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Bullied: You'Re Not Alone - Cyberbully (2013)
Education; Social sciences
Sale, DVD (Region 1 NTSC, 28 mins), $249
Cyberbullying affects all students, regardless of gender, physical difference, or mental ability. The DVD considers the experience of Jacob Zellmann, who was led to the brink of suicide due to cyberbullying....
DVD Find Other To order
Geosynthetics: An Overview of Designs (2009)
Civil Engineering & Building
Sale, DVD (Region 1 NTSC, 45 minutes), $249.00
This programme shows how to use geosynthetics in civil engineering. It covers such topics as walls, slopes, dams, roads, and erosion control.
Bystander Intervention - Diffusion of Responsibility (2008)
DVD (Region 2 PAL, 15 minutes), £39.00
The psychological phenomenon of 'bystander apathy' states that the more bystanders who witness an emergency the less likely it is that any one of them will intervene to help. This programme examines the...
How to Read and Understand a Research Study (2008)
Education; Science
Sale, DVD (Region 1 NTSC, 24 minutes), £219.00
This programme teaches how to understand research studies. It outlines the structure of a scientific paper; differentiates between such research method types as observational studies, true experiments, and...
Presenting William Shakespeare (2007 Video)
Considers William Shakespeare as an actor, poet, and playwright. It examines his life and explores his literary works.
Santiago Calatrava's Sundial Bridge: Angle of Inspiration (2007)
This programme examines the Sundial Bridge in Redding, California, which was designed by Santiago Calatrava. It considers the bridge's innovative architectural design and highlights challenges to its...
Testing Textiles: Testing Fabrics and Products (2005)
Design; Technology
Sale, DVD (Region 1, 30 minutes), £169.00
Featuring footage from industrial textiles testing laboratories, this programme examines the tests that textiles undergo before reaching the market. It looks at such tests as flammability, abrasion, pilling,...
Biological Diversity of Yeasts (2005)
Sale, DVD (Region 1, 45 minutes), $189.00
This programme considers types of yeasts, discusses their biological properties, explains the sexual reproduction and life cycles of yeasts, and looks at yeast ecology.
Anthony Holland Presents Shakespeare (2005 Video)
Actor Antony Holland performs Shakespearean speeches and soliloquies, emphasising the importance of language and text. He offers insights into the characters, settings, and performances and discusses acting...
Period Movement: Early and Late Renaissance (2005 Video)
aka: Period Movement for Actors
Judith Chaffee; Joel Asher
Joel Asher
Dance; Drama
Judith Chaffee conducts a workshop with students on movement for period plays. Working with scenes from Shakespeare's Twelfth Night, and such dances as the pavanne from the early Renaissance and the...
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Blogging can be a great way to position yourself as an authority and create customer loyalty – if you're willing to devote the time necessary.
Follow these 9 Tips for Better Business Blogging and you'll be more successful!
1. Forgo the advertorial dribble. Decades ago some bright entrepreneur realized that if he wrote his sales copy to look like an editorial, he'd get more readers and more sales. Thus we entered into the age of the "advertorial," and it still works in print publications. Where it does NOT work well is on your blog.
Yes, by all means advertise your products and services on your website, but don't be devious about it. Don't make it look like an article when it's really a self-promotion – it annoys your readers and makes them feel let down when you've got them thinking they're going to learn something new, only to discover they've got to pay to get it.
Thus, do not title your blog post, "Revealed: 35 Ways to Drive Traffic" and then reveal nothing about driving traffic except the order link to your new product. Instead, title your blog post, "Revealed: 5 Ways to Drive Traffic," and then GIVE them the 5 ways. At the end, let them know that if they're interested, you reveal 30 additional methods inside your product and give them the link. They'll respect you for this, and because you've given them good information and demonstrated that you know what you're talking about, it will actually INCREASE your sales.
2. Be the authority. This goes along with Tip #1, because when you're dishing about all the latest updates in your industry rather than spending all your time promoting yourself, you become the expert in your prospects' eyes. And who do people want to do business with? The expert. Who can charge more for their products and services? Again, the expert. Offer up the latest news along with plenty of how-to articles and your readers will come to like and trust you, and thus want to do business with you.
3. Educate, inform and inspire. Whenever possible, use real life stories to inspire your readers. Nothing sells like success, and people love to read how others have overcome the same challenges they're having. Whenever you're stuck for a blog post idea, just think, "What do my readers want to know? How can I help my readers and give them value?" Watch the forums for questions that pop up – this is an excellent source for ideas.
4. Show your personality. Building reader loyalty takes time and more than just great information – it also takes personality. Let your own unique style shine through in your posts, and don't be afraid to reveal insights about yourself, and to even be controversial when it's called for. Nothing rallies readers like a good controversy, and nothing makes readers more loyal than if they know you well enough to think of you as a friend.
5. Promote your blog. It would be great if you could write it and people would come, but it's seldom that easy. Use your own network to promote your blog by letting them know each time you've made a new post. Guest write for other blogs to promote your own blog. And always ask people to share your blog with others.
6. Ask readers to reply. The more active your blog is, the more appealing it is to new and old readers alike. Always ask your readers to leave their comments on each of your blog posts. You might even consider running a contest to see who can write the best response, and give a prize to the winner.
Whatever it takes to increase comments is something you should consider. Not only does it make your blog appear that much more active and universally liked – it also provides a sense of ownership to each person who takes the time to leave a comment. The more they comment, the more likely they are to tell others about your blog and to return to your blog in the future. And be sure to respond to your reader's comments so they know they're being heard and appreciated.
7. Don't blog unless you really want to. Blogging is a commitment, and while it might seem exciting and fun in the beginning, writing a new blog post every day or two can get old, especially after the first couple of weeks when you feel like nobody's been reading it. If you're not sure you can keep a blog rolling, consider writing guest posts for other blogs, websites and email newsletters. Every publisher is looking for new material, and being the guest blogger or writer puts you in front of an entirely new audience of prospects each time.
8. If you're guest blogging, write a GREAT resource box. Offer them something intriguing with tremendous value for going to your site, and then deliver above and beyond their expectations. This is an excellent method of increasing your list of prospects and becoming well known in your niche.
9. Interview the big dogs. Interviewing others in your niche not only produces great content – it also provides you with powerful contacts and access to their readers when they mention the interview to their own people. So go ahead and ask others in your niche for interviews – the worst they can do is say no, and you'll be surprised how many quickly say yes. Remember, they want to increase their own readership as well, and it's a tremendous ego boost to be sought out and interviewed as an expert in their industry.
If there is a Tip #10, it is this: Have fun. Whatever it takes to make your blog writing fun is what you should do, since your enjoyment and passion for your topic will shine through in your posts. The more fun you're having writing your blog, the more enjoyable it will be for readers to visit and linger.
So yes, have a blast, add humor, don't be afraid to hold yourself up as an example to your readers, (both good and bad,) put your heart and soul into it, and go for it!


text:
Q: Destiny: I can't see the next story mission This is going to be supernooby.  I'm very new to Destiny.  I rented it, I don't really like it, and I want to beat it and move on.  I'm trying to play the story missions, but after completing one, I don't see another new mission.  I see level requirements being mentioned when I try to read up about this, but do you seriously have to run around trying to level up just to play the story missions?  I appreciate any info.

A: Yes, you are correct. In order to advance in the campaign, you must also advance in levels. To level up one may play-through the campaign missions to receive xp, play in the crucible, complete bounties and strike missions as well as other ways. Unfortunately this game is less of a rent for the weekend and more of a spend weeks grinding for weapons and armour in raids and weeklies, so if that is not your desired play style than this is probably not the game for you.

A: If you're unable to see the next green exclamation mark indicating the next story mission, try travelling to the Reef. Here you have to sit through a cut scene, but then the next mission will be available to you afterwards.

A: It's hard to say without knowing what part you are on. What's the last mission you completed? Do you see and green indicator marks on the planet screen that usually tell you where to go next?
The story is fairly quick to finish in one day and you shouldn't need to worry about leveling up too much in order to complete the story of the game. You will pretty much level up as fast as needed for every following mission until you hit 20 and you're probably done at that point.
The story missions are there pretty much there to get you to level 20 and then you can start doing other things like strikes and raids and crucible to get even better gear that has light levels that get you past level 20 all the way to level 32.

redpajama_set_name: RedPajamaStackExchange


text:
Fifteen and Counting: Iowa to consider legislation against NSA spying
February 14, 2014 Activist Post Uncategorized 0
The movement to thwart NSA spying at the state level continues to grow, with Iowa becoming the 15th state to consider legislation denying support to the spy agency.
On Wednesday, Iowa state Senator Jake Chapman introduced the Fourth Amendment Protection Act. If SF2172 becomes law, the state, it agencies or political subdivisions will not "provide material support for participation with or assistance to, in any form, any federal agency which claims the power, or which purports due to any federal law, regulation, or order, to authorize the collection of electronic data or metadata of any person pursuant to any action not based on a warrant that particularly describes the person, place or thing to be searched or seized."
"When I took the oath of office, I swore to support the Constitution of the United States of America and the Constitution of the great State of Iowa," Chapman said. "This federal agency has usurped our Constitutional rights. My bill affirms our Constitutional rights and protects the citizens of Iowa. We have learned in recent months through investigations and through the media, the NSA is collecting and storing nearly 30 percent of all Americans' call records, they collect and store over 200 million text messages daily, and they are tracking American's through social media, including GPS tracking. We must not trade freedom for security. My bill may not protect all Americans, but it will certainly protect Iowans."
Based on model legislation drafted by a transpartisan coalition organized by the Tenth Amendment Center (TAC) and the Bill of Rights Defense Committee (BORDC) called the OffNow Coalition, the Iowa Fourth Amendment Protection Act blocks state support for the NSA through four avenues.
The first prohibits state and local agencies within their jurisdiction from providing any material support to the NSA. That includes barring government-owned utilities from providing water and electricity. Second, it makes information gathered without a warrant by the NSA and shared with law enforcement inadmissible in state court. Third, it blocks public universities from serving as NSA research facilities or recruiting grounds. And finally, the legislation provides sanctions against corporations that attempt to fill needs not met in the absence of state cooperation.
While the NSA does not currently operate a data or "threat operations" center in Iowa, Tenth Amendment Center national communications director Mike Maharrey said Iowa and states around the country need to pass similar legislation to make NSA expansion more difficult
"We know the NSA is aggressively expanding its physical locations, not just in Utah, but in Texas, Hawaii and other states too," he said. "Since the NSA isn't transparent about its plans, it's essential to not only address where it is today, but work to get the rest of the country to say, 'You're not welcome here either!' If enough states pass this type of legislation, the spy agency will find it increasingly difficult to get the support it needs to violate your rights."
Maharrey said that even without a physical facility to address, the bill will have immediate and dramatic effect in the Hawkeye State, erasing a practical effect of NSA spying – data sharing.
"We know the NSA shares data with state and local law enforcement. We know from a Reuters report that most of this shared data has absolutely nothing to do with national security issues," he said. "This data sharing shoves a dagger into the heart of the Fourth Amendment. This bill would stop that from happening immediately."
The bill would also ultimately end the partnership between Iowa State University and the NSA. The school serves as as an NSA "Center of Academic Excellence." That means the spy agency uses the school as a recruiting ground and a research facility. OffNow spokesperson Shane Trejo said state universities should not support an agency violating the highest law of the land.
"I understand the drive for funding a prestige. But our state universities cannot be allowed to prostitute themselves for a few federal research dollars. As long as the NSA continues to flaunt its constitutional restrains and wantonly violate basic civil liberties, state universities should not partner with them. We just don't aid and abet criminal activity because the price is right," he said.
The legislation rests on a well-established legal principle known as the anti-commandeering doctrine. Simply put, the federal government cannot "commandeer" or coerce states into implementing or enforcing federal acts or regulations – constitutional or not. The anti-commandeering doctrine rests primarily on four Supreme Court cases dating back to 1842. The 1997 case, Printz v. US, serves as the modern cornerstone. The majority opinion deemed commandeering "incompatible with our constitutional system."
We held in New York that Congress cannot compel the States to enact or enforce a federal regulatory program. Today we hold that Congress cannot circumvent that prohibition by conscripting the States' officers directly. The Federal Government may neither issue directives requiring the States to address particular problems, nor command the States' officers, or those of their political subdivisions, to administer or enforce a federal regulatory program. It matters not whether policy making is involved, and no case by case weighing of the burdens or benefits is necessary; such commands are fundamentally incompatible with our constitutional system of dual sovereignty.
Tenth Amendment Center executive director Michael Boldin said he was thrilled to see Iowa join the fight against the NSA, noting that it shows just how quickly the movement has gained momentum.
"It was only a little more than a month ago that Sen. Kelli Ward in Arizona announced she would introduce the Fourth Amendment Protection Act. Now Iowa becomes the 15th state to take action. Take note America – this is a serious movement and we aren't going away," he said. "This isn't going to be easy, but we're not stopping until we win. The NSA has a choice; follow the constitution or get the hell out."
The OffNow coalition is group of grassroots organizations and individuals spanning the political spectrum committed to stopping unconstitutional NSA spying through state and local activism.
If America Continues To Go Down The Same Path As Europe, This Is What Life Will Be Like…
Devout Muslim Group Stones Girl to Death for Belonging to Facebook – 'Tantamount to Adultery'…

redpajama_set_name: RedPajamaCommonCrawl


This season we worked with 34 students in grades K-8, many of whom were returning for their 3rd, 4th, and even 5th seasons at Little Opera. Rather than speak for them, here are a few of their own words about their year with us.
How do you share your ideas and use other students' ideas?
How do you feel about being a student here?
"It's a very good emotional outlet for me, and it helps me cope with what's going on in my life."
What's one thing you learned this year?
"Little Opera made me a lot more confident."
"I didn't know I could sing."
What is your biggest plus about this year?
"Getting better at what I love."
"Little Opera helped me come out of my shell."
""If you look through the clouds of Little Opera, you actually see happiness, joy, and friendship."


text:
Explore the free zone
Explore FEZ
Meet Our Community
FEZ Team
Flexstart 02
Contact us | Klaipėda FEZ
Watch our activity in one place
Inside Klaipeda FEZ: Vakaru Ekspresas' impressions
Dalia Bikauskaite, Vakaru ekspresas
The first impression on arriving at the territory of Klaipeda FEZ, which is mostly located outside the city but also takes up some municipal land, is somewhat puzzling. Buildings, chimneys and manufacturing is everywhere around, yet not a single person is in sight – even though more than 5400 people come to work here every day.
And we arrived at lunchtime as we wanted to meet some of them.
Eimantas Kiudulas, CEO at Klaipeda FEZ, has been commuting to the port city from Vilnius for 18 years because, he explains, his wife wouldn't agree to move. Maybe that's why he says that one day he'd love to see a high-speed train connecting the two cities.
Mr Kiudulas also wants to come to work by bicycle, as he believes that in 2019 electric bicycles will become far more popular in the city.
It was hard to arrange this meeting with Eimantas Kiudulas and ask him to be our guide at Klaipeda FEZ – he is often on a business trip. He says that autumn is the best time for "client hunting". There are few occasions, he explains, where potential FEZ members come to Klaipeda by themselves, so they have to be approached personally.
Klaipeda FEZ workers' impressions
We approached two workers during their lunchbreak. Giedre and Sigita both work at Klaipedos verslo parkas, a company managing three buildings with 30 other businesses that employ almost 1000 people in total. Giedre has been here for 14 years, Sigita 11 years.
Giedre: "I live just outside Klaipeda but it's very convenient to come to work, along the Palanga highway. Of course, it's more complicated if you have to take kids to school. But I try to avoid the city centre, and actually there's no need to go there – once you do, there are always traffic jams. One thing we lack here is probably greenery, but there are some initiatives right now to plant trees. I worked in Melnrage before and there was always this terrible smell from the port. Here, you can sometimes smell tobacco."
Sigita: "We feel good here. There's a restaurant, a conference hall, a recreational area. I worked in the city centre before. There are benefits and there are drawbacks, of course – there you can do errands during the lunchbreak, which is not possible here. We just go out for lunch and come back to the office."
Dovile, a marketing manager at Garant Group, a new company that moved to Klaipeda FEZ three months ago: "It was a bit unusual to start working here: nothing around, only industrial buildings, people busy working, no passers-by. I wouldn't say I miss anything in particular – you go to work because it's your job, after all, and the location is not that important to me. We get along well in our office and don't really meet people from other companies."
Four streets with pavements
Mr Kiudulas recalls 2002, when the area was just empty land. There was a single 1.6 km road, going nowhere, which has now become a street of its own, Pramones Street. There are now four streets in total, all with pavements.
Since 2002, FEZ has seen almost €700 million in investments, and the revenues of all of the FEZ members combined exceed €1 billion. The members come from 16 countries, mostly Scandinavian enterprises. There have been only three cases where investors have discontinued their operations due to changes in the business environment.
FEZ is comprised of two territories – 300 hectares in Klaipeda city and another 70 hectares in Klaipeda region. FEZ both leases and sells the land.
Kiudulas says that the zone's long-term goal is to become an integral part of the city rather than just an industrial park.
"FEZ is not a closed territory, even though outsiders can't enter the industrial buildings freely." He jokes that FEZ does not intend to declare autonomy, no matter how well things are going inside.
Electric buses and bicycles
The average salary at the ten biggest FEZ investors is €1540 brutto, which is twice the Klaipeda city average.
Currently there are 32 investors at FEZ that are developing projects, and 70 more companies are renting premises. This year, Klaipeda FEZ for the first time built a new building without any prospective investors in sight – and most of its premises are already rented. Another building is to be constructed next year. After that, the total area that Klaipeda FEZ can offer to rent will comprise roughly 85,000sqm.
One of the new projects to start in February 2019 is the manufacture of electric buses.
"Klaipeda should be proud of itself, because the world's lightest electric buses are to be manufactured here. The location was determined by a combination of the proximity of the university and FEZ," Kiudulas explains.
He adds that Klaipeda and Palanga municipalities are planning to buy some of the first vehicles produced here. In total, 40 buses will be produced next year, and many more in the future.
Another success story is Glassbel, which is also the only building in the city where the heating is provided by solar panels integrated in the outside walls.
2018 was a good year development-wise: eight new projects are being developed, three of which are new clients. In total, FEZ attracted €72 million in investments this year – which also means roughly 220 new jobs.
Clean environment
Klaipeda FEZ is home to large-scale technology companies. However, Mr Kiudulas claims that the pollution levels are rather minimal. According to him, Pramones Street could hold the title of cleanest street in Klaipeda, as it is cleaned by specific machinery every other day. This is despite the higher traffic flow there.
"FEZ is no park, of course – it's an industrial area, after all. But it is also one of the cleanest areas in the city. The names of the companies that come up when discussing pollution in the city are not chemicals producers, they are, for example, plastics manufacturers and they employ closed-end technologies. They are not responsible for the biggest share of the pollution, yet the names of the materials they process sound like something that is harmful for the environment. But they do less harm than, say, the furniture industry," Kiudulas says.
"The only problematic case was a smell caused by Mestilla, a biofuel producer. But in this case the company reacted immediately, investing more than €300,000 in air filtering technology and the problem was solved."
According to him, not a single chimney at FEZ is emitting smoke – only steam. Kiudulas believes that a far greater share of pollution is caused by households outside the city that are using wood for heating.
"We aim to create more green spaces in the area. I've even seen a couple of roe deer here – I'm pretty sure that once the trees have grown enough the place will become their home."
One farmer has his field in the FEZ territory, where he grows rye. When it's not occupied by industry the land looks much better with crops growing.
Bicycle lanes and further plans
Next year, Klaipeda FEZ is to be connected to the city with a bicycle lane, and since 2016 public transport has also been stopping here. According to Kiudulas, the number of bus routes is to increase in 2019. "There is even a tiny shower for cyclists – although currently it's used more after football games at FEZ," he says.
It is also expected that construction will start soon on the new Statybininku Prospect, a new road that will connect FEZ to the rest of the city.
The biggest companies here have their own canteens, but there is one shared canteen too, where the guest count is approximately 200 people per day.
FEZ has its own fire station, football and basketball fields, and a tennis court. Kiudulas says that all these facilities are full every day after 6pm, and he also mentions plans to build indoor sporting facilities.
Another plan is to build a research centre, which Mr Kiudulas expects will become a Klaipeda FEZ landmark that will differ greatly from all the industrial buildings.
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The season finale of the Summit Avenue Music Series is Sunday, April 14, at 3:00 pm in Martin Luther College's Chapel of the Christ and features pianist Lydia Artymiw and cellist Wilhelmina Smith. Artymiw and Smith will perform the Beethoven Sonata No. 4 in C Major, Op. 102 No. 1; the Schubert "Arpeggione" Sonata; and Prokofiev Sonata in C Major, Op. 119.
Purchase tickets in advance through SummitAvenueMusic.com/tickets or at the New Ulm Chamber of Commerce. Tickets may also be purchased at the door.


Every week, we corral the best wellness stories from around the internet—just in time for your weekend bookmarking. This week: a look at the dilemma of physician burnout, new insights into the teenage brain, and why more older adults are experiencing loneliness.
In this thought-provoking in-depth piece, we learn just how wide, varied, and sometimes banal hallucinations are.
A new study shows that career burnout—a common issue many face today—is affecting a group of people we all heavily depend upon: physicians.
The quest to understand the teenage brain continues. Now Professor Sarah-Jayne Blakemore sheds light on the adolescent mind in her new book, Inventing Ourselves: The Secret Life of the Teenage Brain.
Loneliness is on the rise—and according to recent reports, it's hitting older adults hard. Jamie Ducharme examines the causes behind this alarming spike and what we can do to combat it.


text:
\section{Introduction}
The group structure of elliptic curves (EC) defined over finite fields
has a theoretical interest, but also an interest in applications to cryptography and error-correcting codes. More precisely, those elliptic curves with cyclic groups of rational points are of special interest for various problems (see for example \cite{koblitz1987elliptic}, \cite{Morain1991} and \cite{Kaliski1991}).
In many algorithms that use elliptic curves over finite fields one chooses it at random, thus, knowing such statistics can be useful.

The structure of all possible groups for EC
over a finite field was discovered (independently)
in \cite{SCHOOF1987183}, \cite{ruck1987note},\cite{tsfasman1985group} and \cite{voloch1988note}. However, this result does not
provide immediately the statistics for cyclic EC over a given field or its
extensions, which was explored in \cite{VLADUT199913} and \cite{VLADUT1999354}. See also \cite{shparlinski2012group}, where the realizability of possible groups for EC was studied from a statistical point of view.

There are many other closely related topics which were considered in the literature. In \cite{shparlinski2005orders} and \cite{Murty2007} the order of points on the groups of rational points was studied, in particular lower bounds for the \textit{exponent} (the largest order), and some applications as well. Also, the exponent under base field extensions was examined in \cite{shparlinski2005exponent}. This question is closely related to the question about cyclicity, since cyclicity is equivalent to the fact that the exponent equals the cardinality of the group of rational points.

For both theoretical and practical interests, it is very natural to extend this investigation to abelian varieties of higher dimensions. Many facts are still unknown in this case. Nevertheless, in a recent paper \cite{Rybakov2010}, Rybakov gives a very explicit description of all possible groups of rational points of an abelian variety in a given isogeny class. His result is formulated in terms of the Newton polygon of the variety and the Hodge polygon of the group. In particular, from this description it follows that varieties with cyclic group of rational points always exist in an isogeny class.
In \cite{chantal2014surfaces} it is shown that for abelian surfaces, ``very split" groups of rational points occur with density zero. This is compatible with the general framework of the Cohen–Lenstra heuristics, which predicts that random abelian groups naturally occur with probability inversely proportional to the size of their automorphism groups. Note that the very split groups have many more automorphisms than the cyclic group of the same size.

Another related and widely explored topic is maximal curves, \textit{i.e.}, those with a maximal number of rational points, such curves are very important for applications. Goppa  showed that the applications of EC to coding theory are interesting only if the curve has many rational points (\cite{goppa1983algebraico}). Theoretical and practical bounds were explored and many algorithms to construct it were proposed. Relating ``maximal'' abelian varieties with its structures of rational points does not seem to be a simple task, but with many promising applications.

In cryptography, the Discrete Logarithm Problem is extensively used in such systems and, under some conditions, cyclic groups are suitable for these purposes (\cite{GALBRAITH2005544}). In many applications, security is based on working with big cyclic subgroups of the group of rational points.

Another interesting related topic is counting isomorphism classes of abelian varieties defined over a fixed finite field. In the $1$-dimensional case, this corresponds to compute class numbers of some orders in quadratic number fields, and in the general case, this was treated for example (very recently) in \cite{xue2017counting}. Isomorphism classes can be grouped in isogeny classes. In \cite{DIPIPPO1998426}, DiPippo and Howe developed some techniques and gave an asymptotic formula (with explicit error terms) for the number of isogeny classes of $n$-dimensional abelian varieties over the finite field with $q$ elements, for fixed $n$ and $q$ tending to infinity.

In this paper we give some statistical results concerning cyclic isogeny classes, namely, in Theorem \ref{th:weil_polynomial_criterion} we present a simple criterion to know if an isogeny class is \emph{cyclic}, \textit{i.e.}, all its varieties have a cyclic group of rational points, this criterion is based on the characteristic polynomial of the isogeny class. Further, in Theorems \ref{th:weil_midterm} and \ref{th:weil_q} we give asymptotic lower bounds on the fraction of cyclic isogeny classes when some coefficients on the characteristic polynomial corresponding to these isogeny classes are fixed. Theorem \ref{th:weil_q} needs an extra condition and, we show that for ``good" coefficients, this condition holds true for abelian surfaces. Finally, we prove in Theorem \ref{th:max_surface_cyclic} that under some restrictions, isogeny classes of abelian surfaces defined over finite fields with maximal number of rational points are cyclic and ordinary. Also, we give some examples of families of cyclic isogeny classes of abelian surfaces.

The rest of this paper is organized as follows: in Section \ref{sec:PresProb} we briefly recall some general facts about abelian varieties over finite fields and we state our results more precisely; in Section \ref{sec:GenCase} we prove Theorems \ref{th:weil_polynomial_criterion} to \ref{th:weil_q} and, Section \ref{sec:AbSurf} is devoted to abelian surfaces.

\section{Preliminaries and Statement of the Results}\label{sec:PresProb}
For the general theory of abelian varieties see for example \cite{mumford1970abelian}, and for precise results over finite fields, see \cite{Waterhouse1969}.

Let $q=p^r$ be a power of a prime, and let $k=\Fq$ be a finite field with $q$ elements. Let $A$ be an abelian variety of dimension $g$ over $k$. For an extension field $K$ of $k$, we denote by $\End_K (A)$ the ring of $K$-endomorphism of $A$ and by $\Endo_K (A) =(\End_K (A))\otimes\mathbb{Q}$ its endomorphism algebra, the latter being an invariant of its isogeny class $\mathcal{A}$, we can denote it by $\Endo_K(\mathcal{A})$.
For an integer $n$, denote by $\widehat{n}$ the ratio of $n$ to the product of different prime divisors of $n$, and by $A[n]$ the group of $n$-torsion points of $A$ over $\overline{\Fq}$. Then
\begin{align}
\begin{split}\label{eqn:torsion_points}
A[n]&\cong (\mathbb{Z}/n\mathbb{Z})^{2g}, \qquad p\nmid n\\
A[p]&\cong (\mathbb{Z}/p\mathbb{Z})^i, \qquad 0\leq i \leq g
\end{split}
\end{align}

For a fixed prime $\ell$ ($\neq p$), the Tate module $T_\ell(A)$ is defined by $\lim\limits_{\leftarrow} A[\ell^n]$. The Frobenius endomorphism $F$ of $A$ acts on $T_\ell(A)$ by a semisimple linear operator, and its characteristic polynomial $f_{A}(t)$ is called \emph{Weil polynomial of} $A$. Tate proved in \cite{tate1966} that the isogeny class $\mathcal{A}$ of abelian variety is determined by its characteristic polynomial $f_{\mathcal{A}}(t)$ and, if $\mathcal{A}$ is simple, the center of $\Endo_k(\mathcal{A})$ is isomorphic to the number field $\mathbb{Q}(F)\cong \mathbb{Q}[t]/ (f_{\mathcal{A}}(t))$.
The cardinality of the group of rational points $A(k)$ of $A$ equals $f_{\mathcal{A}}(1)$, and thus, it is an invariant of the isogeny class. For a prime number $\ell$, $v_\ell$ denotes the usual $\ell$-adic valuation. For an abelian group $G$ we denote by $G_\ell$ the $\ell$-primary component of $G$. We denote by $\mathcal{P}$ the set of prime integers and by $\mathcal{P}(n)$ the set of prime divisors of $n$.

For an elliptic curve $E$, we have that $E(\Fq)$ is non-cyclic if and only if there exist a prime $\ell\neq p$ such that $E[\ell]\subseteq E(\Fq)$. This is not the case for higher dimensional varieties, however,  a slightly different criterion holds:
\begin{Lemma}\label{lemma:endom_criterion}
\begin{align*}\label{EndCriterion}
A(\Fq) \text{ is not cyclic } \Longleftrightarrow \exists \varphi\in \End_{\overline{k}}(A), \; N/\ell=\varphi\circ (1-F) \text{ for some prime divisor } \ell|N
\end{align*}
\end{Lemma}

Note that in this case we do not require $\ell\neq p$. Note also that if $A$ is simple, if such $\varphi$ exists, it has to be defined over $k$ and it belongs to the center of its endomorphism ring. This lemma is an easy consequence of the theory of abelian varieties over finite fields and we will prove it here within the proof of Theorem \ref{th:weil_polynomial_criterion}. Lemma \ref{lemma:endom_criterion} implies that abelian varieties with few endomorphism are more likely to be cyclic. The group structure of an abelian variety is not necessarily determined by its endomorphism ring, but from Lemma \ref{lemma:endom_criterion} it follows that the property of being cyclic or not depends on its endomorphism ring, and thus, it is an invariant of it.

We say that $A$ is \emph{cyclic} if the group $A(\Fq)$ is cyclic.
 Counting cyclic varieties in a given isogeny class is the same as counting some fractional ideals in the center of $\Endo_k(A)$. In this paper we will focus on \emph{cyclic} isogeny classes, \textit{i.e.}, those that contain only cyclic varieties.
 We have the following criterion:
\begin{Theorem}\label{th:weil_polynomial_criterion}
Let $\mathcal{A}$ be a $g$-dimensional $\Fq$-isogeny class of abelian varieties corresponding to the Weil polynomial $f_\mathcal{A}(t)$. Then $\mathcal{A}$ is cyclic if and only if  $f'_\mathcal{A}(1)$ is coprime with $\widehat{f_\mathcal{A}(1)}$.
\end{Theorem}
which can be proven by using Rybakov's Theorem (\cite{Rybakov2010}) when $\Endo(A)$ is commutative, and we shall prove it using the Lemma \ref{lemma:endom_criterion} in the general case. Note that we do not require $\mathcal{A}$ to be simple.

Every Weil polynomial has the form
\begin{equation}
f_\mathcal{A}(t)=t^{2g}+a_1 t^{2g-1}+\dots + a_g t^g + a_{g-1} q t^{g-1}+\dots a_1 q^{g-1} t + q^{g} \in \mathbb{Z}[t],
\end{equation}
and for simplicity, we write it as $f_{q,(a_1,\dots,a_{g})}(t)$.

Denote by $\mathcal{S}^g(q)$ the set of $g$-dimensional isogeny classes $\mathcal{A}$ of abelian varieties  defined over the finite field $\Fq$, and by $\mathcal{S}^g_{\mathfrak{c}}(q)$ the subset of cyclic isogeny classes. We write $(a_1,\dots,a_{g})\in \mathcal{S}^g(q)$ if the polynomial $f_{q,(a_1,\dots,a_{g})}(t)$ defines such an isogeny class.

Our first approach is to fix a vector $\mathbf{a}=(a_1,\dots,a_{g-1})\in\mathbb{Z}^{g-1}$ of $g-1$ integers, and define two sets \emph{of possible values of} $a_g$:
\begin{align}
I_{\mathbf{a}}(q)&:=\{z\in\mathbb{Z} |  (a_1,\dots,a_{g-1},z) \in \mathcal{S}^g(q)\},\\
I_{\mathbf{a},\mathfrak{c}}(q)&:=\{z\in\mathbb{Z} | (a_1,\dots,a_{g-1},z) \in \mathcal{S}^g_{\mathfrak{c}}(q)\}.
\end{align}
It is obvious that $I_{\mathbf{a}}(q)$ is finite and non empty from a certain value of $q=q(\mathbf{a})$. Fixing a prime $p$, we are interested on the proportion
\begin{align}
r_{p,\mathbf{a}}(n):=\frac{\sum_{i=1}^n \# I_{\mathbf{a},\mathfrak{c}}(p^i)}{\sum_{i=1}^n \# I_{\mathbf{a}}(p^i)},
\end{align}
when $n$ tends to infinity.

Finally, if we define the polynomial $h(X,\mathbf{a}):=gf_{X,(\mathbf{a},z)}(1)-f'_{X,(\mathbf{a},z)}(1)$ (in fact, it is independent of $z$), and we have
\begin{Theorem}\label{th:weil_midterm}
Given $\mathbf{a}\in\mathbb{Z}^{g-1}$ and a prime $p$, then
\begin{align}
\liminf r_{p,\mathbf{a}} &\geq 1 - \frac{p}{p-1} \left[ \left(1- \prod\limits_{\ell\in \mathcal{P}(h(p,\mathbf{a}))}\left( 1-\frac{1}{\ell^2} \right) \right) \left( 1-\frac{1}{p^{g/2}}\right) +\frac{1}{p^{g/2}} \right].
\end{align}
\end{Theorem}
\begin{Rmk}
It is easy to see that if $p\gg 0$, this bound is approximately
\[
\prod\limits_{\ell\in \mathcal{P}(h(p,\mathbf{a}))}\left( 1-\frac{1}{\ell^2} \right),
\]
which is greater than $1-P(2)=0.54\dots$, where $P(s)$ is the \emph{prime zeta function}, defined as
\begin{align*}
P(s)=\sum_{\ell\in\mathcal{P}} \frac{1}{\ell^s}, \qquad s\in\mathbb{C}.
\end{align*}
\end{Rmk}

Let us follow now a different approach by fixing a vector $\mathbf{b}=(b_1,\dots,b_g)\in\mathbb{Z}^{g}$ of $g$ integers, we want to study when $f_{q,\mathbf{b}}(t)$ defines a cyclic isogeny class over $\Fq$ when $q$ varies over the power of primes. We consider this problem in two ways: (i)$q$ varies over primes and (ii) we fix a prime $p$ and $q$ varies over the powers of $p$.

For our next result, we will need an extra hypothesis, which we make explicit as follows:
\begin{quote}
\textbf{Hyp}($\mathbf{b}$):
there exist integers $\eta(\mathbf{b}), t(\mathbf{b})$ and $s(\mathbf{b})$ with no common factor, such that  the only possible divisors of $f_{q, \mathbf{b}}(1)$ and $f'_{q, \mathbf{b}}(1)$ are divisors of $\eta$ and $tq+s$.
\end{quote}

\begin{Rmk}
The key idea is that these values, in particular $\eta$, are independent of $q$. Such $\eta$ can always be obtained by successive Euclidean division, since we can see $f$ and $f'$ as polynomials in $\mathbb{Q}[q]$; but it is not sure to have a linear polynomial in $q$, with the condition of no common factors. For the $2$-dimensional case, we will see in Section \ref{sec:AbSurf} that this hypothesis holds for a wide range of values of $\mathbf{b}$, and we will explicit them.
\end{Rmk}

Then we are interested on the following proportions:
\begin{gather}
x_{\mathbf{b}}(n)=\frac{\#\{\ell\leq n \text{ prime}|  \mathbf{b}\in \mathcal{S}^g_{\mathfrak{c}}(\ell) \}}{\#\{\ell\leq n \text{ prime}| \mathbf{b}\in \mathcal{S}^g(\ell) \}},\\
y_{p,\mathbf{b}}(n)=\frac{\#\{i\leq n| \mathbf{b} \in \mathcal{S}^g_{\mathfrak{c}}(p^i)\}}{\#\{i\leq n| \mathbf{b}\in \mathcal{S}^g(p^i) \}},
\end{gather}
when $n$ tends to infinity. Note that the above fractions are defined from a big enough $n$. Then our next result is
\begin{Theorem}\label{th:weil_q}
Given $\mathbf{b}_1, \mathbf{b}_2\in\mathbb{Z}^g$ and a prime number $p>2$ such that
\begin{enumerate}
\item {\normalfont\textbf{Hyp}($\mathbf{b}_i$)} holds, with $\eta, t$ and $s$ given,
\item $p$ doesn't divide the last coordinate of $\mathbf{b}_2$, and
\item the multiplicative group $(\mathbb{Z}/\eta(\mathbf{b}_2)\mathbb{Z})^*$ is generated by $p$.
\end{enumerate}
Then
\begin{align*}
1.\quad\liminf x_{\mathbf{b}_1}(n) \geq L_1 \qquad\text{ and, }\qquad 2.\quad\liminf y_{p,\mathbf{b}_2}(n) \geq L_2,
\end{align*}
where
\begin{align*}
L_i=\prod\limits_{\ell\in \mathcal{P}(\eta(\mathbf{b}_i))\setminus \mathcal{P}(t(\mathbf{b}_i) s(\mathbf{b}_i))}\frac{\ell-2}{\ell-1}, \qquad i=1,2.
\end{align*}
\end{Theorem}

\begin{Rmk}\label{rmk:optimHyp}
As we can see, this bound depends on the choose of $(\eta,t,s)$. We can verify that if $(\eta_1,t_1,s_1)$ and $(\eta_2,t_2,s_2)$ verify \textbf{Hyp}($\mathbf{b}$) for some fixed $\mathbf{b}$, then $((\eta_1,\eta_2),t_1,s_1)$ also does. Also, if $(\eta,t_1,s_1)$ and $(\eta,t_2,s_2)$ verify \textbf{Hyp}($\mathbf{b}$), and if there exist a prime $\ell\in \mathcal{P}(t_2 s_2)\setminus \mathcal{P}(t_1 s_1)$ dividing $\eta$, then $(\eta/\ell,t\ell,s\ell)$ also verifies \textbf{Hyp}($\mathbf{b}$). This allows us to optimize the bound in that way. However, in Section \ref{sec:AbSurf} we will see with an example that Theorem \ref{th:weil_q} does not provide necessarily the best bound.
\end{Rmk}

\begin{Ex}
As example, we discuss the case of elliptic curves. Here we have
\begin{align*}
f_{q,a}(1)=(1+a)+q \qquad &\text{and} \qquad f'_{q,a}(1)=2+a,
\end{align*}
thus, \textbf{Hyp}($a$) holds for all values of $a$ if we take
\[
\left \{
  \begin{array}{l}
\eta(a)=2+a, \\
t(a)=1, \\
s(a)=1+a.
  \end{array}
\right.
\]

Also, since $\eta(a)$ and $s(a)$ are always coprimes, the bound for Theorem \ref{th:weil_q} takes the form
\begin{align*}
L=\prod_{\ell\in \mathcal{P}(2+a)} \frac{\ell-2}{\ell-1}.
\end{align*}
Concerning Theorem \ref{th:weil_midterm}, we have that $h(p,\mathbf{a})=p-1$, where $\mathbf{a}$ can be considered empty. This provides a complete answer to the fraction of cyclic isogeny classes defined over some finite extension of $\mathbb{F}_p$, and for $p\gg 0$, the bound takes the form
\begin{align*}
\prod\limits_{\ell\in \mathcal{P}(p-1)}\left( 1-\frac{1}{\ell^2} \right).
\end{align*}
To finish this example, let us say that the question
\begin{center}\textit{
For which values of $q$, all isomorphism classes of EC defined over $\Fq$ are cyclic?}
\end{center}
is answered in Theorem $4.1$ in \cite{VLADUT199913}, and in view of Theorem \ref{th:weil_polynomial_criterion}, this  is equivalent to
\begin{center}\textit{
For which values of $q$, $(\widehat{f_{q,a}(1)},f'_{q,a}(1))=1 \quad\forall a\in [-2\sqrt{q}, 2\sqrt{q}]$?}
\end{center}
\end{Ex}

\section{General case: Proof of Theorems 1-3}\label{sec:GenCase}
In this section we prove Theorems \ref{th:weil_polynomial_criterion} to \ref{th:weil_q}.

\begin{proof}[Proof of Theorem \ref{th:weil_polynomial_criterion}:]
$ $\newline
First we consider the case when $\mathcal{A}$ is simple. For an abelian variety $A\in\mathcal{A}$, the center of its endomorphism algebra $\End_{\Fq}(A)\otimes\mathbb{Q}$ is isomorphic to the number field $K=\mathbb{Q}(\pi)$, where $\pi$ is a root of $f_\mathcal{A}$ and represents the Frobenius endomorphism $F$. For every variety $A$ in $\mathcal{A}$, $\End_{\Fq}(A)\cap K$ is an order in $K$ (thus contained in $\mathcal{O}_K$), and $\mathcal{O}_K$ is the endomorphism ring for some $B\in\mathcal{A}$ by Theorem 3.13 in \cite{Waterhouse1969}. For any $x\in K$, we have that $x\in\mathcal{O}_K$ holds if and only if its (unitary) minimal polynomial has integer coefficients, that is, is an algebraic integer.

Denote by $N$ the value $f_{\mathcal{A}}(1)$, then Lemma \ref{lemma:endom_criterion}
\begin{align*}
A(\Fq) \text{ is not cyclic } \Longleftrightarrow \exists \varphi\in \End_{\overline{\Fq}}(A), \; N/\ell=\varphi\circ (1-F) \text{ for some prime divisor } \ell|N
\end{align*}
follows from the fact that $A(\Fq)$ is not cyclic if and only if $A(\Fq)=\ker (1-F) \subset \ker (N/\ell) =A[N/\ell]$ for some $\ell|N$, since $1-F$ is separable. Since $A$ is simple, if such $\varphi$ exist, it must be in $K$: if $\psi\in \End_{\Fq}(A)$, multiplying the previous equation in both sides and subtracting them, we have $0=(\varphi\psi -\psi\varphi)\circ (1-F)$, thus, $\varphi$ commutes with $\psi$. Similarly, we can show that this endomorphism is defined over $\Fq$.

We consider the minimal polynomial $p_\ell (t)$ of $\frac{N}{\ell(1-\pi)}$. All non-leading coefficients of $p_\ell$ are of the form $\frac{N^{i-1}}{\ell^i}$ times an integer, except for the coefficient of $t^{2g-1}$, which equals $-\frac{f'(1)}{\ell}$, and that the constant term equals $\frac{N^{2g-1}}{\ell^{2g}}$.

If $\mathcal{A}$ is cyclic, then all elements of $\{\frac{N}{\ell(1-\pi)}, \ell\vert N \}$ are algebraic non integers: if not, for some $\ell$, $\frac{N}{\ell(1-\pi)}\in\mathcal{O}_K$ and for some abelian variety $B$ with $\End_{\Fq}(B)\cap K=\mathcal{O}_K$ we will have that $B(\Fq)$ is not cyclic. Therefore $p_\ell$ have some non integer coefficient, either $\ell\nmid f'(1)$, either $v_\ell(N)=1$. In consequence, $gcd(\widehat{N},f'(1))=1$.

If $gcd(\widehat{N},f'(1))=1$, then for any $\ell|N$, either $(\ell,f'(1))=1$, or $v_\ell(N)=1$. In both cases, $p_\ell$ have some non integer coefficient, and thus, $\frac{N}{\ell(1-\pi)}\notin \mathcal{O}_K$. This implies that, for all $A\in\mathcal{A}$, there is no $\varphi\in \End(A)$ with $\frac{N}{\ell}=\varphi\circ (1-F)$: if such $\varphi$ exist, it has to be equal to $\frac{N}{\ell(1-\pi)}$, which is a contradiction.

Now we consider the general case. Then $\mathcal{A}$ is isogenous to a product $\prod \mathcal{A}_i$ of simple isogeny classes, and we have that $f_\mathcal{A} (t)=\prod f_i(t)$, where $f_i(t)$ is the characteristic polynomial of $\mathcal{A}_i$. Suppose $\mathcal{A}$ is not cyclic, we have two possibilities: (i) one of its components is not cyclic or (ii) $\exists \ell$ prime such that $\ell| (f_1(1),f_2(1))$ (WLOG). In (i), by the first part of the theorem, $(\widehat{f_1(1)},f'_1(1))>1$ (WLOG), writing $f'_\mathcal{A} (t)= f'_1(t) \prod_{i>1} f_i(t)+f_1(t) (\prod_{i>1} f_i(t))'$ we have that $(\widehat{ f_\mathcal{A}(1)}, f'_\mathcal{A}(1))>1$. In (ii), writing $f'_\mathcal{A} (t)= f'_1(t) f_2(t) \prod_{i>2} f_i(t) + f_1(t) f'_2(t) \prod_{i>2} f_i(t) + f_1(t) f_2(t) (\prod_{i>2} f_i(t))'$ we have that $(\widehat{ f_\mathcal{A}(1)},f'_\mathcal{A}(1))>1$. Suppose now that $\ell | (\widehat{f_{\mathcal{A}}(1)}, f'_{\mathcal{A}}(1))$, thus, $\ell^2|f_{\mathcal{A}}(1)$ and $\ell | f'_{\mathcal{A}}(1)$. Again, we have two possibilities: (i) $\ell| f_1(1)$ and $\ell| \prod_{i>1} f_i(1)$ or, (ii) $\ell^2| f_1(1)$ and $\ell\nmid \prod_{i>1} f_i(1)$. In (i), $\mathcal{A}$ is not cyclic. In (ii), writing $f'_\mathcal{A} (t)= f'_1(t) \prod_{i>1} f_i(t)+f_1(t) (\prod_{i>1} f_i(t))'$ we see that $\ell |f'_1(1)$, thus, $\mathcal{A}$ is not cyclic by the first part of the theorem. Theorem \ref{th:weil_polynomial_criterion} is thus proved.
\end{proof}

Note that Lemma 3.3.1 in \cite{DIPIPPO1998426} implies that for fixed $\mathbf{a}\in\mathbb{Z}^{g-1}$, $f_{q,(\mathbf{a},z)}(t)$ always defines an ordinary isogeny class from a $q$ big enough when $p\nmid z$, and they are almost all the isogeny classes. This will be important for the proofs of both theorems that follow.

\begin{proof}[Proof of Theorem \ref{th:weil_midterm}:]
$ $\newline
From now on, $\mathbf{a}$ and $p$ are fixed. For simplicity, we denote by $N_i$ and $S_i$ the integers $\# I_{\mathbf{a}}(p^i)$ and the sets of prime divisors of $h(p^i, \mathbf{a})$, with $S=S_1$, respectively. We set $M_i=\max(I_{\mathbf{a}}(q))-\min(I_{\mathbf{a}}(q))$. Also, we write $f_{i,z}(t)$ instead of $f_{p^i,(\mathbf{a},z)}(t)$.

From Lemma 3.3.1 in \cite{DIPIPPO1998426} we deduce that
\begin{align*}
N_i/M_i \rightarrow \frac{p-1}{p} \qquad \text{ and }\qquad M_i/4p^{ic}\rightarrow 1 \qquad \text{ when } i\rightarrow \infty,
\end{align*}
with $c=g/2$.

The following elementary lemma allows us to identify the primes that can simultaneously divide the arithmetic progressions $f_{i,z}(1)$ and $f'_{i,z}(1)$ (as sequences on $z$):

\begin{Lemma}
Let $x_n$ and $y_n$ be two arithmetic progressions of difference $d_x$ and $d_y$ respectively, and let $\ell$ be a prime such that $d_x \not\equiv 0 \pmod{\ell}$. Then $\ell|(x_i,y_i)$ for some $i$ if and only if $d_x y_j \equiv d_y x_j \pmod{\ell}$ for some $j$.
\end{Lemma}

\begin{proof}
First, observe that if the congruence $d_x y_j \equiv d_y x_j \pmod{\ell}$ is verified for some $j$, then it verifies for all $i$:
\begin{align*}
d_x y_i &\equiv d_x(y_i - y_j + y_j) \pmod{\ell} \\
&\equiv d_x y_j + d_x d_y (i-j) \\
&\equiv d_y x_j + d_x d_y (i-j) \\
&\equiv d_y(x_j + d_x (i-j)) \\
&\equiv d_y x_i
\end{align*}
Suppose now that $d_x y_j \equiv d_y x_j
\pmod{\ell}$, then we have to solve the system

\[
\left \{
  \begin{array}{r}
x_0 + d_x i \equiv 0  \pmod{\ell} \\
y_0 + d_y i \equiv 0  \pmod{\ell}
  \end{array}
\right.
\]
We get that $i\equiv -x_0 d_x^{-1}$, and thus, $y_0 + d_y (-x_0 d_x^{-1}) \equiv y_0 -  (d_y x_0) d_x^{-1} \equiv y_0 - d_x y_0 d_x^{-1} \equiv 0 \pmod{\ell}$
\end{proof}

We are in the case where $g(f_{i,z}(1)-z)\equiv f'_{i,z}(1)-gz \pmod{\ell}$, thus, only the primes in $S_i$ can simultaneously divide $f_{i,z}(1)$ and $f'_{i,z}(1)$.

For an integer $n$ and a finite set of integers $Z$, let denote
\begin{align*}
\sigma(n,Z)&=\sum\limits_{i\geq 1} (-1)^{i+1}\sum\limits_{R\subset Z, |R|=i} \left\lceil\frac{n}{\prod_{x\in R} x^2}\right\rceil, \\
\xi(Z)&=\sum\limits_{i\geq 1} (-1)^{i+1}\sum\limits_{R\subset Z, |R|=i} \frac{1}{\prod_{x\in R} x^2}=1-\prod\limits_{x\in Z}\left( 1-\frac{1}{x^2} \right).
\end{align*}
Observe that $\xi$ can be defined for infinite subsets of $\mathcal{P}$ since $\xi(\mathcal{P})$ converges and, if $Z\subseteq Z'$, $\xi(Z) \leq \xi(Z')$.

We will study the sequence $1-r_{p,\mathbf{a}}$, \textit{i.e.}, we will count non cyclic isogeny classes using the lemma above. Note that the number of non cyclic isogeny classes for $p^i$ is at most $\sigma(M_i,S_i)$.\\

For every finite set $T$ of prime numbers that doesn't divides $h(p, \mathbf{a})$ ($T\subset\mathcal{P}\setminus S_1$ finite),
we consider the polynomial
\begin{align*}
g(X)=m(g)X+1, \quad m(g)=\left(\prod\limits_{\ell\in T}\ell-1\right),
\end{align*}
thus, $h(p^i, \mathbf{a}) \equiv h(p, \mathbf{a}) \not\equiv 0 \pmod{\ell}$ for all $\ell\in T$ and for all $i\in g(\mathbb{N})$, hence $S_i\subset \mathcal{P}\setminus T$ for all $i\in g(\mathbb{N})$.

We get
\begin{align*}
1-r_{p,\mathbf{a}}(n) &\leq \frac{\sum \sigma(M_i,S_i)}{\sum N_i} \leq \frac{\sum\limits_{i\notin g(\mathbb{N})} M_i+\sum\limits_{i\in g(\mathbb{N})} \sigma(M_i,S_i)}{\sum N_i} \\
&\approx \frac{p}{p-1}\frac{\sum\limits_{i\notin g(\mathbb{N})} M_i+\sum\limits_{i\in g(\mathbb{N})} \sigma(M_i,S_i)}{\sum M_i} \\
&\approx \frac{p}{p-1}\frac{\sum\limits_{i\notin g(\mathbb{N})} M_i+\sum\limits_{i\in g(\mathbb{N})} M_i\xi(S_i)}{\sum M_i} \\
&\leq\frac{p}{p-1}\frac{\sum\limits_{i\notin g(\mathbb{N})} M_i+\sum\limits_{i\in g(\mathbb{N})} M_i\xi(\mathcal{P}\setminus T)}{\sum M_i} \\
&\approx \frac{p}{p-1}\frac{\sum\limits_{i\notin g(\mathbb{N})} (p^{c})^i+\xi(\mathcal{P}\setminus T)\sum\limits_{i\in g(\mathbb{N})} (p^{c})^{i}}{\sum (p^{c})^i} \\
\end{align*}
where all the sums are taken over $1\leq i\leq n$ and $\approx$ means that both sequences have the same limit. The last sequence converges to
\begin{align*}
\frac{p}{p-1} \left[1+ (\xi(\mathcal{P}\setminus T)-1)\frac{p^{cm(g)}(p^{c}-1)}{p^{c(m(g)+1)}-1}\right].
\end{align*}
As we have the previous reasoning for all finite set $T$ disjoint to $S$, thus,
\begin{align*}
\limsup (1-r_{p,\mathbf{a}}(n)) \leq \frac{p}{p-1}\left[1+(\xi(S)-1)(1-\frac{1}{p^c})\right]=\frac{p}{p-1} \left[ \xi(S) \left( 1-\frac{1}{p^c}\right) +\frac{1}{p^c} \right].
\end{align*}
Finally
\begin{align*}
\liminf r_{p,\mathbf{a}} \geq 1 - \frac{p}{p-1} \left[ \xi(S) \left( 1-\frac{1}{p^c}\right) +\frac{1}{p^c} \right].
\end{align*}
That finish the proof of Theorem \ref{th:weil_midterm}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{th:weil_q}:]
$ $\newline
We write $q$ for $\ell$ varying (i) over the primes or (ii) over powers $p^i$ when $i$ varies over positive integers.

As a consequence of Dirichlet's Theorem or Theorem's hypothesis, the values of $q$ are (asymptotically) evenly distributed among the invertible congruence classes modulo $\eta$, when $q$ varies according to (i) or (ii), respectively.

Set $\mathfrak{j}(q)=tq+s$, then, by \textbf{Hyp}, the condition $(\mathfrak{j},\eta)=1$ would imply that $\mathbf{b}_i \in \mathcal{S}_{\mathfrak{c}}^g(q)$, from a certain value of $q$ when $f_{q,\mathbf{b}_i}(t)$ defines an isogeny class.

Observe that $(\mathfrak{j},\eta)=1$ if and only if $\mathfrak{j}$ is invertible in $\mathbb{Z}/\eta\mathbb{Z}$; then we conclude by using the following lemma.
\end{proof}

\begin{Lemma}\label{lemma_jdistribution}
Let $t,s$ and $n$ be integers with no common factor, then
\begin{align*}
\frac{\# \{ x\in (\mathbb{Z}/n\mathbb{Z})^*, tx+s\in (\mathbb{Z}/n\mathbb{Z})^*\} }{\# (\mathbb{Z}/n\mathbb{Z})^*} =\prod\limits_{\ell\in \mathcal{P}(n)\setminus \mathcal{P}(ts)} \frac{\ell-2}{\ell-1}
\end{align*}
\end{Lemma}
\begin{proof}
Take an $\ell\in \mathcal{P}(n)$ and set $m=\ell^{v_\ell(n)}$, we have that $\# (\mathbb{Z}/m\mathbb{Z})^*=m-m/\ell$. If $\ell\nmid ts$ we have that $\# \{ x\in (\mathbb{Z}/m\mathbb{Z})^*, tx+s\in (\mathbb{Z}/m\mathbb{Z})^*\}=m-2m/\ell$. If $\ell|ts$, for all $x\in(\mathbb{Z}/m\mathbb{Z})^*$ we have that $tx+s\in (\mathbb{Z}/m\mathbb{Z})^*$. We conclude by using the isomorphism $\mathbb{Z}/n\mathbb{Z}\cong\prod_{\ell|n}\mathbb{Z}/\ell^{v_\ell(n)}\mathbb{Z}$
\end{proof}

\section{The case of Abelian Surfaces}\label{sec:AbSurf}
In this section we consider the case of abelian surfaces, where the characteristic polynomial takes the form
\begin{align*}
f_\mathcal{A}(t)=t^{4}+at^3 + b t^2 + a q t + q^2,
\end{align*}
thus, we have $\mathbf{b}=(a,b)$.

\subsection{Theorems applied to Surfaces.} In order to apply Theorem \ref{th:weil_q} to surfaces, we need to verify \textbf{Hyp}($\mathbf{b}$); and this is a consequence of the proposition below and the conditions on $\mathbf{b}$ that follow:
\begin{Prop}\label{lemma_eta}
Let $a, b$ and $q$ be integers, then the common divisors of
\begin{align*}
N&=1+a(q+1)+b+q^2 \text{ and }\\
J&=4+a(q+3)+2b
\end{align*}
are divisors of $\eta(a,b):=(2a+b+2)(a^2-4a+4b-16)$.
\end{Prop}
\begin{proof}

Let $\ell$ be a prime divisor of $N$ and $J$, then it divides
\begin{align*}
(q+3)N-(q+1)J&=(q-1)(q^2+4q-1+b) \text{ and}\\
2N-J&=(q-1)(2q+2+a)
\end{align*}
If $\ell$ divides $q-1$, then $N=q^2-1 + a(q-1) +2a +b +2 \equiv 2a+b+2 (\mod \ell)$.
We set
\begin{align*}
\gamma &=q^2+4q-1+b\\
\theta &=2q+2+a
\end{align*}
So, if $\ell$ doesn't divide $q-1$, then it divides $\gamma$ and $\theta$, and
\begin{align*}
4\gamma=\theta(2q+6-a)+(a^2-4a+4b-16)
\end{align*}
Finally, the possible common divisors are the divisors of $(2a+b+2)(a^2-4a+4b-16)$
\end{proof}
We have $\eta(a,b)$ as defined in the proposition and clearly $f'(1)=aq+4+3a+2b$, thus, we can take
\[
\left\{
\begin{array}{l}
\eta(a,b)=(2a+b+2)(a^2-4a+4b-16),\\
t(a,b)=a,\\
s(a,b)=4+3a+2b.
\end{array}\right.
\]
To finish, and in order to have the condition of no common divisor (with the values previously chosen), the integers in $\mathbf{b}=(a,b)$ must verify:
\begin{align*}
(a,b+2)=1, \qquad\text{ and }\qquad a\equiv 1 \pmod{2}.
\end{align*}

A priori these values are not optimal in the sense of Remark \ref{rmk:optimHyp}, and as mentioned previously, Theorem \ref{th:weil_q} does not necessarily provide the optimal bound. For example, observe that  \textbf{Hyp}($0,0$) holds with $\eta=2, t=1,s=1$, in which case $L_1=L_2=0$, which are clearly very bad bounds. But by applying ``manually'' Theorem \ref{th:weil_polynomial_criterion} we can have better bounds, first we have:
\begin{align*}
f_{q,\mathbf{b}}(1)=1+q^2=(q^2-1)+2 \qquad \text { and } \qquad f'_{q,\mathbf{b}}(1)=4,
\end{align*}
thus, $f_{q,\mathbf{b}}(1)$ is equivalent to $1$ or $2$ modulo $4$, depending if $p$ equals $2$ or not, respectively, thus $L_1=L_2=1$. This is because Theorem \ref{th:weil_q} uses only a weak result:
\begin{align*}
(f(1),f'(1))=1 \qquad\Rightarrow \qquad \text{cyclicity},
\end{align*}
and not the stronger equivalence from Theorem \ref{th:weil_polynomial_criterion}:
\begin{align*}
(\widehat{f(1)},f'(1))=1 \qquad\Leftrightarrow \qquad \text{cyclicity}.
\end{align*}

Concerning Theorem \ref{th:weil_midterm} applied to abelian surfaces, this provides the fraction of cyclic isogeny classes with a fixed trace of Frobenius $a$ and a fixed base field characteristic $p$. We have that $h(p,a)=(p-1)[2(p+1)+a]$. Thus, for $a=-4$ and a Fermat prime $p$, the lower bound for Theorem \ref{th:weil_midterm} is $0.75$ and we cannot improve it. Even for primes of form $2\ell+1$ (with $\ell\gg 0$ a prime number) we have a good bound, around $0.75$.

\subsection{Maximal Surfaces}\label{ssec:MaxSurf}
Now, let us discuss about the other topic mentioned in the Introduction, varieties with many rational points. R\"uck's theorem \cite{Ruck1990} describes polynomials that occurs as Weil polynomials for abelian surfaces with endomorphism algebra being a field:

\begin{Theorem}(R\"uck)
The set of $f_A(t)$ for all abelian varieties $A$ over $\Fq$ of dimension 2 whose algebra $\End(A)\otimes \mathbb{Q}$ is a field is equal to the set of polynomials $f(t)=t^4+at^3+bt^2+aqt+q^2$ where the integers $a$ and $b$ satisfy the conditions
\begin{enumerate}
\item $\vert a \vert < 4\sqrt{q}, \quad 2\vert a\vert\sqrt{q}-2q < b < a^2/4+2q$
\item $\Delta=a^2-4b+8q$ is not a square in $\mathbb{Z}$, and
\item either
\begin{enumerate}
\item $v_p(a)=0, \quad v_p(b)\geq r/2$ and $(b+2q)^2-4qa^2$ is not a square in $\mathbb{Z}_p$,
\item\label{ordinarycase} $v_p(b)=0$, or
\item $v_p(a)\geq r/2, v_p(b) \geq r$, and $f(t)$ has no root in $\mathbb{Z}_p$.
\end{enumerate}
\end{enumerate}
\end{Theorem}

Concerning this type of isogeny classes, we have the following result

\begin{Theorem}\label{th:max_surface_cyclic}
Let $q=p^r$ with $r$ even. Consider the $\Fq$-isogeny classes $\mathcal{A}$ of abelian surfaces
with $\Endo_{\Fq}(\mathcal{A})$ being a field. Then the isogeny class with maximal number of rational points among such classes is ordinary and cyclic, and corresponds to the Weil polynomial
\begin{align*}
f(t)=t^4+at^3+bt^2+aqt+q^2,\quad \text{where} \quad a=4\sqrt{q}-3 \quad \text{and} \quad b=6q-6\sqrt{q}+1.
\end{align*}
\end{Theorem}
\begin{proof}
For a fixed $q$, the number of rational points increases with $a$ and $b$. From the conditions of R\"uck's Theorem we have that the biggest possible values are $a=4\sqrt{q}-3$ and $b=6q-6\sqrt{q}+1$ with $\Delta=5$, and the ordinariness is clear.

Then we have
\begin{align*}
f(1)&=-1-2\sqrt{q}+3q+4(\sqrt{q})^3+q^2 \text{ and}\\
f'(1)&=-3+9q+4(\sqrt{q})^3
\end{align*}
Writing $f(1)$ and $f'(1)$ as functions $N$ and $\mathfrak{j}$ of $x=\sqrt{q}$, respectively, we see that
\begin{align*}
N(x)=x^4+4x^3+3x^2-2x-1,\qquad \mathfrak{j}(x)=4x^3+9x^2-3,
\end{align*}
and thus,
\begin{align*}
(14x^3+49x^2+28x-14)\mathfrak{j}(x)-(56x^2+98x-7)N(x)=35=5\cdot 7
\end{align*}

Finally, we can see case by case, that for $\ell\in \{5,7\}$ we cannot have that $\ell|\mathfrak{j}(x)$ and $\ell^2|N(x)$ simultaneously:
\begin{itemize}
\item for $\ell=5$,
\begin{center}
  \begin{tabular}{ | c | c |c |c |c |c |}
    \hline
    $x \pmod{5}$ & 0 & 1 & 2 & 3 & 4 \\ \hline
    $\mathfrak{j} \pmod{5}$ & 2 & 0 & 0 & 1 & 2 \\ \hline
  \end{tabular}
    \begin{tabular}{ | c | c |c |c |c |c |c |c |c |c |c |}
    \hline
    $x \pmod{25}$ & 1 & 2 & 6 & 7 & 11 & 12 & 16 & 17 & 21 & 22 \\ \hline
    $N \pmod{25}$ & 5 & 5 & 5 & 5 & 5 & 5 & 5 & 5 & 5 & 5 \\ \hline
  \end{tabular}
\end{center}
\item for $\ell=7$,
\begin{center}
  \begin{tabular}{ | c | c |c |c |c |c |c |c |}
    \hline
    $x \pmod{7}$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
    $\mathfrak{j} \pmod{7}$ & 4 & 5 & 3 & 1 & 2 & 2 & 4 \\ \hline
  \end{tabular}
  \end{center}
\end{itemize}
\end{proof}

Surfaces having more number of rational points $N$ than those considered in Theorem \ref{th:max_surface_cyclic} are only (in decreasing order of $N$):
\[
E_{max}^2, \quad E_{max}\times E_{max-1}, \quad E_{max}\times E_{max-2}, \quad E_{max-1}^2 \quad \text{and} \quad E_{max-1}\times E_{max-2},
\]
 where $E_{max}, E_{max-1}$ and $E_{max-2}$ are the elliptic curves defined by its Frobenius traces $2\sqrt{q}, 2\sqrt{q}-1$ and $2\sqrt{q}-2$, respectively. We can easily check, by using Theorem \ref{th:weil_polynomial_criterion}, that $E_{max}$ is always not cyclic, $E_{max-2}$ is always cyclic, and $E_{max-1}$ is cyclic or not, depending on $q\equiv 0,1 \pmod{3}$ or $q\equiv 2 \pmod{3}$, respectively (since $3$ is the only possible prime divisor of $f(1)$ and $f'(1)$); thus, the cyclicity of $E_{max-1}\times E_{max-2}$ depends only on the cyclicity of $E_{max-1}$, since the cardinalities of $E_{max-1}$ and $E_{max-2}$ are relatively prime.

On the other hand, note that for $q=23^3$ the ordinary class with maximal number of rational points, among the classes considered in Theorem \ref{th:max_surface_cyclic}, is given by
\begin{align*}
f(t)=t^4+438t^3+72293t^2+438(23^3)t+23^6
\end{align*}
and is not cyclic since $7|f'(1)$ and $49|f(1)$, so Theorem \ref{th:max_surface_cyclic} holds only for even powers.

\subsection{Families of cyclic isogeny classes}
We finish this section by giving two sequences of cyclic isogeny classes of abelian surfaces in Proposition \ref{prop:seq_cyclic_1} and \ref{prop:seq_cyclic_2}. Basically, we get the cyclicity of these families by choosing appropriate values of $q$ and $\mathbf{b}$.

\begin{Prop}\label{prop:seq_cyclic_1}
Given an integer $b$ and a power of a prime $q=p^r>4$, such that:
\begin{enumerate}
\item $p\nmid b$
\item $8\sqrt{q}-4q\leq b\leq 4$,
\item $N_0=b-3-2q+q^2$ is coprime with $N'_0=2(b-4)$
\end{enumerate}
Let be $s=\prod\limits_{\ell\in\mathcal{P}(N'_0)} (\ell-1)$.
Then for $i\in\mathbb{Z}_{\geq 0}$
\begin{align*}
f_i(t)=t^4-4t^3+b_it^2-4q_it+q_i^2,\quad \text{where} \quad q_i=p^{r+is} \quad \text{and} \quad b_i=b+2q_i,
\end{align*}
defines a ordinary cyclic isogeny class $\mathcal{A}_i$ of abelian surfaces over $\mathbb{F}_{q^i}$.
\end{Prop}
\begin{proof}
Note that $f_i(t)=f_{q_i,(-4,b_i)}(t)$. Every $f_i$ defines an ordinary isogeny class of abelian surfaces since $p\nmid b_i$, and $f_i$ is a Weil polynomial since
\begin{gather*}
8\sqrt{q}p^{is/2}-4qp^{is} \leq 8\sqrt{q}-4q \leq b \qquad\Rightarrow\qquad 8\sqrt{q_i}-2q_i \leq b+2q_i=b_i\\
b_i=b+2q_i \leq 4+2q_i=\frac{(-4)^2}{4}+2q_i.
\end{gather*}

Observe that $f'_{q_i,(-4,b_i)}(1) = 4-4(q_i+3)+2(b+2q_i)=N'_0=f'_{q,(-4,b)}(1)$. Also, we have that
\begin{align*}
f_{i+1}(1) - f_i(1) = (p^s-1)[p^{2(r+is)}(p^s+1)+2p^{r+is}].
\end{align*}
Thus, $(f_{i}(1),N'_0)=1$ implies $(f_{i+1}(1),N'_0)=1$, then if $\mathcal{A}_i$ is cyclic, $\mathcal{A}_{i+1}$ is cyclic as well. We conclude by using the theorem's hypothesis.
\end{proof}

It is not hard to find values $(q,b)$ verifying the last hypothesis. Moreover, for a fixed prime $p$, there are infinite many pairs $(r,b)$ such that $(p^r,b)$ verifies it.

\begin{Prop}\label{prop:seq_cyclic_2}
Given an integer $b$ and an odd prime $p>b+2$ such that $b\not\equiv 2 (\mod 4)$ and
\[
\exists r\geq 1, (p^{2r}-1, b+2)=1,
\]
then for every $i\geq 0$,
\[
f_i(t)=t^4+bt^2+q_i^2
\]
defines an ordinary cyclic isogeny class $\mathcal{A}_i$ of abelian surfaces over $\Fq[i]{}$, where $q_i=p^{r+is}$ and
\begin{align*}
s=\prod\limits_{\ell\in\mathcal{P}(b+2)} (\ell-1).
\end{align*}
\end{Prop}
\begin{proof}

It is easy to check that $f_i(t)$ is a Weil polynomial and it defines an ordinary isogeny class since $p>b$.
Put
\begin{align*}
c=\prod\limits_{\ell\in\mathcal{P}(b+2)}  \ell,
\end{align*}
We have
\begin{align*}
f_i(1)=p^{2(r+is)}-1+(b+2), \; i\in\mathbb{Z}_{\geq 0}.
\end{align*}

Observe that $f_i(t)=f_{q_i,(0,b)}(t)$ and $f'_i(1)=2(b+2)$. For $\mathcal{A}_i$ to be cyclic it is enough to have $(f_i(1),2(b+2))=1$, and since $f_i(1)\equiv b+2 \not\equiv 0 \pmod{4}$, it is enough to have $(f_i(1),b+2)=(f_i(1),c)=1$. The last statement is true since by hypothesis $(f_0(1),c)=1$ and since $c$ divides the difference $f_{i+1}(1)-f_i(1)=p^{2(r+is)}(p^s+1)(p^s-1)$ (using Fermat's little theorem).
\end{proof}

For a fixed $p$, big enough, take a prime $\ell$ such that $2<\ell <p$ and $\ell\nmid p^2 -1$, then set $b=\ell -2$. This gives $b$ and $p$ satisfying the proposition hypothesis.

\section*{Acknowledgement}
I would like to thank my advisor Serge Vl\u adu\c t for very fruitful discussions and for his very useful comments and suggestions.

redpajama_set_name: RedPajamaArXiv


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For Halloween, a reworking of an old drawing of mine.
You know, this was the type of thing I wanted to do for Halloween but could never get together in as good as form as you did. Most excellent!
I hope you meant as a picture and not a costume.


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Software gratuito ou freeware é qualquer programa de computador cuja utilização não implica o pagamento de licenças de uso ou royalties. É importante não confundir o free de freeware com o free de free software, pois no primeiro uso o significado é de gratuito, e no segundo de livre. Um programa licenciado como freeware não é necessariamente um software livre, pode não ter código aberto e pode acompanhar licenças restritivas, limitando o uso comercial, a redistribuição não autorizada, a modificação não autorizada ou outros tipos de restrições. O freeware diferencia-se do shareware, no qual o usuário deve pagar para acessar a funcionalidade completa ou tem um tempo limitado de uso gratuito.

História
O termo freeware foi introduzido por Andrew Fluegelman quando desejava vender um programa chamado PC-Talk que havia criado mas não queria usar métodos tradicionais de distribuição por causa do custo. Fluegelman, na realidade, distribuiu o PC-Talk sob uma licença agora chamada de shareware: o uso atual do termo não é necessariamente compatível com o conceito original.

Critérios de Definição
Um software é gratuito ou freeware quando possui suas funcionalidades completas por tempo ilimitado sem custo monetário. A licença pode restringir o tipo de uso, como uso para fins não lucrativos, não comerciais, uso acadêmico, entre outros. Dessa forma a licença pode ser "gratuito para uso não comercial"

Derivações
Após a disseminação das distribuições como shareware, outros tipos de distribuições com sufixo -ware surgiram, como adware, que denota softwares que possuem propagandas - que também podem ser shareware; ou outros como careware, solicitando que o usuário faça doações a alguma instituição de caridade, crippleware, software com recursos limitados e outras, como beerware, que requer que o usuário pague uma cerveja ao autor.

Licenças Creative Commons
As licenças Creative Commons que permitem que um desenvolvedor defina "Freeware" em um domínio jurídico seguro e internacional, respeitando o caminho. O "compartilhamento" do caso de uso pode ser refinado com cláusulas de restrição como não-comercialidade (CC BY-NC) ou sem derivações (CC BY-ND), veja a descrição das licenças.

Ver também
 GPL
 Código aberto

Ligações externas

Distribuição de programas de computador

Licenças de programas de computador

redpajama_set_name: RedPajamaWikipedia


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> PA Admits Israeli Sanctions Are Cutting Deep
PA Admits Israeli Sanctions Are Cutting Deep
PA prime minister Salam Fayyad says Israel freeze of tax revenues to Ramallah is having a 'devastating impact'
Gabe Kahn.
PA Salam Fayyad Israel Tax Revenues
Salam Fayyad Flash 90
Palestinian Authority Prime Minister Salam Fayyad admitted Sunday Israel's decision to freeze tax revenue transfers to the PA were cutting deep.
Fayyad told reporters Sunday that the sanctions have a "devastating impact" on the economy in PA-run enclaves.
Israel froze the transfer of tax revenues - amounting to roughly $100 million per month - after the PA launched a series of unilateral moves at the United Nations aimed at attaining recognition of a state based on pre-1967 lines at the world body.
Officials in Jerusalem said the moves, which began with a now-moribund application for statehood at the UN Security Council and culminated in UNESCO voting to induct the PA as a 'member state,' were a direct violation of the bilateral Oslo Accords.
All contacts and cooperation with the PA - including the transfer of tax revenues - occur under the auspices of bilateral agreements that proscribe unilateral moves like those the PA is pursuing, Israeli officials note.
PA officials were forced to suspend their unilateral track after UN secretary general Ban Ki-moon withdrew his support for their statehood bid at the world body in the face of promised funding cuts to UN agencies by the United States. The United States underwrites 22 percent of the UN operating budget and 27 percent of its peace keeping operations
Senior economic officials in Ramallah have long said Israel's transfers of tax revenues along with foreign aid are crucial to keeping Fayyad's the cash-strapped administration afloat. Israel's funds freeze to Ramallah coincides with a downturn in foreign donations to the PA – especially from Arab countries coping with the Arab spring.
Last week Fayyad told reporters if Israel did not resume the transfer of tax funds the PA would have to close its doors.
Israeli officials, however, say they are unmoved and that funds will not begin flowing to Ramallah again until PA officials accept a proposal by the Quartet for Middle East Peace for resumed talks without preconditions.
Israeli officials have accepted the Quartet proposal, which calls for a final status agreement by the end of 2012.
Previously, officials in the office of Prime Minister Binyamin Netanyahu said, while the PA shutting down would be "unfortunate" it "would not be the end of the world."


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Dan Froomkin - Press Watch
A section of the US/Mexico border fence. (Getty/AP/Photo Montage by Salon)
Trump: "We are preparing for the military to secure our border"
The president claims he is gearing up to order the U.S. military to police parts of the Mexican-American border
April 4, 2018 4:47AM (UTC)
President Donald Trump said on Tuesday that he is gearing up to order the U.S. military to police parts of the Mexican-American border. The president's remarks intensified his recent rant regarding his desperate desire to see through his campaign promise to tighten border security.
"We have very bad laws for our border, and we are going to be doing some things. I've been speaking with General Mattis. We're going to be doing things militarily," Trump said at the White House, during a meeting with visiting Baltic leaders. "Until we can have a wall and proper security, we're going to be guarding our border with the military. That's a big step. We really haven't done that before — or certainly not very much before."
Trump reportedly reiterated to reporters on Tuesday: "We are preparing for the military to secure our border between Mexico and the United States. I think it's something we have to do."
The president also corroborated his remarks on Twitter:
During the last couple of days, Trump has actively blustered about the need for stricter border control on Twitter, citing an increased urgency after hearing that a group of migrants from Honduras were traveling through Mexico. The story had been reported by his media outlet of choice, Fox News.
Mexican government authorities intervened and have agreed to provide humanitarian visas to some of the 1,200 migrants who were reportedly part of the "caravan," including infants and elderly individuals, according to the New York Times. The caravan is an annual event intended to draw attention to the growing refugee crisis in Central America, the Washington Post reported.
This would not be the first time that troops have been deployed to the border. In 2010, then President Barack Obama's administration sent 1,200 National Guardsmen to support Homeland Security's CBP and ICE amid heightened concerns about drug trafficking.
Advocacy groups highlighted how such patrol is harmful to border communities on social media.
https://twitter.com/RI4A/status/981290747667206145
Trump's latest remarks appear to be another attempt for the president to deliver to his base. Indeed, last week, Trump reportedly suggested in a private meeting with House Speaker Paul Ryan that funding for the construction of the border wall could come from the U.S. military budget, according to CNN. Reports of that plan were met with derision from both sides of the aisle.
"First Mexico was supposed to pay for it, then U.S. taxpayers and now our men and women in uniform? This would be a blatant misuse of military funds and tied up in court for years. Secretary Mattis ought not bother and instead use the money to help our troops rather than advance the president's political fantasies," Senate Minority Leader Chuck Schumer , D-N.Y., said in a statement.
Nicole Karlis is a news writer at Salon. She covers health, science, tech and gender politics. Tweet her @nicolekarlis.
MORE FROM Nicole Karlis
Donald Trump Minority Leader Chuck Schumer Trump Border Wall Trump Immigration Policy
Check out this article! https://www.salon.com2018/04/04/trump-we-are-preparing-for-the-military-to-secure-our-border/


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