The Thurston norm via spun-normal immersions
Daryl Cooper, Stephan Tillmann and William Worden

A theory of transversely oriented spun-normal immersed surfaces in ideally triangulated 3–manifolds is developed in this paper, including linear functionals determining the boundary curves, Euler characteristic and homology class of these immersions. This is used to develop and implement an algorithm to compute the unit ball of the Thurston norm for cusped hyperbolic 3–manifolds of finite volume. As an application of independent interest, we give an upper bound on the minimal entropy of pseudo-Anosov maps of surfaces with number of cusps bounded linearly in genus.

57K32, 57K31, 57K10

§ AN APPLICATION

Our proof of <Ref> relies on the following result of Fried [6], for which we need some notation. Let $\F_\Q$ be the set of rational points on a fibered face of a Thurston norm unit ball. For $x\in \F_\Q$, let $F_x$ be the fiber over $x$, and let $\lambda_x$ be the dilatation of the associated monodromy. Let $L_\F:\F_\Q \to \RR$ be defined by $L_\F(x)=|\chi(F_x)|\log(\lambda_x)$.

The function $L_\F$ extends to a continuous convex function on $\F$ going to infinity toward the boundary of $\F$.

We prove <Ref> by finding a particular fibered link for which the Thurston unit norm ball has a fibered face containing many fibers. The following proposition describes precisely the properties such a link (and its unit norm ball) needs to have.

Let $M=\SS^3\setminus (L=L_1\sqcup L_2\sqcup L_3)$ be the complement of a fibered 3-component link, and let $\F$ be a fibered face of the Thurston norm unit ball for $M$ with vertices $x_1,x_2$ and $x_3$ having the following properties:

(1) $2x_1,2x_2$ and $x_3$ have norm minimizing representatives $S_1=\Sigma_{1,2}$, $S_2=\Sigma_{1,2}$, and $S_3=\Sigma_{1,1}$, respectively.

(2) $S_1$ and $S_2$ each have a single boundary component on $L_1$, and a single boundary component on $L_2$, and no boundary component on $L_3$.

(3) The slopes for $S_1$ and $S_2$ are distinct on $L_1$ and on $L_2$.

(4) $S_3$ has no boundary components on $L_1$ or $L_2$, and has one boundary component on $L_3$.

Then for any $0<\varepsilon<1$, there is a compact subset $K_\varepsilon\subset\F$ such that $\RR^+K_\varepsilon$ contains a fiber $S_{g,n}$ for every $g\ge 2$ and every $n$ satisfying $\varepsilon g + 2\le n\le \frac{g}{\varepsilon}+2$.

Suppose we have such a link $L$ and fibered face $\F$. We first prove the following:

Claim 1: For every integer $g\ge 2$, there are positive integers $p,q$ such that $p+q=g$, $\gcd(p,q)=1$, and $|p-q|\le 4$.

Proof of Claim 1:
Given $g$, choose $p$ and $q$ as follows. If $g=2$, set $p=q=1$. If $g$ is odd, let $p=\frac{g-1}{2}+1$ and let $q=\frac{g-1}{2}$, so that $q=p+1$ and hence $\gcd(p,q)=1$. If $g\equiv 0\mod 4$, choose $p=\frac{g}{2}-1$ and $q=\frac{g}{2}+1$, so that $q=p+2$ and $p$ and $q$ are both odd (since $\frac{g}{2}$ is even). Then $\gcd(p,q)\mid 2$, but since $p$ and $q$ are odd we must have $\gcd(p,q)=1$. If $g\equiv 2\mod 4$, choose $p=\frac{g}{2}-2$ and $q=\frac{g}{2}+2$, so that $q=p+4$. Then $\gcd(p,q)\mid 4$, but $\frac{g}{2}$ is odd, so $p$ and $q$ must also be odd and hence $\gcd(p,q)=1$ is the only possibility, thus concluding the proof of Claim 1.

Fix $0<\varepsilon <1$, and define $K_\varepsilon\subset \F$ as follows:

K_\varepsilon:= \left\{x\in \F \,\,\Big |\,\, x= \frac{px_1+qx_2+rx_3}{|px_1+qx_2+rx_3|},\,\, where \,\,\, \frac{\varepsilon}{6}\le \frac{p}{r},\frac{q}{r},\frac{p}{q}\le \frac{6}{\varepsilon}\right\}

Since $H_2(M,\del M; \RR)\cong \RR^3$ and $x_1,x_2,x_3$ are linearly independent, we may identify $\frac{x_i}{\|x_i\|}$, $i=1,2,3$,  with the standard basis of $\RR^3$, so that $\F$ is the standard 2-simplex. The conditions in the definition of $K_\varepsilon$ say that the projection of the ray determined by $x=px_1+qx_2+rx_3$ to each coordinate plane has slope bounded away from $0$ and $\infty$, so that the intersection of this ray with $\F$ stays a bounded distance (in the $\RR^3$ metric) from the boundary of $\F$. It follows that $K_\varepsilon$ is a compact subset of $\F$.

We first show that $\RR^+\F$ contains a fiber $F=\Sigma_{g,n}$ for every $g\ge 2$ and $n\ge 3$. To this end, fix such a pair $(g,n)$ and choose $p,q$ as in Claim 1. Let $S'=pS_1+qS_2$. Since $S_1$ and $S_2$ have distinct slopes on $L_1$ and $L_2$ and $\gcd(p,q)=1$, $S'$ has exactly two boundary components (one on $L_1$ and one on $L_2$). Since the Thurston norm is linear over the fibered face we have

2+2\cdot \genus(S')-2=|\chi(S')|=\|S'\|=p\|S_1\|+q\|S_2\|=2p+2q \implies \genus(S')=p+q=g

Letting $F=S'+rS_3$ with $r=n-2$, we compute similarly

2\cdot\genus(F)-2+|\del F|=|\chi(F)|=\|F\|=\|S'\|+r\|S_3\|=2p+2q+r

Since the two boundary components of $S'$ are on $L_1$ and $L_2$, and the $r$ boundary components of $rS_3$ are on $L_3$ (i.e., disjoint from the boundary of $S'$), $F$ has a total of $r+2=n$ boundary components, which in turn implies that the genus of $F$ is also $g=p+q$. Since there is only one boundary component on, e.g, $L_1$, $F$ cannot be a multiple of a fiber, and must therefore be a fiber. Since $p,q,r$ are positive, $F$ is in the interior of the face $\F$.

Now let $F$ be as above, and assume that $n$ satisfies $\varepsilon g + 2\le n\le \frac{g}{\varepsilon}+2$. Since $\varepsilon<1$ and $|p-q|\le 4$, we immediately have $\frac{\varepsilon}{6}\le \frac{1}{5}\le \frac{p}{q}\le 5\le \frac{6}{\varepsilon}$. Using $r=n-2$ and $g=p+q$ we have

\varepsilon g + 2\le n\le \frac{g}{\varepsilon}+2 \implies \varepsilon\le\frac{p+q}{r}\le \frac{1}{\varepsilon} \implies \frac{p}{r},\frac{q}{r}\le \frac{1}{\varepsilon}\le \frac{6}{\varepsilon}

Since $|p-q|\le 4$ we have $p\le q+4\le 5q $ and $q\le p+4\le 5p$. This gives

\varepsilon \le \frac{p+q}{r}\le \frac{6p}{r} \implies \frac{\varepsilon}{6}\le \frac{p}{r}

and similarly for $\frac{q}{r}$. It follows that $F\in \RR^+K_\varepsilon$ and the proposition is proved.

We successively deplumb Hopf bands from the Seifert surface for the oriented link L9a46, until we are left with a single Hopf band.

Let $L$ be the census link , pictured in <Ref> (upper left). Then $L$ is fibered, and at least one face of its Thurston norm unit ball for $\SS^3\setminus L$ satisfies the conditions of <Ref>.

First we show that $L$ is fibered. We will use a criterion of Gabai, given in <cit.>, which states that an oriented alternating link $L$ is fibered if and only if the surface $R$ obtained by Seifert's algorithm is a plumbing of Hopf bands. This means that $R$ can be obtained by taking successive Murasugi 4-sums of Hopf bands. Or, equivalently, that we can successively remove Hopf bands from $R$ by desumming, until we obtain a disjoint union of Hopf bands. For a discussion of Murasugi sums see [9]. Note that for a link, Seifert's algorithm depends on the orientations of the components. If we choose the wrong orientation on $L$ we may get a Seifert surface which is not a plumbing of Hopf bands, but it may be that $L$ is fibered with respect to some other orientation. With the orientation shown in the upper left of <Ref>, we obtain the Seifert surface $R$ shown at upper middle-left of the same figure. In general, a Seifert surface is some number of disks joined by twisted bands. If two such bands join the same two Seifert disks and twist in the same direction, then we say they are parallel. We first show the following:

Claim 1: If $b_l$ and $b_r$ are parallel bands of a Seifert surface $R$ that are separated by at most one other band $b'$, then we can deplumb a Hopf band $H$ from $R$ in such a way that either $B_l$ or $B_r$ is removed from $R$.

Proof of Claim 1: Refer to <Ref>. In this figure we remove $b_r$. In general, whether we can remove $b_r$ or $b_l$ depends on the direction of the twist of $b_r$ and $b_l$, and on the position of $b'$. Since we will only need the case shown in <Ref>, we will not discuss the other cases (though they are similar). For our case of interest, we (1) slide $b'$ out of the way (down and to the left), then (2) slide $b_r$ into position to be desummed, then (3) deplumb $b_r$, and finally we (4) slide $b'$ back to its original position.

With the claim proved, we can now show that $L$ is fibered by following the steps of <Ref>. In the first step (after Seifert's algorithm), we identify three parallel bands, and remove two of them (i.e., we do steps $(2\to 3)$ of <Ref> twice). In the next two steps we identify three bands $b_r$, $b_l$, and $b'$, and remove $b_r$ using steps $(1\to 4)$ of <Ref>. Finally, we remove another parallel band, then untwist the last band. We are left with a Hopf band, as desired.

Finally, the claim that the Thurston norm ball for $\SS^3\setminus L$ has a face satisfying the conditions of <Ref> is proved by explicitly computing the norm ball. Using the implementation described in <Ref>, we find that the norm ball is an octahedron, with vertices $x_1,x_2,x_3$ as shown in <Ref>, in which $x_1$ and $x_2$ are the respective vertices on the positive $y$- and $z$-axes, and $x_3$ is the vertex on the positive $x$-axis. The points $2x_1$ and $2x_2$ lying over $x_1$ and $x_2$ are represented by surfaces $S_1$ and $S_2$, both of which are twice-puntured tori. The vertex $x_3$ is represented by a once-punctured torus $S_3$. The boundary slopes of $S_1$ are $[(0,1),(-1,0),(0,0)]$, the boundary slopes of $S_2$ are $[(-1,0),(0,1),(0,0)]$, and the boundary slopes of $S_3$ are $[(0,0),(0,0),(0,1)]$. Since every face has vertices $\pm x_1, \pm x_2, \pm x_3$ for some choice of signs, all faces of the octahedron satisfy conditions (1) though (4) of <Ref>. Since the link is fibered, at least one of these faces is a fibered face, so the hypotheses of <Ref> are satisfied.

Deplumbing a Hopf band.

We are now ready to prove <Ref>.

By <Ref> there is a fibered link satisfying the conditions of <Ref>. For $\varepsilon>0$ fixed, let $c_\varepsilon$ be the maximum of Fried's function $L_\F$ on $K_\varepsilon$. By <Ref>, for every $g\ge 2$ and $n$ satisfying $\varepsilon g + 2\le n\le \frac{g}{\varepsilon}+2$, $\RR^+K_\varepsilon$ contains an element $x$ represented by a fiber $F\cong \Sigma_{g,n}$. Thus

$$l_{g,n}\le \log(\lambda_x)=\frac{L_\F(x)}{|\chi(F)|}\le \frac{c_\varepsilon}{|\chi(F)|}$$

This is the desired upper bound.

§.§ Acknowledgements

The authors thank Eriko Hironaka, Craig Hodgson, Saul Schleimer, Henry Segerman and Mehdi Yazdi for comments and discussions related to this work.
Research of the third author is supported in part under the Australian Research Council's ARC Future Fellowship FT170100316.

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Department of Mathematics, University of California Santa Barbara, CA 93106, USA

School of Mathematics and Statistics, The University of Sydney, NSW 2006, Australia

Department of Mathematics, Rice University, Houston, TX 77005, USA


# Coherence of absolute integral closures

Shravan Patankar  Department of Mathematics, Statistics and Computer Science
University of Illinois at Chicago
Chicago, IL 60607-7045
USA<EMAIL_ADDRESS>https://spatan5.people.uic.edu/

###### Abstract.

We prove that the absolute integral closure $R^{+}$ of an equicharacteristic
zero noetherian complete local domain $R$ is not coherent, provided
$\dim(R)\geq 2$. As a corollary, we give an elementary proof of the mixed
characteristic version of the result due to Asgharzadeh and extend it to
dimension $3$. Furthermore, we apply the methods of Aberbach and Hochster used
to prove the positive characteristic version of this result to study
F-coherent rings and our work naturally suggests a mixed characteristic
analogue of a result of Smith.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Main theorems
  4. 4 Coherence in mixed characteristic
  5. 5 An analogue of Smith’s result
  6. 6 Acknowledgements

##  1\. Introduction

Throughout this article, all rings are assumed to be commutative and contain
an identity element. Recall from [Art71] that the _absolute integral closure_
of a domain $R$, denoted by $R^{+}$, is the integral closure of $R$ inside an
algebraic closure of its fraction field. As Aberbach and Hochster remark in
[AH97, Introduction], although $R^{+}$ is never noetherian (if $\dim(R)\geq
1$), there is a tremendous motivation to study such algebras due to the deep
and seminal results surrounding them: Hochster-Huneke [HH94] and Bhatt’s
[Bha20] results roughly state that $R^{+}$ is (cohomologically) Cohen-Macaulay
in positive and mixed characteristics respectively and Smith’s [Smi94] work
shows that $R^{+}$ has intimate connections with tight closure theory and
rational singularities. We remark that although $R^{+}$ is never Cohen-
Macaulay when $R$ has equicharacteristic zero if $\dim(R)\geq 3$, there has
been some speculation as to whether certain ‘almost’ variants hold [RSS07].

###### Definition 1.1 [Gla89,Def. on p. 44].

A ring $R$ is said to be coherent if every finitely generated ideal of $R$ is
finitely presented.

Clearly, noetherian rings are coherent and coherence is a natural condition to
investigate in non-noetherian (for example, analytic) situations. Hence, it is
natural to ask whether $R^{+}$ is coherent. We show the following:

######  Theorem A.

Let $R$ be an equicharacteric zero (i.e. $\mathbb{Q}\subset R$) noetherian
complete local domain such that $\dim(R)\geq 2$. Then $R^{+}$ is not coherent.

Our key ingredient for the proof comes from the work of Roberts, Singh, and
Srinivas [RSS07]. They show that there are elements of arbitrarily low degree
annihilating the local cohomology of the cone over an elliptic curve in
characteristic zero by constructing the annihilators explicitly in the
absolute integral closure. It follows that the dagger closure of a certain
ideal is non-trivial and a standard argument shows that this cannot happen if
$R^{+}$ is coherent. As a corollary, we recover the mixed characteristic
version of the above result due to Asgharzadeh [Asg17, Theorem 6.6] (who
proved it when $\dim(R)\geq 4$) and extend it to $\dim(R)=3$, answering his
explicit question from the statement of [Asg17, Theorem 6.6].

######  Corollary 1.2.

Let $R$ be a noetherian complete local domain of mixed characteristic. If
$\dim(R)\geq 3$ then $R^{+}$ is not coherent.

In contrast to Asgharzadeh’s proof which uses the direct summand theorem
[And18], or even the methods of Aberbach and Hochster [AH97] who prove the
corresponding positive characteristic statement, our methods are elementary.
In particular, we do not use perfectoid geometry or any of the recent advances
in mixed characteristic commutative algebra and algebraic
geometry.111Shimomoto has given another proof of Asgharzadeh’s theorem (in
private communication, see proof of Theorem 3.6) using the main theorems of
[Bha20]. Similar to Asgharzadeh’s proof, it uses highly non-trivial results
and is unable to achieve our $\dim(R)=3$ improvement. Furthermore, Corollary
1.2 supplements the story of $R^{+}$ in mixed characteristic which has
garnered a tremendous amount of interest in recent years due to [Bha20] and
its applications to mixed characteristic singularities and algebraic geometry
[BMP+20], [TY20], [MS21, Remark 4.11].

The positive characteristic version of the above statements is due to Aberbach
and Hochster [AH97, Section 4] who develop and use the famous homological
properties of perfect rings (see also [BS17, Section 3], [BIM18]). Perhaps as
interesting as our results are the reasons as to why our methods don’t work in
positive characteristic. While similar dagger closure computations work in
positive characteristic 222The computations are much easier since dagger
closure agrees with tight closure for complete local domains by [HH91]., we
crucially use that normal equicharacteristic zero domains are splinters and
hence certain elements in the dagger closure of an ideal are not in its plus
closure. In positive characteristic, dagger closure agrees with plus closure
in many situations and for our example (with the coefficient field of positive
characteristic) by the deep result of [Smi94]. To the best of the author’s
knowledge, the correct analogue of the main result of [Smi94] in mixed
characteristic is not known ([Bha20, Remark 5.5], [MS21, Remark 4.11]).

The proof ideas of our main theorems do not show that the $p$-adic completion
of $R^{+}$, $\widehat{R^{+}}$, is not coherent and we discuss progress towards
it in the fourth section. Although we largely use the methods of Aberbach and
Hochster, new ingredients from mixed characteristic such as the homological
properties of perfectoid rings from [BIM18] and the theory of coherent rings
are required. Our refinement of their methods yields an application to
‘$F$-coherent’ rings and connections to Heitmann’s recent work [Hei21],
homological properties of valuation rings. As mentioned above, our main
theorems are intimately connected to Smith’s result [Smi94]. In the fifth
section, we propose an analogue of Smith’s result in mixed characteristic and
demonstrate the desired containment for the running example in this note and
what is often the first example in tight closure theory, the degree $3$ Fermat
curve. The techniques used in these sections are much more sophisticated in
comparison.

##  2\. Preliminaries

In this section, we recall definitions and lemmas on valuations, closures, and
coherent rings which we need to prove the main theorems.

###### Definition 2.1.

Let $A$ be a domain. A valuation on $A$ is a function $v$ is a from A to
$\mathbb{Q}\cup\infty$ such that

  1. (1)

$v(a)=\infty$ if and only if $a=0$

  2. (2)

$v(ab)=v(a)+v(b)$ for all $a,b\in A$

  3. (3)

$v(a+b)\geq\min(v(a),v(b))$ for all $a,b\in A$

We need the following standard lemma regarding coherent rings and ‘almost
zero’ elements, a similar argument and discussion appears in [AH97,
Proposition 2.3], [AS12, Remark 2.4] and [Shi10, Discussion 3.7, Thm. 3.8]:

######  Lemma 2.2.

Let $(R,m)$ be a noetherian local domain and let $J$ be a finitely generated
proper ideal of $R^{+}$. Then $J$ cannot contain elements which are
arbitrarily small with respect to a valuation on $R^{+}$ positive on $J$.

###### Proof.

Let $S$ be a finite extension of $R$ which contains generators of $J$. Since
$J$ is a proper ideal, there exists $P$, a prime ideal of $S$, containing $J$
and the generators of $J$. There exists a DVR $(V,tV)$ such that $S\subseteq
V\subseteq K$ where $K=Frac(S)$ and $P=S\cap tV$. Note that this implies that
valuations of elements in $R$ but not in $m$ is $0$. We can extend the
valuation to $R^{+}$ by the discussion below. This shows there exists a
valuation positive on $J$. Let $\alpha$ be the minimum of the valuations of
the elements generating $J$. Since $v(a)\geq\alpha$ for all $a$ _generating_
$J$, we have that $v(a)\geq\alpha$ for all elements of $J$ (by properties
listed in Definition 2.1) and hence we are done.
The existence and extension to $R^{+}$ of valuation rings above is well known
and is mentioned without a reference in [Shi10, Discussion 3.7], [HH91, page
237], and [RSS07, page 238]; we give a few more details at the suggestion of
the referee. The DVR $V$ above exists by [HS18, Theorem 6.4.3]. We have
$S\subseteq V\subseteq K$ and hence we have $S^{+}\subseteq V^{+}\subseteq
K^{+}$. $V^{+}$ has a maximal ideal containing the maximal ideal of $V$ by the
Cohen-Seidenberg theorems. $V^{+}$ need not be a valuation domain in general
(it is if $V$ is henselian), however, it is a Prüfer domain by [Bou89, Chapter
VI, §8.6, Prop. 6] and hence localizing at the aforementioned maximal ideal
gives the desired valuation. See also [Ho07, Page 28].

∎

We recall the definitions of dagger closure and plus closure:

###### Definition 2.3 [HH91].

Let $(R,m)$ be a complete local domain and let
$v:R^{+}\xrightarrow{}\mathbb{Q}\cup\infty$ be a valuation positive on $m$.
Let $I$ be an ideal of $R$. The dagger closure of $I$, $I^{\dagger}$ is the
set of all elements $x\in R$ such that there are elements $u\in R^{+}$ of
arbitrarily small valuation with $ux\in IR^{+}$

Dagger closure was defined as an attempt to provide a characteristic free
ideal closure operation with properties similar to tight closure. Hochster-
Huneke [HH91, Page 236] remark that a priori, each valuation may give a
different dagger closure, and show that this is not the case if $R$ has
positive characteristic by proving that dagger closure agrees with tight
closure ([HH91, Theorem 3.1]) for complete local domains (see [BS12,
Proposition 1.8] where it is shown that dagger closure and tight closure agree
for domains essentially of finite type over an excellent local ring). We do
not know if dagger closure depends on the choice of the valuation if $R$ has
equicharacteristic zero or mixed characteristic. Although tight closure has
been studied extensively in positive characteristic, dagger closure continues
to be mysterious333Some progress towards understanding dagger closure has been
made by Stäbler in his thesis work [Sta10]. and we refer the reader to [RSS07,
Questions 4.1, 4.2] for some basic questions about dagger closure that remain
unanswered. The results of this note can be viewed as an application of the
theory of dagger closure.

######  Definition 2.4.

Let $R$ be a domain and let $I$ be an ideal of $R$. An element $a\in R$ is
said to be in the plus closure of $I$, $I^{+}$, if $a\in IR^{+}$.

######  Definition 2.5.

Let $R$ be a noetherian domain. $R$ is said to be a splinter if the inclusion
map $R\rightarrow S$ splits (as a map of $R$-modules) for any module finite
extension $S$ of $R$.

The study of splinters has been a celebrated endeavor with a rich history. The
direct summand conjecture, now a theorem, says precisely that regular rings
are splinters (refer to [DT20] for a survey, basic properties, and additional
references). However we only need to use the following well known and basic
facts ([Ho73, Lemma 2]):

######  Lemma 2.6.

Let $R$ be a normal equicharacteristic zero domain. Then $R$ is a splinter and
hence $I^{+}=I$ for all ideals $I$ of $R$.

###### Proof.

Let $S$ be a module finite extension of $R$. We can quotient $S$ by a minimal
prime and assume it is a domain. Let $d$ be the degree of the extension of the
corresponding fraction fields, i.e. $[Frac(S):Frac(R)]$. We have a trace map
on the fraction fields, $Tr:Frac(S)\rightarrow Frac(R)$. Trace map preserves
integral elements, that is for $s\in S$, $Tr(s)\in Frac(R)$ is integral over
$R$ since $S$ is integral over $R$. Since $R$ is normal, it is integrally
closed in $Frac(R)$ and hence $Tr(s)\in R$. One checks that $Tr(r)=dr$ for any
$r\in R$ and hence $\frac{1}{d}Tr$ gives the desired splitting $S\rightarrow
R$. Let $i\in I^{+}$, then by definition $i\in IR^{+}$ and we can write $i$ as
a finite sum $i=\sum a_{j}r_{j}$ where $a_{j}\in I$ and $r_{j}\in R^{+}$. Let
$S$ be a finite extension of $R$ containing all $r_{j}$, then we have $i\in
IS$. Since $R$ is a splinter, we have $R\rightarrow S$ splits and since $i\in
R$, we have $i\in I$. ∎

We need to use the following standard property of coherent rings ([Gla89, Thm
2.3.2 6]) and recall the proof for the convenience of the reader:

######  Lemma 2.7.

Let $R$ be a coherent ring. Let $I$ be a finitely generated ideal of $R$ and
let $a\in R$. Then the ideal $(I:a)=\\{x|xa\in I\\}$ is finitely generated.

###### Proof.

Let $I=(u_{1},\cdots,u_{r})$ and let $F$ be the free module on generators
$(x_{1},\cdots,x_{r+1})$. Consider the short exact sequence

$0\rightarrow K\rightarrow F\rightarrow(I,a)\rightarrow 0$

where the map from $F$ to $(I:a)$ is defined by $x_{i}\rightarrow u_{i}$ for
$1\leq i\leq r$ and $x_{r+1}\rightarrow a$. Note that $K$ is finitely
generated since $F$ is finitely generated and $(I,a)$ is finitely presented
(as it is a finitely generated ideal of a coherent ring $R$). Let $F^{\prime}$
be the free submodule of $F$ with generators $x_{1},\cdots,x_{r}$. Any element
of $K$ is of the form $u=a_{1}x_{1}+\cdots+a_{r+1}x_{r+1}$ where
$a_{r+1}\in(I:a)$. Thus we have a map $K\rightarrow(I:a)$ defined by
$u\rightarrow a_{r+1}$ and we also have the short exact sequence

$0\rightarrow K\cap F^{\prime}\rightarrow K\rightarrow(I:a)\rightarrow 0$

Hence $(I:a)$ is finitely generated. ∎

##  3\. Main theorems

The following is our key ingredient:

###### Theorem 3.1 [RSS07,Example. 2.4].

Let $k$ be a field of characteristic zero. Let
$A=k[[x,y,z]]/(x^{3}+y^{3}+z^{3})$ or $A=k[x,y,z]/(x^{3}+y^{3}+z^{3})$. Then
$z^{2}$ is inside the dagger closure of $(x,y)$ (with respect to any valuation
$v$ of $A^{+}$ positive on $x$, $y$, and $z$ and $0$ on $k-\\{0\\}$). That is,
there are elements of arbitrarily low valuation in the ideal
$((x,y):_{A^{+}}(z^{2}))$.

###### Proof.

We wish to show that there are elements of arbitrarily low valuation in the
ideal $((x,y):_{A^{+}}(z^{2}))$. Since $k[x,y,z]/(x^{3}+y^{3}+z^{3})$ is a
subring of $k[[x,y,z]]/(x^{3}+y^{3}+z^{3})$, the absolute integral closure of
$k[x,y,z]/(x^{3}+y^{3}+z^{3})$ is a subring of the absolute integral closure
of $k[[x,y,z]]/(x^{3}+y^{3}+z^{3})$. Hence we may work with
$k[x,y,z]/(x^{3}+y^{3}+z^{3})$, as the elements we construct will be inside
the absolute integral closure of $k[x,y,z]/(x^{3}+y^{3}+z^{3})$. We may assume
that $k$ is algebraically closed and by a linear change of coordinates work
with the ring $A=k[x,y,z]/(\theta x^{3}+\theta^{2}y^{3}+z^{3})$ where $\theta$
is a primitive third root of unity. The strategy is to embed $A$ into a copy
of $A$ with ‘lower’ grading via a ‘Frobenius-like’ map and induction takes
care of the rest. Since the embedding is constructed explicitly using
equations with cube roots the elements will turn out to have arbitrarily low
valuation with respect to _any_ valuation $v$ (positive on $x$, $y$, and $z$).
Let $(x_{1},y_{1},z_{1})$ be such that

$\displaystyle x_{1}^{3}=\theta^{\frac{1}{3}}x+\theta^{\frac{2}{3}}y$
$\displaystyle y_{1}^{3}=\theta^{\frac{1}{3}}x+\theta^{\frac{5}{3}}y$
$\displaystyle z_{1}^{3}=\theta^{\frac{1}{3}}x+\theta^{\frac{8}{3}}y$

It follows from direct calculations that $A_{1}=k[x_{1},y_{1},z_{1}]$ is
isomorphic to and contains $A$: $x$, $y$ are in $A_{1}$ as they are a linear
combination of $(x_{1},y_{1},z_{1})$. The element $z$ is in $A_{1}$ as
$(x_{1}y_{1}z_{1})^{3}=\theta x^{3}+\theta^{2}y^{3}=-z^{3}$ so
$-z=x_{1}y_{1}z_{1}$, and $\theta x_{1}^{3}+\theta^{2}y_{1}^{3}+z_{1}^{3}=0$;
one checks that $(0=\theta+\theta^{2}+1)$ factors from the coefficients of
each of $(x,y,z)$ upon expanding. Define $A_{2}$ and $(x_{2},y_{2},z_{2})$
similarly. We have

$x_{1}z^{2}=x_{1}x_{1}^{2}y_{1}^{2}z_{1}^{2}=x_{1}^{3}y_{1}^{2}z_{1}^{2}=(\theta^{\frac{1}{3}}x+\theta^{\frac{2}{3}}y)(y_{1}^{2}x_{1}^{2})\in(x,y)$.

Also,

$x_{2}z^{2}=x_{2}x_{1}^{2}y_{1}^{2}z_{1}^{2}=x_{2}x_{1}^{2}y_{1}^{2}x_{2}^{2}y_{2}^{2}z_{2}^{2}=x_{2}^{3}x_{1}^{2}y_{1}^{2}y_{2}^{2}z_{2}^{2}=(\theta^{\frac{1}{3}}x_{1}+\theta^{\frac{2}{3}}y_{1})x_{1}^{2}y_{1}^{2}y_{2}^{2}z_{2}^{2}\in(x_{1}^{3},y_{1}^{3})\in(x,y)$.

Similarly we get that $x_{n},y_{n},z_{n}$ are in the ideal $((x,y):z^{2})$.

We check that the valuations become arbitrarily low as we iterate this
process. Note that the formulae for $x_{n}$, $y_{n}$, $z_{n}$ imply that they
have positive valuation. Suppose for the sake of contradiction there is a real
number $\delta>0$ such that $v(x_{n}),v(y_{n}),v(z_{n})$ is greater than or
equal to $\delta$. Pick a large positive integer $N$ such that
$(2N+1)\cdot\delta>v(z)$. Since $(x_{n}y_{n}z_{n})^{3}=-z_{n-1}^{3}$ for all
$n$ by construction, we have $v(z_{N-1})=v(x_{N})+v(y_{N})+v(z_{N})>3\delta$.
Similarly,
$v(z_{N-2})=v(x_{N-1})+v(y_{N-1})+v(z_{N-1})>\delta+\delta+3\delta=5\delta$.
Repeating this $N$ times we get
$v(z)=v(x_{1})+v(y_{1})+v(z_{1})>(2N+1)\delta>v(z)$, a contradiction. Hence
there are elements of arbitrarily low valuation in the ideal
$((x,y):_{A^{+}}z^{2})$ for any valuation positive on $x$ and $y$. ∎

######  Remark 3.2.

The above construction without the linear change of coordinates is given in
[Sin09, Example 3.4].

######  Remark 3.3.

A similar statement as Theorem 3.1 can be deduced using (algebro-)geometric
ideas, see [RSS07, Theorem 3.4]. Since $A$ is (the completion of the cone
over) an elliptic curve, there is a multiplication by $n$ endormorphism of
rings of $A$ which has degree $n^{2}$ (in general morphisms of projective
varieties do not correspond to morphisms of the corresponding graded rings,
the statement here uses vanishing theorems of line bundles on abelian
varieties). It follows that there are elements of degree $\frac{1}{n^{2}}$ in
$((x,y):z^{2})$, and by iterating this map one gets that there are elements of
arbitrarily low degree in the same ideal. It is a natural question whether one
can conclude there are elements of arbitrarily low _valuation_ in this ideal
for a valuation $v$ on $R^{+}$ as in Theorem 3.1 above using the
multiplication map, however we do not know the answer 444There seems to be a
subtle change of language in literature in the context of the above dagger
closure example while relating ‘low degree’ and ‘low valuation’ elements.
[RSS07, last paragraph, Page 5] says that “ The examples are
$\mathbb{N}$-graded, and in such cases it is natural to use the valuation
arising from the grading: $v(r)$ is the least integer $n$ such that the $n$-th
degree component of $r$ is nonzero” where as the subsequent work [Sin09, Page
10] mentions “We focus next on an example; for graded domains, we use the
grading in lieu of a valuation.”. A priori, a notion of a degree can only be
extended to $R^{+GR}$ (refer to Section 3, [RSS07]) and hence one would only
get the non-coherence of $R^{+GR}$ this way.

Here is our main theorem:

######  Theorem A.

Let $R$ be an equicharacteric zero (i.e. $\mathbb{Q}\subset R$) noetherian
complete local domain such that $\dim(R)\geq 2$. Then $R^{+}$ is not coherent.

###### Proof.

Cohen Structure Theorem implies $R$ is module finite over a power series ring
over a field. Hence we may assume $R$ itself is isomorphic to
$k[[x_{1},\cdots,x_{{n}}]]$. The ideal $I=((x_{1},x_{2}):_{R^{+}}z^{2})$ where
$z\in R^{+}$ is such that $z^{3}+x_{1}^{3}+x_{2}^{3}=0$, is proper: if it
contained $1$, we would have $z^{2}\in(x_{1},x_{2})R^{+}$. However, $R[z]$ is
normal and as normal rings are splinters (here we are using $R$ has
equicharacteristic zero) we have that $z^{2}\in(x_{1},x_{2})R[z]$. If $R$ is a
ring then $R[[x]]$ is faithfully flat over $R$: it is flat over $R$ as it is
flat over the polynomial ring $R[x]$ (which is free over $R$) and is faithful
as there is a surjection $R[[x]]\rightarrow R$ sending $x$ to $0$. Hence
$R[z]$ is faithfully flat over $k[[x_{1},x_{2}]][z]$ and it follows that
$z^{2}\in(x_{1},x_{2})k[[x_{1},x_{2}]][z]$ which is not true as for example
the class [$\frac{z^{2}}{x_{1}x_{2}}$] spans the degree zero part of the top
local cohomology of $k[[x_{1},x_{2}]][z]$ ([RSS07, Example 2.4]). If $R^{+}$
were coherent this ideal would be finitely generated by Lemma 2.7 (here we are
using coherence). Consider the valuation arising in Lemma 2.2, that is choose
a extension of $R$ containing the generators of $I$ and choose a valuation
dominating a prime ideal containing $I$ and all $x_{i}$. Then there are
elements of arbitrarily low valuation in $I$ by Theorem 3.1, which is a
contradiction by Lemma 2.2. ∎

######  Remark 3.4.

It is interesting to note why the above strategy fails if $k$ has positive
characteristic. Indeed by standard tight closure computations, $z^{2}$ is
inside the tight closure (which is the same as dagger closure in complete
local domains by [HH91]) of $(x_{1},x_{2})$ and hence one gets elements of
arbitrarily low valuation in $((x_{1},x_{2}):_{R^{+}}z^{2})$. However, we are
crucially using the fact that normal rings are splinters in equicharacteristic
zero. One cannot conclude in general that this ideal is proper and in fact it
is not in this case: by a deep result of Smith [Smi94], tight closure is plus
closure for ideals generated by systems of parameters (under mild assumptions)
and hence $z^{2}$ will in fact be inside $(x_{1},x_{2})R^{+}$.

Asgharzadeh’s result follows immediately from (the proof of) Theorem A after
localizing and we extend his results to dimension $3$, answering his question
from [Asg17, Theorem 6.6]:

######  Corollary 3.5.

Let $R$ be a noetherian complete local domain of mixed characteristic $(0,p)$.
If $\dim(R)\geq 3$ then $R^{+}$ is not coherent.

###### Proof.

Cohen Structure Theorem implies that $R$ is module finite over a power series
ring over a DVR, and hence we may assume $R=V[[x_{1},\cdots,x_{n}]]$ where $V$
is a DVR. $R_{p}=R[\frac{1}{p}]$ need not be a complete local domain in
equicharacteristic zero, however, $x_{1},\cdots,x_{n}$ are still algebraically
independent elements in $R_{p}$. Arbitrary localizations of coherent rings are
coherent [Gla89, Thm 2.4.2] and absolute integral closures commute with
localization, hence to show that $R^{+}$ is not coherent it is enough to show
that $R^{+}[\frac{1}{p}]=R_{p}^{+}$ is not coherent. One similarly observes
that there are arbitrarily small elements in
$I=((x_{1},x_{2}):_{R_{p}^{+}}z^{2})$; slight care is required here - $R_{p}$
may not be or contain a two dimensional _power series_ ring over a
characteristic zero field, however it does contain a two dimensional
polynomial ring over a characteristic zero field (as $x_{1}$ and $x_{2}$ are
algebraically independent) and one can repeat the construction and
calculations of Theorem 3.1 for the ring
$V[\frac{1}{p}][x_{1},x_{2},z]/(x_{1}^{3}+x_{2}^{3}+z^{3})$. Choosing a
valuation $v$ positive on $I$ as in the proof of Lemma 2.2, we have that there
are elements of arbitrarily low valuation in $I$ by Theorem 3.1. This ideal is
proper, as if $1\in I$ then as $R_{p}[z]$ is normal it is a splinter and we
have $z^{2}\in(x_{1},x_{2})R_{p}$. The completion of $R_{p}$ is
$V[\frac{1}{p}][[x_{1},\cdots,x_{n},z]]/(x_{1}^{3}+x_{2}^{3}+z^{3})$. It
suffices to show the containment does not hold in the
$(x_{1},\cdots,x_{n})$-adic completion as completion is faithfully flat, for
which one can run the same argument as in Theorem A (which we can apply as
$V[\frac{1}{p}]$ is a characteristic zero field). If $R_{p}^{+}$ was coherent
we would have that the ideal $I$ is finitely generated which is not possible
by Lemma 2.2. ∎

In particular we do not use the direct summand theorem [And18] or any of the
recent advances in mixed characteristic commutative algebra (for example,
[Bha20]). Kazuma Shimomoto has given (in private communication) another proof
of Asgharzadeh’s theorem [Asg17, Theorem 6.6] using the main results of
[Bha20] which we reproduce here, with permission, for the sake of
completeness. Similar to Asgharzadeh’s methods it uses highly non-trivial
results and is unable to achieve our $\dim(R)=3$ improvement. We note that
Shimomoto’s proof allows us to relax the assumptions of Asgharzadeh’s theorem
from a complete local domain to an excellent local domain, since the
corresponding statements from [Bha20] and [BMP+20] have (the weaker)
excellence assumptions. The author thanks Bhargav Bhatt for pointing this out.

######  Theorem 3.6.

Let $R$ be an excellent local domain of mixed characteristic $(0,p)$. If
$\dim(R)\geq 4$ then $R^{+}$ is not coherent.

###### Proof.

(Shimomoto) If $R$ is an excellent local domain the $p$-adic completion of
$R^{+}$, $\widehat{R^{+}}$ is a big balanced Cohen-Macaulay algebra by
[BMP+20, Corollary 2.9]. In general completions of non-noetherian rings may
not be (faithfully) flat over the ring, however, if $R^{+}$ is coherent,
completions with respect to finitely generated ideal are flat by [Has11,
Theorem 3.58]. Hence $\widehat{R^{+}}$ is flat over $R^{+}$. Since $R$ is
local $p$ is in all maximal ideals of $R^{+}$ (by the Cohen-Seidenberg
theorems) and $\widehat{R^{+}}/p\simeq R^{+}/p$ it follows that
$\widehat{R^{+}}$ is faithfully flat over $R^{+}$. This implies $R^{+}$ is big
balanced Cohen-Macaulay, which is not true if $\dim(R)\geq 4$ by [AH97,
Proposition 3.6] or [ST21, Proposition 2.1] (the contradiction is obtained by
a standard trick of localizing at $p$ and using that $R^{+}$ is not Cohen-
Macaulay if $\dim(R)\geq 3$ when $R$ has characteristic zero). This strategy
fails when $dim(R)=3$ because $R^{+}$ itself _is_ Cohen-Macaulay in this
situation by [ST21, Corollary 2.3]. ∎

######  Remark 3.7.

One can observe that the non-coherence of absolute integral closures (in
characteristic zero) relies on the non-triviality of dagger closure and the
fact that normal rings (even non-noetherian ones) are splinters. Our complete
local domain assumption allows us to reduce to the case of the explicit
Theorem 3.1, where we know the dagger closure is non-trivial. Hence it seems
that the complete local domain assumption should not be essential in the
statements of our main results (Theorem A, Corollary 3.5).

######  Remark 3.8.

Note that when $\dim(R)=1$, $R$ is regular if it is normal and finite domain
extensions of normal dimension $1$ domains are flat. Hence $R^{+}$ _is_
coherent as it is a (flat) colimit of normal domains (a flat direct limit of
coherent rings is coherent, [Gla89, Theorem 2.3.3]). The interested reader can
observe that this also follows from the fact that $R^{+}$ is a Prüfer domain.
When $R$ has positive characteristic, Aberbach and Hochster show that $R^{+}$
is not coherent for any noetherian domain $R$ if $\dim(R)\geq 3$ and if $R$ is
a complete local domain containing a field of positive transcendence degree
over a finite field if $\dim(R)=2$. It is not known whether $R^{+}$ (where $R$
is a complete local domain) is coherent in positive and mixed characteristics
when $\dim(R)=2$ in complete generality. It is noteworthy that we are able to
settle the coherence problem in all dimensions (for complete local domains) in
characteristic zero.

##  4\. Coherence in mixed characteristic

Let $R$ be a domain of mixed characteristic $(0,p)$. Throughout this section
we will denote the $p$-adic completion of $R^{+}$ by $\widehat{R^{+}}$. In
mixed characteristic [Bha20, Corollary 5.17] says that $\widehat{R^{+}}$ is
Cohen-Macaulay where $R$ is a complete local domain of mixed characteristic
(where as $R^{+}$ is merely ‘cohomologically Cohen-Macaulay’, i.e.
$H^{i}_{m}(R^{+})=0$ for $0\leq i\leq\dim(R)-1$, $m$ the maximal ideal of
$R^{+}$, and $H^{i}_{m}(R^{+})$ is the local cohomology, see [Bha20, Theorem
5.1]). For clarification on Bhatt’s result and the precise definition(s) of
Cohen-Macaulay, see [BMP+20, Discussion 2.2]. Our methods do not show whether
$\widehat{R^{+}}$ is coherent.

######  Question 4.1.

Let $R$ be a noetherian complete local domain of mixed characteristic. Is the
$p$-adic completion of $R^{+}$, $\widehat{R^{+}}$, coherent?

In the positive characteristic version of our results, Aberbach and Hochster
first use the homological properties of perfect rings to (implicitly) show
that if $\widehat{R^{+}}$ is coherent then it is a G.C.D domain, see proof of
[AH97, Proposition 4.4] and “Thus $a$ and $b$ have a G.C.D in $R_{s}^{+}$”.
Elementary properties of divisor class groups then imply that every complete
local domain has a torsion divisor class group, which is not true if dimension
of the ring is $3$ or more. Using the corresponding homological properties of
perfectoid rings, we are able to obtain the following partial result in mixed
characteristic by mirroring their methods:

######  Theorem 4.2.

Let $R$ be a complete noetherian local domain of mixed characteristic. If
$\widehat{R^{+}}$ is coherent then it is a G.C.D domain.

###### Proof.

Note that $\widehat{R^{+}}$ is a perfectoid ring [BIM18, Example 3.8 (2)]. As
in the proof of Theorem 3.5 we may assume $R$ is a power series ring over a
discrete valuation ring $V[[x_{1},\cdots,x_{n-1}]]$ and hence
$p,x_{1},\cdots,x_{n-1}$ form a s.o.p for $R$. Consider the radical ideal
$m=\sqrt{p,x_{1},\cdots,x_{n-1}}$ where the radical is taken in
$\widehat{R^{+}}$ and hence $m$ is an ideal of $\widehat{R^{+}}$. The ideal
$m\widehat{R^{+}}/\sqrt{p\widehat{R^{+}}}$ of
$\widehat{R^{+}}/\sqrt{p\widehat{R^{+}}}$ is generated up to radical by $n$
elements, hence by [BIM18, Lemma 3.7] flat dimension of $\widehat{R^{+}}/m$
over $\widehat{R^{+}}$ is less than or equal to $n$555The proof of this for
$p$-torsion free perfectoid rings follows immediately from the corresponding
properties of perfect rings.. Note that $m$ is the maximal ideal of
$\widehat{R^{+}}$: since $p\in m$,
$\widehat{R^{+}}/m=R^{+}/\sqrt{p,x_{1},\cdots,x_{n-1}}$ is a field (since
$R^{+}$ is an integral extension of $R$).

Let $I$ be a finitely generated ideal of $\widehat{R^{+}}$. Since
$\widehat{R^{+}}$ is coherent, $\widehat{R^{+}}/I\widehat{R^{+}}$ has a
resolution by finitely generated $\widehat{R^{+}}$ modules by [Gla89,
Corollary 3.5.2]. Since $\widehat{R^{+}}$ is local there is a minimal
resolution of $\widehat{R^{+}}/I\widehat{R^{+}}$ and the $i$th betti number is
given by the vector space dimension over $\widehat{R^{+}}/m\widehat{R^{+}}$ of
$Tor_{i}^{\widehat{R^{+}}}(\widehat{R^{+}}/I\widehat{R^{+}},\widehat{R^{+}}/m)$.
Since the flat dimension of $\widehat{R^{+}}/m$ is less than or equal to $n$,
we have that the projective dimension of $\widehat{R^{+}}/I\widehat{R^{+}}$
over $\widehat{R^{+}}$ is less than or equal to $n$.

The above discussion implies that the projective dimension of finitely
generated ideals of $\widehat{R^{+}}$ is finite. Hence $\widehat{R^{+}}$ is by
definition a (non-noetherian) regular ring [Gla89, Definition on page 200].
Since $R$ is complete and local, $R^{+}$ and consequently $R^{+}/p$ are local.
Hence $\widehat{R^{+}}$ is a local ring. Coherent regular local rings are
G.C.D domains by [Gla89, Corollary 6.2.10]. ∎

######  Remark 4.3.

In the proof above we show that if $\widehat{R^{+}}$ is coherent then it is
also regular and that a coherent regular local ring is a G.C.D domain (just as
in the case of noetherian regular local rings) and in particular that it is an
integral domain. This assertion is not obvious - the main result of Heitmann’s
recent work says that if $R$ is a complete local domain then $\widehat{R^{+}}$
is (surprisingly) an integral domain _regardless_ of our coherence (and
regular) hypothesis [Hei21, Theorem 1.7].

######  Remark 4.4.

As mentioned in Remark 3.8, if $R$ is regular, dimension $1$, and of positive
characteristic, then $R^{+}$ is coherent and it follows from the proof of the
above Theorem 4.2 that projective dimension of $R^{+}/IR^{+}$ (over $R^{+}$)
is $\leq 1$ if $I$ is a finitely generated ideal since $R^{+}$ is a perfect
ring. This may fail if the ideal is not finitely generated: if $R$ is complete
and of dimension $1$ then the global dimension of $R^{+}$ is the projective
dimension of $R^{+}/m$ and is equal to $2$ where $m$ is the maximal ideal of
$R^{+}$ (note that it is not finitely generated). In this situation we may
assume $R$ to be a complete discrete valuation ring and hence $R^{+}$ is a
valuation ring (regardless of the characteristic of $R$, see proof of Lemma
2.2). The result now follows from the theory of valuation rings - the global
dimension of valuation rings was studied by Osofsky [Oso67] and a more modern
and simplified treatment sufficient for the preceding result can be found in
the works of Asgharzadeh [Asg10, Theorem 1.1] and Datta [Dat17, Theorem
4.2.5].

######  Remark 4.5.

In Remark 4.4 we used the fact that absolute integral closures of complete (or
more generally, henselian) valuation rings are valuation rings. This is quite
useful in studying absolute integral closures in dimension $1$, or more
generally, valuation rings. For example, $2$ out of $3$ proofs of ‘Valuation
rings are derived splinters’ by Antieau and Datta [AD21] use the preceding
fact to reduce to the case of absolutely integrally closed valuation rings. As
another example, although by Remark 4.4 the projective dimension of $R^{+}/m$
is $2$ (where $dim(R)=1$, $R$ is complete, and $m$ is the maximal ideal of
$R^{+}$), the $Tor$ dimension of $R^{+}/m$ (over $R^{+}$) is $1$ as we have a
short exact sequence $0\rightarrow m\rightarrow R^{+}\rightarrow
R^{+}/m\rightarrow 0$ and $m$ is a flat because it is a torsion free module
over the valuation ring $R^{+}$. The interested reader can observe that this
answers [AH97, Question 3.7] when $dim(R)=1$.

We expect $\widehat{R^{+}}$ to be not coherent as well. Unfortunately we do
not know how to obtain a contradiction assuming $\widehat{R^{+}}$ is coherent
given Theorem 4.2. One cannot use the same properties of divisor class groups
as in [AH97] as $\widehat{R^{+}}$ is not an integral extension of $R$.
Motivated by the statements above, we would like to include, with permission,
the following related question communicated to us by Kazuma Shimomoto:

######  Question 4.6.

(Shimomoto) Does every noetherian local domain map to a coherent big Cohen-
Macaulay algebra?

When $R$ is excellent of positive characteristic $R^{+}$ is a big Cohen-
Macaulay algebra by [HH94] and not coherent by [AH97] hence one might suspect
that the question has a negative answer. However, the answer is yes for a
class of positive characteristic rings called ‘$F$-coherent’ rings:

###### Definition 4.7 [Shi10,Def. 3.2].

Let $R$ be a ring of positive characteristic. $R$ is said to be ‘$F$-coherent’
if the perfect closure of $R$, $R_{perf}$, is a coherent ring.

It is well known that under mild assumptions666such as $R$ is a quotient of a
Gorenstein ring, see [Shi10, Theorem 3.11] on the ring $R$, $R_{perf}$ is an
‘almost Cohen-Macaulay’ algebra in positive characteristic and if it is
coherent, it follows (refer to [Shi10, Theorem 3.11]) that it is Cohen-
Macaulay, coherence does not allow flexibility of almost mathematics as
indicated by the proofs of our main theorems above (for a more precise version
of this statement see [AS12, Theorem 3.12] which roughly states that
‘coherence and almost Cohen-Macaulay implies (big balanced) Cohen-Macaulay’).
Hence the answer to the above question is yes if the domain is $F$-coherent.
It follows from Kunz’s theorem that regular rings are $F$-coherent ([Shi10,
Proposition 3.3]) and hence so is any purely inseparable extension of a
regular ring (as both have the same perfection). The question of whether these
are all $F$-coherent rings is one of the key open problems in the theory
([Shi10, Question 4]).

######  Remark 4.8.

If $S$ is a module finite purely inseparable extension of a regular $F$-finite
ring $R$ of positive characteristic, $S$ is contained inside
$R^{\frac{1}{p^{e}}}$ for $e$ large enough (take maximum of all $e$ such that
the generators of $S$ as an $R$-module are contained in
$R^{\frac{1}{p^{e}}}$). $R^{\frac{1}{p^{e}}}$ is isomorphic to $R$ as an
algebra and hence $S$ not only has a coherent Cohen-Macaulay algebra, it has a
_noetherian_ Cohen-Macaulay (and in fact, regular) algebra .

######  Remark 4.9.

The proof of Theorem 4.2 offers insights into the characterization of
$F$-coherent rings. The proof and homological properties of perfect rings show
that if $R$ is an excellent noetherian normal $F$-coherent domain then
$R_{perf}$ is a G.C.D domain (see also [Asg17, Corollary 3.7]). Let $(a,b)$ be
a height one ideal of $R$. As $R_{perf}$ is a G.C.D domain $a,b$ have a G.C.D
$d$ in $R_{perf}$. Let $S$ be the normalization of $R[d]$, $S$ is purely
inseparable over $R$ (in general normalizations of $F$-coherent rings are
$F$-coherent by [Shi10, Theorem 3.8]). Standard properties of divisor class
groups (see discussion after [AH97, Lemma 4.3]) imply that there are maps
$Cl(R)\rightarrow Cl(S)$ and $Cl(S)\rightarrow Cl(R)$ with the latter map
multiplication by $[S:R]$. Since $(a,b)$ is trivial in the divisor class group
of $S$ and $S$ is purely inseparable over $R$, it follows that $(a,b)$ is
$p^{\infty}$ torsion in $Cl(R)$. Hence $Cl(R)$ is $p^{\infty}$ torsion for any
excellent normal noetherian $F$-coherent domain $R$. The preceding discussion
shows that the divisor class group of any $F$-coherent ring and any purely
inseparable extensions of it (which is also $F$-coherent) is $p^{\infty}$
torsion under mild conditions on the ring. We do not know if it is reasonable
to expect a converse. Under similar assumptions, if every purely inseparable
extension of a ring $R$ has a $p^{\infty}$ torsion divisor class group, or if
the perfection of the ring is a G.C.D domain, is the ring $F$-coherent?
$p^{\infty}$ torsion in divisor class groups has historically often shown up
as a separate case which requires new methods to deal with in problems in
$F$-singularity theory and algebraic geometry. See for example [PS20] and
[Kol21].

##  5\. An analogue of Smith’s result

As remarked in Remark 3.4, our methods are closely related to the deep result
of Smith ([Smi94]), hence we would like to ask if the following analogous
result holds in mixed characteristic:

######  Question 5.1.

Let $I$ be a parameter ideal of a mixed characteristic complete local domain
$R$ and fix a valuation $v$ on $R^{+}$ positive on $I$. Does the dagger
closure of $I$ (with respect to $v$) coincide with $I\widehat{R^{+}}\cap R$?

We illustrate the question in a series of propositions below using the running
example in our note, $R=\mathbb{Z}_{p}[[x,y,z]]/(x^{3}+y^{3}+z^{3})$. First we
explain why $z^{2}$ is not in the plus closure of $I=(x,y)$ (i.e. $IR^{+}\cap
R$) and that $z^{2}$ is in fact in the dagger closure of $I$. This is implicit
in the proofs of our main theorems in Section $3$ and explains why we ask for
the “$\widehat{R^{+}}$-closure of $I$” (i.e. $I\widehat{R^{+}}\cap R$) to
coincide with the dagger closure instead of the usual plus closure of $I$.
Then we show that $z^{2}$ is in $(p^{n},x,y)R^{+}$ for every $n$ by exploiting
the fact that $R$ is the (completion of) the cone over an abelian variety -
the $p^{n}$ power maps on the abelian variety lift to ring endomorphisms of
$R$. Then we use [Bha20] to show that $z^{2}$ is in $I\widehat{R^{+}}$.

######  Proposition 5.2.

Let $R=\mathbb{Z}_{p}[[x,y,z]]/(x^{3}+y^{3}+z^{3})$ and $I=(x,y)$, then
$z^{2}$ is not in the plus closure of $I$.

###### Proof.

If $z^{2}$ was in the plus closure of $I$, as absolute integral closures
commute with localization, after localizing at $p$ we have that
$z^{2}\in(x,y)R[\frac{1}{p}]^{+}$. This implies $z^{2}\in(x,y)R[\frac{1}{p}]$
as $R[\frac{1}{p}]$ is a normal equi-characteristic zero domain and hence a
splinter. Finally, $z^{2}\notin(x,y)R[\frac{1}{p}]$ as the containment can be
checked after $(x,y)$-adic completion (which is faithfully flat) and the
$I$-adic completion of $R[\frac{1}{p}]$ is
$\mathbb{Q}_{p}[[x,y,z]]/(x^{3}+y^{3}+z^{3})$ where the containment does not
hold. ∎

######  Proposition 5.3.

Let $R=\mathbb{Z}_{p}[[x,y,z]]/(x^{3}+y^{3}+z^{3})$ and $I=(x,y)$, then
$z^{2}$ is in the dagger closure of $I$.

###### Proof.

The reader can observe that the same argument as in Theorem 3.1 works: the
absolute integral closure of polynomial ring
$\mathbb{Z}_{p}[x,y,z]/(x^{3}+y^{3}+z^{3})$ is inside the absolute integral
closure of $R$, hence we may assume
$R=\mathbb{Z}_{p}[x,y,z]/(x^{3}+y^{3}+z^{3})$. We may also enlarge $R$ and
assume it contains $\theta^{\frac{1}{3}}$ where $\theta$ is the primitive
third root of unity. The construction in Theorem 3.1 give a finite extension
of $R$ isomorphic to itself. In the new extension $((x,y):z^{2})$ contains
elements of lower valuation and by iterating this process the calculation
performed at the end of Theorem 3.1 shows that there are elements of arbitrary
low valuation in the colon ideal $((x,y):_{R^{+}}z^{2})$. ∎

######  Proposition 5.4.

Let $R=\mathbb{Z}_{p}[[x,y,z]]/(x^{3}+y^{3}+z^{3})$, then $z^{2}$ is in
$(p^{n},x,y)R^{+}$ for all $n\geq 1$.

###### Proof.

Let $S=\mathbb{Z}_{p}[x,y,z]/(x^{3}+y^{3}+z^{3})$. Note that $S$ is the
homogeneous coordinate ring of an elliptic curve $Proj(S)=E$ over
$\mathbb{Z}_{p}$. In general morphisms of projective varieties do not
correspond to morphisms of the corresponding graded rings, however for abelian
varieties, the $p^{n}$ power maps $[p^{n}]$ lift to module-finite graded
_ring_ endomorphisms $m_{p^{n}}$ of the homogenous coordinate ring $S$, see
discussion after [RSS07, (3.4.1)], paragraph starting with “Thus we have a map
of graded rings from $R$ to itself…”. The maps $[p^{n}]:E\rightarrow E$ induce
multiplication by $p^{n}$ maps on
$H^{1}(E,\mathcal{O}_{E})\simeq\mathbb{Z}_{p}$, this is a standard fact about
abelian schemes and we refer the reader to [BMP+20, Example 4.14] for a proof.
Since the degree zero part of $H^{2}_{(x,y,z)}(S)$ is isomorphic to
$H^{1}(E,\mathcal{O}_{E})$ and by construction of $m_{p^{n}}$ is defined by
the same Čech complex,
$m_{p^{n}}:(H^{2}_{(x,y,z)}(S))_{0}\rightarrow(H^{2}_{(x,y,z)}(S))_{0}$ is
also the multiplication by $p^{n}$ map.

Note that $\sqrt{(x,y)}=\sqrt{(m_{p^{n}}(x),m_{p^{n}}(y))}=(x,y,z)$ is the
irrelevant ideal. The class $[\frac{z^{2}}{xy}]$ spans
$(H^{2}_{(x,y,z)}(S))_{0}$ and let $[\frac{\alpha}{m_{p^{n}}(x)m_{p^{n}}(y)}]$
be another generator of $(H^{2}_{(x,y,z)}(S))_{0}$ obtained by computing the
Čech complex with respect to $(m_{p^{n}}(x),m_{p^{n}}(y))$. Hence we have
$[\frac{m_{p^{n}}(z^{2})}{m_{p^{n}}(x)m_{p^{n}}(y)}]=[p^{n}\frac{\alpha}{m_{p^{n}}(x)m_{p^{n}}(y)}]$
as classes in $(H^{2}_{(x,y,z)}(S))_{0}$. Unwinding the definition of
$H^{2}_{(x,y,z)}(S)$ using the Čech complex we have
$\frac{m_{p^{n}}(z^{2})}{m_{p^{n}}(x)m_{p^{n}}(y)}=p^{n}\frac{\alpha}{m_{p^{n}}(x)m_{p^{n}}(y)}+\frac{\beta_{1}}{m_{p^{n}}(y)}+\frac{\beta_{2}}{m_{p^{n}}(x)}$
for some $\beta_{1},\beta_{2}\in S$ and cross multiplying by
$m_{p^{n}}(x)m_{p^{n}}(y)$ we have
$m_{p^{n}}(z^{2})\in(p^{n},m_{p^{n}}(x),m_{p^{n}}(y))$ for all $n$. Since
$m_{p^{n}}:S\rightarrow S$ is a module finite morphism, we have a copy of $S$,
$T$ in $S^{+}$ such that the inclusion $S\rightarrow T$ is the same as
$m_{p^{n}}:S\rightarrow S$. Hence $z^{2}\in(p^{n},x,y)S^{+}$. Since $S$ is
contained in $R$ we have $S^{+}\subset R^{+}$ and hence
$z^{2}\in(p^{n},x,y)R^{+}$. ∎

The anonymous referee kindly pointed out to us that this proposition also
follows from the fact that since $p$ is a non-zero divisor on $R^{+}/(x,y)$ by
[Bha20] the derived $p$ adic completion of $R^{+}/(x,y)$ coincides with itself
and hence is in particular $p$-adically separated.

######  Proposition 5.5.

Let $(R,m)$ be an excellent local domain of mixed characteristic $(0,p)$ and
let $(p,x,y)$ denote a system of parameters. Then
$\cap_{n}(p^{n},x,y)R^{+}\subset(x,y)\widehat{R^{+}}$.

###### Proof.

We will be using the main theorems of [Bha20], that is by [Bha20, Corollary
5.11] we know that $(p^{n},x,y)$ is a regular sequence on $R^{+}$ for every
$n$. The reader is encouraged to look at [HM21, Lemma 3.2] where this
proposition is proved for a two element ideal $(p,g)$ and does not use
[Bha20], since Cohen-Macaulayness of $R^{+}$ in dimension $2$ follows from
elementary commutative algebra. Let $\alpha\in\cap_{n}(p^{n},x,y)R^{+}$. We
have $\alpha=a_{1}x+b_{1}y+c_{1}p$ for some $(a_{1},b_{1},c_{1})\in R^{+}$ and
we also have that $c_{1}p\in\cap_{n}(p^{n},x,y)$. So
$c_{1}p=a_{2}x+b_{2}y+c_{2}p^{2}$ for some $(a_{2},b_{2},c_{2})\in R^{+}$. We
will now use that $(p,x,y)$ is a regular sequence to show that we may choose
$(a_{2},b_{2})$ to be divisible by $p$. Since
$c_{1}p=a_{2}x+b_{2}y+c_{2}p^{2}$, we have that $(a_{2}x+b_{2}y)=0$ as an
element in $R^{+}/p$. Hence $b_{2}y=0$ as an element of $R^{+}/(p,x)$ and as
$(p,x,y)$ form a regular sequence on $R^{+}$ we have $b_{2}=gx+hp$ for some
$(g,h)\in R^{+}$ and hence $c_{1}p=(a_{2}+gy)x+hpy+c_{2}p^{2}$. Since $x$ is a
non-zero divisor on $R^{+}/p$ we have $(a_{2}+gy)=jp$ for some $j\in R^{+}$
and hence $\alpha=a_{1}x+b_{1}y+(a_{2}+gy)x+hpy+c_{2}p^{2}$. Thus we may
choose $a_{2}$ and $b_{2}$ to be divisible by $p$. We repeat this process to
get $(a_{i},b_{i})$: we have $c_{2}p^{2}\in\cap_{n}(p^{n},x,y)$ and hence we
may write $c_{2}p^{2}=a_{3}x+b_{3}y+c_{3}p^{3}$ and by the same argument with
$p$ replaced by $p^{2}$ we may assume $a_{3},b_{3}$ to be divisible by
$p^{2}$.
Continuing this way we have that $(a_{i},b_{i})\in p^{i-1}R^{+}$ and hence the
sums $(a_{1}+a_{2}+...),(b_{1}+b_{2}+...)$ converge and are well defined
elements of $\widehat{R^{+}}$. Finally, we have that
$\alpha=a_{1}x+b_{1}y+a_{2}x+b_{2}y+...=x(a_{1}+a_{2}+...)+y(b_{1}+b_{2}+...)\in(x,y)\widehat{R^{+}}$.
∎

######  Corollary 5.6.

Let $R=\mathbb{Z}_{p}[[x,y,z]]/(x^{3}+y^{3}+z^{3})$ and $I=(x,y)$. Then
$z^{2}$ is not in $I^{+}=IR^{+}\cap R$ but is in the dagger closure of $I$,
$I^{\dagger}$, and the “$\widehat{R^{+}}$-closure of $I$”,
$I\widehat{R^{+}}\cap R$.

###### Proof.

Follows from the propositions above. ∎

######  Remark 5.7.

Note that valuations on $R$ (positive on $I$) extend to those on
$\widehat{R^{+}}$ by [Hei21, Lemma 1.3]. Given this, one can quickly conclude
that a negative answer to Question 5.1 for any finitely generated ideal $I$ of
$R^{+}$ (not just one generated by system of parameters) implies that the
answer to Question 4.1 is negative, that is $\widehat{R^{+}}$ is not coherent
(as any element of $I\widehat{R^{+}}$ will have valuation at least the minimum
of valuations of generators of $I$). This is analogous to the fact that if
absolute integral closures were coherent in positive characteristic then tight
closure would be equal to plus closure and hence would commute with
localization. Brenner and Monsky’s paper [BM10] which settled the long
standing question of whether tight closure localizes (in the negative)
immediately implies that $R^{+}$ is not coherent in positive characteristic
for their choice of $R$ (see [Shi10, Proposition 4.5] and [AH97, Proposition
2.3]).

##  6\. Acknowledgements

The author thanks Mohsen Asgharzadeh, Bhargav Bhatt, Rankeya Datta, Arnab
Kundu, Vaibhav Pandey, Kazuma Shimomoto, and his advisor Kevin Tucker for
conversations, mentorship, encouragement, useful suggestions, or a reading of
the manuscript, and especially Kazuma Shimomoto for the permission to include
Question 4.6 and his proof of Theorem 3.6. The anonymous referee suggested
improvements in exposition and pointed out mistakes and serious gaps in a
former version of this article (especially in what is now Section $5$) which
we are grateful for.

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