1/p]$ via the canonical map $R \rightarrow {\overline{R}}[1/p]$.

Proof

Since $J$ is derived $p$-complete, we can apply [12, Theorem 5.4] for $R/J$. Then we have

$$\begin{aligned} J_{{{\,\textrm{perfd}\,}}}[1/p] = \ker ({\overline{R}}[1/p] \rightarrow \overline{(R/J)_{{{\,\textrm{perfd}\,}}}}[1/p]) = \ker ({\overline{R}}[1/p] \rightarrow \widehat{((R/J)[1/p])^u}) \end{aligned}$$

and the last term is equal to $(J[1/p])_{{{\,\textrm{sp}\,}}}$ because of the definition of the spectral radical and of the uniform completion [13, Remark 5.7]. $\square$

Lemma 4.4

Let $I$ and $J$ be derived $p$-complete ideals of a perfectoid ring $R$ of characteristic $0$. Then we have

$$\begin{aligned} \widetilde{(I \cap J)_{{{\,\textrm{perfd}\,}}}} = \widetilde{I_{{{\,\textrm{perfd}\,}}}} \cap \widetilde{J_{{{\,\textrm{perfd}\,}}}}. \end{aligned}$$

Proof

The intersection $I \cap J$ is derived $p$-complete since it is the kernel of the map of derived $p$-complete rings $R \rightarrow (R/I) \times (R/J)$. Therefore, the assertion follows from Lemma 4.3 and the fact that the map $(-)_{{{\,\textrm{sp}\,}}}$ on ideals of a perfectoid Tate ring commutes with intersections. $\square$

Lemma 4.5

Let $J$ be a perfectoid ideal of a perfectoid ring $R$. Then the $p$-saturation ${\widetilde{J}}$ of $J$ in $R$ is also a perfectoid ideal.

Proof

Note that an element $f$ of $R$ is in ${\widetilde{J}}$ if and only if the image of $f$ in $R/J$ is $p$-power torsion. That is, $R/{\widetilde{J}}$ is the quotient of $R/J$ by its $p$-power torsion part $(R/J)[p^\infty ]$. By [7, 2.1.3] or [10, Proposition 2.19], we see that ${\widetilde{J}}$ is a perfectoid ideal. $\square$

Lemma 4.6

Let $I$ and $J$ be ideals of a $t$-adically separated ring $R$ for some element $t \in R$. Let us denote by ${\overline{\mathfrak {a}}}^t$ the (topological) closure of an ideal $\mathfrak {a}$ of $R$ with respect to the $t$-adic topology of $R$. Then we have

$$\begin{aligned} {\overline{I}}^t \cap {\overline{J}}^t \subseteq \sqrt{\overline{I \cap J}^t}. \end{aligned}$$

Proof

Let $f$ be an element of ${\overline{I}}^t \cap {\overline{J}}^t$. Choose sequences $(f_n) \subset I$ and $(g_n) \subset J$ such that $f = \lim _n f_n = \lim _n g_n$ in the $t$-adically separated ring $R$. Then $(f_n g_n)$ is a sequence in $IJ \subseteq I \cap J$ and $f^2 = (\lim _n f_n) (\lim _n g_n) = \lim _n f_n g_n$ is in $\overline{I \cap J}^t$. $\square$

Proposition 4.7

Let $R$ be a perfectoid ring of characteristic $0$ and let $\mathfrak {q}$ be a $p^\flat$-adically closed prime ideal of $R^\flat$. Then the untilt $\mathfrak {q}^\sharp$ is a prime ideal of $R$.

Proof

In view of the canonical isomorphism $R/p R \cong R^\flat /p^\flat R^\flat$, the assertion is clear if $p^\flat \in \mathfrak {q}$ (or equivalently, if $p \in \mathfrak {q}^\sharp$). Hence we may assume that $p^\flat \notin \mathfrak {q}$. We proceed as in the proof of [9, Theorem 4.16]. So, let $I$ and $J$ be ideals of $R$ with $IJ \subseteq \mathfrak {q}^\sharp$ and it suffices to show that either $I$ or $J$ is contained in $\mathfrak {q}^\sharp$.

By Lemma 3.3, the untilt $\mathfrak {q}^\sharp$ is a perfectoid ideal of $R$. Then $\mathfrak {q}^\sharp$ is a radical ideal by the reduced property of perfectoid rings [7, 2.1.3]. Since every radical ideal is the intersection of the prime ideals containing of it, the inclusion $IJ \subseteq \mathfrak {q}^\sharp$ implies $I \cap J \subseteq \mathfrak {q}^\sharp$. Since any perfectoid ideal of $R$ is $p$-adically closed in $R$, we have

$$\begin{aligned} \sqrt{\overline{I \cap J}^p} \subseteq \mathfrak {q}^\sharp \end{aligned}$$

where $\overline{I \cap J}^p$ is the (topological) closure of $I \cap J$ with respect to the $p$-adic topology of $R$. By Lemma 4.6, we have

$$\begin{aligned} {\overline{I}}^p \cap {\overline{J}}^p \subseteq \mathfrak {q}^\sharp . \end{aligned}$$

By Lemma 4.4, we see that

$$\begin{aligned} \widetilde{({\overline{I}}^p)_{{{\,\textrm{perfd}\,}}}} \cap \widetilde{({\overline{J}}^p)_{{{\,\textrm{perfd}\,}}}} = \widetilde{({\overline{I}}^p \cap {\overline{J}}^p)_{{{\,\textrm{perfd}\,}}}}. \end{aligned}$$

Since the map $(-)^\flat$ from subsets of $R$ to subsets of $R^\flat$ preserves inclusions and intersections, this means that

$$\begin{aligned} (\widetilde{({\overline{I}}^p)_{{{\,\textrm{perfd}\,}}}})^\flat \cap (\widetilde{({\overline{J}}^p)_{{{\,\textrm{perfd}\,}}}})^\flat \subseteq (\widetilde{\mathfrak {q}^\sharp })^\flat . \end{aligned}$$

By the definition of the tilting operation (3), if $f \in (\widetilde{\mathfrak {q}^\sharp })^\flat$, then $p^n f^\sharp \in \mathfrak {q}^\sharp$ for some $n \ge 0$. Using the fact that $p^n = ((p^\flat )^n)^\sharp \in R$ and Theorem 3.5, we have $(p^\flat )^n f \in \mathfrak {q}^{\sharp \flat } = \mathfrak {q}$ and thus $f \in \mathfrak {q}$ because of the assumption that $p^\flat \notin \mathfrak {q}$. This shows that

$$\begin{aligned} (\widetilde{({\overline{I}}^p)_{{{\,\textrm{perfd}\,}}}})^\flat \cap (\widetilde{({\overline{J}}^p)_{{{\,\textrm{perfd}\,}}}})^\flat \subseteq \mathfrak {q}. \end{aligned}$$

Since $\mathfrak {q}$ is a prime ideal of $R^\flat$, we see that either

$$\begin{aligned} (\widetilde{({\overline{I}}^p)_{{{\,\textrm{perfd}\,}}}})^\flat \subseteq \mathfrak {q}\text { or } (\widetilde{({\overline{J}}^p)_{{{\,\textrm{perfd}\,}}}})^\flat \subseteq \mathfrak {q}. \end{aligned}$$

On the other hand, Lemma 4.5 ensures that $\widetilde{({\overline{I}}^p)_{{{\,\textrm{perfd}\,}}}}$ and $\widetilde{({\overline{J}}^p)_{{{\,\textrm{perfd}\,}}}}$ are perfectoid ideals of $R$. Applying the inclusion-preserving map $(-)^\sharp$ and using Theorem 3.5, we conclude that

$$\begin{aligned} I \subseteq \widetilde{({\overline{I}}^p)_{{{\,\textrm{perfd}\,}}}} \subseteq \mathfrak {q}^\sharp \text { or } J \subseteq \widetilde{({\overline{J}}^p)_{{{\,\textrm{perfd}\,}}}} \subseteq \mathfrak {q}^\sharp , \end{aligned}$$

as desired. $\square$

5 The perfectoid spectrum

We will show that the one-to-one correspondence in Theorem 3.5 induces a homeomorphism between a certain subspace of the spectrum of R and a certain subspace of the spectrum of $R^{\flat }$.

Definition 5.1

For a perfectoid ring R of characteristic 0, the perfectoid spectrum of R is the subspace

$$\begin{aligned} {{\,\text {Spec}}}_{{{\text {perfd}\,}}}(R) :=\{\, \mathfrak {p}\in {{\,\text {Spec}\,}}(R)\mid \mathfrak {p}~\text{ is } \text{ a } \text{ perfectoid } \text{ ideal }\,\} \end{aligned}$$

of ${{\,\textrm{Spec}\,}}(R)$ endowed with the subspace topology induced from the Zariski topology on ${{\,\textrm{Spec}\,}}(R)$. The perfectoid spectrum of $R^{\flat }$ is the subspace

$$\begin{aligned} {{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat }) :=\{\, \mathfrak {p}\in {{\,\textrm{Spec}\,}}(R^{\flat })\mid \mathfrak {p}~\text {is}~p^{\flat }\text {-adically closed}\,\} \end{aligned}$$

of ${{\,\textrm{Spec}\,}}(R^{\flat })$, again with the subspace topology induced from the Zariski topology.

We recall a variant of the notion of the topological spectrum from [9].

Definition 5.2

(cf. [9, Definition 2.18]) The topological spectrum of a Tate ring A is the subspace

$$\begin{aligned} {{\,\text {Spec}}}_{{{\text {Top}\,}}}(A) :=\{\mathfrak {p}\in {{\,\text {Spec}\,}}(A)\mid \mathfrak {p}~\text{ is } \text{ spectrally } \text{ reduced }\} \end{aligned}$$

of ${{\,\textrm{Spec}\,}}(A)$, where the topology on ${{\,\textrm{Spec}\,}}_{{{\,\textrm{Top}\,}}}(A)$ is the subspace topology induced from the Zariski topology on ${{\,\textrm{Spec}\,}}(A)$.

We note that, by [9, Remark 2.19], for any seminorm $\Vert \cdot \Vert$ defining the topology on A such that there exists a topologically nilpotent unit $t\in A$ multiplicative with respect to $\Vert \cdot \Vert$, the topological spectrum of A defined as above is equal to the topological spectrum of the seminormed ring $(A, \Vert \cdot \Vert )$ in the sense of [9, Definition 2.18].

First, we record the following observations about the perfectoid spectrum and the topological spectrum.

Lemma 5.3

Let R be a perfectoid ring of characteristic 0, let $\pi$ be a perfectoid element of R and let $\pi ^{\flat }$ be an element of $R^{\flat }$ such that $\pi ^{\flat \sharp }$ is a unit multiple of $\pi$ in R (by [5, Lemma 3.9], such an element $\pi ^\flat$ always exists). Then ${{\,\textrm{Spec}\,}}_{{{\,\textrm{Top}\,}}}(R[1/\pi ])$ is equal to the open subspace $D(\pi )\cap {{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R)$ of ${{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R)$ and ${{\,\textrm{Spec}\,}}_{{{\,\textrm{Top}\,}}}(R^{\flat }[1/\pi ^{\flat }])$ is equal to the open subspace $D(\pi ^{\flat })\cap {{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat })$ of ${{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat })$.

Proof

By Corollary 3.6 a prime ideal of $R^{\flat }$ is $\pi ^{\flat }$-adically closed if and only if it is $p^{\flat }$-adically closed. Hence the assertion follows from Lemma 2.7 and Proposition 2.8. $\square$

Combining Theorem 3.5 and Proposition 4.7, we obtain a correspondence (Theorem 5.5 below) between perfectoid prime ideals of a perfectoid ring $R$ of characteristic 0 and $p^\flat$-adically closed prime ideals of $R^\flat$. This correspondence is indeed a homeomorphism, not only a bijection. To prove this, we record the following lemma.

Lemma 5.4

Let R be a perfect(oid) ring of characteristic p and let I, J be ideals of R. If J is a radical ideal, then so is ${\overline{J}}^{I}$, the closure of J with respect to the I-adic topology on R.

Proof

The quotient $R/{\overline{J}}^{I}$ is the I-adic completion of the perfect ring R/J, so the lemma follows from [8, Lemma 3.6]. $\square$

Theorem 5.5

Let R be a perfectoid ring of characteristic 0. Then the tilting map (3)

$$\begin{aligned} \mathfrak {p} \mapsto \mathfrak {p}^{\flat } = \{f = (f^{(n)})_n \in R^\flat \mid f^\sharp \in \mathfrak {p}\} \end{aligned}$$

induces a homeomorphism

$$\begin{aligned} {{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R) \xrightarrow {\simeq } {{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat }), \end{aligned}$$

whose inverse is given by the untilting map $\mathfrak {q}\mapsto \mathfrak {q}^\sharp$ defined in (5).

Proof

It is readily seen that the map $J\mapsto J^{\flat }$ takes (p-adically closed) prime ideals of R to prime ideals of $R^{\flat }$. By Proposition 4.7, the map $I\mapsto I^\sharp$ also takes ($p^{\flat }$-adically closed) prime ideals of $R^{\flat }$ to (perfectoid) prime ideals of R. By Theorem 3.5, the two maps $J \mapsto J^{\flat }$ and $I\mapsto I^\sharp$ are inverse to each other, so we obtain a bijection ${{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R) \xrightarrow {\simeq } {{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat })$.

To see that this bijection is a homeomorphism, note that a perfectoid prime ideal $\mathfrak {p}$ of R contains an ideal J if and only if it contains the perfectoid ideal $({\overline{J}}^{p})_{{{\,\textrm{perfd}\,}}}$: Indeed, if $J \subseteq \mathfrak {p}$, we have ${\overline{J}}^{p}\subseteq \mathfrak {p}$ since perfectoid ideals are p-adically closed, and then the map of semiperfectoid rings

$$\begin{aligned} R/{\overline{J}}^{p} \rightarrow R/\mathfrak {p} \end{aligned}$$

factors through $R/({\overline{J}}^{p})_{{{\,\textrm{perfd}\,}}}$ by the universal property of perfectoidization. Hence every closed subset of ${{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R)$ is of the form

$$\begin{aligned} \mathcal {V}_{{{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R)}(J)= \{\mathfrak {p}\in {{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R)\mid \mathfrak {p}\supseteq J\} \end{aligned}$$

for some perfectoid ideal J of R.

Using Lemma 5.4, every closed subset of ${{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat })$ is of the form $\mathcal {V}_{{{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat })}(I)$ for some $p^{\flat }$-adically closed radical ideal $I$ of $R^{\flat }$. Since the mutually inverse maps $J \mapsto J^{\flat }$ and $I \mapsto I^\sharp$ are inclusion-preserving, we see that

$$\begin{aligned} (\mathcal {V}_{{{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R)}(J))^{\flat } = \mathcal {V}_{{{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat })}(J^{\flat }) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {V}_{{{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat })}(I))^\sharp = \mathcal {V}_{{{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R)}(I^\sharp ) \end{aligned}$$

for any perfectoid ideal J of R and any $p^{\flat }$-adically closed radical ideal I of $R^{\flat }$. It follows that the map

$$\begin{aligned} {{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R) \rightarrow {{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat }), \mathfrak {p} \mapsto \mathfrak {p}^{\flat }, \end{aligned}$$

is a homeomorphism and its inverse map is $\mathfrak {q}\mapsto \mathfrak {q}^\sharp$. $\square$

Let us discuss the relationship between the above theorem and the main result of [9]. Recall that the Berkovich spectrum $\mathcal {M}(A)$ of a seminormed ring $(A, \Vert \cdot \Vert )$ (introduced for Banach rings in [3]) is the topological space of bounded multiplicative seminorms $\phi :A\rightarrow \mathbb {R}_{\ge 0}$ on A, endowed with the weakest topology making each of the maps

$$\begin{aligned} \mathcal {M}(A) \rightarrow \mathbb {R}_{\ge 0}, \phi \mapsto \phi (f), \end{aligned}$$

for $f\in A$, continuous.

We note that if there is a topologically nilpotent unit $t\in A$ which is multiplicative with respect to the seminorm on A, then, by [9, Lemma 2.11], for every continuous multiplicative seminorm $\phi :A\rightarrow \mathbb {R}_{\ge 0}$, there exists $s \in (0,1)$ such that $\phi ^{s} \in \mathcal {M}(A)$ (in fact, if the seminorm $\phi$ satisfies $\phi (t) = \Vert t\Vert$, then $\phi \in \mathcal {M}(A)$, see the proof of loc. cit.).

If A is a perfectoid Tate ring, viewed as a uniform Banach ring by choosing a power-multiplicative norm defining the topology on A, then we have a canonical map

$$\begin{aligned} \mathcal {M}(A) \rightarrow \mathcal {M}(A^{\flat }), \phi \mapsto \phi ^{\flat }, \end{aligned}$$

where the seminorm $\phi ^{\flat }:A^\flat \rightarrow \mathbb {R}_{\ge 0}$ is defined by $\phi ^{\flat }(f) :=\phi (f^\sharp )$ for all $f\in A^{\flat }$. By [15, Theorem 3.3.7(c)] (which is an analogue for Berkovich spectra of [17, Corollary 6.7(iii)]), this map is a homeomorphism $\mathcal {M}(A) \xrightarrow {\simeq } \mathcal {M}(A^{\flat })$, for every perfectoid Tate ring A; we denote by $\phi \mapsto \phi ^\sharp$ its inverse.

The following proposition shows that the untilting operation $(-)^\sharp$ introduced in this paper (5) generalizes the analytically defined untilting operation (denoted by the same symbol) which was introduced in the context of perfectoid Tate rings in [9, Section 4].

Proposition 5.6

Let R be a perfectoid ring of characteristic 0, let $\pi$ be a perfectoid element of R and choose an element $\pi ^{\flat }$ of $R^{\flat }$ such that $\pi ^{\flat \sharp }$ is a unit multiple of $\pi$ in R (this is always possible by [5, Lemma 3.9]). Let I be a $p^{\flat }$-adically closed (or, equivalently, $\pi ^{\flat }$-adically closed, see Corollary 3.6) radical ideal of $R^{\flat }$ such that $R^{\flat }/I$ is $\pi ^{\flat }$-torsion-free. Endow the perfectoid Tate ring $R[1/\pi ]$ (respectively, $(R[1/\pi ])^{\flat }=R^{\flat }[1/\pi ^{\flat }]$²) with a power-multiplicative norm $\Vert \cdot \Vert$ defining its topology such that the element $\pi \in R[1/\pi ]$ (respectively, $\pi ^{\flat } \in (R[1/\pi ])^{\flat }$) is a multiplicative element with respect to $\Vert \cdot \Vert$ (note that such a power-multiplicative norm always exists by [15, Remark 2.8.18] in conjunction with [9], Remark 2.13). Then we have

$$\begin{aligned} I^\sharp = \bigcap _{\begin{array}{c} \phi \in \mathcal {M}(R^{\flat }[1/\pi ^{\flat }]) \\ \ker (\phi ) \supseteq I \end{array}}\ker (\phi ^\sharp ) \cap R. \end{aligned}$$

Proof

By [9, Proposition 4.9], for any spectrally reduced ideal $\mathcal {I}$ of $(R[1/\pi ])^\flat$, the map

$$\begin{aligned} \mathcal {I}\mapsto \bigcap _{\begin{array}{c} \phi \in \mathcal {M}(R^{\flat }[1/\pi ^{\flat }]) \\ \ker (\phi ) \supseteq \mathcal {I} \end{array}} \ker (\phi ^\sharp ) \end{aligned}$$

is inverse to the map $\mathcal {J}\mapsto \mathcal {J}^{\flat }$ from the set of spectrally reduced ideals of $R[1/\pi ]$ to the set of spectrally reduced ideals (or, equivalently, closed radical ideals, see [9, Remark 2.21]) of $(R[1/\pi ])^{\flat }=R^{\flat }[1/\pi ^{\flat }]$. But for an ideal I of $R^{\flat }$ and $\phi \in \mathcal {M}(R^{\flat }[1/\pi ^{\flat }])$, we have $\ker (\phi ) \supseteq I[1/\pi ^{\flat }]$ if and only if $\ker (\phi ) \supseteq I$, and by Lemma 2.7 a radical ideal J of R is perfectoid if and only if the ideal $J[1/\pi ]$ of $R[1/\pi ]$ is spectrally reduced.

Consequently,

$$\begin{aligned} I \mapsto \bigcap _{\begin{array}{c} \phi \in \mathcal {M}(R^{\flat }[1/\pi ^{\flat }]) \\ \ker (\phi ) \supseteq I \end{array}}\ker (\phi ^\sharp )\cap R \end{aligned}$$

is inverse to the map $J \mapsto J^{\flat }$ from the set of perfectoid ideals of R to the set of $p^{\flat }$-adically closed radical ideals of $R^{\flat }$: Indeed, the tilt of the right hand side is equal to the inverse image of $I[1/\pi ^\flat ]$ via $R^\flat \rightarrow \overline{R^\flat }[1/\pi ^\flat ]$ by the above arguments and the inverse image is $I$ because $R^\flat /I$ is $\pi ^\flat$-torsion-free. By Theorem 3.5, the map $I \mapsto I^\sharp$ is also inverse to $J \mapsto J^{\flat }$. The assertion follows. $\square$

Remark 5.7

Note that the space ${{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat })$ of $p^\flat$-adically closed (or, by Corollary 3.6, of $\pi ^{\flat }$-adically closed) prime ideals of $R^\flat$ can be written as

$$\begin{aligned} {{\,\textrm{Spec}\,}}_{{{\,\textrm{Top}\,}}}(R^\flat [1/\pi ^\flat ]) \cup {{\,\textrm{Spec}\,}}(R^\flat /(\pi ^\flat ))={{\,\textrm{Spec}\,}}_{{{\,\textrm{Top}\,}}}(R^\flat [1/\pi ^\flat ]) \cup {{\,\textrm{Spec}\,}}(R/(\pi )). \end{aligned}$$

If we restrict the tilting map $(-)^\flat$ to ${{\,\textrm{Spec}\,}}_{{{\,\textrm{Top}\,}}}(R[1/\pi ]) = D_R(\pi ) \cap {{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R)$, the homeomorphism induces a homeomorphism of topological spectra ${{\,\textrm{Spec}\,}}_{{{\,\textrm{Top}\,}}}(R[1/\pi ]) \cong {{\,\textrm{Spec}\,}}_{{{\,\textrm{Top}\,}}}(R^\flat [1/\pi ^\flat ]) \cong {{\,\textrm{Spec}\,}}_{{{\,\textrm{Top}\,}}}((R[1/\pi ])^\flat )$ by Lemma 5.3.

Moreover, by Proposition 5.6, the analogous restriction of the inverse map $(-)^\sharp$ is the same as the map $(-)^\sharp$ defined in [9, Section 4], and thus Theorem 5.5 provides a new proof of the main result of [9] (loc. cit., Theorem 4.16): For a perfectoid Tate ring A of characteristic 0, choose a topologically nilpotent unit $\pi$ of A such that $\pi ^{p}$ divides p in $A^{\circ }$ and then apply the above arguments to the perfectoid ring $R=A^{\circ }$ and its perfectoid element $\pi \in A^{\circ }$. We note that this new proof of loc. cit., Theorem 4.16, is more algebraic in nature in that, unlike the previous proof, it does not make use of the homeomorphism of Berkovich spectra

$$\begin{aligned} \mathcal {M}(R[1/\pi ]) \simeq \mathcal {M}(R^{\flat }[1/\pi ^{\flat }]) \end{aligned}$$

from [15, Theorem 3.3.7(c)], or the homeomorphism of adic spectra

$$\begin{aligned} {{\,\textrm{Spa}\,}}(R[1/\pi ], R[1/\pi ]^{\circ }) \simeq {{\,\textrm{Spa}\,}}(R^{\flat }[1/\pi ^{\flat }], R^{\flat }[1/\pi ^{\flat }]^{\circ }) \end{aligned}$$

from [17, Corollary 6.7(iii)].

Remark 5.8

(cf. [9, 14]) There exist perfectoid rings R of characteristic 0 such that ${{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R)$ is not equal to ${{\,\textrm{Spec}\,}}(R)$ and ${{\,\textrm{Spec}\,}}_{{{\,\textrm{perfd}\,}}}(R^{\flat })$ is not equal to ${{\,\textrm{Spec}\,}}(R^{\flat })$. An example is as follows.

Let $K$ be a perfectoid field with pseudo-uniformizer $\pi$ and let $X$ be an infinite totally disconnected compact Hausdorff space. Consider the Banach ring $C(X, K)$ of continuous $K$-valued functions on $X$ whose topology is defined by uniform convergence on all of $X$. Then $C(X, K)$ is a perfectoid Tate ring whose pseudo-uniformizer is the constant function $x \mapsto \pi$ which we also denote $\pi$ and $C(X, K)^\circ =C(X, K^\circ )$ is a perfectoid ring. Moreover, any closed prime ideal of $C(X, K)$ is maximal by [14, Theorem 29] but $C(X, K)$ has a prime ideal $\mathfrak {p}$ which is not a maximal one by the example after the proof of [14, Theorem 30]. Thus $\mathfrak {p}$ is a non-closed ideal of $C(X, K)$, in which case

$$\begin{aligned} \mathfrak {p}^\circ :=\mathfrak {p}\cap C(X, K^\circ ) \end{aligned}$$

is a prime ideal of the perfectoid ring $C(X, K^\circ )$ which is not closed with respect to the subspace topology induced from $C(X, K^\circ )$ (if $\mathfrak {p}^\circ$ was closed, then the quotient $C(X, K^\circ )/\mathfrak {p}^\circ$ would be $\pi$-adically separated and then the quotient Tate ring $C(X, K)/\mathfrak {p}=(C(X, K^\circ )/\mathfrak {p}^\circ )[1/\pi ]$ would be Hausdorff, which is a contradiction).

If the perfectoid field $K$ is of characteristic $0$ and contains $\mathbb {Q}_p$,

we can choose $p$ as $\pi$ and then $C(X, K^\circ )$ is equipped with the $p$-adic topology. This shows that $C(X, K^\circ )/\mathfrak {p}^\circ$ is not $p$-adically complete and, in particular, $\mathfrak {p}^\circ$ is not perfectoid.

Similarly, in the above situation we can let $\mathfrak {q}$ be a non-closed prime ideal in the perfect(oid) Tate ring $C(X, K^\flat )$ and then $\mathfrak {q}^\circ :=\mathfrak {q}\cap C(X, K^{\flat \circ })$ is a prime ideal in the perfect(oid) ring $C(X, K^{\flat \circ })$ of characteristic $p$ which is not $p^\flat$-adically closed.

As an application, we show that a perfectoid ring (not a perfectoid Tate ring) is an integral domain if and only if its tilt is an integral domain. This contains (and is equivalent to) a previous result about perfectoid Tate rings [9, Corollary 4.17].

Corollary 5.9

Let R be a perfectoid ring of characteristic 0, let $\pi$ be a perfectoid element in R and let $\pi ^{\flat }\in R^{\flat }$ be such that $\pi ^{\flat \sharp }$ is a unit multiple of $\pi$ in R. Then $R/R[\pi ^{\infty }]$ is an integral domain if and only if $R^{\flat }/R^{\flat }[(\pi ^{\flat })^\infty ]$ is an integral domain. In particular, $R$ is an integral domain if and only if its tilt $R^\flat$ is an integral domain.

Proof

By [7, 2.1.3], the $\pi ^{\flat }$-torsion-free quotient $R^{\flat }/R^{\flat }[(\pi ^{\flat })^\infty ]$ is the tilt of the perfectoid ring $R/R[\pi ^{\infty }]$. Hence it suffices to prove the assertion in the case when R is $\pi$-torsion-free (and thus $R^{\flat }$ is $\pi ^{\flat }$-torsion-free). Since the zero ideal of R is a perfectoid ideal, the assertion follows from Theorem 5.5. $\square$

Remark 5.10

Actually, the above corollary is a direct consequence of the previous result [9, Corollary 4.17]. In fact, as we saw in the proof of Corollary 5.9, it suffices to prove the corollary in the case when R is $\pi$-torsion-free (and $R^{\flat }$ is $\pi ^{\flat }$-torsion-free). In this case, we have injective maps $R\hookrightarrow R[1/\pi ]$ and $R^{\flat }\hookrightarrow R^{\flat }[1/\pi ^{\flat }]$ and we can apply the previous result to conclude. However, our proof does not rely on the results of [9]. In particular, we obtain a new proof of loc. cit., Corollary 4.17, which is more algebraic in nature in the sense that it does not use homeomorphisms between Berkovich spectra or adic spectra.