Perfectoid multiplier/test ideals in regular rings and bounds on symbolic powers

TL;DR

This paper introduces a new mixed characteristic analog of multiplier and test ideals using perfectoid algebras, and applies it to establish uniform bounds on symbolic powers of radical ideals in regular rings.

Contribution

It develops a novel mixed characteristic framework for multiplier and test ideals and extends uniform symbolic power bounds to all excellent regular rings.

Findings

Established subadditivity of the new ideal.

Derived uniform bounds on symbolic powers.

Extended results from equal characteristic to mixed characteristic.

Abstract

Using perfectoid algebras, we introduce a mixed characteristic analog of the multiplier ideal, respectively test ideal, from characteristic zero, respectively $p > 0$, in the case of a regular ambient ring. We prove several properties about this ideal such as subadditivity. We then use these techniques to derive a uniform bound on the growth of symbolic powers of radical ideals in all excellent regular rings. The analogous result was shown in equal characteristic by Ein-Lazarsfeld-Smith and Hochster-Huneke.

Equations

$$Q^{(mh)}\subseteq Q^m.$$

$$\mathcal{J}(R,{\mathfrak{a}}^t)\subseteq \mathcal{J}(R,{\mathfrak{b}}^t)$$

$$\mathcal{J}(R,{\mathfrak{a}}^{t^{\prime }})\subseteq \mathcal{J}(R,{\mathfrak{a}}^t).$$

$$\mathcal{J}(R,{\mathfrak{a}}^{tn})=\mathcal{J}(R,({\mathfrak{a}}^n{)}^t).$$

$$\mathcal{J}({\mathfrak{a}}^s{\mathfrak{b}}^t)\subseteq \mathcal{J}({\mathfrak{a}}^s)\cdot \mathcal{J}({\mathfrak{b}}^t).$$

$$\mathcal{J}({\mathfrak{a}}^{tn})\subseteq \mathcal{J}({\mathfrak{a}}^t{)}^n.$$

$$\begin{array}[]{rl}(\mathcal{J}/\tau)(R,\mathfrak{a}^{t})=&\operatorname{{Ann}}_{R}\{\eta\in H^{d}_{\mathfrak{m}}(R)\;|\;\text{ $\eta$'s image in $H^{d}_{\mathfrak{m}}(B)$ is ``annihilated'' by $\mathfrak{a}^{t}$}\}.\end{array}$$

$$\text{uptau}(A,{\mathfrak{a}}^t)={\mathrm{Ann}}_A\{\eta \in H_{\mathfrak{m}}^d(A)|\eta \text{ is “almost” annihilated by }{\mathfrak{a}}^t\}$$

$$\text{uptau}(A,{\mathfrak{a}}^t)\subseteq \Gamma (Y,{\mathcal{O}}_Y(\lceil K_{Y/X}-tG\rceil ))$$

$$\text{uptau}(A,{\mathfrak{a}}^t)\cdot A[1/p]\subseteq \mathcal{J}(A[1/p],(\mathfrak{a}\cdot A[1/p]{)}^t).$$

$$M{\otimes }_A^{\mathbf{L}}S\cong M{\otimes }_{A/p^k}^{\mathbf{L}}(A/p^k{\otimes }_A^{\mathbf{L}}S)\cong M{\otimes }_{A/p^k}^{\mathbf{L}}S/p^k$$

$$\underset{m}{\underleftarrow{\lim }}(F_{\bullet }^{(m)}{\otimes }_{A}S)\cong \underset{m}{\underleftarrow{\lim }}(F_{\bullet }{\otimes }_A(S/p^m))\cong F_{\bullet }{\otimes }_{A}S.$$

$$0\to {\underset{m}{\underleftarrow{\lim }}}^1{\mathcal{H}}^{-2}(F_{\bullet }^{(m)}{\otimes }_{A}S)\to {\mathcal{H}}^{-1}(\underset{m}{\underleftarrow{\lim }}(F_{\bullet }^{(m)}{\otimes }_{A}S))\to \underset{m}{\underleftarrow{\lim }}{\mathcal{H}}^{-1}(F_{\bullet }^{(m)}{\otimes }_{A}S)\to 0.$$

$${\mathcal{H}}^{-i}(F_{\bullet }^{(m)}{\otimes }_{A}S)\cong {\mathcal{H}}^{-i}(F_{\bullet }^{(m)}{\otimes }_{A/p^m}(S/p^m))\cong {\mathrm{Tor}}_i^{A/p^m}(M/p^m,S/p^m)$$

$$R_{g_1,\dots ,g_n}={\hat{\underrightarrow{\lim }}}_l{S}_l,\text{where }{S}_l:={\mathcal{O}}_Y^{+}(U(\frac{T_{g_1}-g_1,\dots ,T_{g_n}-g_n}{p^l})).$$

$$M{\otimes }_{R/t^h}^{\mathbf{L}}S/t^h\cong M{\otimes }_{R/t^{h-1}}^{\mathbf{L}}(R/t^{h-1}{\otimes }_{R/t^h}^{\mathbf{L}}S/t^h)\cong M{\otimes }_{R/t^{h-1}}^{\mathbf{L}}S/t^{h-1}.$$

$$S_l/p^{1/p}={\mathcal{O}}_{Y^{♭}}^{+}(U(\frac{f_1,\dots ,f_n}{(p^{♭}{)}^l}))/(p^{♭}{)}^{1/p}.$$

$$B:=R_n^{♭}[u_1^{1/p^{\infty }},\dots ,u_n^{1/p^{\infty }}]/(u_i^{1/p^k}(p^{♭}{)}^{l/p^k}-f_i^{1/p^k},\forall i,k).$$

$$C:=R_n^{♭}[u_1,\dots ,u_n]/(u_1(p^{♭}{)}^l-f_1,\dots ,u_n(p^{♭}{)}^l-f_n),$$

$$f_i=T_{g_i}^{♭}-g_i^{♭}\text{ in }{R}_n^{♭}/(p^{♭}{)}^{1/p}=R_n/p^{1/p}$$

$$C/(p^{♭}{)}^{1/p}\cong R^{♭}/(p^{♭}{)}^{1/p}[(T_{g_1}^{♭}{)}^{1/p^{\infty }},\dots ,(T_{g_n}^{♭}{)}^{1/p^{\infty }}][u_1,\dots ,u_n]/(T_{g_1}^{♭}-g_1^{♭},\dots ,T_{g_n}^{♭}-g_n^{♭}).$$

$$A_{\infty }=\hat{\underrightarrow{\lim }}{R}_{g_1,\dots ,g_n},$$

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\{\eta \in H_{\mathfrak{m}}^d(A)|c^{1/p^e}\eta =0\in H_{\mathfrak{m}}^d(A_{\infty })\text{ for all e}\}=0.$$

$$c^{1/p^e}(x_1\cdots x_d{)}^w\in (x_1^{w+1},\dots ,x_d^{w+1})A_{\infty }.$$

$$\frac{A}{(x_1,\dots ,x_d{)}^{p^e}}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\begin{array}[]{r}0^{\diamondsuit\mathfrak{a}^{t}}_{H_{\mathfrak{m}}^{d}(A)}=\left\{\eta\in H_{\mathfrak{m}}^{d}(A)\;|\;\forall e>0,p^{1/p^{\infty}}f^{1/p^{e}}\eta=0\text{ in }H_{\mathfrak{m}}^{d}(A_{\infty})\text{ for all $f\in\mathfrak{a}^{\lceil tp^{e}\rceil}$ }\right\},\end{array}$$

$$\begin{array}[]{rl}0^{\diamondsuit[f_{1},\dots,f_{n}]^{t}}_{H_{\mathfrak{m}}^{d}(A)}=\Big{\{}\eta\in H_{\mathfrak{m}}^{d}(A)\;|&\forall e>0,p^{1/p^{\infty}}g\eta=0\text{ in }H_{\mathfrak{m}}^{d}(A_{\infty})\\ &\text{for all $g=\prod_{i=1}^{a}f_{j_{i}}^{1/p^{e}}$ where $a\geq tp^{e}$}\Big{\}},\end{array}$$

$${\text{uptau}}^{♯}({\mathfrak{a}}^t)={\text{uptau}}^{♯}(A,{\mathfrak{a}}^t)={\mathrm{Ann}}_A(0_{H_{\mathfrak{m}}^d(A)}^{♢{\mathfrak{a}}^t})$$

$${\text{uptau}}^{♯}([f_1,\dots ,f_n{]}^t)={\text{uptau}}^{♯}(A,[f_1,\dots ,f_n{]}^t)={\mathrm{Ann}}_A(0_{H_{\mathfrak{m}}^d(A)}^{♢[f_1,\dots ,f_n{]}^t}).$$

Full text

The first named author was supported in part by NSF Grant #1836867/1600198 and NSF CAREER Grant DMS #1252860/1501102.

The second named author was supported in part by the NSF FRG Grant DMS #1265261/1501115 and NSF CAREER Grant DMS #1252860/1501102.

1. Introduction

In this paper, we prove the following result on the uniform bound on the growth of symbolic powers of ideals.

Main Theorem** (Theorem 7.4).**

Suppose that $R$ is a Noetherian regular ring with reduced formal fibers (e.g. $R$ is excellent). If $Q\subseteq R$ is a prime ideal of height $h$, then for all $m>0$ we have

[TABLE]

where $Q^{(mh)}$ is the $mh$ symbolic power of $Q$.111$Q^{(mh)}=Q^{mh}R_{Q}\cap R$, i.e., the elements of $R$ which vanish generically to order $mh$ at $Q$.

When $R$ is finite type over $\mathbb{C}$, this result was proved as an application of multiplier ideals by Ein–Lazarsfeld–Smith [ELS01]. Shortly later, Hochster–Huneke [HH02] used tight closure theory to prove the result when $R$ contains a field. Also see [Swa00] where it was first shown that there is a linear containment relation between symbolic and ordinary powers. Our contribution to the Main Theorem is the mixed characteristic case, which answers the question of Hochster–Huneke in [HH02, Section 5].

Since regular local rings are UFDs, every height one prime $Q$ is principal and hence $Q^{(m)}=Q^{m}$. Thus the Main Theorem can be viewed as a strengthening and generalization of this classical fact to primes of higher codimension. Moreover, starting with the aforementioned results, the question of the growth of symbolic powers has been of central importance in commutative algebra and its applications to algebraic geometry over the past few decades, see for example [HH07, HKV09, BDH*+*09, BH10, DDG*+*18].

The main ideas of our proof come from the recent solution of the direct summand conjecture and its derived variant [And16a, Bha18]: we introduce a mixed characteristic analog of the multiplier ideal or test ideal using perfectoid algebras, prove many properties of it, and finally (and analogously to the strategy of [ELS01], see also [Har05]) use those properties to deduce the Main Theorem above.

1.1. Multiplier and test ideals

Suppose that $R$ is an equal characteristic regular domain satisfying mild geometric assumptions222For example, of essentially finite type over a field, or complete, or $F$-finite in characteristic $p>0$.. Further suppose that $\mathfrak{a}\subseteq R$ is an ideal and $t\in\mathbb{R}_{\geq 0}$ a formal exponent for $\mathfrak{a}$. In this setting we can form the test ideal $\tau(R,\mathfrak{a}^{t})$ in characteristic $p>0$ or the multiplier ideal $\mathcal{J}(R,\mathfrak{a}^{t})$ in characteristic [math].

This is an ideal of $R$ which measures the singularities of $V(\mathfrak{a})\subseteq\operatorname{{Spec}}R$, scaled by $t$. Roughly speaking, for relevant values of $t$, the multiplier or test ideal of $(R,\mathfrak{a}^{t})$ is smaller/deeper than that of $(R,\mathfrak{b}^{t})$ if $V(\mathfrak{a})$ has the same dimension as $V(\mathfrak{b})$ and is more singular than $V(\mathfrak{b})$. Crucially for the applications to symbolic powers, the multiplier or test ideal satisfies the following list of properties, see for example [HH90, HY03, Laz04]. We state them for the multiplier ideal $\mathcal{J}(R,\mathfrak{a}^{t})$ but they also hold for the test ideal $\tau(R,\mathfrak{a}^{t})$.

• (A)

Basic containments: If $\mathfrak{a}\subseteq\mathfrak{b}$ is a containment of ideals, then

[TABLE]

and if $t<t^{\prime}$ then

[TABLE]

• (B)

Unambiguity of exponent: For any positive integer $n$,

[TABLE]

• (C)

Not too small: $\mathfrak{a}\subseteq\mathcal{J}(R,\mathfrak{a})$.

• (D)

Not too big: If $\mathfrak{a}$ is prime of height $h$, $\mathcal{J}(R,(\mathfrak{a}^{(lh)})^{{1\over l}})\subseteq\mathfrak{a}$.

• (E)

Subadditivity: If $\mathfrak{b}$ is another ideal and if $s\geq 0$ is another real number, then

[TABLE]

In particular we have

[TABLE]

Combining these results, the application to the growth of symbolic powers follows from a clever asymptotic construction of multiplier ideals [ELS01], see also [Har05]. We aim to do the same thing in mixed characteristic.

Very roughly, the multiplier ideal and test ideal of a regular local ring $(R,\mathfrak{m})$ of dimension $d$ can be defined in the following way:

[TABLE]

Here $B$ and “annihilated” are made precise as follows:

$B$** in characteristic zero: **

$B={\bf R}\pi_{*}\mathcal{O}_{Y}$ where $\pi:Y\xrightarrow{\ \ }\operatorname{{Spec}}R$ is a log resolution.333$\pi:Y\xrightarrow{\ \ }\operatorname{{Spec}}R$ is proper birational, $Y$ is regular and $\mathfrak{a}\cdot\mathcal{O}_{Y}$ defines a SNC divisor. We define the $\eta$ which are “annihilated” as follows: write $\mathfrak{a}\cdot\mathcal{O}_{Y}=\mathcal{O}_{Y}(-G)$ and consider $\eta$ whose image in $\mathbb{H}^{d}_{\mathfrak{m}}({\bf R}\pi_{*}\mathcal{O}_{Y}(\lfloor tG\rfloor))$ is zero.

$B$** in characteristic $p>0$: **

$B=R^{1/p^{\infty}}$, the perfection of $R$. We define the $\eta$ which are “annihilated” as follows: those $\eta$ such that $c^{1/p^{e}}(\mathfrak{a}^{\lceil tp^{e}\rceil})^{1/p^{e}}\eta=0\in H^{d}_{\mathfrak{m}}(R^{1/p^{\infty}})$ for some $c\neq 0$ and all $e>0$.

The real power of both the multiplier and test ideal (and their related circles of ideas) are the associated vanishing theorems that accompany them. In characteristic zero this is Kawamata–Viehweg vanishing [Kaw82, Vie82], see for instance [EV92]. In characteristic $p>0$, Serre vanishing combined with the Frobenius morphism plays an analogous role.

Our goal in this article is to develop a theory of the test ideal in mixed characteristic regular local rings and prove it satisfies properties (A) through (E) above.

1.2. Perfectoid test ideals

Now let us assume that $A$ is a complete regular local ring of mixed characteristic $(0,p)$. In this situation, for every fixed element $g\in A$, André constructed an $A$-algebra $A_{\infty}$ that is an integral perfectoid algebra almost faithfully flat over $A$ mod powers of $p$ and such that $g\in A$ has a compatible system of $p$-power roots in $A_{\infty}$ (in this case, $g^{1/p^{e}}$ will be declared compatible). This ingenious construction is crucial in the solution of the mixed characteristic case of the direct summand conjecture [And16a, Bha18] and the existence of big Cohen-Macaulay algebras [And16a, HM17, Shi17, And18].

In this article, we iterate André’s construction to obtain a huge extension $A\xrightarrow{\ \ }A_{\infty}$ that is almost faithfully flat over $A$ and such that all elements of $A$ have a compatible systems of $p$-power roots in $A_{\infty}$. We will use this $A_{\infty}$ as the $B$ to replace $R^{1/p^{\infty}}$ in the definition of the test ideal in characteristic $p>0$ (or as a replacement for the ${\bf R}\pi_{*}\mathcal{O}_{Y}$ in the definition of the multiplier ideal in characteristic [math]). Inspired by this, let $\mathfrak{a}\subseteq A$ be an ideal, we define the perfectoid test ideal of $(A,\mathfrak{a}^{t})$ to be

[TABLE]

There are different ways to interpret this $\mathfrak{a}^{t}$ action in our setting, and at least in some proofs, it is convenient to define our analog of the test ideal with respect to a sequence of elements $\{f_{1},\ldots,f_{n}\}$ that generate $\mathfrak{a}$. See Section 3 for more details of these definitions.

We then show that the above perfectoid test ideals satisfy the analogs of properties (A) through (E) above in Section 3, Section 3, Section 3, Section 3, Theorem 5.11, Theorem 4.4. Putting these together, and defining asymptotic perfectoid test ideals similar to how asymptotic multiplier ideals were introduced in [ELS01], we obtain our Main Theorem.

Beyond this, it is also natural to compare our perfectoid test ideals with multiplier ideals in characteristic 0. We obtain the following result, which is essentially a corollary of our proof of property (D) in mixed characteristic.

Theorem** (Theorem 6.1, Theorem 6.3).**

Suppose that $(A,\mathfrak{a}^{t})$ is a pair, where $A$ is a complete regular local ring of mixed characteristic $(0,p)$. First suppose that $\pi:Y\xrightarrow{\ \ }X=\operatorname{{Spec}}A$ is a proper birational map with $Y$ normal and such that $\mathfrak{a}\cdot\mathcal{O}_{Y}=\mathcal{O}_{Y}(-G)$. Then

[TABLE]

where the object on the right would be the multiplier ideal if $Y$ is a log resolution.

Furthermore, since $A[1/p]$ has characteristic zero, we can form the multiplier ideal $\mathcal{J}(A[1/p],(\mathfrak{a}\cdot A[1/p])^{t})$. We have:

[TABLE]

We also expect that the characteristic zero statement is an equality, but we do not know how to show this.

Acknowledgements:

The authors would like to thank Bhargav Bhatt, Raymond Heitmann, Kiran Kedlaya, Tiankai Liu, Stefan Patrikis, and Peter Scholze for valuable conversations. We thank Rankeya Datta for comments on a previous draft. Finally, we thank all the referees for numerous comments on previous versions – their feedback has substantially improved the paper.

2. Perfectoid algebras and André’s construction

Throughout this paper we will use the language of (integral) perfectoid algebras and almost mathematics as in [Sch12], [GR03], [Bha18], [And18]. We will work over a fixed perfectoid field $K=\widehat{\mathbb{Q}_{p}(p^{1/p^{\infty}})}$ and its ring of integers $K^{\circ}=\widehat{\mathbb{Z}_{p}[p^{1/p^{\infty}}]}$. We collect some definitions from [Sch12, Section 5], [Bha18, Section 1.4], [And18, Section 2.2]. Additionally, we use the notation ${\mathcal{H}}^{j}(\bullet)$ to denote the $j$th cohomology of a complex.

A perfectoid $K$-algebra is a Banach $K$-algebra $R$ such that the subring of power-bounded elements $R^{\circ}\subseteq R$ is bounded and the Frobenius is surjective on $R^{\circ}/p$. A $K^{\circ}$-algebra $S$ is called integral perfectoid if it is $p$-adically complete, $p$-torsion free, and the Frobenius induces an isomorphism $S/p^{1/p}\xrightarrow{\sim}S/p$. If $R$ is a perfectoid $K$-algebra, then the subring of power-bounded elements $R^{\circ}$ is integral perfectoid, and if $S$ is integral perfectoid, then $S[1/p]$ perfectoid, see [Sch12, Theorem 5.2]. Unless otherwise stated, almost mathematics in this paper will always be measured with respect to the ideal $(p^{1/p^{\infty}})\subseteq K^{\circ}$.

Remark 2.1*.*

In [Bha18], there is an extra condition in the definition of integral perfectoid algebra: one requires that $S=S_{*}:=\{x\in S[1/p]\hskip 5.0pt|\hskip 5.0ptp^{1/p^{n}}x\in S\text{ for all }n\}$. If we impose this extra condition then [Sch12, Theorem 5.2] says the two categories are equivalent. In particular, $S_{*}=S[1/p]^{\circ}$ is integrally closed in $S[1/p]=S_{*}[1/p]$. Since passing from $S$ to $S_{*}$ is harmless for all our purposes (they are almost isomorphic to each other), we will assume that $S$ is integrally closed in $S[1/p]$ for all integral perfectoid algebras in the remainder of the article.

We recall the following definitions.

• $\circ$

A map $A\xrightarrow{\ \ }S$ such that $S$ is a $K^{\circ}$-algebra is almost flat if $\operatorname{{Tor}}_{i}^{A}(M,S)$ is almost zero (i.e., annihilated by $(p^{1/p^{\infty}})$) for all $A$-modules $M$ and all $i>0$. By taking syzygies and degree shifting, it suffices that $\operatorname{{Tor}}_{1}^{A}(M,S)$ is almost zero for all $A$-modules $M$.

• $\circ$

A map $R\xrightarrow{\ \ }S$ of $K^{\circ}$-algebras is almost faithfully flat if it almost flat, and such that if $M\otimes_{R}S$ is almost zero then $M$ is almost zero.

The goal of this section is to explain the following:

Theorem 2.2**.**

Let $(A,\mathfrak{m})$ be a complete regular local ring of mixed characteristic $(0,p)$ and dimension $d$. Then there exists a map $A\xrightarrow{\ \ }A_{\infty}$ to an integral perfectoid $K^{\circ}$-algebra $A_{\infty}$ such that:

1. (a)

All elements of $A$ have a compatible system of $p$-power roots in $A_{\infty}$. 2. (b)

$A\xrightarrow{\ \ }A_{\infty}$* is almost flat. In particular, $A\hookrightarrow A_{\infty}$ is injective, and nonzero elements of $A$ are nonzerodivisors in $A_{\infty}$.* 3. (c)

If $M\otimes_{A}A_{\infty}$ is almost zero, then $M=0$.555We caution the reader that the term “almost faithfully flat” is only defined when we consider maps of $K^{\circ}$-algebras while our base ring $A$ here is not defined over $K^{\circ}$ (e.g., saying $M$ is almost zero does not usually make sense here since $M$ is just an $A$-module). This is the reason we treat the properties $(b)$ and $(c)$ separately in the statement.

In order to prove Theorem 2.2, we need the following fact about almost flat maps which we believe is known to experts (this is the “almost” analog of [BMS16, Remark 4.31]). We include an elementary proof since we cannot find a good reference.

Lemma 2.3**.**

Let $A\xrightarrow{\ \ }S$ be a map of $p$-adically complete and $p$-torsion free rings such that $A$ is Noetherian and $S$ is a $K^{\circ}$-algebra. If $A/p^{k}\xrightarrow{\ \ }S/p^{k}$ is almost flat for all $k>0$, then $A\xrightarrow{\ \ }S$ is almost flat.

Proof.

We want show $\operatorname{{Tor}}_{1}^{A}(M,S)$ is almost zero for all $A$-modules $M$. We can assume that $M$ is finitely generated by taking direct limit. By considering $0\xrightarrow{\ \ }\Gamma_{(p)}M\xrightarrow{\ \ }M\xrightarrow{\ \ }\overline{M}\xrightarrow{\ \ }0$, we only need to handle the case when $M$ is annihilated by $p^{k}$ for some $k$ and the case when $M$ is $p$-torsion free.

Case 1:

If $M$ is annihilated by $p^{k}$, then we have

[TABLE]

and hence $\operatorname{{Tor}}_{1}^{A}(M,S)\cong\operatorname{{Tor}}_{1}^{A/p^{k}}(M,S/p^{k})$ is almost zero.

Case 2:

Now we assume that $M$ is $p$-torsion free. Let $F_{\bullet}\xrightarrow{\ \ }M\xrightarrow{\ \ }0$ be a free resolution of $M$ with each term of $F_{\bullet}$ a finite free $A$-modules (since $A$ is Noetherian and $M$ is finitely generated). We set $F_{\bullet}^{(m)}=F_{\bullet}\otimes A/p^{m}\cong M/p^{m}$ since $M$ is $p$-torsion free. Since $S$ is $p$-adically complete, we have

[TABLE]

Since $\{F_{\bullet}^{(m)}\otimes_{A}S\}_{m}$ forms a tower of chain complexes with surjective transition maps, we have a Milnor exact sequence [Wei94, Theorem 3.5.8]:

[TABLE]

Since $S/p^{m}$ is almost flat over $A/p^{m}$, we know that

[TABLE]

is almost zero for every $i>0$. This together with the above discussion shows that $\operatorname{{Tor}}_{1}^{A}(M,S)={\mathcal{H}}^{-1}(F_{\bullet}\otimes_{A}S)={\mathcal{H}}^{-1}(\varprojlim_{m}(F_{\bullet}^{(m)}\otimes_{A}S))$ is almost zero. ∎

To prove Theorem 2.2, we also need an ingenious construction of certain integral perfectoid algebras due to André [And16a, Section 2.5], see also [Bha18, Secton 2]. Below we recall André’s construction.

2.1. André’s technique of adjoining $p$-power roots

Let $R$ be an integral perfectoid $K^{\circ}$-algebra and $g_{1},\dots,g_{n}$ be a finite set of elements of $R$. We want to construct an almost faithfully flat extension $R\xrightarrow{\ \ }R_{g_{1},\dots,g_{n}}$ of integral perfectoid $K^{\circ}$-algebras such that each $g_{i}$ admits a compatible system of $p$-power roots in $R_{g_{1},\dots,g_{n}}$.

We consider the integral perfectoid algebra $R_{n}=R\langle T_{g_{1}}^{1/p^{\infty}},\dots,T_{g_{n}}^{1/p^{\infty}}\rangle$ where the $T_{g_{i}}$ are indeterminates, and we let $Y=\operatorname{Spa}(R_{n}[\frac{1}{p}],R_{n})$ be the associated perfectoid space. We set $R_{g_{1},\dots,g_{n}}$ to be the integral perfectoid ring of functions on the Zariski closed subset of $Y$ defined by the ideal $(T_{g_{1}}-g_{1},\dots,T_{g_{n}}-g_{n})$. Explicitly, we have

[TABLE]

Here $U(\frac{T_{g_{1}}-g_{1},\dots,T_{g_{n}}-g_{n}}{p^{l}})$ denotes the rational subset \big{\{}y\in Y\;\big{|}\;|T_{g_{i}}(y)-g_{i}(y)|\leq|p^{l}|,\forall i\big{\}} and the completion is $p$-adic. Each $S_{l}$ is integral perfectoid. Using the explicit description of $R_{g_{1},\dots,g_{n}}$ (which does not depend on the order of $g_{1},\dots,g_{n}$) we have a map $R_{g_{1},\dots,g_{n}}\xrightarrow{\ \ }R_{g_{1},\dots,g_{n},h_{1},\dots,h_{m}}$. Therefore $\{R_{g_{1},\dots,g_{n}}\}$, where $g_{1},\dots,g_{n}$ runs through all the finite sets of elements of $A$, naturally form a directed system of integral perfectoid $K^{\circ}$-algebras.

Note that in $R_{g_{1},\dots,g_{n}}$, we have $T_{g_{i}}=g_{i}$ for each $i$ since $T_{g_{i}}-g_{i}$ is divisible by $p^{l}$ for all $l$ in $R_{g_{1},\dots,g_{n}}$. Thus, each $g_{i}$ has a compatible system of $p$-power roots $g_{i}^{1/p^{k}}=T_{g_{i}}^{1/p^{k}}$ in $R_{g_{1},\dots,g_{n}}$. In the case that we only have one element $g=g_{1}$, the next lemma is [Bha18, Theorem 2.3] (which was originally proved by André in [And16a, Section 2.5] under a slightly different setup). A similar argument works for any finite sets of elements $g_{1},\dots,g_{n}$ and we give details below.

Lemma 2.4** (André).**

For each $l>0$, the map $R/p^{h}\xrightarrow{\ \ }S_{l}/p^{h}$ is almost faithfully flat for all $h>0$. Consequently, $R/p^{h}\xrightarrow{\ \ }R_{g_{1},\dots,g_{n}}/p^{h}$ is almost faithfully flat.

Proof.

We first claim that it is enough to show $R/p^{1/p}\xrightarrow{\ \ }S_{l}/p^{1/p}$ is almost faithfully flat. This follows from the more general fact:

Claim 2.5**.**

Let $R\xrightarrow{\ \ }S$ be a map of $K^{\circ}$-algebras (resp. $K^{\flat\circ}$-algebras). Suppose $R$, $S$ are both $t$-torsion free for some $0\neq t\in K^{\circ}$ (resp. $K^{\flat\circ}$). If $S/t$ is almost faithfully flat over $R/t$, then $S/t^{h}$ is almost faithfully flat over $R/t^{h}$ for all $h\geq 1$.

Proof.

We first prove $\operatorname{{Tor}}_{1}^{R/t^{h}}(S/t^{h},M)$ is almost zero for all $R/t^{h}$-modules $M$. By considering $0\xrightarrow{\ \ }tM\xrightarrow{\ \ }M\xrightarrow{\ \ }M/t\xrightarrow{\ \ }0$, it is enough to prove this when $M$ is annihilated by $t^{h-1}$. Since $t$ is a nonzerodivisor on both $R$ and $S$, we have

[TABLE]

Therefore we are done by induction on $h$.

We next note that if $N$ is an $R/t^{h}$-module and $S/t^{h}\otimes_{R/t^{h}}N$ is almost zero, then $S/t\otimes_{R/t}{N/t}$ is almost zero. Hence $N/t$ is almost zero since $S/t$ is almost faithfully flat over $R/t$. But this implies $N=N/t^{h}$ is almost zero because $N/t^{h}$ has a finite filtration with each factor a quotient of $N/t$. This finishes the proof. ∎

By Scholze’s approximation lemma [Sch12, Corollary 6.7], there exists $f_{1},\dots,f_{n}$ in $R_{n}^{\flat}=R^{\flat}\langle(T_{g_{1}}^{\flat})^{1/p^{\infty}},\dots,(T_{g_{n}}^{\flat})^{1/p^{\infty}}\rangle$ such that

• $\circ$

$f_{i}^{\sharp}\equiv T_{g_{i}}-g_{i}$ mod $p^{1/p}$ for every $i$.

• $\circ$

$U(\frac{T_{g_{1}}-g_{1},\dots,T_{g_{n}}-g_{n}}{p^{l}})=U(\frac{f_{1}^{\sharp},\dots,f_{n}^{\sharp}}{p^{l}})$ as rational subsets of $Y$.

Therefore $S_{l}=\mathcal{O}_{Y}^{+}(U(\frac{f_{1}^{\sharp},\dots,f_{n}^{\sharp}}{p^{l}}))$ whose tilt is $\mathcal{O}_{Y^{\flat}}^{+}(U(\frac{f_{1},\dots,f_{n}}{(p^{\flat})^{l}}))$ by [Sch12, Theorem 6.3 (ii)]. In particular, we have

[TABLE]

Thus proving $S_{l}/p^{1/p}$ is almost faithfully flat over $R/p^{1/p}$is the same as proving $\mathcal{O}_{Y^{\flat}}^{+}(U(\frac{f_{1},\dots,f_{n}}{(p^{\flat})^{l}}))/(p^{\flat})^{l}$ is $(p^{\flat})^{1/p^{\infty}}$-almost faithfully flat over $R^{\flat}/(p^{\flat})^{1/p}$. Now by [Sch12, Lemma 6.4, third equation in the proof of (i)], we know that $\mathcal{O}_{Y^{\flat}}^{+}(U(\frac{f_{1},\dots,f_{n}}{(p^{\flat})^{l}}))$ is $(p^{\flat})^{1/p^{\infty}}$-almost isomorphic to the $p^{\flat}$-adic completion of

[TABLE]

Hence it is enough to show that $R^{\flat}/(p^{\flat})^{1/p}\xrightarrow{\ \ }B/(p^{\flat})^{1/p}$ is almost faithfully flat.

At this point, we note that $B$ is the perfection of

[TABLE]

and by our choice of $f_{i}$,

[TABLE]

for some $g^{\flat}_{i}\in R_{n}^{\flat}$ (which amounts to choose a compatible sequence $\{g_{i,k}\}$ such that $g_{i,k}^{p^{k}}=g_{i}$ in $R_{n}/p^{1/p}$). Thus we have

[TABLE]

It follows that $C/(p^{\flat})^{1/p}$ is free over $R^{\flat}/(p^{\flat})^{1/p}$ and that $(p^{\flat})^{1/p},u_{1}(p^{\flat})^{l}-f_{1},\dots,u_{n}(p^{\flat})^{l}-f_{n}$ is a regular sequence on $R_{n}^{\flat}[u_{1},\dots,u_{n}]$. In particular, $(p^{\flat})^{1/p}$ is a nonzerodivisor on $C$.666We are using the fact that if $y_{1},\dots,y_{n}$ is a regular sequence on a (possibly non-Noetherian, non-local) ring $R$, then $y_{1}$ is always a nonzerodivisor on $R/(y_{2},\dots,y_{n})$. By induction it comes down to the case $n=2$, where one can check directly [BH93, Paragraph before Proposition 1.1.6]. By Section 2.1, $R^{\flat}\xrightarrow{\ \ }C$ is almost faithfully flat mod any power of $(p^{\flat})^{1/p}$. Since $R^{\flat}\xrightarrow{\ \ }B=C_{\text{perf}}$ is the direct limit of the maps $R^{\flat}\xrightarrow{\text{Frob}^{e}}R^{\flat}\xrightarrow{\ \ }C$ and the latter is almost faithfully flat mod any power of $(p^{\flat})^{1/p}$ ($\text{Frob}^{e}$ is an isomorphism), $R^{\flat}\xrightarrow{\ \ }B$ is almost faithfully flat mod any power of $(p^{\flat})^{1/p}$ as desired. ∎

2.2. Proof of Theorem 2.2 and its consequences

We now prove our main result in this section. We need the following construction from [Bha18].

Lemma 2.6** (Proposition 5.2 in [Bha18]).**

Let $(A,\mathfrak{m})$ be a complete regular local ring of mixed characteristic $(0,p)$ and dimension $d$. Then there exists a map $A\xrightarrow{\ \ }R$ such that $R$ admits the structure of an integral perfectoid $K^{\circ}$-algebra, and such that

1. (1)

$A\xrightarrow{\ \ }R$* is almost flat.* 2. (2)

If $M\otimes_{A}R$ is almost zero, then $M=0$.

If we compare Section 2.2 with Theorem 2.2, the latter satisfies one extra condition (that all elements of $A$ have compatible system of $p$-power roots). Our strategy of the proof is to start with $R$ as in Section 2.2, apply the construction in Section 2.1 to $R$ for all the finite sets of elements $g_{1},\dots,g_{n}$, and then take a (huge) completed direct limit. We will show that the conditions $(1)$ and $(2)$ in Section 2.2 are preserved under these constructions. Below we give details.

Proof of Theorem 2.2.

We first construct $A\xrightarrow{\ \ }R$, where $R$ is an integral perfectoid $K^{\circ}$-algebra as in Section 2.2. For any finite set of elements $g_{1},\dots,g_{n}$ of $A$, we have $R\xrightarrow{\ \ }R_{g_{1},\dots,g_{n}}$ as in Section 2.1. Since $\{R_{g_{1},\dots,g_{n}}\}$ form a directed system, we set

[TABLE]

where the direct limit is taken over all the finite sets of elements of $A$ and the completion is $p$-adic. Then $A_{\infty}$ is an integral perfectoid $K^{\circ}$-algebra such that all elements of $A$ have a compatible system of $p$-power roots in $A_{\infty}$. It remains to prove that $A_{\infty}$ satisfies properties $(b)$ and $(c)$ in Theorem 2.2.

Proving (b)

Since $A_{\infty}/p^{k}=\varinjlim(R_{g_{1},\dots,g_{n}}/p^{k})$ and $R_{g_{1},\dots,g_{n}}/p^{k}$ is almost flat over $R/p^{k}$ by Section 2.1, $R/p^{k}\xrightarrow{\ \ }A_{\infty}/p^{k}$ is almost flat for all $k>0$. Since $A\xrightarrow{\ \ }R$ is almost flat by Section 2.2, it follows that $A/p^{k}\xrightarrow{\ \ }A_{\infty}/p^{k}$ is almost flat for all $k>0$. This implies $A\xrightarrow{\ \ }A_{\infty}$ is almost flat by Section 2.

Proving (c)

Suppose that $M\otimes_{A}A_{\infty}$ is almost zero but $M\neq 0$. We can choose $A/I$ a nonzero cyclic $A$-submodule of $M$. Since $A_{\infty}$ is almost flat over $A$, $A/I\otimes_{A}A_{\infty}\xrightarrow{\ \ }M\otimes_{A}A_{\infty}$ is an almost injection. Thus $A/I\otimes_{A}A_{\infty}$ is almost zero, and it follows that $A/\mathfrak{m}\otimes_{A}A_{\infty}$ is almost zero since $A/I\otimes_{A}A_{\infty}\twoheadrightarrow A/\mathfrak{m}\otimes_{A}A_{\infty}$. But $R/\mathfrak{m}R\cong A/\mathfrak{m}\otimes_{A}R$ is not almost zero by Section 2.2 (2), and since $R/p\xrightarrow{\ \ }A_{\infty}/p$ is almost faithfully flat, it follows that $A/\mathfrak{m}\otimes_{A}A_{\infty}\cong R/\mathfrak{m}R\otimes_{R/p}A_{\infty}/p$ is not almost zero. This is a contradiction. ∎

Corollary 2.7**.**

Let $(A,\mathfrak{m})$ be a complete regular local ring of mixed characteristic $(0,p)$ and dimension $d$. Let $A\xrightarrow{\ \ }A_{\infty}$ be a map as in Theorem 2.2. Then we have

1. (1)

If $x_{1},\dots,x_{d}$ is a system of parameters of $A$, then it is an almost regular sequence on $A_{\infty}$, i.e., $(x_{1},\ldots,x_{i})A_{\infty}:_{A_{\infty}}x_{i+1}\over(x_{1},\ldots,x_{i})A_{\infty}$ is almost zero for every $0\leq i\leq d-1$ and $A_{\infty}\over(x_{1},\ldots,x_{d})A_{\infty}$ is not almost zero.777Recall that $(x_{1},\ldots,x_{i})A_{\infty}:_{A_{\infty}}x_{i+1}$ denotes the set of elements of $A_{\infty}$ that multiply $x_{i+1}$ into $(x_{1},\ldots,x_{i})A_{\infty}$. 2. (2)

$A_{\infty}\otimes\operatorname{Hom}_{A}(N,M)$* is almost isomorphic to $\operatorname{Hom}_{A_{\infty}}(N\otimes A_{\infty},M\otimes A_{\infty})$ for all finitely generated $A$-modules $N$ and all $A$-modules $M$.*

Proof.

$(1)$: $x_{1},\dots,x_{d}$ is a regular sequence on $A$ because $A$ is regular. Since $A\xrightarrow{\ \ }A_{\infty}$ is almost flat by Theorem 2.2 $(b)$, it follows that $(x_{1},\ldots,x_{i})A_{\infty}:x_{i+1}\over(x_{1},\ldots,x_{i})A_{\infty}$ is almost zero. Moreover, $A_{\infty}\over(x_{1},\ldots,x_{d})A_{\infty}$ is not almost zero by Theorem 2.2 $(c)$.

$(2)$: Let $A^{\oplus l}\xrightarrow{\ \ }A^{\oplus n}\xrightarrow{\ \ }N\xrightarrow{\ \ }0$ be a presentation of $N$, then we have $A_{\infty}^{\oplus l}\xrightarrow{\ \ }A_{\infty}^{\oplus n}\xrightarrow{\ \ }N\otimes A_{\infty}\xrightarrow{\ \ }0$. We look at the following commutative diagram:

[TABLE]

The second line is exact, and the first line is almost exact because $A_{\infty}$ is almost flat over $A$ by Theorem 2.2 $(b)$. The conclusion follows by chasing the diagram. ∎

Remark 2.8*.*

We record here some very recent progress towards constructing integral perfectoid (almost) big Cohen-Macaulay algebras. For simplicity we will keep our notations: $A$ is a complete regular local ring of mixed characteristic $(0,p)$.

1. (a)

In Bhatt’s unpublished note [Bha17, Corollary 9.4.7], it is proved that there exists an integral perfectoid $A$-algebra $A_{\infty,\infty}$ that is almost flat over $A$ mod $p^{k}$, and such that $A_{\infty,\infty}$ is absolutely integrally closed: each monic polynomial has a root in $A_{\infty,\infty}$. In particular every element of $A_{\infty,\infty}$ admits a compatible system of $p$-power roots. It follows from the same proof of Theorem 2.2 that $A_{\infty,\infty}$ satisfies the conclusion of Theorem 2.2. Bhatt’s construction is similar to ours in spirit (i.e., iterate André’s construction and take certain completed direct limit). 2. (b)

Shimomoto [Shi17, Main Theorem 2] and André [And18, Theorem 3.1.1] proved that one can construct an integral perfectoid $A$-algebra $B$ that is a big Cohen-Macaulay $A^{+}$-algebra. Here $A^{+}$ denotes the absolute integral closure of $A$: the integral closure of $A$ in an algebraic closure of its fraction field. Since $B$ is an $A^{+}$ algebra, elements of $A$ have compatible system of $p$-power roots in $B$, and $B$ is (honestly) faithfully flat over $A$ because $A$ is regular and $B$ is big Cohen-Macaulay. Therefore $B$ also satisfies the conclusion of Theorem 2.2. The construction of $B$ is quite difficult and relies on André’s perfectoid Abhyankar lemma [And16b].

Setting 2.9**.**

Throughout the rest of this article. We fix our notations as follows:

• $\circ$

$(A,\mathfrak{m})$ will always denote a complete regular local ring of mixed characteristic $(0,p)$ and dimension $d$.

• $\circ$

$A_{\infty}$ will always denote a fixed integral perfectoid $K^{\circ}$-algebra that satisfies the conclusion of Theorem 2.2. The existence of such $A_{\infty}$ follows from Theorem 2.2 (see also Section 2).

• $\circ$

$p^{1/p^{\infty}}z=0$ means $z$ is almost zero. More precisely, this means $p^{1/p^{k}}z=0$ for all positive integers $k$, or equivalently, $z$ is $(p^{1/p^{\infty}})$-torsion where $(p^{1/p^{\infty}})\subseteq K^{\circ}$ is the ideal that we use to measure almost mathematics.

We caution the reader that, although $A_{\infty}$ is reduced (since it is integral perfectoid), taking $p^{e}$-th roots in $A_{\infty}$ is generally not unique because we are working in mixed characteristic. In particular, elements of $A$ may have many compatible systems of $p$-power roots in $A_{\infty}$. This will be addressed carefully throughout this paper.

We end this section with the following lemma.

Lemma 2.10**.**

Let $c\neq 0$ be an element of $A$, and fix any compatible system of $p$-power roots $\{c^{1/p^{e}}\}_{e=1}^{\infty}$ in $A_{\infty}$. Then

[TABLE]

Proof.

Recall that $H_{\mathfrak{m}}^{d}(A)$ is the injective hull of $A/\mathfrak{m}$ (since $A$ is Gorenstein [BH93, Proposition 3.5.4]), and by the Čech complex description of local cohomology [BH93, page 130], the socle element (up to multiplication by a unit) of $H_{\mathfrak{m}}^{d}(A)$ can be expressed as $\frac{1}{x_{1}\cdots x_{d}}$ where $x_{1},\dots,x_{d}$ is a regular system of parameters of $A$. Since the injective hull is an essential extension, we get that $\frac{1}{x_{1}\cdots x_{d}}$ lies in the left hand side if the latter is not zero. Thus we have $\frac{c^{1/p^{e}}}{x_{1}\cdots x_{d}}=0$ in $H_{\mathfrak{m}}^{d}(A_{\infty})$ for every $e$. This means for every $e$, there exists a $w$ depending on $e$ such that

[TABLE]

Since $x_{1},\dots,x_{d}$ is an almost regular sequence on $A_{\infty}$ by Section 2, this implies $p^{1/p^{e}}c^{1/p^{e}}\in(x_{1},\dots,x_{d})A_{\infty}$888In general, $z(x_{1}\cdots x_{d})^{w}\in(x_{1}^{w+t},\dots,x_{d}^{w+t})A_{\infty}$ implies $p^{1/p^{\infty}}z\in(x_{1}^{t},\dots,x_{d}^{t})A_{\infty}$. The condition implies $x_{1}^{w}(z(x_{2}\cdots x_{d})^{w}-ax_{1}^{t})\in(x_{2}^{w+t},\dots,x_{d}^{w+t})A_{\infty}$ for some $a\in A_{\infty}$. So $p^{1/p^{\infty}}(z(x_{2}\cdots x_{d})^{w}-a_{1}x_{1}^{t})\in(x_{2}^{w+t},\dots,x_{d}^{w+t})A_{\infty}$ and thus $p^{1/p^{\infty}}z(x_{2}\cdots x_{d})^{w}\in(x_{1}^{t},x_{2}^{w+t},\dots,x_{d}^{w+t})A_{\infty}.$ Note that we have dropped the exponent of $x_{1}$ at the expense of multiplying by $p^{1/p^{\infty}}$. Do the same thing for $x_{2},\dots,x_{d}$ consecutively, we have $p^{1/p^{\infty}}z\in(x_{1}^{t},\dots,x_{d}^{t})A_{\infty}.$ This fact will be used in Section 5.for every $e$ and thus $pc\in(x_{1},\dots,x_{d})^{p^{e}}A_{\infty}$ for every $e$. Now we consider the following commutative diagram:

[TABLE]

where the vertical maps send $1$ to $pc$. Since $pc\in(x_{1},\dots,x_{d})^{p^{e}}A_{\infty}$, we have $\phi=0$. On the other hand, $\phi$ is an almost injection because $A_{\infty}$ is almost flat over $A$ by Theorem 2.2. Thus $\frac{A_{\infty}}{((x_{1},\dots,x_{d})^{p^{e}}:pc)A_{\infty}}$ is almost zero, so $\frac{A}{(x_{1},\dots,x_{d})^{p^{e}}:pc}=0$ by Theorem 2.2. Therefore $pc\in(x_{1},\dots,x_{d})^{p^{e}}A$ for every $e$, which is a contradiction. ∎

3. Perfectoid test ideals

Our goal in this section is to introduce a mixed characteristic ideal which is analogous to the multiplier ideal which appeared in higher dimensional algebraic geometry and the test ideal which appeared independently in characteristic $p>0$ commutative algebra [Laz04, HH90]. We recall the reader that we are using the notations as in Section 2.

Definition 3.1**.**

Fix an ideal $\mathfrak{a}\subseteq A$ and a real number $t\geq 0$. We define

[TABLE]

where $f^{1/p^{e}}$ denotes all possible $p^{e}$-th roots of $f$ in $A_{\infty}$ that are part of a compatible system of $p$-power roots. If we also fix a sequence of elements $\{f_{1},\dots,f_{n}\}$ generating $\mathfrak{a}$ and for each $i$, one fixed compatible system of $p$-power roots $\{f_{i}^{1/p^{e}}\}_{e=1}^{\infty}$ for $f_{i}$ (these data we denote by $[f_{1},\dots,f_{n}]$), then we set

[TABLE]

In the above product, $f_{j_{i}}^{1/p^{e}}$ runs over our chosen elements in $\{f_{1}^{1/p^{e}},\dots,f_{n}^{1/p^{e}}\}$ (note that $j_{i}$ belongs to the set $\{1,\dots,n\}$, and repeats are allowed). We then define

[TABLE]

and

[TABLE]

We will usually write ${\uptau}^{\sharp}([\underline{f}]^{t})$ for ${\uptau}^{\sharp}([f_{1},\dots,f_{n}]^{t})$ when $f_{1},\dots,f_{n}$ is clear from the context. In the case that $\mathfrak{a}=(f)$ is principal, we will write ${\uptau}^{\sharp}(f^{t})$ for ${\uptau}^{\sharp}(\mathfrak{a}^{t})$. We emphasize that a priori this is different from ${\uptau}^{\sharp}([f]^{t})$. This is because in the former one considers all possible $p^{e}$-th roots of elements in $(f)$ in $A_{\infty}$ that are part of a compatible system of $p$-power roots, while in the latter one we fix a single compatible system of $p$-power roots of $f$.

Remark 3.2*.*

A key part of the definition of $0^{\diamondsuit[f_{1},\dots,f_{n}]^{t}}_{H_{\mathfrak{m}}^{d}(A)}$ and ${\uptau}^{\sharp}([f_{1},\dots,f_{n}]^{t})$ is that we are not considering $p^{e}$-th roots of elements like $f_{1}+f_{2}$, and this is important, because unlike in characteristic $p>0$, taking $p^{e}$-th roots is not additive in mixed characteristic. Another key point is that in the notation $[f_{1},\dots,f_{n}]^{t}$ we control directly which $p^{e}$-th roots we are taking.

The following properties are straightforward from the definition.

Proposition 3.3**.**

Fix $\{f_{1},\ldots,f_{n}\}$ a sequence of generators of an ideal $\mathfrak{a}\subseteq A$ and for each $f_{i}$ fix a compatible system of $p$-power roots of $f_{i}$ in $A_{\infty}$ (in order to define ${\uptau}^{\sharp}([f_{1},\ldots,f_{n}]^{t})$). Then we have

1. (a)

For every $t\geq 0$, ${\uptau}^{\sharp}([f_{1},\ldots,f_{n}]^{t})\subseteq{\uptau}^{\sharp}(\mathfrak{a}^{t}).$ 2. (b)

If $t^{\prime}>t$, then ${\uptau}^{\sharp}(\mathfrak{a}^{t^{\prime}})\subseteq{\uptau}^{\sharp}(\mathfrak{a}^{t})$ and ${\uptau}^{\sharp}([f_{1},\ldots,f_{n}]^{t^{\prime}})\subseteq{\uptau}^{\sharp}([f_{1},\ldots,f_{n}]^{t})$. 3. (c)

If $\mathfrak{a}\subseteq\mathfrak{b}$, and $f_{1},\dots,f_{n},f_{n+1},\dots,f_{m}$ is a fixed set of generators of $\mathfrak{b}$ (we should also fix a compatible system of $p$-power roots of each $f_{i}$), then for every $t\geq 0$, ${\uptau}^{\sharp}(\mathfrak{a}^{t})\subseteq{\uptau}^{\sharp}(\mathfrak{b}^{t})$ and ${\uptau}^{\sharp}([f_{1},\ldots,f_{n}]^{t})\subseteq{\uptau}^{\sharp}([f_{1},\ldots,f_{m}]^{t})$.

One of the key properties of multiplier and test ideals is the fact that small positive perturbations of the exponent do not change the ideal. We do not know if this is true for ${\uptau}^{\sharp}$.

Question 3.4*.*

Is it true that ${\uptau}^{\sharp}(\mathfrak{a}^{t})={\uptau}^{\sharp}(\mathfrak{a}^{t+\epsilon})$ or ${\uptau}^{\sharp}([f_{1},\ldots,f_{n}]^{t})={\uptau}^{\sharp}([f_{1},\ldots,f_{n}]^{t+\epsilon})$ for all $\epsilon\ll 1$?

Due to this, we make the following definition. This will be our real definition of perfectoid test ideals.

Definition 3.5**.**

Fix $\{f_{1},\ldots,f_{n}\}$ a sequence of generators of an ideal $\mathfrak{a}\subseteq A$ and for each $f_{i}$ fix a compatible system of $p$-power roots of $f_{i}$ (in order to define ${\uptau}^{\sharp}([f_{1},\ldots,f_{n}]^{t})$). We define ${\uptau}(\mathfrak{a}^{t})={\uptau}(A,\mathfrak{a}^{t})$ to be the union or sum of $\{{\uptau}^{\sharp}(\mathfrak{a}^{t^{\prime}})\}$ for all $t^{\prime}>t$. Since ${\uptau}^{\sharp}(\mathfrak{a}^{t^{\prime}})\subseteq{\uptau}^{\sharp}(\mathfrak{a}^{t})$ for all $t^{\prime}>t$, by the Noetherian property of $A$, this is ${\uptau}^{\sharp}(\mathfrak{a}^{t+\epsilon})$ when $\epsilon\ll 1$. Similarly, we define ${\uptau}([f_{1},\ldots,f_{n}]^{t})={\uptau}(A,[f_{1},\ldots,f_{n}]^{t})$ to be ${\uptau}^{\sharp}([f_{1},\ldots,f_{n}]^{t+\epsilon})$ for $\epsilon\ll 1$.

It follows from the definition and Section 3 that we have:

[TABLE]

As with multiplier ideals and test ideals, in the notation ${\uptau}(\mathfrak{a}^{t})$, $\mathfrak{a}^{t}$ is a formal object. This can cause confusion when $t$ is an integer since, for instance, $\mathfrak{a}^{2}$ makes sense on its own. Thus it is natural to ask whether ${\uptau}^{\sharp}((\mathfrak{a}^{n})^{t})={\uptau}^{\sharp}(\mathfrak{a}^{nt})$ and whether ${\uptau}((\mathfrak{a}^{n})^{t})={\uptau}(\mathfrak{a}^{nt})$. We do not know how to do this with our definition of ${\uptau}^{\sharp}$, but as we shall see it is not difficult to show this for ${\uptau}(\mathfrak{a}^{t})$ and ${\uptau}([f_{1},\ldots,f_{n}]^{t})$. This will be crucial for our later purposes (and suggests that ${\uptau}$ is a better definition than ${\uptau}^{\sharp}$).

Lemma 3.6**.**

In the definition of $0^{\diamondsuit\mathfrak{a}^{t}}_{H_{\mathfrak{m}}^{d}(A)}$ and $0^{\diamondsuit[f_{1},\dots,f_{n}]^{t}}_{H_{\mathfrak{m}}^{d}(A)}$, one may restrict to $e\gg 0$.

Proof.

Indeed, restricting the $e$ to those $\gg 0$ results in fewer conditions and hence a larger subset of $H^{d}_{\mathfrak{m}}(A)$. On the other hand suppose that $\eta$’s image in $H^{d}_{\mathfrak{m}}(A_{\infty})$ is annihilated by $p^{1/p^{\infty}}f^{1/p^{e}}$ for all $f\in\mathfrak{a}^{\lceil tp^{e}\rceil}$ where $f^{1/p^{e}}$ is part of a compatible system of $p$-power roots, for $e\geq e_{0}$. Fix now some $n\geq 0$ and $f\in\mathfrak{a}^{\lceil tp^{n}\rceil}$. Then for $e\geq e_{0}$,

[TABLE]

since $f^{p^{e}}\in\mathfrak{a}^{p^{e}\lceil tp^{n}\rceil}\subseteq\mathfrak{a}^{\lceil tp^{e+n}\rceil}$. This shows that $\eta\in 0^{\diamondsuit\mathfrak{a}^{t}}_{H_{\mathfrak{m}}^{d}(A)}$. A similar argument works for $0^{\diamondsuit[f_{1},\dots,f_{n}]^{t}}_{H_{\mathfrak{m}}^{d}(A)}$ and we omit the details. ∎

Proposition 3.7**.**

For all positive integers $n$, ${\uptau}((\mathfrak{a}^{n})^{t})={\uptau}(\mathfrak{a}^{nt})$.

Proof.

Fix an $\varepsilon>0$ so that ${\uptau}((\mathfrak{a}^{n})^{t})={\uptau}^{\sharp}((\mathfrak{a}^{n})^{t+\varepsilon})$ and ${\uptau}(\mathfrak{a}^{nt})={\uptau}^{\sharp}(\mathfrak{a}^{nt+\varepsilon})$. By definition, any ${\uptau}^{\sharp}((\mathfrak{a}^{n})^{t+\varepsilon^{\prime}})$ where $\varepsilon^{\prime}\leq\varepsilon$ also computes ${\uptau}((\mathfrak{a}^{n})^{t})$. We first show that $0^{\diamondsuit(\mathfrak{a}^{n})^{t+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}\supseteq 0^{\diamondsuit\mathfrak{a}^{nt+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}$. Note that $0^{\diamondsuit\mathfrak{a}^{nt+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}$ consists of $\eta\in H^{d}_{\mathfrak{m}}(A)$ whose images in $H_{\mathfrak{m}}^{d}(A_{\infty})$ are annihilated by $p^{1/p^{\infty}}f^{1/p^{e}}$ with $f\in\mathfrak{a}^{\lceil(nt+\varepsilon)p^{e}\rceil}$ and $f^{1/p^{e}}$ part of a compatible system of $p$-power roots of $f$, while $0^{\diamondsuit(\mathfrak{a}^{n})^{t+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}$ consists of $\eta$ that are annihilated by $p^{1/p^{\infty}}g^{1/p^{e}}$, where $g\in(\mathfrak{a}^{n})^{\lceil(t+\varepsilon)p^{e}\rceil}$ and $g^{1/p^{e}}$ part of a compatible system of $p$-power roots of $g$. Since $\lceil(nt+\varepsilon)p^{e}\rceil\leq n\lceil(t+\varepsilon)p^{e}\rceil$, one sees that $(\mathfrak{a}^{n})^{\lceil(t+\varepsilon)p^{e}\rceil}\subseteq\mathfrak{a}^{\lceil(nt+\varepsilon)p^{e}\rceil}$ and so $0^{\diamondsuit(\mathfrak{a}^{n})^{t+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}\supseteq 0^{\diamondsuit\mathfrak{a}^{nt+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}$. Thus ${\uptau}((\mathfrak{a}^{n})^{t+\varepsilon})\subseteq{\uptau}(\mathfrak{a}^{nt+\varepsilon})$.

Conversely, note that

[TABLE]

for $e\gg 0$. Thus

[TABLE]

and so arguing as above, and using Section 3, we see that

[TABLE]

where the last equality follows from our choice of $\varepsilon$. This finishes the proof. ∎

We next prove the analog result for ${\uptau}([f_{1},\dots,f_{n}]^{t})={\uptau}([\underline{f}]^{t})$.

Proposition 3.8**.**

Fix $\{\underline{f}\}=\{f_{1},\ldots,f_{n}\}$ a sequence of generators of an ideal $\mathfrak{a}\subseteq A$ and for each $f_{i}$ fix a compatible system of $p$-power roots of $f_{i}$ to define ${\uptau}^{\sharp}([\underline{f}]^{t})$. Set $\underline{f}^{\bullet n}$ to be the set of degree $n$ monomials in the $f_{i}$, and we use the product of the fixed compatible system of $p$-power roots of $f_{i}$ to build compatible system of $p$-power roots for $\underline{f}^{\bullet n}$. Then for all positive integers $n$ and real numbers $t\geq 0$, ${\uptau}([\underline{f}^{\bullet n}]^{t})={\uptau}([\underline{f}]^{nt})$.

Proof.

Fix an $\varepsilon>0$ so that ${\uptau}([\underline{f}^{\bullet n}]^{t})={\uptau}^{\sharp}([\underline{f}^{\bullet n}]^{t+\varepsilon})$ and ${\uptau}([\underline{f}]^{nt})={\uptau}^{\sharp}([\underline{f}]^{nt+\varepsilon})$. By definition, any ${\uptau}^{\sharp}([\underline{f}^{\bullet n}]^{t+\varepsilon^{\prime}})$ where $\varepsilon^{\prime}\leq\varepsilon$ also computes ${\uptau}([\underline{f}^{\bullet n}]^{t})$. As in the proof of Section 3, the containment $0^{\diamondsuit[\underline{f}^{\bullet n}]^{t+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}\supseteq 0^{\diamondsuit[\underline{f}]^{nt+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}$ follows from the fact that $\lceil(nt+\varepsilon)p^{e}\rceil\leq n\lceil(t+\varepsilon)p^{e}\rceil$.

Next we show $0^{\diamondsuit[\underline{f}^{\bullet n}]^{t+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}\subseteq 0^{\diamondsuit[\underline{f}]^{nt+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}$. If $a/p^{e}\geq nt+\varepsilon$, then we can write $a=bn+c$ such that $b=\lfloor a/n\rfloor$ and $0\leq c\leq n-1$. Pick $e\gg 0$ such that $c/p^{e}\leq\varepsilon/2$, since $\frac{bn+c}{p^{e}}\geq nt+\varepsilon$, we must have $bn/p^{e}\geq nt+\varepsilon/2$ and thus $b/p^{e}\geq t+\varepsilon/2n$. Therefore if $\eta$ is annihilated by $p^{1/p^{\infty}}g$ in $H_{\mathfrak{m}}^{d}(A_{\infty})$ for all $g=\prod^{b}g_{k}^{1/p^{e}}$ with $g_{k}^{1/p^{e}}=\prod^{n}f_{j}^{1/p^{e}}$ and $b/p^{e}\geq t+\varepsilon/2n$, then it is annihilated by $p^{1/p^{\infty}}f$ with $f=\prod^{a}f_{j}^{1/p^{e}}$ and $a/p^{e}\geq nt+\varepsilon$. This proves that

[TABLE]

where again the last equality follows from our choice of $\varepsilon$. This finishes the proof. ∎

Our next goal is to show that $\mathfrak{a}\subseteq{\uptau}([f_{1},\dots,f_{n}])$ for any set of generators $\{f_{1},\dots,f_{n}\}$ of $\mathfrak{a}$ (and any fixed set of compatible system of $p$-power roots $\{f_{i}^{1/p^{e}}\}_{e=1}^{\infty}$ in order to define ${\uptau}([f_{1},\dots,f_{n}])$). It would follow that $\mathfrak{a}\subseteq{\uptau}(\mathfrak{a})$ by ($\dagger$ ‣ 3).

Proposition 3.9**.**

Fix $\{\underline{f}\}=\{f_{1},\ldots,f_{n}\}$ a set of elements of $A$, and for each $f_{i}$ fix a compatible system of $p$-power roots of $f_{i}$ in $A_{\infty}$ to define ${\uptau}([\underline{f}])$. Then we have $(f_{1},\dots,f_{n})\subseteq{\uptau}([\underline{f}])={\uptau}([\underline{f}]^{1})$. In particular, $\mathfrak{a}\subseteq{\uptau}(\mathfrak{a})$ for any ideal $\mathfrak{a}\subseteq A$.

Proof.

It is enough to show that $(f_{1},\dots,f_{n})$ annihilates $0^{\diamondsuit[\underline{f}]^{1+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}$ for $0<\varepsilon\ll 1$. Fix an $\eta\in 0^{\diamondsuit[\underline{f}]^{1+\varepsilon}}_{H_{\mathfrak{m}}^{d}(A)}$. Our hypothesis implies that $(pf_{i})^{1/p^{\infty}}(f_{i}\eta)=0$ in $H_{\mathfrak{m}}^{d}(A_{\infty})$. Applying Section 2 to $c=pf_{i}$, we have $f_{i}\eta=0$ in $H_{\mathfrak{m}}^{d}(A)$, i.e., $f_{i}$ annihilates $\eta$. ∎

Corollary 3.10**.**

For $0\neq f\in A$, we have ${\uptau}([f])={\uptau}^{\sharp}([f])=(f)$.

Proof.

We know $(f)\subseteq{\uptau}([f])\subseteq{\uptau}^{\sharp}([f])$ by Section 3 and (missing), thus it suffices to show that ${\uptau}^{\sharp}([f])={\uptau}^{\sharp}([f]^{1})\subseteq(f)$. But if $f\eta=0$, then $p^{1/p^{\infty}}f\eta=0$ in $H_{\mathfrak{m}}^{d}(A_{\infty})$ and so $\{\eta\in{H^{d}_{\mathfrak{m}}(A)}\;|\;f\eta=0\}\subseteq 0^{\diamondsuit[f]^{1}}_{H_{\mathfrak{m}}^{d}(A)}$, therefore the result follows by applying Matlis duality. ∎

Corollary 3.11**.**

Fix $\{\underline{f}\}=\{f_{1},\ldots,f_{n}\}$ a sequence of generators of an ideal $\mathfrak{a}\subseteq A$, and for each $f_{i}$ fix a compatible system of $p$-power roots of $f_{i}$ to define ${\uptau}([\underline{f}])$. Then we have ${\uptau}([\underline{f}]^{0})={\uptau}(\mathfrak{a}^{0})=A$.

Proof.

Since ${\uptau}([\underline{f}]^{0})\subseteq{\uptau}(\mathfrak{a}^{0})\subseteq A$, it suffices to prove that ${\uptau}([\underline{f}]^{0})=A$, that is, $0^{\diamondsuit[\underline{f}]^{\epsilon}}_{H_{\mathfrak{m}}^{d}(A)}=0$ when $\epsilon\ll 1$. Suppose that $\eta\in 0^{\diamondsuit[\underline{f}]^{\epsilon}}_{H_{\mathfrak{m}}^{d}(A)}=0^{\diamondsuit[\underline{f}]^{1/p^{e}}}_{H_{\mathfrak{m}}^{d}(A)}$ for all $e\gg 0$, then we have $p^{1/p^{e}}f_{i}^{1/p^{e}}\eta=0$ in $H_{\mathfrak{m}}^{d}(A_{\infty})$ for all $e\gg 0$ and all $i$. Applying Section 2 to $c=pf_{i}$, we have $\eta=0$. ∎

4. The subadditivity theorem

The goal in this section is to prove the subadditivity for ${\uptau}([\underline{f}]^{t})$. This is in analogous to [DEL00] and [HY03]. We do not know how to prove the subadditivity property for ${\uptau}(\mathfrak{a}^{t})$. This is the main reason that we need to work with ${\uptau}([\underline{f}]^{t})$ in our later applications. We start by introducing the mixed perfectoid test ideals.

Definition 4.1**.**

Let $\{f_{1},\dots,f_{n}\}$ and $\{g_{1},\dots,g_{m}\}$ be fixed sets of elements of $A$. We also fix a compatible system of $p$-power roots $\{f_{i}^{1/p^{e}}\}_{e=1}^{\infty}$, $\{g_{j}^{1/p^{e}}\}_{e=1}^{\infty}$ for all $f_{i}$ and $g_{j}$ in $A_{\infty}$. Let $t,s\geq 0$ be two real numbers.

[TABLE]

We define ${\uptau}^{\sharp}([\underline{f}]^{t}[\underline{g}]^{s})=\operatorname{{Ann}}_{A}0^{\diamondsuit[\underline{f}]^{t}[\underline{g}]^{s}}_{H_{\mathfrak{m}}^{d}(A)}$, and ${\uptau}([\underline{f}]^{t}[\underline{g}]^{s})={\uptau}^{\sharp}([\underline{f}]^{t+\epsilon}[\underline{g}]^{s+\epsilon})$ for $\epsilon\ll 1$.

Remark 4.2*.*

For $\epsilon\ll 1$, we have ${\uptau}([\underline{f}]^{t}[\underline{f}]^{s})={\uptau}^{\sharp}([\underline{f}]^{t+\epsilon}[\underline{f}]^{s+\epsilon})={\uptau}^{\sharp}([\underline{f}]^{t+s+\epsilon})={\uptau}([\underline{f}]^{t+s})$ (compare with the proof of Section 3).

Before we prove our subadditivity theorem, we recall some notations which appear frequently in the study of Artinian modules in commutative algebra, as well as facts about $H^{d}_{\mathfrak{m}}(R)$. These are well known to experts in commutative algebra, but we do not know of a good reference.

Remark 4.3* (Annihilators of submodules of $H^{d}_{\mathfrak{m}}(A)$).*

We will show, in our setting, that if $M\subseteq H^{d}_{\mathfrak{m}}(A)$ is a submodule and $J=\operatorname{{Ann}}_{A}M$, then

[TABLE]

Note here (and in the proof below) we slightly abuse the notation of annihilators to select the $J$-torsion of a module. We hope this will not cause substantial confusion.

Now we verify (1). Recall that because $A$ is regular $H^{d}_{\mathfrak{m}}(A)$ is isomorphic to $E$, the injective hull of the residue field. Next observe that $E$ is Artinian (as are all its submodules). Because $A$ is complete and so isomorphic to the Matlis dual of $E$, the submodules $M$ of $E$ are Matlis dual to the quotients of $A$. Now, the annihilator $J$ of $M$ is equal to the annihilator of its Matlis dual, and hence the Matlis dual of $M$ is $A/J$. On the other hand, $M=\operatorname{Hom}_{A}(A/J,E)$ is the submodule of $E$ that $J$ annihilates.

We are ready to prove our subadditivity theorem. Our proof is inspired from the proof of subadditivity for test ideals in characteristic $p>0$ given by S. Takagi from [Tak06, Theorem 2.4]. The essential reason that the theorem holds is because $A_{\infty}$ is almost flat over $A$ by Theorem 2.2.

Theorem 4.4** (Subadditivity).**

With notation as in Section 4, we have ${\uptau}^{\sharp}([\underline{f}]^{t}[\underline{g}]^{s})\subseteq{\uptau}^{\sharp}([\underline{f}]^{t}){\uptau}^{\sharp}([\underline{g}]^{s})$ so also ${\uptau}([\underline{f}]^{t}[\underline{g}]^{s})\subseteq{\uptau}([\underline{f}]^{t}){\uptau}([\underline{g}]^{s})$. In particular,

[TABLE]

for all $t\in\mathbb{R}_{\geq 0}$ and all $n\in\mathbb{N}$, where we define ${\uptau}([\underline{f}^{\bullet n}]^{t})$ as in Section 3.

Proof.

We first claim that it is enough to show that

[TABLE]

To see this claim, if $z\in\operatorname{{Ann}}_{H_{\mathfrak{m}}^{d}(A)}{\uptau}^{\sharp}([\underline{f}]^{t}){\uptau}^{\sharp}([\underline{g}]^{s})$ then ${\uptau}^{\sharp}([\underline{f}]^{t})z\subseteq\operatorname{{Ann}}_{H_{\mathfrak{m}}^{d}(A)}{\uptau}^{\sharp}([\underline{g}]^{s})=0^{\diamondsuit[\underline{g}]^{s}}_{H_{\mathfrak{m}}^{d}(A)}$ and thus $z\in\{\eta\in H^{d}_{\mathfrak{m}}(A)\;|\;{\uptau}^{\sharp}([\underline{f}]^{t})\eta\subseteq 0^{\diamondsuit[\underline{g}]^{s}}_{H_{\mathfrak{m}}^{d}(A)}\}\subseteq 0^{\diamondsuit[\underline{f}]^{t}[\underline{g}]^{s}}_{H_{\mathfrak{m}}^{d}(A)}$. Therefore

[TABLE]

and thus ${\uptau}^{\sharp}([\underline{f}]^{t}[\underline{g}]^{s})\subseteq{\uptau}^{\sharp}([\underline{f}]^{t}){\uptau}^{\sharp}([\underline{f}]^{s})$ as desired.

Next we prove (missing). Suppose that $\eta\in{H_{\mathfrak{m}}^{d}(A)}$ satisfies ${\uptau}^{\sharp}([\underline{f}]^{t})\eta\ \subseteq 0^{\diamondsuit[\underline{g}]^{s}}_{H_{\mathfrak{m}}^{d}(A)}$. By definition we know that $p^{1/p^{\infty}}g\eta\cdot{\uptau}^{\sharp}([\underline{f}]^{t})=0$ in $H_{\mathfrak{m}}^{d}(A_{\infty})$ for all $g=\prod_{i=1}^{b}g_{k_{i}}^{1/p^{e}}$ with $b\geq sp^{e}$. This means $p^{1/p^{\infty}}g\eta\in\operatorname{{Ann}}_{H_{\mathfrak{m}}^{d}(A_{\infty})}({\uptau}^{\sharp}([\underline{f}]^{t})A_{\infty})$. Since $H_{\mathfrak{m}}^{d}(A_{\infty})\cong H_{\mathfrak{m}}^{d}(A)\otimes A_{\infty}$ (which follows from the Čech complex description of local cohomology [BH93, page 130] as $d=\dim A$), we know from Section 2 that $\operatorname{{Ann}}_{H_{\mathfrak{m}}^{d}(A_{\infty})}({\uptau}^{\sharp}([\underline{f}]^{t})A_{\infty})=\operatorname{Hom}_{A_{\infty}}(A_{\infty}/{\uptau}^{\sharp}([\underline{f}]^{t})A_{\infty},H_{\mathfrak{m}}^{d}(A_{\infty}))$ is almost isomorphic to

[TABLE]

Therefore $p^{1/p^{\infty}}g\eta\in A_{\infty}\otimes 0^{\diamondsuit[\underline{f}]^{t}}_{H_{\mathfrak{m}}^{d}(A)}$, which means for every $k$ we can write

[TABLE]

where $\eta_{i}\in 0^{\diamondsuit[\underline{f}]^{t}}_{H_{\mathfrak{m}}^{d}(A)}$ and $a_{i}\in A_{\infty}$. So for all $f=\prod_{i=1}^{a}f_{j_{i}}^{1/p^{e}}$ with $a\geq tp^{e}$,

[TABLE]

for all $k$, $k^{\prime}$. Thus we know $p^{1/p^{\infty}}fg\cdot\eta=0$ for all $f=\prod_{i=1}^{a}f_{j_{i}}^{1/p^{e}}$ and all $g=\prod_{i=1}^{b}g_{k_{i}}^{1/p^{e}}$ such that $a\geq tp^{e}$ and $b\geq sp^{e}$. Hence $\eta\in 0^{\diamondsuit[\underline{f}]^{t}[\underline{g}]^{s}}_{H_{\mathfrak{m}}^{d}(A)}$ as desired.

Finally, (missing) follows from Section 3, Section 4 and the inclusion ${\uptau}^{\sharp}([\underline{f}]^{t}[\underline{g}]^{s})\subseteq{\uptau}^{\sharp}([\underline{f}]^{t}){\uptau}^{\sharp}([\underline{g}]^{s})$ we just proved applied to $\underline{f}=\underline{g}$, and $t,s$ both equal to $t+\epsilon$ for $\epsilon\ll 1$, plus an induction on $n$. ∎

We could also define the mixed characteristic perfectoid test ideal for ${\uptau}^{\sharp}(\mathfrak{a}^{t})$ in an analogous way:

[TABLE]

where $f^{1/^{e}}$ and $g^{1/p^{e}}$ denote all possible parts of a compatible system of $p$-power roots of $f$ and $g$ respectively. We then define ${\uptau}^{\sharp}(\mathfrak{a}^{t}\mathfrak{b}^{s})=\operatorname{{Ann}}_{A}0^{\diamondsuit\mathfrak{a}^{t}\mathfrak{b}^{s}}_{H_{\mathfrak{m}}^{d}(A)}$ and ${\uptau}(\mathfrak{a}^{t}\mathfrak{b}^{s})={\uptau}^{\sharp}(\mathfrak{a}^{t+\epsilon}\mathfrak{b}^{s+\epsilon})$ for $\epsilon\ll 1$. In fact, working with this definition, one can also prove ${\uptau}^{\sharp}(\mathfrak{a}^{s}\mathfrak{b}^{t})\subseteq{\uptau}^{\sharp}(\mathfrak{a}^{s}){\uptau}^{\sharp}(\mathfrak{b}^{t})$ following a very similar argument as in Theorem 4.4. The problem is that, it is not clear to us whether ${\uptau}(\mathfrak{a}^{t}\mathfrak{a}^{s})={\uptau}(\mathfrak{a}^{t+s})$, and hence the second conclusion of the subadditivity theorem does not seem to work for ${\uptau}(\mathfrak{a}^{t})$.

5. Comparison with the blowup

The goal of this section is to prove Theorem 5.11, that is, ${\uptau}((I^{(hl)})^{1/l})\subseteq I$ for every radical ideal $I\subseteq A$ of height $h$ and every positive integer $l$. This follows from our core result Section 5. Theorem 5.11 implies immediately that ${\uptau}([\underline{f}]^{1/l})\subseteq I$ for every fixed generating set $\{\underline{f}\}$ of $I^{(hl)}$ (and every fixed compatible system of $p$-power roots of $\underline{f}$) since we always have ${\uptau}([\underline{f}]^{1/l})\subseteq{\uptau}((I^{(hl)})^{1/l})$ by (missing).

Our key idea is to study how information about our perfectoid test ideal can be obtained by blowing up a finitely generated ideal $J$. The situation is easier if $\sqrt{J}$ contains $p$ as then the blowup of $JA_{\infty}$ is admissible since it is trivial outside of $V(p)$. This allows us to use Scholze’s vanishing theorem for perfectoid spaces [Sch12, Proposition 6.14] which tells us that passing to the blowup is essentially harmless, up to almost mathematics and factoring. In the case that $J$ does not contain a power of $p$, however, we use a similar strategy to the one in [Bha18, section 6]: we need to pass to certain enlargement of $A_{\infty}$ where a multiple of $p$ is contained in $JB$. Throughout this section, we will mostly work with ${\uptau}(\mathfrak{a}^{t})$, but as discussed earlier, the final result Theorem 5.11 holds for ${\uptau}([\underline{f}]^{t})$ as well simply because of (missing).

We start by proving a series of four crucial lemmas (Section 5, Section 5, Section 5 and Section 5) that allow us to handle the case that the ideal we are blowing up does not contain a power of $p$. The reader who is only interested in the case when the ideal we blow up contains a power of $p$ may wish to jump directly to Section 5 where the main result of the section is proven. First we need a definition.

Definition 5.1**.**

We define

[TABLE]

where as usual $f^{1/p^{h}}$ denotes all $p^{h}$-th roots of $f$ that are part of a compatible system of $p$-power roots of $f$ in $A_{\infty}$. We use $0^{\diamondsuit\mathfrak{a}^{t}}_{[l,\infty]}$ (resp. $0^{\diamondsuit\mathfrak{a}^{t}}_{[\infty,h]}$) if we allow $h$ (resp. $l$) to range over all positive integers. Under this definition we have $0^{\diamondsuit\mathfrak{a}^{t}}_{H_{\mathfrak{m}}^{d}(A)}=0^{\diamondsuit\mathfrak{a}^{t}}_{[\infty,\infty]}$.

Lemma 5.2**.**

We have $0^{\diamondsuit\mathfrak{a}^{t}}_{H_{\mathfrak{m}}^{d}(A)}=0^{\diamondsuit\mathfrak{a}^{t}}_{[k,k]}$ for all $k\gg 0$.

Proof.

We have containments $\dots\subseteq 0^{\diamondsuit\mathfrak{a}^{t}}_{[l+1,\infty]}\subseteq 0^{\diamondsuit\mathfrak{a}^{t}}_{[l,\infty]}\subseteq\cdots$. Since $H_{\mathfrak{m}}^{d}(A)$ is Artinian, $0^{\diamondsuit\mathfrak{a}^{t}}_{H_{\mathfrak{m}}^{d}(A)}=0^{\diamondsuit\mathfrak{a}^{t}}_{[\infty,\infty]}=0^{\diamondsuit\mathfrak{a}^{t}}_{[l,\infty]}$ for all $l\gg 0$. Now we fix such an $l\gg 0$, it follows from the proof of Section 3 that we have containments $\dots\subseteq 0^{\diamondsuit\mathfrak{a}^{t}}_{[l,h+1]}\subseteq 0^{\diamondsuit\mathfrak{a}^{t}}_{[l,h]}\subseteq\cdots.$ Thus by the Artinian property of $H_{\mathfrak{m}}^{d}(A)$ again, $0^{\diamondsuit\mathfrak{a}^{t}}_{[l,\infty]}=0^{\diamondsuit\mathfrak{a}^{t}}_{[l,h]}$ for all $h\gg 0$. Now take $k\geq\max\{l,h\}$. We have $0^{\diamondsuit\mathfrak{a}^{t}}_{H_{\mathfrak{m}}^{d}(A)}=0^{\diamondsuit\mathfrak{a}^{t}}_{[k,k]}$. ∎

The next lemma is a slight generalization of [HM17, Lemma 3.2]. We recall that, for any element $g\in A_{\infty}$, $A_{\infty}\langle\frac{p^{n}}{g}\rangle$ denotes the integral perfectoid algebra, which is the ring of bounded functions on the rational subset $\{x\in X\hskip 5.0pt|\hskip 5.0pt|p^{n}|\leq|g(x)|\}$ where $X=\operatorname{Spa}(A_{\infty}[1/p],A_{\infty})$ is the perfectoid space associated to $(A_{\infty}[1/p],A_{\infty})$.

Lemma 5.3**.**

Let $I=(p^{c},y_{1},\dots,y_{s})$ be an ideal of $A$ (that contains a power of $p$). Let $g=p^{m}g_{0}\in A$ where $p\nmid g_{0}$, and consider the map $A\xrightarrow{\ \ }A_{\infty}\xrightarrow{\ \ }A_{\infty}\langle\frac{p^{b}}{g}\rangle$ for every positive integer $b$. Suppose the image of $z\in A_{\infty}$ is contained in $IA_{\infty}\langle\frac{p^{b}}{g}\rangle$. If $cp^{a}+m<b$, then for every $g^{1/p^{a}}$ that is part of a compatible system of $p$-power roots of $g$ in $A_{\infty}$, we have $p^{1/p^{a}}g^{1/p^{a}}z\in IA_{\infty}$.

In particular, if the image of $z$ is contained in $IA_{\infty}\langle\frac{p^{b}}{g}\rangle$ for all $b>0$, then $p^{1/p^{a}}g^{1/p^{a}}z\in IA_{\infty}$ for all $a>0$.

Proof.

We fixed a compatible system of $p$-power roots of $g$, call it $\{g^{1/p^{e}}\}_{e=1}^{\infty}$ that contains the particular $g^{1/p^{a}}$. Since $A_{\infty}\langle\frac{p^{n}}{g}\rangle$ is almost isomorphic to the $p$-adic completion of $A_{\infty}[(\frac{p^{n}}{g})^{\frac{1}{p^{\infty}}}]$ by [Sch12, Lemma 6.4], we have

[TABLE]

for some $t>a$. The image of $p^{1/p^{t}}z$ inside $A_{\infty}[(\frac{p^{b}}{g})^{1/p^{\infty}}]/p^{c}=\widehat{A_{\infty}[(\frac{p^{b}}{g})^{1/p^{\infty}}]}/p^{c}$ is contained in the ideal $(y_{1},\dots,y_{s})$. Therefore we can write

[TABLE]

where $f_{0},f_{1},\dots,f_{s}\in A_{\infty}[(\frac{p^{b}}{g})^{1/p^{\infty}}]$. Since this is a finite sum, there exist integers $k$ and $h$ such that $f_{0},f_{1},\dots,f_{s}$ are elements in $A_{\infty}[(\frac{p^{b}}{g})^{1/p^{k}}]$ of degree in $(\frac{p^{b}}{g})^{1/p^{k}}$ bounded by $p^{k}h$.

Next we claim that multiplying by $g_{0}^{h}$ in (missing) will clear all the denominators of the $f_{i}$. This is because every $g^{1/p^{e}}$ (that is part of the compatible system of $p$-power roots of $g$) has the form $p^{m/p^{e}}g_{0}^{1/p^{e}}$ for a certain $g_{0}^{1/p^{e}}\in A_{\infty}$ (that is part of a compatible system of $p$-power roots of $g_{0}$). To see this, simply observe that

[TABLE]

Since $A_{\infty}$ is integrally closed in $A_{\infty}[1/p]$, we have $\frac{g^{1/p^{e}}}{(p^{1/p^{e}})^{m}}\in A_{\infty}$ whose $p^{e}$-th power is $g_{0}$. One checks that after multiplying by $g_{0}^{h}$ to (missing) we get:

[TABLE]

From this we know:

[TABLE]

where $h_{0},h_{1},\dots,h_{s}\in A_{\infty}$. Rewriting this we have

[TABLE]

Since $p\nmid g_{0}$, $g_{0}$ is a nonzerodivisor on $A/p$. This implies $g_{0}^{h-(1/p^{a})}$ is an almost nonzerodivisor on $A_{\infty}/p^{(b-m)/p^{a}}$ since $A\xrightarrow{\ \ }A_{\infty}$ is almost flat by Theorem 2.2. Hence $p^{1/p^{t}}g_{0}^{1/p^{a}}z-p^{c}h_{0}-y_{1}h_{1}-\cdots-y_{s}h_{s}$ is annihilated by $(p^{1/p^{\infty}})$ in $A_{\infty}/p^{(b-m)/p^{a}}$. In particular, since $t>a$, we know

[TABLE]

Finally, note that $b>cp^{a}+m$ and thus $p^{(b-m)/p^{a}}$ is a multiple of $p^{c}$, and $g^{1/p^{a}}$ is a multiple of $g_{0}^{1/p^{a}}$. Therefore we have

[TABLE]

This finishes the proof. ∎

The main technical statement which allows us to pass to the enlargement of $A_{\infty}$ is contained below.

Lemma 5.4**.**

Let $p,x_{1},\dots,x_{d-1}$ be a system of parameters of $A$. For all $\epsilon\ll 1$ we have

[TABLE]

where $z\in A$, $c\in\mathbb{N}$, $f^{1/p^{e}}\in A_{\infty}$ runs over all possible $p^{e}$-th roots of $f$ that are part of a compatible system of $p$-power roots of $f$.

Proof.

We first prove the containment $``\subseteq"$. This works for any $\epsilon>0$. Suppose $\frac{z}{p^{c}x_{1}^{c}\cdots x_{d-1}^{c}}\in 0^{\diamondsuit\mathfrak{a}^{t+\epsilon}}_{H_{\mathfrak{m}}^{d}(A)}$, then $\frac{p^{1/p^{\infty}}f^{1/p^{e}}z}{p^{c}x_{1}^{c}\cdots x_{d-1}^{c}}=0$ in $H_{\mathfrak{m}}^{d}(A_{\infty})$ for all $f\in\mathfrak{a}^{\lceil(t+\epsilon)p^{e}\rceil}$ and all $f^{1/p^{e}}$ that are part of a compatible system of $p$-power roots. This means for every $l>0$,

[TABLE]

for some $w$ (which depends on $c$, $e$ and $l$). But since $p,x_{1},\dots,x_{d-1}$ is an almost regular sequence on $A_{\infty}$ by Section 2, this implies

[TABLE]

for all $f\in\mathfrak{a}^{\lceil(t+\epsilon)p^{e}\rceil}$ and all $l>0$. Hence its image in $A_{\infty}\langle\frac{p^{b}}{g}\rangle$ is contained in $(p^{c},x_{1}^{c},\dots,x_{d-1}^{c})A_{\infty}\langle\frac{p^{b}}{g}\rangle$ for all $0\neq g\in\mathfrak{a}$ and all $b>0$. Thus $p^{1/p^{\infty}}f^{1/p^{e}}z\in(p^{c},x_{1}^{c},\dots,x_{d-1}^{c})A_{\infty}\langle\frac{p^{b}}{g}\rangle$.

Next we prove the other containment $``\supseteq"$. We take $\varepsilon_{0}\ll 1$ such that $0^{\diamondsuit\mathfrak{a}^{t+\varepsilon_{0}}}_{H_{\mathfrak{m}}^{d}(A)}$ computes $0^{\diamondsuit\mathfrak{a}^{t+\epsilon}}_{H_{\mathfrak{m}}^{d}(A)}$. We note that $\varepsilon_{0}$ depends only on $\mathfrak{a}$ and $t$, and for every $\varepsilon_{1}<\varepsilon_{0}$, $0^{\diamondsuit\mathfrak{a}^{t+\varepsilon_{1}}}_{H_{\mathfrak{m}}^{d}(A)}$ also computes $0^{\diamondsuit\mathfrak{a}^{t+\epsilon}}_{H_{\mathfrak{m}}^{d}(A)}$. We choose $k\gg 0$ such that $0^{\diamondsuit\mathfrak{a}^{t+\varepsilon_{0}}}_{[k,k]}=0^{\diamondsuit\mathfrak{a}^{t+\varepsilon_{0}}}_{H_{\mathfrak{m}}^{d}(A)}$ by Section 5, and $\frac{\varepsilon_{0}}{2}p^{k}\geq t+\varepsilon_{0}$. We also observe that $k$ depends on $\mathfrak{a}$, $t$, $\varepsilon_{0}$ (and hence only depends on $\mathfrak{a}$ and $t$). We will show that $0^{\diamondsuit\mathfrak{a}^{t+\varepsilon_{0}}}_{[k,k]}=0^{\diamondsuit\mathfrak{a}^{t+\varepsilon_{0}}}_{H_{\mathfrak{m}}^{d}(A)}=0^{\diamondsuit\mathfrak{a}^{t+\varepsilon_{0}/2}}_{H_{\mathfrak{m}}^{d}(A)}$ contains

[TABLE]

where $f^{1/p^{2k}}$ runs over all possible $p^{{2k}}$-th roots of $f$ that are part of a compatible system of $p$-power roots of $f$. This will establish the $``\supseteq"$ because the object in (missing) (when applied to $\epsilon=\varepsilon_{0}/2$) is larger than the object in the statement of Section 5 (since it requires fewer conditions).

So select an arbitrary $\frac{z}{p^{c}x_{1}^{c}\cdots x_{d-1}^{c}}$ in the set in (missing), we have

[TABLE]

for all $b>0$. By Section 5, we get

[TABLE]

for all $f\in\mathfrak{a}^{\lceil(t+\varepsilon_{0}/2)p^{2k}\rceil}$, all $f^{1/p^{2k}}$ part of a compatible system of $p$-power roots of $f$, all $0\neq g\in\mathfrak{a}$, and all $g^{1/p^{2k}}$ part of a compatible system of $p$-power roots of $g$.

Finally, for every $\tilde{f}\in\mathfrak{a}^{\lceil(t+\varepsilon_{0})p^{k}\rceil}$, and every $\tilde{f}^{1/p^{k}}$ part of a compatible system of $p$-power roots, we can write $\tilde{f}^{1/p^{k}}=\tilde{f}^{\frac{p^{k}-1}{p^{2k}}}\tilde{f}^{1/p^{2k}}$, where $\tilde{f}^{1/p^{2k}}$ is the $p^{k}$-th root of $\tilde{f}^{1/p^{k}}$ in the compatible system. We claim that $\tilde{f}^{p^{k}-1}\in\mathfrak{a}^{\lceil(t+\varepsilon_{0}/2)p^{2k}\rceil}$, this is because

[TABLE]

by our choice of $k$. Now apply (6) to $g=\tilde{f}\in\mathfrak{a}$ (and we use $\tilde{f}^{1/p^{2k}}$ as part of the compatible system of $p$-power roots of $g$) and $f=\tilde{f}^{p^{k}-1}$, we find that $p^{1/p^{k}}\tilde{f}^{1/p^{k}}z\in(p^{c},x_{1}^{c},\dots,x_{d-1}^{c})A_{\infty}$. Thus $\frac{z}{p^{c}x_{1}^{c}\cdots x_{d-1}^{c}}$ is annihilated by $p^{1/p^{k}}\tilde{f}^{1/p^{k}}$ in $H_{\mathfrak{m}}^{d}(A_{\infty})$ for every $\tilde{f}^{1/p^{k}}$ part of a compatible system of $p$-power roots of $\widetilde{f}$ with $\tilde{f}\in\mathfrak{a}^{\lceil(t+\varepsilon_{0})p^{k}\rceil}$. Hence

[TABLE]

as desired. ∎

Lemma 5.5**.**

With the notations as in Section 5, we have

[TABLE]

Here we set $X_{\infty}^{b,g}=\operatorname{Spa}(A_{\infty}\langle\frac{p^{b}}{g}\rangle[1/p],A_{\infty}\langle\frac{p^{b}}{g}\rangle)$ to be the perfectoid space associated to $(A_{\infty}\langle\frac{p^{b}}{g}\rangle[1/p],A_{\infty}\langle\frac{p^{b}}{g}\rangle)$.

Proof.

This is true by Section 5 and utilizing the fact that $A_{\infty}\langle\frac{p^{b}}{g}\rangle$ is almost isomorphic to $\mathbf{R}\Gamma(X_{\infty}^{b,g},A_{\infty}\langle\frac{p^{b}}{g}\rangle)$ with respect to $(p^{1/p^{\infty}})$ by Scholze’s vanishing theorem of perfectoid spaces [Sch12, Proposition 6.14]. ∎

We need to recall one more fact well known to experts.

Lemma 5.6**.**

Suppose that $B$ is an integral perfectoid algebra and that $J\subseteq B$ is a finitely generated ideal containing a power of $p$. Set $X=\operatorname{Spa}(B[1/p],B)$ to be the perfectoid space associated to $(B[1/p],B)$. Then the map of ringed spaces

[TABLE]

factors through the blowup of $J$.

Proof.

This is described in footnote #8 in [Bha18, Proof of Proposition 6.2] and as pointed out there is implicit in the description of adic spaces found in [GR04, 14.8]. ∎

We are ready to prove our core lemma in this section.

Lemma 5.7**.**

Let $\pi:Y\xrightarrow{\ \ }X=\operatorname{{Spec}}A$ be the blowup of some ideal $J\subseteq A$ such that $Y$ is normal and that $\mathfrak{a}\subseteq\sqrt{J}$. Suppose that $E$ on $Y$ is a Weil divisor with $\pi(E)\subseteq V(J)$. Fix $t\in\mathbb{R}_{\geq 0}$ and suppose that for every $e>0$ and every $f\in\mathfrak{a}^{\lceil tp^{e}\rceil}$,

[TABLE]

Then ${\uptau}(\mathfrak{a}^{t})\subseteq\Gamma(Y,\mathcal{O}_{Y}(K_{Y/X}-E))\subseteq A$.

The strategy of the proof, at least in the case that $J$ contains a power of $p$, is to show we can factor the map $A\xrightarrow{\ \ }{\bf R}\Gamma(Y,\mathcal{O}_{Y})\xrightarrow{\cdot f^{1/p^{e}}}{\bf R}\Gamma(X_{\infty},\mathcal{O}_{X_{\infty}})$ through ${\bf R}\Gamma(Y,\mathcal{O}_{Y}(E))$ where $X_{\infty}=\operatorname{Spa}(A_{\infty}[1/p],A_{\infty})$. After this, we use the fact that ${\bf R}\Gamma(X_{\infty},\mathcal{O}_{X_{\infty}})$ is almost quasi-isomorphic to $A_{\infty}$ [Sch12, Proposition 6.14]. Thus our overall strategy is similar to (and inspired by) the proof of the derived direct summand conjecture, [Bha18]. Roughly speaking, the idea is that our divisor condition forces $f^{1/p^{e}}\mathcal{O}_{Y}(E)$ to pullback to something contained in $\mathcal{O}_{X_{\infty}}$, at least up to some issues of integrality. Below we give details.

Proof.

Let $J=(z_{1},\dots,z_{m})$, write $Y=\operatorname{{Proj}}A\oplus JT\oplus J^{2}T^{2}\oplus\cdots$ and let $U_{1},\dots,U_{m}$ be an affine cover of $Y$ with $U_{j}=Y\setminus V(z_{j}T)\cong\operatorname{{Spec}}A[\frac{z_{1}}{z_{j}},\frac{z_{2}}{z_{j}},\dots,\frac{z_{m}}{z_{j}}]$. We fix an $e$ and an element $f^{1/p^{e}}\in A_{\infty}$ such that $f^{1/p^{e}}$ is part of a compatible system of $p$-power roots of $f$ with $f\in\mathfrak{a}^{\lceil(t+\epsilon)p^{e}\rceil}\subseteq\mathfrak{a}^{\lceil tp^{e}\rceil}$.

Let $h\in A_{\infty}[z_{j}^{-1}]$ be such that $h\in f^{1/p^{e}}\mathcal{O}_{Y}(E)(U_{j})\subseteq f^{1/p^{e}}A[z_{j}^{-1}]$ for some $j$ (note we are using the $\pi(E)\subseteq V(J))$. Since $\operatorname{{div}}_{Y}(f)\geq p^{e}E$, we have $h^{p^{e}}\in\mathcal{O}_{Y}(U_{j})$. For any such fixed $h$, we claim the following:

Claim 5.8**.**

There exists $h^{\prime}\in\overline{(z_{1},\dots,z_{m})A_{\infty}}=\overline{JA_{\infty}}$ (where $\overline{\bullet}$ denotes integral closure of an ideal) such that, if $Y^{\prime}_{\infty}\xrightarrow{\ \ }\operatorname{{Spec}}A_{\infty}$ is the blow up of $(z_{1},\dots,z_{m},h^{\prime})$ in $\operatorname{{Spec}}A_{\infty}$ and $\rho^{\prime}:Y^{\prime}_{\infty}\xrightarrow{\ \ }Y$ is the induced affine map (see Appendix A), then we have that

[TABLE]

Proof of Claim.

By construction, we can write $h={f^{1/p^{e}}w\over z_{j}^{p^{l}}}$ for some integer $l$ and some $w\in A$. Since

[TABLE]

we know that there exists some $d\gg 0$ such that $fw^{p^{e}}z_{j}^{p^{d}-p^{e+l}}\in(z_{1},\ldots,z_{m})^{p^{d}}$. Fixing a compatible system of $p$-power roots of $w$ and $z_{j}$ in $A_{\infty}$, we note that

[TABLE]

Thus $f^{1/p^{d}}w^{1/p^{d-e}}z_{j}^{(p^{d}-p^{e+l})/p^{d}}\in\overline{(z_{1},\dots,z_{m})A_{\infty}}$. We set $h^{\prime}=f^{1/p^{d}}w^{1/p^{d-e}}z_{j}^{(p^{d}-p^{e+l})/p^{d}}$, and let $Y^{\prime}_{\infty}$ be the blow up. We have

[TABLE]

This finishes the proof of the Claim. ∎

Because the module $f^{1/p^{e}}\mathcal{O}_{Y}(E)(U_{j})$ is finitely generated over $\mathcal{O}_{Y}(U_{j})$ for every $j$, we collect the generators for all $1\leq j\leq m$ and we call them $h_{1},\dots,h_{k}$ (there are implicit $j$s we are suppressing). For each $h_{i}$ we construct $h_{i}^{\prime}\in\overline{JA_{\infty}}$ as in Section 5. Let $Y_{\infty}$ be the blow up of $(z_{1},\dots,z_{m},h_{1}^{\prime},\dots,h_{k}^{\prime})$ of $\operatorname{{Spec}}A_{\infty}$. Since each $h_{i}^{\prime}$ is in the integral closure of $(z_{1},\dots,z_{m})A_{\infty}$, the inverse image of the $\{U_{j}\}$ forms an affine cover of $Y_{\infty}$ by Appendix A. We then have a factorization $Y_{\infty}\xrightarrow{\rho}Y\xrightarrow{\ \ }X$ with $\rho$ affine, and for each $j$ we have a natural map $\mathcal{O}_{Y}(U_{j})\xrightarrow{\ \ }\mathcal{O}_{Y_{\infty}}(\rho^{-1}U_{j})$.

Claim 5.9**.**

With notation as above, the canonical map

[TABLE]

factors through $\mathcal{O}_{Y}(E)$.

Proof of claim.

By Section 5 and construction, we know that the $\mathcal{O}_{Y}(U_{j})$-generators of $f^{1/p^{e}}\mathcal{O}_{Y}(E)(U_{j})$ are contained in $\mathcal{O}_{Y^{\prime}_{\infty}}(\rho^{\prime-1}U_{j})$. Hence

[TABLE]

for every $1\leq j\leq m$, which proves the claim. ∎

The case when $J$ contains a power of $p$

At this point we are essentially done in the case that $J$ contains a power of $p$. We note that by Section 5 applied to $B=A_{\infty}$ and $X=X_{\infty}=\operatorname{Spa}(A_{\infty}[1/p],A_{\infty})$, we have a factorization $(X_{\infty},\mathcal{O}_{X_{\infty}}^{+}=A_{\infty})\xrightarrow{\ \ }Y_{\infty}\xrightarrow{\ \ }\operatorname{{Spec}}A_{\infty}$ because $Y_{\infty}$ is the blow up of $A_{\infty}$ at a finitely generated ideal $(z_{1},\dots,z_{m},h_{1}^{\prime},\dots,h_{k}^{\prime})$ that contains a power of $p$ (by our hypotheses). Therefore we have a commutative diagram (we abuse notation and use $\Gamma_{\mathfrak{m}}(Y,\bullet)$ to denote the functor $\Gamma_{\mathfrak{m}}(\Gamma(Y,\bullet))$):

[TABLE]

Here the existence of the dotted arrows follows from Section 5. Since, by [Sch12, Proposition 6.14], ${\bf R}\Gamma(X_{\infty},\mathcal{O}_{X_{\infty}}^{+})$ is almost isomorphic to $A_{\infty}$, the map $\phi$ is an almost isomorphism. Hence elements in $0^{\diamondsuit\mathfrak{a}^{t+\epsilon}}_{H_{\mathfrak{m}}^{d}(A)}$ are precisely those $\eta\in H_{\mathfrak{m}}^{d}(A)$ whose image is almost zero in ${\mathcal{H}}^{d}{\bf R}\Gamma_{\mathfrak{m}}(X_{\infty},\mathcal{O}_{X_{\infty}}^{+})$, when we vary over all $e>0$ and all $f\in\mathfrak{a}^{\lceil(t+\epsilon)p^{e}\rceil}$ (and all $f^{1/p^{e}}$ that is part of a compatible system of $p$-power roots of $f$) in the above diagram. But this is the case if $\eta$ has trivial image in ${\mathcal{H}}^{d}{\bf R}\Gamma_{\mathfrak{m}}(Y,\mathcal{O}_{Y}(E))$ by the commutative diagram. Therefore we have

[TABLE]

However by local and Grothendieck duality, see for instance [Har66], the Matlis dual of ${\mathcal{H}}^{d}\mathbf{R}\Gamma_{\mathfrak{m}}(Y,\mathcal{O}_{Y}(E))$ is ${\mathcal{H}}^{-d}({\bf R}\Gamma(Y,\omega_{Y}(-E)[d])$ and so the Matlis dual of the map $H^{d}_{\mathfrak{m}}(A)\xrightarrow{\ \ }{\mathcal{H}}^{d}\mathbf{R}\Gamma_{\mathfrak{m}}(Y,\mathcal{O}_{Y}(E))$ is $A\cong\omega_{A}\leftarrow\Gamma(Y,\omega_{Y}(-E))$. It follows that

[TABLE]

Here we take $K_{X}=0$ and $K_{Y}=K_{Y/X}$ as described in and before Appendix A. This proves the case when $J$ contains a power of $p$.

The case when $J$ may not contain a power of $p$

We now handle the general case. This is the place that we need to use Section 5 (which relies on the technical Section 5 and Section 5).

Let $Y_{\infty}^{b,g}\xrightarrow{\ \ }\operatorname{{Spec}}A_{\infty}\langle\frac{p^{b}}{g}\rangle$ be the blowup of the ideal $(z_{1},\dots,z_{m},h_{1}^{\prime},\dots,h_{k}^{\prime})A_{\infty}\langle\frac{p^{b}}{g}\rangle$, where $0\neq g\in\mathfrak{a}$. So we have a commutative diagram:

[TABLE]

Now, for each $0\neq g\in\mathfrak{a}\subseteq\sqrt{J}$, for some $l>0$ we have $g^{l}\in J$ and so $p^{lb}$ is contained inside $J\cdot A_{\infty}\langle\frac{p^{b}}{g}\rangle\subseteq(z_{1},\dots,z_{m},h_{1}^{\prime},\dots,h_{k}^{\prime})A_{\infty}\langle\frac{p^{b}}{g}\rangle$. Therefore by Section 5 applied to $B=A_{\infty}\langle\frac{p^{b}}{g}\rangle$, for every $b$ and every $0\neq g\in\mathfrak{a}$ we have a factorization

[TABLE]

where we use $X_{\infty}^{b,g}=\operatorname{Spa}(A_{\infty}\langle\frac{p^{b}}{g}\rangle[1/p],A_{\infty}\langle\frac{p^{b}}{g}\rangle)$ to denote the perfectoid space associated to $(A_{\infty}\langle\frac{p^{b}}{g}\rangle[1/p],A_{\infty}\langle\frac{p^{b}}{g}\rangle)$.

The above discussion shows that for every $f^{1/p^{e}}$ part of a compatible system of $p$-power roots of $f$ with $f\in\mathfrak{a}^{\lceil(t+\epsilon)p^{e}\rceil}$, every $0\neq g\in\mathfrak{a}$ and every positive integer $b$, we have the following commutative diagram:

[TABLE]

Again, the key point is that we have the dotted arrows in the above diagram, because we proved that the map $\mathcal{O}_{Y}\xrightarrow{\ \ }\mathcal{O}_{Y_{\infty}}\xrightarrow{\cdot f^{1/p^{e}}}\mathcal{O}_{Y_{\infty}}$ factors through $\mathcal{O}_{Y}(E)$ by Section 5 (up to pushforward by affine morphisms that we omit from the notation).

By Section 5, $\frac{z}{p^{c}x_{1}^{c}\cdots x_{d-1}^{c}}\in 0^{\diamondsuit\mathfrak{a}^{t+\epsilon}}_{H_{\mathfrak{m}}^{d}(A)}$ if and only if the image of $z$ is annihilated by $(p^{1/p^{\infty}})$ under the natural map induced from the top left to the right bottom of the above diagram

[TABLE]

for every $f^{1/p^{e}}$, every $0\neq g\in\mathfrak{a}$ and every $b>0$. But this is clearly the case if $z$ has trivial image in ${\mathcal{H}}^{0}\left(\frac{A}{(p^{c},x_{1}^{c},\dots,x_{d-1}^{c})}\otimes^{\mathbf{L}}\mathbf{R}\Gamma(Y,\mathcal{O}_{Y}(E))\right)$. Thus we have

[TABLE]

is contained in $0^{\diamondsuit\mathfrak{a}^{t+\epsilon}}_{H_{\mathfrak{m}}^{d}(A)}$. Now, the above is precisely

[TABLE]

because \varinjlim_{c}{\mathcal{H}}^{0}\big{(}\frac{A}{(p^{c},x_{1}^{c},\dots,x_{d-1}^{c})}\otimes^{\mathbf{L}}\mathbf{R}\Gamma(Y,\mathcal{O}_{Y}(E))\big{)}\cong{\mathcal{H}}^{d}\mathbf{R}\Gamma_{\mathfrak{m}}(Y,\mathcal{O}_{Y}(E)). Note that here the transition maps are $\frac{A}{(p^{c},x_{1}^{c},\dots,x_{d-1}^{c})}\xrightarrow{\cdot px_{1}\cdots x_{d-1}}\frac{A}{(p^{c+1},x_{1}^{c+1},\dots,x_{d-1}^{c+1})}$, which follows from the Čech complex characterization of $\mathbf{R}\Gamma_{\mathfrak{m}}(\bullet)$, see [BH93, page 130]. Again by Matlis, local and Grothendieck duality, $\operatorname{{Ann}}_{A}\left(\ker\left(H_{\mathfrak{m}}^{d}(A)\xrightarrow{\ \ }{\mathcal{H}}^{d}\mathbf{R}\Gamma_{\mathfrak{m}}(Y,\mathcal{O}_{Y}(E))\right)\right)=\Gamma(Y,\mathcal{O}_{Y}(K_{Y/X}-E))\subseteq A.$ Therefore we have

[TABLE]

which proves the lemma. ∎

We come to our main result of the section. We first remind our reader of the general definition of a symbolic power of a radical ideal.

Definition 5.10**.**

If $Q\subseteq A$ is a prime ideal, then the $n$th symbolic power of $Q$, denoted $Q^{(n)}$ is defined to be $Q^{n}A_{Q}\cap A$.

Suppose that $I\subseteq A$ is a radical ideal. Suppose $I=Q_{1}\cap\dots\cap Q_{t}$ is a decomposition of $I$ into minimal primes of $I$. In this case the $n$th symbolic power of $I$, denoted $I^{(n)}$, is defined to be the intersection:

[TABLE]

Theorem 5.11**.**

If $I\subseteq A$ is a radical ideal such that each prime component has height $\leq h$ then we have

[TABLE]

for every $l$.

Proof.

Let $\pi:Y\xrightarrow{\ \ }\operatorname{{Spec}}A$ be the normalization of the blowup of $I\subseteq A$. In particular, $\pi$ is the blowup of some $J=\overline{I^{n}}$ by Appendix A. Since $A$ is regular, we let $D=\sum_{i}D_{i}$ denote the union of components of the inverse image of $V(I)$ which dominate components of $V(I)$. If we write $I=Q_{1}\cap\dots\cap Q_{t}$ a primary decomposition, then over the localization $A_{Q_{i}}$, we are simply blowing up a power the maximal ideal $Q_{i}A_{Q_{i}}$ in a regular local ring. It follows that there is exactly one $D_{i}$ lying over each $V(Q_{i})$.

Next notice that $I^{(lh)}$ is contained in $\sqrt{J}$ since neither symbolic powers or integral closures change the vanishing locus. Furthermore, elements of

[TABLE]

vanish to order at least $p^{e}h$ on each $D_{i}$ by construction, and so we may apply Section 5 with $E:=hD$. It is then enough to show that

[TABLE]

Thus we must compute the exceptional divisor $K_{Y/X}$. Since regular local rings are pseudo rational by [LT81, Section 4], we see that $H^{d}_{\mathfrak{m}}(A)\xrightarrow{\ \ }H^{d}_{\mathfrak{m}}(Y,\mathcal{O}_{Y})$ injects, and so by local and Grothendieck duality, the Matlis dual map $\Gamma(Y,\mathcal{O}_{Y}(K_{Y/X}))\xrightarrow{\ \ }A$ surjects. It follows that $K_{Y/X}\geq 0$.

We now write

[TABLE]

and compute the integers $a_{i}$. Since $D_{i}$ is the only exceptional divisor dominating a component $V(Q_{i})\subseteq V(I)$, this can be done after localizing at $Q_{i}$ and so the statement reduces to computing the relative canonical divisor of the blowup of a regular local ring of dimension $h_{i}\leq h$ at its maximal ideal. At that point we see that $a_{i}=h_{i}-1\leq h-1$ by Appendix A. It follows immediately that $\Gamma(Y,\mathcal{O}_{Y}(K_{Y/X}-E))_{Q_{i}}\subseteq\Gamma(Y,\mathcal{O}_{Y}(-D))_{Q_{i}}=Q_{i}$ and so $\Gamma(Y,\mathcal{O}_{Y}(K_{Y/X}-E))\subseteq I$ as desired. ∎

6. Relation with multiplier ideals

We have defined ${\uptau}(\mathfrak{a}^{t})={\uptau}(A,\mathfrak{a}^{t})\subseteq A$ and have shown it satisfies at least some formal properties similar to those of the multiplier ideal [Laz04] (in this section we will always write ${\uptau}(A,\mathfrak{a}^{t})$ to clarify which ring we are working with). On the other hand, $A[1/p]$ is a ring of equal characteristic [math] and so there exists a log resolution of $(\operatorname{{Spec}}A[1/p],\mathfrak{a}\cdot A[1/p])$ and so we can compute its multiplier ideal. Thus it is natural to compare ${\uptau}(A,\mathfrak{a}^{t})\cdot A[1/p]$ with $\mathcal{J}(A[1/p],(\mathfrak{a}\cdot A[1/p])^{t})$.

Theorem 6.1**.**

For a pair $(A,\mathfrak{a}^{t})$, we have

[TABLE]

Proof.

First let $J\subseteq A[1/p]$ be an ideal whose blowup produces a log resolution of $(A[1/p],(\mathfrak{a}\cdot A[1/p])^{t})$ [Tem08]. Because a log resolution principalizes $\mathfrak{a}$, the blowup of $\mathfrak{a}\cdot J$ also produces the same log resolution of $(A[1/p],(\mathfrak{a}\cdot A[1/p])^{t})$. Since we may choose this log resolution to be an isomorphism outside of $\mathfrak{a}\cdot A[1/p]$ (since $A[1/p]$ is regular), we may assume that $\mathfrak{a}\cdot A[1/p]\subseteq\sqrt{J}$. Consider $J^{\prime}=J\cap A$ and notice that $\mathfrak{a}\subseteq(\mathfrak{a}\cdot A[1/p])\cap A\subseteq\sqrt{J}\cap A\subseteq\sqrt{J^{\prime}}$. Let $\pi:Y\xrightarrow{\ \ }X=\operatorname{{Spec}}A$ be the normalized blowup of $\mathfrak{a}\cdot J^{\prime}$ (the blowup of $\overline{(\mathfrak{a}\cdot J^{\prime})^{n}}$ for some $n>0$ by Appendix A). Write $\mathfrak{a}\cdot\mathcal{O}_{Y}=\mathcal{O}_{Y}(-G)$ and let $E=\lfloor tG\rfloor$. Finally write $U=\operatorname{{Spec}}A[1/p]\subseteq\operatorname{{Spec}}A$ and $V=\pi^{-1}(U)$. Now observe that $\pi|_{V}:V\xrightarrow{\ \ }U$ is the blowup of $J$ and hence the log resolution we started with. It follows that the hypotheses of Section 5 are satisfied and so ${\uptau}(A,\mathfrak{a}^{t})\subseteq\Gamma(Y,\mathcal{O}_{Y}(K_{Y/X}-\lfloor tG\rfloor))$. But now by definition, $\Gamma(V,\mathcal{O}_{Y}(K_{Y/X}-\lfloor tG\rfloor))=\mathcal{J}(A[1/p],(\mathfrak{a}\cdot A[1/p])^{t})$ and the result follows. ∎

We expect that the other containment should hold as well, namely:

Conjecture 6.2**.**

${\uptau}(A,\mathfrak{a}^{t})\cdot A[1/p]=\mathcal{J}(A[1/p],(\mathfrak{a}\cdot A[1/p])^{t})$**

We do not know how to prove it unfortunately, and it is certainly related to the question of localizing ${\uptau}$ which we also do not know how to handle.

Alternatively, in mixed characteristic one can define the multiplier ideal $\mathcal{J}(A,\mathfrak{a}^{t})$ valuatively. This is equivalent to defining

[TABLE]

where $Y\xrightarrow{\ \ }X:=\operatorname{{Spec}}A$ runs over all proper birational maps with $Y$ normal and such that $\mathfrak{a}\cdot\mathcal{O}_{Y}=\mathcal{O}_{Y}(-G_{Y})$. Note it is also not clear whether this definition commutes with localization (since the intersection is infinite). With this definition, we have the following.

Theorem 6.3**.**

Suppose that $(A,\mathfrak{a}^{t})$ is a pair, then

[TABLE]

Proof.

This will follow from Section 5 but we must argue that we only need consider projective birational $\pi:Y\xrightarrow{\ \ }X$ that are blowups of some ideal $J$ such that $\mathfrak{a}\subseteq\sqrt{J}$. Moving from proper to projective maps (and hence blowups) is simply Chow’s Lemma. Since $A$ is regular local and hence a UFD, one may always choose $J$ so that $V(J)$ is the locus over which $\pi$ is not an isomorphism (cf. [Har77, Chapter II, Exercise 7.11(c)]). Thus it suffices to show that we can restrict our attention to $\pi$ that are an isomorphism outside of $V(\mathfrak{a})$.

To do this, first notice that our multiplier ideal can also be written as

[TABLE]

Next consider an arbitrary projective birational map $\pi:Y\xrightarrow{\ \ }X$ with $Y$ normal. Fix some prime divisor $D$ on $Y$ with corresponding discrete valuation $v$ and set $W=\pi(E)$ to be the center of $v$. If $W$ is not contained in $V(\mathfrak{a})$, then the coefficient of $\lfloor tG_{Y}\rfloor-K_{Y/X}$ along $E$ is negative (since $K_{Y/X}$ is effective, see the proof of Theorem 5.11), and so $E$ imposes no condition on the elements $f$ which make up $\mathcal{J}(A,\mathfrak{a}^{t})$. Thus, we only need to consider $E$ whose centers are contained in $V(\mathfrak{a})$. Next observe that the $E$ coefficient of $\lfloor tG_{Y}\rfloor-K_{Y/X}$ only depends on the valuation $v$, it does not depend on the particular choice of $Y$. Therefore, it suffices to show that if $v$ is a divisorial valuation whose center is contained in $V(\mathfrak{a})$, then there is a projective birational map $\pi:Y\xrightarrow{\ \ }X$, with $Y$ normal and $\pi$ an isomorphism outside of $V(\mathfrak{a})$ such that the valuation ring of $v$ is the local ring of some prime divisor on $Y$.

However, [Art86, Section 5] (see also [Abh56, Proposition 3]), shows that by repeatedly blowing up the center of the valuation $v$ one eventually obtains a birational model which realizes the valuation ring of $v$ as the localization at a generic point of a prime divisor. Compositions of such blowups are isomorphisms away from the center of $v$, which is contained in $V(\mathfrak{a})$.

The Theorem then follows by Section 5. ∎

7. Asymptotic perfectoid test ideals

We let $\{\mathfrak{a}_{n}\}_{n=1}^{\infty}$ be a graded sequence of ideals, i.e., $\mathfrak{a}_{n}\mathfrak{a}_{m}\subseteq\mathfrak{a}_{m+n}$ for all $m,n$. By analogy with for instance [ELS01], it would be natural to define the $n$-th asymptotic perfectoid test ideal of the graded sequence $\{\mathfrak{a}_{n}\}$ as the ideal

[TABLE]

However, since we don’t know the subadditivity theorem for ${\uptau}(\mathfrak{a}^{t})$, this version of asymptotic test ideal will not give us the desired result on symbolic powers of ideals. Instead we have to build the definition using ${\uptau}([\underline{f}]^{t})$. Now since everything depends on the choice of the generating set $\{\underline{f}\}$ and a choice of compatible system of $p$-power roots of $\{\underline{f}\}$, we must proceed carefully.

Let $\{\mathfrak{a}_{n}\}_{n=1}^{\infty}$ be a graded sequence of ideals of $A$. We will be mostly interested in the situation that $\mathfrak{a}_{n}=I^{(n)}$, the $n$-th symbolic power of a radical ideal $I\subseteq A$. We define a generating set of each $\mathfrak{a}_{n}$ inductively as follows:

First, we let $\{\underline{f}_{(1)}\}=\{f_{(1),1},\dots,f_{(1),n_{1}}\}$ be a fixed generating set of $\mathfrak{a}_{1}$, and we also fix a compatible system of $p$-power roots for each $f_{(1),i}$ in $A_{\infty}$ (that we will use to define ${\uptau}([\underline{f}_{(1)}]^{t})$). Now, suppose that a generating set $\{\underline{f}_{(s)}\}$ of $\mathfrak{a}_{s}$ and a compatible system of $p$-power roots of $\underline{f}_{(s)}$ have been chosen for all $s<m$. We let

[TABLE]

be a generating set of $\mathfrak{a}_{m}$ satisfying the condition that it contains all possible $f_{(s),i}f_{(t),j}$, where $s+t=m$ and $f_{(s),i}$, $f_{(t),j}$ are part of the chosen generating set of $\mathfrak{a}_{s}$ and $\mathfrak{a}_{t}$. Moreover, we fix a compatible system of $p$-power roots for each $f_{(m),k}$, such that, if $f_{(m),k}=f_{(s),i}f_{(t),j}$, then we use the product of the compatible system of $p$-power roots of $f_{(s),i}$ and $f_{(t),j}$, i.e., we let

[TABLE]

It might happen that $f_{(s),i}f_{(t),j}=f_{(m),k}=f_{(s^{\prime}),i^{\prime}}f_{(t^{\prime}),j^{\prime}}$ for $s+t=m=s^{\prime}+t^{\prime}$, but the product of the chosen compatible system of $p$-power roots of $f_{(s),i}$ and $f_{(t),j}$ is not the same as the product of the chosen compatible system of $p$-power roots of $f_{(s^{\prime}),i^{\prime}}$ and $f_{(t^{\prime}),j^{\prime}}$. In this case, we simply allow $f_{(m),k}$ to appear multiple times in the generating set, but we use different compatible system of $p$-power roots, one coming from $f_{(s),i}f_{(t),j}$ and the other coming from $f_{(s^{\prime}),i^{\prime}}f_{(t^{\prime}),j^{\prime}}$.

We have defined a generating set $\{\underline{f}_{(m)}\}$ for each $\mathfrak{a}_{m}$ in the graded sequence, as well as a compatible system of $p$-power roots for each element $f_{(m),i}$ appearing in the generating set. Now we give our definition of asymptotic perfectoid test ideal.

Definition 7.1** (Asymptotic perfectoid test ideals).**

We define

[TABLE]

By Section 3,

[TABLE]

where as in the notation of Section 3, $\underline{f}_{(ln)}^{\bullet m}$ denotes the set of all degree $m$ monomials in $\{\underline{f}_{(ln)}\}=\{f_{(ln),1},\dots,f_{(ln),n_{ln}}\}$. More importantly, by our choice of the generating set $\{\underline{f}_{(mln)}\}$ (and the way we fix the compatible system of $p$-power roots of elements in the generating set), we have

[TABLE]

by Section 3. Therefore we have

[TABLE]

Proposition 7.2**.**

We have ${\uptau}_{\infty}([\underline{f}_{(mn)}])\subseteq{\uptau}_{\infty}([\underline{f}_{(n)}])^{m}$ for all $n,m\in\mathbb{N}$.

Proof.

By the above discussion, we can choose $l$ sufficiently large and divisible such that ${\uptau}_{\infty}([\underline{f}_{(mn)}])={\uptau}([\underline{f}_{(lmn)}]^{\frac{1}{l}})$ and ${\uptau}_{\infty}([\underline{f}_{(n)}])={\uptau}([\underline{f}_{(lmn)}]^{\frac{1}{ml}})$. Now we have

[TABLE]

where the second equality follows from Section 3, and the only inclusion is by the subadditivity property Theorem 4.4. This completes the proof. ∎

Theorem 7.3**.**

Let $I\subseteq A$ be a radical ideal such that each prime component has height $\leq h$. Then we have $I^{(hn)}\subseteq I^{n}$ for all $n\in\mathbb{N}$.

Proof.

Let $\{\mathfrak{a}_{n}\}=\{I^{(n)}\}$ be the graded sequence of ideals. We select a generating set $\{\underline{f}_{(n)}\}$ for each $\mathfrak{a}_{n}$ in this graded sequence as well as a compatible system of $p$-power roots for each element $f_{(n),i}$ appearing in the generating set as in the discussion before Section 7, and we form the asymptotic perfectoid test ideal ${\uptau}_{\infty}([\underline{f}_{(n)}])$ as in Section 7.

Since $\{\underline{f}_{(hn)}\}$ is a generating set of $\mathfrak{a}_{hn}=I^{(hn)}$, by Section 3 we have

[TABLE]

where the last containment follows from Section 7. But by Section 7, we know that

[TABLE]

for all $n$. Therefore we are done if we can show that ${\uptau}_{\infty}([\underline{f}_{(h)}])\subseteq I$. However,

[TABLE]

for some sufficiently large and divisible $l$, and since $\{\underline{f}_{(lh)}\}$ is a generating set for $\mathfrak{a}_{lh}=I^{(lh)}$, we know from (missing) and Theorem 5.11 that

[TABLE]

This finishes the proof. ∎

Theorem 7.4**.**

Let $R$ be a Noetherian regular ring with reduced formal fibers (e.g. $R$ is excellent) and let $I\subseteq R$ be a radical ideal such that each minimal prime of $I$ has height $\leq h$. Then for every integer $n>0$,

[TABLE]

Proof.

Since the formation of symbolic powers commutes with localization, it is enough to prove $I^{(hn)}\subseteq I^{n}$ after localizing at each prime ideal of $R$ and so we may assume that $R$ is local. Since $I^{(hn)}\widehat{R}\subseteq(I\widehat{R})^{(hn)}$ and $I\widehat{R}$ is still a radical ideal, see [GD65, 7.6.7(ii)], and each minimal prime has height $\leq h$, see [GD65, 7.1.10] or [Mat89, Corollary on page 251]. If we can show that $(I\widehat{R})^{(hn)}\subseteq(I\widehat{R})^{n}=I^{n}\widehat{R}$, then it follows that $I^{(hn)}\subseteq I^{n}\widehat{R}\cap R=I^{n}$. Hence we may assume that $A:=\widehat{R}$ is a complete regular local ring. In the case that $A$ is of equal characteristic, the result is already known by [HH02, Theorem 1.1 (a)] (also see [ELS01, Theorem A]). If $A$ has mixed characteristic, then we are done by Theorem 7.3. This completes the proof. ∎

8. An example

We can compute this perfectoid test ideal in a simple case.

Example 8.1** (SNC pair).**

Consider $A=W(k)\llbracket x_{1},\ldots,x_{d-1}\rrbracket$ for $k$ some perfect field of characteristic $p>0$ and let $f=p^{a_{0}}x_{1}^{a_{1}}x_{d-1}^{a_{2}}\cdots x_{d}^{a_{d-1}}$ for some integers $a_{i}$. Suppose that $A_{\infty}$ contains a fixed copy of $A_{0}:=A[p^{1/p^{\infty}},x_{1}^{1/p^{\infty}},\dots,x_{d-1}^{1/p^{\infty}}]$, this follows for instance if one constructs $A_{\infty}$ via the $R$ from [Bha18, Proposition 5.2] as stated in Section 2.2. In particular for our compatible system of $p$-power roots of $f$, we fix the ones given in $A_{0}$ via products of roots of monomials.

We claim

[TABLE]

Since in the definition of ${\uptau}([f]^{t})$, we are building the $+\epsilon$ variant of the perfectoid test ideal, we may work with a fixed $t+\epsilon=b/p^{e}$. Consider $A^{\prime}=A[p^{1/p^{e}},x_{1}^{1/p^{e}},\dots,x_{d-1}^{1/p^{e}}]$ and observe it is also regular and contains $f^{b/p^{e}}$.

Since $A^{\prime}\subseteq A_{0}$, we have a factorization

[TABLE]

Also note that we can factor the map $A\xrightarrow{\cdot f^{b/p^{e}}}A_{\infty}$ as

[TABLE]

Since $A^{\prime}$ is regular, we have by Section 2 that

[TABLE]

By using this and the argument of Section 3, we have that

[TABLE]

It then follows from local duality that

[TABLE]

where $\Phi$ is the generator of $\operatorname{Hom}_{A}(A^{\prime},A)$ as an $A^{\prime}$-module (for example, it can be taken to be the map which sends the monomial basis $p^{a_{0}/p^{e}}x_{1}^{a_{1}/p^{e}}\cdots x_{d-1}^{a_{d-1}/p^{e}}\mapsto p^{(a_{0}-p^{e}+1)/p^{e}}x_{1}^{(a_{1}-p^{e}+1)/p^{e}}\cdots x_{d-1}^{(a_{d-1}-p^{e}+1)/p^{e}}$ if that term makes sense in $A$ and zero otherwise). But this image is precisely $(p^{\lfloor a_{0}t\rfloor}\cdot x_{1}^{\lfloor a_{1}t\rfloor}\cdots x_{d-1}^{\lfloor a_{d-1}t\rfloor})$ as desired (at this point, it is the same computation as the one for the test ideal).

9. Further questions

We record some open questions regarding the results herein.

Question 9.1*.*

Fix $\{f_{1},\ldots,f_{n}\}$ a sequence of generators of an ideal $\mathfrak{a}\subseteq A$ and for each $f_{i}$ fix a compatible system of $p$-power roots of $f_{i}$ in $A_{\infty}$ (in order to define ${\uptau}^{\sharp}([f_{1},\ldots,f_{n}]^{t})$). Is any inclusion in (missing):

[TABLE]

an equality?

Another fundamental question left open in this paper is:

Question 9.2*.*

Is ${\uptau}([\underline{f}]^{t})$ or ${\uptau}(\mathfrak{a}^{t})$ independent of the choice of $A_{\infty}$?

We also ask how our object behaves with respect to localization. Note that it is still an open question whether or not the formation of the classical characteristic $p>0$ test ideal commutes with localization.

Question 9.3*.*

If $Q\in\operatorname{{Spec}}A$ is a prime containing $p$ and $\mathfrak{a}$, is it true that

[TABLE]

[TABLE]

Appendix A Blowups

In this appendix we briefly recall (and in some cases prove) facts about blowups of ideals. These are well known but we record them here for ease of the reader. Note, we are working with potentially non-Noetherian rings in most cases.

Setting A.1**.**

Throughout this section, $R$ will be a reduced ring and $J\subseteq R$ will be a finitely generated ideal. We let $X=\operatorname{{Spec}}R$ and let $Y\xrightarrow{\ \ }X$ be the blowup of $J$ in $X$. In particular, set $S=R\oplus JT\oplus JT^{2}\oplus\dots$ where the $T$ serve as a dummy variable to help distinguish degree, and thus $Y=\operatorname{{Proj}}S$.

Lemma A.2**.**

If $J=(z_{1},\ldots,z_{m})$, then the complements $U_{i}$ of $V(z_{i}T)\subseteq Y$ form an affine cover of $Y$ with $U=\operatorname{{Spec}}R[z_{1}/z_{i},\ldots,z_{m}/z_{i}]$.

In the above $R[z_{1}/z_{i},\ldots,z_{m}/z_{i}]$ is viewed as the subring of elements of $S[(z_{i}T)^{-1}]$ of the form $gT^{n}/(z_{i}T)^{n}$ as in [Sta18, Tag 052P].

Proof.

Note any homogeneous prime of $S$ does not contain some $z_{i}$ and so this follows from for instance [Sta18, Tag 0804]. ∎

Lemma A.3**.**

Suppose that $f\in R$ is integral over $J$. Define $J^{\prime}=J+(f)$ and let $Y^{\prime}\xrightarrow{\ \ }X$ be the blowup of $J^{\prime}$. Then $Y^{\prime}\xrightarrow{\ \ }X$ factors through $Y$ and $Y^{\prime}$ is a partial normalization of $Y$ generated locally by adding a single integral element to the rings defining the affine charts $U_{i}$.

Proof.

Write $f^{n}+a_{1}f^{n-1}+\dots+a_{n}=0$ with $a_{i}\in J^{i}$. Now write $J=(z_{1},\ldots,z_{m})$ and form the Rees algebra $S$ as above. Let $S^{\prime}=R\oplus J^{\prime}T\oplus J^{\prime 2}T\oplus\dots\supseteq S$. We will first prove that the $U_{i}^{\prime}=Y^{\prime}\setminus V(z_{i}T)$ form an open cover of $Y^{\prime}$ (in particular, we do not need $V(fT)$). Suppose that $Q\subseteq S^{\prime}$ is a homogeneous prime ideal containing all of the $z_{i}T$ but not $fT$. Obviously $Q$ contains $0=f^{n}T^{n}+a_{1}f^{n-1}T^{n}+\dots+a_{n}T^{n}$ also note that $Q$ contains $a_{n}T^{n}$ since $a_{n}T^{n}\in(z_{1},\dots,z_{n})^{n}T^{n}$. But then since $Q$ does not contain $fT$, $Q$ must contain

[TABLE]

But $Q$ also contains $a_{n-1}T^{n-1}$ as before and so continuing in this way, we eventually deduce that $fT\in Q$, a contradiction. Thus we have shown that $\{U_{i}\}$ form an open cover of $\operatorname{{Proj}}S^{\prime}=Y^{\prime}$.

On the other hand, each $U_{i}^{\prime}=\operatorname{{Spec}}R[z_{1}/z_{i},\dots,z_{m}/z_{i},f/z_{i}]$ and $y=f/z_{i}$ satisfies the monic polynomial equation

[TABLE]

where each $a_{j}/z_{i}^{j}\in R[z_{1}/z_{i},\dots,z_{m}/z_{i}]$ by construction. The lemma follows. ∎

Next we recall a partial converse to the previous Lemma.

Lemma A.4**.**

Suppose additionally to Appendix A that $R$ is normal, and that the normalization $\mu:Y^{\prime}\xrightarrow{\ \ }Y$ is finite over $Y$. Then $\pi:Y^{\prime}\xrightarrow{\ \ }X$ is the blowup of $\overline{J^{n}}$ for some $n>0$ where $\overline{\bullet}$ denotes the integral closure of the ideal.

Proof.

Write $J=(z_{1},\dots,z_{m})$ and consider the ring $R_{i}:=R[z_{1}/z_{i},\dots,z_{m}/z_{i}]$ defining an affine chart $U_{i}$ on $Y$. Suppose that $x\in\mathcal{O}_{Y^{\prime}}(\mu^{-1}U_{i})$, and hence $x$ is integral over $R_{i}$. It follows that $x$ satisfies some integral equation

[TABLE]

with $f_{j}=f_{j}(z_{1}/z_{i},\dots,z_{m}/z_{i})\in R_{i}$. Note that we can pick a sufficiently large $h$ such that $f_{j}z_{i}^{h}\in J^{h}$ for all $j$ (i.e., clearing all the denominators of $f_{j}$). It follows that $f_{j}z_{i}^{hj}\in J^{hj}\subseteq R$ for all $j$. Multiplying by $z^{hl}$ we get

[TABLE]

Now, $xz_{i}^{h}$ is in $R$ since it is integral over $R$ and $R$ is normal. Since $f_{j}z_{i}^{hj}\in J^{hj}$ for all $j$, we also have $xz_{i}^{h}\in\overline{J^{h}}$ and thus $x\in R[\frac{\overline{J^{h}}}{z_{i}^{h}}]$. We can do this for the finitely many generators of each chart, and pick $h\gg 0$ that works for all these generators. It follows that there exists $h\gg 0$ such that $\mathcal{O}_{Y^{\prime}}(\mu^{-1}U_{i})\subseteq R[\frac{\overline{J^{h}}}{z_{i}^{h}}]$ for every $i$. But then $\mathcal{O}_{Y^{\prime}}(\mu^{-1}U_{i})=R[\frac{\overline{J^{h}}}{z_{i}^{h}}]$ because the latter is integral over $R_{i}$ and $\mathcal{O}_{Y^{\prime}}(\mu^{-1}U_{i})$ is the integral closure of $R_{i}$. Therefore $Y^{\prime}$ is the blow up of $\overline{J^{h}}$ as desired. ∎

Remark A.5*.*

Another way to prove this when $R$ is normal, Noetherian and excellent is to consider the Rees algebra $S$, and observe that the normalization $S^{\prime}$ of $S$ is

[TABLE]

see for instance [HS06, Proposition 5.2.1]. It easily follows that $\operatorname{{Proj}}S^{\prime}$ is the normalization of $\operatorname{{Proj}}S$ [PS10, 6.C.9 Exercise]. Since $S$ is excellent, $S^{\prime}$ is finite over $S$ and hence Noetherian. We thus see that $S^{\prime n}$, the $n$th Veronese of $S^{\prime}$, is generated in degree $1$ for $n$ sufficiently divisible [Bou98, Chapter III, §1.3, Proposition 3]. But $\operatorname{{Proj}}S^{\prime n}\cong\operatorname{{Proj}}S^{\prime}$ is the blowup of $\overline{J^{n}}$.

Finally, we now move to blowups in Noetherian regular local rings. First we recall some notation, suppose that $\pi:Y\xrightarrow{\ \ }X=\operatorname{{Spec}}A$ is a finite type birational map between normal Noetherian integral schemes where $X$ is regular (or at least Gorenstein). We also fix a choice of a dualizing complex \omega_{A}^{{{\,\begin{picture}(1.0,1.0)(-1.0,-2.0)\circle*{2.0}\end{picture}\ }}} on $A$. Since $A$ is Gorenstein and integral, this complex has cohomology only in a single degree (which we select to be $-\dim X$), and that cohomology is a line bundle which is denoted by $\omega_{X}$. We then define the dualizing complex \omega_{Y}^{{{\,\begin{picture}(1.0,1.0)(-1.0,-2.0)\circle*{2.0}\end{picture}\ }}} on $Y$ to be \pi^{!}\omega_{X}^{{{\,\begin{picture}(1.0,1.0)(-1.0,-2.0)\circle*{2.0}\end{picture}\ }}} where we have sheafified our dualizing complex on $A$. We also set \omega_{Y}:={\mathcal{H}}^{-\dim X}\omega_{Y}^{{{\,\begin{picture}(1.0,1.0)(-1.0,-2.0)\circle*{2.0}\end{picture}\ }}} and observe that this is not necessarily a line bundle.

By a canonical divisor on $X$ we mean any Weil divisor $K_{X}$ on $X$ such that $\mathcal{O}_{X}(K_{X})\cong\omega_{X}$. Since $X$ is Gorenstein, $\mathcal{O}_{X}(K_{X})$ is a line bundle and hence $K_{X}$ is Cartier. Likewise a canonical divisor on $Y$ is any Weil divisor $K_{Y}$ so that $\mathcal{O}_{Y}(K_{Y})\cong\omega_{Y}$.

Lemma A.6**.**

There exist canonical divisors $K_{Y}$ and $K_{X}$ that agree where $\pi$ is an isomorphism. Furthermore, for any choice of $K_{X}$, there is such a compatible choice of $K_{Y}$.

Our proof also holds if $X$ is not necessarily Gorenstein but only normal with a dualizing complex.

Proof.

First notice that even though $\omega_{Y}$ is not a line bundle, $\omega_{Y}$ is still a reflexive rank-1 sheaf, and so there exists a $K_{Y}$ with $\mathcal{O}_{Y}(K_{Y})=\omega_{Y}$. Consider the divisor $\pi_{*}K_{Y}$ on $X$ obtained by throwing away any irreducible component of $K_{Y}$ that is mapped to a subscheme of codimension $\geq 2$. This divisor agrees with $K_{Y}$ wherever $\pi$ is an isomorphism, which is a set $U$ whose complement has codimension $\geq 2$ on $X$. In particular, $\mathcal{O}_{U}(\pi_{*}K_{Y})\cong\omega_{X}|_{U}$. Thus $\mathcal{O}_{X}(\pi_{*}K_{Y})$ is a reflexive sheaf that agrees with $\omega_{X}$ outside a set of codimension $\geq 2$, and so $\mathcal{O}_{X}(\pi_{*}K_{Y})\cong\omega_{X}$, cf. [Har94]. Setting $K_{X}=\pi_{*}K_{Y}$ proves the first part of the lemma.

Now suppose that $K_{X}^{\prime}$ is another choice of canonical divisor. Since $\mathcal{O}_{X}(K_{X}^{\prime})\cong\omega_{X}\cong\mathcal{O}_{X}(K_{X})$, we see that $K_{X}^{\prime}\sim K_{X}$ and so there exists some element $f$ of the fraction field $K(A)$ so that $K_{X}^{\prime}=K_{X}+\operatorname{{div}}_{X}(f)$. We then set $K_{Y}^{\prime}=K_{Y}+\operatorname{{div}}_{Y}(f)$ and observe that $K_{Y}^{\prime}$ and $K_{X}^{\prime}$ agree where $\pi$ is an isomorphism. ∎

Definition A.7** (Relative canonical divisor).**

Choose $K_{Y}$ and $K_{X}$ as in Appendix A. We define the relative canonical divisor $K_{Y/X}:=K_{Y}-\pi^{*}K_{X}$, and observe it is exceptional and also independent of the choice of $K_{Y}$ and $K_{X}$. Note that if one chooses $\omega_{X}\cong\mathcal{O}_{X}$, then one may take $K_{X}=0$ and so $K_{Y}=K_{Y/X}$ may be chosen to be exceptional.

Lemma A.8**.**

Suppose that $(R,\mathfrak{m},k)$ is a regular local Noetherian ring of dimension $d$ and that $Y\xrightarrow{\ \ }X=\operatorname{{Spec}}R$ is the blowup of $\mathfrak{m}$. Then $Y$ is regular, has prime exceptional divisor $E$ with $\mathfrak{m}\mathcal{O}_{Y}=\mathcal{O}_{Y}(-E)$ and $K_{Y/X}=(d-1)E$.

Proof.

This is well known, but because we do not know of a reference where it is phrased in this language outside of the context of varieties over a field, we include a quick geometric proof. Equivalent commutative algebra statements can be found for example in [HV85, HSV87, TW89].

A direct computation shows that the exceptional divisor $E\cong\mathbb{P}^{d-1}_{k}$ lives in the regular scheme $Y$. The same computation also shows that $\mathcal{O}_{X}(-E)|_{E}=\mathcal{O}_{E}(1)$. Because we know999The adjunction formula still works in this generality, simply take the Grothendieck dual of the short exact sequence $0\xrightarrow{\ \ }\mathcal{O}_{Y}(-E)\xrightarrow{\ \ }\mathcal{O}_{Y}\xrightarrow{\ \ }\mathcal{O}_{E}\xrightarrow{\ \ }0$ which yields $0\xrightarrow{\ \ }\omega_{Y}\xrightarrow{\ \ }\omega_{Y}(E)\xrightarrow{\ \ }\omega_{E}\xrightarrow{\ \ }0$. that $(K_{Y}+E)|_{E}=K_{E}$ and that $\mathcal{O}_{E}(K_{E})=\mathcal{O}_{E}(-d)$, if we write $K_{Y}=nE$, then $(K_{Y}+E)|_{E}=(nE+E)|_{E}=K_{E}$ and so $-(n+1)=-d$ and thus $n=d-1$ as claimed. ∎