# On the finiteness of local homology modules

Ali Fathi1 1 Department of Mathematics, Zanjan Branch, Islamic Azad
University, Zanjan, Iran<EMAIL_ADDRESS>and  Alireza Hajikarimi2 2
Department of Mathematics, Mobarakeh Branch, Islamic Azad University, Isfahan,
Iran<EMAIL_ADDRESS>
###### Abstract.

Let $R$ be a commutative Noetherian ring and $\mathfrak{a}$ an ideal of $R$.
Let $M$ be a finitely generated $R$-module and $N$ an Artinian $R$-module. The
concept of filter coregular sequence is introduced to determine the infimum of
the integers $i$ such that the generalized local homology
$\textrm{H}^{\mathfrak{a}}_{i}(M,N)$ is not finitely generated as an
$\widehat{R}^{\mathfrak{a}}$-module, where $\widehat{R}^{\mathfrak{a}}$
denotes the $\mathfrak{a}$-adic completion of $R$. In particular, it is shown
that $\textrm{H}^{\mathfrak{a}}_{i}(M,N)$ is a finitely generated
$\widehat{R}^{\mathfrak{a}}$-module for all $i\in\mathbb{N}_{0}$ if and only
if $(0:_{N}\mathfrak{a}+\textrm{Ann}_{R}(M))$ has finite length whenever $R$
is a complete semi-local ring.

###### Key words and phrases:

Local homology, Tor functor, finiteness, filter coregular sequence

###### 2010 Mathematics Subject Classification:

13D07, 13E05, 13C15

##  1\. Introduction

Throughout this paper, $R$ is a commutative Noetherian ring with nonzero
identity. Let $\mathfrak{a}$ be an ideal of $R$ and let $M,N$ be two
$R$-modules. We denote the $\mathfrak{a}$-adic completion of $N$ by
$\Lambda_{\mathfrak{a}}(N)$. The $\mathfrak{a}$-adic completion functor
$\Lambda_{\mathfrak{a}}(\cdot)$ is an additive covariant functor on the
category of $R$-modules. In [2] N. T. Cuong and T. T. Nam defined the $i$th
local homology $\operatorname{H}^{\mathfrak{a}}_{i}\left(N\right)$ of $N$ with
respect to $\mathfrak{a}$ by

$\operatorname{H}^{\mathfrak{a}}_{i}\left(N\right)=\underset{n\in\mathbb{N}}{\varprojlim}\operatorname{Tor}^{R}_{i}\left(R/\mathfrak{a}^{n},N\right).$

Also, they showed that
$L^{\mathfrak{a}}_{i}(N)\cong\operatorname{H}^{\mathfrak{a}}_{i}\left(N\right)$
when $N$ is Artinian, where $L^{\mathfrak{a}}_{i}(\cdot)$ denotes the $i$th
left derived functor of $\Lambda_{\mathfrak{a}}(\cdot)$. Note that since the
tensor functor is not left exact and the inverse limit is not right exact on
the category of $R$-modules, the functor of the $\mathfrak{a}$-adic completion
$\Lambda_{\mathfrak{a}}(\cdot)$ is neither left nor right exact. Therefore, in
general, $L_{0}^{\mathfrak{a}}(\cdot)\ncong\Lambda_{\mathfrak{a}}(\cdot)$ (
and so, in general,
$L^{\mathfrak{a}}_{i}(\cdot)\ncong\operatorname{H}^{\mathfrak{a}}_{i}\left(\cdot\right)$).

Similarly, the $i$th generalized local homology
$\operatorname{H}^{\mathfrak{a}}_{i}(M,N)$ of $M,N$ with respect to
$\mathfrak{a}$ is defined by

$\operatorname{H}^{\mathfrak{a}}_{i}(M,N)=\underset{n\in\mathbb{N}}{\varprojlim}\operatorname{Tor}^{R}_{i}\left(M/\mathfrak{a}^{n}M,N\right);$

see [11] for basic properties and more details. We use filter coregular
sequences to study the finiteness of (generalized) local homology modules.

The set of all prime ideals of $R$ containing the $\operatorname{Ann}_{R}(N)$
is denoted by $\operatorname{Cosupp}_{R}(N)$. Also, we denote the set $\\{x\in
N:\mathfrak{a}x=0\\}$ by $(0:_{N}\mathfrak{a})$. A sequence
$x_{1},\dots,x_{n}$ of elements of $\mathfrak{a}$ is called a filter coregular
$N$-sequence (of length $n$) in $\mathfrak{a}$ whenever

$\operatorname{Cosupp}_{R}\big{(}(0:_{N}(x_{1},\dots,x_{i-1})R)/x_{i}(0:_{N}(x_{1},\dots,x_{i-1})R)\big{)}\subseteq\operatorname{Max}(R)$

for all $1\leq i\leq n$, where $\operatorname{Max}(R)$ denotes the set of
maximal ideals of $R$. In the remainder of this section, let $M$ be finitely
generated and $N$ be Artinian. We prove that, if there exists a filter
coregular $N$-sequence in $\mathfrak{a}$ of infinite length, then each filter
coregular $N$-sequence in $\mathfrak{a}$ can be extended to a filter coregular
$N$-sequence in $\mathfrak{a}$ of infinite length. Now, assume all filter
coregular $N$-sequences in $\mathfrak{a}$ have finite length. Then all maximal
filter coregular $N$-sequences in $\mathfrak{a}$ are of the same length, and
this common length is denoted by $\operatorname{f-width}(\mathfrak{a},N)$. We
set $\operatorname{f-width}(\mathfrak{a},N)=\infty$ whenever there exists a
filter coregular $N$-sequence in $\mathfrak{a}$ of infinite length. We use
filter coregular sequences to determine the first local homology modules that
is not Artinian as an $\widehat{R}^{\mathfrak{a}}$-module. More precisley, if
$L$ is a finitely generated $R$-module such that
$\operatorname{Supp}_{R}(L)=\operatorname{Supp}_{R}(M/\mathfrak{a}M)$, then we
prove that

$\displaystyle\operatorname{f-width}(\mathfrak{a}+\operatorname{Ann}_{R}(M),N)$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{i}\left(L,N\right)\right)\nsubseteq\operatorname{Max}(R)\\}$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{Tor}^{R}_{i}\left(L,N\right)\textrm{
is not a finitely generated }R\textrm{-module}\\}$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{H}^{\mathfrak{a}}_{i}(M,N)\textrm{
is not a finitely generated }\widehat{R}^{\mathfrak{a}}\textrm{-module}\\};$

see Theorem 2.10 and Remark 2.11. In particular, we have the following special
cases

$\displaystyle\operatorname{f-width}(\operatorname{Ann}_{R}(M),N)$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{Tor}^{R}_{i}\left(M,N\right)\textrm{
is not a finitely generated }R\textrm{-module}\\}$

and

$\displaystyle\operatorname{f-width}(\mathfrak{a},N)$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{H}^{\mathfrak{a}}_{i}\left(N\right)\textrm{
is not a finitely generated }\widehat{R}^{\mathfrak{a}}\textrm{-module}\\}.$

Finally, we show, in Corollary 2.13, that if
$\operatorname{H}^{\mathfrak{a}}_{i}(M,N)$ is a finitely generated
$\widehat{R}^{\mathfrak{a}}$-module (resp. if
$\operatorname{Tor}^{R}_{i}\left(M,N\right)$ has finite length) for all
$i\in\mathbb{N}_{0}$, then $(0:_{N}\mathfrak{a}+\operatorname{Ann}_{R}(M))$
(resp. $(0:_{N}\operatorname{Ann}_{R}(M))$) has finite length. The converse
statements also hold whenever $R$ is a semi-local ring which is complete with
respect to its Jacobson radical.

##  2\. Main Results

Let $M$ be a finitely generated $R$-module and $\mathfrak{a}$ an ideal of $R$.
We recall that a sequence $x_{1},\dots,x_{n}$ of elements of $R$ is called a
filter regular $M$-sequence (of length $n$) whenever one of the following
equivalent conditions

  * •

$\operatorname{Supp}_{R}(0:_{M/(x_{1},\dots,x_{i-1})M}x_{i}R)\subseteq\operatorname{Max}(R)$,
or

  * •

$x_{i}\notin\bigcup_{\mathfrak{p}\in\operatorname{Ass}_{R}(M/(x_{1},\dots,x_{i-1})M)\setminus\operatorname{Max}(R)}\mathfrak{p}$,
or

  * •

$(0:_{M/(x_{1},\dots,x_{i-1})M}x_{i}R)$ has finite length

is satisfied for all $1\leq i\leq n$. If, in addition, $x_{1},\dots,x_{n}$
belong to $\mathfrak{a}$, then we say that $x_{1},\dots,x_{n}$ is a filter
regular $M$-sequence in $\mathfrak{a}$. There is a filter regular $M$-sequence
in $\mathfrak{a}$ of infinite length if and only if $M/\mathfrak{a}M$ has
finite length. Also, if $\dim_{R}(M/\mathfrak{a}M)>0$, then all maximal filter
regular $M$-sequences in $\mathfrak{a}$ have the same (finite) length and this
common length is denoted by $\operatorname{f-depth}(\mathfrak{a},M)$. We set
$\operatorname{f-depth}(\mathfrak{a},M)=\infty$ whenever
$\dim(M/\mathfrak{a}M)\leq 0$. We denote the set of all prime ideals of $R$
containing $\mathfrak{a}$ by $\operatorname{V}\left(\mathfrak{a}\right)$. If
$N$ is a finitely generated $R$-module such that
$\operatorname{Supp}_{R}(N)=\operatorname{V}\left(\mathfrak{a}\right)$, then

$\displaystyle\operatorname{f-depth}(\mathfrak{a},M)$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{Ext}^{i}_{R}\left(N,M\right)\textrm{is
not Artinian }\\}$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{H}^{i}_{\mathfrak{a}}\left(M\right)\textrm{is
not Artinian }\\}.$

In particular, $\operatorname{H}^{i}_{\mathfrak{a}}\left(M\right)$ is Artinian
for all $i\in\mathbb{N}_{0}$ if and only if $M/\mathfrak{a}M$ has finite
length; see [4, Sec. 3] or [14] for more details (see also [6] when $R$ is a
local ring).

In this paper, the dual of filter regular sequence is established to study the
finiteness of (generalized) local homology modules.

######  Definition 2.1.

Let $M$ be an $R$-module. We say that a prime ideal $\mathfrak{p}$ of $R$ is
an attached prime of $M$, if there exists a submodule $N$ of $M$ such that
$\mathfrak{p}=\operatorname{Ann}_{R}(M/N)$. We denote by
$\operatorname{Att}_{R}(M)$ the set of attached primes of $M$. Also we denote
$\operatorname{V}\left(\operatorname{Ann}_{R}(M)\right)$ by
$\operatorname{Cosupp}_{R}(M)$. It is clear that
$\operatorname{Att}_{R}(M)\subseteq\operatorname{Cosupp}_{R}(M)$; see [12] for
basic properties.

When an $R$-module $M$ has a secondary representation in the sense of [8], our
definition of $\operatorname{Att}_{R}(M)$ coincides with that of Macdonald [1,
Exercise 7.2.5]. In particular, the set of attached primes of an Artinian
$R$-module is a finite set.

The following lemmas and propositions are needed in the proof of our main
theorem.

Let $N$ be an $R$-module and $\mathfrak{a}$ an ideal of $R$. We use
$(0:_{N}\mathfrak{a})$ to denote the submodule $\\{x\in N:\mathfrak{a}x=0\\}$
of $N$.

######  Definition 2.2.

Let $N$ be an $R$-module. A sequence $x_{1},\dots,x_{n}$ of elements of $R$ is
called a filter coregular $N$-sequence (of length $n$) whenever

$\operatorname{Cosupp}_{R}\big{(}(0:_{N}(x_{1},\dots,x_{i-1})R)/x_{i}(0:_{N}(x_{1},\dots,x_{i-1})R)\big{)}\subseteq\operatorname{Max}(R)$

for all $1\leq i\leq n$, where $\operatorname{Max}(R)$ denotes the set of all
maximal ideals of $R$. If, in addition, $x_{1},\dots,x_{n}$ belong to an ideal
$\mathfrak{a}$, then we say that $x_{1},\dots,x_{n}$ is a filter coregular
$N$-sequence in $\mathfrak{a}$.

######  Lemma 2.3.

Let $N$ be an Artinian $R$-module. Then the following statements are
equivalent.

  1. (i)

$N$ has finite length.

  2. (ii)

$\operatorname{Cosupp}_{R}(N)\subseteq\operatorname{Max}(R)$.

  3. (iii)

$\operatorname{Att}_{R}(N)\subseteq\operatorname{Max}(R)$.

###### Proof.

Assume that $N$ has finite length. Since $N$ is finitely generated, we have
$\operatorname{Cosupp}_{R}(N)=\operatorname{Supp}_{R}(N)$. Also, the
Artinianness of $N$ implies that
$\operatorname{Supp}_{R}(N)\subseteq\operatorname{Max}(R)$, and so
$\operatorname{Cosupp}_{R}(N)\subseteq\operatorname{Max}(R)$. This proves the
implication (i)$\Rightarrow$(ii). The implication (ii)$\Rightarrow$(iii) is
clear. Finally, to prove the implication (iii)$\Rightarrow$(i), suppose that
$\operatorname{Att}_{R}(N)\subseteq\operatorname{Max}(R)$. Then, by [1,
Proposition 7.2.11], we have
$\bigcap_{\mathfrak{q}\in\operatorname{Att}_{R}(N)}\mathfrak{q}=\sqrt{\operatorname{Ann}_{R}(N)},$
and so
$\big{(}\bigcap_{\mathfrak{q}\in\operatorname{Att}_{R}(N)}\mathfrak{q}\big{)}^{n}N=0$
for some positive integer $n$. Since $\operatorname{Att}_{R}(N)$ is a finite
set, $N$ is annihilated by the multiplication of finitely many maximal ideals
and so has finite length. ∎

######  Proposition 2.4.

Let $x_{1},\dots,x_{n}$ be elements of $R$ and $N$ an Artinian $R$-module.
Then the following statements are equivalent.

  1. (i)

$x_{1},\dots,x_{n}$ is a filter coregular $N$-sequence.

  2. (ii)

$(0:_{N}(x_{1},\dots,x_{i-1})R)/x_{i}(0:_{N}(x_{1},\dots,x_{i-1})R)$ has
finite length for all $1\leq i\leq n$.

  3. (iii)

$\operatorname{Att}_{R}\left((0:_{N}(x_{1},\dots,x_{i-1})R)/x_{i}(0:_{N}(x_{1},\dots,x_{i-1})R)\right)\subseteq\operatorname{Max}(R)$
for all $1\leq i\leq n$.

  4. (iv)

$x_{i}\notin\bigcup_{\mathfrak{p}\in\operatorname{Att}_{R}\left(0:_{N}(x_{1},\dots,x_{i-1})R\right)\setminus\operatorname{Max}(R)}\mathfrak{p}$
for all $1\leq i\leq n$.

###### Proof.

The statements (i), (ii) and (iii) are equivalent by Lemma 2.3. Set
$N_{i-1}=(0:_{N}(x_{1},\dots,x_{i-1})R)$ for all $1\leq i\leq n$. Then
$x_{i}\notin\bigcup_{\mathfrak{p}\in\operatorname{Att}_{R}(N_{i-1})\setminus\operatorname{Max}(R)}\mathfrak{p}$
if and only if
$\operatorname{V}\left(x_{i}R\right)\cap\operatorname{Att}_{R}(N_{i-1})\subseteq\operatorname{Max}(R)$.
Also, by [12, Proposition 2.13], we have
$\operatorname{Att}_{R}(N_{i-1}/x_{i}N_{i-1})=\operatorname{V}\left(x_{i}R\right)\cap\operatorname{Att}_{R}(N_{i-1})$,
and so the statements (iii) and (iv) are also equivalent. ∎

######  Lemma 2.5.

Let $M,N$ and $L$ be $R$-modules.

  1. (i)

If $0\rightarrow M\rightarrow N\rightarrow L\rightarrow 0$ is an exact
sequence, then
$\operatorname{Cosupp}_{R}(N)=\operatorname{Cosupp}_{R}(M)\cup\operatorname{Cosupp}_{R}(L)$.

  2. (ii)

If $M{\rightarrow}N{\rightarrow}L$ is an exact sequence, then
$\operatorname{Cosupp}_{R}(N)\subseteq\operatorname{Cosupp}_{R}(M)\cup\operatorname{Cosupp}_{R}(L)$.

###### Proof.

To proof (i) see [12, Proposition 2.9(4)]. Now, assume
$M\stackrel{{\scriptstyle f}}{{\rightarrow}}N\stackrel{{\scriptstyle
g}}{{\rightarrow}}L$ is exact. It follows from the previous part and the exact
sequence $0\rightarrow{\rm Im}f\rightarrow N\rightarrow{\rm Im}g\rightarrow 0$
that

$\operatorname{Cosupp}_{R}(N)=\operatorname{Cosupp}_{R}({\rm
Im}f)\cup\operatorname{Cosupp}_{R}({\rm Im}g).$

Again by using (i), we obtain $\operatorname{Cosupp}_{R}({\rm
Im}f)=\operatorname{Cosupp}_{R}(M/{\rm
Ker}f)\subseteq\operatorname{Cosupp}_{R}(M)$ and
$\operatorname{Cosupp}_{R}({\rm Im}g)\subseteq\operatorname{Cosupp}_{R}(L)$.
These prove (ii). ∎

######  Proposition 2.6.

Let $M,N$ be $R$-modules and $x_{1},\dots,x_{n}$ elements of $R$. Then, for
each $i\in\mathbb{N}_{0}$,

(2.1)
$\displaystyle\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{i}\left(M,(0:_{N}(x_{1},\dots,x_{n})R)\right)\right)\subseteq\left(\bigcup_{j=i}^{i+n}\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{j}\left(M,N\right)\right)\right)$
$\displaystyle\cup\left(\bigcup_{k=1}^{n}\bigcup_{j=i+2}^{i+2+n-k}\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{j}\left(M,\frac{(0:_{N}(x_{1},\dots,x_{k-1})R)}{x_{k}(0:_{N}(x_{1},\dots,x_{k-1})R)}\right)\right)\right)$
.

If, in addition, $x_{1},\dots,x_{n}$ belong to $\operatorname{Ann}_{R}(M)$,
then

(2.2)
$\displaystyle\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{i}\left(M,N\right)\right)\subseteq\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{i-n}\left(M,(0:_{N}(x_{1},\dots,x_{n})R)\right)\right)$
$\displaystyle\cup\left(\bigcup_{k=1}^{n}\
\bigcup_{j=i+1-k}^{i+2-k}\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{j}\left(M,\frac{(0:_{N}(x_{1},\dots,x_{k-1})R)}{x_{k}(0:_{N}(x_{1},\dots,x_{k-1})R)}\right)\right)\right)$
.

###### Proof.

We prove the inclusions by induction on $n$. The following commutative diagram
with exact rows

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(0:_{N}x_{1}R)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x_{1}}$$\scriptstyle{x_{1}}$$\scriptstyle{x_{1}}$$\textstyle{x_{1}N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{x_{1}N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\textstyle{N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N/x_{1}N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

induces the following commutative diagram

$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{i}(0:_{N}x_{1}R)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{i}(N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x_{1}^{(i)}}$$\scriptstyle{x_{1}^{(i)}}$$\scriptstyle{x_{1}}$$\textstyle{T_{i}(x_{1}N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{i-1}(0:_{N}x_{1}R)\cdots}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{i+1}(N/x_{1}N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{i}(x_{1}N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{i}}$$\textstyle{T_{i}(N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{i}(N/x_{1}N)\cdots}$

with exact rows, where
$T_{i}(\cdot)=\operatorname{Tor}^{R}_{i}\left(M,\cdot\right)$ and
$x_{1}^{(i)}=\operatorname{Tor}^{R}_{i}\left({\rm id}_{M},x_{1}\right)$. Lemma
2.5(ii) implies that

$\displaystyle\operatorname{Cosupp}_{R}\left(T_{i}(0:_{N}x_{1}R)\right)\subseteq\operatorname{Cosupp}_{R}\left(T_{i}(N)\right)\cup\operatorname{Cosupp}_{R}\left(T_{i+1}(x_{1}N)\right)$
$\displaystyle\subseteq\operatorname{Cosupp}_{R}\left(T_{i}(N)\right)\cup\operatorname{Cosupp}_{R}\left(T_{i+1}(N)\right)\cup\operatorname{Cosupp}_{R}\left(T_{i+2}(N/x_{1}N)\right)$

for all $i\in\mathbb{N}_{0}$. This proves the inclusion (2.1) in the case when
$n=1$. Now assume, inductively, that $n>1$ and inclusion (2.1) holds for
smaller values of $n$. If we replace $N$ by $(0:_{N}x_{1}R)$, then, by the
inductive hypothesis for elements $x_{2},\dots,x_{n}$, we have

$\displaystyle\operatorname{Cosupp}_{R}\left(T_{i}{(0:_{N}(x_{1},\dots,x_{n})R)}\right)\subseteq\left(\bigcup_{j=i}^{i+n-1}\operatorname{Cosupp}_{R}\left(T_{j}{(0:_{N}x_{1}R)}\right)\right)$
$\displaystyle\cup\left(\bigcup_{k=2}^{n}\bigcup_{j=i+2}^{i+2+n-k}\operatorname{Cosupp}_{R}\left(T_{j}\left(\frac{(0:_{N}(x_{1},\dots,x_{k-1})R)}{x_{k}(0:_{N}(x_{1},\dots,x_{k-1})R)}\right)\right)\right)$

(note that if we set $y_{1}=x_{2},\dots,y_{n-1}=x_{n}$ and $l=k-1$, then
$1\leq l\leq n-1$ and $i+2\leq j\leq i+2+n-1-l$ yield $2\leq k\leq n$ and
$i+2\leq j\leq i+2+n-k$). Now, we combine this inclusion with the previous
inclusion to obtain (2.1).

Now assume $x_{j}M=0$ for all $1\leq j\leq n$, and we prove, by induction on
$n$, that inclusion (2.2) holds. Thus, the multiplication by $x_{j}$ on
$T_{i}(N)$ is also zero for all $i\in\mathbb{N}_{0}$ and all $1\leq j\leq n$.
The triangle in above diagram commutes, and so
$\operatorname{Im}x_{1}^{(i)}\subseteq\operatorname{Ker}f_{i}$ for all
$i\in\mathbb{N}_{0}$. Therefore Lemma 2.5 implies

$\operatorname{Cosupp}_{R}\left(\operatorname{Im}x^{(i)}_{1}\right)\subseteq\operatorname{Cosupp}_{R}\left(\operatorname{Ker}f_{i}\right)\subseteq\operatorname{Cosupp}_{R}\left(T_{i+1}(N/x_{1}N)\right).$

Again by Lemma 2.5(ii), we have from the exactness of rows

$\operatorname{Cosupp}_{R}\left(T_{i}(x_{1}N)\right)\subseteq\operatorname{Cosupp}_{R}\left(\operatorname{Im}x^{(i)}_{1}\right)\cup\operatorname{Cosupp}_{R}\left(T_{i-1}(0:_{N}x_{1})\right)$

and

$\operatorname{Cosupp}_{R}\left(T_{i}(N)\right)\subseteq\operatorname{Cosupp}_{R}\left(T_{i}(x_{1}N)\right)\cup\operatorname{Cosupp}_{R}\left(T_{i}(N/x_{1}N)\right).$

These inclusions yield

$\displaystyle\operatorname{Cosupp}_{R}\left(T_{i}(N)\right)\subseteq\operatorname{Cosupp}_{R}\left(T_{i-1}(0:_{N}x_{1}R)\right)\cup\operatorname{Cosupp}_{R}\left(T_{i}\left({N}/{x_{1}N}\right)\right)$
$\displaystyle\cup\operatorname{Cosupp}_{R}\left(T_{i+1}\left({N}/{x_{1}N}\right)\right)$
.

Hence, inclusion (2.2) holds in the case when $n=1$. Next, suppose,
inductively, that $n>1$ and the desired inclusion has been proved for smaller
values of $n$. If we use $(0:_{N}x_{1}R)$ and $i-1$ instead of $N$ and $i$
respectively, then the inductive hypothesis for elements $x_{2},\dots,x_{n}$
implies that

$\displaystyle\operatorname{Cosupp}_{R}\left(T_{i-1}(0:_{N}x_{1}R)\right)\subseteq\operatorname{Cosupp}_{R}\left(T_{i-n}{(0:_{N}(x_{1},\dots,x_{n})R)}\right)$
$\displaystyle\cup\left(\bigcup_{k=2}^{n}\
\bigcup_{j=i+1-k}^{i+2-k}\operatorname{Cosupp}_{R}\left(T_{j}\left({\frac{(0:_{N}(x_{1},\dots,x_{k-1})R)}{x_{k}(0:_{N}(x_{1},\dots,x_{k-1})R)}}\right)\right)\right)$
.

Combining two last inclusions yield (2.2), and this completes the inductive
step. ∎

######  Corollary 2.7.

Let $M,N$ be two $R$-modules and $x_{1},\dots,x_{n}$ a filter coregular
$N$-sequence in $\operatorname{Ann}_{R}(M)$. Then

(2.3)
$\displaystyle\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{i}\left(M,N\right)\right)\subseteq\operatorname{Max}(R)$

for all $i<n$, and

(2.4)
$\displaystyle\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{n}\left(M,N\right)\right)\cup\operatorname{Max}(R)=$
$\displaystyle\operatorname{Cosupp}_{R}\left(M\otimes_{R}(0:_{N}(x_{1},\dots,x_{n})R)\right)\cup\operatorname{Max}(R).$

###### Proof.

For each $1\leq k\leq n$, since
$\operatorname{Cosupp}_{R}\left(\frac{(0:_{N}(x_{1},\dots,x_{k-1})R)}{x_{k}(0:_{N}(x_{1},\dots,x_{k-1})R)}\right)\subseteq\operatorname{Max}(R)$,
we have

(†)
$\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{i}\left(M,\frac{(0:_{N}(x_{1},\dots,x_{k-1})R)}{x_{k}(0:_{N}(x_{1},\dots,x_{k-1})R)}\right)\right)\subseteq\operatorname{Max}(R)$

for all $i\in\mathbb{N}_{0}$. Hence inclusion (2.3) is an immediate
consequence of inclusion (2.2).

Now, we prove the equation (2.4). If set $i=0$ in (2.1), then it follows from
(${\dagger}$) and (2.3) that

$\operatorname{Cosupp}_{R}\left(M\otimes_{R}(0:_{N}(x_{1},\dots,x_{n})R)\right)\subseteq\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{n}\left(M,N\right)\right)\cup\operatorname{Max}(R).$

Conversely, if set $i=n$ in (2.2), then, by (${\dagger}$), we have

$\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{n}\left(M,N\right)\right)\subseteq\operatorname{Cosupp}_{R}\left(M\otimes_{R}(0:_{N}(x_{1},\dots,x_{n})R)\right)\cup\operatorname{Max}(R).$

These inclusions yield the equation (2.4). ∎

Let $\mathfrak{a}$ be an ideal of $R$ and let $M,N$ be two $R$-modules. Let
$n$ be a fixed integer, and for each $i$, we set
$X_{i}=\operatorname{Tor}^{R}_{n}\left(M/\mathfrak{a}^{i}M,N\right)$. If
$i\leq j$, then the natural homomorphism
$\pi_{ij}:{M/\mathfrak{a}^{j}M}\rightarrow{M/\mathfrak{a}^{i}M}$ induces the
homomorphism $\varphi_{ij}:X_{j}\rightarrow X_{i}$. Since
$\\{M/\mathfrak{a}^{i}M,\,\pi_{ij}:{M/\mathfrak{a}^{j}M}\rightarrow{M/\mathfrak{a}^{i}M}\\}_{i\leq
j}$ is an inverse system, $\\{X_{i},\,\varphi_{ij}:X_{j}\rightarrow
X_{i}\\}_{i\leq j}$ is also an inverse system. We define

$X=\\{(x_{i})_{i}\in\prod_{i}X_{i}:x_{i}=\varphi_{ij}(x_{j})\textrm{ whenever
}i\leq j\\}$

and, for each $i$, assume $\varphi_{i}:X\rightarrow X_{i}$ is the restriction
of natural projection $p_{i}:\prod_{i}X_{i}\rightarrow X_{i}$ to $X$. Then
$\\{X,\,\varphi_{i}:X\rightarrow X_{i}\\}_{i}$ is the inverse limit of the
inverse system $\\{X_{i},\,\varphi_{ij}:X_{j}\rightarrow X_{i}\\}_{i\leq j}$
(see [13, Chapter 2] for more details about the inverse system and inverse
limit). This inverse limit is called $n$-th generalized local homology of
$M,N$ with respect to $\mathfrak{a}$ and is denoted by
$\operatorname{H}^{\mathfrak{a}}_{n}(M,N)$; see [2, 3, 10, 11] for more
details and basic properties. Now, we show that the $R$-module $X$ also has an
$\widehat{R}^{\mathfrak{a}}$-module structure, where

$\widehat{R}^{\mathfrak{a}}=\\{(r_{i}+\mathfrak{a}^{i})_{i}\in\prod_{i}R/\mathfrak{a}^{i}:r_{j}-r_{i}\in\mathfrak{a}^{i}\textrm{
whenever }i\leq j\\}$

denotes the $\mathfrak{a}$-adic completion of $R$. For each
$(r_{i}+\mathfrak{a}^{i})_{i}\in\widehat{R}^{\mathfrak{a}}$ and
$(x_{i})_{i}\in X$, we define

$(r_{i}+\mathfrak{a}^{i})_{i}(x_{i})_{i}=(r_{i}x_{i})_{i}.$

First note that, since $\mathfrak{a}^{i}X_{i}=0$ for all $i$, in view of the
definitions of $\widehat{R}^{\mathfrak{a}}$ and $X$, we have

$\displaystyle\varphi_{ij}(r_{j}x_{j})-r_{i}x_{i}=r_{j}\varphi_{ij}(x_{j})-r_{i}x_{i}=r_{j}(\varphi_{ij}(x_{j})-x_{i})+(r_{j}-r_{i})x_{i}=0$

for all $i\leq j$. Therefore $(r_{i}x_{i})_{i}\in X$. Also, for each other
element $(r^{\prime}_{i}+\mathfrak{a}^{i})_{i}$ in
$\widehat{R}^{\mathfrak{a}}$, if
$(r_{i}+\mathfrak{a}^{i})_{i}=(r^{\prime}_{i}+\mathfrak{a}^{i})_{i}$, then we
have $r_{i}-r^{\prime}_{i}\in\mathfrak{a}^{i}$, and so
$r_{i}x_{i}=r^{\prime}_{i}x_{i}$. Therefore the above definition is well-
defined and it is easy to see that $X$ is an
$\widehat{R}^{\mathfrak{a}}$-module with this multiplication. Finally, let
$\alpha:R\rightarrow\widehat{R}^{\mathfrak{a}}$ be the natural homomorphism of
rings given by $\alpha(r)=\widehat{r}=(r+\mathfrak{a}^{i})_{i}$. Since $X$ has
an $\widehat{R}^{\mathfrak{a}}$-module structure, it has an $R$-module
structure via $\alpha$. But, if $r\in R$ and $x=(x_{i})_{i}\in X$, then we
have

$r\cdot
x=\alpha(r)x=(r+\mathfrak{a}^{i})_{i}(x_{i})_{i}=(rx_{i})_{i}=r(x_{i})_{i}=rx.$

Therefore this $R$-module structure of $X$ coincides with the original
$R$-module structure of $X$.

######  Lemma 2.8 ([11, Proposition 2.3]).

Let $\mathfrak{a}$ be an ideal of $R$. For each $i\in\mathbb{N}_{0}$ and
$R$-modules $M$ and $N$, the generalized local homology
$\operatorname{H}^{\mathfrak{a}}_{i}(M,N)$ is an $\mathfrak{a}$-separated
$R$-module, i.e.,

$\bigcap_{t\in\mathbb{N}}\mathfrak{a}^{t}\operatorname{H}^{\mathfrak{a}}_{i}(M,N)=0.$

######  Lemma 2.9 ([9, Theorem 8.4]).

Let $\mathfrak{a}$ be an ideal of $R$ such that $R$ is complete with respect
to the $\mathfrak{a}$-adic topology and $M$ an $R$-module. If
$M/\mathfrak{a}M$ is a finitely generated $R$-module and $M$ is
$\mathfrak{a}$-separated (i.e.,
$\bigcap_{t\in\mathbb{N}}\mathfrak{a}^{t}M=0$), then $M$ is a finitely
generated $R$-module.

Now, we are ready to state and prove the main result of this paper. Assume
$\mathfrak{a}$ is an ideal of $R$, $M$ is a finitely generated $R$-module and
$N$ is an Artinian $R$-module. Among the other things, we show that the least
integer $i$ such that the local homology modules
$\operatorname{H}^{\mathfrak{a}}_{i}\left(N\right)$ is not Artinian as an
$\widehat{R}^{\mathfrak{a}}$-module is equal by the length of maximal filter
coregular $N$-sequences in $\mathfrak{a}$.

######  Theorem 2.10.

Let $\mathfrak{a}$ be an ideal of $R$, $M$ a finitely generated $R$-module and
$N$ an Artinian $R$-module. Then, for each $n\in\mathbb{N}$, the following
statements are equivalent.

  1. (i)

There is a filter coregular $N$-sequence in $\mathfrak{a}$ of length $n$.

  2. (ii)

Any filter coregular $N$-sequence in $\mathfrak{a}$ of length less than $n$
can be extended to a filter coregular $N$-sequence in $\mathfrak{a}$ of length
$n$.

  3. (iii)

$\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{i}\left(R/\mathfrak{a},N\right)\right)\subseteq\operatorname{Max}(R)$
for all $i<n$.

  4. (iv)

If $\operatorname{Supp}_{R}(M)=\operatorname{V}\left(\mathfrak{a}\right)$,
then
$\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{i}\left(M,N\right)\right)\subseteq\operatorname{Max}(R)$
for all $i<n$.

  5. (v)

If $\operatorname{Supp}_{R}(M)=\operatorname{V}\left(\mathfrak{a}\right)$,
then $\operatorname{Tor}^{R}_{i}\left(M,N\right)$ is finitely generated for
all $i<n$.

  6. (vi)

If $\operatorname{Ann}_{R}(M)\subseteq\mathfrak{a}$, then
$\operatorname{H}^{\mathfrak{a}}_{i}(M,N)$ is a finitely generated
$\widehat{R}^{\mathfrak{a}}$-module for all $i<n$.

###### Proof.

(iii)$\Rightarrow$(ii). Assume that
$\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{i}\left(R/\mathfrak{a},N\right)\right)\subseteq\operatorname{Max}(R)$
for all $i<n$, and suppose, for the sake of contradiction, that
$x_{1},\dots,x_{m}$ is a maximal filter coregular $N$-sequence in
$\mathfrak{a}$ of length $0\leq m<n$. The maximality of $x_{1},\dots,x_{m}$
yields

$\mathfrak{a}\subseteq\bigcup_{\mathfrak{p}\in\operatorname{Att}_{R}\left(0:_{N}(x_{1},\dots,x_{m})R\right)\setminus\operatorname{Max}(R)}\mathfrak{p}.$

Since $\operatorname{Att}_{R}\left(0:_{N}(x_{1},\dots,x_{m})R\right)$ is a
finite set, it follows from the Prime Avoidance Theorem that
$\mathfrak{a}\subseteq\mathfrak{p}$ for some
$\mathfrak{p}\in\operatorname{Att}_{R}\left(0:_{N}(x_{1},\dots,x_{m})R\right)\setminus\operatorname{Max}(R)$.
Hence, by equation (2.4) in Corollary 2.7 and hypothesis, we have

$\displaystyle\mathfrak{p}\in\operatorname{V}\left(\mathfrak{a}\right)\cap\operatorname{Att}_{R}(0:_{N}(x_{1},\dots,x_{m})R)$
$\displaystyle=\operatorname{Att}_{R}(R/\mathfrak{a}\otimes_{R}(0:_{N}(x_{1},\dots,x_{m})R))$
$\displaystyle\subseteq\operatorname{Cosupp}_{R}(R/\mathfrak{a}\otimes_{R}(0:_{N}(x_{1},\dots,x_{m})R))$
$\displaystyle\subseteq\operatorname{Cosupp}_{R}(\operatorname{Tor}^{R}_{m}\left(R/\mathfrak{a},N\right))\cup\operatorname{Max}(R)$
$\displaystyle\subseteq\operatorname{Max}(R),$

which is a contradiction.

(ii)$\Rightarrow$(iv). We argue by induction on $n$. Assume that
$\operatorname{Supp}_{R}(M)=\operatorname{V}\left(\mathfrak{a}\right)$ and any
filter coregular $N$-sequence in $\mathfrak{a}$ of length less than $n$ can be
extended to a filter coregular $N$-sequence in $\mathfrak{a}$ of length $n$.
If $n=1$, then, by hypothesis, there exists $x_{1}\in\mathfrak{a}$ such that
$\operatorname{Cosupp}_{R}(N/x_{1}N)\subseteq\operatorname{Max}(R)$. Since
$\operatorname{Supp}_{R}(M)=\operatorname{V}\left(\mathfrak{a}\right)$, we
have $x_{1}^{t}\in\operatorname{Ann}_{R}(M)$ for some $t\in\mathbb{N}$. Also,
by [12, Proposition 2.13], we have

$\operatorname{Att}_{R}(N/x^{t}_{1}N)=\operatorname{Att}_{R}(N/x_{1}N)\subseteq\operatorname{Max}(R).$

Thus $x^{t}_{1}$ is a filter coregular $N$-sequence in
$\operatorname{Ann}_{R}(M)$, and so
$\operatorname{Cosupp}_{R}(M\otimes_{R}N)\subseteq\operatorname{Max}(R)$ by
inclusion (2.3) in Corollary 2.7. Now assume, inductively, that $n>1$ and the
result has been proved for smaller values of $n$. Set
$T_{i}(\cdot)=\operatorname{Tor}^{R}_{i}\left(M,\cdot\right)$ for all
$i\in\mathbb{N}_{0}$. By hypothesis, we can extend the filter coregular
$N$-sequence $x^{t}_{1}$ to a filter coregular $N$-sequence
$x^{t}_{1},x_{2},\dots,x_{n}$ in $\mathfrak{a}$. Hence $x_{2},\dots,x_{n}$ is
a filter coregular $(0:_{N}x^{t}_{1}R)$-sequence in $\mathfrak{a}$ of length
$n-1$. Therefore, by the inductive hypothesis,

$\operatorname{Cosupp}_{R}(T_{i}{(0:_{N}x^{t}_{1}R)})\subseteq\operatorname{Max}(R)$

for all $i<n-1$. Also, since $x^{t}_{1}M=0$, inclusion (2.2) in Proposition
2.6 yields

$\displaystyle\operatorname{Cosupp}_{R}\left(T_{i}(N)\right)\subseteq\operatorname{Cosupp}_{R}\left(T_{i-1}(0:_{N}x^{t}_{1}R)\right)\cup\operatorname{Cosupp}_{R}\left(T_{i}\left({N}/{x^{t}_{1}N}\right)\right)$
$\displaystyle\cup\operatorname{Cosupp}_{R}\left(T_{i+1}\left({N}/{x^{t}_{1}N}\right)\right)$

(set $n=1$ in inclusion (2.2)). By the definition of filter coregular
$N$-sequences,
$\operatorname{Cosupp}_{R}\left({N}/{x^{t}_{1}N}\right)\subseteq\operatorname{Max}(R)$,
and hence
$\operatorname{Cosupp}_{R}\left(T_{i}\left({N}/{x^{t}_{1}N}\right)\right)\subseteq\operatorname{Max}(R)$
for all $i\in\mathbb{N}_{0}$. Therefore the above inclusions yield
$\operatorname{Cosupp}_{R}\left(T_{i}(N)\right)\subseteq\operatorname{Max}(R)$
for all $i<n$.

The implication (iv)$\Rightarrow$(iii) is clear, and so (ii), (iii) and (iv)
are equivalent. Also, (i)$\Rightarrow$(iii) is an immediate consequence of
inclusion (2.3) in Corollary 2.7, and (ii)$\Rightarrow$(i) is clear. Hence the
statements (i), (ii), (iii) and (iv) are equivalent.

(iv)$\Leftrightarrow$(v). Since $M$ is finitely generated and $N$ is Artinian,
$\operatorname{Tor}^{R}_{i}\left(M,N\right)$ is Artinian for all
$i\in\mathbb{N}_{0}$. Hence, for each $i\in\mathbb{N}_{0}$,
$\operatorname{Tor}^{R}_{i}\left(M,N\right)$ is finitely generated (or
equivalently has finite length) if and only if
$\operatorname{Att}_{R}(\operatorname{Tor}^{R}_{i}\left(M,N\right))\subseteq\operatorname{Max}(R)$;
see Lemma 2.3.

(i)$\Leftrightarrow$(vi). Assume that
$\operatorname{Ann}_{R}(M)\subseteq\mathfrak{a}$. We prove, by induction on
$n$, that (i) and (vi) are equivalent. We first assume that $n=1$. Since
$M\otimes_{R}N$ is Artinian, we have

$\operatorname{H}^{\mathfrak{a}}_{0}(M,N)\cong\Lambda_{\mathfrak{a}}(M\otimes_{R}N)\cong(M\otimes_{R}N)/\mathfrak{a}^{s}(M\otimes_{R}N)\cong\operatorname{Tor}^{R}_{0}\left(M/\mathfrak{a}^{s}M,N\right)$

for sufficiently large integer $s$. Also, since
$\operatorname{Supp}_{R}(M/\mathfrak{a}^{s}M)=\operatorname{V}\left(\mathfrak{a}\right)$,
the equivalence of (i) and (v) implies that
$\operatorname{H}^{\mathfrak{a}}_{0}(M,N)$ is a finitely generated $R$-module
or equivalently it is a finitely generated $\widehat{R}^{\mathfrak{a}}$-module
if and only if $\mathfrak{a}$ contains a filter coregular element on $N$ (note
that since $\operatorname{H}^{\mathfrak{a}}_{0}(M,N)$ is
$\mathfrak{a}$-torsion, its submodules as an $R$-module and as an
$\widehat{R}^{\mathfrak{a}}$-module are same; see [5, Lemma 1.3]). Now,
assume, inductively, that $n>1$ and the result has been proved for smaller
values of $n$. Since $N$ ia Artinian, there exists $t\in\mathbb{N}$ such that
$\mathfrak{a}^{t}N=\mathfrak{a}^{s}N$ for all $s\geq t$. Thus we have
$\Lambda_{\mathfrak{a}}(N)\cong N/\mathfrak{a}^{t}N$. Assume either (i) or
(vi) holds. Since $n>1$, it follows from the equivalence of (i) and (v), and
also the equivalence of (vi) and (v) in the case $n=1$ that
$\Lambda_{\mathfrak{a}}(N)\cong\operatorname{Tor}^{R}_{0}\left(R/\mathfrak{a}^{t},N\right)$
is a finitely generated $R$-module. Now, the exact sequence

$0\rightarrow\mathfrak{a}^{t}N\rightarrow
N\rightarrow\Lambda_{\mathfrak{a}}(N)\rightarrow 0$

of Artinian $R$-modules induces the long exact sequences

$\displaystyle\cdots\rightarrow\operatorname{H}^{\mathfrak{a}}_{i+1}(M,\Lambda_{\mathfrak{a}}(N))\rightarrow\operatorname{H}^{\mathfrak{a}}_{i}(M,\mathfrak{a}^{t}N)\rightarrow\operatorname{H}^{\mathfrak{a}}_{i}(M,N)\rightarrow\operatorname{H}^{\mathfrak{a}}_{i}(M,\Lambda_{\mathfrak{a}}(N))\rightarrow\cdots$

(see [11, Proposition 2.4]) and

$\displaystyle\cdots$
$\displaystyle\rightarrow\operatorname{Tor}^{R}_{i+1}\left(R/\mathfrak{a},\Lambda_{\mathfrak{a}}(N)\right)\rightarrow\operatorname{Tor}^{R}_{i}\left(R/\mathfrak{a},\mathfrak{a}^{t}N\right)\rightarrow\operatorname{Tor}^{R}_{i}\left(R/\mathfrak{a},N\right)$
$\displaystyle\rightarrow\operatorname{Tor}^{R}_{i}\left(R/\mathfrak{a},\Lambda_{\mathfrak{a}}(N)\right)\rightarrow\cdots.$

Since $\Lambda_{\mathfrak{a}}(N)$ is Artinian, it is linearly compact by [7,
3.10], and so [10, Theorem 3.6] yields

$\operatorname{H}^{\mathfrak{a}}_{i}(M,\Lambda_{\mathfrak{a}}(N))\cong
H_{i}\left(\Lambda_{\mathfrak{a}}(M\otimes_{R}F_{\bullet})\right),$

where $F_{\bullet}$ is a free resolution of $\Lambda_{\mathfrak{a}}(N)$. By
hypothesis, $\Lambda_{\mathfrak{a}}(N)$ is finitely generated, and so we can
assume that every component of $F_{\bullet}$ is finitely generated. Also,
$\Lambda_{\mathfrak{a}}(\cdot)$ is exact on the category of finitely generated
$R$-modules, and hence it commutes with the homological functor in this
category. Therefore

$\operatorname{H}^{\mathfrak{a}}_{i}(M,\Lambda_{\mathfrak{a}}(N))\cong\Lambda_{\mathfrak{a}}\left(H_{i}\left(M\otimes_{R}F_{\bullet}\right)\right)\cong\Lambda_{\mathfrak{a}}\left(\operatorname{Tor}^{R}_{i}\left(M,\Lambda_{\mathfrak{a}}(N)\right)\right).$

Since $\operatorname{Tor}^{R}_{i}\left(M,\Lambda(N)\right)$ is Artinian, we
obtain

$\Lambda_{\mathfrak{a}}\left(\operatorname{Tor}^{R}_{i}\left(M,\Lambda_{\mathfrak{a}}(N)\right)\right)\cong\operatorname{Tor}^{R}_{i}\left(M,\Lambda_{\mathfrak{a}}(N)\right)/\mathfrak{a}^{r}\operatorname{Tor}^{R}_{i}\left(M,\Lambda_{\mathfrak{a}}(N)\right)$

for sufficiently large integer $r$. Hence
$\operatorname{H}^{\mathfrak{a}}_{i}(M,\Lambda_{\mathfrak{a}}(N))$ is a
finitely generated $R$-module for all $i\in\mathbb{N}_{0}$. Also, the above
isomorphism shows that
$\operatorname{H}^{\mathfrak{a}}_{i}(M,\Lambda_{\mathfrak{a}}(N))$ is
$\mathfrak{a}$-torsion, and so
$\operatorname{H}^{\mathfrak{a}}_{i}(M,\Lambda_{\mathfrak{a}}(N))$ is a
finitely generated $\widehat{R}^{\mathfrak{a}}$-module for all
$i\in\mathbb{N}_{0}$ by [5, Lemma 1.3]. Now, for each $i\in\mathbb{N}_{0}$, it
follows from the first long exact sequence that
$\operatorname{H}^{\mathfrak{a}}_{i}(M,N)$ is a finitely generated
$\widehat{R}^{\mathfrak{a}}$-module if and only if
$\operatorname{H}^{\mathfrak{a}}_{i}(M,\mathfrak{a}^{t}N)$ is a finitely
generated $\widehat{R}^{\mathfrak{a}}$-module. Also, for each
$i\in\mathbb{N}_{0}$, it follows from the second long exact sequence that
$\operatorname{Tor}^{R}_{i}\left(R/\mathfrak{a},N\right)$ is finitely
generated if and only if
$\operatorname{Tor}^{R}_{i}\left(R/\mathfrak{a},\mathfrak{a}^{t}N\right)$ is
finitely generated. Thus, to prove the equivalence of (i) and (vi), in view of
the equivalence of (i) and (v), we can replace $N$ by $\mathfrak{a}^{t}N$ and
assume, in addition, that $\mathfrak{a}N=N$. Therefore
$\operatorname{V}\left(\mathfrak{a}\right)\cap\operatorname{Att}_{R}(N)=\emptyset$,
and so
$\mathfrak{a}\nsubseteq\bigcup_{\mathfrak{p}\in\operatorname{Att}_{R}(N)}\mathfrak{p}$.
Let
$x_{1}\in\mathfrak{a}\setminus\bigcup_{\mathfrak{p}\in\operatorname{Att}_{R}(N)}\mathfrak{p}$.
Then
$\operatorname{V}\left(x_{1}R\right)\cap\operatorname{Att}_{R}(N)=\emptyset$,
and so $N=x_{1}N$. The exact sequence

$0\rightarrow(0:_{N}x_{1}R)\rightarrow N\stackrel{{\scriptstyle
x_{1}}}{{\longrightarrow}}N\rightarrow 0$

induces the long exact sequence

$\cdots\rightarrow\operatorname{H}^{\mathfrak{a}}_{i+1}(M,N)\stackrel{{\scriptstyle
x_{1}}}{{\longrightarrow}}\operatorname{H}^{\mathfrak{a}}_{i+1}(M,N)\rightarrow\operatorname{H}^{\mathfrak{a}}_{i}(M,(0:_{N}x_{1}R))\rightarrow\cdots.$

We first assume that (i) holds. By the equivalence of (i) and (ii), we can
extend $x_{1}$ to a filter coregular $N$-sequence of length $n$, say
$x_{1},x_{2},\dots,x_{n}$. Hence $x_{2},\dots,x_{n}$ is a filter coregular
$(0:_{N}x_{1}R)$-sequence in $\mathfrak{a}$, and so, by the inductive
hypothesis, $\operatorname{H}^{\mathfrak{a}}_{i}(M,(0:_{N}x_{1}R))$ is a
finitely generated $\widehat{R}^{\mathfrak{a}}$-module for all $i<n-1$. It
follows from the above long exact sequence that
$\operatorname{H}^{\mathfrak{a}}_{i}(M,N)/x_{1}\operatorname{H}^{\mathfrak{a}}_{i}(M,N)$
and consequently its quotient
$\operatorname{H}^{\mathfrak{a}}_{i}(M,N)/(\mathfrak{a}\widehat{R}^{\mathfrak{a}})\operatorname{H}^{\mathfrak{a}}_{i}(M,N)$
is a finitely generated $\widehat{R}^{\mathfrak{a}}$-module for all $i<n$.
Also, by Lemma 2.8, we have

$\bigcap_{t\in\mathbb{N}}(\mathfrak{a}\widehat{R}^{\mathfrak{a}})^{t}\operatorname{H}^{\mathfrak{a}}_{i}(M,N)=\bigcap_{t\in\mathbb{N}}({\mathfrak{a}}^{t}\widehat{R}^{\mathfrak{a}})\operatorname{H}^{\mathfrak{a}}_{i}(M,N)=\bigcap_{t\in\mathbb{N}}{\mathfrak{a}}^{t}\operatorname{H}^{\mathfrak{a}}_{i}(M,N)=0.$

Hence, by Lemma 2.9, $\operatorname{H}^{\mathfrak{a}}_{i}(M,N)$ is a finitely
generated $\widehat{R}^{\mathfrak{a}}$-module for all $i<n$. Conversely,
assume that $\operatorname{H}^{\mathfrak{a}}_{i}(M,N)$ is a finitely generated
$\widehat{R}^{\mathfrak{a}}$-module for all $i<n$. It follows from the above
long exact sequence that
$\operatorname{H}^{\mathfrak{a}}_{i}(M,(0:_{N}x_{1}R))$ is a finitely
generated $\widehat{R}^{\mathfrak{a}}$-module for all $i<n-1$, and so, by the
inductive hypothesis, there is a filter coregular $(0:_{N}x_{1}R)$-sequence in
$\mathfrak{a}$ of length $n-1$, say $x_{2},\dots,x_{n}$. Therefore
$x_{1},x_{2},\dots,x_{n}$ is a filter coregular $N$-sequence. This completes
the inductive step. ∎

######  Remark 2.11.

Let $\mathfrak{a}$ be an ideal of $R$ and $N$ an Artinian $R$-module. If there
exists a filter coregular $N$-sequence in $\mathfrak{a}$ of infinite length,
then, by the equivalence of the statements (i) and (ii) in Theorem 2.10, any
filter coregular $N$-sequence in $\mathfrak{a}$ can be extended to a filter
coregular $N$-sequence in $\mathfrak{a}$ of arbitrary length, and in this case
we set $\operatorname{f-width}(\mathfrak{a},N)=\infty$. Now assume all filter
coregular $N$-sequences in $\mathfrak{a}$ have finite length. Again, by the
equivalence of the statements (i) and (ii) in Theorem 2.10, we can extend any
filter coregular $N$-sequence in $\mathfrak{a}$ to a maximal one, and all
maximal filter coregular $N$-sequences in $\mathfrak{a}$ are of the same
length which we denote this common length by
$\operatorname{f-width}(\mathfrak{a},N)$. Moreover, if $M$ is a finitely
generated $R$-module such that
$\operatorname{Supp}_{R}(M)=\operatorname{V}\left(\mathfrak{a}\right)$, then,
by Theorem 2.10, we have

(2.5) $\displaystyle\operatorname{f-width}(\mathfrak{a},N)$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{Cosupp}_{R}\left(\operatorname{Tor}^{R}_{i}\left(M,N\right)\right)\nsubseteq\operatorname{Max}(R)\\}$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{Tor}^{R}_{i}\left(M,N\right)\textrm{
is not a finitely generated }R\textrm{-module}\\}$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{H}^{\mathfrak{a}}_{i}\left(N\right)\textrm{
is not a finitely generated }\widehat{R}^{\mathfrak{a}}\textrm{-module}\\}$

(we note that
$\operatorname{H}^{\mathfrak{a}}_{i}(R,N)=\operatorname{H}^{\mathfrak{a}}_{i}\left(N\right)$
and the statements of Theorem 2.10 are independence of the choice of $M$).
Also, for an arbitrary finitely generated $R$-module $L$, since
$\operatorname{H}^{\mathfrak{a}+\operatorname{Ann}_{R}(L)}_{i}(L,N)\cong\operatorname{H}^{\mathfrak{a}}_{i}(L,N)$,
if we replace $\mathfrak{a}$ by $\mathfrak{a}+\operatorname{Ann}_{R}(L)$ in
Theorem 2.10, then the equivalence of (ii) and (vi) in Theorem 2.10 yields

(2.6)
$\displaystyle\operatorname{f-width}(\mathfrak{a}+\operatorname{Ann}_{R}(L),N)$
$\displaystyle=\inf\\{i\in\mathbb{N}_{0}:\operatorname{H}^{\mathfrak{a}}_{i}(L,N)\textrm{
is not a finitely generated }\widehat{R}^{\mathfrak{a}}\textrm{-module}\\}.$

Finally, since
$\operatorname{V}\left(\mathfrak{a}\right)=\operatorname{V}\left(\sqrt{\mathfrak{a}}\right)$,
by equation (2.5), we obtain
$\operatorname{f-width}(\mathfrak{a},N)=\operatorname{f-width}(\sqrt{\mathfrak{a}},N).$

######  Proposition 2.12.

Let $\mathfrak{a}$ be an ideal of $R$ and $N$ an Artinian $R$-module. If
$\operatorname{f-width}(\mathfrak{a},N)=\infty$, then $(0:_{N}\mathfrak{a})$
has finite length; and the converse statements also hold whenever $R$ is a
semi-local ring which is complete with respect to its Jacobson radical.

###### Proof.

We first assume that $\operatorname{f-width}(\mathfrak{a},N)=\infty$, and
$x_{1},x_{2},x_{3},\dots$ is a filter coregular $N$-sequence of infinite
length in $\mathfrak{a}$. There is the following descending chain of
submodules of $N$

$(0:_{N}x_{1}R)\supseteq(0:_{N}(x_{1},x_{2})R)\supseteq(0:_{N}(x_{1},x_{2},x_{3})R)\supseteq\cdots.$

Hence, $(0:_{N}(x_{1},\dots,x_{n-1})R)=(0:_{N}(x_{1},\dots,x_{n})R)$ for some
$n\in\mathbb{N}$, and so $x_{n}(0:_{N}(x_{1},\dots,x_{n-1})R)=0$. It follows
that $(0:_{N}(x_{1},\dots,x_{n-1})R)$ has finite length because
$(0:_{N}(x_{1},\dots,x_{n-1})R)/x_{n}(0:_{N}(x_{1},\dots,x_{n-1})R)$ has
finite length by definition. Therefore
$(0:_{N}\mathfrak{a})\subseteq(0:_{N}(x_{1},\dots,x_{n-1})R)$ has finite
length.

Conversely, assume that $R$ is a complete semi-local ring and that
$(0:_{N}\mathfrak{a})$ has finite length. Hence

$\operatorname{Cosupp}_{R}(0:_{N}\mathfrak{a})=\operatorname{Supp}_{R}(0:_{N}\mathfrak{a})\subseteq\operatorname{Max}(R).$

On the other hand, for each $i\in\mathbb{N}_{0}$,
$\mathfrak{a}+\operatorname{Ann}_{R}(N)\subseteq\operatorname{Ann}_{R}\operatorname{Tor}^{R}_{i}\left(R/\mathfrak{a},N\right)$.
Therefore, in view of [12, Proposition 2.12], we have

$\displaystyle\operatorname{Cosupp}_{R}(\operatorname{Tor}^{R}_{i}\left(R/\mathfrak{a},N\right))$
$\displaystyle\subseteq\operatorname{V}\left(\mathfrak{a}+\operatorname{Ann}_{R}(N)\right)$
$\displaystyle=\operatorname{V}\left(\mathfrak{a}\right)\cap\operatorname{Cosupp}_{R}(N)$
$\displaystyle=\operatorname{Cosupp}_{R}(0:_{N}\mathfrak{a})\subseteq\operatorname{Max}(R)$

for all $i\in\mathbb{N}_{0}$. It follows, by Theorem 2.10, that
$\operatorname{f-width}(\mathfrak{a},N)=\infty$. ∎

######  Corollary 2.13.

Let $\mathfrak{a}$ be an ideal of $R$, $M$ a finitely generated $R$-module and
$N$ an Artinian $R$-module. Then

  1. (i)

If $\operatorname{Tor}^{R}_{i}\left(M,N\right)$ has finite length for all
$i\in\mathbb{N}_{0}$, then $(0:_{N}\operatorname{Ann}_{R}(M))$ has finite
length.

  2. (ii)

If $\operatorname{H}^{\mathfrak{a}}_{i}(M,N)$ is a finitely generated
$\widehat{R}^{\mathfrak{a}}$-module for all $i\in\mathbb{N}_{0}$, then
$(0:_{N}\mathfrak{a}+\operatorname{Ann}_{R}(M))$ has finite length. In
particular, $(0:_{N}\mathfrak{a})$ has finite length whenever
$\operatorname{H}^{\mathfrak{a}}_{i}\left(N\right)$ is a finitely generated
$\widehat{R}^{\mathfrak{a}}$-module for all $i\in\mathbb{N}_{0}$.

Moreover, when $R$ is a complete semi-local ring, then the converse statements
also hold.

###### Proof.

It follows by equations (2.5), (2.6) and Proposition 2.12. ∎

## References

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