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Former church built by stonemasons in 1917 for sale near Young for $180,000
Glynis Quinlan 31 March 2018 29 March 2018
This quaint former church built in 1917 is on the market 11 km from Young. All photos supplied by Aston & Joyce Pty Ltd.
A quaint former church built by local stonemasons in 1917 is for sale near Young for just $180,000 and offers the attractions of pressed tin ceilings, arched windows and stained timber floorboards.
Located around two hours' drive from Canberra on the 'wine trail' between Cowra and Young, the former church offers a slice of history within a lovely rural setting.
Already partly converted, the former Woodonga Uniting Church building at 1123 Olympic Highway North is at lock up stage and includes a recently constructed addition that is in keeping with the original church building.
According to listing agent Danny Joyce of stock and property agency Aston & Joyce Pty Ltd, the new owners will have the opportunity to complete the project to their own taste.
"There is potential to continue to make it a top-class dwelling or by thinking outside the square, creating a business interest such as a B&B, function centre or ecotourism," Mr Joyce said.
Development approval is in place for the residence but the new owner will need to obtain an occupancy certificate after completing the church conversion.
"The original church building features pressed tin ceilings, stained timber floorboards and arched windows, while the new addition consists of two bedrooms with provisions for built-in robes and a three-way bathroom with provision for bath and shower," Mr Joyce said.
The property is fully-fenced and includes a rainwater storage tank, a shed and secure parking.
Mr Joyce said there has been a lot of interest in the property because people are fascinated with former church buildings.
The lovely stone walls of the church building still include the original stone laid to mark the official dedication of the building by Mrs W. Robinson in November 1918.
Mr Joyce said that the 924-square metre property would have been subdivided from a farm in the early 1900s in order for the church to be built to serve the local community.
Following its decommissioning in more recent years, the church was bought and destined for a church conversion. However, the current owner became too busy with business interests to complete the work and decided to put the property on the market.
For more information or to inspect the property contact Mr Joyce on 0428 436 266 or go to the website by clicking here.
Tags Aston & Joyce Pty Ltd church conversion Danny Joyce former church historic property property for sale real estate wine trail young


text:
Articles > SPAAAAAAACE 5/13/13 - Chris Hadfield Comes Home
SPAAAAAAACE 5/13/13 - Chris Hadfield Comes Home
By Typical Michael | 13th May, 2013 | 8:23 pm | SPAAAAAAACE
Chris Hadfield is one of the coolest guys ever:
http://www.guardian.co.uk/science/2013/ ... ce-station
The first Space Rock Star. He has done a lot to really humanize astronauts, I think. They are friggin awesome, but still real people. And they like to do real things. For me, this really showed emotion. The guy just loves space!
"In the old days, the astronaut corps was almost a silent priesthood. No one knew much about them. And their operations in space were a black box," said Kevin Fong, director of University College London's centre for space medicine. "We've seen a transition, a breaking down of the barriers, between people who experience space and those who want to experience space vicariously."
Welcome home, Commander Hadfield!
http://www.cbc.ca/news/technology/story ... earth.html
Check it live! Godspeed!
Watch some of the biggest solar flares of the year:
http://www.space.com/21118-biggest-sola ... tions.html
Here is more about solar flares, like what causes them, and what they can cause.
http://www.space.com/21121-solar-flares ... ation.html
Scientists classify strong solar flares into one of three categories: C, M or X (with A and B classes, too, for weaker eruptions). There's a tenfold increase in power from one class to the next, so an X flare is 10 times stronger than an M flare, and 100 times more powerful than a C.
Gany's Galaxy
Here, Gany tells us about these dastardly...or lonely? Rogue Planets!
Alright everybody, it's time to go rogue!
Rogue planets are planets which don't orbit any star. They just fly through space alone, without "the man" keeping them down.
There are thought to be two ways for a planet to have gone rogue. The first is that the planet originally formed as part of a solar system, orbiting a star, but then due to gravitational interactions in the early, more turbulent solar system, the planet was flung out into interstellar space, left to wander its path alone, like some poor lonely black-sheep of the solar system family. This could happen in the early stages of a solar system, when the orbits of planetary bodies are chaotic and irregular, before they settle down to a more ordered state. The other, less depressingly scary way, is that the planet was formed in a cluster, in much the same way as a star but never obtained enough mass to ignite fusion. This type of object is known as a brown-dwarf, an almost-but-not-quite star. Brown-dwarfs are like giant Jupiters, mainly composed of hydrogen, but not massive enough to ignite fusion in its core, and therefore they don't shine.
Rogue planets can be either Jovian (gas giants) or terrestrial (rocky, like Earth). Generally, these planets have a similar orbit to the stars, orbiting the centre of the galaxy, although their orbits tend to be highly eccentric, especially the ones turned rogue by ejection. The rogue planets would be dark, and very cold, without a host star to provide heat energy. It is possible, however, that some terrestrial rogue planets may maintain a type of atmosphere. A rogue planet doesn't necessarily have to remain lonely forever, though. If it happens to enter a solar system, the orphan planet might get adopted by the star. It may get caught by the gravitational effects of the objects in the system and assume a more regular orbit about the star. Rogue planets had been predicted for decades, but the first direct observations were made in 2011, and since then at least ten have been identified and confirmed. It is thought, due to recent observations, that there are probably more rogue planets than stars in our galaxy.
So the question is, "could Earth ever go rogue?". And the answer is "yes, it totally could, but not for a few 10's of millions of years, so don't worry!". The orbits of the bodies in our solar system are well understood, but only for the short term. (Note: "the short term" for an astronomer is like 10 million years). After a certain point, the rather tenuous order in our system becomes uncertain and unpredictable. Small and gradual gravitational influences build up over time and it becomes impossible to predict the orbits accurately after about 10 million years or so. For example, take Jupiter and Mercury. The perihelion of both planets (perihelion is when the planet is closest to the Sun), precesses at a rate of about 1.5 degrees every 1000 years. If the perihelion of both planets synced up eventually, and both planets were at their closest to the Sun at the same time, the gravitational effect from Jupiter on Mercury could accumulate and eventually just pull Mercury off course altogether. This of course would have an effect on Venus and Earth etc. and while that would not be enough to eject Earth (probably), it is a good example of how easily the balance of orbits could be disturbed. Considering the amount of objects on our solar system and the gravitational influence that they all have on each other, and including the possibility of the influence of outside objects such as asteroids, the possibility of Earth going rogue is unlikely in the short term, and possible in the long term.
But, like I said, don't worry! We have 10 million years of safety (well, safety against expulsion from the solar system at least, in space you are never truly safe), and by the time anything does happen we should be able to thwart our demise by escaping in our super-advanced warp-enabled FTL-travelling starships, right? Right! Better get on that shit astro-engineers!
If you want to read more about rogue planets:
http://www.universetoday.com/94656/rogu ... her-stars/
http://en.wikipedia.org/wiki/Rogue_planets
Share on facebook and twitter, or Gany will use psychic power to destroy a life-bearing planet.
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redpajama_set_name: RedPajamaCommonCrawl


Mauricio Pochettino is hoping to gain some measure of revenge on Arsenal when the two sides face off at the Emirates on Sunday afternoon.
Pochettino's side were five points clear of Arsene Wenger's team last season with three games to play, and on the final day, Spurs only needed a draw to finish above their city rivals for the first time since 1995.
But an inexplicable 5-1 defeat to relegated Newcastle coupled with Arsenal's defeat of Aston Villa, meant Spurs ended up behind the Gunners in the final Premier League standings, in third place.
Ahead of this weekend's eagerly-anticipated north London derby, Pochettino admits he can't explain his side's capitulation.
"I still feel bad about last season," he said. "It was so easy to be above Arsenal in that moment.
Spurs's winning run at the tail end of the campaign came with two draws, at home to West Brom and away to Chelsea, ahead of losing to Southampton and Newcastle.
Pochettino added: "It's true that we felt very disappointed after the Chelsea game but with two games ahead and only one point (needed) to be above them, that was very frustrating and disappointing.


text:
TodayTopicsWritersPodcastsMagazineMoreSearchAccount
Magazine: 7 March 2020
Issue: 7 March 2020
No sacred cows
My night in A&E left me worried about the NHS's coronavirus response
From magazine issue: 7 March 2020
CommentsShare
I wish I shared the Prime Minister's confidence about the ability of the NHS to cope with coronavirus. 'I have no doubt that with the help of the NHS and its incomparable staff, this country will get through it – and beat it,' he said on Sunday. Not if my experience in A&E last week is anything to go by.
I wasn't keen to visit A&E in the midst of the current crisis, obviously, and had it been my own health I was worried about I would have stayed in bed. But it was Sasha's, my 16-year-old daughter. She was pushed down some stairs at a party (not deliberately) and one of her fake fingernails got bent back, ripping her actual nail off its nail bed. She was complaining about the pain and it looked infected so I called 111 and asked what to do. Answer: take her to A&E. So off we went to Chelsea and Westminster.
We arrived at about 8.30 p.m. on a Sunday and the waiting area was quite crowded so Sasha and I reconciled ourselves to a long stay. At about 10 p.m. a Chinese-looking man walked in, approached the receptionist and said: 'I think I've got coronavirus.' He said this quite loudly, so everyone in the room could hear, and I exchanged a nervous glance with a middle-aged woman opposite. 'What makes you think that?' asked the sceptical receptionist. 'Have you been in any of the following countries in the past two weeks?' He then rattled off a list of 'hot zones' and, sure enough, this man had been in one of them two days earlier. He was also suffering from a bad cold, judging from how often he sneezed. The receptionist told him to leave, call 111 and follow the advice.
The man left, having spent no more than five minutes in the waiting area, and everyone breathed a sigh of relief. Then, about half an hour later, he reappeared, this time wearing a mask. Why the hell had he come back? Surely that couldn't have been the advice he'd been given?
'At least he's wearing a mask,' I thought, at which point his mobile rang and he took the mask off and started jabbering into the receiver. As the conversation became more animated, he began to pace up and down. He then sat down opposite the water cooler, still talking on the phone, with his head no more than three feet from the 'clean' plastic cups.
The reaction of the man sitting next to him was wonderfully British. Rather than get up straight away and flee to the other side of the room, he started patting his pockets and shaking his head, as if he'd forgotten some vital document in his car. He then got up, left the waiting area entirely, only to reappear five minutes later and take a seat as far away from Coronavirus Man as possible.
The middle-aged woman sitting opposite me slowly pulled her jumper up so it covered her nose and mouth, but she made a pretence of shivering ostentatiously before doing so, as if her only motive was to keep warm. It goes without saying that the waiting room was absolutely sweltering. My daughter, who was possibly the only person there oblivious to the unfolding drama — because she was on Snapchat — then asked me to get her a glass of water.
'Er, I don't think that's a good idea, darling,' I whispered. 'Why not?' she said. 'I'm really thirsty.' 'Just trust me, OK?' The woman opposite stifled a laugh beneath her jumper.
When Sasha and I were eventually seen at about midnight, I alerted the doctor to the fact that there was a man wandering around the waiting room who might have coronavirus.
'Oh really,' he said, looking at me with disdain. 'And what makes you think he has Covid-19? Is it because he's of Asian ancestry?'
'No,' I replied. 'It's because he went up to the receptionist and said "I think I've got coronavirus." '
At this point the doctor looked a little alarmed and disappeared to investigate. He returned a few minutes later and told me there was nothing to worry about. 'That man does not have Covid-19,' he declared.
I n-odded politely, but inside I was thinking: 'Tell at a glance, can you? That's some pretty impressive diagnostic skills you've got there. You should be working for the World Health Organisation.'
I'm a big fan of Chelsea and Westminster, which saved my son Ludo's life when he was a newborn, and I've raised money for the hospital in the past. But I'm not convinced the NHS is properly prepared for the looming emergency.
A Chinese-looking man walked into the waiting room and said quite loudly: 'I think I've got coronavirus'
Toby Young is associate editor of The Spectator.
Written byToby Young
Toby Young is the co-author of What Every Parent Needs to Know and the co-founder of several free schools. In addition to being an associate editor of The Spectator, he is an associate editor of Quillette. Follow him on Twitter @toadmeister
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Hidden Article


Anita and Olayemi look like a royal couple straight out of a fairytale novel. Their pre-wedding shoot ahead of their wedding is absolutely beautiful, we expect nothing less from Jide Odukoya Photography.
The lovebirds met 7 years ago and with each passing year, they've fallen even more in love with each other. BellaNaija Weddings wishes the lovely couple a life of immense joy and unwavering love. Looking forward to #Alfreds2017!
We met through a mutual friend on July 16th 2010 and we connected instantly, God bless that day! We became very close friends and started dating in 1 year. I love everything about Yemi, he's the most loyal friend anybody can ask for, he taught me how to love, he's patient and caring. Always pushing me to be a better person.
The first 4 years of our relationship, Yemi was in Lagos and I was in Abuja, the distance wasn't easy but he was very devoted to making us work.
It's been 7 years of genuine friendship and love, I can't wait to share the rest of my life with you, to be my second self and best earthly companion.
July 16th 2010, you brightened up my morning, I knew from that moment I'd share my life with you. Talking with you, laughing with you, going through tough times with you, it's been 7 years of bliss. Putting my heart in your hands has been the best decision I ever made. Every part of our lives is perfect because of you.
Yemi proposed 9th July 2016. OMG it was so beautiful!
He was Lagos and was going to travel out of the country for a course, he asked me to come see him before he travelled. I left Abuja to Lagos, and he said we should go see a movie with some of his friends we usually hang out with at a private cinema.
We got there, watched a movie and at the end of the movie, I started seeing my pictures on the screen with music playing in the background and Yemi came to kneel down with a ring! I was so surprised because I never suspected a thing!
I knelt down too out of confusion! Next thing I see my friends there out of the blues, like where did you guys come out from?! I couldn't even read what was being displayed on screen again. Everyone wanted a picture and was screaming, he asked me to be his wife and I said YES! That day was so beautiful, I felt so special and I'm just grateful to God for the love we share.
beauty queen!c as she fine pass all the mbgn contestants.
all these dresses for pre-wedding photoshoot…oga o..but they are fine sha..
She don sell her makate. We all need to sell our makate to move our economy forward.
Her dresses are really beautiful… Wishing them all the best..
but why guys are taking now too long before putting the ring?? 7 years? wow..
This is beautiful….. Msmakor wears Msmakor #Alfreds2017. Absolutely amazing. Wishing you both all the love and joy?
Congratulations Olayemi God bless ur marriage.
All these mutual friends I that have are not doing their work, imagine every wedding story I have read on bellanaija.com, all met through their mutal friends, am getting new ones asap.


text:
\section{Introduction}

As a generalization of the notion of local cohomology modules,
R. Takahashi, Y. Yoshino, and T. Yoshizawa \cite{TakYoYo},
introduced local cohomology modules with respect to a pair
of ideals. This paper is concerned about this new notion of
local cohomology and Serre subcategories. For notations and
terminologies not given in this paper, if necessary,
the reader is referred to \cite{TakYoYo} and \cite{AghMel}.
\\Throughout this paper, $R$ is denoted a commutative
noetherian ring with non-zero identity, $I$ , $J$ are denoted
two ideals of $R$, and $M$ is denoted an arbitrary $R$-module.
The $(I,J)$-torsion submodule $\Gamma_{I,J}(M)$ of $M$ is a
submodule of $M$ consists of all elements $x$ of $M$ with
Supp$(Rx)\subseteq W(I,J)$, in which
$$W(I,J)=\{~\mathfrak{p}\in\textmd{Spec}(R)\mid I^n\subseteq
\mathfrak{p}+J \textmd{\ \ for an integer} \ n\geq1\}.$$
For an integer $n$, the $n$-th local cohomology functor
$\lc^{n}_{I,J}$ with respect to $(I,J)$ is the $n$-th
right derived functor of $\Gamma_{I,J}$.
The $R$-module $\lc^{n}_{I,J}(M)$ is called the $n$-th
local cohomology module of $M$ with respect to $(I,J)$.
In the case $J=0$, $\lc^{n}_{I,J}$ coincides with
the ordinary functor $\lc^{n}_{I}$.
Also, we are concerned with the following set of ideals of $R$:
$$\tilde{W}(I,J)=\{~\mathfrak{a}\trianglelefteq R\mid
I^n\subseteq \mathfrak{a}+J \textmd{\ \ for an integer} \ n\geq0\}.$$
A class of $R$-modules is called a Serre subcategory
(or Serre class) of the category of $R$-modules when it is
closed under taking submodules, quotients and extensions.
Always, $\mathcal{S}$ stands for a Serre class.
\\According to the third Huneke$^,$s problem on local cohomology
\cite[Conjecture 4.3]{Hu}, one of the main problem in commutative
algebra is finiteness of the socle of local cohomology modules
on a local ring. Solving this problem, gives an answer to the
finiteness of the set of associated primes of local cohomology modules.
\\On this area, some remarkable attempts have been done, e.g. see
\cite{HuSh}, \cite{Ly}, \cite{Ly1}, \cite{MarVa}, and
\cite{AsgTo}.\\
In section 2, we give a positive answer to the Huneke$^,$s
conjecture more general for an arbitrary Serre subcategory
$\mathcal{S}$, instead of the category of finitely generated
modules. Let $R$ be a noetherian (not necessary local) ring.
Let $R/\mathfrak{m}\in\mathcal{S}$ for all $\mathfrak{m}\in\textmd{Max}(R)$.
For any minimax $R$-module $M$ of krull dimension less than 3,
we show that
$\Ext^{j}_R\big(R/\mathfrak{m},\lc^{i}_{I}(M)\big)\in\mathcal{S}$
for any $\mathfrak{m}\in\textmd{Max}(R)\cap\V(I)$ and all $i,j\geq0$.
In particular $\Hom_R\big(R/\mathfrak{m},\lc^{i}_{I}(M)\big)\in\mathcal{S}$
for any $\mathfrak{m}\in\textmd{Max}(R)\cap\V(I)$ and all $i\geq0$
(see Theorem \ref{2.14}).\\
We get the same result for local cohomology modules with
respect to a pair of ideals, but in local case
(see Theorem \ref{2.18}).\\
In section 3, we obtain some results on cofiniteness
and artinianness of local cohomology with respect to a
pair of ideals.\\
M. Aghapournahr and L. Melkersson in \cite{AghMel},
obtained some conditions in which the ordinary local
cohomology $\lc^{i}_{I}(M)$ belongs to $\mathcal{S}$
for all $i<n$ (from below).
In section 4, as a complement of this work, we study some
conditions in which the local cohomology $\lc^{i}_{I,J}(M)$
belongs to $\mathcal{S}$ for all $i>n$ (from top).
For a ZD-module $M$ of finite krull dimension, we show that if
the integer $n\in\mathbb{N}$ is such that
$\lc^{i}_{I,J}(M)\in\mathcal{S}$ for all $i>n$, then the modules
$\lc^{i}_{I,J}(M)/\mathfrak{a}^{j}\lc^{i}_{I,J}(M)\in\mathcal{S}$ for any $\mathfrak{a}\in\tilde{W}(I,J)$, all $i\geq n$, and all $j\geq0$
(see Theorem \ref{4.4}).
Replacing $\mathcal{S}$ with some familiar
Serre subcategories such as zero modules, finite modules,
and artinian modules (resp.), we show that a necessary
and sufficient condition for $\lc^{n}_{I,J}(M)$ to be zero,
inite, and artinian (resp.), is the existence of an integer
$m\in\mathbb{N}_0$ such that $I^m\lc^{n}_{I,J}(M)$ is zero,
finite, and artinian (resp.) (see Corollary \ref{4.6}).\\
More generally, for a finite $R$-module $M$ if
$n\in\mathbb{N}_0$ is such that $\lc^{i}_{I,J}(M)$ belongs to
$\mathcal{S}$ for all $i>n$ and $\mathfrak{b}$ is an ideal of $R$
such that $\lc^{n}_{I,J}(M/\mathfrak{b} M)$ belongs to $\mathcal{S}$,
then $\lc^{n}_{I,J}(M)/{\mathfrak{b}}\lc^{n}_{I,J}(M)$ belongs to
$\mathcal S$ (see Theorem \ref{4.7}).
As a consequence of this theorem, we obtain a similar
result as Corollary \ref{4.6}, for the finite module $M$
(see Corollary \ref{4.10}).
Other consequence of this theorem is concerned about
finiteness of $\lc^{n}_{I,J}(M)$ where $n=\textmd{cd}(I,J,M)$
or $n=\textmd{dim}M\geq1$ (see Corollary \ref{4.11}).\\
Finally, these results motivate us to introduce
the concept of Serre cohomological dimension of $M$ with
respect to $(I,J)$, $\textmd{cd}_{\mathcal{S}}(I,J,M)$, as the
supremum of non-negative integers $i$ such that
$\lc^{i}_{I,J}(M)\not\in\mathcal{S}$. We give some
characterizations to $\textmd{cd}_{\mathcal{S}}(I,J,M)$
(Corollary \ref{4.20}) and as a main result of its properties,
we show that for an integer $r\in\mathbb{N}_0$,
$\lc^{j}_{I,J}(R)\in\mathcal{S}$ for all $j>r$ iff
$\lc^{j}_{I,J}(M)\in\mathcal{S}$ for all $j>r$
and any finite $R$-module $M$
(see Corolary \ref{4.23})

\maketitle

\section{The membership of the
$\textmd{Hom}_R\big(-,\lc^{t}_{I,J}(M)\big)$ in Serre classes}

In this section, Proposition \ref{2.5} plays a main
role to obtain our results. For this purpose we need
the following Lemmas.

\begin{lem}\label{2.1}
For a Serre class $\mathcal{S}$, we have $\mathcal{S}\neq{0}$ if and only if
$R/\mathfrak{m}\in\mathcal{S}$ for some $\mathfrak{m}\in\Max(R)$.

\begin{proof}
$(\Rightarrow)$ Let $L\in\mathcal{S}$ and $0\neq x\in L$. Then $(0:_Rx)\subseteq\mathfrak{m}$
for some $\mathfrak{m}\in$ Max$(R)$. Now, since $Rx\in\mathcal{S}$, the assertion follows from
the natural epimorphism $Rx\cong R/(0:_Rx)\rightarrow R/\mathfrak{m}$.\\
$(\Leftarrow)$ It is obvious.
\end{proof}
\end{lem}

\begin{lem}\label{2.2}
Let $\mathcal{FL}$ be the class of finite length $R$-modules. Then
$\mathcal{FL}\subseteq\mathcal{S}$ if and only if $R/\mathfrak{m}\in\mathcal{S}$ for all
$\mathfrak{m}\in\Max(R)$.
\end{lem}
\begin{proof}
$(\Rightarrow)$ It is obvious.\\
$(\Leftarrow)$ Let $N\in\mathcal{FL}$ and $\ell:=\ell_R(N)$. So, consider the
following chain of $R$-submodules of $N$:
$$0=N_0\subseteq N_1\subseteq\cdots\subseteq N_\ell=N$$
in which, for all $1\leq j\leq \ell$, $N_j/N_{j-1}\cong R/\mathfrak{m}$  for some
$\mathfrak{m}\in$ Max$(R)$. Now, the assertion can be followed by induction on $\ell$.
\end{proof}

\begin{cor}\label{2.3}
Let $(R,\mathfrak{m})$ be a local ring and $\mathcal{S}\neq0$. Then
$\mathcal{FL}\subseteq\mathcal{S}$.
\end{cor}

\begin{exam}\label{2.4}
\emph{In general case, it not true that $R/\mathfrak{m}\in\mathcal{S}$
for all $\mathfrak{m}\in\Max(R)$. To see this, let $R$ be a non-local
ring and $\mathfrak{m}\in$ Max$(R)$.
Let  $I$ be an ideal of $R$ such that $I\nsubseteq\mathfrak{m}$.
Let $\mathcal{S}$ be the
class of $I$-cofinite minimax $R$-modules,
(see \cite[Corollary 4.4]{Mel1}).
Then Supp$(R/\mathfrak{m})\nsubseteq V(I)$ and so
$R/\mathfrak{m}\not\in\mathcal{S}$.
For example,
let $R:=\mathbb{Z}[x]$, $\mathfrak{m}:=(x-1)R$, and $I:=xR$.}
\end{exam}

\begin{prop}\label{2.5}
For a noetherian ring $R$, we have
\begin{enumerate}

  \item[\rm{(}i\rm{)}] If $R/\mathfrak{m}\in\mathcal{S}$, for any $\mathfrak{m}\in$ Max$(R)$, and
  $M$ is a finite or an artinian $R$-module, then $\Ext^j_R(R/\mathfrak{m},M)\in\mathcal{S}$
  for any $\mathfrak{m}\in$ Max$(R)$ and all $j\geq0$.

  \item[\rm{(}ii\rm{)}] If $R/\mathfrak{m}\in\mathcal{S}$, for any $\mathfrak{m}\in$ Max$(R)$, and $M$
  is a minimax $R$-module, then $\Ext^j_R(R/\mathfrak{m},M)\in\mathcal{S}$
  for any $\mathfrak{m}\in$ Max$(R)$ and all $j\geq0$.

  \item[\rm{(}iii\rm{)}] If $(R,\mathfrak{m})$ be a local ring, $\mathcal{S}\neq0$, and $M$
  be a minimax $R$-module, then $\Ext^j_R(R/\mathfrak{m},M)\in\mathcal{S}$ for all $j\geq0$.

\end{enumerate}

\begin{proof}
(i) Let $\mathfrak{m}\in$ Max$(R)$ and $j\geq0$. Since $\Ext^j_R(R/\mathfrak{m},M)$ is finite and is
annihilated by $\mathfrak{m}$, hence $\Ext^j_R(R/\mathfrak{m},M)$ has finite length. Now, the result
follows from Lemma \ref {2.2}.\\
(ii) Since $M$ is minimax, there exists a short exact sequence
$$0\rightarrow N\rightarrow M\rightarrow A\rightarrow0,$$
where $N$ is a finite module and $A$ is an artinian module.
This induces the exact sequence
$$\cdots\rightarrow\Ext^j_R(R/\mathfrak{m},N)\rightarrow\Ext^j_R(R/\mathfrak{m},M)\rightarrow
\Ext^j_R(R/\mathfrak{m},A)\rightarrow\cdots .$$
Now, the assertion follows from part (i).\\
(iii) Apply Lemma \ref {2.1} and part (ii).
\end{proof}
\end{prop}

\begin{cor}\label{2.6}
Let $R/\mathfrak{m}\in\mathcal{S}$ for all $\mathfrak{m}\in$ Max$(R)$. Let $t\in\mathbb{N}_0$
be such that $\lc^{t}_{I,J}(M)$ is a minimax $R$-module. Then
$\Ext^j_R\big(R/\mathfrak{m},\lc^{t}_{I,J}(M)\big)\in\mathcal{S}$ for all $i\geq0$.
\end{cor}

\begin{lem}\label{2.7}
Let $\mathfrak{a}\in\tilde{W}(I,J)$. Let $X$ be an $R$-module. Then
  $$\big( 0 :_X\mathfrak{a}\big) = \big(0:_{\Gamma_{\mathfrak{a}}(X)}\mathfrak{a}\big) =
  \big(0:_{\Gamma_{\mathfrak{a},J}(X)}\mathfrak{a}\big) = \big(0:_{\Gamma_{I,J}(X)}\mathfrak{a}\big),$$
      in particular, for any ideal $\mathfrak{b}$ of $R$ with $\mathfrak{b}\supseteq\ I$, we have
 \begin{enumerate}

  \item[\rm{(}i\rm{)}] $\big( 0 :_X\mathfrak{b}\big) = \big(0:_{\Gamma_{\mathfrak{b}}(X)}\mathfrak{b}\big) =
  \big(0:_{\Gamma_{\mathfrak{b},J}(X)}\mathfrak{b}\big) = \big(0:_{\Gamma_{I,J}(X)}\mathfrak{b}\big) =
  \big(0:_{\Gamma_{I}(X)}\mathfrak{b}\big)$.

  \item[\rm{(}ii\rm{)}] $\big(0:_X\mathfrak{b}\big) = \big(0:_{\Gamma_{\mathfrak{b},J}(X)}\mathfrak{b}\big)
  \subseteq \big(0:_{\Gamma_{\mathfrak{b},J}(X)}I\big) \subseteq
  \big(0:_{\Gamma_{I,J}(X)}I\big) = \big( 0 :_{X}I\big)$.

\end{enumerate}

\begin{proof}
All proofs are easy and we leave them to the reader.
\end{proof}
\end{lem}

\begin{prop}\label{2.8}
Let $\mathfrak{a}\in\tilde{W}(I,J)$ and $t\in\mathbb{N}_0$. Consider the natural
homomorphism
$$\psi : \Ext^t_R\big(R/\mathfrak{a},M\big)\longrightarrow \Hom_R\big(R/\mathfrak{a},
\lc^{t}_{I,J}(M)\big).$$

\begin{enumerate}

  \item[\rm{(}i\rm{)}] If $\Ext^{t-j}_R\big(R/\mathfrak{a},\lc^{j}_{I,J}(M)\big)
  \in\mathcal{S}$ for all $j<t$, then \emph{Ker} $\psi\in\mathcal{S}$.

  \item[\rm{(}ii\rm{)}] If $\Ext^{t+1-j}_R\big(R/\mathfrak{a},\lc^{j}_{I,J}(M)\big)
  \in\mathcal{S}$ for all $j<t$, then \emph{Coker} $\psi\in\mathcal{S}$.

  \item[\rm{(}iii\rm{)}] If $\Ext^{n-j}_R\big(R/\mathfrak{a},\lc^{j}_{I,J}(M)\big)
  \in\mathcal{S}$ for $n=t,\ t+1$ and for all $j<t$, then \emph{Ker} $\psi$
  and \emph{Coker} $\psi$ both belong to $\mathcal{S}$.
  Thus $\Ext^t_R\big(R/\mathfrak{a},M\big)\in\mathcal{S}$ iff
  $\Hom_R\big(R/\mathfrak{a},\lc^{t}_{I,J}(M)\big)\in\mathcal{S}$.

\end{enumerate}
\begin{proof}
Let $\textmd{F} (-) = \Hom_R\big(R/\mathfrak{a},-\big)$ and
$\textmd{G} (-) = {\Gamma_{I,J}\big(-\big)}$.
By Lemma \ref {2.7}, $\textmd{FG} =\textmd{F}$.
Now, the result can be followed by \cite[Proposition 3.1]{AghMel1}.
\end{proof}
\end{prop}

The next result can be a generalization of main results of
\cite{AsgTo}, \cite{KhaSal}, \cite{BroLash},
\cite{DivMaf}, \cite{LoSahYas}, \cite{BahNag},
and \cite{AsaKhaSal}.

\begin{thm} \label {2.9}
Let $\mathfrak{a}\in\tilde{W}(I,J)$ and $~t\in\mathbb{N}_0$ be such that
$\Ext^{t}_R\big(R/\mathfrak{a},M\big)\in\mathcal{S}$
and $\Ext^{n-j}_R\big(R/\mathfrak{a},\lc^{j}_{I,J}(M)\big)$ $\in\mathcal{S}$ for $n=t,\ t+1$
and for all $j<t$. Then for any submodule $N$ of $\lc^{t}_{I,J}(M)$ such that
$\Ext^1_R\big(R/\mathfrak{a},N\big)\in\mathcal{S}$, we have

\begin{enumerate}

  \item[\rm{(}i\rm{)}] $\Hom_R\big(R/\mathfrak{a},\lc^{t}_{I,J}(M)/N\big)\in\mathcal{S}$.

  \item[\rm{(}ii\rm{)}] $\Hom_R\big(L,\lc^{t}_{I,J}(M)/N\big)\in\mathcal{S}$
  for any finite $R$-module $L$ with $\Supp(L)\subseteq\textmd{V}(\mathfrak{a})$.

  \item[\rm{(}iii\rm{)}] $\Hom_R\big(R/\mathfrak{p},\lc^{t}_{I,J}(M)/N\big)\in\mathcal{S}$
  for any $\mathfrak{p}\in\textmd{V}(\mathfrak{a})$.
\end{enumerate}
All the statements are hold for $\mathfrak{a}=I$.
\begin{proof}
(i) Apply Proposition \ref{2.8} and the exact sequence
$$0\rightarrow N\rightarrow\lc^{t}_{I,J}(M)\rightarrow\lc^{t}_{I,J}(M)/N\rightarrow0.$$
For (ii) and (iii) apply \cite[Theorem 2.10]{AhSad}.
\end{proof}
\end{thm}

\begin{cor}\label{2.10}
Let $(R,\mathfrak{m})$ be a local ring and $\mathfrak{a}\in\tilde{W}(I,J)$.
Let $t\in\mathbb{N}_0$ be such that
$\Ext^{t}_R(R/\mathfrak{a},M)\in\mathcal{S}$ and
$\Ext^{n-j}_R\big(R/\mathfrak{a},\lc^{j}_{I,J}(M)\big)\in\mathcal{S}$
for $n=t,\ t+1$ and all $j<t$.
Then for any submodule $N$ of $\lc^{t}_{I,J}(M)$ such that
$\Ext^1_R\big(R/\mathfrak{a},N\big)\in\mathcal{S}$, we have
$\Hom_R\big(R/\mathfrak{a},\lc^{t}_{I,J}(M)/N\big)\in\mathcal{S}$,
specially $\Hom_R\big(R/\mathfrak{m},\lc^{t}_{I,J}(M)/N\big)\in\mathcal{S}$.
\end{cor}

\begin{exam} \label{2.11}
\emph{In Theorem \ref{2.9},
the assumption $\Ext^{t}_R\big(R/\mathfrak{a},M\big)\in\mathcal{S}$ is necessary.
To see this, let $(R,\mathfrak{m})$ be a local Gorenstein ring of positive dimension $d$,
and $\mathcal{S}=0$. Then $\lc^{i}_{\mathfrak{m}}(R)=0$ for $i<d$. But we have
$\Ext^{d}_R\big(R/\mathfrak{m},R\big)\cong\Hom_R\big(R/\mathfrak{m},\lc^{d}_{\mathfrak{m}}(R)\big)\cong
\Hom_R\big(R/\mathfrak{m},\emph{E}(R/\mathfrak{m})\big)\cong R/\mathfrak{m}\neq0$}.
\end{exam}

One of the main result of this paper is the following theorem which is a
generalization of \cite[Theorem 2.12]{AsgTo}

\begin{thm} \label {2.12}
Let $R/\mathfrak{m}\in\mathcal{S}$ for all $\mathfrak{m}\in$ $\emph{Max}(R)$. Let $M$ be a minimax
$R$-module and $t\in\mathbb{N}_0$ be such that
$\Ext^{n-j}_R\big(R/\mathfrak{m},\lc^{j}_{I,J}(M)\big)$ $\in\mathcal{S}$ for
$n=t,\ t+1$, all $j<t$, and all $\mathfrak{m}\in$ $\emph{Max}(R)$.
Then for any submodule $N$ of $\lc^{t}_{I,J}(M)$ such that
$\Ext^1_R\big(R/\mathfrak{m},N\big)\in\mathcal{S}$, we have
$\Hom_R\big(R/\mathfrak{m},\lc^{t}_{I,J}(M)/N\big)\in\mathcal{S}$,
for any $\mathfrak{m}\in\emph{W}_\emph{Max}(I,J) := \emph{Max}(R)\cap\emph{W}(I,J)$.
\end{thm}
\begin{proof}
The assertion follows from Proposition \ref{2.5} (ii) and Theorem \ref{2.9}.
\end{proof}

\begin{cor}\label{2.13}
Let $(R,\mathfrak{m})$ be a local ring, $\mathcal{S}\neq0$ and $M$ be a minimax $R$-module.
Let $~t\in\mathbb{N}_0$ be such that $\Ext^{n-j}_R\big(R/\mathfrak{m},\lc^{j}_{I,J}(M)\big)$
$\in\mathcal{S}$ for $n=t,\ t+1$ and for all $j<t$.
Then for any minimax submodule $N$ of $\lc^{t}_{I,J}(M)$,
we have $\Hom_R\big(R/\mathfrak{m},\lc^{t}_{I,J}(M)/N\big)\in\mathcal{S}.$
\end{cor}
\begin{proof}
Apply Proposition \ref{2.5} and Theorem \ref{2.12}.
\end{proof}

The following familiar conjecture is due to Huneke \cite{Hu}.\\
\textbf{Conjecture}. Let $(R,\mathfrak{m},k)$ be a regular local ring and $I$ be an ideal of $R$.
For all $n$, soc$\big(\lc^{n}_{I}(R)\big)$ is finitely generated.\\
As we mentioned in the introduction, the following theorem which is one of
the main result of this section, can be a positive answer to Huneke$^,$s
conjecture. In fact the following theorem proves a generalization of the conjecture
for an arbitrary noetherian ring $R$ (not necessary regular local one), a minimax
$R$-module $M$ of krull dimension less than 3, and a Serre class $\mathcal{S}$.

\begin{thm} \label {2.14}
Let $R$ be a noetherian ring and $M$ be a minimax $R$-module
of krull dimension less than \emph{3}.
Let $R/\mathfrak{m}\in\mathcal{S}$ for any $\mathfrak{m}\in\emph{Max}(R)$.
Then $\Ext^{j}_R\big(R/\mathfrak{m},\lc^{i}_{I}(M)\big)\in\mathcal{S}$ for any $\mathfrak{m}\in\emph{Max}(R)\cap\V(I)$ and all $i,j\geq0$.
In particular $\Hom_R\big(R/\mathfrak{m},\lc^{i}_{I}(M)\big)\in\mathcal{S}$
for any $\mathfrak{m}\in\emph{Max}(R)\cap\V(I)$ and all $i\geq0$.
\end{thm}

\begin{proof}
Let $\mathfrak{m}\in\textmd{Max}(R)\cap\V(I)$. By Proposition \ref{2.5} (ii)
and the Grothendieck$^,$s vanishing theorem there is
nothing to prove for cases $i=0$ and $i>2$.
Now, assume that $0<i\leq2$. If dim $M=2$ and $i=2$, then
the result follows from \cite[Corollary 3.3]{DivEsm}
and Proposition \ref{2.5} (i).
Also, in the case $i=1$, the result is obtained from
\cite[Theorem 2.3 ]{AghTahVah}, by replacing $s:=j$,
$t:=1$, and $N:=R/\mathfrak{m}$, and Proposition \ref{2.5} (ii).
Finally if dim$M\leq1$, we can obtain the desired
result in similar way.
\end{proof}

\begin{cor} \label {2.15}
Let $R$ be a noetherian ring of dimension less than \emph{3}
and let $M$ be a minimax $R$-module. Let $R/\mathfrak{m}\in\mathcal{S}$
for any $\mathfrak{m}\in\emph{Max}(R)$. Then
$\Ext^{j}_R\big(R/\mathfrak{m},\lc^{i}_{I}(M)\big)\in\mathcal{S}$ for any
$\mathfrak{m}\in\emph{Max}(R)\cap\V(I)$ and all $i,j\geq0$. In particular for the class of finite $R$-modules.
\end{cor}

\begin{cor}\label{2.16}
Let $R$ be a noetherian ring. Let $M$ be a minimax $R$-module
of krull dimension less than \emph{3}. Then the Bass numbers of
$\lc^{i}_{I}(M)$ are finite for all $i\geq0$, in particular it is true
when $\dim R\leqslant 2$.
\end{cor}

One can generalize Theorem \ref{2.14} for local cohomology
modules with respect to a pair of ideals, but in local case.
To do this, we need the
following lemma.

\begin{lem}\label{2.17}
Let $(R,\mathfrak{m})$ be a local ring and $M$ be a minimax $R$-module
of finite krull dimension d. Then $\lc^{d}_{I,J}(M)$ is artinian.
\begin{proof}
Let $N$ be a finite submodule of $M$ such that $A:=M/N$
is artinian. Since $\textmd{dim}~N\leq d$, by \cite[Theorem 2.1]{ChWa},
$\lc^{d}_{I,J}(N)$ is artinian. Also, the exact sequence
$0\rightarrow N\rightarrow M \rightarrow A\rightarrow0$
induces the following exact sequence
$$0\rightarrow\Gamma_{I,J}(N)\rightarrow\Gamma_{I,J}(M)\rightarrow \Gamma_{I,J}(A)
\rightarrow \lc^{1}_{I,J}(N)\rightarrow \lc^{1}_{I,J}(M) \rightarrow0,$$
and $\lc^{i}_{I,J}(N)\cong\lc^{i}_{I,J}(M)$ for all $i\geq2$.
By the exactness of the above sequence, it is easy to see that $\lc^{d}_{I,J}(M)$ is
artinian.
\end{proof}
\end{lem}

\begin{thm}\label{2.18}
Let $(R,\mathfrak{m})$ be a local ring and $M$ be a minimax $R$-module
of krull dimension less than \emph{3}. Let $\mathcal{S}\neq0$.
Then $\Ext^{j}_R\big(R/\mathfrak{m},\lc^{i}_{I,J}(M)\big)\in\mathcal{S}$
for all $i,j\geq0$.
\begin{proof}
Apply Lemma \ref{2.17}, Proposition \ref{2.5}, and
the same method of the proof of Theorem \ref{2.14}.
\end{proof}
\end{thm}

The next ressults can be useful for finiteness of Bass numbers of
local cohomology modules over a noetherian ring and modules
of krull dimension 3.
Recall that an $R$-module $M$ is called locally minimax if
$M_\mathfrak{m}$ is minimax for any $\mathfrak{m}\in$ Max$(R)$ (see \cite{AghMel2}).

\begin{thm} \label {2.19}
Let $R$ be a noetherian ring and $M$ be an $R$-module
of krull dimension \emph{3}.
Let $R/\mathfrak{m}\in\mathcal{S}$ for any $\mathfrak{m}\in\emph{Max}(R)$.
Then $\Ext^{j}_R\big(R/\mathfrak{m},\lc^{i}_{I}(M)\big)\in\mathcal{S}$
for any $\mathfrak{m}\in\emph{Max}(R)\cap V(I)$ and all $i,j\geq0$,
if one of the following conditions holds:
\begin{enumerate}

  \item[\rm{(}i\rm{)}] $M$ and $\lc^{1}_{I}(M)$ are minimax;

  \item[\rm{(}ii\rm{)}] $M$ is finite and $\lc^{1}_{I}(M)$ is
  locally minimax;

  \item[\rm{(}iii\rm{)}] $M$ and $I^{m}\lc^{2}_{I}(M)$ are minimax
  for some $m\in\mathbb{N}_0$;

  \item[\rm{(}iv\rm{)}] $M$ is minimax, $I^{m}\lc^{2}_{I}(M)$ is
  locally minimax, and $\Hom_{R}\big(R/I,I^{m}\lc^{2}_{I}(M)\big)$ is
  finite for some $m\in\mathbb{N}_0$;

  \item[\rm{(}v\rm{)}] $(R,\mathfrak{m})$ is local, $M$ is minimax, and $I^{m}\lc^{2}_{I}(M)$ is locally minimax for some $m\in\mathbb{N}_0$.
\end{enumerate}

\begin{proof}
Let $\mathfrak{m}\in\emph{Max}(R)\cap\V(I)$.

(i) By \cite[Corollary 3.3]{DivEsm}, $\lc^{3}_{I}(M)$ is artinian.
Thus the assertion is true for $i=0,1,3$, by Proposition \ref{2.5}.
Also, for $i=2$, we apply \cite[Theorem 2.3]{AghTahVah}, by replacing
$s:=j$, $t:=2$, and $N:=R/\mathfrak{m}$.

(ii) By \cite[Theorem 2.3]{BahNag} and \cite[Theorem 2.6]{AghMel2},
$\lc^{1}_{I}(M)$ is minimax. Now, the assertion follows from part (i).

(iii) By the short exact sequence
$$0\rightarrow I^m\lc^{2}_{I}(M)\rightarrow
\lc^{2}_{I}(M)\rightarrow \lc^{2}_{I}(M)/I^m\lc^{2}_{I}(M)
\rightarrow0,$$
and \cite[Theorem 3.1]{AsgTo}, for the class of minimax $R$-modules,
we get $\lc^{2}_{I}(M)$ is minimax. Now, for $i=0,2,3$, the claim
follows from \cite[Corollary 3.3]{DivEsm} and Proposition \ref{2.5}.
In the case $i=2$, apply \cite[Theorem 2.3 ]{AghTahVah},
by replacing $s:=j$, $t:=1$, $N:=R/\mathfrak{m}$ and Proposition \ref{2.5}.

(iv) Apply \cite[Theorem 2.6]{AghMel2} and part (iii).

(v) By proof of Lemma \ref{2.17}, we may assume that $M$
is finite. Now, the assertion follows from
\cite[Propositions 3.4 , 2.2]{AghMel2} and part (iii).
\end{proof}
\end{thm}

\begin{cor}\label{2.20}
Let $R$ be a noetherian ring and $M$ be a minimax $R$-module
of krull dimension $d\leq\emph{3}$. Then the Bass numbers of $\lc^{i}_{I}(M)$
are finite for all $i\geq0$, if one of the following conditions holds:
\begin{enumerate}

  \item[\rm{(}i\rm{)}] $M$ and $\lc^{1}_{I}(M)$ are minimax;

  \item[\rm{(}ii\rm{)}] $M$ is finite and $\lc^{1}_{I}(M)$ is
  locally minimax;

  \item[\rm{(}iii\rm{)}] $M$ and $I^{m}\lc^{2}_{I}(M)$ are minimax
  for some $m\in\mathbb{N}_0$;

  \item[\rm{(}iv\rm{)}] $M$ is minimax, $I^{m}\lc^{2}_{I}(M)$ is
  locally minimax, and $\Hom_{R}\big(R/I,I^{m}\lc^{2}_{I}(M)\big)$ is
  finite for some $m\in\mathbb{N}_0$;

  \item[\rm{(}v\rm{)}] $M$ is minimax, and $I^{m}\lc^{2}_{I}(M)$ is locally minimax, for some $m\in\mathbb{N}_0$.
\end{enumerate}
In particular the statments are true when $\dim R\leqslant 3$.
\end{cor}
\begin{proof}
For part (v), apply localization and Theorem \ref{2.19} (v).
\end{proof}

\section{Cofiniteness and Artinianness of $\lc^{i}_{I,J}(M)$}

In this section, we need the concept of $(I,J)$-cofinite modules and $\cd(I,J,M)$.
An $R$-module $M$ is called $(I,J)$-cofinite if Supp$(M)\subseteq W(I,J)$
and $\Ext^i_R(R/I,M)$ is a finite $R$-module, for all $i\geq0$.
Also $~\cd(I,J,M)={\sup\big\{i\in\mathbb{N}_0}\mid H^i_{I,J}(M)\neq0\big\}$
(see \cite{TehTa} and \cite{ChWa}, resp.)

\begin{thm}\label{3.1}
Let $(R,\mathfrak{m})$ be a local ring and $M$ be a finite $R$-module.
Let $t\in\mathbb{N}$ be an integer such that
$\Supp(\lc^{i}_{I,J}(M))\subseteq \{ \mathfrak{m} \}$ for all $i<t$.
Then, for all $i< t$, the $R$-module $\lc^{i}_{I,J}(M)$
 is artinian and $I$-cofinite.
\begin{proof}
We do this by induction on $t$. When $t=1$, it is obvious that
$\lc^{0}_{I,J}(M)$ is artinian $I$-cofinite $R$-module,
since it is a finite module with support in $\{\mathfrak{m}\}$. Now, suppose that
$t\geq2$ and the case $t-1$ is settled.
By \cite[Corollary 1.13]{TakYoYo}, we may assume that $M$ is
$(I,J)$-torsion free, and so $I$-torsion free $R$-module.
Therefore, by \cite[Lemma 2.1.1]{BroSh}, there exists
$x\in I\setminus\bigcup_{\mathfrak{p}\in \Ass(M)}\mathfrak{p}$ such that
$0\rightarrow M^{{.x}\atop{\longrightarrow}} M\rightarrow M/xM \rightarrow0$
is exact. Now, by the exact sequence
$$\lc^{t-2}_{I,J}(M)\rightarrow\lc^{t-2}_{I,J}(M/xM)\rightarrow
\lc^{t-1}_{I,J}(M)^{{.x}\atop{\longrightarrow}}\lc^{t-1}_{I,J}(M),$$
we get the following exact sequence
$$\lc^{t-2}_{I,J}(M)\rightarrow\lc^{t-2}_{I,J}(M/xM)\rightarrow
\big(0:_{\lc^{t-1}_{I,J}(M)}x\big)\rightarrow0.$$
Thus by inductive hypothesis $\lc^{t-2}_{I,J}(M/xM)$ is
artinian $I$-cofinite . So $\big(0:_{\lc^{t-1}_{I,J}(M)}x\big)$
is artinian $I$-cofinite. As Supp$(\lc^{t-1}_{I,J}(M))\subseteq
 \{ \mathfrak{m} \}$, hence $\lc^{t-1}_{I,J}(M)$ is $I$-torsion.
 Now, the assertion follows from \cite[Proposition 4.1]{Mel1}.
\end{proof}
\end{thm}

\begin{thm} \label {3.2}
Let $M$ be an $R$-module such that $\Ext^{i}_R(R/I,M)$ is finite
for all $i\geq0$. Let $~t\in\mathbb{N}_0$ be such that
$\lc^{i}_{I,J}(M)$ is $(I,J)$-cofinite for all $i\neq t$.
Then $\lc^{t}_{I,J}(M)$ is $(I,J)$-cofinite.
\begin{proof}
Apply \cite[Theorem 3.11]{AhSad}.
\end{proof}
\end{thm}

\begin{thm} \label {3.3}
Let $M$ be a finite $R$-module. If
$~\emph{cd}(I,J,M)\leq1$, then
$~\lc^{i}_{I,J}(M)$ is $(I,J)$-cofinite for all $i\geqslant 0$.
\begin{proof}
When $i=0$ and $i\geq2$, the claim is true, since $\Gamma_{I,J}(M)$
is finite and $\cd(I,J,M)\leq1$.  For $i=1$, apply Theorem \ref{3.2}.
\end{proof}
\end{thm}

\begin{thm} \label {3.4}
Let $(R,\mathfrak{m})$ be a local ring and $M$ be a finite $R$-module with $\dim M=n$.
Then $\lc^{n}_{I,J}(M)$ is artinian and $I$-cofinite. In fact,
$\Ext^i_R\big(R/I,\lc^{n}_{I,J}(M)\big)$ has finite length for all $i$.
\begin{proof}
The assertion follows from \cite[Theorem 2.1]{ChWa}, \cite[Theorem 2.3]{Ch} and
\cite[Theorem 3]{DelMar}.
\end{proof}
\end{thm}

\begin{thm} \label {3.5}
Let $(R,\mathfrak{m})$ be a local ring and $M$ be a finite $R$-module.
If $\dim M\leqslant 2$, then $\lc^{i}_{I,J}(M)$ is $(I,J)$-cofinite
for all $i\geqslant 0$, in particular it is true when $\dim R\leqslant 2$.
\begin{proof}
Apply Theorems \ref{3.4} , {3.2}.
\end{proof}
\end{thm}

\section{Upper bounds of $\lc^{i}_{I,J}(M)$}

In this section, we introduce the concept of Serre cohomological dimension of
$M$ with respect to a pair of ideals $(I,J)$, but first, we
characterize the membership of $\lc^{i}_{I,J}(M)$ in a Serre class from
upper bound.

\begin{thm}\label{4.1}
Let $n\in\mathbb{N}_0$ and $M$ be a finite $R$--module.
Then the following statements are equivalent:

\begin{enumerate}
  \item[\rm{(}i\rm{)}] $\lc^{i}_{I,J}(M)$ is in $\mathcal S$ for all $i> n$.

  \item[\rm{(}ii\rm{)}] $\lc^{i}_{I,J}(N)$ is in $\mathcal S$ for all $i> n$
  and for any finite $R$-module $N$ such that $\Supp_R(N)\subseteq\Supp_R(M)$.

  \item[\rm{(}iii\rm{)}] $\lc^{i}_{I,J}(R/\mathfrak{p})$ is in $\mathcal S$ for all
  $\mathfrak{p}\in\Supp_R(M)$ and for all $i> n$.

  \item[\rm{(}iv\rm{)}] $\lc^{i}_{I,J}(R/\mathfrak{p})$ is in $\mathcal S$ for all
  $\mathfrak{p}\in\Min\Ass_R(M)$ and for all $i> n$.

\end{enumerate}
\end{thm}
\begin{proof}
Apply the method of the proof of Theorem 3.1 in \cite{AghMel}
to $\lc^{i}_{I,J}$ .
\end{proof}

\begin{cor}\label{4.2}
Let $M$ , $N$ be finite $R$-modules such that $~\Supp(N)\subseteq\Supp(M)$ and
$n\in\mathbb{N}_0$. If $\lc^{i}_{I,J}(M)\in\mathcal S$ for all $i> n$, then
$\lc^{i}_{I,J}(N)\in\mathcal S$ for all $i> n$.
\end{cor}

\begin{lem}\label{4.3}
For any $R$-module $N$, the following statements are fulfilled.

\begin{enumerate}

  \item[\rm{(}i\rm{)}] $\Gamma_{\mathfrak{a}}(N)\subseteq\Gamma_{\mathfrak{a},J}(N)
  \subseteq\Gamma_{I,J}(N)$ for any $\mathfrak{a}\in\tilde{W}(I,J)$.

  \item[\rm{(}ii\rm{)}] $\Gamma_{I,J}(N)=0$ if and only if $\Gamma_{\mathfrak{a},J}(N)=0$
  for any $\mathfrak{a}\in\tilde{W}(I,J)$.

  \item[\rm{(}iii\rm{)}] $\Gamma_{I,J}(N)=N$ if and only if there exists
  $\mathfrak{a}\in\tilde{W}(I,J)$ such that $\Gamma_{\mathfrak{a},J}(N)=N$.

  \item[\rm{(}iv\rm{)}] If there exists $\mathfrak{a}\in\tilde{W}(I,J)$ such that
  $\Gamma_{\mathfrak{a}}(N)=N$, then $\Gamma_{I,J}(N)=N$.

  \item[\rm{(}v\rm{)}] $\Gamma_{I,J}(N)=\cup_{\mathfrak{a}\in\tilde{W}(I,J)}\Gamma_{\mathfrak{a}}(N)=
  \cup_{\mathfrak{a}\in\tilde{W}(I,J)}\Gamma_{\mathfrak{a},J}(N)$

\end{enumerate}

\begin{proof}
All these statements follow easily from the definitions. We will only prove
the statement (v). Since $\Gamma_{\mathfrak{b}}(N)\subseteq\Gamma_{\mathfrak{b},J}(N)$, for
any ideal $\mathfrak{b}$ of $R$, so by \cite[Theorem 3.2]{TakYoYo} and part(i),
we get\\
$\Gamma_{I,J}(N)\subseteq\cup_{\mathfrak{a}\in\tilde{W}(I,J)}\Gamma_{\mathfrak{a}}(N)\subseteq
\cup_{\mathfrak{a}\in\tilde{W}(I,J)}\Gamma_{\mathfrak{a},J}(N)\subseteq\Gamma_{I,J}(N).$
\end{proof}
\end{lem}

In \cite{DivEsm}, authors introduced the concept of ZD-modules.
An $R$-module $M$ is said to be a ZD-module (zero-divisor module) if for every
submodule $N$ of $M$, the set of zero divisors of $M/N$ is a union of finitely
many prime ideals in Ass$_R(M/N)$. By \cite[Example 2.2]{DivEsm}, the class of
ZD-modules contains modules with finite support, finitely generated, Laskerian,
weakly Laskerian, linearly compact, Matlis reflexive and minimax $R$-modules.

Now, we are in position to prove two other main results of this paper
(Theorem \ref{4.4} and Theorem \ref{4.7}), which the first one can be
considered as a generalization of \cite[Theorem 3.1]{AsgTo}.

\begin{thm}\label{4.4}
Let $M$ be a ZD-module of finite Krull dimension. Let $n\in\mathbb{N}$ be such that
$\lc^{i}_{I,J}(M)\in\mathcal{S}$ for all $i>n$. Then
$\lc^{i}_{I,J}(M)/\mathfrak{a}^{j}\lc^{i}_{I,J}(M)\in\mathcal{S}$ for any
$\mathfrak{a}\in\tilde{W}(I,J)$, all $i\geq n$, and all $j\geq0$.
\begin{proof}
It is enough to verify the assertion for just $i=n$ and $j=1$.
To do this, we use induction on
$d$:=dim$M$. When $d=0$, the result follows from \cite[Theorem 3.2]{TakYoYo} and
Grothendieck$^,$s Vanishing theorem.\\
Next, we assume that $d>0$ and the claim is true for
all $R$-modules of dimension less than $d$. By \cite[Theorem 1.3 (4)]{TakYoYo}, we have
$\lc^{j}_{I,J}(M)\cong \lc^{j}_{I,J}\big(M/\Gamma_{I,J}(M)\big)$ for all $j>0$.
Also, $M/\Gamma_{I,J}(M)$ has dimension not exceeding $d$, and is an $(I,J)$-torsion-free
$R$-module. Therefore we may assume that $\Gamma_{I,J}(M)=0$ and so, by Lemma \ref{4.3},
$\Gamma_{\mathfrak{a}}(M)=0$. By \cite[Lemma 2.4]{DivEsm}, $x\in \mathfrak{a}\setminus\bigcup_{\mathfrak{p}\in \textmd{Ass}(M)}\mathfrak{p}$.
Now, the $R$-module $M/xM$ is ZD-module of dimension $d-1$.
Considering the exact sequence
$$0\rightarrow M^{{.x}\atop{\longrightarrow}} M\rightarrow M/xM \rightarrow0$$
induces a long exact sequence of local cohomology modules, which shows that
$\lc^{i}_{I,J}(M/xM)\in\mathcal{S}$ for all $i>n$.
Thus by inductive hypothesis $\lc^{n}_{I,J}(M/xM)/\mathfrak{a}\lc^{n}_{I,J}(M/xM)\in\mathcal{S}$.
Now, the exact sequence $$\lc^{n}_{I,J}(M)^{{.x}\atop{\longrightarrow}}
\lc^{n}_{I,J}(M)^{{\alpha}\atop{\longrightarrow}} \lc^{n}_{I,J}(M/xM)^{{\beta}\atop
{\longrightarrow}}\lc^{n+1}_{I,J}(M),$$ induces the following exact sequences
$$\lc^{n}_{I,J}(M)^{{.x}\atop{\longrightarrow}}\lc^{n}_{I,J}(M)\rightarrow
N:=\Image\alpha\rightarrow0,$$
$$0\rightarrow N\rightarrow \lc^{n}_{I,J}(M/xM)\rightarrow K:=\Image\beta\rightarrow0.$$
Therefore, the following two sequences

\begin{enumerate}
\item[]$(\ast)$ $~~$ $~~$ $~\lc^{n}_{I,J}(M)/\mathfrak{a}\lc^{n}_{I,J}(M)^{{.x}\atop
{\longrightarrow}}\lc^{n}_{I,J}(M)/\mathfrak{a}\lc^{n}_{I,J}(M)\rightarrow N/\mathfrak{a} N\rightarrow0,$
\end{enumerate}

\begin{enumerate}
\item[]$(\ast\ast)$ $~~$ $~~$ ${\textmd{Tor}^{R}_{1}(R/\mathfrak{a},K)}\rightarrow N/\mathfrak{a} N
\rightarrow\lc^{n}_{I,J}(M/xM)/\mathfrak{a}\lc^{n}_{I,J}(M/xM)\rightarrow K/\mathfrak{a} K\rightarrow0$
\end{enumerate}
are both exact. Since $x\in \mathfrak{a}$ and from the exact sequence $(\ast)$, we get
$N/\mathfrak{a} N\cong \lc^{n}_{I,J}(M)/\mathfrak{a}\lc^{n}_{I,J}(M)$.
On the other hand, by \cite[Lemma 2.1]{AsgTo}, we have
Tor$^R_1(R/\mathfrak{a},K)\in\mathcal{S}$. Therefore $N/\mathfrak{a} N\in\mathcal{S}$, by $(\ast\ast)$,
as required.
\end{proof}
\end{thm}

The following result is an immediate consequence of the above theorem.

\begin{cor}\label{4.5}
Let $M$ be a ZD-module of finite Krull dimension. Let $n\in\mathbb{N}$ be such that
$\lc^{i}_{I,J}(M)\in\mathcal{S}$ for all $i>n$. Then $\lc^{n}_{I,J}(M)\in\mathcal{S}$
if and only if there exist $\mathfrak{a}\in\tilde{W}(I,J)$ and $m\in\mathbb{N}_0$ such that
$\mathfrak{a}^m\lc^{n}_{I,J}(M)\in\mathcal{S}$.
\begin{proof}
$(\Rightarrow)$ It is obvious.\\
$(\Leftarrow)$ Apply the short exact sequence $0\rightarrow \mathfrak{a}^m\lc^{n}_{I,J}(M)\rightarrow
\lc^{n}_{I,J}(M)\rightarrow \lc^{n}_{I,J}(M)/\mathfrak{a}^m\lc^{n}_{I,J}(M)\rightarrow0$ and
Theorem \ref{4.3}.
\end{proof}
\end{cor}

Applying Corollary \ref{4.5}, for some familiar Serre classes of modules, we get some
results as follows.

\begin{cor}\label{4.6}
Let $M$ be a ZD-module of finite Krull dimension and let $n\in\mathbb{N}$.
Then the following statements are fulfilled.

\begin{enumerate}
  \item[\rm{(}i\rm{)}] If $\lc^{i}_{I,J}(M)$ is finite for all $i>n$, then
  $\lc^{n}_{I,J}(M)$ is finite if and only if there exist $\mathfrak{a}\in\tilde{W}(I,J)$
  and $m\in\mathbb{N}_0$ such that $\mathfrak{a}^m\lc^{n}_{I,J}(M)$ is finite
  if and only if there exist $\mathfrak{a}\in\tilde{W}(I,J)$
  and $m\in\mathbb{N}_0$ such that
  $\lc^{n}_{I,J}(M)/(0:_{\lc^{n}_{I,J}(M)}\mathfrak{a}^m)$ is finite .

  \item[\rm{(}ii\rm{)}] If $\lc^{i}_{I,J}(M)$ is artinian for all $i>n$, then
  $\lc^{n}_{I,J}(M)$ is artinian if and only if there exist
  $\mathfrak{a}\in\tilde{W}(I,J)$ and $m\in\mathbb{N}_0$
  such that $\mathfrak{a}^m\lc^{n}_{I,J}(M)$ is artinian.

  \item[\rm{(}iii\rm{)}] If $\lc^{i}_{I,J}(M)=0$ for all
  $i>n$, then $\lc^{n}_{I,J}(M)={\mathfrak{a}}^j\lc^{n}_{I,J}(M)$ for any
  $\mathfrak{a}\in\tilde{W}(I,J)$ and all $j\geq0$.
  Thus $\lc^{n}_{I,J}(M)=0$ if and only if there exist $\mathfrak{a}\in\tilde{W}(I,J)$
  and $m\in\mathbb{N}_0$ such that ${\mathfrak{a}}^m\lc^{n}_{I,J}(M)=0$.

\end{enumerate}
In particular for $n=\emph{dim}M$ and $\emph{cd}(I,J,M)$
\begin{proof}
Apply Corollary \ref{4.5} and \cite[Theorem 3.1]{AghMel2}.
\end{proof}

\end{cor}

Although the following theorem is seemed to be similar to Theorem \ref{4.4}, but it is
more useful and general than \ref{4.4} for finite $R$-modules.

\begin{thm}\label{4.7}
Let $M$ be a finite $R$ module and $n\in\mathbb{N}_0$ be such that $\lc^{i}_{I,J}(M)$
belongs to $\mathcal S$ for all $i> n$. If $\mathfrak{b}$ is an ideal of $R$ such that
 $\lc^{n}_{I,J}(M/{\mathfrak{b}}M)$ belongs to $\mathcal S$,
then the module $\lc^{n}_{I,J}(M)/{\mathfrak{b}}\lc^{n}_{I,J}(M)$ belongs to
 $\mathcal S$.
\end{thm}
\begin{proof}
Let $\mathfrak{b}=(b_1,\dots,b_r)$ and consider the map
$ f : M^r \to M $, defined by
$f(x_1,\dots,x_r)=\sum_1^r b_ix_i$.
Then $\Image f=\mathfrak{b} M$ and $\Coker f=M/\mathfrak{b} M$.
Since $\h^i_{I,J}(M)$ is in $\mathcal S$ for all $i>n$ and
$\Supp(\Ker f)\subseteq\Supp(M)$, it follows from Theorem \ref{4.1}
that $\h^{n+1}_{I,J}(\Ker f)$ is also in $\mathcal S$.
By hypothesis $\h^n_{I,J}(\Coker f)$ belongs to $\mathcal S$.
Hence by \cite[Corollary 3.2]{Mel1} $\Coker{\h^n_{I,J}(f)}$, which equals to
$\h^n_{I,J}(M)/{{\mathfrak{b}}\h^n_{I,J}(M)}$, is in $\mathcal S$.
\end{proof}

\begin{cor}\label{4.8}
Let $M$ be a finite $R$-module and $n\in\mathbb{N}$ be such that
$\lc^{i}_{I,J}(M)\in\mathcal{S}$ for all $i> n$. Then
$\lc^{n}_{I,J}(M)/{\mathfrak{a}}\lc^{n}_{I,J}(M)\in\mathcal S$ for any
$\mathfrak{a}\in\tilde{W}(I,J)$, in particular for $\mathfrak{a}=I$.
\begin{proof}
Let $\mathfrak{a}\in\tilde{W}(I,J)$. Since $M/{\mathfrak{a}}M$ is $\mathfrak{a}$-torsion $R$-module,
thus the assertion follows from Lemma \ref{4.3} (iv), \cite[Corollary 1.13]{TakYoYo}
and Theorem \ref{4.7}.
\end{proof}
\end{cor}

\begin{cor}\label{4.9}
Let $M$ be a finite $R$-module. Let $n\in\mathbb{N}$ be such that
$\lc^{i}_{I,J}(M)\in\mathcal{S}$ for all $i>n$. Then
$\lc^{n}_{I,J}(M)\in\mathcal{S}$ if and only if there exists
$m\in\mathbb{N}_0$ such that $\mathfrak{a}^m\lc^{n}_{I,J}(M)\in\mathcal{S}$.
\end{cor}

\begin{cor}\label{4.10}
Let $M$ be a finite $R$-module. Let $\mathfrak{a}\in\tilde{W}(I,J)$ and
$n\in\mathbb{N}$.
\begin{enumerate}

  \item[\rm{(}i\rm{)}] If $\lc^{i}_{I,J}(M)$ is finite for all $i>n$, then
  $\lc^{n}_{I,J}(M)$ is finite if and only if there exist $\mathfrak{a}\in\tilde{W}(I,J)$
  and $m\in\mathbb{N}_0$ such that $\mathfrak{a}^m\lc^{n}_{I,J}(M)$ is finite
  if and only if there exist $\mathfrak{a}\in\tilde{W}(I,J)$
  and $m\in\mathbb{N}_0$ such that
  $\lc^{n}_{I,J}(M)/(0:_{\lc^{n}_{I,J}(M)}\mathfrak{a}^m)$ is finite

  \item[\rm{(}ii\rm{)}] If $\lc^{i}_{I,J}(M)$ is artinian for
  all $i>n$, then $\lc^{n}_{I,J}(M)$ is artinian if and only if there exists
  $m\in\mathbb{N}_0$ such that ${\mathfrak{a}}^m\lc^{n}_{I,J}(M)$ is artinian.

  \item[\rm{(}iii\rm{)}] If $\lc^{i}_{I,J}(M)=0$ for all
  $i>n$, then $\lc^{n}_{I,J}(M)={\mathfrak{a}}^j\lc^{n}_{I,J}(M)$ for all $j\in\mathbb{N}_0$.
  Thus $\lc^{n}_{I,J}(M)=0$ if and only if there exists $m\in\mathbb{N}_0$ such that
  ${\mathfrak{a}}^m\lc^{n}_{I,J}(M)=0$, in particular for $n=\dim M$ and $\cd(I,J,M)$.

  \item[\rm{(}iv\rm{)}] If $R$ is local and $\lc^{i}_{I,J}(M)$ is finite for all $i>n$,
  then $\lc^{n}_{I,J}(M)={\mathfrak{a}}^j\lc^{n}_{I,J}(M)$ for all $j\geq0$.

\end{enumerate}
\begin{proof}
Apply Corollary \ref{4.9}, \cite[Proposition 1]{LotPay}, and
\cite[Theorem 3.1]{AghMel2}.
\end{proof}
\end{cor}

The following result is more useful whenever $R$ is a local ring and
$I$ is a proper ideal. (see \cite[Lemma 2.1]{Bro})

\begin{cor}\label{4.11}
Let $M$ be a non-zero ZD-module with $d:=\dim M$ and $~t:=\cd(I,J,M)$.
Let $\mathfrak{a}\in\tilde{W}(I,J)$ be such that ${\mathfrak{a}}^m\subseteq\emph{Jac}(R)$ for some
$m\in\mathbb{N}_0$.

\begin{enumerate}

  \item[\rm{(}i\rm{)}] If $t\geq1$, then ${\mathfrak{a}}^j\lc^{t}_{I,J}(M)$
  is not finite for all $j\geq0$.

  \item[\rm{(}ii\rm{)}] If $d\geq1$, then $\lc^{d}_{I,J}(M)$ is finite
  if and only if $~\lc^{d}_{I,J}(M)=0$.

  \item[\rm{(}iii\rm{)}] If $d\geq2$ and $\lc^{d}_{I,J}(M)$ is finite,
  then $\lc^{d-1}_{I,J}(M)/{\mathfrak{a}}^j\lc^{d-1}_{I,J}(M)$ has finite length
  for all $j\geq0$.

\end{enumerate}
In particular when $R$ is local ring and $I\neq R$.
\begin{proof}
(i), (ii) Apply Corollary \ref{4.6} (iii) and Nakayama$^,$s Lemma.\\
(iii) Apply part (ii) and Theorem \ref{4.4} for the class of finite
length.
\end{proof}
\end{cor}

\begin{cor}\label{4.12}
Let $M$ be a non-zero finite $R$-module and set
$t:=\emph{sup}\big\{i\geq1\mid\lc^{i}_{I,J}(M)
\emph{~is not finite}\big\}$,
$n:=\emph{cd}(I,J,M)$, and $r:=\emph{dim}M/JM$.

\begin{enumerate}

  \item[\rm{(}i\rm{)}] If $\mathfrak{a}\in\tilde{W}(I,J)$ is such that
  ${\mathfrak{a}}^m\subseteq\emph{Jac}(R)$ for some $m\in\mathbb{N}_0$, then $n\geq1$
  if and only if $n=t$.

  \item[\rm{(}ii\rm{)}] Let $(R,\mathfrak{m})$ be a local ring and $r\geq1$
  be an integer such that $\lc^{r}_{I,J}(M)$ is finite. Let
  $\mathfrak{a}\in\tilde{W}(I,J)$ be such that ${\mathfrak{a}}^m+J\subseteq\mathfrak{m}$ for some
  $m\in\mathbb{N}_0$. Then $~\lc^{r}_{I,J}(M)=0$.

  \item[\rm{(}iii\rm{)}] If $(R,\mathfrak{m})$ is a local ring and $I+J$ is an
  $\mathfrak{m}$-primary ideal, then $r=n=t$.

\end{enumerate}

\begin{proof}
Apply Corollaries \ref{4.9}, \ref{4.10}, and \cite[Theorems 4.3,4.5]{TakYoYo}.
\end{proof}
\end{cor}

\begin{cor}\label{4.13}
Let $M$ be a finite $R$-module and $n\in\mathbb{N}$.
If $\lc^{i}_{I,J}(M)$ is artinian for all $i> n$. Then
$\lc^{n}_{I,J}(M)/{\mathfrak{a}}\lc^{n}_{I,J}(M)$ is artinian for any
$\mathfrak{a}\in\tilde{W}(I,J)$.
\end{cor}
\begin{proof}
Note that $\lc^{n}_{I,J}(M/{\mathfrak{a}}M)= 0$ for any $\mathfrak{a}\in\tilde{W}(I,J)$
and all $n\ge 1$. Now, apply Theorem \ref{4.7} for the class of artinian modules.
\end{proof}

\begin{rem}\label{4.14}
\emph{In \ref{4.13} we have to assume that $n\ge 1$.
Take an ideal $I$ in a ring $R$ such that $R/I$ is not artinian. Let $J=0$ and
 $M=R/I$. Then $\lc^{i}_{I,J}(M)= 0$ for $i\ge 1$, and $\G_{I}(M)= M$.
On the other hand $M/{I}M\cong M$. Thus
  $\lc^{0}_{I,J}(M)/{I}\lc^{0}_{I,J}(M)$ is not artinian}.
\end{rem}

Properties of Serre classes of modules and the previous results motivate us to
introduce the following definition as a generalization of the concept of
cohomological dimension. (see \cite{ChWa}).

\begin{defn}\label{4.15}
\emph{Let $I$ , $J$ be two ideals of $R$ and let $M$ be an $R$-module.
For a Serre subcategory $\mathcal{S}$ of the category of $R$-modules, we define
Serre cohomological dimension of $M$ with respect to $(I,J)$, by
$${\textmd{cd}}_{\mathcal{S}}(I,J,M)={\textmd{sup}}\big\{{i\in\mathbb{N}_0}\mid
\lc^{i}_{I,J}(M)\not\in\mathcal{S}\big\},$$
if this supremum exists, and $\infty$ otherwise.\\
It is easy to see that
${\textmd{cd}}_{\mathcal{S}}(I,J,M)={\textmd{inf}}\big\{{n\in\mathbb{N}_0}\mid
\lc^{i}_{I,J}(M)\in\mathcal{S} ~\textmd{~for ~all} ~~i>n\big\}$}.
\end{defn}

\begin{rem} \label{4.16}
\emph{For an arbitrary Serre class $\mathcal{S}$ , we have
${\textmd{cd}}_{\mathcal{S}}(I,J,M)\leq{\emph{cd}}(I,J,M)$, and if
in Definition 4.16, we let $\mathcal{S}:=0$ then we have
$${\textmd{cd}}_{\mathcal{S}}(I,J,M)={\textmd{sup}}\big\{{i\in\mathbb{N}_0}\mid
\lc^{i}_{I,J}(M)\neq0\big\}=\textmd{cd}(I,J,M).$$
Also, if $\mathcal{S}$ is the class of Artinian $R$-modules, we get
$${\textmd{cd}}_{\mathcal{S}}(I,J,M)={\textmd{sup}}\big\{{i\in\mathbb{N}_0}\mid
\lc^{i}_{I,J}(M)~ {\textmd{is not Artinian $R$-module}}\big\}.$$
We denote it by $\q(I,J,M)$}.
\end{rem}

\begin{prop} \label{4.17}
Let $M$ be a ZD-module of finite krull dimension or finite $R$-module. Let
$\mathcal{S}$ be a Serre class and $n:=\emph{cd}_{\mathcal{S}}(I,J,M)\geq1$.
Then ${\mathfrak{a}}^j\lc^{n}_{I,J}(M)\not\in\mathcal{S}$ for any
$\mathfrak{a}\in\tilde{W}(I,J)$ and all $j\geq0$,
in particular ${\mathfrak{a}}^j\lc^{n}_{I,J}(M)\neq0$
for any $\mathfrak{a}\in\tilde{W}(I,J)$ and all $j\geq0$.
\begin{proof}
Apply Corollaries \ref{4.5} , \ref{4.9}.
\end{proof}
\end{prop}

It is well known that if $\mathfrak{a}$ is an ideal of $R$ and $M$ , $N$ are finite $R$-modules
with $\textmd{Supp}(N)\subseteq \textmd{Supp}(M)$, then
${\textmd{cd}}(\mathfrak{a},N)\leq {\textmd{cd}}(\mathfrak{a},M)$. The next result is a generalization
of this fact.

\begin{prop} \label{4.18}
Let $M$ , $N$ be finite $R$-modules such that $\emph{Supp}(N)\subseteq \emph{Supp}(M)$.
Then ${\emph{cd}}_{\mathcal{S}}(I,J,N)\leq {\emph{cd}}_{\mathcal{S}}(I,J,M)$.
\end{prop}
\begin{proof}
Apply Corollary \ref{4.2}.
\end{proof}

\begin{cor} \label{4.19}
Let $M$ , $N$ be finite $R$-modules such that $\emph{Supp}(N)\subseteq \emph{Supp}(M)$.
Then  ${\emph{cd}}(I,J,N)\leq {\emph{cd}}(I,J,M)$.
\end{cor}

\begin{cor} \label{4.20}
For a finite $R$-module $M$, there exist the following equalities.
\begin{align*}
\emph{cd}_{\mathcal{S}}(I,J,M)&={\emph{max}~}\big\{\emph{cd}_
{\mathcal{S}}(I,J,R/\mathfrak{p})\mid\mathfrak{p}\in\ {\emph{Ass}}(M)\big\}\\
&={\emph{max}~}\big\{\emph{cd}_{\mathcal{S}}(I,J,R/\mathfrak{p})\mid
\mathfrak{p}\in\ {\emph{Min Ass}}(M)\big\}\\ &={\emph{max}~}\big\{\emph{cd}_
{\mathcal{S}}(I,J,R/\mathfrak{p})\mid\mathfrak{p}\in\ {\emph{Supp}}(M)\big\}\\
&={\emph{max}~}\big\{\emph{cd}_{\mathcal{S}}(I,J,R/\mathfrak{p})\mid
\mathfrak{p}\in\ {\emph{Min Supp}}(M)\big\}\\ &={\emph{max}~}\big\{
\emph{cd}_{\mathcal{S}}(I,J,N)\mid N ~\emph{is a finite submodule } M \big\}\\
&={\emph{max}~}\big\{i\geq0\mid H^i_{I,J}(R/\mathfrak{p})\not\in\mathcal{S},
~\emph{for some} ~\mathfrak{p}\in{\emph{Ass}}(M)\big\}\\
&=\emph{min~}\big\{{n\geq0}\mid\lc^{i}_{I,J}(R/\mathfrak{p})\in\mathcal{S}
~\emph{~for ~all} ~~i>n ~\emph{and  ~all} ~~i>n\big\}.\end{align*}
\end{cor}

\begin{proof}
Apply Corollary \ref{4.1} and Proposition \ref{4.19}.
\end{proof}

\begin{cor} \label{4.21}
Let $M$ , $N$ be finite $R$-modules such that
$\emph{Supp}(N)\subseteq \emph{Supp}(M)$.
Then  $\q(I,J,N)\leq \q(I,J,M)$.
\end{cor}

\begin{prop} \label{4.22}
Let $0\rightarrow L\rightarrow M\rightarrow N\rightarrow0$
be an exact sequence of finite $R$-modules. Then
$\emph{cd}_{\mathcal{S}}(I,J,M)={\emph{max}}~\big\{\emph{cd}_
{\mathcal{S}}(I,J,L)~,~\emph{cd}_{\mathcal{S}}(I,J,N)\big\}$
\end{prop}
\begin{proof}
Let $t:=\textmd{cd}_{\mathcal{S}}(I,J,M)$ and $s:={\textmd{max}~}\big\{
\emph{cd}_{\mathcal{S}}(I,J,L)~,~\emph{cd}_{\mathcal{S}}(I,J,N)\big\}$.
By, Corollary \ref{4.2} , we have $t\geq s$. Now, let $t>s$. Then by
the following exact sequence
$$\cdots\rightarrow H^t_{I,J}(L)\rightarrow H^t_{I,J}(M)\rightarrow
H^t_{I,J}(N)\rightarrow\cdots,$$
we get $H^t_{I,J}(M)\in\mathcal{S}$
which is a contradiction with $\textmd{cd}_{\mathcal{S}}(I,J,M)=t$.
\end{proof}

\begin{cor} \label{4.23}
For a noetherian ring $R$ there exists the following equality.
$$\emph{cd}_{\mathcal{S}}(I,J,R)={\emph{sup}~}\big\{\emph{cd}_
{\mathcal{S}}(I,J,N)\mid ~N~ {\textmd{is a finite $R$-module}} \big\}.$$
In particular, for $r\in\mathbb{N}_0$ the following statements are equivalent:
\begin{enumerate}

\item[(i)] $H^j_{I,J}(R)\in\mathcal{S}$ for all $j>r$.

\item[(ii)] $H^j_{I,J}(M)\in\mathcal{S}$ for all $j>r$ and all finite $R$-module $M$.
\end{enumerate}
\end{cor}

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


text:
Category Archives: Muslims
Babar Ahmad: 7 years without charge
Babar Ahmad
From: http://uk.news.yahoo.com/terror-suspect-interview-ruling-due-031218373.html
" Babar Ahmad, a 37-year-old British Muslim, is being held under controversial extradition laws as he fights removal to the US, where he is wanted for allegedly raising funds for Chechen and Afghan insurgents over the internet.
He strongly denies any involvement with terrorism."
From: http://www.freebabarahmad.com/the-story/who-is-babar-ahmad
Babar Ahmad is a 37 year old British Muslim and the longest detained-without-charge British detainee held as part of the global 'war on terror'.
Babar Ahmad was born in London in May 1974, and, until his imprisonment in 2004, lived in Tooting, South West London. His parents migrated to the UK from Pakistan in the early 1960's. His father is a retired civil servant and his mother a retired Science teacher.
He went to a reputable public school where he won academic prizes and obtained outstanding results at both GCSE and A-Level. He then went to university and obtained a Master's degree in Engineering from the University of London. Before his imprisonment in August 2004, he was working in the IT department at Imperial College London.
He is well-known locally in Tooting as a 'caring and helpful' member of the community through his years of youth work.
In December 2003 Babar was arrested at his London home under anti-terror legislation. By the time he reached the police station Babar had sustained at least 73 forensically recorded injuries, including bleeding in his ears and urine. Six days later he was released without charge.
Babar then filed a formal complaint that he had been subjected to horrific physical, sexual and religious abuse by the arresting police officers. In March 2009 the Metropolitan Police finally admitted in the Royal Courts of Justice in London that they did indeed carry out the Islamophobic and brutal assault on Babar Ahmad in December 2003. Moreover, they paid Babar Ahmad £60,000 compensation for damages. However, the Metropolitan Police have still offered no apology for the actions of their officers.
In August 2004 Babar was re-arrested in London and taken to prison pursuant to an extradition request from the US under the controversial, no-evidence-required, Extradition Act 2003. The US has alleged that in the 1990s Babar was a supporter of "terrorism". Babar denies any involvement in terrorism. He has now been in prison for over five years even though he has not been charged in the UK.
On 03 June 2011, following a trial lasting five weeks at Southwark Crown Court, all four police officers charged with assaulting Babar Ahmad were acquitted. The jury took only 45 mins to reach their verdict and requested to meet the officers to shake their hands following the conclusion of the trial.
After the acquittal, Judge Geoffrey Rivlin QC, the Recorder of Westminster and Deputy High Court Judge said, 'I express the hope that his ordeal as a man in detention in this country for a number of years without trial is brought to an end as soon as possible…It is no concern of this court as to which, but it is a matter of concern and I would have thought should be a matter of concern to the public at large, quite apart from Mr Ahmad, that here is a man who has been in custody for literally years without knowing what his fate is to be.'
On 22 June 2011, in a report released by the House of Parliament Joint Committee on Human Rights (JCHR) urged the UK government to change the law so that Babar Ahmad perpetual threat of extradition is ended without further delay.
Babar's family, friends and campaigners have mounted a high profile campaign for his release. He appeared in the news when it was revealed that the police had bugged his prison visits with his MP, Sadiq Khan (Labour-Tooting). His final appeal against extradition is at The European Court of Human Rights in Strasbourg, which is due to decide on it in late 2011. If extradited he faces the rest of his natural life in solitary confinement in a maximum security US 'Supermax' prison.
(Click here for statements from friends, work colleagues and family)
Posted by Escapees Blog on 2012-01-11 in Corruption, Courts, Crime, Establishment bullshit, Government, Judges, Law, Madness, Muslims, Police State, Polital Prisoner, Prisons, Protest, Security, Terrorists, The War on Terrror, Torture, US
The Real Truth of Wars – Dr. Dahlia Wasfi's
I've a worse than crap Xmas Day – though, infinitely better than most people. I haven't been murdered, bombed on, shot at, etc., etc. .
From: http://youtu.be/6_RvhvGZVkk
We have an obligation to every last victim of this illegal aggression because all of this carnage has been done in our name. Since World War II, 90% of the casualties of war are unarmed civilians. 1/3 of them children. Our victims have done nothing to us. From Palestine to Afghanistan to Iraq to Somalia to wherever our next target may be, their murders are not collateral damage, they are the nature of modern warfare. They don't hate us because of our freedoms. They hate us because every day we are funding and committing crimes against humanity. The so-called "war on terror" is a cover for our military aggression to gain control of the resources of western Asia.
This is sending the poor of this country to kill the poor of those Muslim countries. This is trading blood for oil. This is genocide, and to most of the world, we are the terrorists. In these times, remaining silent on our responsibility to the world and its future is criminal. And in light of our complicity in the supreme crimes against humanity in Iraq and Afghanistan, and ongoing violations of the U.N. Charter in International Law, how dare any American criticize the actions of legitimate resistance to illegal occupation.
Our so-called enemies in Afghanistan, Iraq, Palestine, our other colonies around the world, and our inner cities here at home, are struggling against the oppressive hand of empire, demanding respect for their humanity. They are labeled insurgents or terrorists for resisting rape and pillage by the white establishment, but they are our brothers and sisters in the struggle for justice. The civilians at the other end of our weapons don't have a choice, but American soldiers have choices, and while there may have been some doubt 5 years ago, today we know the truth. Our soldiers don't sacrifice for duty-honor-country, they sacrifice for Kellogg Brown & Root.
They don't fight for America, they fight for their lives and their buddies beside them, because we put them in a war zone. They're not defending our freedoms, they're laying the foundation for 14 permanent military bases to defend the freedoms of Exxon Mobil and British Petroleum.
They're not establishing democracy, they're establishing the basis for an economic occupation to continue after the military occupation has ended. Iraqi society today, thanks to American "help" is defined by house raids, death squads, check-points, detentions, curfews, blood in the streets, and constant violence. We must dare to speak out in support of the Iraqi people, who resist and endure the horrific existence we brought upon them through our bloodthirsty imperial crusade. We must dare to speak out in support of those American war-resisters, the real military heroes, who uphold their oath to defend the constitution of the United States against all enemies, foreign and domestic, including those terrorist cells in Washington DC more commonly known as the Legislative, Executive & Judicial branches.
"If There Is No Struggle, There Is No Progress"
Frederick Douglass said
"Those who profess to favor freedom, and yet deprecate agitation, are people who want crops without plowing the ground. They want rain without thunder and lightning. They want the ocean without the awful roar of its many waters. This struggle may be a moral one, or it may be a physical one, and it may be both … but it must be a struggle. Power concedes nothing without a demand. It never did and it never will."
Every one of us, every one of us must keep demanding, keep fighting, keep thundering, keep plowing, keep speaking, keep struggling until justice is served. NO justice, NO peace.
http://phoenixaquua.blogspot.com/2011/05/truth-hurts-lies-kill-real-truth-of….
Posted by Escapees Blog on 2011-12-28 in Big Business, Corporations, Corruption, Crime, Genocide, Madness, Money, Muslims, Oil, Parasites, Police brutality, Police State, Rape, Terrorists, Torture, US, War, War Crimes
Detained without charge in the United Kingdom for 7 years
Still think that you live in a free country?
From: http://www.indymedia.org.uk/en/2011/10/487372.html
"Babar Ahmad, a British citizen detained for 7 years without charge is sought by the U.S. for extradition. The Free Babar Ahmad Campaign need 100,000 signatures by 10th November 2011 deadline, to necessarily force the issue of his extradition before Parliament. Please sign the online petition and ask your friends to do the same by clicking here http://www.freebabarahmad.com/ ."
Posted by Escapees Blog on 2011-10-25 in Corruption, Courts, Crime, Establishment bullshit, Government, Judges, Law, Madness, Muslims, New World Order, Police State, Polital Prisoner, Prisons, Terrorists, The War on Terrror, Torture, US
US assassinates it own citizens with no due process
Dead (wo)men can't protest their innocence.
"Glenn Greenwald reports on the US assassination of Anwar al-Awlaki, a US citizen in Yemen, who had not been charged with (or convicted of) any crime. Al-Awlaki was "far from any battlefield," and no judge or jury considered any accusations that had been levelled against him, nor did he have the opportunity to face his accusers nor offer a defense. Whether or not al-Awlaki was a terrorist (something no court can determine now), this sets a new precedent: the US can assassinate its own citizens on presidential order without any due process or accountability."
From: http://boingboing.net/2011/09/30/new-reality-us-assassinates-it-own-citizens-with-no-due-process.html
Posted by Escapees Blog on 2011-10-02 in Corruption, Crime, Government, Judges, Law, Madness, Muslims, New World Order, Obama, Police State, Security, Terrorists, The War on Terrror, Thought Crimes, US
FBI Teaches Agents: 'Mainstream' Muslims Are 'Violent, Radical'
The FBI is teaching its counterterrorism agents that "main stream" [sic] American Muslims are likely to be terrorist sympathizers; that the Prophet Mohammed was a "cult leader"; and that the Islamic practice of giving charity is no more than a "funding mechanism for combat."
At the Bureau's training ground in Quantico, Virginia, agents are shown a chart contending that the more "devout" a Muslim, the more likely he is to be "violent." Those destructive tendencies cannot be reversed, an FBI instructional presentation adds: "Any war against non-believers is justified" under Muslim law; a "moderating process cannot happen if the Koran continues to be regarded as the unalterable word of Allah."
These are excerpts from dozens of pages of recent FBI training material on Islam that Danger Room has acquired. In them, the Constitutionally protected religious faith of millions of Americans is portrayed as an indicator of terrorist activity.
"There may not be a 'radical' threat as much as it is simply a normal assertion of the orthodox ideology," one FBI presentation notes. "The strategic themes animating these Islamic values are not fringe; they are main stream."
The FBI isn't just treading on thin legal ice by portraying ordinary, observant Americans as terrorists-in-waiting, former counterterrorism agents say. It's also playing into al-Qaida's hands.
Focusing on the religious behavior of American citizens instead of proven indicators of criminal activity like stockpiling guns or using shady financing makes it more likely that the FBI will miss the real warning signs of terrorism. And depicting Islam as inseparable from political violence is exactly the narrative al-Qaida spins — as is the related idea that America and Islam are necessarily in conflict. That's why FBI whistleblowers provided Danger Room with these materials.
Over the past few years, American Muslim civil rights groups have raised alarm about increased FBI and police presence in Islamic community centers and mosques, fearing that their lawful behavior is being targeted under the broad brush of counterterrorism. The documents may help explain the heavy scrutiny.
They certainly aren't the first time the FBI has portrayed Muslims in a negative light during Bureau training sessions. As Danger Room reported in July, the FBI's Training Division has included anti-Islam books, and materials that claim Islam "transforms [a] country's culture into 7th-century Arabian ways." When Danger Room confronted the FBI with that material, an official statement issued to us claimed, "The presentation in question was a rudimentary version used for a limited time that has since been replaced."
But these documents aren't relics from an earlier era. One of these briefings, titled "Strategic Themes and Drivers in Islamic Law," took place on March 21.
The Islam briefings are elective, not mandatory. "A disclaimer accompanied the presentation stating that the views expressed are those of the author and do not necessarily reflect the views of the U.S. government," FBI spokesman Christopher Allen tells Danger Room.
"The training materials in question were delivered as Stage Two training to counterterrorism-designated agents," Allen adds. "This training was largely derived from a variety of open source publications and includes the opinion of the analyst that developed the lesson block."
Not all counterterrorism veterans consider the briefings so benign. "Teaching counterterrorism operatives about obscure aspects of Islam," says Robert McFadden, who recently retired as one of the Navy Criminal Investigative Service's al-Qaida-hunters, "without context, without objectivity, and without covering other non-religious drivers of dangerous behavior is no way to stop actual terrorists."
Still, at Quantico, the alleged connection between Islam and violence isn't just stipulated. It's literally graphed.
An FBI presentation titled "Militancy Considerations" measures the relationship between piety and violence among the texts of the three Abrahamic faiths. As time goes on, the followers of the Torah and the Bible move from "violent" to "non-violent." Not so for devotees of the Koran, whose "moderating process has not happened." The line representing violent behavior from devout Muslims flatlines and continues outward, from 610 A.D. to 2010. In other words, religious Muslims have been and always will be agents of aggression.
Training at Quantico isn't designed for intellectual bull sessions or abstract theory, according to FBI veterans. The FBI conducts its training so that both seasoned agents and new recruits can sharpen their investigative skills.
In this case, the FBI's Allen says, the counterterrorism agents who received these briefings have "spent two to three years on the job." The briefings are written accordingly. The stated purpose of one, about allegedly religious-sanctioned lying, is to "identify the elements of verbal deception in Islam and their impacts on Law Enforcement." Not "terrorism." Not even "Islamist extremism." Islam.
What's more, the Islamic "insurgency" is all-encompassing and insidious. In addition to outright combat, its "techniques" include "immigration" and "law suits." So if a Muslim wishes to become an American or sues the FBI for harassment, it's all just part of the jihad.
On Tuesday, the leaders of the Senate Committee on Homeland Security, Joe Lieberman (I-Connecticut) and Susan Collins (R-Maine), warned that law enforcement lacks "meaningful standards" to prevent anti-Islam material from seeping into counterterrorism training. Some FBI veterans suspect the increased pressure on American Muslims has a lot to do with the kind of training that Quantico offers.
"Seeing the materials FBI agents are being trained with certainly helps explain why we've seen so many inappropriate FBI surveillance operations broadly targeting the Muslim-American community, from infiltrating mosques with agents provocateur to racial- and ethnic-mapping programs," Mike German, a former FBI agent now with the American Civil Liberties Union, tells Danger Room after being shown the documents. "Biased police training can only result in biased policing." (Full disclosure: This reporter's wife works for the ACLU.)
The chief of the Training Division, Assistant FBI Director Thomas Browne, came into his current job in January. His official biography lists no terrorism expertise beyond serving as a coordinator for a bureau "Domestic Terrorism Program" in Tennessee sometime in the last decade.
It is unclear what vetting process the FBI used to approve these briefings; if any Muslim scholars contributed to them; and what criteria Quantico uses to determine Islamic expertise. "The development of effective training is a constantly evolving process," says FBI spokesman Allen. "Sometimes the training is adapted for long-term use. This particular training segment was delivered a single time and not used since."
Several of these briefings were the work of a single author: an FBI intelligence analyst named William Gawthrop. In 2006, before he joined the Bureau, he gave an interview to the website WorldNetDaily, and discussed some of the themes that made it into his briefings, years later. The Prophet "Muhammad's mindset is a source for terrorism," Gawthrop told the website, which would later distinguish itself as a leader of the "birther" movement, a conspiracy theory that denies President Obama's American citizenship.
At the time, Gawthrop's major suggestion for waging the war on terrorism was to attack what he called "soft spots" in Islamic faith that might "induce a deteriorating cascade effect upon the target." That is, to discredit Islam itself and cause Muslims to abandon their religion. "Critical vulnerabilities of the Koran, for example, are that it was uttered by a mortal," he said. Alas, he lamented, he faced the bureaucratic obstacle of official Washington's "political taboo of linking Islamic violence to the religion of Islam," according to the website.
Back then, however, Gawthrop didn't work for the FBI. He had recently stepped down from a position with the Defense Department's Counterintelligence Field Activity. That agency came under withering criticism during the Bush administration for keeping a database about threats to military bases that included reports on peaceful antiwar protesters and dovish Church groups. It is unclear how Gawthrop came to work for the FBI.
Through an intermediary, Gawthrop told Danger Room that he was unavailable for comment before our deadline.
The FBI didn't always conflate terrorism with Islam. "I never saw that," says Ali Soufan, one of the FBI's most distinguished counterterrorism agents and author of the new memoir The Black Banners, who retired from the bureau in 2005. "Sometimes, toward the end of my time, I started noticing it with different entities outside the FBI. You started feeling like they had a problem with Islam-as-Islam, because of the media. But that was a few people, and was usually hidden behind closed doors."
Soufan, a Muslim, has interrogated members of al-Qaida and contributed to rolling up one of its cells in Yemen after 9/11. But by the logic of the FBI's training materials, Soufan's religious practices make him a potential terrorist.
McFadden, the former NCIS counterterrorist, has a lot of respect for his FBI colleagues, who he believes are ill-served by these Islam briefings. "These are earnest special agents and police officers who want to know how do their job better," McFadden says.
Too often, McFadden says, counterterrorism training becomes simultaneously over-broad and ignorant. "Instead of looking for indicators of nefarious behavior, you have a sweeping generalization of things like, for instance, the Hawala system," McFadden explains. "It's a system that most of the developing world and expatriates from it use to move money around, including terrorists. But you can't say the whole hawala system is about terrorism, just like you can't say that Islam as a whole has anything to do with bad behavior."
McFadden, a Catholic, believes that obsessing over obscure Koranic verses is as useful a guide to terrorist behavior as "diving into the rite of exorcism" is to understanding Catholicism.
On April 6, barely two weeks after the "Islamic Motivations for 'Suicide' Bombers" briefing at Quantico, FBI Director Robert Mueller defended the bureau's budget before a congressional committee. Among his major points: the FBI needs cooperation from American Muslims to stop the next terrorist attack.
"Since September 11th, every one of our 56 field offices and the leadership of those offices have had outreach to the Muslim community," Mueller said. "We need the support of that community … our business is basically relationships." That is exactly the opposite message sent in the training rooms of Quantico, where the next generation of FBI counterterrorism is shaped.
From: http://www.wired.com/gadgetlab/2011/09/no-flash-windows-8-metro/
Posted by Escapees Blog on 2011-09-16 in Establishment bullshit, FBI, Government, Madness, Muslims, Police State, Terrorists, The War on Terrror, Thought Crimes, US


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