text:
Tag: science experiments
Co-op Chemistry, by Cheryl
Mentos and diet soda, in cold weather
Science With Friends
My kids love doing science experiments! I really do too, but with everything else we do at home, the experiments sometimes get put on the back burner. I wanted to be sure that my eight-year-old science enthusiast had a year full of science experiments. To be sure I followed my plan, I signed up to teach a class for our co-op. I have never had so much fun with a group of seven, eight, and nine-year-olds!
We have spent the last two months studying the basics of chemistry. We have covered atoms; the periodic table; mixtures; four kinds of reactions; four types of evidence of reactions; polymers, and more.
Experiments we have done so far:
Mixing things found around the house to check for reactions: baking soda, lemon juice, vinegar, salt water, egg whites, and milk. As we mixed items, we recorded what happened in a chart. The kids loved baking soda with vinegar and baking soda with lemon juice. Milk and lemon juice was another fun reaction. We looked for bubbles or precipitation in these experiments.
We also spent a week studying the pH of various household liquids by mixing them with red cabbage juice. Acids turned our juice from purple to pink, bases turned it green/blue, and neutral items did not change the color. We also neutralized the acids by adding bases to watch the color turn back to purple. The kids had a great time mixing things back and forth to watch the color changes.
We tested mixtures. Our first project was to make a cake. We made a nearly homogeneous mixture as we stirred the mix, eggs, oil, and water then we added giant drops of frosting to turn it into a heterogeneous mixture. I think this was their favorite experiment because we ate it! While that cooked we tried mixing oil and water based liquids to see what would happen.
We checked for starch in food items with iodine.
The second favorite experiment was making gooey putty. Mix equal parts liquid starch and white glue to create a polymer that is fun to play with!
Our final chemistry experiment was the "Mentos in diet soda" reaction. We read about the reason for the reaction, discussed what we thought would happen with different types of soda and then we went outside with eight bottles (four types of soda) and two flavors of Mentos. We had a soda fountain show! Our soda did not shoot as high as Mythbusters' did. Our hypothesis is that the soda and Mentos (which sat in my car all night in the freezing cold) do not react as strongly when cold. We plan to test it again when it is warmer outside.
The kids have been introduced to some important foundational concepts of chemistry as well as the idea of the scientific method in experimentation. We are only halfway through our class. I cannot wait to get started on physics with these kids!
Cheryl–Cheryl is a singing, dancing, baking, homeschooling mom of three. She has danced her whole life and taught ballet and theatre for most of her adult life. Her favorite pastime has always been cooking and baking, and as a Pampered Chef Independent Consultant she gets to share that love with others. Home educating her three children has been and continues to be one of her greatest learning experiences! It is an adventure she is ready to continue.
Author mizsocratesPosted on March 15, 2014 Categories ChemistryTags Chemistry, Cheryl, co-ops, March, recent, Science, science experiments1 Comment on Co-op Chemistry, by Cheryl
Science and History: Hand in Hand, by Lynne
Teaching Science at Home
I am a history nerd. I've always enjoyed history classes, historical fiction, and historical documentaries. I never considered myself a science nerd, though. As I've said before, The Well Trained Mind by Jessie Wise and Susan Wise Bauer changed my life. Of course, I knew that science and history did not happen independently of one another. And, of course, I knew that history influenced and inspired scientific discovery. I also knew that certain events in history often discouraged scientific advancement, such as when Mongol invaders destroyed libraries and universities during their rampages. What I didn't know until using the four year history and science cycle described in the book was how much sense the history of science would make when studying science in conjunction with history.
"We divide the four years of science into subjects that roughly correspond to the history periods. First graders, who are studying the ancients, learn about those things the ancients could see — animal life, the human body, and plants. . . Second graders collect facts about the earth and sky, a division designed to go along with the medieval-early Renaissance period, when Copernicus and Tycho Brahe observed the heavens." (The Well-Trained Mind pp. 157- 158. 2009. W. W. Norton & Co.) The book goes on to describe how third graders learn about chemistry while learning about great chemists of the early modern period, like Robert Boyle. Fourth graders learn about physics and technology while they are studying the history of the modern age and all its exciting scientific and technological developments.
These cycles of history and science together are then repeated again in the logic and rhetoric stages, building on the facts learned during the grammar stage.
Before we started homeschooling, I had been afterschooling my kids in history by having us all listen to The Story of the World by Susan Wise Bauer in the car as we drove all around. My younger son was in Kindergarten at the time, and my older son was in first grade — the year students study ancient history. The next year we were homeschooling full time, so I decided to move ahead with history and study the medieval period. Therefore, we didn't do life sciences the first time around in our four year cycle. We went straight into earth science and astronomy.
I was amazed by the resources my public library had on these topics, so I didn't even purchase a curriculum for science that year. We read dozens of library books about weather, geology, the solar system, etc. We also did experiments from Janice Van Cleave's 201 Awesome, Magical, Bizarre & Incredible Experiments. We made a rain gauge and kept weather journals. We went to the planetarium and learned about the constellations. We searched for different kinds of rocks and labeled them. We even put together our own models of the solar system. As we learned about the rotation of the planets in science, we were learning in history class about Hans Lippershey inventing the telescope and Galileo being excommunicated for his radical scientific views. How cool is that?
In our second year of homeschooling, we were studying chemistry in science and the early modern era in history. Now, you could not have prepared me for the enthusiasm my older son would have for chemistry! I was not a fan of chemistry in high school, so I never imagined that my kids could like it so much. I purchased this beautiful deck of element cards, and my older son would sit with them, and pour over them, and read information from them to me. We even used them to build a giant periodic table on our sunroom floor. In addition to the Janice van Cleave experiment book, we also used Real Science 4 Kids Chemistry Pre-Level I. This book was a great introduction to basic concepts in Chemistry, and my kids really liked how atoms were drawn with arms to show how they linked to other atoms to form molecules. The atoms were drawn with the same number of arms as they had available spaces in their outer electron shell. Such an easy way for elementary kids to understand the concept of how atoms could join with other atoms! There were numerous advancements in science during the early modern period, and it was interesting to see how the advancements in science instigated changes in world politics and history. The study of chemistry and physical properties of matter were at the heart of these advancements.
Our third year of homeschooling corresponded to the fourth year of the history/science cycle- the Modern Age. This year, we used Real Science for Kids Physics Level I. We also used the corresponding lab book. I liked how the lab book had students conduct their experiments by using the scientific method. We learned about electricity and simple machines. The kids took a class at the Metroparks about how light and sound waves work. We learned that there was such a thing as nuclear physics. We had interesting conversations about the historical implications of nuclear physics, such as the devastation caused by atom bombs, and (an event my kids remember) the damage caused by earthquakes to the nuclear reactors in Japan. We also had some laughs when a physics discussion would be prompted by a family TV night watching The Big Bang Theory.
This year, we are in the Logic Stage and have gone back to the first year of our cycle. We are studying Biology, using Pandia Press R.E.A.L. Science Odyssey Biology 2. This is a fascinating book, with lots of hands on experiments and many chances to look at things under a microscope. We are also taking a second look at ancient history. So much was happening in life sciences during the ancient times. Egyptians were mummifying bodies, Hippocrates was busy establishing the study of medicine in Greece, and people in India were busy working on a classification system.
Obviously, during all periods of history, all kinds of science was happening in all sorts of areas. The study of chemistry wasn't limited to the early modern period any more than astronomy was limited to medieval times. People have studied the world around them since the dawn of time. You can't have biology or physics without chemistry. It's all interconnected. So, even though this four year cycle breaks down the different disciplines of science to correspond with different eras in history, it doesn't limit us to these four designations. If you are studying modern times, you are going to learn about the newest discoveries of the latest satellite or Mars Rover, right along with your physics lessons. The four year cycle just provides a consistent backbone to pursue historical AND scientific scholarship simultaneously. When my kids had to temporarily go back to public school this fall, giving up this four year cycle was, honestly, the biggest regret I had concerning their academics. I am relieved we get to continue on this path, and so are the kids.
Lynne–Lynne has enjoyed homeschooling her two sons for the past three years, after their brief stint in the local public school. Her older son is a humorous fellow with high functioning autism who thrives in a home education environment. Her younger son is a sensitive soul with a great deal of patience. The boys, Mom, and Dad, along with the two guinea pigs, live in Northeast Ohio. Lynne holds a Master's Degree in French Language and Literature. She is also a Harry Potter fanatic, enjoys line dancing and Zumba, spends hours scrapbooking, and loves organic vegetables. You can visit her soon-to-be revitalized blog at www.daysofwonderhomeschool.blogspot.com.
Author mizsocratesPosted on March 9, 2014 Categories ScienceTags Chemistry, History, Lynne, March, RS4K, Science, science experiments, science history, Susan Wise Bauer, The Well-Trained Mind, WTM Science1 Comment on Science and History: Hand in Hand, by Lynne
sandboxtosocrates Blog at WordPress.com.


Please click, print, and sign these petition letters (2 for your Senators [S.878] and 1 for your Representative [H.Res.248 & H.R.2002]) with your neighbors, friends and family. Mail the stack to FAPA Headquarters. THANK YOU!
On March 16, 2018, President Trump has signed Taiwan Travel Act (HR 535) into law!
To all our members, supporters and Taiwanese American all over the states, thank you for your hard work and support. This wouldn't happen if it weren't for you.
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Rob arrived at Bromley after living in the greater Toronto Area for the past 12 years. Rob is married to Jill and together they have 2 children, Isaac and Nevaeh. Rob received an undergraduate degree at Tyndale University College and Seminary, and his masters at Wycliffe College at the University of Toronto. Rob has served as a youth pastor, children's pastor, an associate pastor and now as a senior pastor.


Times of India – COLOMBO: Sri Lanka's sports minister called on the International Cricket Council to establish "simple, clear rules" Wednesday after skipper Dinesh Chandimal was punished for ball tampering "Laws governing the offence / act of altering the condition of the ball are somewhat vague and lack clarity," Faiszer Musthapha said in a statement.
COLOMBO: Sri Lanka's sports minister called on the International Cricket Council to establish "simple, clear rules" Wednesday after skipper Dinesh Chandimal was punished for ball tampering "Laws governing the offence / act of altering the condition of the ball are somewhat vague and lack clarity," Faiszer Musthapha said in a statement.
"I invite the ICC to revisit the applicable laws and ensure that a set of simple, clear and certain rules and procedures are put in place.
"He did not elaborate but the statement came after Chandimal, coach Chandika Hathurusingha and manager Asanka Gurusinha were sanctioned following a tampering scandal.
His teammates David Warner and Cameron Bancroft were also banned.Du Plessis said tougher sanctions would make players think twice before putting something in their mouth and then using saliva to shine the ball.


Dialogs work by selecting or typing in keywords or very short versions of what is actually being said.
The first video game about Dialog: Keywords was released on December 17, 1988.
The keywords may occasionally be depicted in form of icons/images.


text:
Students look at housing issue
It's important for the Island's education system to instill civic mindedness in students, preparing them to become good citizens and think of social justice and equality. As a final assignment for a Grade 12 Global Issues class at Colonel Gray High, teacher Andrew Rowe did just that.
For their final presentations of the year, the students were asked to research a real world issue and relate it back locally. The topics ranged from global warming and pollution, to hunger, immigration and Internet privacy.
Connor Crosby and Young Hao Yu chose to focus on PEI's affordable housing crisis. Under the leadership of Mr. Rowe, they surveyed 260 people to better understand the housing issue on Prince Edward Island. They compared Victoria, B.C. - which has a vacancy rate of 0.8 per cent - with Prince Edward Island - which has a vacancy rate of 0.3 per cent - and considered whether some of the solutions under review in Victoria could work here.
The students examined possible solutions such as government loans to families and construction companies, reserving assisted housing for low income families, building new public housing, and regulating Air BnB type rentals. They analyzed the pros and cons of each idea.
"I just think it's terrible that people don't have the resources to live a good life," Crosby told his classmates, 50 per cent of whom immigrated to Canada in the past five years. "We need to start these projects right away so we can start to see a drastic change immediately."
Young says he believes the lack of affordable housing is the most important issue facing Islanders.
"How can we worry about anything else when people don't have a place to lay their head at night?"
After carefully weighing the pros and cons of the options and the results of their survey, Crosby and Yu concluded that government should invest heavily in new public housing.
Social Development and Housing Minister Ernie Hudson, along with other provincial and municipal government representatives and members from Prince Edward Island's three other political parties, attended the students' presentation to hear Crosby and Yu's ideas and share their own ideas and feedback.
"Islanders are struggling to find affordable, accessible and appropriate housing and we need to work collaboratively and creatively to find solutions," Minister Hudson said.
"A good idea is a good idea no matter where it comes from."
"It is great to see our Island education system encouraging students. This can only lead to strong leadership for our province in the future."
Minister Hudson thanked the students for their work and asked to have a copy of the presentation emailed to him so he could share it with his department and fellow cabinet ministers.
"Housing is certainly the number one priority for us right now," Hudson said.
For more information on the Housing Strategy
Tweets by AgInfoPEI
Department of Social Development and Housing
2nd Floor, Jones Building
11 Kent Street
DeptSDH@gov.pe.ca

redpajama_set_name: RedPajamaCommonCrawl


Fr. Thielo Ramirez, parochial vicar at Ss. Simon and Jude Cathedral shares his vocation story.
MaryRuth Hackett, Ph.D. introduces us to the new Parenting Smarts podcast.
Joe Cady, director of Adult Catechesis at St. Timothy's, instructor at Kino Catechetical Institute and University of Mary, Tempe explains aspects of the faith many cradle Catholics may have missed in their formation.
Tony Gutierrez, editor of The Catholic Sun, gives us a glimpse into the lives of the saints featured in this month's edition.
Fr. Michael Passo, FSSP, pastor, Mater Misercordiae in Phoenix talks about his vocation, the Latin Mass and Mater Misercordiae Mission.
Recent graduates from the Tepeyac Leadership Initiative share their experiences.


text:
Q: Compiler fails due to recursion in MSVS2017 I am using compile time features of boost::hana::string for generating xaml markup strings for a GUI.
I am using VS2017 configured as C++17 language, but there is some moment where strings are too long that I received the next message from compiler:
 fatal error C1202: recursive type or function dependency context too complex

If I delete a few characters from the string, everything is ok, so I guess is length dependant.
Next code is working right in coliru under gcc, but it fails miserabily in MSVS2017:
NOTE: Please, take into account that I have removed all the configurabled and templated variables in the example to focus just in the problem. Of course, it is not worthy to use boost::hana like sample code, is just a minimal working source.
#include <boost/hana/string.hpp>

constexpr auto markup_ribbon()
{
    return (BOOST_HANA_STRING(
        "<?xml version='1.0' encoding='iso-8859-1'?>"
        "<Border VerticalAlignment='Top'>"
        "<Border.Background>"
            "<LinearGradientBrush StartPoint='0,0.2' EndPoint='0.8,0.8'>"
                "<GradientStop Offset='0' Color='#cd0000'/>"
                "<GradientStop Offset='0.75' Color='#fc999e'/>"
                "<GradientStop Offset='1' Color='#e4ecf7'/>"
            "</LinearGradientBrush>"
        "</Border.Background>"
        "<StackPanel Orientation='Vertical' TextBlock.FontFamily='Tahoma'>"
            "<StackPanel.Resources>"
                "<Style TargetType='Path'>"
                    "<Setter Property='Fill' Value='Blue'/>"
                    "<Style.Triggers>"
                        "<Trigger Property='IsMouseOver' Value='True'>"
                            "<Trigger.Setters>"
                                "<Setter Property='Fill' Value='Red'/>"
                            "</Trigger.Setters>"
                        "</Trigger>"
                    "</Style.Triggers>"
                "</Style>"
            "</StackPanel.Resources>"
            "<Grid>"
                "<Grid.ColumnDefinitions>"
                    "<ColumnDefinition Width='300'/>"
                    "<ColumnDefinition Width='*'/>"
                    "<ColumnDefinition Width='*'/>"
                    "<ColumnDefinition Width='50'/>"
                    "<ColumnDefinition Width='200'/>"
                "</Grid.ColumnDefinitions>"
                "<Image Margin='10' HorizontalAlignment='Left' VerticalAlignment='Center' Source='1'/>"
                "<TextBlock Grid.Column='1' HorizontalAlignment='Left' VerticalAlignment='Center' FontSize='44' FontWeight='Bold' Foreground='#ffffff'>ASDF2021</TextBlock>"
                "<Image Grid.Column='2' HorizontalAlignment='Left' VerticalAlignment='Top' Source='3'/>"
                "<TextBlock Grid.Column='3' HorizontalAlignment='Right' VerticalAlignment='Center'><Hyperlink TextDecorations='' Tag='back' ToolTip='Back to administration screen. Back to administration screen.'><Path Stroke='Purple' StrokeThickness='2' Data='M12.981,9.073V6.817l-12.106,6.99l12.106,6.99v-2.422c3.285-0.002,9.052,0.28,9.052,2.269c0,2.78-6.023,4.263-6.023,4.263v2.132c0,0,13.53,0.463,13.53-9.823C29.54,9.134,17.952,8.831,12.981,9.073z'/></Hyperlink></TextBlock>"
                "<TextBlock Grid.Column='4' Margin='0,0,30,0' HorizontalAlignment='Right' VerticalAlignment='Center' FontSize='18' FontWeight='Bold'><Hyperlink Tag='exit' TextDecorations='' ToolTip='Close current session and go back to user screen. Good Bye, good bye.'>Close current session</Hyperlink></TextBlock>"
            "</Grid>"
            "<Border Height='3' Background='#9ebbdd'/>"
            "<Border Height='3' Background='White'/>"
        "</StackPanel>"
        "</Border>")
        ).c_str();
}

#include <iostream>

constexpr auto markup = markup_ribbon();

int main()
{
    std::cout << markup << std::endl;
}

I understand that problem is compiler+recursion related, not boost::hana's problem at all, but maybe is there some tweak, hack or something that I may adjust in IDE so this works?

A: The compiler is giving up because you are creating a template with too many parameters. The representation of a Hana string is hana::string<'<', '?', 'x', 'm',...> where each character is a non-type template parameter. This is generated by some tricks inside the BOOST_HANA_STRING macro.
I suggest looking at the following presentation for some insight on how to deal with large compile-time strings: Constexpr All the Things - Jason Turner, Ben Deane
Aside from that, here is a basic example of one possible approach that you could possibly put in your own macro depending on how you are using these.
static constexpr char name[] = "really long string";

template <char const* str>
struct my_string {
    static constexpr char const* value = str;
};

auto name_s = my_string<name>{};

static_assert(name_s.value[3] == 'l', "");

redpajama_set_name: RedPajamaStackExchange


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TOOWOOMBA-based Broadspectrum subsidiary, Easternwell, has been awarded a five-year contract to provide well servicing rigs and associated services for Shell's QGC coal seam gas business in the Surat Basin.
Broadspectrum Managing Director and Chief Executive Officer, Nick Miller, said the new contract with Shell's QGC business reinforced his company's ability to deliver technical, innovative solutions to the resources sector.
Easternwell is currently recruiting a number of positions on their website.
"We look forward to continuing to work with Shell's QGC business to grow our operations in regional Queensland, providing a boost for local economic activity," he said.
The contract will result in a boost for the local economy over five years including a 20 per cent increase in local employment and a nine per cent increase in local procurement.
It will also provide an opportunity to increase Indigenous employment by 50 per cent and double the level of Indigenous procurement.
Under the contract, Easternwell will provide its next-generation Advantage Light Rig platform, which features several improvements including safer operations, efficiency outcomes, and the capability to increase productivity by 30 per cent in the field.
Broadspectrum's chief executive, resources, Tom Quinn, said the ALRx had been designed and built from the company's Toowoomba hub, and showcases homegrown innovation.


P.S. If you prefer bite-sized video chunks, then check out this video demo page instead. Videos are broken down by specific features.
A mindcrumb is a daily journal entry for capturing your up-to-date thoughts on a specific journey toward learning a certain skill or embarking on a new life experience.
Use it to de-clutter your mind, track your progress, and systematically experiment with new strategies to advance on that journey. Mindcrumbs are smart and they will get smarter the more you keep up with your journaling routine.
How to Create a Mindcrumb?
In order to input your mindcrumbs you'll first need to create a new journey. Click on the "Start a New Journey" button in the "My Journeys" tab above. If you already have a journey, then just navigate there by clicking on the same "My Journeys" tab.
Click on the "Enter a New Mindcrumb" text box on your journey page in order to create a new mindcrumb. You can include rich text, links, images, videos, and tags (see next section) in your mindcrumb.
How to Use Tags for Systematic Filtering?
One recurring #sticking_issue that has been frustrating me lately is figuring out how to stay on beat when executing spinning moves.
The first paragraph in the example above will be labeled with the #sticking_issue tag. The second paragraph will be labeled with the #milestone and #habit tags. As a result, filtering through the #habit tag will include the second paragraph along with similarly tagged paragraphs in other mindcrumbs. You can perform such filtering by clicking the magnifier icon next to the "All Mindcrumbs" tab or by clicking on any visible #tags when browsing through a journey page.
How to Edit the Milestones Timeline?
Mindcrumb paragraphs with the #milestone tag will automatically appear on the milestones timeline.
No one can see your mindcrumbs unless you mark your journey as public. You can change the privacy of your journeys any time by visiting your profile page. Simply click on the user icon in the menu bar above.


text:
Megan Thee Stallion Announces "B.I.T.C.H" Release Date With Cover Art Reveal
16 January 2020 MIXTAPESfolk, hea, morning, new, No, Rap, Social Media, summer
Megan Thee Stallion will be ending the month with a bang.
. In January, only the folks seriously tuned into what was buzzing in the rap world had heard of her but . It's crazy how fast these things can happen. The star has been supplementing her highly-successful Fever mixtape for months, introducing us to Hot Girl Summer and linking up with some of the biggest names in the business. and now, it looks like she's getting back on her grind, inching us toward her upcoming debut album with the release of "B.I.T.C.H," the first solo single of the year from Meg.
Taking to social media this morning, the Houston rapper informed everybody to set their alarms for January 24, because that's when she'll be dropping her next single. Titled "B.I.T.C.H," the H-Town Hottie has a barking dog on her cover artwork and .
. Will you be checking for the new track at the end of the month?
Corey Jorgenson
All Three R. Kelly Lawyers Are Abandoning His Case
Roddy Ricch Says Lauren London Smiling After Nipsey's Win Was "One Of The Greatest Moments" Of Grammys
« Rubi Rose Always Keeps Weed, $500 Perfume, & Her Student ID In Her Bag
Doja Cat Almost Died On A Jet Ski In Cabo: "The Sharks Were En Route" »


text:
By Kit O'Connell Kit O'Connell @KitOConnell
The only openly Anti-Semitic Alex Jones,' is now staunchly Zionist.
AUSTIN, Texas — Infowars, an alternative multimedia outlet notorious for peddling hoaxes and anti-Semitic ideas, is now toeing a far-right, extremist line on U.S.-Israel relations.
The site's shifting political viewpoint is another sign of the transformation that Alex Jones, the site's founder, has undergone as he's transitioned from an offbeat talk show host on public access TV to a Washington insider with intimate access to President-elect Donald Trump.
Despite his modest beginnings, Jones has built an expansive and profitable media empire that includes multiple websites, TV and radio programs, documentaries, and lucrative advertising deals that promote products ranging from Infowars-branded male "vitality" supplements to doomsday preparation supplies.
Jones amassed a following by vociferously opposing status quo politics and the mainstream media narrative, even when it meant propagating blatantly false hoaxes such as Sandy Hook "trutherism" or #Pizzagate. During the 2016 election cycle, Jones and Infowars were enthusiastic supporters of Trump's candidacy, and the president-elect favored Jones with an exclusive interview last month, just weeks after his election.
In addition to peddling conspiracy theories, bucking the mainstream, and joining Team Trump, Infowars and its publisher frequently veer into blatant anti-Semitism. In an October episode of his radio show, Jones claimed Chicago Mayor Rahm Emanuel and his brother, Ezekiel, who are Jewish, are part of a mythical "Jewish mafia." Earlier that same month, when Trump referenced an anti-Semitic conspiracy theory about global finance during a stump speech in Florida, media analysts cited Jones as one of the sources of his beliefs.
However, Infowars sometimes has been justifiably critical of Israel's foreign policy and treatment of the indigenous Palestinian population. During Israel's brutal 2014 assault on Gaza that ultimately killed over 2,200 people, many of them children, Infowars published an opinion piece calling U.S. support for Israel the "moral failure of the West."
In the July 2014 op-ed, Paul Craig Roberts, who served as assistant secretary of the Treasury for Economic Policy under President Ronald Reagan, wrote:
"One would think that the Great Moral West would be discussing sanctions on Israel and on Washington's stooge government in Kiev, which is bombing civilian homes, apartment complexes, and infrastructure … But the Great Moral West only aids the perpetrators of death and destruction, not the victims."
Also in July 2014, Paul Joseph Watson, Infowars' editor-at-large, criticized the invasion of Gaza and its high death toll in a video satirically titled "Israel is Only Defending Itself."
"Even if you think butchering kids is a legitimate form of self-defense, can we at least refrain from applauding, cheering, and celebrating the murder of children?" Watson asked.
However, Infowars' tone on Israel has changed dramatically in recent months, coinciding with Jones' support for Trump. After the United States refused to block a U.N. resolution condemning Israel's expansion of illegal settlements on Palestinian land, Infowars republished a bizarre diatribe by blogger Michael Snyder which claimed that the United States is under a supernatural curse for opposing the Israeli government's apartheid agenda. Snyder wrote:
"Because Barack Obama has cursed Israel at the United Nations, America is now under a curse. Friday's stunning betrayal of Israel at the UN Security Council is making headlines all over the planet, but the truth is that what Obama has just done is far more serious than most people would dare to imagine."
Contrary to Snyder's claims that Obama betrayed Israel, the president actually helped Israel cement a record-breaking $38 billion dollar military aid package in the final months of his administration.
In a video Jones published on his YouTube channel on Jan. 3, he claims the United States' failure to veto the resolution is part of a plot by billionaire investor George Soros to "bring down" the United States and Israel. Although Soros does have links to numerous foreign NGOs and has been accused of supporting regime change in Ukraine and elsewhere, he also frequently stars in right-wing conspiracy theories that tie him to everything from the Black Lives Matter movement to labor rights protests.
In the video, Jones claims that documents published by WikiLeaks "prove" that the Obama administration and George Soros are funding Muslim attacks on the West and its allies, and that the failure to support illegal Israeli settlements is a part of that plot. Yet he fails to provide examples, sources, or facts to support these statements.
While anti-Semitic conspiracy theories and support for Israeli war crimes may seem like a strange combination, they actually closely align Jones and Infowars with the incoming administration. Steve Bannon, the notoriously racist, sexist, anti-Semitic but pro-Israel founder of Breitbart News, is the president-elect's chief strategist, and, last month, Trump selected David Friedman, his pro-settlement, anti-peace process bankruptcy lawyer, as the next ambassador to Israel.
Watch "Israel is Only Defending Itself" from Paul Joseph Watson:
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Social Media Censorship Reaches New Heights as Twitter Permanently Bans Dissent
Mnar Muhawesh speaks with journalist Daniel McAdams about being permanently banned from Twitter, social media censorship and more.
Media Backs Billionaire President as Chileans Protest Privatization
As millions gather in Santiago and whole families march against privatization and rising inequality, the corporate media is busy defending Chile's billionaire president.
Rage Against the War Machine: An Interview with Peace Activist Cindy Sheehan
An interview with anti-war campaigner Cindy Sheehan on how liberals learned to love her, her clash with the Women's March and more.
How Holes in the Burning Saudi Oil Fields Narrative Could Draw the US Into a War With Iran
It is inconceivable to the United States that a "rag-tag" force like Yemen's Houthis might, with home-made weapons, outdo the hundred-billion-dollar might and sophistication of the state-of-the-art systems Washington has lavished upon the Saudis.
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Much of the attention for the Polymath network has to do with the connection established between KYC providers, smart contract developers, legal experts, and token investors for their security token. They recently announced a partnership with SeriesOne, which will create an "end to end" blockchain solution that will hopefully reduce the difficulties associated with security tokens issuance.
The partnership was announced originally on Tuesday by SeriesOne, who said that the prospective token issuers would have an easier time in the creation of compliant digital securities. Considering the evolving regulatory space with the SEC, compliance with the current rules is crucial to avoid substantial fees or even a shutdown.


text:
\section{Introduction and Preliminaries} \label{sec:intro}
In this paper, we will be interested in fields ${\mathbb F}$ that have characteristic $0$.
Thus, one can assume that  ${\mathbb Q}\subseteq {\mathbb F}\subseteq {\mathbb C}$, where ${\mathbb Q}$ is the field of rational numbers and ${\mathbb C}$ is the field of complex numbers.  An element ${\alpha} \in {\mathbb C}$ is said to be algebraic over  ${\mathbb F}$, if ${\alpha}$ is a
root of a polynomial $f(x) \in {\mathbb F}[x]$. The polynomial $f(x)$ is
said to be the minimal polynomial of ${\alpha}$ over ${\mathbb F}$, if ${\alpha}$ is a root
of $f(x)$,  $f(x)$ is monic  and  is irreducible in
${\mathbb F}[x]$. In this paper,  ${\mathbb M}_n({\mathbb F})$ will denote the set of all $n \times n$ matrix over ${\mathbb F}$.  All  vector symbols will denote column vectors and they will be written in bold face. Also, the vector of all $1$'s will be denoted by ${\mathbf e}$ and a square matrix with all entries $1$, will be denoted by ${\mathbf J}$. Then ${\mathbf J} = {\mathbf e}
{\mathbf e}^t$, where ${\mathbf e}^t$ denotes the transpose of ${\mathbf e}$. The symbol ${\mathbf 0}$ will denote either a vector or a matrix having all entries zero.

Recall that for any $A \in {\mathbb M}_n({\mathbb F})$, a celebrated result, commonly known as the Cayley-Hamilton Theorem, states that the matrix $A$ satisfies its own characteristic polynomial. That is, if $\varsigma_A(x) = \det(x I - A)$ is the characteristic polynomial of $A$, then $\varsigma_A(A)$ as an element of ${\mathbb F}[A]$, equals ${\mathbf 0}$. Let $S=\{f(x) \in {\mathbb F}[x] | f(A)={\mathbf 0}\}$. Then  $S$ is  an ideal in ${\mathbb F}[x]$ and $S = \langle p(x) \rangle$ for some monic polynomial $p(x) \in {\mathbb F}[x]$.  By definition, $p(x)$ divides $\varsigma_A(x)$ and for any $B \in {\mathbb F}[A]$, there exists a unique
polynomial $g(x) \in {\mathbb F}[x]$, with $\deg(g(x)) < \deg (p(x))$ such that $B = g(A).$ The polynomial $p(x) $  is called the minimal polynomial of $A$ and is  denoted by $p_A(x)$.

We are now ready to state a few results from matrix theory and
abstract algebra.  For proofs and notations related with these results
the reader is advised to refer to the book Abstract Algebra by Dummit $\&$
Foote~\cite{D:F} and Linear Algebra by Hoffman $\&$ Kunze~\cite{H:K}.

\begin{lemma}[Hoffman $\&$ Kunze, Pages~$204, 231$ \cite{H:K}] \label{lem:aa3}
Let $A$ be a square matrix.
\begin{enumerate}
\item  \label{lem:aa3:1} Then A
 is diagonalizable if and only if its minimal polynomial is separable.
\item  \label{lem:aa3:2} Let A be a matrix with distinct eigenvalues. Then a
matrix $B$ commutes
with $A$ if and only if $B$ is a polynomial in $A$.
\end{enumerate}
\end{lemma}

Before stating the next result,  recall that for a monic polynomial $f(x) = x^n - c_{n-1}x^{n-1} - c_{n-2}x^{n-2} - \cdots - c_1
x - c_0 \in {\mathbb F}[x]$, its  companion matrix, denoted
${\mathcal C}(f)$, is defined as
$$ {\mathcal C}(f) =
    \begin{bmatrix}
    0 & 1 & 0 & \dots & 0 & 0 \\
    0 & 0 & 1 & \dots & 0 & 0 \\
        \vdots &\vdots & \ddots & \ddots & \vdots & \vdots \\
     0 & 0 & 0 &\dots & 0 & 1\\
     c_0 & c_1 & c_2 & \dots & c_{n-2} & c_{n-1} \\
     \end{bmatrix}.$$
For example, the $n \times n$ matrix $$W_n=
\begin{bmatrix}
    0 & 1 & 0 & \dots & 0 & 0 \\
    0 & 0 & 1 & \dots & 0 & 0 \\
        \vdots &\vdots & \ddots & \ddots & \vdots & \vdots \\
     0 & 0 & 0 &\dots & 0 & 1\\
     1 & 0 & 0 & \dots & 0 & 0 \\
     \end{bmatrix}$$ is the companion matrix of the polynomial $x^n -1.$ Note that $W_n$ is a $0,1$-circulant matrix and $x^n - 1$ is its minimal polynomial. It is well known (for example, see Davis~\cite{davis}) that every
circulant matrix is a polynomial in $W_n$. Due to the above property, the matrix $W_n$ is called the fundamental circulant matrix. The next result also appears in~\cite{H:K}.

\begin{lemma}[Hoffman $\&$ Kunze, Page~$230$ \cite{H:K}] \label{lem:la}
Let ${\mathcal C}(f)$ be the companion matrix of $f(x) =  x^n - c_{n-1}x^{n-1} - c_{n-2}x^{n-2} - \cdots - c_1
x - c_0 \in {\mathbb F}[x]$ . Then \begin{enumerate}
\item \label{lem:la:1} $f(x)$ is both the minimal and the
characteristic polynomial of ${\mathcal C}(f)$.
\item \label{lem:la:2} all eigenvalues of the companion matrix
${\mathcal C}(f)$ are distinct if and only if ${\mathcal C}(f)$ is diagonalizable.
\end{enumerate}
\end{lemma}

The next result is also well known. The proof  can be easily obtained by using basic results in abstract algebra and it also appears in~\cite{D:F}.

\begin{theorem}\label{thm:a3}
Let $p_A(x)$ be  the minimal polynomial of $A\in {\mathbb M}_n({\mathbb F})$.
\begin{enumerate}
\item \label{thm:a3:2} Let $g(x) \in {\mathbb F}[x]$ and let $h(x) =
\gcd(g(x),p_A(x))$. Then $\langle g(A) \rangle =\langle h(A) \rangle.$
\item \label{thm:a3:1}
If $q(x)$ is a non-constant factor of $p_A(x)$ in ${\mathbb F}[x]$ then
${\mathbb F}[A]/\langle q(A)\rangle \cong {\mathbb F}[x]/\langle q(x)\rangle$. In
particular, if $q(x)$ is irreducible and $q({\alpha}) = 0$ for some ${\alpha} \in
{\mathbb C}$ then ${\mathbb F}[A]/\langle q(A)\rangle \cong {\mathbb F}[x]/\langle q(x) \rangle
\cong  {\mathbb F}({\alpha}).$ That is, ${\mathbb F}[A]/\langle q(A) \rangle$ is a field.
\end{enumerate}
\end{theorem}

As a corollary of Theorem~\ref{thm:a3}, one has the following result. To state the result, recall that a pair of $n\times n$ matrices $(A,B)$ over ${\mathbb F}$ is said to represent an extension field ${\mathbb K}$ if ${\mathbb K} \cong
{\mathbb F}[A]/\langle B \rangle$, where $\langle B \rangle$ is an ideal in  ${\mathbb F}[A]$
generated by $B$.

\begin{cor}\label{c1} Let ${\alpha} \in {\mathbb C}$. Then the matrix pair $(A,B)$ represents ${\mathbb F}({\alpha})$, a field, if and only if ${\alpha}$ is an eigenvalue of $A$, $q(x)$ is the minimal polynomial of ${\alpha}$ over ${\mathbb F}$ and
$\langle B\rangle =\langle q(A) \rangle$ in ${\mathbb F}[A]$.
\end{cor}

That is, suppose that $q(x)$ is the minimal polynomial of an eigenvalue ${\alpha}$ of $A$. Then the matrix pair $(A,B)$ represents an extension
${\mathbb K}={\mathbb F}({\alpha})$ of ${\mathbb F}$ if and only if  $\langle B \rangle = \langle
q(A)\rangle$. Hence with an abuse of the language, we may say that the matrix
$A$ represent ${\mathbb K}$ to mean that ${\mathbb K}={\mathbb F}({\alpha})$, whenever ${\alpha}$ is an eigenvalue of
$A$. Also, it is well known that the choice of  $\alpha$ is not unique.
Therefore, depending on the
choice of $\alpha$ the corresponding matrices that represent ${\mathbb K}$ can
vary. This issue becomes significant when we search for a smallest representation (in terms of order).

We are now ready to explain the motivation for our study.
Let $G$ be a finite group and let $n\in {\mathbb Z}^+$. A matrix representation of $G$ is a homomorphism from $G$ into $GL_n({\mathbb F})$, where $GL_n({\mathbb F})$ is the group of
invertible $n\times n$ matrices with entries from ${\mathbb F}$.
The representation is called faithful if the image of the homomorphism is
isomorphic to $G$.  A similar question arises whether an extension field of ${\mathbb F}$ has a representation in ${\mathbb M}_n({\mathbb F})$. For example, let  ${\alpha}$ be an algebraic number over ${\mathbb F}$ with $q(x) \in {\mathbb F}[x] $ as  its minimal polynomial.
Then, using Lemma~\ref{lem:la} and Corollary~\ref{c1}, we see that ${\mathbb F}({\alpha}) \cong {\mathbb F}[{\mathcal C}(q))]$, where ${\mathcal C}(q)$ is the companion matrix of $q(x)$. This leads
to the following natural questions:
\begin{enumerate}
\item does there
exist a matrix $A$ in ${\mathbb M}_n({\mathbb F})$  with some  specified properties such that
${\mathbb F}[A]\cong {\mathbb F}({\alpha})$?
\item if it exists, what is the smallest possible positive integer $n$?
\end{enumerate}

For example, fix a positive integer $n$ and consider ${\zeta}_n$, a primitive $n$-th
root of unity. Then the polynomial $\Phi_n(x)$ over ${\mathbb Q}$, called the $n$-th
cyclotomic polynomial,  is the minimal polynomial of ${\zeta}_n$ and hence is irreducible over ${\mathbb Q}$. In this case, is
it possible to find a matrix $A$ which is either circulant over ${\mathbb Q}$ or is a
$0,1$-companion matrix of $\Phi_n(x)$ such that  ${\mathbb Q}[A]\cong {\mathbb Q}({\zeta}_n)$?  It
can easily be checked that this is true only
when $n=1\;$ or $n=2$. For  $n> 2$, such a result is not true. To understand this, recall that  $x^n-1=\prod\limits_{d|n}\Phi_d(x)$, where $\Phi_d(x) \in {\mathbb Z}[x]$ is the minimal polynomial of ${\zeta}_d$ in ${\mathbb Q}[x]$ and for any two integers $s,t$, the notation $s | t$ means that $s$ divides $t$.
Consequently, from Corollary~\ref{c1}, it follows that for each divisor $d$
of $n$,
\begin{equation}\label{eqn:zetad}
{\mathbb Q}[W_n] / \langle \Phi_d(W_n) \rangle \cong {\mathbb Q}({\zeta}_d).
\end{equation}  In particular,
${\mathbb Q}[W_n] / \langle \Phi_n(W_n) \rangle \cong {\mathbb Q}({\zeta}_n).$ That is, in this case, the pair $(W_n,\Phi_n(W_n))$ represents the field
${\mathbb Q}({\zeta}_n)$.

To proceed further, we need the following definitions and notations.
A directed graph (in short, digraph) is an ordered pair $X=(V,E)$ that
consists of two sets
$V$, the vertex set, and $E$, the edge set, where $V$ is non-empty and  $E
\subset V \times V$. If $e=(u, v) \in E$ with $u \ne v$ then the edge $e$ is
said to be incident from  $u$ to $v$ or $u$ is said to be the
initial vertex and $v$ the terminal vertex of $e$. An edge $e = (u,u)$ is
called a loop. A digraph is called a  graph if $(u,v) \in E$ whenever $(v,u) \in E$, for any two elements $u,v \in V$.  If $u, v \in V$, then an edge between $u$ and $v$ in the graph $X$ is denoted by $e=\{u,v\}$ and in this case, we say that $e$ is incident with $u$ and $v$ or the vertex $u$ is adjacent to the vertex $v$, or vice-versa.   For any finite set $S$, let $|S|$ denote the number of elements in $S$. Then a graph/digraph is said to be finite, if  $|V| $ (called the
order of $X$)  is finite. A graph is called simple if it has no loops.

Let $X=(V,E)$ be a graph. Then the degree of a vertex $v\in V$, denoted
$d(v)$, is the number of edges incident with it. In case $v$ is  a vertex of a
digraph $X$, one defines in-degree of $v$, denoted $d^+(v)$, as the number
of edges that have $v$ as a terminal vertex and out-degree of $v$, denoted
$d^-(v)$, as the number of edges that have $v$ as an initial vertex.  The simple
graph $X$ that has an edge for each pair of vertices is called a complete
graph, denoted $K_n$, where $n$ is the number of vertices of $X$. A graph with no edge is called a null graph. The  cycle graph on $n$
vertices, say $u_1, u_2, \ldots, u_n$, denoted $C_n$,
 is a simple graph in which  $\{u_i,u_j\}$ is an edge
if and only if $i-j\equiv \pm 1 \pmod n$. A graph on $n$ vertices, say $u_1, u_2,
\ldots, u_n$, denoted $X_n$,  is called a path graph,
if for each $i, 1
\le i \le n-1$, the set $\{ u_i, u_{i+1}\}$ is an edge.
A graph (digraph) $X$ is said to be $k$- regular if
$d(v) = k$ ($d^+(v) = d^-(v) = k$) for all $v \in V$. Unless specified otherwise, all the graphs in this paper are assumed to be finite and simple.

Let $X=(V,E)$ be a digraph. Then  the adjacency matrix of $X$, denoted $A = [a_{ij}]$ is a square matrix of order $|V|$ with $a_{ij}=1$ whenever $(i,j) \in E$ and $0$, otherwise. In case $X$ is a graph then it can be easily seen that $A$ is a symmetric matrix.

Now, let $A$ be the adjacency matrix of a connected $k$-regular graph $X$ on
$n$ vertices. Then, it is well known that $k$ is a simple eigenvalue of
$A$. Thus, the minimal polynomial of $A$ is of the form $(x-k) q(x)$ for
some $q(x) \in {\mathbb Z}[x]$. Note that $k$ is a simple eigenvalue of $A$ implies
that $q(k) \ne 0$ and for any other eigenvalue ${\alpha}$ of $A, \; q({\alpha}) =
0$. Then with $q(x)$ as defined, we state the following well known result. We
present the proof for the sake of completeness.

\begin{lemma}[Hoffman~\cite{hof}] \label{lem:regular:J}
Let $X$ be a connected $k$-regular graph on $n$ vertices with
minimal polynomial $(x-k) q(x) \in {\mathbb Z}[x]$. Then the matrix ${\mathbf J}$ equals $ \dfrac{n}{q(k)} q(A)$.
\end{lemma}

\begin{proof}
As $X$ is a $k$-regular graph, its adjacency matrix $A$  satisfies $A {\mathbf e} = k {\mathbf e}$
and hence
\begin{equation} \label{eqn:J} {\mathbf J} A = A
{\mathbf J} = k {\mathbf J} \; {\mbox{ and }} \; q(A) {\mathbf e} = q(k) {\mathbf e}. \end{equation}
Also, the eigenvectors of $A$ can be chosen to form an orthonormal basis
$\mathcal B$ of ${\mathbb R}^n$. Hence $\dfrac{1}{\sqrt{n}} {\mathbf e} \in {\mathcal B}$ and
thus,
for any vector ${\mathbf x} \in {\mathcal B}, {\mathbf x} \ne {\mathbf e}$, ${\mathbf x}^t {\mathbf e} = 0$. Therefore,
${\mathbf J} {\mathbf x} = {\mathbf 0}$ and using Equation~(\ref{eqn:J}), ${\mathbf J} \dfrac{1}{\sqrt{n}} {\mathbf e} =
\dfrac{n}{\sqrt{n}} {\mathbf e} = \bigl(\dfrac{n}{q(k)} q(k) \bigr)
\dfrac{1}{\sqrt{n}} {\mathbf e}= \dfrac{n}{q(k)} q(A) \dfrac{1}{\sqrt{n}}
{\mathbf e}.$
Also, $q({\lambda}) = 0$ for any eigenvalue ${\lambda} \ne k$ of $A$ implies that $q(A) {\mathbf x} =
q({\alpha}) {\mathbf x} = {\mathbf 0}$. That is, $\dfrac{n}{q(k)} q(A) {\mathbf x} = {\mathbf 0}$.

Thus, the image of two matrices ${\mathbf J}$ and
$\dfrac{n}{q(k)} q(A)$ on a basis of ${\mathbb R}^n$ are same and hence the
two matrices are equal. Therefore  ${\mathbf J} = \dfrac{n}{q(k)} q(A).$
\end{proof}

The next corollary is an immediate consequence of
Theorem~\ref{thm:a3} and
Lemma~\ref{lem:regular:J}.

\begin{cor}\label{cor:regular}
Let $A$ be the adjacency matrix of a connected k-regular graph $X$
on $n$ vertices.  Then ${\mathbb F}[A]/\langle {\mathbf J} \rangle \cong {\mathbb F}[x]/\langle
q(x)\rangle$.
\end{cor}

\begin{proof}
Since $q(x)$ is a factor of the minimal polynomial $(x-k)q(x)$
of $A$, using Theorem~\ref{thm:a3} and Lemma~\ref{lem:regular:J}, one has
${\mathbb F}[x]/\langle q(x)\rangle \cong {\mathbb F}[A]/\langle q(A) \rangle \cong {\mathbb F}[A] /
\left \langle \dfrac{n}{q(k)}q(A) \right \rangle = {\mathbb F}[A] / \langle
{\mathbf J} \rangle.
$
\end{proof}

A.J. Hoffman $\&$ M.H. McAndrew~\cite{hoff} extended Lemma~\ref{lem:regular:J} to digraphs and is stated below. Note that in their paper, regular digraph were referred as strongly regular digraph.

\begin{lemma}[Hoffman $\&$McAndrew(1965)~\cite{hoff}]\label{lem:hoff}
Let $A$ be the adjacency matrix of a digraph $X$. Then there exists a polynomial $g(x) \in {\mathbb Q}[x]$ such that ${\mathbf J} = g(A)$ if and only if $X$ is strongly connected and  regular.
\end{lemma}

In this paper, we determine the smallest order circulant matrix representation for all
subfields of cyclotomic fields.
We also determine smallest order $0,1$-companion matrix representation for cyclotomic
fields in a restricted sense. We only consider those $0,1$-companion matrices
which have $\zeta_n$ as an eigenvalue.
We begin with a review of some facts related  with the representation of cyclotomic fields and their
subfields by circulant matrices in Section~\ref{sec:represent:cyclotomic}.
The results about representations of $p$-th cyclotomic field, $p$ a prime, by
$0,1$-circulant matrices are given in Section~\ref{sec:prime}. In section
\ref{sec7} we present results on the size of the smallest order $0,1$-companion matrix representations of cyclotomic fields.

\section{Representation of cyclotomic fields and their subfields}
\label{sec:represent:cyclotomic}

We start this section with a result about the irreducible factors of the
minimal polynomial of a
companion matrix. The proof of this  result can be easily obtained using the
theory of minimal polynomials and Lemmas~\ref{lem:aa3} and
\ref{lem:la}. Hence we omit the proof.

\begin{lemma} \label{lem:a9}
Let $f(x) \in {\mathbb F}[x]$ be a  monic separable polynomial with irreducible
factors $q_1(x), q_2(x), \ldots, q_k(x)$ in ${\mathbb F}[x]$.  Suppose   $A \in
{\mathbb M}_n({\mathbb F})$ commutes with the companion matrix ${\mathcal C}(f)$.
\begin{enumerate}\item  Then $A=g({\mathcal C}(f))$ for some $g(x)\in {\mathbb F}[x]$.
\item Let ${\alpha}_i$ be a root of $q_i(x)$ and let $\chi_{q_i,g}(x)$ be the minimal
polynomial of $g({\alpha}_i)$ over ${\mathbb F}$. Then the minimal polynomial of $A$ is
the maximal square-free factor of $\prod\limits_{i=1}^k \chi_{q_i,g}(x)$.

In particular, the number of irreducible factors of the minimal polynomial of
$g(A)$ in ${\mathbb F}[x]$ are at most the number of irreducible factors of $f(x)$
in ${\mathbb F}[x]$.
\end{enumerate}
\end{lemma}

Recall that for a fixed positive integer $n$,
$\;\deg(\Phi_n(x)) = \varphi(n)$, where $\varphi(n)$ denotes the
well known Euler-totient function.  The function $\varphi(n)$ also gives the  number of integers between $1$ and $n$ that are coprime to $n$.
We omit the proof of the next result as it directly follows from Lemma~\ref{lem:a9}, Theorem \ref{thm:a3}.\ref{thm:a3:1} and the fact that $x^n-1=\sum\limits_{d|n}\Phi_d(x)$.

\begin{theorem}\label{thm:a10}
Fix a positive integer $n$ and let $A=g(W_n )$, for some  $g(x)\in {\mathbb Q}[x]$ with $1\leq \deg(g(x))\leq n-1$.  Also, for each divisor $d$ of $n$, let
$\chi_{\Phi_d,g}(x)$ be the minimal polynomial of $g({\zeta}_d)$ over ${\mathbb Q}$. Then
\begin{enumerate}
\item \label{thm:a10:1}
$p_A(x)$, the minimal polynomial of $A$,  is the maximal square free factor of
$\prod\limits_{d|n}\chi_{\Phi_d,g}(x)$ and
$\deg(\chi_{\Phi_d,g}(x))$ divides $\deg(\Phi_d(x))$.
\item \label{thm:a10:2} the number of irreducible factors of $p_A(x)$
is at most the number of divisors of $n$.
\item \label{thm:a10:3}  ${\mathbb Q}\bigl(g(\zeta_d)\bigr) \cong {\mathbb Q}[A]/ \langle
\chi_{\Phi_d,g}(A) \rangle.$
\end{enumerate}
Furthermore, if $n$ is a prime,
say $p$,  then the number of irreducible factors of $p_A(x)$ is exactly two.
One of the factors is of degree $1$ and the degree of the other factor is a
divisor of $\varphi(p)=p-1$.
\end{theorem}

\begin{proof}
Proofs of Part~\ref{thm:a10:1},~\ref{thm:a10:2} and~\ref{thm:a10:3} are direct consequence of  Lemma~\ref{lem:a9} and Theorem~\ref{thm:a3}. For the last statement, note that $n$ is prime and hence it has exactly two factors, namely $1$ and $n$. Hence,  using Lemma~\ref{lem:a9}, it is sufficient to prove that $p_A(x)$ has at least two irreducible factors.

As $A = g(W_n)$, the eigenvalues of $A$ are $g({\zeta}_n^i)$ for $0 \le i \le n-1$. Now observe that for $i=0$, $g({\zeta}_n^0) = g(1)\in {\mathbb Q}$ as $g(x) \in {\mathbb Q}[x]$. Therefore, $(x-g(1))$ is an irreducible factor of $p_A(x)$. Also, for some $i, 1 \le i \le n-1$, if $g({\zeta}_n^i)\neq g(1)$  then the minimal polynomial of $g({\zeta}_n^i)$ is  another irreducible factor of $p_A(x)$. Hence, $p_A(x)$  has at least two irreducible factors.

Thus, we need to show that $g({\zeta}_n^i)= g(1)$ cannot hold true for
all $i\in\{1,2\ldots, n-1\}$. On the contrary, assume that
$g({\zeta}_n^i)= g(1)$ for all $i, 1 \le i \le n-1$. Define, $h(x) =
g(x) - g(1) \in {\mathbb Q}[x]$. Then $h(x)$ has $n$ distinct zeros,
${\zeta}_n^i, \; 0 \le i \le n-1$. This contradicts the definition of
$h(x)$ and the assumption that $\deg(g(x)) \le n-1$ as a polynomial
$f(x)$ has at most $\deg(f(x))$ zeros over ${\mathbb C}$.
\end{proof}

Theorem~\ref{thm:a10} establishes that, any field which is represented by an
$n\times n$ circulant matrix is a subfield of the $d$-th cyclotomic field
for some $d$ that divides $n$. It also describes
the correspondence between the set of all circulant matrices and the set
of all subfields of cyclotomic fields. It can also be concluded that the minimal
polynomial of every circulant matrix other than the scalar matrix
has at least two irreducible factors. One also concludes the next result and hence the proof is omitted.

\begin{cor}\label{cor:5.2}
Let $A$ be an $n\times n$ circulant matrix. Then $A$ represents a field ${\mathbb F}$ over ${\mathbb Q}$ if and only if ${\mathbb F}$ is a subfield of ${\mathbb Q}[\zeta_d]$, for some $d$ dividing $n$.
\end{cor}

The next result gives the smallest positive integer $d$ for which a field ${\mathbb L}$ over ${\mathbb Q}$ (as a subfield of ${\mathbb Q}[{\zeta}_d]$) is represented by a circulant matrix.

\begin{cor}\label{cor:5.3}
Let ${\mathbb L}$ be a finite extension of ${\mathbb Q}$. If $d$ is the smallest positive integer such that ${\mathbb L}$ is a subfield of $\mathbb{Q}(\zeta_d)$ then the smallest circulant matrix representation of ${\mathbb L}$ is of order $d$.
\end{cor}

\begin{proof}
Since ${\mathbb L}$ is a subfield of $\mathbb{Q}(\zeta_d)$ there exists $g(x)\in {\mathbb Q}[x]$ such that ${\mathbb L}={\mathbb Q}\bigl(g({\zeta}_d)\bigr)$.
Thus, by Theorem~\ref{thm:a10}.\ref{thm:a10:3},  ${\mathbb L}$ is
represented by
$A=g(W_d)$.

Now, assume that there exists a $d'\times d'$ circulant matrix
$B= h(W_{d'})$ that represents ${\mathbb L}$, for some $d'<d$. Let
$\chi_{\Phi_{d',h}}(x) \in {\mathbb Q}[x]$ be the minimal polynomial of $h({\zeta}_{d'})$. Then, using Theorem~\ref{thm:a10}.\ref{thm:a10:3}, one has ${\mathbb L}\cong
{\mathbb Q}[B]/\langle \chi_{\Phi_{d',h}}(B) \rangle \cong {\mathbb Q}[h(\zeta_{d'})]$.
That is,  ${\mathbb L}$ is a subfield of ${\mathbb Q}[\zeta_{d'}]$ as well. This contradicts
 our hypothesis that $d$ was the smallest positive integer such that
${\mathbb L}$ was a subfield of $\mathbb{Q}[\zeta_d]$. Hence, one has the required
result.
\end{proof}

Let $n,m $ and $a$ be positive integers with $n = 2^a m$ and $a \ge 1$.
Then it is known that ${\mathbb Q}[\zeta_{m}] \cong {\mathbb Q}[{\zeta}_n]$, whenever $m$ is odd and $a=1$.
Let $n=2^a\cdot m$ for some odd positive integer $m$. Then, using Corollaries~\ref{cor:5.2} and~\ref{cor:5.3}, the smallest representation of
${\mathbb Q}[{\zeta}_n]$ is of order $n$, whenever $a\ne 1$ and its order is
$\frac{n}{2}$, whenever $a=1$. The following theorem gives a $0,1$-symmetric
and circulant matrix
representation of order $n$ for  the largest subfield of ${\mathbb Q}[\zeta_n]$.

\begin{theorem}\label{thm:sym:represent}
Let ${\delta}_n = \zeta_n+\zeta_n^{-1}$. \begin{enumerate} \item
\label{thm:sym:represent:1} Then $\mathbb{Q}[{\delta}_n]$ has a symmetric $0,1$-circulant matrix representation of order $n$. \item
\label{thm:sym:represent:2} Let ${\mathbb L}$ be a subfield of $\mathbb{Q}[{\delta}_n]$. Then there exists a symmetric circulant matrix that represents ${\mathbb L}$.
\end{enumerate}
\end{theorem}

\begin{proof}
Proof of part~\ref{thm:sym:represent:1}: Let ${\mathbb K}=\mathbb{Q}[\zeta_n]$.
Then ${\mathbb Q}[{\delta}_n]$ is a subfield of ${\mathbb K}$ and $\zeta_n$ is a zero of the polynomial $x^2-{\delta}_n x+1 \in {\mathbb Q}[{\delta}_n][x]$. So, $[{\mathbb K}:{\mathbb Q}[{\delta}_n]]=2$.
As $A = W_n + W_n^{-1}$, $A$ is a symmetric, $0,1$-circulant matrix. Also ${\delta}_n$ is an eigenvalue of $W_n+W_n^{n-1}=W_n + W_n^{-1} = A$. Thus, by Theorem~\ref{thm:a10}, $A$ represents ${\mathbb Q}[{\delta}_n]$. This completes the proof of Part~\ref{thm:sym:represent:1}.

Proof of part~\ref{thm:sym:represent:2}: Since ${\mathbb L}$ is a subfield of
${\mathbb Q}[{\delta}_n]$, there exists a polynomial $g(x) \in {\mathbb Q}[x]$ such that ${\mathbb L} =
{\mathbb Q}[g({\delta}_n)]$. So ${\mathbb L}={\mathbb Q}[h({\zeta}_n)]$ where $h(x)=g(x+x^{n-1})$. Hence ${\mathbb L}$ can be
represented by $h(W_n)=g(W_n+W_n^{n-1})=g(A)$. Clearly, $h(W_n)$ is a symmetric, circulant matrix. Thus, the required result follows.
\end{proof}

We end this subsection, by a remark that gives an improvement on the order of the matrix $A$ of Theorem~\ref{thm:sym:represent}, whenever $n$ is an even integer. It is important to note that the representation given in the next remark, need not be a circulant representation. To do this, one uses a well known result that relates the eigenvalues of a cycle graph with the eigenvalues of a path graph.

\begin{rem}[Bapat, page~$27$, \cite{bapat}] \label{rem:even:cycle}
Let $n$ be an even positive integer and let $A$ denote the adjacency
matrix of the cycle $C_n$. Then the following results are well known:
\begin{enumerate}
\item $2$ and $-2$ are eigenvalues of  $A$.
\item Let $B_m$ be the adjacency matrix of the path $X_m$, on $m$ vertices.
 Then the set of eigenvalues of $B_{n/2 -
1}$ and the set of distinct eigenvalues of $A$, different from $2$ and $-2$,
are equal.
\end{enumerate}
Thus, the subfields of ${\mathbb Q}[{\delta}_n]$ can also be represented by
 $g(B_{n/2-1})$, for some polynomial $g(x) \in {\mathbb Q}[x]$.
\end{rem}

\subsection{Representations of prime order}\label{sec:prime}

Let $p$ be a prime and let ${\mathbb K}$ be a subfield of ${\mathbb Q}[{\zeta}_p]$. Then it is shown
in this subsection that there exists  a zero-one circulant matrix
$A$ of order $p$ such that the pair $(A,{\mathbf J})$ represents ${\mathbb K}$.
To do this, we define Cayley graphs/digraphs.

\begin{definition}\label{def:cay}
Let $G$ be a group and let $S$ be a non-empty subset of $G$ that does not
contain the identity element of $G$. Then the {\em Cayley digraph/graph}\index{Cayley digraph} associated
with the pair $(G,S)$, denoted ${\mbox{Cay}}(G,S)$, has the set $G$ as its
vertex set and for any two vertices $x,y \in G, \; (x,y)$ is an edge if $x
y^{-1} \in S$.
\end{definition}

Observe that ${\mbox{Cay}}(G,S)$ is a graph if and only if $S$ is closed with
respect to inverse ($S = S^{-1} = \{ s^{-1} : s \in S \}$). Also, the graph is
$k$-regular if $S$ has $k$ elements. The set $S$ is called the connection set of
the graph and it can be easily verified that the graph ${\mbox{Cay}}(G,S)$ is
connected if and only if $G = \langle S \rangle$. We also recall that a  digraph
is called a {\em circulant digraph}\index{circulant digraph} if its adjacency matrix is a circulant
matrix. The next lemma, due to Biggs,  states that every circulant digraph can
be obtained as a Cayley digraph.

\begin{lemma}[Biggs~\cite{biggs}]\label{lem:cgroup}
Consider ${\mathbb Z}_n$ as a cyclic group of order $n$. Then every Cayley
digraph ${\mbox{Cay}}({\mathbb Z}_n,S)$ is a circulant digraph. Conversely,
every circulant digraph on $n$ vertices is ${\mbox{Cay}}({\mathbb Z}_n,S)$, for some non-empty subset $S$ of ${\mathbb Z}_n$.
\end{lemma}

We now  state a result due to Turner~\cite{turner} that relates the isomorphism of two circulant graphs of
prime order with their eigenvalues.

\begin{lemma}[Turner~\cite{turner}] \label{lem:turner}
Let $X_1$ and $X_2$ be  two circulant graphs of prime order. Then they are
isomorphic if and only if they have the same set of eigenvalues. Or
equivalently,  their connection sets are equivalent.
\end{lemma}

Before proving  a couple of results, we recall the following facts. These facts
are not stated in the present form but they can be obtained from
the results stated on  Pages~$554,577$ of Dummit $\&$ Foote~\cite{D:F}.

\begin{fact}[Dummit $\&$ Foote, Pages~$554, 577$~\cite{D:F}] \label{fact:unique}
Let $p$ be a prime. Then \begin{enumerate} \item \label{fact:unique:1} the
polynomial $f(x) = 1 + x + \cdots + x^{p-1}$ is irreducible over ${\mathbb Q}$.
\item \label{fact:unique:2} the Galois group of ${\mathbb Q}[{\zeta}_p] $ over ${\mathbb Q}$ is
isomorphic to ${\mathbb Z}_p^*$, a cyclic group of order $p-1$. \item
\label{fact:unique:3}  for each divisor $d$ of  $p-1$, ${\mathbb Z}_p^*$ has a
unique subgroup of order $d$ and there exists a unique subfield of ${\mathbb Q}[{\zeta}_p]$ whose degree of extension over ${\mathbb Q}$ is $d$.
\end{enumerate}
\end{fact}

\begin{lemma}\label{lem:a13}
Let  $p$ be a  prime number and let $k$  be any factor of $p-1$.
Then the edge set of  $K_p=({\mathbb Z}_p,E)$, the complete graph on $p$ vertices, can be
partitioned into $k$  subsets $E_0,E_1, \ldots, E_{k-1}$ such
that the digraphs $X_i=({\mathbb Z}_p,E_i)$, for $0 \le i \le k-1$, are $r$-regular circulant digraphs, where $r=\frac{p-1}{k}$. Moreover, the digraphs $X_i$ and $X_j$, for $0 \le i < j \le k-1$, are isomorphic.
\end{lemma}

\begin{proof}
 Let ${\alpha}$ be a generator of ${\mathbb Z}_p^*$. Then $H=\langle \alpha^{k}\rangle = \{1,
{\alpha}^k, \ldots, {\alpha}^{k(r-1)}\}$ is
a subgroup of ${\mathbb Z}_p^*$ having $r$ elements and let $H_j=\alpha^j H$, for  $j = 0,
1, \ldots, k-1$, be the cosets of $H$ in ${\mathbb Z}_p^*$ with $H_0 = H$.
It is important to note that $H_j$, as a subset of ${\mathbb Z}_p$, generates ${\mathbb Z}_p$, for
each $j = 0,1,\ldots, k-1$. Let us now define
a digraph $X_j$ by having ${\mathbb Z}_p$ as its vertex set  and  any two vertices $x, y \in {\mathbb Z}_p$, $(x,y)$ is an edge in $X_j$ if and only if $y-x \in H_j$. Then  it is easy to verify that $X_j$ is an $r$-regular Cayley digraph, ${\mbox{Cay}}({\mathbb Z}_p,H_j)$. Also, observe that if we define   $A_j=\sum\limits_{h\in H_j}W_p^h$, for
$ 0 \le j \le k-1$, then $A_j$ is a $0,1$-circulant matrix and is the adjacency matrix of $X_j$.

Since the cosets $H_j$, for $0 \le j \le k-1$, are disjoint, one has
obtained $k$ disjoint digraphs that are $r$-regular and this completes the
proof of the first part.

We now need to show that the $k$ digraphs,
$X_j$, for $ 0 \le j \le k-1$, are mutually isomorphic. We
will do so by proving that the digraphs $X_0$ and $X_j$ are isomorphic,
for $1 \le j \le k-1$.

Let us define a map $\psi: \; V(X_0)\rightarrow V(X_j)$ by $\psi(s)={\alpha}^{j}s$
for each $s \in V(X_0)$. Then it can be easily verified that $\psi$ is one-one
and onto. Thus, we just need to show that $\psi\bigl( (x,y)
\bigr) $ is an edge in $X_j$ if and only if  $(x,y) $ is an edge in $X_0$.
Or equivalently, we need to show that
$\psi(y) - \psi(x) \in H_j$ if and only if  $y-x \in H_0=H$. And this holds true
as $$y-x \in H \Leftrightarrow {\alpha}^j(y-x)\in H_j \Leftrightarrow ({\alpha}^j y-{\alpha}^j x)\in H_j \Leftrightarrow \psi(y)-\psi(x)\in H_j.$$ This completes the proof of the lemma.
\end{proof}

Before coming to the main result of this section, we have the following remark.
\begin{rem}\label{rem:hj} Let $p$ be a prime and let the cyclic group ${\mathbb Z}_p^* = \langle {\alpha} \rangle$ and its cosets $H_j = {\alpha}^j H$, for $0 \le j \le k-1$, be defined as in Lemma~\ref{lem:a13}. Then
\begin{enumerate}
\item \label{rem:hj:1}  using  Fact~\ref{fact:unique}.\ref{fact:unique:3},  the Cayley digraphs, $X_0, X_1, \ldots, X_{k-1}$,  constructed in the proof of Lemma~\ref{lem:a13} are unique.
\item \label{rem:hj:2} for a fixed $j$, $0 \le j \le k$, we observe the following.
\begin{enumerate}
\item \label{rem:hj:2:1} For each $h \in H, h H_j = H_j$. That is, for each $t, 0 \le t \le r-1$, ${\alpha}^{t k} H_j = H_j$. That is, ${\alpha}^{j} H = {\alpha}^{j + t k} H$, for all $t, 0 \le t \le r-1$.
\item \label{rem:hj:2:2} Let ${\zeta}_p$ be a primitive $p$-th root of unity. Then, for each  $t, 0 \le t \le r-1$, $$\sum\limits_{h \in H} \bigl({\zeta}_p^{{\alpha}^j}\bigr)^h = \sum\limits_{s \in H_j} {\zeta}_p^s = \sum\limits_{h \in H} \bigl({\zeta}_p^{{\alpha}^{j+ t k}}\bigr)^h.$$
\end{enumerate}
\end{enumerate}
\end{rem}
We are now ready to state and prove the main result of this section.

\begin{theorem} \label{thm:a14}
Let $p$ be a prime and let ${\mathbb L}$ be a subfield  of  ${\mathbb Q}[\zeta_p]$. Then
there exists a circulant digraph on $p$ vertices whose adjacency matrix
represents ${\mathbb L}$.
\end{theorem}

\begin{proof}
Let $ r =[{\mathbb L}: \mathbb{Q}]$. Then $r$ divides $p-1 = [\mathbb{Q}[\zeta_p]:
 \mathbb{Q}]$, as ${\mathbb Q} \subset {\mathbb L} \subset \mathbb{Q}[\zeta_p]$.  Let
$k=\frac{p-1}{r}$.   We now claim  the existence of a $0,1$-circulant matrix
$A$ of order $p$ whose minimal polynomial has an irreducible factor of degree
$r$.

As $k$ divides $p-1$,  Lemma~\ref{lem:a13}, gives us a collection, say $X_0, X_1,
\ldots, X_{k-1}$,  of $r$-regular circulant digraphs on $p$ vertices that are
mutually isomorphic. Let $A$ be the adjacency matrix of $X_0$. Then, using the
definition of $X_0$, its adjacency matrix  $A=\sum\limits_{h\in H} W_p^h$. Thus, $A$ is
a circulant matrix and hence diagonalizable. Thus, we just need to find the
eigenvalues of $A$ to get the minimal polynomial of $A$.  By  definition,  the eigenvalues of $A$ are $\lambda_i=
\sum\limits_{h\in H} ({\zeta}_p^i)^{h}$, for $0 \le i \le p-1$. Observe
that $|{\lambda}_i| \le r$ and ${\lambda}_i = r$ if and only if $i = 0$. Fix an $i \in \{1, 2, \ldots, p-1\}$. Then, $i \in H_j$, for some coset $H_j$, $0 \le j \le k-1$, of ${\mathbb Z}_p^*$. Therefore, using Remark~\ref{rem:hj}.\ref{rem:hj:2}, we see that
${\lambda}_{{\alpha}^j} = {\lambda}_{{\alpha}^{j+k}} = \cdots = {\lambda}_{{\alpha}^{j+ (r-1) k}}$, for each $j \in \{0, 1, 2, \ldots, k-1\}$. That is, for each fixed $j \in \{0, 1, \ldots, k-1\}$ and  $s, t \in H_j$, ${\lambda}_s = {\lambda}_t$.
Thus,  $A$ has exactly $k$ distinct eigenvalues other than  the eigenvalue $r$.
Also, note that $A$ is  a circulant matrix of order $p$, a prime.
Therefore, by Theorem~\ref{thm:a10}.\ref{thm:a10:2}, the minimal
polynomial of $A$ factors into two distinct irreducible factors. One of
the factor is  $x-r$, corresponding to the simple eigenvalue $r$ of $A$ and the
other must contain all the distinct eigenvalues of $A$, different from $r$.
Hence, the minimal polynomial of $A$ equals $(x-r) \prod\limits_{i=1}^{k}
(x - {\lambda}_i) = (x-k) q(x) \in {\mathbb Q}[x].$

As $\deg(q(x)) = k$, the $0,1$-circulant matrix $A$ represents a subfield, say
${\mathbb K}$, of ${\mathbb Q}[{\zeta}_p]$ such that $[{\mathbb K}: {\mathbb Q}[{\zeta}_p] ] = k.$ Thus, the proof of the
claim is complete.

Now, using Fact~\ref{fact:unique}.\ref{fact:unique:3}, the subfield ${\mathbb K}$ is
indeed the subfield ${\mathbb L}$.
\end{proof}

We have seen that if $p$ is a prime then ${\mathbb Q}[{\zeta}_p]$ has a unique subfield
for each divisor $d$ of $p-1$. But all the real subfields of ${\mathbb Q}[{\zeta}_p]$
are also subfields of $\mathbb{Q}(\zeta_p+\zeta_p^{-1})$. This observation
leads to the last result of this section.

\begin{cor}
Let $p$ be a prime number. Then every real subfield of
$\mathbb{Q}[\zeta_p]$ has a symmetric $0,1$-circulant matrix
representation of order $p$.
\end{cor}

Let $p$ be a prime and consider the digraph $X_0$ in the proof of
Theorem~\ref{thm:a14}. Since $p$ is a prime, it can be easily verified that
$X_0$ is a strongly connected regular digraph.   Hence, using
Lemma~\ref{lem:hoff}, one immediately obtains  the following result and hence
the proof is omitted.

\begin{cor} \label{a15}
Let  ${\mathbb L}$ be  a subfield  of $\mathbb{Q}[\zeta_p]$. Then there exists a
$0,1$-circulant matrix $A$ of order $p$ such that $(A,{\mathbf J})$ represents
${\mathbb L}$.
\end{cor}

\section{Smallest $0,1$- Companion Matrices Whose Minimal Polynomial is Divisible
by $\Phi_n(x)$}
\label{sec7}

In this section, for a fixed positive integer $n$, our objective is
to find a $0,1$-companion matrix of the smallest order that
represents ${\mathbb Q}[{\zeta}_n]$. Let $\alpha$ denote the generic element such
that ${\mathbb Q}[\alpha]= {\mathbb Q}[\zeta_n]$. Using Corollary~\ref{c1}, this is
equivalent to finding the smallest $0,1$-companion matrix whose
minimal polynomial is divisible by the minimal polynomial of
$\alpha$. As there are infinitely many choices for  $\alpha$, we
restrict ourselves to $\alpha={\zeta}_n$. Hence,  we search for a
polynomial $f(x) \in {\mathbb Z}[x]$ of least degree such that $\Phi_n(x)$
divides $f(x)$ and ${\mathcal C}(f)$, the companion matrix of $f(x)$, is a
matrix with entries $0$ and $1$.  A similar study was made by  Filaseta $\&$ Schinzel~\cite{M:A} and Steinberger~\cite{J:S}, where they looked at polynomials with integer coefficients that are divisible by $\Phi_n(x)$.

Let $f(x) = x^n- a_{n-1}x^{n-1}-\dots -a_1x-a_0 \in {\mathbb Z}[x]$. Then ${\mathcal C}(f)$ is a
$0,1$-matrix if and only if for each $i, 0 \le i \le n-1, \; a_i\in \{0,1\}$.
Since $gcd(x^k,\Phi_n(x)) = 1$ for all $k\geq 1$, without loss of generality, we
can assume that  $a_0=1$. By definition, $\Phi_n(x)$ divides $g(x) = x^n-1$.
Hence ${\mathcal C}(g)$ is a $0,1$-matrix of order $n$ that represents ${\mathbb Q}[\zeta_n]$.
In order to determine whether there exists a matrix of smaller order, we define
a set ${\mathcal A}_n$ as \begin{eqnarray}
{\mathcal A}_n = \{ f(x) \in {\mathbb Z}[x] & : & \Phi_n(x) \mid f(x), f(x) \equiv x^m - a_{m-1}
x^{m-1} - \cdots a_1 x-1 \nonumber \\ && \hspace{.5in}  m < n, a_i \in \{0, 1\}
{\mbox{ for }} 1 \le i < m \} \label{defn:an}
\end{eqnarray}
and try to find the polynomial of least degree in ${\mathcal A}_n$.

Let $f(x) \in {\mathcal A}_n$. Then $f(x)$ has at least three terms as $m < n$. Hence,
$f(1)\ne 0$. Now, let $p$ be a prime. Then $\deg(f(x)) > \deg(\Phi_p(x)) =p-1$.
Thus, ${\mathcal A}_p $ is an empty set.  This is stated as our next result.

\begin{lemma}\label{lem:prime}
Let $p$ be a prime number. Then ${\mathcal A}_p$ is an empty set.
\end{lemma}

\begin{rem}\label{rem:Ap}
Let $p$ be a prime. Then, starting with the field ${\mathbb Q}[\zeta_p]$, a $0,1$-matrix representing ${\mathbb Q}[\zeta_p]$ of least order is $W_p$, the companion matrix of $x^p - 1$. But, it can be easily seen that if there exists a $0,1$-matrix $A \in {\mathbb M}_{\ell}({\mathbb C})$ representing ${\mathbb Q}[{\alpha}] \cong {\mathbb Q}[\zeta_p]$ then $\ell \ge p-1$. Thus, it may be possible to get a $0,1$-matrix $A \in {\mathbb M}_{p-1}({\mathbb C})$ such that $A$ represents  ${\mathbb Q}[{\alpha}] \cong {\mathbb Q}[\zeta_p]$.
\end{rem}

We now state a well known result about cyclotomic polynomials which enables us  to consider only  square-free positive integers $n$, where recall that  a positive integer $n$  is said to be square free if the decomposition of $n$ into primes
does not have any repeated factors.

\begin{lemma} [Prasolov, Page:93~\cite{V:P}] \label{lem:cyclo}
Let p be a prime number and let $n$ be a positive integer. Then
$$\Phi_{pn}(x)=
\begin{cases} \Phi_n(x^p), & \mbox{ if } p\mid n,\\
\frac{\Phi_n(x^p)}{\Phi_n(x)}, & \mbox{ if } p\nmid n.
\end{cases}$$
In particular, if   $n=p_1^{a_1} \cdots p_k^{a_k}$ is a prime
factorization of $n$ into distinct primes $p_1, p_2, \ldots, p_k$ and if
$n_0 = p_1 p_2 \cdots p_k$ then $\Phi_n(x)=\Phi_{n_0}(x^{n/n_0})$.
\end{lemma}

Steinberger~\cite{J:S} pointed out that
the problem of finding polynomials divisible by $\Phi_n(x)$ is
equivalent to finding polynomials divisible by $\Phi_{n_0}(x),$
where $n_0$ is the maximum square-free factor of $n$. Following is
a similar assertion in the current context. We give the proof for the sake of
completeness.

\begin{lemma}\label{lem:squarefree}
Let $n = p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ be a factorization of $n$
into distinct primes $p_1, p_2, \ldots, p_k$ and let $n_0=p_1p_2 \cdots p_k$.
Then
$$\min\{\deg(f(x)): f(x) \in {\mathcal A}_n\}=\frac{n}{n_0}\min\{\deg(f(x)): f(x)
\in {\mathcal A}_{n_0}\}.$$
\end{lemma}

\begin{proof}
Let $f(x)\in {\mathcal A}_{n_0}$. Then by  Lemma~\ref{lem:cyclo},  $f(x^{n/n_0})\in
{\mathcal A}_n$.

Conversely, suppose $f(x)\in {\mathcal A}_n$. Then $\Phi_n(x)$
divides $f(x)$ and therefore using Gauss lemma on polynomials [see~Dummit $\&$ Foote, Page~$304$ \cite{D:F}] and  Lemma~\ref{lem:cyclo}, $$f(x) = \Phi_n(x)
g(x) = \Phi_{n_0}(x^{n/n_0}) g(x) {\mbox{ for some }} g(x) \in {\mathbb Z}[x].$$ We now
group the terms of
$g(x)$ such that $g(x)=\sum\limits_{i=0}^{\frac{n}{n_0}-1}g_i(x^{n/n_0})
x^i$, where $g_i(x^{n/n_0})$ is a polynomial in $x^{n/n_0}$ (collect the
terms containing the exponents that are equivalent to $i \pmod {n/n_0}$).
Therefore,
$$f(x) = \sum_{i=0}^{\frac{n}{n_0}-1} \Phi_{n_0}(x^{n/n_0}) g_i(x^{n/n_0})
x^i = \sum_{i=0}^{\frac{n}{n_0}-1} f_i(x) x^i \; {\mbox{(say)}}.$$ That
is, the polynomials $f_i(x)$, for $0 \le i \le n/n_0 -1$, are divisible by
$\Phi_n(x) = \Phi_{n_0}(x^{n/n_0})$. Let the polynomial $f_j(x) x^j$ contain
$x^m$, the leading term of $f(x)$. Then $f(x) \in {\mathcal A}_n$ implies that
$$\Phi_{n_0}(x^{n/n_0}) g_j(x^{n/n_0})
x^j = x^m-x^{r_\ell}-x^{r_{\ell-1}}-\cdots-x^{r_1}, \; {\mbox{ with }} j
\le  r_1 < r_2 < \cdots < r_\ell.$$
 As $\gcd(\Phi_n(x),x^{r_1})=1$, the
polynomial $h(x)=x^{m-r_1}-x^{r_\ell-r_1}-\cdots-x^{r_2-r_1}-1$ is expressible
as a polynomial in $x^{n/n_0}$ and is divisible by $\Phi_{n_0}(x^{n/n_0})$.
Hence, one obtains a polynomial
$h_1(y)=y^{m'}-y^{r'_\ell}-\cdots-y^{r'_2}-1\in {\mathcal A}_{n_0}$ such that
$h(x)=h_1(x^{n/n_0})$ and $\frac{n}{n_0}\deg(h_1)\leq \deg(f)=m$. Thus, the
desired result follows.
\end{proof}

\begin{rem}\label{rem:squarefree}
Lemma~\ref{lem:squarefree} implies that in order to determine
$\min\{\deg(f(x)): f(x) \in {\mathcal A}_n\}$,  it is sufficient to solve the same problem
in ${\mathcal A}_{n_0}$, where $n_0$ equals the product of  all the prime factors
of $n$, a square-free positive integer. Henceforth, $n$  will be a square-free
positive integer.
\end{rem}

Lemma~\ref{lem:prime} together with Lemma~\ref{lem:squarefree} leads to
our next result.

\begin{lemma}\label{lem:pk}
Let $p$ be a prime and let $n=p^k$ for some $k\in{\mathbb Z}^+$. Then ${\mathcal A}_n$ is an empty set.
\end{lemma}

Thus, we will be interested only in those positive integers $n$ that has at
least two prime factors. In this case, it will be shown (see
Corollary~\ref{cor:pq} on Page~\pageref{cor:pq}) that the set ${\mathcal A}_n$ in
non-empty. To start with, note that  \begin{equation}
\varphi(n)\leq \min\{\deg(f(x)) : \; f(x) \in {\mathcal A}_n \} < n. \label{eqn:bound:an}
\end{equation}

Using a small observation, we improve the lower-bound in
Equation~(\ref{eqn:bound:an}) as follows.

\begin{lemma}\label{lem:lower}
Let $n$ be a positive integer. Then
\begin{equation}
\label{eqn:bound}\max \{\varphi(n), \lceil\frac{n}{2} \rceil \} <
\min\{\deg(f(x)) : \; f(x) \in
{\mathcal A}_n \} < n.\end{equation}
\end{lemma}

\begin{proof}
The lemma is immediate from Equation~(\ref{eqn:bound:an}) if we can show that
$\deg(f(x)) >\lceil\frac{n}{2} \rceil$. Suppose
$f(x)=x^m-x^{k_{\ell}}-x^{k_{\ell-1}}-\dots-x^{k_1}-1 \in {\mathcal A}_n$ with $0<
k_1<k_2< \cdots < k_\ell<m$. As $ \Phi_n(x)$ divides $f(x)$,  $f({\zeta}_n) = 0$.
Now, let if possible, $m\leq\frac{n}{2}$. Then $n - 2m \ge 0$ and using the fact
that ${\zeta}_n = {\zeta}_{2n}^2 = \cos(2 \pi/n) + i \sin(2 \pi/n)$ and ${\zeta}_{2n}^n =
-1$, we get
\begin{eqnarray*}
0 = f({\zeta}_n) &=&
-{\zeta}_n^m+\zeta_n^{k_\ell}+\zeta_n^{k_{\ell-1}}+\cdots+\zeta_n^{k_1}+1 =
-{\zeta}_{2n}^{2m}+\zeta_{2n}^{2 k_\ell} + \zeta_{2n}^{2 k_{\ell-1}} + \cdots +
\zeta_{2n}^{2 k_1}+1 \\
&=& 1+{\zeta}_{2n}^{n-2m+2k_{\ell}} +  {\zeta}_{2n}^{n-2m+2k_{\ell -1}} + \cdots +
{\zeta}_{2n}^{n-2m+2k_1}+ {\zeta}_{2n}^{n-2m} \\
&=& 1 + \sum_{j=0}^\ell \left( \cos\left(\frac{(n-2m + 2 k_j) \pi}{n}\right) + i
 \sin\left(\frac{(n-2m + 2 k_j) \pi}{n}\right)\right),
\end{eqnarray*}
 where $k_0=0$. Now using the choice of $k_j$'s, one gets $n - 2m + 2 k_j < n$
for each $j=0, 1,2, \ldots, \ell$. Hence, $\sum\limits_{j=0}^\ell
\sin(\frac{(n-2m + 2 k_j) \pi}{n})$ cannot be zero. Thus, we have arrived at a
contradiction and therefore the required result follows.
\end{proof}

This section is arranged as follows: the first subsection is devoted to
characterizing ${\mathcal A}_n$ in terms of certain subsets of roots of unity.  In
Subsection~\ref{sec2:2:1}, the exact value of $\min\{\deg(f(x)): f(x) \in
{\mathcal A}_n\}$ is obtained whenever $n$ has exactly two prime factors. The last
subsection, namely Subsection~\ref{sec2:2:2}, gives a bound on
$\min\{\deg(f(x)): f(x) \in {\mathcal A}_n\}$ whenever $n$ has $3$ or more prime factors.

\subsection{Characterization of ${\mathcal A}_n$ by One-Sums}
 \label{sec:An-Bn}

For a fixed positive integer $n$,  let $U_n = \{k : 1 \le k \le n, \gcd(k,n) =
1\}$ and  $R_n = \{{\zeta}_n^k : \; 0 \le k \le n-1
\}$. Then $| U_n| = \varphi(n)$, $R_n$ contains all the $n$-th roots of unity
and $\{{\zeta}_n^k : k \in U_n \}$ contains all the primitive $n$-th roots of unity.
 The following result can be found in Apostol~\cite{T:A}.

\begin{lemma}[Apostol~\cite{T:A}]\label{lem:mu}
Let $n$ be a positive integer. Then $\sum\limits_{k \in U_n}  {\zeta}_n^k =
\mu(n)$, where $$ \mu(n)=
\begin{cases} 0, & {\mbox{ if }} \; n \; {\mbox{ is not square free,}} \\
 1, &  {\mbox{ if }} \; n \; {\mbox{ has even number of prime factors,}}\\
-1, & {\mbox{ if }} \; n \; {\mbox{ has odd number of prime factors.}}
\end{cases}$$
\end{lemma}

Fix a positive integer $n$ and let $T \subset R_n$. Let $\sigma(T) =
\sum\limits_{{\alpha} \in T} {\alpha}$, denote the sum of all the elements of $T$. In
particular, recall that $\sigma (R_n)
= 0$. We now define a subset ${\mathcal B}_n $ of $R_n$ by
 \begin{equation}\label{defn:bn}
 {\mathcal B}_n = \{ T \subset R_n\setminus \{1\} : \; \sigma(T) = 1 \}.
\end{equation}
Then the next result gives a bijection
between the sets ${\mathcal A}_n$ and ${\mathcal B}_n$. This correspondence is useful in
constructing  members of ${\mathcal A}_n$.

\begin{theorem}\label{thm:An-Bn}
Let ${\mathcal A}_n$ and ${\mathcal B}_n$ be defined as above. Then there exists a bijection
 between  ${\mathcal A}_n$ and
${\mathcal B}_n$ such that $x^m-1-x^{k_1}-x^{k_2}-\cdots -x^{k_l}\in {\mathcal A}_n$ corresponds to
$\{\zeta^{n-m},\zeta^{n-m+k_1},\cdots ,\zeta^{n-m+k_l}\}\in {\mathcal B}_n$.
\end{theorem}

 \begin{proof}
Let $f(x)=x^m-1-x^{k_1}-x^{k_2}-\dots-x^{k_\ell} \in {\mathcal A}_n$ with $1\le
k_1<k_2< \cdots < k_\ell<m< n$. As $f(x) \in {\mathcal A}_n,  f({\zeta}_n) = 0$ and hence
$\zeta_n^m=1+\zeta_n^{k_1}+\zeta_n^{k_2}+\cdots+\zeta_n^{k_\ell}.$
Or equivalently, $T = \{\zeta_n^{n-m},\zeta_n^{n - m +
k_1},\ldots,\zeta_n^{n-m+k_\ell}\}\in {\mathcal B}_n \; {\mbox{ as }} \; \sigma(T) = 1.$

Conversely, let $T = \{{\zeta}_n^{k_0}, {\zeta}_n^{k_0+k_1},
{\zeta}_n^{k_0+k_2}, \ldots, {\zeta}_n^{k_0+k_\ell} \} \in {\mathcal B}_n$,
where  $1\le k_0< k_0+k_1<k_0+k_2< \cdots < k_0+k_\ell < n$. Then
$\sum\limits_{i=1}^{\ell} {\zeta}_n^{k_0+k_i} + {\zeta}_n^{k_0} = 1$, or equivalently,
 ${\zeta}_n^{n-k_0} = 1 + {\zeta}_n^{k_1} + \cdots + {\zeta}_n^{k_\ell}=0.$ Thus,  $f(x) =
x^{n-k_0} - x^{k_\ell}- x^{k_{\ell-1}}
\cdots - x^{k_1} - 1\in {\mathcal A}_n$ and  the required result
follows.
 \end{proof}

Theorem~\ref{thm:An-Bn} leads to the following important remark.

\begin{rem}\label{rem:min:degree}
Fix a positive integer $n$ and let $f(x)$ be a polynomial of least degree
in ${\mathcal A}_n$. Then $\deg(f(x)) = n- k_0$, where $k_0$ is obtained as
follows: ``for each element $T$ of ${\mathcal B}_n$, let $k_T$ be the least
positive integer such that ${\zeta}_n^{k_T} \in T.$ Then $k_0 = \max\{k_T : T
\in {\mathcal B}_n\}$".
\end{rem}

We now observe the following. Let $n$ be a positive integer
and let $d$ be the product
of an even number of distinct prime divisors of $n$. Also, let us write $
{\zeta}_n^{n/d} = {\zeta}_d.$ Then using Lemma~\ref{lem:mu},  $\{ {\zeta}_d^k : \; k \in
U_d\} \in {\mathcal B}_n$.
Observe that $U_d=\{1,1+k_1,1+k_2,\dots,1+k_\ell = d-1\}$ for some $k_i$'s
satisfying $1 \le k_1 < k_2 < \cdots < k_\ell = d-2$.
Therefore, ${\zeta}_d^{d-1}= {\zeta}_d^{-1} = 1 + {\zeta}_d^{k_1} + \cdots +
{\zeta}_d^{k_{l-1}}+
{\zeta}_d^{d-2}$ and hence
$$f(x) = x^{\frac{n}{d}(d-1)}
-x^{\frac{n}{d}(d-2)}-x^{\frac{n}{d}(k_{\ell-1})}-\dots -x^{\frac{n}{d}(k_1)}
-1 \in {\mathcal A}_n$$ is the corresponding polynomial.
Note that $\deg(f(x)) = n - \frac{n}{d}.$ This observation leads to the
first part of the following result. The second part directly follows from
the first part and hence the proof is omitted.

\begin{cor}\label{cor:pq}
Let $n = p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ be a factorization of $n$
into distinct primes and let $d$ be the product of an even number of
distinct prime divisors of $n$. If $U_d=\{1,1+k_1,1+k_2,\ldots,
1+k_\ell\}$ with $1 \le k_1 < k_2 < \cdots < k_\ell = d-2$, then
\begin{enumerate}
\item \label{cor:pq:1}$\Phi_n(x)$ divides the polynomial
$f(x) = x^{\frac{n}{d}(d-1)} -x^{\frac{n}{d}(d-2)}-x^{\frac{n}{d}(k_{\ell-1})}
-\dots -x^{\frac{n}{d}(k_1)}-1.$
\item \label{cor:pq:2} $\min\{ \deg(f(x)) : \; f(x) \in {\mathcal A}_n
\} \le n- \dfrac{n}{p_1 p_2}$, where $p_1$ and $p_2$ are the two smallest
prime divisors of $n$.
\end{enumerate}
\end{cor}

\subsection{Integers with Two Prime Factors} \label{sec2:2:1}

Let $n = p_1^{a_1} p_2^{a_2}$ be the factorization of $n$ as product of two
distinct primes $p_1$ and $p_2$. Then it is shown that the upper bound obtained
in Corollary~\ref{cor:pq} is indeed attained. That is,
$\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{n} \} = \frac{n}{p_1 p_2}\left\{p_1
p_2-1\right\}$.

Before proceeding further,
recall that for any positive integer $n$ and non-negative integer $m$, the
{\em Ramanujan's sum}\index{Ramanujan's sum}  is defined as $c_n(m) =
\sum\limits_{k \in U_n} ({\zeta}_n^k)^m.$ The next lemma is  a well known result
related with the Ramanujan's sum (for results related with Ramanujan's sum and coefficients of cyclotomic polynomials, see  Moree $\&$ Hommerson~\cite{mor}).

\begin{lemma}[Moree $\&$ Hommerson~\cite{mor}]\label{lem:ramanujan}
Fix  positive integers $m$ and $n$.  Then, for each divisor   $d$ of $n$,
$c_n(d)=\mu(\frac{n}{d})\frac{\varphi(n)}{\varphi(\frac{n}{d})}$.
Furthermore, $c_n(m)=c_n(d)$ whenever $\gcd(m,n)=d$.
\end{lemma}

Let $m < n$ be a positive integer. Then Ramanujan's sum is used to assign a number to  a polynomial
$g(x)=\sum\limits_{i=0}^m a_i x^i \in {\mathbb Q}[x]$ via the sum $\sum_{k\in U_n} g({\zeta}_n^k)$, denoted $S_g$. Then
\begin{equation}\label{eqn:sg}
S_g=\sum_{i=0}^m a_i c_n(i)=a_0\varphi(n)+
\sum\limits_{d \mid n} \left( \sum\limits_{i\in U_d}a_{ni/d}
\right)\mu(d)\frac{\varphi(n)}{\varphi(d)}. \end{equation}
Since, $\Phi_n(x)$ divides $f(x)$, $f({\zeta}_n^k) = 0$
for each $k \in U_n$. Thus, the next result is immediate  and hence the proof is
omitted.

\begin{lemma}\label{lem:Sf}
Let  $n$ be a positive integer. Then for each $f(x) \in {\mathcal A}_n$, $S_f = 0$.
\end{lemma}

Therefore, for any  $f(x) \in {\mathbb Q}[x]$, $S_f = 0$ gives a necessary
condition for $\Phi_n(x)$ to divide $f(x)$.
The next result is the main result of this subsection and it is shown that
$\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{p_1 p_2} \} =
p_1 p_2-1$, whenever $p_1$ and $p_2$ are distinct primes. This result together
with Lemma \ref{lem:squarefree}  implies that if $p_1$ and $p_2$ are distinct
primes and $n=p_1^{a_1} p_2^{a_2}$, for some positive integers $a_1$ and $a_2$,
then $\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{n} \} = \frac{n}{p_1 p_2}
\left\{p_1 p_2-1\right\}.$

\begin{theorem} \label{thm:p1p2}
Let $p_1$ and $p_2$ be two distinct primes. Then
$$\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{p_1 p_2} \} = p_1 p_2-1.$$
\end{theorem}

\begin{proof}
Let $n = p_1 p_2$. Then using a contrapositive argument, we will first show that
$\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{n} \} \ge
n-1$.  Let $f(x) \in {\mathcal A}_{n}$ be the
polynomial of least degree with $\deg(f(x)) < n-1$. Then
Theorem~\ref{thm:An-Bn} gives the existence of a subset $T = \{ {\zeta}_n^{k_1},
{\zeta}_n^{k_2}, \ldots, {\zeta}_n^{k_\ell}\}$ of ${\mathcal B}_n$ with $2 \le k_1 < k_2 < \cdots
< k_\ell$ that corresponds to $f(x)$. Define $g(x)
= \sum\limits_{i=1}^\ell x^{k_i} - 1. $ Then $g(x) \in {\mathbb Z}[x]$ and $g({\zeta}_n) =
0$. Thus, for all $k \in U_n, g({\zeta}_n^k)=0$ and $S_g = 0$.

Now, for each divisor $d$ of $n$, define $N_d = \{i \frac{n}{d}  : \; i\in U_d\} \cap \{k_1, k_2,
\ldots, k_\ell\}$. Then, using Equation~(\ref{eqn:sg}), one has $0 = S_g =
\sum\limits_{d \mid n} |N_d| \; \mu(d) \;\frac{\varphi(n)}{\varphi(d)} -
\varphi(n).$ Or equivalently, $\varphi(n) = \sum\limits_{d \mid n} |N_d| \;
\mu(d) \;\frac{\varphi(n)}{\varphi(d)}.$ Therefore, using
$\mu(p_i) = -1$ for $i=1,2$ and $\mu(p_1 p_2) = 1$, one gets
$$\frac{|N_n|}{\varphi(n)}=1+ \frac{|N_{p_1}|}{\varphi(p_1)}+
\frac{|N_{p_2}|}{\varphi(p_2)}$$
as $|N_1|=0$.
But observe that $|N_n| < \varphi(n)$ as $k_1 \ge 2$. That is, the left hand
side of the above identity is less than $1$ which contradicts the expression
that appears on the right hand side. Thus, our assumption is not valid and
hence $\min\{ \deg(f(x)) :
f(x) \in {\mathcal A}_{n} \} \ge n-1$.

As $n = p_1 p_2$, Corollary~\ref{cor:pq}.\ref{cor:pq:2} implies that
${\mathcal A}_n$ is non-empty and hence
$\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{n} \} \le n-1$. Thus, the
required result follows.
 \end{proof}

\subsection{Even Integers with $3$ or more Prime Factors}
\label{sec2:2:2}

In this subsection, we improve the bound
given in  Corollary~\ref{cor:pq} for all even positive integers that have more
than $2$ prime factors.

\begin{theorem}\label{thm:main}
Let $ p_1<p_2<\cdots <p_k$ be odd primes and let $n= 2 p_1 p_2\cdots p_k$. Then
$\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{n} \}\leq n-v$ where $$v=
\begin{cases}
\frac{n}{2} \cdot \frac{p_1+p_2}{p_1p_2}, & {\mbox{ if }}\; 2p_1>p_2,\\
\frac{3n}{2p_2},  & {\mbox{ if }}\; 2p_1<p_2<3p_1,\\
\frac{n}{2p_1}, & {\mbox{ if }}\; 3p_1<p_2.
\end{cases}$$
\end{theorem}

\begin{proof}
Let $f_0(x)$ be the polynomial of least degree in ${\mathcal A}_n$.  We will find numbers
$v_1$ and $v_2$, as lower bounds for $n-\deg(f_0(x))$ and take
$v=max\{v_1,v_2\}$. The value of $v_1=\frac{n}{2 p_1}$ is a direct application
of
Corollary~\ref{cor:pq}  as $2$ and $ p_1$ are the smallest two prime
divisors of $n$. Now, let us compute $v_2$.

To get the value of $v_2$, consider  $T = \{ {\zeta}_n^{n r/{2 p_1}} : \; r \in
U_{2 p_1} \} \cup \{ {\zeta}_n^{n \ell/{p_2}} : \; \ell \in U_{p_2} \}.$ Then using
Lemma~\ref{lem:mu},
$ \sum\limits_{z \in T} z = \sum\limits_{r \in U_{2 p_1}}  {\zeta}_n^{n r/{2
p_1}} + \sum\limits_{\ell \in U_{p_2}} {\zeta}_n^{n \ell/{p_2}} = 1 + (-1) =
0.$ Multiplying both sides by ${\zeta}_n^{n/{2 p_2}}$ and observing that
$\left({\zeta}_n^{n/{2 p_2}}\right)^{p_2} = -1 $ (as $p_2$ is an odd prime), one gets
\begin{eqnarray}
0 &=& \sum_{r \in U_{2 p_1}}  {\zeta}_n^{n r/(2 p_1) + n/(2 p_2)} + \sum_{\ell
\in U_{p_2}} {\zeta}_n^{n \ell/{p_2} + n/(2 p_2)} \nonumber \\
&=&  \sum_{r \in U_{2 p_1}} {\zeta}_n^{n r/(2 p_1) + n/(2 p_2)} + \sum_{\ell
\in U_{p_2},  2 \ell < p_2 -1 } {\zeta}_n^{n \ell/{p_2} + n/(2 p_2)}
+{\zeta}_n^{\frac{n}{2}}
\nonumber\\
&& \hspace{.5in} + \sum_{\ell \in U_{p_2},  2 \ell > p_2 - 1 } {\zeta}_n^{n
\ell/{p_2} + n/(2
p_2)}  \nonumber \\
&=&  \sum_{r \in U_{2 p_1}} {\zeta}_n^{n r/(2 p_1) + n/(2 p_2)} + \sum_{\ell
\in U_{p_2},  2 \ell < p_2 -1 } {\zeta}_n^{n \ell/{p_2} + n/(2 p_2)} -1\nonumber \\
&& \hspace{.5in} + \sum_{\ell \in U_{p_2},  2 \ell > p_2 - 1 } {\zeta}_n^{n
\ell/{p_2} + n/(2
p_2)}. \label{eqn:v2}
\end{eqnarray}
Thus, Equation~(\ref{eqn:v2}) implies
$$T' = \{ {\zeta}_n^{n r/(2 p_1) + n/(2 p_2)} : r \in U_{2 p_1}\} \cup \{ {\zeta}_n^{n
\ell/{p_2} + n/(2
p_2)} : \ell \in U_{p_2}\setminus \{(p_2-1)/2\} \}  \; \in {\mathcal B}_n.$$ That is,
$v_2 =\min\{ r : {\zeta}_n^r \in T' \} =
\begin{cases}
\frac{n}{2} \left( \frac{p_1+p_2}{p_1 p_2} \right), & {\mbox{ if }}\; 2p_1>p_2,
\\
\frac{3n}{2p_2}, & {\mbox{ if }}\; 2p_1<p_2.
\end{cases}$

Hence, using Remark~\ref{rem:min:degree}  the required result follows.

\end{proof}

\subsection{When $n$ is Even and $\Phi_n(x)$ is Flat}
In this subsection, the upper bound for $\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{n} \}$ is improved further  whenever $n$ is even  and the cyclotomic polynomial $\Phi_n(x)$ is {\em flat}\index{flat polynomial}. To do so, recall that the {\em height}\index{height of a polynomial} of a polynomial in ${\mathbb Z}[x]$ is the largest absolute value of its coefficients and a polynomial is said to be {\em flat} if its height is $1$.
 Let $A(n)$ be the height of $\Phi_n(x)$.
It is known that for all $n < 105$,
$\Phi_n(x)$ is flat and height of $\Phi_{105}(x)$ is $2$. In
fact, the height of $\Phi_n(x)$ is unbounded [see Emma Lehmer~\cite{lehmer}].

Let $k$ be the number of distinct odd
prime factors of $n$. For square-free $n$, this number $k$ is
called the  {\em order}\index{order of a cyclotomic polynomial} of the cyclotomic polynomial
$\Phi_n(x)$.  It is known that all cyclotomic polynomials of order
$1$ and order $2$ are flat.
Gennady Bachman~\cite{gb} gave the first infinite family of flat cyclotomic
polynomials of order three
 and this family was expanded by Kaplan~\cite{kap1}. In \cite{kap2}, Kaplan gave
some flat
polynomials of order four. It is unknown whether there
are any flat cyclotomic polynomials of order greater than four.

Fix a positive integer $k$ and let  $n= 2 p_1 p_2 \cdots p_k$, for distinct odd primes $p_1<p_2<\cdots < p_k$.
Let  $\Phi_n(x)=\prod\limits_{i=1}^{\varphi(n)}(x-x_i) =\sum\limits_{t=0}^{\varphi(n)}(-1)^t e_t x^{\varphi(n)-t}$, where  $x_1, x_2,\ldots, x_{\varphi(n)}$ are distinct roots of $\Phi_n(x)$ and
$e_t=\sum\limits_{1\leq i_1<i_2<\dots<i_t\leq \varphi(n)}\prod\limits_{j=1}^tx_{i_j}$. Then it is known that $e_t=e_{\varphi(n)-t}$, for $0\leq t\leq \varphi(n)$ and  $e_0=1$ [see Thangadurai~\cite{tan}]. Further,  by  Newton-Girard formulas
\begin{equation}\label{eqn:sum}
me_m=e_{m-1}c_n(1)-e_{m-2}c_n(2)+\dots + (-1)^me_1c_n(m-1)+(-1)^{m-1}c_n(m).
\end{equation}
where $c_n(m)$ is the Ramanujan's sum defined in Page~\pageref{lem:ramanujan}. In particular, using Lemma~\ref{lem:mu} $e_1=c_n(1)= \mu(n) = -1$.

Now let $k$ be an even integer. That is, $n$ is product of odd number of distinct primes.   Then for any positive integer $m<p_1$,
\begin{equation*}
c_n(m)=
\begin{cases}
- 1, &
{\mbox{ if }} \; m \; {\mbox{ is odd,}} \\
1, &
{\mbox{ if }} \; m \; {\mbox{ is even. }}
\end{cases}
\end{equation*}
Now, using Equation~(\ref{eqn:sum}) recursively, it is easy to show that $e_2=\cdots=e_{p_1-1}=0$ and $e_{p_1}=1$.

With these observations, we have
\begin{equation}\label{eq:Phi}
\Phi_n(x)= \begin{cases} x^{\varphi(n)}-x^{\varphi(n)-1}\pm \dots -x+1, &
{\mbox{ if }} \; k \; {\mbox{ is odd,}} \\
x^{\varphi(n)}+x^{\varphi(n)-1}-x^{\varphi(n)-p_1}\pm \dots -x^{p_1}+x+1, &
{\mbox{ if }} \; k \; {\mbox{ is even.}}
\end{cases}
\end{equation}

From now on, we consider only flat cyclotomic polynomials. Then $\Phi_n(x)=f_1(x)-f_2(x)$, for some $0,1$-polynomials $f_1(x)$ and $f_2(x)$. Observe that  the representation of $\Phi_n(x)$ as difference of two $0,1$-polynomials is unique. Also,
$ \Phi_n({\zeta}_n)=0$ implies that $  f_1({\zeta}_n)-f_2({\zeta}_n)=0$ and hence $ f_1({\zeta}_n)+{\zeta}_n^{n/2}\cdot f_2({\zeta}_n)=0.$  That is, $ \Phi_n(x)$ divides  $f_1(x)+x^{n/2}f_2(x)$.

Let $\Phi^T_n(x)= f_1(x)+x^{n/2}f_2(x)$. Then $\Phi^T_n(x)$ is a
$0,1$-polynomial and $\Phi_n^T({\zeta}_n) = 0$.
And from Equation~(\ref{eq:Phi}), we have
\begin{equation}
 \deg(\Phi^T_n(x))=\begin{cases} \phi(n)-1+\frac{n}{2},\;\; \mbox{whenever }  k \mbox{ is odd},\\
\phi(n)-p_1+\frac{n}{2},\;\; \mbox{whenever }    k \mbox{ is even.}
                    \end{cases}
\end{equation}

We now construct a polynomial
$\Phi_n^*(x)\in {\mathcal A}_n$ from $\Phi^T_n(x)$  as follows. Let the degree of $\Phi_n^T(x)$ be $D$. Consider the monomials
in $\Phi_n^T(x)$ having exponent strictly between $D-n/2$ and $n/2$. If $x^b$ is the monomial
with smallest exponent among these, then $\Phi_n^*(x)=x^{b+n/2}+x^b-\Phi_n^T(x)$.
Since $n$ is even, $\Phi_n(x)$ divides $x^{n/2} + 1$ and hence $\Phi_n(x)^* \in {\mathcal A}_n$. Also, the monomial $x^b$ comes from the polynomial $f_1(x)$ and therefore
\begin{equation}\label{eqn:Phi}
\deg(\Phi_n^*(x))= \begin{cases}
\frac{n}{2}+\varphi(n), & {\mbox{ if }} k  {\mbox{ is odd,}} \\
\frac{n}{2}+\varphi(n)-1, & {\mbox{ if }} k  {\mbox{ is even.}}
\end{cases}
\end{equation}

 Since $\Phi_n^*(x)\in {\mathcal A}_n$, using Equation~(\ref{eqn:Phi}), the following result follows and hence the proof is omitted.

\begin{lemma}\label{lem:flat}
Let $n= 2 p_1 p_2\cdots p_k$ be the factorization of  $n$ into
odd primes  $p_1<p_2<\cdots <p_k$. Suppose that the cyclotomic polynomial $\Phi_n(x)$ is flat. Then
$$\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{n} \}\leq
\begin{cases}
\frac{n}{2}+\varphi(n), & {\mbox{\; if }}  k {\mbox{ is odd}}, \\
\frac{n}{2}+\varphi(n)-1, & {\mbox{ if }} k {\mbox{ is even}}.\\
\end{cases}$$
\end{lemma}

\begin{rem}
In general, we are not able to give exact comparison between the bounds obtained in Theorem~\ref{thm:main} and the bound in Lemma~\ref{lem:flat}. But it can be checked that whenever $3 p_1 < p_2$ then the bound in Lemma~\ref{lem:flat} is better than the bound in Theorem~\ref{thm:main}.
\end{rem}

\section*{Conclusion}

In this paper, we have tried to study the representations of subfields of a cyclotomic field with the help of circulant  and $0,1$-companion matrices. In particular, the following results have been obtained.

\begin{enumerate}
\item  A subfield of a cyclotomic field is representable by some circulant
matrix and conversely every circulant matrix represents a
subfield of a cyclotomic field.

\item Every real subfield of ${\mathbb Q}[\zeta_n]$ is representable by
a polynomial in the adjacency matrix of $C_n$, the cyclic graph.
Consequently, every real subfield of ${\mathbb Q}[\zeta_n]$ has integer symmetric
circulant matrix representation.

\item  Let $p$ be a prime and let ${\mathbb K}$ be a subfield ${\mathbb Q}[{\zeta}_p]$. Then a $0,1$
circulant matrix $A$ of order $p$ is obtained such that $(A,{\mathbf J})$ represents
${\mathbb K}$.

\item Let $n= p^k$ for some prime $p$. Then the smallest $0,1$-companion matrix having ${\zeta}_n$ as an eigenvalue  is $W_n$, the companion matrix of $x^n - 1$.

\item Let $n=p_1^{a_1} p_2^{a_2}$ be the prime factorization of $n$ as product of distinct primes. Then $\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{n} \} = \frac{n}{p_1p_2}(p_1p_2-1)$.

\item Let $n$ be a positive integer having $3$ or
more prime factors. Then  $\min\{ \deg(f(x)) : f(x) \in {\mathcal A}_{n} \} \le \frac{n}{p_1p_2}(p_1p_2-1)$, where $p_1$ and $p_2$ are the smallest two distinct primes dividing $n$. Furthermore, if  $n$ is even then this  upper bound is improved in Theorem~\ref{thm:main} and Lemma~\ref{lem:flat}.

It will be nice to improve the bounds obtained in this paper. Also, it will be nice to get examples where the bounds are attained.
\end{enumerate}

redpajama_set_name: RedPajamaArXiv