# homology of conductor

Mohsen Asgharzadeh<EMAIL_ADDRESS>
###### Abstract.

We study some homological properties of the conductor ideal.

###### Key words and phrases:

annihilator; conductor ideal; integral closure; homological methods; parameter
sequence

###### 2010 Mathematics Subject Classification:

Primary 13B22; 13D45.

###### Contents

  1. 1 Introduction
  2. 2 $\mathfrak{C}$ as a test module
  3. 3 $\mathfrak{C}$ and cohomological annihilator
  4. 4 Ikeda’s problem

##  1\. Introduction

The main theme of this paper is about of the ideal of definition of the normal
locus:

$\operatorname{Var}(\mathfrak{C}):=\\{\mathfrak{p}\in\operatorname{Spec}(R):R_{\mathfrak{p}}\emph{
is not normal}\\},$

if it is exists. Under some mild conditions, for example the analytically
unramified condition, it is closed in Zariski topology. Namely, and up to
radical, we are interested in
$\mathfrak{C}:=\mathfrak{C}_{R}:=\operatorname{Ann}_{R}(\frac{\overline{R}}{R})$
and we refer it the conductor ideal. There are some forgotten conjectures in
this area:

######  Conjecture 1.1.

(See [13, Page 51]). Suppose $R$ has no nilpotent. Then
$\frac{\overline{R}}{R}$ is $h$-reduced.

We record [1] to see some partial answers. The other one was asked by Ikeda
[11]:

######  Problem 1.2.

Let $\underline{x}$ be a system of parameter. When is
$\mathfrak{C}\nsubseteq\underline{x}R$?

Also, it stated as an open problem in the book of Huneke-Swanson when the ring
is Cohen-Macaulay, one may deal with the middle (see Conjecture 4.1). We
checked Problem 1.2 in a lot of isolated examples, some of them are presented
in Section 4. The best answer to this is due to Ikeda over any complete
Gorenstein integral domain. We present a new proof of this:

Theorem A. Adopt one of the following assumptions:

  1. i)

Suppose $R$ is quasi-Gorenstein.

  2. ii)

Suppose $0pt(R)=1$.

  3. iii)

Suppose $R$ is a two-dimensional Buchsbaum ring.

  4. iv)

Suppose $\mathfrak{m}=\mathfrak{C}+yR$ for some $y$.

Then $\mathfrak{C}\nsubseteq\underline{x}R$ where $\underline{x}$ is a system
of parameter.

We mention that there are several non-trivial examples for which
$\mathfrak{m}=\mathfrak{C}+yR$. Next, we deal with:

######  Conjecture 1.3.

$\mathfrak{C}_{R}$ can be used as a test object for any homological properties
of rings.

Theorem B. The following assertions are valid.

  1. i)

If $\operatorname{pd}_{R}(\mathfrak{C})<\infty$, then $R$ is normal.

  2. ii)

If $\operatorname{id}_{R}(\mathfrak{C})<\infty$, then $R$ is normal. In
particular, $R$ is Gorenstein.

  3. iii)

Suppose $R$ is $(S_{2})$. If
$\operatorname{Ext}_{R}^{i}(\mathfrak{C},\mathfrak{C})=0$ for some $i>0$, then
$\mathfrak{C}$ is free. In particular, $R$ is normal.

  4. iv)

If $\mathfrak{C}^{i\otimes}$ is torsion-free for some $i>1$ and $R$ is
$(S_{2})$, then $\mathfrak{C}$ is free. In particular, $R$ is normal.

  5. v)

If $R$ is one-dimensional, and $\operatorname{Ext}_{R}^{i}(\mathfrak{C},R)=0$
for some $i>0$ then $R$ is Gorenstein.

Among other things, part i) uses the Roberts’ intersection theorem. It is not
so adjective any use of Bass’ conjecture in the second item. Also, part iii)
is related to a conjecture of Auslander-Reiten. Recall that part iv) is a
special case of a conjecture by Vasconcelos, which is widely open. We think
the use of $(S_{2})$-condition is superficial. In this regard, we replace it
with the seminormailty or even with quasi-Buchsbaum condition. Also, we
present the Gorenstein analogue of i), see Observation 2.8. Part v) extends a
result of Ulrich [22] and Rego [20, main result], as they worked with
$\operatorname{Ext}_{R}^{1}(\mathfrak{C},R)=0$. It may be nice to note that
Rego used the compactified Jacobian of a singular curve. In addition, we
present a higher dimensional analogue. Reflexivity of $\overline{R}$ seems
more subtle, as we connect to initial Bass’ numbers. In this regard, let $R$
be a 2-dimensional Buchsbaum ring which is $(R_{1})$. In Fact 2.9 we show
$\overline{R}$ is reflexive iff $\mu^{1}(\mathfrak{m},R)\leq 1$.

Commuting annihilator of modules is a difficult task, and this become more
subtle if the module under consideration is involved in the integral closure,
as it is difficult to compute $\overline{R}$ too. In this regard, there are
several (co-)homological approaches to approximates the annihilator. Namely,
Vaconcelos’ homological annihilator denoted by $\operatorname{h-Ann}(-)$, and
Bridger-Roberts’ cohomological annihilator denoted by
$\operatorname{Ann}^{h}(-)$. For the precise definitions, see Section 3. We
note these (co-)homological approaches are not necessarily equal to
$\operatorname{Ann}_{R}(-)$. In Section 3 we show:

Theorem C. The following assertions are valid.

  1. i)

Let $R$ be a 2-dimensional Gorenstein ring. Then
$\operatorname{h-Ann}(\frac{\overline{R}}{R})=\operatorname{Ann}_{R}(\frac{\overline{R}}{R})$.

  2. ii)

Let $R$ be Cohen-Macaulay. Then
$\mathfrak{C}_{R}\subseteq\operatorname{Ann}^{h}(\overline{R})$. If $R$ is not
normal, then
$\operatorname{rad}\mathfrak{C}_{R}\neq\operatorname{rad}\operatorname{Ann}^{h}(\overline{R})$.

  3. iii)

Let $R$ be as ii). Then
$\mathfrak{C}_{R}\subseteq\operatorname{Ann}\operatorname{Tor}^{R}_{+}(\overline{R},R/\underline{x}R)=0$.

  4. iv)

Let $R$ be a 2-dimensional analytically unramified ring satisfying $(R_{1})$.
Then $\mathfrak{C}_{R}=\operatorname{Ann}(H^{1}_{\mathfrak{m}}(R))$.

  5. v)

If $R$ is Gorenstein and $\overline{R}$ is Cohen-Macaulay, then
$\mathfrak{C}_{R}$ is maximal Cohen-Macaulay.

Concerning iii), it is true in slightly more general setting than be Cohen-
Macaulay. Also, it is trivial if $i>1$ or even $\overline{R}$ is Cohen-
Macaulay, and maybe there is a related result from literature. Namely, over
complete 1-dimensional integral domains, and for any maximal Cohen-Macaulay
module $M$, Wang proved that
$\mathfrak{C}_{R}\operatorname{Tor}^{R}_{1}(M,-)=0$, see [27, 3.1+1.5].
Concerning v), and when the ring is standard graded over a field, it is proved
in [9] independently by Eisenbud and Ulrich.

We follow the book of Huneke-Sawnson [21] for all unexplained notations. Also,
see the books [6, 16] as classical references.

##  2\. $\mathfrak{C}$ as a test module

Let $(R,\mathfrak{m})$ be a commutative noetherian local ring, and let
$\overline{R}$ be the normalization of $R$ in the total ring of fraction.
Also, $\underline{x}$ is a system of parameter for $R$.

######  Hypothesis 2.1.

We assume in addition that the ring is analytically unramified.

Recall that Serre’s $(S_{n})$ condition indicates
$0pt(R_{\mathfrak{p}})\geq\min\\{n,\operatorname{ht}(\mathfrak{p})\\}$.
Serre’s $(R_{n})$ condition means $R_{\mathfrak{p}}$ is regular for all prime
ideal $\mathfrak{p}$ of height at most $n$.

######  Fact 2.2.

(Serre). $R$ is normal iff it satisfies $(S_{2})$ and $(R_{1})$.

By $\operatorname{pd}_{R}(-)$ we mean the projective dimension of $(-)$.

######  Lemma 2.3.

Suppose $R$ is $(S_{2})$. If $\operatorname{pd}_{R}(\mathfrak{C})<\infty$,
then $R$ is normal.

###### Proof.

Let $\mathfrak{p}$ be in $\operatorname{Spec}(R)$ of height one. Recall that
$\mathfrak{C}=\operatorname{Hom}_{R}(\overline{R},R)$. Since $R$ is
analytically unramified, $\overline{R}$ is finitely generated as an
$R$-module. In particular,

$\operatorname{Hom}_{R}(\overline{R},-)_{\mathfrak{p}}=\operatorname{Hom}_{R_{\mathfrak{p}}}(\overline{R}_{\mathfrak{p}},-_{\mathfrak{p}})=\operatorname{Hom}_{R_{\mathfrak{p}}}(\overline{R_{\mathfrak{p}}},-_{\mathfrak{p}}).$

From this,
$(\mathfrak{C}_{R})_{\mathfrak{p}}=\mathfrak{C}_{R_{\mathfrak{p}}}$. So,
projective dimension of $(\mathfrak{C}_{R})_{\mathfrak{p}}$ is finite. Recall
that

$\operatorname{pd}_{R_{\mathfrak{p}}}(\frac{R_{\mathfrak{p}}}{(\mathfrak{C}_{R})_{\mathfrak{p}}})\leq
0pt(R_{\mathfrak{p}})\leq\dim(R_{\mathfrak{p}})=1,$

and consequently,
$(\mathfrak{C}_{R})_{\mathfrak{p}}=\mathfrak{C}_{R_{\mathfrak{p}}}$ is free.
In particular, it should be a principal ideal. Recall from [21, Ex. 12.4] that
$\mathfrak{C}_{R_{\mathfrak{p}}}$ is not subset of any proper principal ideal.
In sum, $\mathfrak{C}_{R_{\mathfrak{p}}}={R_{\mathfrak{p}}}$. This implies
${R_{\mathfrak{p}}}$ is normal. By dimension consideration,
${R_{\mathfrak{p}}}$ is regular. Following Fact 2.2, $R$ is normal. ∎

Let us drop the $(S_{2})$ condition.

######  Observation 2.4.

If $\operatorname{pd}_{R}(\mathfrak{C})<\infty$, then $R$ is normal.

###### Proof.

We processed by induction on $d:=\dim R$. Recall that $R$ is analytically
unramified. By definition, its completion with respect to $\mathfrak{m}$-adic
topology is reduced, see [21]. Zero-dimensional reduced rings are regular.
This proves the claim for $d=0$. Now, suppose $d>0$ and assume inductively
that $R$ is locally normal over punctured spectrum. Recall that both
assumptions $\operatorname{pd}_{R}(\mathfrak{C})<\infty$ and analytically
unramified behaved well with respect to localization. It turns out that
$\mathfrak{C}_{R_{\mathfrak{p}}}=R_{\mathfrak{p}}$ for all
$\mathfrak{p}\in\operatorname{Spec}(R)\setminus\\{\mathfrak{m}\\}$, i.e.,
$\frac{{R}}{\mathfrak{C}}$ is of finite length. Suppose on the way of
contradiction that $\frac{{R}}{\mathfrak{C}}\neq 0$. Thanks to Lemma 2.3 $R$
is not $(S_{2})$. Since $\frac{{R}}{\mathfrak{C}}$ is nonzero, of finite
length and of finite projective-dimension, and in the light of Roberts’
intersection theorem [19, 6.2.4], we see $R$ is Cohen-Macaulay. In particular,
$R$ is $(S_{2})$. This contradiction shows $\frac{{R}}{\mathfrak{C}}=0$, so
$R$ is normal. ∎

By $\operatorname{id}_{R}(-)$ we mean the injective dimension of $(-)$.

######  Observation 2.5.

If $\operatorname{id}_{R}(\mathfrak{C})<\infty$, then $R$ is normal and in
particular, $R$ is Gorenstein.

###### Proof.

By Bass’ conjecture, which is a theorem, $R$ is Cohen-Macaulay (see [17] and
[19]). In particular, $R$ is $(S_{2})$. Let
$\mathfrak{p}\in\operatorname{Spec}(R)$ be of height one. Recall that
$(\mathfrak{C}_{R})_{\mathfrak{p}}=\mathfrak{C}_{R_{\mathfrak{p}}}$ and that
$\mathfrak{C}_{R_{\mathfrak{p}}}$ is integrally closed, as
$\dim({R_{\mathfrak{p}}})=1$ and $\mathfrak{C}_{R_{\mathfrak{p}}}$ is a common
ideal of ${R_{\mathfrak{p}}}$ and $\overline{R_{\mathfrak{p}}}$ (see [21, Ex.
12.2]). Due to Serre’s criteria we may assume
${R_{\mathfrak{p}}}\neq\overline{R_{\mathfrak{p}}}$. This yields that
$\mathfrak{C}_{R_{\mathfrak{p}}}$ is primary to the maximal ideal, and
integrally closed. This in turn implies that $\mathfrak{C}_{R_{\mathfrak{p}}}$
is full. Also
$\operatorname{id}_{R_{\mathfrak{p}}}(\mathfrak{C}_{\mathfrak{p}})<\infty$. In
view of [14, Theorem 2.2] we observe that ${R_{\mathfrak{p}}}$ is regular.
Again, by Serre’s criteria, $R$ is normal. Then $\mathfrak{C}=R$ has finite
injective dimension. So, $R$ is Gorenstein. ∎

Let $I$ be an ideal generated by $f_{1},\ldots,f_{r}$ of $R$. For an
$R$-module $M$, we have the Čech complex

(2.1) ${C}^{\bullet}(I;M):0\longrightarrow M\longrightarrow\prod_{1\leq j\leq
r}M_{f_{j}}\longrightarrow\prod_{1\leq j_{1}<j_{2}\leq
r}M_{f_{j_{1}}f_{j_{2}}}\longrightarrow\cdots\longrightarrow M_{f_{1}\cdots
f_{r}}\longrightarrow 0.$

Then its $i$-th cohomology is denoted by $H^{i}_{I}(M)$.

######  Observation 2.6.

Suppose $\operatorname{Ext}_{R}^{i}(\mathfrak{C},\mathfrak{C})=0$ for some
$i>0$ where $R$ is one of the following:

  1. i)

$R$ is $(S_{2})$

  2. ii)

$R$ is quasi-Buchsbaum***here, and elsewhere we only need
$\mathfrak{m}H^{1}_{\mathfrak{m}}(R)=0$., or

  3. iii)

$R$ is seminormal.

Then $\mathfrak{C}$ is free and in particular $R$ is normal.

###### Proof.

i) Suppose first that $R$ is $(S_{2})$ Let $\mathfrak{p}$ be in
$\operatorname{Spec}(R)$ of height one. Let $A:={R_{\mathfrak{p}}}$. Recall
that $(\mathfrak{C}_{R})_{\mathfrak{p}}=\mathfrak{C}_{A}$. We look at the
following exact sequence taken from [4]

$\operatorname{Tor}_{2}^{A}(\operatorname{Tr}(\operatorname{Syz}_{i}(\mathfrak{C}_{A})),\mathfrak{C}_{A})\rightarrow\operatorname{Ext}^{i}_{A}(\mathfrak{C}_{A},A)\otimes_{A}\mathfrak{C}_{A}\rightarrow\operatorname{Ext}^{i}_{A}(\mathfrak{C}_{A},\mathfrak{C}_{A})\rightarrow\operatorname{Tor}_{1}^{A}(\mathfrak{C}_{A},\operatorname{Tr}(\operatorname{Syz}_{i}(\mathfrak{C}_{A})))\rightarrow
0,$

where $\operatorname{Syz}_{i}(-)$ is the $i^{th}$-syzygy of a module $(-)$ and
$\operatorname{Tr}(\operatorname{Syz}_{i}(\mathfrak{C}_{A}))$ stands for the
Auslander’s transpose. In order to define it, we first set
$(-)^{\ast}:=\operatorname{Hom}_{A}(-,A)$ and take $A^{n}\to
A^{m}\to\operatorname{Syz}_{i}(\mathfrak{C}_{A})\to 0$ be a finite type
presentation. Now, by taking dual, $\operatorname{Tr}$ is defined by the
following cokernel:

$0\longrightarrow\operatorname{Syz}_{i}(\mathfrak{C}_{A})^{\ast}\longrightarrow(A^{\ast})^{m}\longrightarrow(A^{\ast})^{n}\longrightarrow\operatorname{Tr}(\operatorname{Syz}_{i}(\mathfrak{C}_{A}))\longrightarrow
0.$

From this
$\operatorname{Tor}_{1}^{A}(\mathfrak{C}_{A},\operatorname{Tr}(\operatorname{Syz}_{i}(\mathfrak{C}_{A})))=0$.
By shifting

$\operatorname{Tor}_{2}^{A}(A/\mathfrak{C}_{A},\operatorname{Tr}(\operatorname{Syz}_{i}(\mathfrak{C}_{A})))=0\quad(\ast)$

Without loss of generality $\mathfrak{C}_{A}$ is integrally closed and primary
to maximal ideal of $A$. This along with $(\ast)$ implies that
$\operatorname{pd}(\operatorname{Tr}(\operatorname{Syz}_{i}(\mathfrak{C}_{A})))<2$
(see [7, 3.3]). Since $(\operatorname{Syz}_{i}(\mathfrak{C}_{A}))^{\ast}$ is
the second syzygy of
$\operatorname{Tr}(\operatorname{Syz}_{i}(\mathfrak{C}_{A}))$ we deduce that
$\operatorname{Syz}_{i}(\mathfrak{C}_{A})^{\ast}$ is free. Thanks to [8, Lemma
3.9] we know that $\operatorname{Syz}_{i}(\mathfrak{C}_{A})$ is free. Then
$\operatorname{pd}(\mathfrak{C}_{A})<\infty$. In view of Observation 2.3 we
deduce that $A$ is normal and via the dimension consideration, $A$ becomes
regular. We proved that $R$ satisfies $(R_{1})$. According to Fact 2.2, $R$ is
normal. In particular, $\mathfrak{C}=R$ is free.

ii) Suppose now that $R$ is quasi-Buchsbaum. We processed by induction on
$d:=\dim R$. Recall that $R$ is analytically unramified. This proves the claim
for $d=0$. Now, suppose $d>0$ and assume inductively that $R$ is locally
normal over punctured spectrum. It turns out that $\frac{{R}}{\mathfrak{C}}$
(resp. $\frac{\overline{R}}{R}$) is of finite length. Suppose on the way of
contradiction that $\frac{{R}}{\mathfrak{C}}\neq 0$. Thanks to the first case,
and without loss of generality, we assume in addition that $R$ is not
$(S_{2})$. In fact, following part i) we may and do assume in addition that
$d>1$, as 1-dimensional reduced rings are $(S_{2})$. This allows us to deal
with the annihilator of $H^{1}_{\mathfrak{m}}(R)$ as follows. Namely, from
$0\to R\to\overline{R}\to\frac{\overline{R}}{R}\to 0$ we deduce
$0=H^{0}_{\mathfrak{m}}(\overline{R})\to
H^{0}_{\mathfrak{m}}(\frac{\overline{R}}{R})\to H^{1}_{\mathfrak{m}}(R)\to
H^{1}_{\mathfrak{m}}(\overline{R})=0.$ From this
$H^{0}_{\mathfrak{m}}(\frac{\overline{R}}{R})=\frac{\overline{R}}{R}$.
Consequently,
$H^{1}_{\mathfrak{m}}(R)=H^{0}_{\mathfrak{m}}(\frac{\overline{R}}{R})=\frac{\overline{R}}{R}$,
so
$\mathfrak{C}_{R}=\operatorname{Ann}(H^{1}_{\mathfrak{m}}(R))=\operatorname{Ann}(\oplus_{X}k)$
where $X$ is an index set. Since $\mathfrak{C}_{R}\neq R$, $X$ is non-empty.
It follows that $\mathfrak{C}=\mathfrak{m}$. Since
$\operatorname{Ext}^{i}_{R}(\mathfrak{m},\mathfrak{m})=0$ we deduce
$\operatorname{Ext}^{i+1}_{R}(R/\mathfrak{m},\mathfrak{m})=0$. In other words,
$\mu^{i+1}(\mathfrak{m})=0$. We know from [18, Theorem II] that
$i+1\not\in[0pt{\mathfrak{m}}=1,\operatorname{id}_{R}(\mathfrak{m})]$. Thus
$\operatorname{id}_{R}(\mathfrak{m})<\infty$, i.e., $R$ is regular a
contradiction. Thus, $\frac{{R}}{\mathfrak{C}}=0$, and the claim follows.

iii) We processed by induction on $d:=\dim R$. Recall that $R$ is analytically
unramified. Since zero-dimensional reduced rings are regular, we proved the
claim for $d=0$. Now, suppose $d>0$ and assume inductively that $R$ is locally
normal over punctured spectrum. Recall that all three assumptions,
seminormality, $\operatorname{Ext}_{R}^{i}(\mathfrak{C},\mathfrak{C})=0$ and
analytically unramified behaved well with respect to localization. It turns
out that $\mathfrak{C}_{R_{\mathfrak{p}}}=R_{\mathfrak{p}}$ for all
$\mathfrak{p}\in\operatorname{Spec}(R)\setminus\\{\mathfrak{m}\\}$, i.e.,
$\frac{{R}}{\mathfrak{C}}$ is of finite length. Suppose on the way of
contradiction that $\frac{{R}}{\mathfrak{C}}\neq 0$. Then $\mathfrak{C}$ is
primary to a maximal ideal of normalization. In particular, $R$ is not
regular. By seminormality, we know $\mathfrak{C}$ is a radical ideal in
$\overline{R}$. Thus, $\mathfrak{C}=\mathfrak{m}_{\overline{R}}$. Then
$\mathfrak{C}=\mathfrak{C}\cap R=\mathfrak{m}_{\overline{R}}\cap
R=\mathfrak{m}$, see [16, page 62]. Since
$\operatorname{Ext}^{i}_{R}(\mathfrak{m},\mathfrak{m})=0$ we deduce a
contradiction. Thus, $\frac{{R}}{\mathfrak{C}}=0$, and the claim follows. ∎

######  Definition 2.7.

  1. i)

By $(G_{n})$ we mean that $R_{\mathfrak{p}}$ is Gorenstein for any
$\mathfrak{p}$ of height at most $n$.

  2. ii)

Recall from [25] that a ring $R$ is called quasi-normal if it is $(S_{2})$ and
$(G_{1})$.

By $\operatorname{Gdim}_{R}(-)$ we mean the G-dimension of $(-)$.

######  Observation 2.8.

Suppose $R$ is $(S_{2})$ and $\operatorname{Gdim}_{R}(\mathfrak{C})<\infty$.
Then $R$ is quasi-normal.

###### Proof.

This is similar to Observation 2.5, and we leave the straightforward
modification to the reader. ∎

We say a module $M$ is reflexive if the natural map $\delta_{M}:M\to
M^{\ast\ast}$ is an isomorphism. One may ask the reflexivity problem of the
$R$-module $\overline{R}$.

######  Fact 2.9.

i) Suppose $\dim R=1$ then $\overline{R}$ is reflexive.

ii) Suppose $R$ is $(S_{2})$. Then $\overline{R}$ is reflexive.

iii) Let $R$ be 2-dimensional Buchsbaum ring which is $(R_{1})$. If
$\overline{R}$ is reflexive, then $\mu^{1}(\mathfrak{m},R)\leq 1$.

iv) Let $R$ be 2-dimensional Buchsbaum ring which is $(R_{1})$ and
$\mu^{1}(\mathfrak{m},R)\leq 1$. Then $\overline{R}$ is reflexive.

###### Proof.

i) Following [21, Ex. 12.13] we know $\mathfrak{C}^{\ast}=\overline{R}$. But,
$\mathfrak{C}=\operatorname{Hom}_{R}(\overline{R},R)$. Combining these
$\overline{R}^{\ast\ast}=\mathfrak{C}^{\ast}=\overline{R},$ it follows the
natural map $\overline{R}\to\overline{R}^{\ast\ast}$ is an isomorphism, and so
$\overline{R}$ is reflexive.***Alternatively, use the fact that any dual
module over Cohen-Macaulay rings is reflexive.

ii) Let $\mathfrak{p}\in\operatorname{Spec}(R)$ be so that
$0ptR_{\mathfrak{p}}\geq 2$. Recall that the map
$R_{\mathfrak{p}}\longrightarrow\overline{R}_{\mathfrak{p}}$ is finite, and in
view of [6, 1.2.26(b)]
$0pt_{{R}_{\mathfrak{p}}}(\overline{R}_{\mathfrak{p}})=0pt_{\overline{R}_{\mathfrak{p}}}(\overline{R}_{\mathfrak{p}})\geq
2$. Recall from i) that the claim is clear in dimension one, so:

  1. (a)

$\overline{R}_{\mathfrak{p}}$ is reflexive for all $\mathfrak{p}$ with
$0ptR_{\mathfrak{p}}\leq 1$, and

  2. (b)

$0pt(\overline{R})_{\mathfrak{p}}\geq 2$ for all $\mathfrak{p}$ with
$0ptR_{\mathfrak{p}}\geq 2$.

From this, and in the light of [6, Proposition 1.4.1(b)], $\overline{R}$ is
reflexive.

iii) Suppose first that $R$ is Cohen-Macaulay. Then $0pt(R)=2$ and
$\mu^{1}(\mathfrak{m},R)=0<1$. Now, suppose $R$ is 2-dimensional Buchsbaum
ring which is $(R_{1})$ and not Cohen-Macaulay. Recall that
$H^{1}_{\mathfrak{m}}(R)=H^{0}_{\mathfrak{m}}(\frac{\overline{R}}{R})=\frac{\overline{R}}{R}$,
so
$\mathfrak{C}_{R}=\operatorname{Ann}(H^{1}_{\mathfrak{m}}(R))=\operatorname{Ann}(\oplus_{n}k)$
where $n$ is an integer. Since $\mathfrak{C}_{R}\neq R$, $n>0$ and
consequently, $\mathfrak{C}_{R}=\mathfrak{m}$. In particular, there is an
exact sequence

$0\longrightarrow R\longrightarrow\overline{R}\longrightarrow
k^{n}\longrightarrow 0\quad(\ast)$

Recall that $0pt(R)=1$. This shows that $(\frac{{R}}{\mathfrak{m}})^{\ast}=0$
and $\operatorname{Ext}_{R}^{1}(\frac{{R}}{\mathfrak{m}},R)=k^{t}$ where
$t:=\mu^{1}(\mathfrak{m},R)\neq 0$. By dual $(\ast)$, we have

$\begin{CD}0@>{}>{}>\overline{R}^{\ast}@>{}>{}>{R}^{\ast}@>{}>{}>\operatorname{Ext}_{R}^{1}(k^{n},R)=k^{nt}@>{}>{}>\operatorname{Ext}_{R}^{1}(R,R)=0\\\
\end{CD}$

Another duality yields that

$0\longrightarrow{R}^{\ast\ast}\longrightarrow\overline{R}^{\ast\ast}\longrightarrow\operatorname{Ext}_{R}^{1}(k^{nt},R)=k^{nt^{2}}\longrightarrow\operatorname{Ext}_{R}^{1}(R^{\ast},R)=0.$

Recall by work of Eilenberg-Maclane that the biduality is a functor. This
means by applying it to $M\to N$ yields that

$\begin{CD}M^{\ast\ast}@>{}>{}>N^{\ast\ast}\\\
\delta_{M}@A{}A{}A\delta_{N}@A{}A{}A\\\ <EMAIL_ADDRESS>\end{CD}$

There is a map $g$ completes the following depicted diagram:

$\begin{CD}0@>{}>{}>R^{\ast\ast}@>{}>{}>\overline{R}^{\ast\ast}@>{}>{}>k^{nt^{2}}@>{}>{}>0\\\
\delta_{{R}}@A{}A{}A\delta_{\overline{R}}@A{}A{}A\exists g@A{}A{}A\\\
0@>{}>{}>R@>{}>{}>\overline{R}<EMAIL_ADDRESS>\end{CD}$

If $\delta_{\overline{R}}:\overline{R}\to\overline{R}^{\ast\ast}$ is an
isomorphism, it shows $g$ is an isomorphism, and so $t=1$.

iv) As the claim is clear when $R$ is Cohen-Macaulay, we may assume
$\mu^{1}(\mathfrak{m},R)=1$. Recall that there is the following diagram:

$\setcounter{MaxMatrixCols}{11}\begin{CD}R^{\ast\ast}@>{}>{}>\overline{R}^{\ast\ast}@>{}>{}>k^{n}@>{}>{}>0@>{}>{}>0\\\
\delta_{{R}}@A{}A{}A\delta_{\overline{R}}@A{}A{}Ag@A{}A{}Ah_{4}@A{}A{}Ah_{5}@A{}A{}A\\\
R@>{}>{}>\overline{R}@>{}>{}>k^{n}<EMAIL_ADDRESS>\end{CD}$

Recall that $h_{2}:=\delta_{\overline{R}}$ and $h_{4}$ are injective and
$h_{1}:=\delta_{{R}}$ is surjective, then by a version of 5-lemma $h_{3}:=g$
is injective. But, any injective endomorphism of a finite dimensional vector
space is indeed an isomorphism. Thus, $g$ is an isomorphism. In view of

$\begin{CD}0@>{}>{}>R^{\ast\ast}@>{}>{}>\overline{R}^{\ast\ast}@>{}>{}>k^{n}@>{}>{}>0\\\
\cong @A{}A{}A\delta_{\overline{R}}@A{}A{}Ag@A{}A{}A\\\
0@>{}>{}>R@>{}>{}>\overline{R}@>{}>{}>k^{n}@>{}>{}>0\\\ \end{CD}$

we deduce $\delta_{\overline{R}}$ is an isomorphism. So, the claim follows. ∎

######  Corollary 2.10.

The ring $R$ is quasi-normal, provided all of the following three assertions
hold:

  1. i)

$\dim R=2$

  2. ii)

$\operatorname{Gdim}_{R}(\mathfrak{C})<\infty$, and

  3. iii)

$\overline{R}$ is reflexive.

###### Proof.

According to the argument of Observation 2.8, $R$ satisfies $(G_{1})$. Suppose
on the way of contradiction that the ring is not $(S_{2})$. So, $0pt(R)=1$.
Thanks to Auslader-Bridger formula, $\mathfrak{C}$ is totally reflexive. It
turns out that its dual is also totally reflexive. But,
$\mathfrak{C}^{\ast}=\overline{R}^{\ast\ast}\stackrel{{\scriptstyle
iii)}}{{=}}\overline{R}$. Again, duo to Auslader-Bridger formula we have

$1=0pt_{R}(R)=\operatorname{Gdim}_{R}(\overline{R})+0pt_{R}(\overline{R})=\operatorname{Gdim}_{R}(\overline{R})+0pt_{\overline{R}}(\overline{R})=2,$

a contradiction. Thus the ring satisfies $(S_{2})$. By Observation 2.8, $R$ is
quasi-normal. ∎

Here, we recall that $M^{(n+1)\otimes}:=M^{n\otimes}\otimes_{R}M$. Vasconcelos
[23] asked:

######  Question 2.11.

Let $R$ be a local domain and $M$ be torsion-free. Is there an integer $e$
guaranteeing that if $M$ is not free, then the tensor power $M^{e\otimes}$ has
nontrivial torsion?

See [2] for some partial answers. Here, we prove it for conductor ideal:

######  Observation 2.12.

Suppose $\mathfrak{C}^{i\otimes}$ is torsion-free for some $i>1$, where $R$ is
one of the following:

  1. i)

$R$ is $(S_{2})$

  2. ii)

$R$ is quasi-Buchsbaum, or

  3. iii)

$R$ is seminormal.

Then $\mathfrak{C}$ is free and in particular $R$ is normal.

###### Proof.

i) We proceed by induction on $i$. The case $i=2$ follows from the following
argument, and avoid its repetition. So, we may assume $i>2$ and inductively we
assume that the desired claim holds for $i-1$. Let $\mathfrak{p}$ be in
$\operatorname{Spec}(R)$ of height one. Let $A:={R_{\mathfrak{p}}}$. Recall
that $(\mathfrak{C}_{R})_{\mathfrak{p}}=\mathfrak{C}_{A}$. We apply
$\mathfrak{C}^{(i-1)\otimes}\otimes_{R}-$ to $0\to\mathfrak{C}\to R\to
R/\mathfrak{C}\to 0$ and deduce the exact sequence

$0=\operatorname{Tor}_{1}^{R}(R,\mathfrak{C}^{(i-1)\otimes})\to\operatorname{Tor}_{1}^{R}(\frac{R}{\mathfrak{C}},\mathfrak{C}^{(i-1)\otimes})\to\mathfrak{C}^{i\otimes}.$

Since
$\operatorname{Tor}_{1}^{R}(\frac{R}{\mathfrak{C}},\mathfrak{C}^{(i-1)\otimes})$
is both torsion and torsion-free it should be zero. Then
$\operatorname{Tor}_{1}^{A}(\frac{A}{\mathfrak{C}_{A}},\mathfrak{C}_{A}^{(i-1)\otimes})=0$.
First assume $\mathfrak{C}\nsubseteq\mathfrak{p}$. This leads to the equality
$\mathfrak{C}_{\mathfrak{p}}=A$ which is free as an $A$-module. So, we may
assume that $\mathfrak{C}\subseteq\mathfrak{p}$. By dimension consideration,
$\mathfrak{C}_{R_{\mathfrak{p}}}$ primary to the maximal ideal, and is
integrally closed, as $\dim({A})=1$ and $\mathfrak{C}_{A}$ is a common ideal
of ${A}$ and $\overline{A}$ (see [21, Ex. 12.2]). In view of [7, 3.3], we know
$\mathfrak{C}_{A}^{(i-1)\otimes}$ is free. Thanks to an inductive argument,
$\mathfrak{C}_{A}$ is free. In sum, in both cases, $\mathfrak{C}_{A}$ is free.
This implies $A$ is normal. By dimension consideration, $A$ is regular. In
other words, $R$ satisfies $(R_{1})$. Again, by Serre’s criteria, $R$ is
normal. This yields that $\mathfrak{C}=R$, which is free.

The second and third cases are similar to the argument presented in
Observation 2.6, and we leave the straightforward modifications to the reader.
∎

The following completes the proof of Theorem B) from introduction.

######  Observation 2.13.

If $R$ is one-dimensional. The following are equivalent:

  1. i)

$\operatorname{Ext}_{R}^{i}(\mathfrak{C},R)=0$ for some $i>0$,

  2. ii)

$R$ is Gorenstein,

  3. iii)

$\operatorname{Ext}_{R}^{i}(\overline{R},R)=0$ for all $i>0$,

  4. iv)

$\operatorname{Ext}_{R}^{i}(\overline{R},R)=0$ for some $i>0$.

###### Proof.

$i)\Rightarrow ii)$: The case $i=1$ is in [20, main theorem]. Without loss of
generality assume that $i>1$. After completion if needed, we may assume $R$
possess a canonical module $\omega_{R}$. Set
$(-)^{+}:=\operatorname{Hom}_{R}(-,\omega_{R})$. Recall that $R$ is Cohen-
Macaulay, so $\operatorname{Hom}(\omega_{R},\omega_{R})=R$. We denote
$(-)^{\vee}:=\operatorname{Hom}_{R}(-,E_{R}(\frac{R}{\mathfrak{\mathfrak{m}}}))$,
and apply Matlis duality to see
$\operatorname{Ext}^{i}_{R}(\mathfrak{C},\omega_{R}^{+})^{\vee}=\operatorname{Tor}_{i-\dim
R}^{R}(\mathfrak{C},\omega_{R})$ (see [15, Lemma 3.5(2)]), as dimension $R$ is
one, and via a shifting, we have

$\operatorname{Tor}_{i-1}^{R}(\mathfrak{C},\omega_{R})=\operatorname{Tor}_{i}^{R}(\frac{R}{\mathfrak{C}},\omega_{R})=0\quad(\ast)$

There is nothing to prove if $\mathfrak{C}=R$, as it implies that $R$ is
normal, and by a dimension-consideration regular, and in particular
Gorenstein. Thus, without loss of generality we may assume $\mathfrak{C}$ is a
proper ideal. Thanks to [21, Ex. 2.11] $\mathfrak{C}$ has a regular element.
Consequently, $\mathfrak{C}$ is $\mathfrak{m}$-primary, and it is integrally
closed, as $\dim({R})=1$ and $\mathfrak{C}$ is a common ideal of ${R}$ and
$\overline{R}$ (see [21, Ex. 12.2]). This along with $(\ast)$ implies that
$\operatorname{pd}(\omega_{R})<i$ (see [7, 3.3]). By Auslander-Buchsbaum
formula, $\omega_{R}$ is free and consequently, $R$ is Gorenstein.

$ii)\Rightarrow iii)$ In view of Auslander-Bridger formula
$\operatorname{Gdim}_{R}(\overline{R})=0$, and the claim follows.

$iii)\Rightarrow iv)$ Trivial.

$iv)\Rightarrow i)$ Since $\overline{R}$ is principal ideal domain, and
$0\neq\mathfrak{C}\lhd\overline{R}$, it should be principal, and so
$\mathfrak{C}\cong\overline{R}$. ∎

The following extends Observation 2.8:

######  Corollary 2.14.

If $\operatorname{Ext}_{R}^{i}(\mathfrak{C},R)=0$ for some $i>0$ then $R$
satisfies $(G_{1})$. In particular, $R$ is quasi-normal if $R$ is $(S_{2})$.

##  3\. $\mathfrak{C}$ and cohomological annihilator

By homological annihilator we mean:

######  Definition 3.1.

(Vasconcelos).
$\operatorname{h-Ann}_{R}(M):=\prod_{i=0}^{d}\operatorname{Ann}_{R}\operatorname{Ext}^{i}_{R}(M,R)$.

If $R$ is Gorenstein, then
$\operatorname{Ann}(M)^{\operatorname{Gdim}(M)-\operatorname{grade}(M)+1}\subseteq\operatorname{h-Ann}(M)\subseteq\operatorname{Ann}(M),$
see e.g. [3]. There are some of interests to find situations for which the
equality holds.

######  Observation 3.2.

Let $R$ be a 2-dimensional Gorenstein ring. Then
$\operatorname{h-Ann}(\frac{\overline{R}}{R})=\operatorname{Ann}_{R}(\frac{\overline{R}}{R})$.

###### Proof.

Recall that
$\operatorname{Ass}(\operatorname{Hom}_{R}(\frac{\overline{R}}{R},R))=\operatorname{Ass}(R)\cap\operatorname{Supp}(\frac{\overline{R}}{R})=\operatorname{Ass}(R)\cap\operatorname{Var}(\mathfrak{C}_{R})=\emptyset$.
From this, $\operatorname{Hom}_{R}(\frac{\overline{R}}{R},R)=0$. Also,
$\overline{R}$ is maximal Cohen-Macaulay. It yields,
$\operatorname{Gdim}_{R}(\overline{R})=0$. From $0\to
R\to\overline{R}\to\frac{\overline{R}}{R}\to 0$ we know
$\operatorname{Gdim}_{R}(\frac{\overline{R}}{R})\leq 1$. So,
$\operatorname{Ext}^{2}_{R}(\frac{\overline{R}}{R},R)=0$. Also, it yields,

$0=\operatorname{Hom}_{R}(\frac{\overline{R}}{R},R)\to\operatorname{Hom}_{R}({\overline{R}},R)\to\operatorname{Hom}_{R}({{R}},R)\to\operatorname{Ext}^{1}_{R}(\frac{\overline{R}}{R},R)\to\operatorname{Ext}^{1}_{R}({\overline{R}},R)=0.$

This gives
$\operatorname{Ext}^{1}_{R}(\frac{\overline{R}}{R},R)=\frac{{R}}{\mathfrak{C}}$.
We deduce that

$\operatorname{h-Ann}(\frac{\overline{R}}{R}):=\prod_{i=0}^{2}\operatorname{Ann}_{R}\operatorname{Ext}^{i}_{R}(\frac{\overline{R}}{R},R)=\operatorname{Ann}_{R}\frac{{R}}{\mathfrak{C}}=\mathfrak{C}=\operatorname{Ann}_{R}(\frac{\overline{R}}{R}),$

as claimed. ∎

######  Definition 3.3.

(Bridger-Roberts). By cohomological annihilator we mean
$\operatorname{Ann}^{h}_{R}(M):=\prod_{i=0}^{\dim
R-1}\operatorname{Ann}_{R}H^{i}_{\mathfrak{m}}(M)$.

######  Observation 3.4.

Let $R$ be a Gorenstein ring. Then

  1. i)

$\mathfrak{C}_{R}\subseteq\operatorname{Ann}^{h}(\overline{R})$,

  2. ii)

Suppose the ring is not normal. Then
$\operatorname{rad}\mathfrak{C}_{R}\neq\operatorname{rad}\operatorname{Ann}^{h}(\overline{R})$.

An ideal is called semiconductor if it defines the same locus as
$\mathfrak{C}_{R}$.

###### Proof.

i) Without loss of generality, we may assume the ring is complete. Recall by
local duality that
$H^{d-j}_{\mathfrak{m}}(\overline{R})^{\vee}=\operatorname{Ext}^{j}_{R}({\overline{R}},R).$
This gives
$\operatorname{Ann}(H^{d-j}_{\mathfrak{m}}(\overline{R}))=\operatorname{Ann}(\operatorname{Ext}^{j}_{R}({\overline{R}},R))$.
From $0\to R\to\overline{R}\to\frac{\overline{R}}{R}\to 0$ we know for any $j$
that

  1. $(\ast)_{j=1}$

:
$\operatorname{Ext}^{1}_{R}(\frac{\overline{R}}{R},R)\to\operatorname{Ext}^{1}_{R}({\overline{R}},R)\to\operatorname{Ext}^{1}_{R}({{R}},R)=0$,

  2. $(+)_{j>1}$

:
$0=\operatorname{Ext}^{j-1}_{R}({R},R)\to\operatorname{Ext}^{j}_{R}(\frac{\overline{R}}{R},R)\to\operatorname{Ext}^{j}_{R}({\overline{R}},R)\to\operatorname{Ext}^{1}_{R}({{R}},R)=0$.

From $(\ast)_{j=1}$ we have
$\mathfrak{C}_{R}\subseteq\operatorname{Ann}(\operatorname{Ext}^{1}_{R}(\frac{\overline{R}}{R},R))\subseteq\operatorname{Ann}(\operatorname{Ext}^{1}_{R}({\overline{R}},R))$.
From $(+)_{j>1}$ we have
$\mathfrak{C}_{R}\subseteq\operatorname{Ann}(\operatorname{Ext}^{J}_{R}(\frac{\overline{R}}{R},R))=\operatorname{Ann}(\operatorname{Ext}^{1}_{R}({\overline{R}},R))$.
This gives $\mathfrak{C}_{R}\subseteq\operatorname{Ann}^{h}(\overline{R})$.

ii) Suppose $\dim R=1$. Since $\mathfrak{C}_{R}$ is nonzero, proper, it is
$\mathfrak{m}$-primary, and so
$\operatorname{rad}\mathfrak{C}_{R}=\mathfrak{m}$. But,
$\operatorname{rad}\operatorname{Ann}^{h}(\overline{R})=\operatorname{rad}\operatorname{Ann}(H^{0}_{\mathfrak{m}}(\overline{R}))=R$.
Now, suppose $\dim R>1$ and suppose on the way of contradiction that

$\operatorname{rad}(\mathfrak{C}_{R})=\operatorname{rad}(\operatorname{Ann}^{h}(\overline{R}))\quad(\dagger)$

  1. $Fcat$:

(Bridger, see [5]). Set
$\gamma(M):=\bigcap_{i>0}\operatorname{rad}(\operatorname{Ann}_{R}\operatorname{Ext}^{i}_{R}(M,R))$
and suppose $\operatorname{Gdim}(M)<\infty$. Then $\gamma(M)$ contains an
$M$-sequence of length $k$ iff $M$ is $k$-torsion-free.

Now, since $\overline{R}$ is $(S_{2})$, it is reflexive and so 2-torsionless.
Thanks to the above fact, $\gamma(\overline{R})$ contains a regular
$\overline{R}$-sequence of length two. In view of $(\dagger)$ we deduce
$\mathfrak{C}_{R}$ contains a regular $\overline{R}$-sequence of length two.
In particular, its height is bigger then one. But, the $(S_{2})$ condition of
$R$ implies $\operatorname{ht}(\mathfrak{C})=1$, a contradiction. ∎

As may the reader see, part $i)$ works for Cohen-Macaulay rings with canonical
module. It may be nice to drop the Cohen-Macaulay assumption:

######  Corollary 3.5.

Let $R$ be so that there is a Cohen-Macaulay ring $G$ between
$R\subseteq\overline{R}$. Then
$\mathfrak{C}_{R}\subseteq\operatorname{Ann}^{h}(\overline{R})$.

###### Proof.

We may assume the ring is complete. Recall that

$\mathfrak{C}_{R}:=\operatorname{Ann}_{R}(\frac{\overline{R}}{R})\subseteq\operatorname{Ann}_{R}(\frac{\overline{R}}{G})\subseteq\operatorname{Ann}_{G}(\frac{\overline{R}}{G})=\operatorname{Ann}_{G}(\frac{\overline{G}}{G})=\mathfrak{C}_{G}\quad(\ast)$

Also, by the independence theorem for local cohomology we have

$\operatorname{Ann}^{h}_{G}(\overline{G})\cap
R=\operatorname{Ann}^{h}_{R}(\overline{R})\quad(+)$

From $(\ast)$ and Observation 3.4 we have
$\mathfrak{C}_{R}\subseteq\mathfrak{C}_{G}\subseteq\operatorname{Ann}^{h}_{G}(\overline{G}).$
Thus, combining with $(+)$ we deduce
$\mathfrak{C}_{R}\subseteq\operatorname{Ann}^{h}_{G}(\overline{G})\cap
R=\operatorname{Ann}^{h}_{R}(\overline{R})$ as claimed. ∎

The next result is trivial if $i>1$ or $\overline{R}$ is Cohen-Macaulay:

######  Corollary 3.6.

Let $R$ be a Cohen-Macaulay ring and $\underline{x}$ be a full parameter
sequence. Then
$\mathfrak{C}_{R}\subseteq\operatorname{Ann}(\operatorname{Tor}^{R}_{i}(\overline{R},R/\underline{x}R))=0$
for all $i>0$.

###### Proof.

We may assume the ring is complete. Let $F_{\bullet}$ be the Koszul complex
with respect to $\underline{x}$, and set $M:=\overline{R}$. Then $F_{\bullet}$
is a complex of finite free modules with homology of finite length. By [6, Ex.
8.1.6] $\operatorname{Ann}^{h}(M)$ annihilates
$H^{+}(F_{\bullet}\otimes_{R}M)$. Since
$\mathfrak{C}_{R}\subset\operatorname{Ann}^{h}(\overline{R})$ we get the
desired claim. ∎

Let k be a field and $\underline{m}=(m_{1},\ldots,m_{n})\in\mathbb{N}_{0}^{n}$
with $m_{1}+\ldots+m_{n}=d$, we denote by $P_{n,d,\underline{m}}$ the pinched
Veronese ring in $n$ variables formed by removing the monomial generator
$x_{1}^{m_{1}}\dots x_{n}^{m_{n}}$ from the d-Veronese subring of
$k[[x_{1},\ldots,x_{n}]].$ The rings $P_{n,d,\underline{m}}$ are rarely Cohen-
Macaulay. Despite this:

######  Example 3.7.

Let $R:=P_{n,d,\underline{m}}$. Then
$\mathfrak{C}_{R}\subseteq\operatorname{Ann}^{h}(\overline{R})$.

###### Proof.

Recall from [12, Theorem 2.3] that $\overline{P_{n,d,\underline{m}}}$ is the
Veronese ring with the exception of normal rings $P_{n,d,d}$ and
$P_{2,2,(1,1)}$. By [12, Theorem 1.1] we observe that $\overline{R}$ is Cohen-
Macaulay. So, the desired claim follows by Corollary 3.5. ∎

Suppose $R$ is 2-dimensional and Cohen-Macaulay. Since
$0pt(\operatorname{Hom}_{R}({\overline{R}},R))\geq\min\\{2,0pt({R})\\}$, it
follows that $\mathfrak{C}_{R}$ is maximal Cohen-Macaulay. This extends to
higher dimensional as follows:

######  Observation 3.8.

If $R$ is Gorenstein and $\overline{R}$ is Cohen-Macaulay, then
$\mathfrak{C}_{R}$ is maximal Cohen-Macaulay.

###### Proof.

Without loss of generality we assume that $d:=\dim R>1$ and we assume $R$ is
not normal. We observe that
$\operatorname{Ext}^{1}_{R}(\frac{\overline{R}}{R},R)=\frac{{R}}{\mathfrak{C}}$.
From $0\to R\to\overline{R}\to\frac{\overline{R}}{R}\to 0$ we know

$d-1=\dim(\frac{\overline{R}}{R})\geq 0pt_{R}(\frac{\overline{R}}{R})\geq
d-1.$

Thus, $\frac{\overline{R}}{R}$ is Cohen-Macaulay. This gives Cohen-
Macaulayness of
$\operatorname{Ext}^{1}_{R}(\frac{\overline{R}}{R},R)\cong\frac{{R}}{\mathfrak{C}}$
which is of dimension $d-1$. In view of
$0\to\mathfrak{C}\to\overline{R}\to\frac{\overline{R}}{\mathfrak{C}}\to 0$ we
observe $\mathfrak{C}$ is maximal Cohen-Macaulay.∎

######  Example 3.9.

i) Let $G$ be the set of all quadric monomials in $\\{X,Y,Z\\}$. Let
$R:=k[[G\setminus\\{XY\\}]]$. Since $R$ is Gorenstein (see [12, Theorem A])
and $\overline{R}=k[[G]]$ is Cohen-Macaulay, we see $\mathfrak{C}$ is maximal
Cohen-Macaulay.

ii) The Gorenstein assumption of $R$ is needed. Indeed, Let
$R:=k[[s^{2},s^{3},st,t]]$. Here, we claim that $\mathfrak{C}=\mathfrak{m}$:
First, we compute it. Since $s=s^{3}/s^{2}$ it belong to fraction field of
$R$. Also, it is integral over $R$, as $f(s)=0$ where $f[X]:=X^{2}-s^{2}$.
From this, $\overline{R}=k[[s,t]]$. Now, we compute its conductor:

  1. a)

In order to show $t\in\mathfrak{C}$ we note that $t.\overline{R}\subseteq R$.
Indeed, $\\{t.1,t.s,t.t\\}\subset R.$

  2. b)

In order to show $s^{2}\in\mathfrak{C}$ we note that
$s^{2}.\overline{R}\subseteq R$. Indeed, $\\{s^{2}.1,s^{2}.s,s^{2}.t\\}\subset
R.$

  3. c)

In order to show $s^{3}\in\mathfrak{C}$ we note that
$s^{3}.\overline{R}\subseteq R$. Indeed, $\\{s^{3}.1,s^{3}.s,s^{3}.t\\}\subset
R.$

  4. d)

In order to show $st\in\mathfrak{C}$ we note that $st.\overline{R}\subseteq
R$. Indeed, $\\{st.1,st.s,st.t\\}\subset R.$

Thus, $\mathfrak{C}=\mathfrak{m}_{R}$. Since $\overline{R}$ is generated as an
$R$-module by $\\{1,t\\}$ and that $\mathfrak{m}t\subseteq R$, we deduce that
$\frac{\overline{R}}{R}=\frac{{R}}{\mathfrak{m}}$. Since
$0pt(\mathfrak{m})=1$, we deduce that $\mathfrak{C}_{R}$ is not maximal Cohen-
Macaulay.

The following completes the proof of Theorem C) from introduction.

######  Observation 3.10.

Let $R$ be a 2-dimensional analytically unramified ring satisfying $(R_{1})$.
Then $\mathfrak{C}_{R}=\operatorname{Ann}(H^{1}_{\mathfrak{m}}(R))$.

###### Proof.

From $0\to R\to\overline{R}\to\frac{\overline{R}}{R}\to 0$ we deduce
$0=H^{0}_{\mathfrak{m}}(\overline{R})\to
H^{0}_{\mathfrak{m}}(\frac{\overline{R}}{R})\to H^{1}_{\mathfrak{m}}(R)\to
H^{1}_{\mathfrak{m}}(\overline{R})=0.$ According to the $(R_{1})$-condition,
we see the ring is normal over the punctured spectrum. From this
$H^{0}_{\mathfrak{m}}(\frac{\overline{R}}{R})=\frac{\overline{R}}{R}$.
Consequently,
$H^{1}_{\mathfrak{m}}(R)=H^{0}_{\mathfrak{m}}(\frac{\overline{R}}{R})=\frac{\overline{R}}{R}$,
so $\mathfrak{C}_{R}=\operatorname{Ann}(H^{1}_{\mathfrak{m}}(R))$ as claimed.
∎

Let $M$ be an $R$-module. Vasconselos proved that $M$ is projective provided
$M\otimes_{R}\overline{R}$ is $\overline{R}$-projective and that
$M\otimes_{R}R/\mathfrak{C}$ is $R/\mathfrak{C}$-projective. Two natural
questions arises:

  1. $1)$

: Suppose $\operatorname{Gdim}_{\overline{R}}(M\otimes_{R}\overline{R})=0$ and
that
$\operatorname{Gdim}_{R/\mathfrak{C}}(M\otimes_{R}\frac{R}{\mathfrak{C}})=0$.
Is $\operatorname{Gdim}_{R}(M)=0$?

  2. $2)$

: Suppose $\operatorname{pd}_{\overline{R}}(M\otimes_{R}\overline{R})<\infty$
and that
$\operatorname{pd}_{R/\mathfrak{C}}(M\otimes_{R}\frac{R}{\mathfrak{C}})<\infty$.
Is $\operatorname{pd}_{R}(M)<\infty$?

We show non of these true:

######  Example 3.11.

  1. $1)$

: Let $R:=k[[s^{3},s^{4},s^{5}]]$ and $M:=\mathfrak{m}=\mathfrak{C}$. Then
$\operatorname{Gdim}_{R}(M)\neq 0$ but
$\operatorname{Gdim}_{\overline{R}}(M\otimes_{R}\overline{R})=\operatorname{Gdim}_{R/\mathfrak{C}}(M\otimes_{R}\frac{R}{\mathfrak{C}})=0$.

  2. $2)$

: Let $R:=k[[s^{2},s^{3},st,t]]$ and $M:=k$. Then:
$\operatorname{pd}_{R}(M)=\infty$ but
$\operatorname{pd}_{\overline{R}}(M\otimes_{R}\overline{R})<\infty$ and that
$\operatorname{pd}_{R/\mathfrak{C}}(M\otimes_{R}\frac{R}{\mathfrak{C}})<\infty$.

##  4\. Ikeda’s problem

######  Conjecture 4.1.

Suppose $R$ is analytically unramified. Then
$\mathfrak{C}\nsubseteq\underline{x}R$ where $\underline{x}$ is a regular
sequence.

We only use the following weakened property of quasi-Gorenstein:

######  Lemma 4.2.

Let $(R,\mathfrak{m})$ be d-dimensional. If
$\operatorname{id}_{R}(H^{d}_{\mathfrak{m}}(R))<\infty,$ then
$H^{d}_{\mathfrak{m}}(R)$ is injective.

###### Proof.

Easy. ∎

The following drop some conditions from [11], for example the Cohen-Macaulay
condition:

######  Lemma 4.3.

Suppose $R$ is analytically unramified and quasi-Gorenstein. Then
$\operatorname{Hom}_{R}(\mathfrak{C},H^{d}_{\mathfrak{m}}(R))=H^{d}_{\mathfrak{m}}(\overline{R})$,
where $d:=\dim R$.

###### Proof.

Since $\overline{R}$ is finitely generated and $H^{d}_{\mathfrak{m}}(R)$ is
injective, we have the following functorial isomorphisms:

$\begin{array}[]{ll}\operatorname{Hom}_{R}(\mathfrak{C},H^{d}_{\mathfrak{m}}(R))&=\operatorname{Hom}_{R}(\operatorname{Hom}_{R}(\overline{R},R),H^{d}_{\mathfrak{m}}(R))\\\
&\cong\overline{R}\otimes\operatorname{Hom}_{R}(R,H^{d}_{\mathfrak{m}}(R))\\\
&=\overline{R}\otimes H^{d}_{\mathfrak{m}}({R})\\\
&\stackrel{{\scriptstyle(\ast)}}{{\cong}}H^{d}_{\mathfrak{m}}(\overline{R}),\end{array}$

where $(\ast)$ follows because $d$ is the cohomological dimension. ∎

The following extends and simplifies a result of Ikeda:

######  Observation 4.4.

Suppose $R$ is quasi-Gorenstein. Then $\mathfrak{C}\nsubseteq\underline{x}R$
where $\underline{x}$ is a system of parameter.

###### Proof.

Suppose $\mathfrak{C}\subseteq\underline{x}R$ for some system of parameter
$\underline{x}$. We extend it to a sequence of length $d$. By canonical
element conjecture which is a theorem, see [6, Section 9.3], there is a
natural map
$\rho_{\overline{R}}:\frac{\overline{R}}{\underline{x}\overline{R}}\to
H^{d}_{\mathfrak{m}}(\overline{R})$ with the property that
$\alpha:=\rho_{\overline{R}}(1+\underline{x}\overline{R})\neq 0$. Now, look at
$\alpha\in
H^{d}_{\mathfrak{m}}(\overline{R})\stackrel{{\scriptstyle\ref{injh}}}{{=}}\operatorname{Hom}_{R}(\mathfrak{C},H^{d}_{\mathfrak{m}}(R))$.
In particular, there is a nonzero $f:\mathfrak{C}\to H^{d}_{\mathfrak{m}}(R)$
so that $\alpha=\phi(f)$ where
$\phi:H^{d}_{\mathfrak{m}}(\overline{R})\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\operatorname{Hom}_{R}(\mathfrak{C},H^{d}_{\mathfrak{m}}(R))$.
Also, the assignment $m\otimes(M\stackrel{{\scriptstyle
g}}{{\longrightarrow}}N)\mapsto g(m)$ gives the evaluation map
$\varPhi_{M,N}:M\otimes\operatorname{Hom}(M,N)\longrightarrow N$. Here, we set

  1. i)

$\varPhi_{1}:=\varPhi_{\mathfrak{C},H^{d}_{\mathfrak{m}}(R)}:\mathfrak{C}\otimes\operatorname{Hom}(\mathfrak{C},H^{d}_{\mathfrak{m}}(R))\longrightarrow
H^{d}_{\mathfrak{m}}(R)$,

  2. ii)

$\varPhi_{2}=\varPhi_{\overline{R},R}\otimes_{R}\operatorname{id}_{\frac{{R}}{\underline{x}{R}}}:[\operatorname{Hom}(\overline{R},R)\otimes\overline{R}\longrightarrow
R]\otimes_{R}\frac{{R}}{\underline{x}{R}}$.

We have the following commutative diagram

$\textstyle{\mathfrak{C}\otimes_{R}H^{d}_{\mathfrak{m}}(\overline{R})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varPhi_{1}}$$\textstyle{H^{d}_{\mathfrak{m}}(R)}$$\textstyle{\mathfrak{C}\otimes_{R}\frac{\overline{R}}{\underline{x}\overline{R}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{id}_{\mathfrak{C}}\otimes\rho_{\overline{R}}}$$\scriptstyle{\varPhi_{2}}$$\textstyle{\frac{{R}}{\underline{x}{R}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{{R}}}$

Recall that $f:\mathfrak{C}\to H^{d}_{\mathfrak{m}}(R)$ so that
$\alpha=\phi(f)$ is nonzero. Take $y\in\mathfrak{C}$ so that $f(y)\neq 0$.
Since $\mathfrak{C}\overline{R}\subseteq R$ we deduce that $ry\in R$ and so
$\varPhi_{2}(y\otimes(r+\underline{x}\overline{R}))=ry+\underline{x}R\in\frac{{R}}{\underline{x}{R}}$.
Now, recall that $y\in\mathfrak{C}\subseteq\underline{x}R$. This gives

$\rho_{R}\circ\varPhi_{2}((y\otimes(1+\underline{x}\overline{R}))=\rho_{R}(y+\underline{x}R)=0.$

But,
$\varPhi_{1}\circ(\operatorname{id}_{\mathfrak{C}}\otimes\rho_{\overline{R}})((y\otimes(1+\underline{x}\overline{R}))=\varPhi_{1}(y\otimes\alpha)=\varPhi_{1}(y\otimes\phi(f))=f(y)\neq
0.$∎

The following removes dimension restriction from [21, 12.4] and [11].

######  Observation 4.5.

Suppose $R$ is analytically unramified. Then $\mathfrak{C}\nsubseteq{x}R$
where ${x}\in\mathfrak{m}$.

###### Proof.

Suppose $\mathfrak{C}\subseteq{x}R$ for some ${x}\in\mathfrak{m}$. Since $R$
is analytically unramified, it is reduced. It is easy to see $\overline{R}/R$
is analytically unramified, and so reduced. Due to [21, Ex. 2.11] we know
$\mathfrak{C}$ has a regular element. Thus, $\operatorname{grade}({x}R,R)=1$.
Consequently, $x$ is regular. Suppose $\mathfrak{C}=(a_{1},\ldots,a_{n})$ and
let $b_{i}$ so that $a_{i}=b_{i}x$. Let $J:=(b_{1},\ldots,b_{n})$. Then
$\mathfrak{C}=xJ$. Thus,
$J\overline{R}=(1/x)\mathfrak{C}\overline{R}=(1/x)\mathfrak{C}=J$. In other
words, $J$ is a common ideal of $\\{R,\overline{R}\\}$. So,
$J\subseteq\mathfrak{C}$. Therefore, $J=xJ$. By Nakayama’s lemma $J=0$. This
is a contradiction, since $\mathfrak{C}\neq 0$. ∎

######  Corollary 4.6.

Suppose $0pt(R)=1$. Then $\mathfrak{C}\nsubseteq\underline{x}R$ where
$\underline{x}$ is a regular sequence.

######  Example 4.7.

Let $R:=k[[s^{2},s^{3},st,t]]$. Then $\mathfrak{C}\nsubseteq\underline{x}R$
where $\underline{x}$ is a parameter sequence.

###### Proof.

Thanks to Example 3.9 we know $(t,s^{2},s^{3},st)=\mathfrak{C}$, and the
equality follows as the ring is not normal. Suppose on the way of
contradiction that $\mathfrak{C}\subseteq\underline{x}R$. Then
$\mathfrak{m}\subseteq\underline{x}R$. This implies the ring is regular, a
contradiction. ∎

The previous example extends as follows:

######  Observation 4.8.

Suppose $R$ is a two-dimensional Buchsbaum ring which is $(R_{1})$. Then
$\mathfrak{C}\nsubseteq\underline{x}R$ where $\underline{x}$ is a system of
parameter.

###### Proof.

Recall from Observation 3.10 that
$\frac{\overline{R}}{R}=H^{1}_{\mathfrak{m}}(R)=\oplus_{X}k$ where $X$ is an
index set. Without loss of generality we may assume that $R$ is not $(S_{2})$,
as otherwise $\mathfrak{C}=R$ which is not subset of $\underline{x}R$. This
implies $X$ is non-empty. Suppose on the way of contradiction that
$\mathfrak{C}\subseteq\underline{x}R$. Then
$\mathfrak{m}\subseteq\underline{x}R$. This implies the ring is regular, and
so normal. Thus $R=\mathfrak{C}\subseteq\underline{x}R$, a contradiction. ∎

######  Corollary 4.9.

Suppose $R$ is a complete two-dimensional seminormal ring which is $(R_{1})$.
Then $R$ is Buchsbaum.

###### Proof.

We may and do assume that the ring is not $(S_{2})$, as otherwise it is Cohen-
Macaulay and so Buchsbaum. Recall from $(R_{1})$ condition that $\mathfrak{C}$
is $\mathfrak{m}$-primary. By seminormality, we know $\mathfrak{C}$ is a
radical ideal in $\overline{R}$. Thus,
$\mathfrak{C}=\mathfrak{m}_{\overline{R}}$. Then
$\mathfrak{C}=\mathfrak{C}\cap R=\mathfrak{m}_{\overline{R}}\cap
R=\mathfrak{m}$. As
$\mathfrak{C}=\operatorname{Ann}(H^{1}_{\mathfrak{m}}(R))$, it turns out that
$R$ is Buchsbaum. ∎

######  Example 4.10.

Let $R:=k[[X^{2},Y^{2},XY,X^{2}Y,XY^{2}]]$. Then

  1. i)

$R$ is seminomal,

  2. ii)

$\mathfrak{C}\nsubseteq\underline{x}R$ where $\underline{x}$ is a parameter
sequence.

###### Proof.

i) This is well-known.

ii) It is easy to see $\overline{R}=k[[X,Y]]$. Let $J:=(XY,X^{2}Y,XY^{2})$. In
order to show $J\subseteq\mathfrak{C}$ we note by the routine argument of
Example 3.9(ii) that $J\overline{R}\subseteq R$. Suppose on the way of
contradiction that $\mathfrak{C}\subseteq(a,b)R$ where $a,b$ is a parameter
sequence. Then $\mathfrak{m}\subseteq\langle
J,X^{2},Y^{2}\rangle\subseteq\langle
a,b,X^{2},Y^{2}\rangle\subseteq\mathfrak{m}$. From this we see $\langle
a,b,X^{2},Y^{2}\rangle=\mathfrak{m}$. But, $5=\mu(\mathfrak{m})\leq 4$, a
contradiction. ∎

We close this section by presenting some higher dimensional examples.

######  Example 4.11.

Let $G$ be the set of all quadric monomials in $\\{X,Y,Z,W\\}$. Let
$R:=k[[G\setminus\\{ZW\\}]]$. Then $\mathfrak{C}\nsubseteq\underline{x}R$
where $\underline{x}$ is a parameter sequence.

###### Proof.

It is easy to see $\overline{R}=k[[G]]$, the 2-Veronese subring of
$k[[X,Y,Z,W]]$. Let $J:=\langle G\setminus\\{ZW,Z^{2},W^{2}\\}\rangle_{R}$. In
order to show $J\subseteq\mathfrak{C}$, we note by the routine argument
presented in Example 3.9(ii), that $J\overline{R}\subseteq R$. Suppose on the
way of contradiction that $\mathfrak{C}\subseteq(a,b,c,d)R$ where $a,b,c,d$ is
a parameter sequence. Then

$\mathfrak{m}\subseteq\langle J,Z^{2},W^{2}\rangle\subseteq\langle
a,b,c,d,Z^{2},W^{2}\rangle\subseteq\mathfrak{m}.$

From this we see $9=\mu(\mathfrak{m})\leq 6$, a contradiction. ∎

The following completes the proof of Theorem A) from introduction.

######  Observation 4.12.

Let $R$ be analytically unramified containing a field and suppose
$\mathfrak{m}=\mathfrak{C}+yR$ for some $y$. Then
$\mathfrak{C}\nsubseteq\underline{x}R$ where $\underline{x}$ is a system of
parameter.

###### Proof.

Without loss of generality we assume the ring is complete. In particular, $R$
a reduced ring. Suppose on the way of contradiction that there is some system
of parameter $\underline{x}:=x_{1},\ldots,x_{d}$ so that
$\mathfrak{C}\subseteq\underline{x}R$. Then
$\mathfrak{m}=(x_{1},\ldots,x_{d},y)$. Let $S:=k[[X_{1},\ldots,X_{d},X]]$ and
consider the natural projection $\pi:S\twoheadrightarrow R$. Since $R$ is
reduced we see $\ker(\pi)$ is radical and of height one. Since $S$ is regular,
$\ker(\pi)=(f)$ is principal. From this $R$ is hypersurface, in particular
quasi-Gorenstein. This is in contradiction with Observation 4.4. ∎

Here, we present some non-trivial examples for which
$\mathfrak{m}=\mathfrak{C}+xR$:

######  Example 4.13.

Let $G$ be the set of all cubic monomials in $\\{X,Y,Z\\}$. Let
$R:=k[[G\setminus\\{X^{2}Y\\}]]$. Then $\mathfrak{m}=\mathfrak{C}+X^{3}R$. In
particular, $\mathfrak{C}\nsubseteq\underline{x}R$ where $\underline{x}$ is a
parameter sequence.

###### Proof.

As $X^{2}Y=\frac{(XY^{2})XYZ}{Y^{2}Z}$ we see $X^{2}Y\in Q(R)$. Let
$f(t):=t^{2}-(X^{3})(XY^{2})\in R[t]$. Since $f(X^{2}Y)=0$, we see
$\overline{R}=k[[G]]$. Let $J:=\langle
G\setminus\\{X^{2}Y,X^{3}\\}\rangle_{R}$. In order to show
$J\subseteq\mathfrak{C}$ we show $\overline{R}J\subseteq R$. To this end, we
show $X^{2}YJ\subseteq R$. Let us check this by looking at each seven
generators of $J$:

  1. 1)

$(X^{2}Y)Z^{2}Y=XY^{2}(Z^{2}X)\in J\subseteq R,\quad\quad\quad 2)\ \
(X^{2}Y)Y^{3}=(XY^{2})Y^{3}\in J\subseteq R,$

  2. 3)

$(X^{2}Y)ZXY=(Y^{2}X)ZX^{2}\in J\subseteq R,\quad\quad\ \ 4)\ \
(X^{2}Y)Z^{3}=(ZX^{2})Z^{2}Y\in J\subseteq R,$

  3. 5)

$(X^{2}Y)ZY^{2}=(Y^{3})ZX^{2}\in J\subseteq R,\quad\quad\ \ \ \ \ \ \ 6)\ \
(X^{2}Y)Z^{2}X=(X^{3})Z^{2}Y\in J\subseteq R,$

  4. 7)

$(X^{2}Y)ZX^{2}=(X^{3})XYZ\in J\subseteq R$.

Thus, $\mathfrak{m}\subseteq
J+X^{3}R\subseteq\mathfrak{C}+X^{3}R\subseteq\mathfrak{m}$. This proves the
first part. The particular case is in Observation 4.12. ∎

######  Example 4.14.

Let $k$ be a field, $n>1$ and $\\{X_{1},\ldots,X_{n}\\}$ be indeterminate. Let

$R:=k[[X_{1},\ldots,X_{n-1},X_{n}(X_{1},\ldots,X_{n})]]\subseteq
k[[X_{1},\ldots,X_{n}]].$

Then $\mathfrak{m}=\mathfrak{C}+X_{n}^{2}R$. In particular,
$\mathfrak{C}\nsubseteq\underline{x}R$ where $\underline{x}$ is a parameter
sequence.

###### Proof.

As $X_{n}=\frac{X_{n}X_{1}}{X_{1}}$ we see $X_{n}\in Q(R)$. Let
$f(t):=t^{2}-X_{n}^{2}\in R[t]$. Since $f(X_{n}^{2})=0$ we can deduce that
$\overline{R}=k[[X_{1},\ldots,X_{n}]]$. Let $J:=\langle
X_{1},\ldots,X_{n-1},X_{1}X_{n},\ldots,X_{n-1}X_{n}\rangle_{R}$. In order to
show $J\subseteq\mathfrak{C}$ we show $\overline{R}J\subseteq R$. To this end,
we show $X_{n}J\subseteq R$. Let us check this by looking at each $2n-2$
generators of $J$:

  1. $\bullet_{j}$

$(X_{n})X_{j}\in J\subseteq R$ for all $1\leq j<n$

  2. $\bullet_{i}$

$(X_{n})X_{i}X_{n}=X_{i}X_{n}^{2}\in J\subseteq R$ for all $1\leq i<n$.

Thus, $\mathfrak{m}\subseteq
J+X_{n}^{2}R\subseteq\mathfrak{C}+X_{n}^{2}R\subseteq\mathfrak{m}$. This
proves the first part. The particular case is in Observation 4.12. ∎

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