text:
Home News UFC is about ...
UFC is about to get a huge boost with casual sports fans thanks to $1.5 billion deal with ESPN
The UFC is leaving Fox Sports and heading to ESPN as part of a five-year deal worth $1.5 billion that could potentially bring more eyes into mixed martial arts than ever before.
Vaughn Ridley/Getty Images
The UFC is moving from Fox Sports to ESPN in a five-year deal worth $1.5 billion.
The move will likely introduce a new audience to the world of mixed martial arts and will give ESPN plenty of opportunities to promote its new product.
Fights will appear on both ESPN and the network's over-the-top service ESPN+ starting in 2019.
ESPN and the UFC have agreed to a five-year deal worth $1.5 billion that will bring mixed martial arts to the Worldwide Leader in Sports and quite possibly its biggest audience ever.
The deal is set to begin in early 2019, and includes both the previously announced agreement to bring fights to ESPN's new over-the-top offering ESPN+ as well as live shows on the cable network. Thirty UFC Fight Night events will appear on ESPN platforms per year, with ten main cards on the network and the other 20 available on ESPN+. ESPN networks will also reportedly air the preliminary bouts of pay-per-view events.
While interest in UFC has declined somewhat of late – with some of the sport's biggest stars leaving for boxing, wrestling, or waiting out suspensions – the deal could serve as a nice kickstart to bring a new audience of committed fight fans to the ESPN+ service.
More importantly, with ESPN's 24-hour news cycle fueled by debate shows and Stephen A. Smith hot takes, there will be plenty of opportunities for the network to promote interest in their latest acquisition.
For the UFC, the deal is an opportunity to reach a much wider audience. Moving from Fox Sports, who decided to turn down an extension with the promotion in favor of buying the rights to WWE's "SmackDown Live," the UFC will now be a much more significant part of the lives of casual sports fans. Additionally, with the promotion's current star problem, the reliability of ESPN's check coming through every month eases some of the pressure to generate new stars that can sell PPV rates that reach the heights of Conor McGregor.
It feels like a win-win deal for both the network and the fight promotion, but we won't know how the deal is working out until broadcasts start in 2019.


text:
#if WINDOWS_PHONE
using System.Windows;
using System.Windows.Interactivity;
#else
using Windows.UI.Xaml;
#endif

namespace WpWinNl.Behaviors
{
  public class WidthPercentageBehavior : Behavior<FrameworkElement>
  {
    private FrameworkElement visualParent;

    protected override void OnAttached()
    {
      AssociatedObject.Loaded += AssociatedObjectLoaded;

      base.OnAttached();
    }

    private void AssociatedObjectLoaded(object sender, RoutedEventArgs e)
    {
      visualParent = AssociatedObject.GetVisualParent();
      if (visualParent != null)
      {
        visualParent.SizeChanged += VisualParentSizeChanged;
      }
    }

    private void VisualParentSizeChanged(object sender, SizeChangedEventArgs e)
    {
      ApplyNewWidthPercentage(WidthPercentage);
    }

    private void ApplyNewWidthPercentage(int newPercentage)
    {
      if (visualParent != null)
      {
        AssociatedObject.Width = (double) newPercentage/100*visualParent.ActualWidth;
      }
    }

    #region WidthPercentage

    /// <summary>
    /// WidthPercentage Property name
    /// </summary>
    public const string WidthPercentagePropertyName = "WidthPercentage";

    public int WidthPercentage
    {
      get { return (int)GetValue(WidthPercentageProperty); }
      set { SetValue(WidthPercentageProperty, value); }
    }

    /// <summary>
    /// WidthPercentage Property definition
    /// </summary>
    public static readonly DependencyProperty WidthPercentageProperty = DependencyProperty.Register(
        WidthPercentagePropertyName,
        typeof(int),
        typeof(WidthPercentageBehavior),
        new PropertyMetadata(100, WidthPercentageChanged));

    /// <summary>
    /// WidthPercentage property changed callback.
    /// </summary>
    /// <param name="d">The depency object (i.e. the behavior).</param>
    /// <param name="e">The property event args <see cref="DependencyPropertyChangedEventArgs"/> instance containing the event data.</param>
    public static void WidthPercentageChanged(DependencyObject d, DependencyPropertyChangedEventArgs e)
    {
      var behavior = d as WidthPercentageBehavior;
      var newValue = (int)e.NewValue;
      if (behavior != null)
      {
        behavior.ApplyNewWidthPercentage(newValue);
      }
    }

    #endregion
  }
}

redpajama_set_name: RedPajamaGithub


text:
Obama's Gun Speech Gag Order Is Retaliation Against Group That Sued The Gov't Last Month
By: Sean Davis
The Obama administration's latest anti-gun salvo isn't about reducing gun violence or stopping the export of dangerous weapons, it's about pure retaliation against a non-profit that sued the government in federal court last month.
Defense Distributed, a pro-Second Amendment non-profit organization that provides blueprints, plans, and machinery to fabricate or finish firearm components, has been at war with the State Department for nearly two years.
According to a complaint it filed in federal court last month, the organization began to make its data, compiled entirely from publicly available information, available for free on the Internet in December of 2012. Just a few months later, in May of 2013, Defense Distributed received a letter from the State Department alleging that the group was illegally exporting technical data. The federal government then demanded, in contravention of long-standing policy dating back to 1984, that Defense Distributed submit its proposed speech to the federal government for pre-approval.
You read that right: the Obama administration demanded that a pro-Second Amendment non-profit submit its proposed speech, consisting of publicly available information, to the federal government for approval prior to publishing.
For nearly two years, Defense Distributed tried to comply with the State Department's demands. The organization submitted its data for pre-approval. Then it waited. And waited. And waited. After asserting a lawless requirement that the government must pre-approve speech by Defense Distributed, the government refused to do anything but sit on its hands.
After nearly two years of trying in vain to comply with these new restrictions on free speech, Defense Distributed filed suit against the federal government on May 6, 2015. This is no rinky-dink lawsuit. According to court filings, the lead counsel for Defense Distributed is Alan Gura, the same attorney who successfully argued the landmark Heller case, in which the Supreme Court struck down as unconstitutional several gun bans in Washington, D.C. By all appearances, Defense Distributed is bringing out the big guns in its legal fight against the Obama administration's gun speech gag rule.
On June 3, just four weeks after Defense Distributed filed its complaint in federal court, the State Department suddenly decided to propose a new rule giving it the authority to pre-approve speech related to publicly available firearm plans. The State Department's play here is obvious: it hopes to promulgate a new rule making its previous anti-speech efforts superficially legal in order to short-circuit Defense Distributed's court case. If that were to happen, the non-profit would then have to file a new and separate suit alleging the unconstitutionality of the new rule.
Only in Washington, D.C. in 2015 can you blatantly violate the law, get called to court for it, and then unilaterally and retroactively change the law you violated in order to avoid accountability.
Even worse, the federal government issued its proposed regulation before it ever even responded to Defense Distributed's complaint. In fact, the government and all of the officials named in the suit have yet to file a single response to the complaint with the court. Why? Because they're hoping to run out the clock on the case. If the State Department can promulgate its lawless prior restraint on free speech before a court can rule on the merits of Defense Distributed's case, then the case is moot. And while a high-profile constitutional court case like this one can drag on for years, the State Department only needs to wait 60 days from the notice of its proposed rule before it can issue a final rule neutering Defense Distributed's case against it.
At issue are two definitions under a 1976 law governing the import and export of defense articles. Long story short, the Obama administration proposes to throw aside more than three decades of federal precedent, going all the way back to 1984, in order to target Defense Distributed.
First, the administration proposes to change the definition of "technical data" in such a way that mere public discussion of technical gun data–virtually identical to what you can find on any gunowner forum on the Internet–will be treated as a violation of federal weapons export laws. Second, the administration proposes to treat software itself as a "defense article," a move that would put open source computer code in the same category as nuclear warheads and intercontinental ballistic missiles.
The State Department's proposed June 3 rule even includes a bogus "public domain" exception that would make George Orwell proud. In one section, the proposed rule states that any technical data in the public domain–such as information that can be found on a publicly available website or at a public library–is totally kosher:
§ 120.11 Public domain.
(a) Except as set forth in paragraph (b) of this section, unclassified information and software are in the public domain, and are thus not technical data or software subject to the ITAR, when they have been made available to the public without restrictions upon their further
dissemination[.]
Problem solved, right? Not so fast.
In the very next section, the federal government informs us that information can only be considered in the "public domain" if the federal government, via unelected State Department bureaucrats, has issued a prior ruling declaring the information to be in the public domain. In other words, even if speech is obviously in the public domain, it doesn't count as being in the public domain unless the federal government says so:
(b) Technical data or software, whether or not developed with government funding, is not in the public domain if it has been made available to the public without authorization from: (1) The Directorate of Defense Trade Controls; (2) The Department of Defense's Office of Security Review; (3) The relevant U.S. government contracting entity with authority to allow the technical data or software to be made available to the public; or (4) Another U.S. government official with authority to allow the technical data or software to be made available to the public.
This is not how representative government is supposed to work. Our system was not designed so that one branch of the federal government could lawlessly harass its citizens and then turn around and unilaterally enact new laws by administrative fiat when citizens finally decided to take their grievances to court. The system was not designed to give unelected government bureaucrats the right to promulgate new laws on the fly in order to avoid court sanctions. Nowhere in the Constitution is there a grant of authority to the federal government to demand pre-approval of political speech prior to its public dissemination.
If Congress wishes to pass a law banning fabrication of firearms by non-federal firearms licensees, and if the president wishes to sign that law, that is their prerogative. It is not, however, the prerogative of unaccountable bureaucrats to rewrite laws just so they can get around pesky court cases alleging government violation of those very laws.
The State Department did not respond to repeated requests for comment for this article. A court hearing on Defense Distributed's request for a preliminary injunction against the government is set for July 6 in Austin, TX.
Sean Davis is CEO and co-founder of The Federalist. He previously worked as an economic policy adviser to Gov. Rick Perry, as CFO of Daily Caller, and as chief investigator for Sen. Tom Coburn. He was named by The Hill as one of the top congressional staffers under the age of 35 for his role in spearheading the enactment of the law that created USASpending.gov. Sean received a BBA in finance from Texas Tech University and an MBA in finance and entrepreneurial management from the Wharton School. He can be reached via e-mail at sean@thefederalist.com.
Sean Davis
Visit on Twitter @seanmdav
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<# } ); #> wpDiscuz

redpajama_set_name: RedPajamaCommonCrawl


text:
Ruby Crowned Kinglet Photo: Ronald Newhouse
End of the Year Message from Jon Hayes
2019, What a Year!
Birds in This Story
Sialia currucoides
http://www.audubon.org/sites/default/files/MOUBLU_1.songnum1_UTkc.mp3?site=nm&ni...
Nucifraga columbiana
http://www.audubon.org/sites/default/files/CLANUT_1.callsnum1_UTkc_1.mp3?site=nm...
Coccyzus americanus
http://www.audubon.org/sites/default/files/YEBICU_1.c-c-c-kowlpoutbursts3example...
Empidonax traillii
http://www.audubon.org/sites/default/files/WILFLY_1.songnum1_NYle_1.mp3?site=nm&...
As I look back on 2019 I am amazed at the incredible achievements our staff, volunteers, and supporters accomplished for birds and people in just one year. We moved the needle in real and impactful ways and it was thanks to each and every one of you who believe birds and the places they need are worth saving.
As a way of saying thank you, we've picked the 5 things we're most proud of this year and we'd like to share them with you, our Audubon New Mexico family, without whom none of this work would be possible.
We secured a historic water lease – Just before the Thanksgiving holiday we learned that we would be receiving a long-awaited permit to lease 40 acre feet of water on the Rio Chama. This is the first time any private entity in the state has been granted a permit for in-stream flow as a beneficial use of river water. This precedent setting policy innovation will provide a new strategy for water managers to ensure that the Rio Grande and other rivers in New Mexico continue to function in a changing climate, providing habitat for some of our most at risk species like Southwestern Willow Flycatcher and Western Yellow-billed Cuckoo.
We held the grand opening of the David Jay Henderson Pavilion – We were honored to have Senator Tom Udall join us to celebrate our former director's legacy with the opening of this new building in his honor. This multi-purpose event facility will enable us to double our summer camp offerings and other on-site education opportunities while also providing a space for community members to gather and hold events. In fact Governor Michelle Lujan Grisham has already made use of the facility for the signing of a shared stewardship agreement with the US Forest Service.
We released Audubon's Survival by Degrees report – Our newest report on the impact of climate change on North American birds shows that 389 species of birds (including Clark's Nutcracker, Mountain Bluebird, and Sandhill Crane) are vulnerable to extinction by the end of the century if we keep releasing carbon into the atmosphere at our current rates. But this sobering report also provides hope as the number of vulnerable species could be reduced by over 75% if we take steps to stop carbon emissions now. This was front page news in the Santa Fe New Mexican and it provided us the opportunity to brief our congressional delegation on the impacts of climate change and what we can do to address this crisis.
We reached over 5,500 future conservationists – Our education staff does incredible things every day, working tirelessly to educate the future stewards of our planet about the wonders of the natural world and our responsibility to protect it. This year alone, they have had 4,800 direct points of contact with students, brought over 2,000 students (largely from Title 1 schools) to the Randall Davey Audubon Center & Sanctuary to experience nature firsthand, and trained 50 teachers from Valencia, Bernalillo, Santa Fe, and Rio Arriba counties in how to deliver hands-on, outdoor science learning. We consider this an investment in the future of the planet and the places we love.
We opened our Rio Grande office to accommodate our growing team – This year we welcomed a number of new staff to our crew of incredible conservationists who are helping us turn your support into conservation outcomes on-the-ground and at the Capital. In order to have the greatest impact possible; Audubon New Mexico has, for the first time ever, opened an office in Albuquerque to strengthen our connection to that community and the river that runs through it.
With such a momentous year behind us, it's hard to believe we can top it in 2020, but rest assured that the team of environmental leaders we've assembled at Audubon New Mexico are up to the task. This is an exciting time to work in conservation in New Mexico as we have the opportunity to make big changes to the way we manage our resources and landscapes.
Working together, we can prevent the worst effects of climate change, we can protect the quality of our air, soils and water, and we can ensure that our rivers are resilient in the face of an uncertain future. We can and will do these things because that is how we guarantee New Mexico remains a place where birds and people prosper and thrive for generations to come.
Thank you for your continued support and Happy Holidays!!


In Berlin, project spaces come and go faster than postgrad American expats. Yet every once in a while, a space seems to have some staying power, and a decidedly different mission than the 'hey-let's-get-a-bunch-of-twenty-something-friends- together-and-put-experimental-art-on-the- peeling-walls-to-rock / techno / electroclash-music' thing so common in this city over the past 25 years.
Savvy Contemporary is one of those special spaces. Founded a full five years ago, this 'laboratory of form-ideas' (Savvy's self-appointed subtitle) takes a unique position: in a city whose art production so often grapples with an Eastern / Western Europe narrative, Savvy's exhibitions and ambitions go global. Each show represents a 'trialogue' between the curators, a North American or European artist, and another artist from Africa, South America, Asia or Australia.
Past exhibitions, usually bimonthly, were often this simple and straightforward – in 2013, GhostBusters II (Haunted by Heroes) showed artists Kara Lynch and Délio Jasse presented by curators Nadine Siegert and Storm Janse Von Rensburg. More recent projects, however, have expanded the triad into a United Nations-like conversation. For most of 2014, for example, the exhibition and research project Giving Contours to Shadows (partially in cooperation with the Neuer Berliner Kunstverein) mounted an impressively international group exhibition as well as a series of talks and symposia on how the Western canon has dominated historical discussions. The show took Africa as a point of departure, juxtaposing artists like Nigeria-born Otobong Nkanga with Western positions such as that of French-born Alexandre Singh. A series of satellite projects has even been running in African venues, such as the Nairobi Arts Trust / Centre for Contemporary Art of East Africa and the Nest, both in Nairobi, all year.
Savvy's global approach makes perfect sense considering its founder's origins: Bonaventure Ndikung (who contributed an article to the October issue of ArtReview) comes from Cameroon. A sharp dresser (snazzy suits, hats and splashy accessories set him far apart from the usual Berlin hipster) as well as a sharp thinker, Ndikung works as a biochemist by day, but oversees an increasingly multipronged endeavour at the Savvy mothership – assisted by coartistic director Elena Agudio and a troupe of young, friendly volunteers hailing from countries as broadly strewn as the Dominican Republic and Russia. In summer 2013, the organisation moved from a small storefront on Richardstrasse in Neukölln's homey Rixdorf neighbourhood into a two-storey, 400sqm monolithic red-brick power station up the street – apparently the result of Ndikung's tenacity and negotiation skills (the space hadn't been used in decades, and part of it may still become luxury lofts, but hopefully not too soon).
About the same time, Savvy won a €30,000 grant from the Berlin Culture Ministry – additional funding comes from donations, a friends' association, project-specific grants and occasionally renting out the venue. The move has allowed both physical and conceptual expansion: Giving Contours to Shadows, in part funded by the German Federal Cultural Foundation, would have been much more difficult in the smaller old space; an ambitious performance-art programme (curated by Goldsmiths alumna Chiara Cartuccia) now has more room to breathe and evolve. Savvy might still have to close for a week or two in winter when the interior spaces are simply too cold for comfort, but its lightflooded main exhibition space and reading room / library / project archive – which doubles as a stage for discussions on topics like postcolonialism, book readings and film screenings – still attract viewers and researchers. The former storefront space has morphed into a coworking space called Savvy Desk; there's even a residency for artists and scholars.
It's humbling to consider how much goes on in this 'laboratory' – whose nonprofit status perhaps gives it a certain advantage in mounting projects and shows that official, more institutional (read: bureaucratic) channels for non-Western art in Berlin don't always have the agility to address. What's more astonishing is just how much heart Savvy has. Stop by on a weekend and you'll catch groups of volunteers working and chatting; at openings, bands or DJs perform in the basement and the crowds spill into the back courtyard. In an increasingly transient city that often seems to have become a caricature of its own coolness, Savvy stays smart, independent, postcool. Let's hope for at least another five years.


Most applications for graduate study are made through the University's Graduate Admissions Office. You will have the opportunity to name Fitzwilliam as your College of choice.
If you are applying through a Department or Faculty, such as the Judge Business School or Faculty of Education, application procedures may be different. Whatever your route of application, please keep the College Graduate Office informed of your progress.
The University Graduate Admissions Office will send all their applicants information on how to log in to the self-service account. This is where you will track the progress of your University application and make any changes to your contact details.
The College needs to use the information stored in this account to contact you. Please ensure that the contact information is kept up to date. If you do not have access to this service, please inform the College Graduate Office of any changes.
A guide to the Cambridge College system for graduate students is available on the University website.
Colleges only consider applications after a conditional offer has been made by the University. The Fitzwilliam Graduate Officer will contact you by email if the College is able to confirm membership. Once membership is confirmed your papers will not be sent on to any other colleges and the College Membership condition will be completed.
At this stage, we will need some information from you. Please fill in the form below as soon as possible. Do not wait until you know you have funding or until you have met any other condition.
Keep the College Graduate Office informed of your progress toward meeting the conditions of your offer. Further information on University conditions, and the offer booklet, can be found on the Graduate Admissions - Your Offer pages of the University website.
Apply for Accommodation. Fitzwilliam graduate students can live in College accommodation for the duration of their course. Find out more on our accommodation pages. All graduate students of the University, must live within 10 miles of the centre of Cambridge, unless they have specific permission from the University. You can discuss these arrangements with the Graduate Admissions Office.


text:
Q: Boundary in Powell method Scipy Let's minimise the function
f =lambda x: (x+1)**2

using Powell method in scipy
If we use
scipy.optimize.minimize(f, 1, method='Powell', bounds=None)
the return is
   direc: array([[1.]])
     fun: array(0.)
 message: 'Optimization terminated successfully.'
    nfev: 20
     nit: 2
  status: 0
 success: True
       x: array(-1.)

i.e. the minimum is at -1 as it should. If we provide the bounds
scipy.optimize.minimize(f, 1, method='Powell', bounds=[(0,2)])

the return is again
   direc: array([[1.]])
     fun: array(0.)
 message: 'Optimization terminated successfully.'
    nfev: 20
     nit: 2
  status: 0
 success: True
       x: array(-1.)

which is now WRONG! The correct answer should be 0. It is like if the bounds are not taken int account. I am using scipy '1.4.1' and python 3.7.6.
Someone has any clue?

A: With scipy 1.4.x the Powell method cannot handle constraints nor bounds, as you can see here. Updating to scipy 1.5.x, it can handle bounds, see here:
In [11]: scipy.optimize.minimize(f, x0=1.0, method='Powell', bounds=[(0.0,2.0)])
Out[11]:
   direc: array([[1.64428414e-08]])
     fun: array(1.)
 message: 'Optimization terminated successfully.'
    nfev: 103
     nit: 2
  status: 0
 success: True
       x: array([2.44756652e-12])

A: I want to add that I have noticed as well that the Powell method tends to explore out of bounds parameters.

redpajama_set_name: RedPajamaStackExchange


text:
\section{Some definitions}
\begin{enumerate}
\item
Let $p$ be an odd prime. Let $\zeta_p$ be a root of the polynomial equation $X^{p-1}+X^{p-2}+\dots+X+1=0$.
Let $K_p$ be the $p$-cyclotomic field $K_p=\Q(\zeta_p)$. The ring of integers of $K_p$ is $\Z[\zeta_p]$.
Let $K_p^+$ be the maximal totally real subfield of $K_p$.
Let $v$ be a primitive root $\modu p$ and $\sigma: \zeta_p\rightarrow \zeta_p^v$ be a $\Q$-isomorphism of $K_p$.
Let $G_p$ be the Galois group of $K_p/\Q$.
Let ${\bf F}_p$ be the finite field of cardinal $p$ with no null part  ${\bf F}_p^*$.
Let $\lambda=\zeta_p-1$. The prime ideal of $K_p$ lying over $p$ is $\pi=\lambda \Z[\zeta_p]$.
\item
Suppose that $p$ is irregular and that $p$ does not divide  the class number of $K_p^+$.
Let $C_p$ be the $p$-class group of $K_p$.
Let $r$ be the rank of $C_p$.
Let $\Gamma$ be a  subgroup of order $p$ of $C_p$ annihilated by $\sigma-\mu\in {\bf F}_p[G_p]$ with $\mu\in{\bf F}_p^*$. Then $\mu\equiv v^{2m+1}\modu p$ with a natural integer $m,\quad 1\leq m\leq \frac{p-3}{2}$.
\item
An  integer  $A\in \Z[\zeta_p]$ is said singular if $A^{1/p}\not\in K_p$ and if  there exists an   ideal $\mathbf a$  of $\Z[\zeta_p]$ such that
$A \Z[\zeta_p]=\mathbf a^p$.
There exists  singular integers $A$  with $A \Z[\zeta_p] =\mathbf a^p$ where $\mathbf a$ is a {\bf not} principal  ideal of $\Z[\zeta_p]$
verifying
\begin{equation}\label{e512101}
\begin{split}
& Cl(\mathbf a)\in \Gamma,\\
& \sigma(A)=A^\mu\times\alpha^p,\\
\end{split}
\end{equation}
with $\alpha\in K_p$ and $\pi^{2m+1} \ \|\ A-1$.
Moreover, the number $A$ verifies \begin{equation}\label{e512103}
A\times\overline{A}=D^p,
\end{equation}
for some integer $D\in O_{K_p^+}$.
\end{enumerate}
\section{On Kummer and Stickelberger relation}
\begin{enumerate}
\item
Here we fix a notation for the sequel.
Let $v$ be a primitive root $\modu p$. For every integer $k\in\Z$   then  $v^k$ is understood $\modu p$ so $1\leq v^k\leq p-1$.
 If $k<0$ it is to be understood as $v^kv^{-k}\equiv 1\modu p$.

\item
Let $q\not=p$ be an odd prime.
Let $\zeta_q$ be a root of the minimal polynomial equation $X^{q-1}+X^{q-2}+\dots+X+1=0$.
Let $K_q=\Q(\zeta_q)$ be the $q$-cyclotomic field.
The ring of integers of $K_q$ is $\Z[\zeta_q]$.
Here we fix a notation for the sequel.
Let $u$ be a primitive root $\modu q$. For every integer $k\in\Z$   then  $u^k$ is understood $\modu q$ so $1\leq u^k\leq q-1$.
 If $k<0$ it is to be understood as $u^ku^{-k}\equiv 1\modu q$.
Let $K_{pq}=\Q(\zeta_p,\zeta_q)$. Then $K_{pq}$ is the compositum $K_pK_q$.
The ring of integers of $K_{pq}$ is $\Z[\zeta_{pq}]$.
\item
Let $\mathbf q$ be a prime ideal of $\Z[\zeta_p]$ lying over the prime $q$.
Let $m=N_{K_p/\Q}(\mathbf q)= q^f$ where $f$ is the smallest integer such that $q^f\equiv 1\modu p$.
If $\psi(\alpha)=a$ is the image of $\alpha\in \Z[\zeta_p]$ under the natural map
$\psi: \Z[\zeta_p]\rightarrow \Z[\zeta_p]/\mathbf q$, then for
$\psi(\alpha)=a\not\equiv 0$ define a character $\chi_{\mathbf q}^{(p)}$ on ${\bf F}_m=\Z[\zeta_p]/\mathbf q$ by
\begin{equation}
\chi_{\mathbf q}^{(p)}(a)=\{\frac{\alpha}{\mathbf q}\}_p^{-1}=\overline{\{\frac{\alpha}{\mathbf q}\}}_p,
\end{equation}
where $\{\frac{\alpha}{\mathbf q}\}=\zeta_p^c$ for some natural integer $c$,  is the $p^{th}$ power residue character $\modu \mathbf q$.
We define
\begin{displaymath}
g(\mathbf q)=\sum_{x\in{\bf F}_m}(\chi_{\mathbf q}^{(p)}(x)\times
\zeta_q^{Tr_{{\bf F}_m/{\bf F}_q}(x)})\in \Z[\zeta_{pq}],
\end{displaymath}
and $\mathbf G(\mathbf q)= g(\mathbf q)^p$.
It follows that $G(\mathbf q)\in \Z[\zeta_{pq}]$.
Moreover $\mathbf G(\mathbf q)=g(\mathbf q)^p\in \Z[\zeta_p]$, see for instance Mollin \cite{mol} prop. 5.88 (c) p. 308
or Ireland-Rosen \cite{ire} prop. 14.3.1 (c) p. 208.
\end{enumerate}
The Stickelberger's relation is classically:
\begin{thm}\label{t12201}

In $\Z[\zeta_p]$ we have the ideal decomposition
\begin{equation}\label{e512121}
\mathbf G(\mathbf q)\Z[\zeta_p]=\mathbf q^{S},
\end{equation}
with $S=\sum_{t=1}^{p-1} t\times \varpi_t^{-1}$
where  $\varpi_t\in Gal(K_p/\Q)$ is given by $\varpi_t: \zeta_p\rightarrow \zeta_p^t$.
\end{thm}
See for instance Mollin \cite{mol} thm. 5.109 p. 315 and Ireland-Rosen \cite{ire} thm. 2. p.209.
\subsection{On the structure of $\mathbf G(\mathbf q)$.}
In this subsection we are studying carefully the structure of $g(\mathbf q)$ and $\mathbf G(\mathbf q)$.
\begin{lem}\label{l512151}
If $q\not\equiv 1\modu p$ then $g(\mathbf q)\in \Z[\zeta_p]$.
\begin{proof}$ $
\begin{enumerate}
\item
Let $u$ be a primitive root $\modu q$. Let $\tau :\zeta_q\rightarrow \zeta_q^u$ be a $\Q$-isomorphism generating $Gal(K_q/\Q)$.
The isomorphism $\tau$ is extended to a $K_p$-isomorphism of $K_{pq}$ by
$\tau:\zeta_q\rightarrow \zeta_q^u,\quad \zeta_p\rightarrow \zeta_p$.
Then  $g(\mathbf q)^p=\mathbf G(\mathbf q)\in \Z[\zeta_p]$ and so
\begin{displaymath}
\tau(g(\mathbf q))^p=g(\mathbf q)^p,
\end{displaymath}
and it follows that there exists a natural integer $\rho$ with $\rho<p$ such that
\begin{displaymath}
\tau(g(\mathbf q))= \zeta_p^\rho\times  g(\mathbf q).
\end{displaymath}
Then $N_{K_{pq}/K_p}(\tau(g(\mathbf q)))=\zeta_p^{(q-1)\rho}\times N_{K_{pq}/K_p}(g(\mathbf q))$ and so  $\zeta_p^{\rho(q-1)}=1$.
\item
If $q\not\equiv 1\modu p$, it implies that $\zeta_p^\rho=1$ and so that $\tau(g(\mathbf q))=g(\mathbf q)$ and thus that $g(\mathbf q)\in \Z[\zeta_p]$.
\end{enumerate}
\end{proof}
\end{lem}
Let us note in the sequel $g(\mathbf q)=\sum_{i=0}^{q-2} g_i\times \zeta_q^i$ with $g_i\in \Z[\zeta_p]$.
\begin{lem}\label{l512152}
If $q\equiv 1\modu p$ then $g_0=0$.
\begin{proof}
Suppose that $g_0\not=0$ and search for a contradiction:
we start of
\begin{displaymath}
\tau(g(\mathbf q))= \zeta_p^\rho\times  g(\mathbf q).
\end{displaymath}
We have $g(\mathbf q)=\sum_{i=0}^{q-2} g_i\times \zeta_q^i$  and so
$\tau(g(\mathbf q))=\sum_{i=0}^{q-2}  g_i\times \zeta_q^{i u}$,
therefore
\begin{displaymath}
\sum_{i=0}^{q-2} (\zeta_p^\rho g_i)\times \zeta_q^i=\sum_{i=0}^{q-2}  g_i\times \zeta_q^{i u},
\end{displaymath}
thus $g_0=\zeta_p^\rho \times g_0$ and so $\zeta_p^\rho=1$ which
implies that $\tau(g(\mathbf q))=g(\mathbf q)$ and so $g(\mathbf q)\in \Z[\zeta_p]$.
Then $\mathbf G(\mathbf q)=g(\mathbf q)^p$ and so Stickelberger relation leads to
$g(\mathbf q)^p \Z[\zeta_p] =\mathbf q^S$ where $S=\sum_{t=1}^{p-1}t\times\varpi_t^{-1}$.
Therefore $\varpi_1^{-1}(\mathbf q) \ \|\ \mathbf q^S$ because $q$ splits totally in $K_p/\Q$
 and $\varpi_t^{-1}(\mathbf q)\not=\varpi_{t^\prime}^{-1}(\mathbf q)$ for $t\not=t^\prime$. This case is not possible because the first member
$g(\mathbf q)^p$ is a $p$-power.
\end{proof}
\end{lem}
\begin{lem}\label{l512152}
If $q\equiv 1\modu p$ then
\begin{equation}\label{e512151}
\begin{split}
& \mathbf G(\mathbf q) = g(\mathbf q)^p,\\
&g(\mathbf q)=\zeta_q +\zeta_p^\rho\zeta_q^{u^{-1}}+\zeta_p^{2\rho}\zeta_q^{u^{-2}}+\dots \zeta_p^{(q-2)\rho}\zeta_q^{u^{-(q-2)}},\\
& g(\mathbf q)^p \Z[\zeta_p] =\mathbf q^S.\\
\end{split}
\end{equation}
\begin{proof} $ $
\begin{enumerate}
\item
We start of $\tau(g(\mathbf q))=\zeta_p^\rho\times g(\mathbf q)$ and so
\begin{equation}\label{e512152}
\sum_{i=1} ^{q-2}g_i \zeta_q^{ui}=\zeta_p^\rho\times\sum_{i=1}^{q-2} g_i \zeta_q^i,
\end{equation}
which implies that $g_i=g_1\zeta_p^\rho $ for  $u\times i\equiv 1\modu q$ and so $g_{u^{-1}}=g_1\zeta_p^\rho$ (where $u^{-1}$ is to be understood by
$u^{-1}\modu q$,  so $1\leq u^{-1}\leq q-1)$.
\item
Then
$\tau^2(g(\mathbf q))=\tau(\zeta_p^{\rho} g(\mathbf q))=\zeta_p^{2\rho} g(\mathbf q)$.
Then
\begin{displaymath}
\sum_{i=1} ^{q-2}g_i \zeta_q^{u^2i}=\zeta_p^{2\rho}\times (\sum_{i=1}^{q-2} g_i \zeta_q^i),
\end{displaymath}
which implies that $g_i=g_1\zeta_p^{2\rho}$ for $u^2\times i\equiv 1\modu q$ and so $g_{u^{-2}}=g_1\zeta_p^{2\rho}$.
\item
We continue up to
$\tau^{(q-2)\rho}(g(\mathbf q))=\tau^{q-3}(\zeta_p^\rho g(\mathbf q))=\dots=\zeta_p^{(q-2)\rho} g(\mathbf q)$.
Then
\begin{displaymath}
\sum_{i=1} ^{q-2}g_i \zeta_q^{u^{q-2}i}=\zeta_p^{(q-2)\rho}\times(\sum_{i=1}^{q-2} g_i \zeta_q^i),
\end{displaymath}
which implies that $g_i=g_1\zeta_p^{(q-2)\rho}$ for $u^{q-2}\times i\equiv 1\modu q$ and so $g_{u^{-(q-2)}}=g_1\zeta_p^{(q-2)\rho}$.
\item
Observe that $u$ is a primitive root $\modu q$ and so $u^{-1}$ is a primitive root $\modu q$.
Then it follows that
$g(\mathbf q) =g_1\times (\zeta_q +\zeta_p^\rho\zeta_q^{u^{-1}}+\zeta_p^{2\rho}\zeta_q^{u^{-2}}+\dots \zeta_p^{(q-2)\rho}\zeta_q^{u^{-(q-2)}})$.
Let $U=\zeta_q +\zeta_p^\rho\zeta_q^{u^{-1}}+\zeta_p^{2\rho}\zeta_q^{u^{-2}}+\dots \zeta_p^{(q-2)\rho}\zeta_q^{u^{-(q-2)}}$.
\item
We prove now that $g_1\in \Z[\zeta_p]^*$.
From Stickelberger relation $g_1^p \times U^p =\mathbf q^{S}$.
From $S=\sum_{i=1}^{p-1}\varpi_t^{-1}\times t$ it follows that
$\varpi_t^{-1}(\mathbf q)^t\ \|\ \mathbf q^{S}$
and so that $g_1\not\equiv 0\modu \varpi_t^{-1}(\mathbf q)$
because $g_1^p$ is a $p$-power,
which implies that
$g_1\in \Z[\zeta_p]^*$. We can take $g_1=1$ because $\mathbf G(\mathbf q)$ is defined modulo the unit group $\Z[\zeta_p]^*$.
\item
From Stickelberger,  $g(\mathbf q)^p \Z[\zeta_p]=\mathbf q^S$,
which achieves the proof.
\end{enumerate}
\end{proof}
\end{lem}
\paragraph{Remark:}
From
\begin{equation}
\begin{split}
&g(\mathbf q)=\zeta_q +\zeta_p^\rho\zeta_q^{u^{-1}}+\zeta_p^{2\rho}\zeta_q^{u^{-2}}+\dots +\zeta_p^{(q-2)\rho}\zeta_q^{u^{-(q-2)}},\\
&\Rightarrow \tau(g(\mathbf q))=\zeta_q^u +\zeta_p^\rho\zeta_q+\zeta_p^{2\rho}\zeta_q^{u^{-1}}+\dots +\zeta_p^{(q-2)\rho}\zeta_q^{u^{-(q-3)}},\\
&\Rightarrow\zeta^\rho\times g(\mathbf q)=\zeta^\rho\zeta_q +\zeta_p^{2\rho}\zeta_q^{u^{-1}}+\zeta_p^{3\rho}\zeta_q^{u^{-2}}+\dots +\zeta_p^{(q-1)\rho}
\zeta_q^{u^{-(q-2)}}\\
\end{split}
\end{equation}
and we can verify directly that $\tau(g(\mathbf q))=\zeta_p^\rho \times g(\mathbf q)$ for this expression of $g(\mathbf q)$, observing that $q-1\equiv 0\modu p$.
\begin{lem}\label{l12161}
Let $S=\sum_{t=1}^{p-1} \varpi_t^{-1}\times t$ where $\varpi_t$ is the $\Q$-isomorphism
given by $\varpi_t:\zeta_p\rightarrow \zeta_p^t$ of $K_p$.
Let $v$ be a primitive root $\modu p$. Let $\sigma$ be the $\Q$-isomorphism of $K_p$ given by  $\zeta_p\rightarrow\zeta_p^v$.
Let $P(\sigma)=\sum_{i=0}^{p-2} \sigma^i\times v^{-i}\in\Z[G_p]$.
Then $S=P(\sigma)$.
\begin{proof}
Let us consider one term $\varpi_t^{-1} \times t$. Then $v^{-1}=v^{p-2}$ is a primitive root $\modu p$ because $p-2$ and $p-1$ are coprime and so there exists one and one $i$ such that
$t=v^{-i}$. Then $\varpi_{v^{-i}}:\zeta_p\rightarrow \zeta_p^{v^{-i}}$ and so $\varpi_{v^{-i}}^{-1}:\zeta_p\rightarrow\zeta_p^{v^i}$
and so $\varpi_{v^{-i}}^{-1}=\sigma^i$ (observe that $\sigma^{p-1}\times v^{-(p-1)}=1$), which achieves the proof.
\end{proof}
\end{lem}
\paragraph{Remark} : The previous lemma is  a verification of  the consistency of results in Ribenboim \cite{rib} p. 118, of Mollin \cite {mol} p. 315 and of Ireland-Rosen p. 209 with our computation. In the sequel we use Ribenboim notation more adequate for the factorization in ${\bf F}_p[G]$.
In that case the Stickelberger's relation is connected with the Kummer's relation on Jacobi resolvents, see for instance
Ribenboim, \cite{rib} (2A) b. p. 118 and (2C) relation (2.6) p. 119.

\begin{lem}\label{l512162}$ $

\begin{enumerate}
\item
$g(\mathbf q)$ defined in relation (\ref{e512151}) is the  Jacobi resolvent $<\zeta_p^\rho,\zeta_q>$.
\item
$\rho=-v$.
\end{enumerate}
\begin{proof}$ $
\begin{enumerate}
\item
Apply formula of Ribenboim \cite{rib} (2.2) p. 118 with
$p=p, q=q, \zeta =\zeta_p,\quad \rho=\zeta_q, \quad n=\rho,\quad u=i,\quad m=1$ and $h=u^{-1}$
(where the left members notations $p, q, \zeta,\rho, n, u, m$ and $h$ are the Ribenboim notations).
\item
We start of
$<\zeta_p^\rho,\zeta_q>=g(\mathbf q)$.
Then $v$ is a primitive root $\modu p$, so there exists  a natural integer $l$ such that
$\rho \equiv v^l\modu p$.
By conjugation $\sigma^{-l}$ we get
$<\zeta_p,\zeta_q>=g(\mathbf q)^{\sigma^{-l}}$.
Raising to $p$-power
$<\zeta_p,\zeta_q>^p=g(\mathbf q)^{p\sigma^{-l}}$.
From lemma \ref{l12161} and Stickelberger relation
$<\zeta_p,\zeta_q>^p=\mathbf q^{P(\sigma)\sigma^{-l}}$.
From Kummer's relation (2.6) p. 119 in Ribenboim \cite{rib}, we get
$<\zeta_p,\zeta_q>^p=\mathbf q^{P_1(\sigma)}$ with $P_1(\sigma)=\sum_{j=0}^{p-2}\sigma^j v^{(p-1)/2-j}$.
Therefore
$\sum_{i=0}^{p-2} \sigma^{i-l}v^{-i}=\sum_{j=0}^{p-2}\sigma^j v^{(p-1)/2-j}$.
Then
$i-l\equiv j\modu p$ and $-i\equiv \frac{p-1}{2}-j\modu p$ (or  $i\equiv j-\frac{p-1}{2}\modu p$)
imply that
$j-\frac{p-1}{2}-l\equiv j\modu p$,
so
$l+\frac{p-1}{2}\equiv 0\modu p$,
so
$l\equiv -\frac{p-1}{2}\modu p$,
and
$l\equiv \frac{p+1}{2}\modu p$,
thus
$\rho\equiv v^{(p+1)/2}\modu p$ and finally $\rho= -v$.
\end{enumerate}
\end{proof}
\end{lem}
\paragraph{Remark}: The previous lemma allows to verify the consistency of our computation with Jacobi resultents used in Kummer (see Ribenboim
p. 118-119).
\begin{lem}\label{l512161}
$g(\mathbf q)\equiv -1\modu \pi$.
\begin{proof}
From $g(\mathbf q)=\zeta_q +\zeta_p^\rho\zeta_q^{u^{-1}}+\zeta_p^{2\rho}\zeta_q^{u^{-2}}+\dots +\zeta_p^{(q-2)\rho}\zeta_q^{u^{-(q-2)}}$, we see that
$g(\mathbf q)\equiv \zeta_q +\zeta_q^{u^{-1}}+\zeta_q^{u^{-2}}+\dots +\zeta_q^{u^{-(q-2)}}\modu \pi$.
From $u^{-1}$ primitive root $\modu p$ it follows  that
$1+\zeta_q +\zeta_q^{u^{-1}}+\zeta_q^{u^{-2}}+\dots +\zeta_q^{u^{-(q-2)}}=0$, which leads to the result.
\end{proof}
\end{lem}
\subsection{Singular integers and Kummer-Stickelberger's relation}
We know that the group of ideal classes of the cyclotomic field is generated by the ideal classes of prime ideals of degree $1$, see for instance
Ribenboim, \cite{rib} (3A) p. 119.
From now we suppose  that the prime ideal $\mathbf q$ of $\Z[\zeta_p]$ has a class $Cl(\mathbf q)\in \Gamma$ where $\Gamma$ is
a subgroup of order $p$ of $C_p$ previously defined,  with a singular integer $A$  given  by $A \Z[\zeta_p] = \mathbf q^p$ and
$N_{K_p/\Q}(\mathbf q)=q\equiv 1\modu p$.
\begin{thm}\label{l512163}$ $

$(\frac{g(\mathbf q}{\overline{g(\mathbf q})})^{p^2}=(\frac{A^2}{D^p})^{P(\sigma)}$.
\begin{proof}
We start of
$\mathbf G(\mathbf q)\Z[\zeta_p]= g(\mathbf q)^p \Z[\zeta_p]= \mathbf q^S$.
Raising to $p$-power we get
$g(\mathbf q)^{p^2} \Z[\zeta_p]= \mathbf q^{pS}$.
But $A \Z[\zeta_p] = \mathbf q^p$, so
$g(\mathbf q)^{p^2} \Z[\zeta_p]= A^{S}$, so
 $g(\mathbf q)^{p^2}\times  \zeta_p^w\times \eta= A^{S},\quad \eta\in \Z[\zeta+\zeta^{-1}]^*$ where $w$ is a natural number.
Therefore, by complex conjugation, we get
$ \overline{g(\mathbf q)}^{p^2}\times\zeta_p^{-w}\times  \eta= \overline{A}^{S}.$
Then
$ (\frac{g(\mathbf q)}{\overline{g(\mathbf q)}})^{p^2}\times\zeta_p^{2w}=(\frac{A}{\overline{A}})^{S}$.
From $A\times\overline{A}=D^p$ we obtain
$(\frac{g(\mathbf q}{\overline{g(\mathbf q}})^{p^2}\times\zeta_p^{2w}=(\frac{A^2}{D^p})^{S}$.
We have $\pi^{2m+1}\ \|\ A-1$  for $m\geq 1$ and $\pi^{p+1}\ |\ D^p-1$ and so $(\frac{A^{2}}{D^p})^{P(\sigma)}\equiv 1\modu \pi^{2m+1}$.
From $g(\mathbf q)\equiv -1\modu \pi$ it follows that $\overline{g(\mathbf q)}\equiv -1\modu\pi$ and so that  $\frac{g(\mathbf q)}{\overline{g(\mathbf q)}}\equiv 1\modu \pi$ and so
$(\frac{g(\mathbf q)}{\overline{g(\mathbf q)}})^p\equiv 1\modu \pi^p$ and
thus $w=0$,
so $(\frac{g(\mathbf q)}{\overline{g(\mathbf q)}})^{p^2}=
(\frac{A^2}{D^p})^{P(\sigma)}$.
\end{proof}
\end{thm}
\begin{lem}\label{l512171}
\begin{equation}
P(\sigma)=\sum_{i=0}^{p-2}\sigma^i\times v^{-i}=v^{-(p-2)}\times \{\prod_{k=0,\ k\not=1}^{p-2}(\sigma-v^{k})\}+p\times R(\sigma),
\end{equation}
where $R(\sigma)\in\Z[G_p]$ with $deg(R(\sigma))<p-2$.
\begin{proof}
Let us consider the polynomial $R_0(\sigma)=P(\sigma)-v^{-(p-2)}\times \{\prod_{k=0,\ k\not=1}^{p-2}(\sigma-v^{k})\}$  in ${\bf F}_p[G_p]$.
Then $R_0(\sigma)$ is of degree smaller than $p-2$ and  the two
polynomials $\sum_{i=0}^{p-2} \sigma^iv^{-i} $ and  $\prod_{k=0,\ k\not=1}^{p-2}(\sigma-v^{k})$ take a null value in ${\bf F}_p[G_p]$  when $\sigma$ takes  the $p-2$ different  values  $\sigma=v^k$ for $k=0,\dots, p-2,\quad k\not= 1$. Then $R_0(\sigma)=0$ in ${\bf F}_p[G_p]$ which leads to the result in
$\Z[G_p]$.
\end{proof}
\end{lem}
Let us note in the sequel
\begin{equation}\label{e12201}
T(\sigma)=v^{-(p-2)}\times\prod_{k=0,\ k\not=1}^{p-2}(\sigma-v^{k}).
\end{equation}
\begin{thm}\label{l512164}$ $

$A^{P(\sigma)}\equiv 1\modu\pi^{2(p-1)}$.
\begin{proof}$ $
\begin{enumerate}
\item
From $A\times \overline{A}=D^p$ we know that $D\equiv 1\modu\pi$.
Let us consider $D\equiv 1+d\lambda^n\modu \pi^{n+1}$ where $d\in \Z,\quad d\not\equiv 0\modu p$ and $n\geq 1$. From  $D\in \Z[\zeta_p+\zeta_p^{-1}]$, we derive that $n$ is even and $n\geq 2$.
Then $\sigma(D)\equiv 1+d\sigma(\lambda)^n\equiv 1+d(\zeta^v-1)^n\equiv 1+d((\lambda+1)^v-1)^n$
$\equiv 1+d v^n\lambda^n\modu\pi^{n+1}$.
Also $D^{v^n}=(1+d\lambda^n)^{v^n}\equiv 1+dv^n\lambda^n\modu\pi^{n+1}$.
Then $D^{\sigma-v^n}\equiv 1\modu\pi^{n+1}$.
But $D^{\sigma-v^n}\in K_p^+$ and so $\pi^{n^{\prime}}\ \|\ D^{\sigma-v^n}-1$ with $n^\prime$ even and $n^\prime>n$.
 Using the factorization of $T(\sigma)$ given, we can climb  starting $\pi$-adically of $n^\prime$,  step by step, in the $\pi$-adic approximations to get  finally
$D^{T(\sigma)}\equiv 1\modu\pi^{p-1}$. Then $D^{pT(\sigma)}\equiv 1\modu \pi^{2(p-1)}$. From lemma \ref{l512171}, $p P(\sigma)= pT(\sigma)+p^2 \times
R(\sigma)$
where $R(\sigma)\in\Z[G_p]$ and so
$D^{pP(\sigma)}\equiv 1\modu\pi^{2(p-1)}$.
\item
We have $g(\mathbf q)\equiv -1\modu \pi$ in $\Z[\zeta_{pq}]$ and so
$g(\mathbf q)^p\equiv -1\modu\pi^{p}$ in $\Z[\zeta_{pq}]$.
Therefore $g(\mathbf q)^p-1= \lambda^p\times  a$ with
$a=\sum_{i=0}^{p-2}\sum_{j=0}^{q-2} a_{ij} \zeta_p^i\zeta_q^j,\quad a_{ij}\in \Z$.
But $g(\mathbf q)^p\in \Z[\zeta_p]$ so $a\in K_p$ and so $a_{ij}=0$ when $j\not=0$.
Therefore $a=\sum_{i=0}^{p-2} a_{i0}\zeta_p^i,\quad a_{i0}\in\Z$ and so $g(\mathbf q)^p\equiv 1\modu\pi^p$ in $\Z[\zeta_p]$.
Therefore $(\frac{g(\mathbf q}{\overline{g(\mathbf q)}})^{p^2}\equiv 1\modu\pi^{2(p-1)}$ in $K_{p}$.
We have seen also that $D^{pP(\sigma)}\equiv 1\modu\pi^{2(p-1)}$ in $\Z[\zeta_p]$. Therefore from theorem \ref{l512163} $A^{2P(\sigma)}\equiv 1\modu \pi^{2(p-1)}$ and
from  $A\equiv 1\modu \pi^{2m+1}$ it follows that $A^{P(\sigma)}\equiv 1\modu \pi^{2(p-1)}$.
\end{enumerate}
\end{proof}
\end{thm}
\paragraph{Remark:} if $C\in\Z[\zeta_p]$ is a semi-primary number  with  $C\equiv 1\modu\pi^2$
we can only assert in general that $C^{P(\sigma)}\equiv  1\modu\pi^{p-1}$. For the singular numbers $A$ considered here we assert more:
$A^{P(\sigma)}\equiv 1\modu\pi^{2(p-1)}$.  We shall use this $\pi$-adic improvement in the sequel.
\clearpage
\section{Congruences caracterizing the $p$-class group}
\begin{enumerate}
\item
In this section we shall give a  polynomial $Q(\sigma)\in\Z[G_p]$ explicitly computable in a straightforward way    which {\it caracterize}
the structure of the $p$-class group of $K_p$.
\item
We know  that the    $p$-class group $C_p=\oplus_{k=1}^{r} \Gamma_k$ where $\Gamma_k$ are groups of order $p$ annihilated by $\sigma-\mu_k,
\quad \mu_k \equiv v^{2m_k+1}\modu p,\quad 1\leq m_k\leq\frac{p-3}{2}$.
Let us consider the singular numbers $A_k,\quad k=1,\dots,r$, with $\pi^{2m_k+1} \ \|\ A_k-1$  defined  in  lemmas \ref{l512171} and \ref{l512165}.
From Kummer, the group of ideal classes of $K_p$ is generated by the classes of prime ideals of degree $1$ (see for instance Ribenboim \cite{rib} (3A) p. 119).
\end{enumerate}
\begin{lem}\label{l512165}
\begin{equation}\label{e512172}
P(\sigma)\times (\sigma-v)= T(\sigma)\times(\sigma-v)+pR(\sigma)\times (\sigma-v)=p\times Q(\sigma),
\end{equation}
where $Q(\sigma)=\sum_{i=1}^{p-2}\delta_i\times \sigma^i\in\Z[G_p]$ is given by
\begin{equation}
\begin{split}
& \delta_{p-2}= \frac{v^{-(p-3)}-v^{-(p-2)}v}{p},\\
& \delta_{p-3}= \frac{v^{-(p-4)}-v^{-(p-3)}v}{p},\\
& \vdots\\
& \delta_1 = \frac{1-v^{-1}v}{p},\\
\end{split}
\end{equation}
with $-p<\delta_i\leq 0$.
\begin{proof}
We start of  the relation in $\Z[G_p]$
\begin{displaymath}
P(\sigma)\times(\sigma-v)= v^{-(p-2)}\times \prod_{k=0}^{p-2} (\sigma-v^k)+p\times R(\sigma)\times(\sigma-v)=p\times Q(\sigma),
\end{displaymath}
with $Q(\sigma)\in\Z[G_p]$ because  $\prod_{k=0}^{p-2} (\sigma-v^k)=0$ in ${\bf F}_p[G_p]$ and so
$\prod_{k=0}^{p-2} (\sigma-v^k)=p\times R_1(\sigma)$  in $\Z[G_p]$.
Then  we identify in $\Z[G_p]$ the  coefficients in the relation
\begin{displaymath}
\begin{split}
&(v^{-(p-2)}\sigma^{p-2}+v^{-(p-3)}\sigma^{p-3}+\dots+v^{-1}\sigma+1)\times(\sigma-v)=\\
&p\times (\delta_{p-2}\sigma^{p-2}+\delta_{p-3}\sigma^{p-3}+\dots+\delta_1\sigma+\delta_0),\\
\end{split}
\end{displaymath}
where $\sigma^{p-1}=1$.
\end{proof}
\end{lem}
\paragraph{Remark:}
\begin{enumerate}
\item
Observe that, with our notations,  $\delta_i\in \Z,\quad i=1,\dots,p-2$, but generally $\delta_i\not\equiv 0\modu p$.
\item
We see also that $-p< \delta_i\leq 0$.
Observe also that $\delta_0=\frac{v^{-(p-2)}-v}{p}=0$.
\end{enumerate}
We shall show that the polynomial $Q(\sigma)$ caracterize in a straightforward  way the $p$-class group of $K_p$.
\begin{thm}\label{t512171}
Let $p$ be an odd prime. Let $v$ be a primitive root $\modu p$.
For $k=1,\dots,r$ rank of the $p$-class group of $K_p$  then
\begin{equation}\label{e512191}
Q(v^{2m_k+1})=\sum_{i=1}^{p-2} v^{(2m_k+1)\times i}\times(\frac{v^{-(i-1)}-v^{-i}\times v}{p})\equiv 0\modu p,
\end{equation}
(or an other formulation  $\prod_{k=1}^{r}(\sigma-v^{2m_k+1})$ divides $Q(\sigma)$ in ${\bf F}_p[G_p]$).
\begin{proof}$ $
\begin{enumerate}
\item
Let us fix $A$ one the singular numbers $A_k$ with $\pi^{2m+1}\  \|\ A-1$ equivalent to
$\pi^{2m+1}\ \|\  (\frac{A}{\overline{A}} -1)$,
equivalent to
\begin{displaymath}
\frac{A}{\overline{A}} =1+\lambda^{2m+1}\times a,\quad a\in K_p, \quad v_{\pi}(a)=0.
\end{displaymath}
Then raising to $p$-power we get
$(\frac{A}{\overline{A}})^p=(1+\lambda^{2m+1}\times a)^p\equiv 1+p\lambda^{2m+1} a\modu\pi^{p-1+2m+2}$ and so
$\pi^{p-1+2m+1}\ \|\  (\frac{A}{\overline{A}})^p -1.$
\item
From theorem \ref{l512164} we get
\begin{displaymath}
(\frac{A}{\overline{A}})^{P(\sigma)\times(\sigma-v)} =(\frac{A}{\overline{A}})^{pQ(\sigma)}\equiv 1\modu\pi^{2(p-1)}.
\end{displaymath}
We have shown that
\begin{displaymath}
(\frac{A}{\overline{A}})^p=1+\lambda^{p-1+2m+1}b,\quad b\in K_p,\quad v_\pi(b)=0,
\end{displaymath}
then
\begin{equation}\label{e512281}
(1+\lambda^{p-1+2m+1}b)^{Q(\sigma)}\equiv 1\modu  \pi^{2(p-1)}.
\end{equation}
\item
But $1+\lambda^{p-1+2m+1}b\equiv 1+p\lambda^{2m+1} b_1\modu \pi^{p-1+2m+2}$ with $b_1\in\Z,\quad b_1\not\equiv 0\modu p$.
There exists a natural integer $n$ not divisible by $p$ such that
\begin{displaymath}
(1+p\lambda^{2m+1} b_1)^n\equiv 1+p\lambda^{2m+1}\modu\pi^{p-1+2m+2}.
\end{displaymath}
Therefore
\begin{equation}\label{e512192}
(1+p \lambda^{2m+1}b_1)^{n Q(\sigma)}\equiv (1+p \lambda^{2m+1})^{Q(\sigma)}\equiv 1\modu  \pi^{p-1+2m+2}.
\end{equation}
\item
Show that the possibility of climbing up the  step  $\modu \pi^{p-1+2m+2}$  implies that $\sigma-v^{2m+1}$ divides $Q(\sigma)$ in ${\bf F}_p[G_p]$:
we have $(1+p\lambda^{2m+1} )^\sigma=1+p\sigma(\lambda^{2m+1}) = 1+p(\zeta^v-1)^{2m+1} =1+p((\lambda+1)^v-1)^{2m+1}
\equiv 1+pv^{2m+1}\lambda^{2m+1}\modu\pi^{p-1+2m+2}.$
In an other part $(1+p\lambda^{2m+1})^{v^{2m+1}}\equiv 1+p\lambda^{2m+1} v^{2m+1}\modu\pi^{p-1+2m+2}$.
Therefore
\begin{equation}\label{e512282}
(1+p\lambda^{2m+1})^{\sigma-v^{2m+1}}\equiv 1\modu\pi^{p-1+2m+2}.
\end{equation}
\item
By euclidean division of $Q(\sigma)$ by $\sigma-v^{2m+1}$ in ${\bf F}_p[G_p]$, we get
\begin{displaymath}
Q(\sigma) = (\sigma-v^{2m+1}) Q_1(\sigma)+R
\end{displaymath}
with $R\in{\bf F}_p.$
From congruence (\ref{e512192}) and (\ref{e512282})  it follows that  $(1+p\lambda^{2m+1})^R\equiv 1\modu\pi^{p-1+2m+2}$ and so
that $1+pR\lambda^{2m+1}\equiv 1\modu\pi^{p-1+2m+2}$ and finally
that $R=0$.
Then in ${\bf F}_p$ we have $Q(\sigma)=(\sigma-v^{2m+1})\times Q_1(\sigma)$ and so  $Q(v^{2m+1})\equiv 0\modu p$, or explicitly
\begin{displaymath}
\begin{split}
& Q(v^{2m+1})=v^{(2m+1)(p-2)}\times \frac{v^{-(p-3)}-v^{-(p-2)}v}{p}+v^{(2m+1)(p-3)}\times\frac{v^{-(p-4)}-v^{-(p-3)}v}{p}+\dots\\
&+v^{2m+1}\times \frac{1-v^{-1}v}{p}\equiv 0\modu p,\\
\end{split}
\end{displaymath}
which achieves the proof.
\end{enumerate}
\end{proof}
\end{thm}
\paragraph{Remarks:}
\begin{enumerate}
\item
Observe   that it is the $\pi$-adic theorem  \ref{l512164} connected to Kummer-Stickelberger which allows to obtain this result.
\item
It is easy to verify the consistency of relation (\ref{e512191}).
The little following  MAPLE program allows to verify that this formula is consistent with the table of irregular primes and Bernoulli numbers in
Washington, \cite{was} p. 410. We see that  $Q(v^{2m+1})\equiv 0\modu p$ when $B_{p-1-2m}\equiv 0\modu p$ where $B_{p-1-2m}$ is an even Bernoulli number.
\begin{verbatim}
restart;
> # p       : prime  in paper
> # delta_i :  in paper
> # Q_k     :   is Q in paper
> # a_2     : is 2a in Washington table of irregular primes p. 410
> # v       : primitive root mod p
> p:=3:
> p_max:=157:
> while p < p_max do :
>   p:=nextprime(p):
>   for v from 2 to p-2 do:
>   primitive:=1;
>   for i from 1 to p-2 do:
>   if (v^i mod p) =1 then primitive:=0:fi:
>   od:
>   if primitive = 1 then break:fi:
>   od:
> #***************
> for k from 3 to p-2 by 2 do:
> a2:=p-k;
>  Q_k:=0:
>  for i from 1 to p-2 do:
>   delta_i:=v^(-(i-1)) mod p;
>   delta_i:=delta_i-(v^(-i) mod p)*v:
>   #print(`delta_i=`,delta_i):
>   delta_i:=delta_i/p;
>   #print(`p=`,p,`v=`,v,`delta_i=`,delta_i):
>   Q_k:=Q_k+(v^(k*i) mod p)*delta_i :
>  od:
> Q_k:=Q_k mod p:
> if Q_k=0 then print(`p=`,p,`v=`,v,`a2=`,a2);fi;
> od:
> od:
> print (`****`);
                        p=, 37, v=, 2, a2=, 32
                        p=, 59, v=, 2, a2=, 44
                        p=, 67, v=, 2, a2=, 58
                        p=, 101, v=, 2, a2=, 68
                        p=, 103, v=, 5, a2=, 24
                        p=, 131, v=, 2, a2=, 22
                        p=, 149, v=, 2, a2=, 130
                        p=, 157, v=, 5, a2=, 110
                        p=, 157, v=, 5, a2=, 62
                                 ****

\end{verbatim}
\end{enumerate}
An  immediate consequence is an explicit  criterium for $p$ to be a regular prime:
\begin{cor}\label{512301}
Let $p$ be an odd prime. Let $v$ be a primitive root $\modu p$.
If the congruence
\begin{equation}\label{e512301}
\sum_{i=1}^{p-2} X^{i-1}\times(\frac{v^{-(i-1)}-v^{-i}\times v}{p})\equiv 0\modu p
\end{equation}
has no solution $X$ in $\Z$ then the prime $p$ is regular.
\end{cor}

redpajama_set_name: RedPajamaArXiv


Shropshire Fire and Rescue Service has received the official blessing of both the Archbishop of Canterbury and the Archbishop of Westminster in celebration of its diamond jubilee year.
Telford Fire Station has received a brand new fire appliance to deal with emergency call outs to fires and road crashes.
Shropshire Fire and Rescue Service recently held one of their most successful Festival of Carols at St Mary Magdalene Church, Bridgnorth, with around 400 people attending.
Shropshire women got "a taste" of life as a firefighter during awareness training sessions held by Shropshire Fire and Rescue Service which is urging more women to join up.
Kitchen fires "devastate" lives – that is the message going out to the Shropshire public as part of a Christmas fire safety campaign launched by Shropshire Fire and Rescue Service.
Fire safety in Shropshire's rural business regions has been given a major boost with the introduction of specialist community fire safety officers.
Shropshire Fire and Rescue Service celebrated its 60th anniversary at the Brigade's annual Christmas Festival of Carols.
Minsterley firefighter Bob Morris is hanging up his helmet after a 38 year career with Shropshire Fire and Rescue Service.


text:
SAFETY VIDEOS READY FOR LMS INTEGRATION
Video Title Remember Charlie (1 year LMS license) Safety - Everyone's Responsibility (1 year LMS license)
Remember Charlie (1 year LMS license) - $8,500.00 USD Safety - Everyone's Responsibility (1 year LMS license) - $8,500.00 USD
We now offer our best seller videos "Remember Charlie", full length 56-minutes, and "Safety, Everyone's Responsibility", full length 48-minutes, both available in English and Spanish, for integration within your own Learning Management System (LMS). Deliverables in .mp4 format.
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Remember Charlie
Charlie's journey from death's door following a refinery explosion, to his position as the world's leading presenter on why we all should follow safety procedures precisely. Charlie's story has become a phenomenon in the OHS world. It tells of Charlie's journey from death's door following a refinery explosion, to his position as the world's leading presenter on why we all should follow safety procedures precisely.
Remember Charlie - Keynote Safety Speaker Charlie Morecraft from Charles Morecraft on Vimeo.
Safety Everyone's Responsibility
This presentation highly motivates workers and management to work as a team and not as adversaries, ultimately creating a safer work environment. Once you've seen Charlie, you'll ask him to return. Safety – Everyone's Responsibility is the powerful follow-up to Remember Charlie. This presentation highly motivates workers and management to work as a team and not as adversaries, ultimately creating a safer work environment.
Safety, Everyone's Responsibility with Keynote Safety Speaker Charlie Morecraft from Charles Morecraft on Vimeo.


text:
Could a knowledgeable passenger have saved Helios Flight 522?
Helios Flight 522 crashed when the crew unknowingly suffered hypoxia. Although the masks in the passenger cabin deployed, the crew were unaware of this and didn't realise the extent of the problem.
If I was a passenger on a plane at cruise altitude and the oxygen masks deployed, I know that there is only about 10 minutes oxygen available. I would expect the plane to immediately descend to a lower altitude.
If I don't notice any change in attitude and there is no passenger announcement after a few minutes, would it help to alert a flight attendant to check that the pilots are aware of the issue? Or would it be best to assume that 'they know what they are doing' and just keep quiet?
emergency passenger
David Glickman
David GlickmanDavid Glickman
$\begingroup$ Well, the purser was more knowledgeable than a passenger would and he failed to save the plane… $\endgroup$ – Jan Hudec Mar 29 '17 at 18:16
It's never bad to inform a crew member about anything you think may be abnormal.
In this special situation, as you know that the oxygen masks deploy automatically if the cabin altitude raises to high, you expect the aircraft to descend in order to allow the pax to breathe normally again. If this doesn't happen it would be possible that the crew deployed the masks manually for no reason or the automatism faulty detected a low cabin pressure. In both cases I would expect an announcement that the oxygen masks are not needed and can be put down safely.
If neither of those options happens something is definitely wrong. Either the pilots already suffered hypoxia or they didn't notice the passenger masks drop at all. In my opinion the cabin crew should be trained to know what's normal procedure if masks drop, but I would definitely tell a flight attendant about my concerns if nothing happens after two minutes or so.
You always have to remember: In high altitudes you become unconscious after less than 30 secs without the mask, and as the masks can only produce a limited amount of oxygen it's a really time critical issue to reach the flight-deck and initiate a descend if the pilots are already out of service!
pcfreakxxpcfreakxx
$\begingroup$ I believe the cabin crew has mobile oxygen tanks with masks which would allow people to move around whilst still be able to breathe. $\endgroup$ – ROIMaison Mar 28 '17 at 11:28
$\begingroup$ You don't die! Not that fast, really. You become unusably dizzy and incapable of thinking. You become unable to realize that you should put on the mask. But you really don't die that fast. $\endgroup$ – Vladimir F Mar 28 '17 at 11:38
$\begingroup$ Can you actually do anything about it these days though? I thought that since the reinforced doors you can't get into the pilots cabin unless they let you in so if they are both out of commission you're SOL? $\endgroup$ – DRF Mar 28 '17 at 11:51
$\begingroup$ What I was trying to say is that they didn't notice that the pax oxygen masks dropped (automatically or faulty) and therefor do not react via an announcement $\endgroup$ – pcfreakxx Mar 28 '17 at 13:14
$\begingroup$ @njzk2 - you cannot hold your breath. You need to exhale. Pressure outside of your body is so low that your lungs might be damaged [citation needed]. $\endgroup$ – Peter M. - stands for Monica Mar 28 '17 at 14:19
First and foremost you should read the accident report which you can find in full here. I will pull some excerpts from that below.
Helios Flight 522 crashed when the crew unknowingly suffered hypoxia. Although the masks in the passenger cabin deployed, the crew were unaware of this and didn't realize the extent of the problem.
This is the one of the debated/interesting aspects of the accident. The crew would have been alerted by alarms to the depressurization event
...Non-identification of the warnings and the reasons for the activation of the warnings (Cabin Altitude Warning Horn, Passenger Oxygen Masks Deployment indication, Master Caution).
Many might say they did in fact know there was a problem but for some unknown reason they did nothing about it. Generally speaking "Don Oxygen Masks" is the first item on the emergency checklist following the mishap of the South Dakota LearJet crash in 1999. One can conclude from the report that its possible the pilots assumed the cabin pressure system was in the proper place (auto) when in fact it was set to manual (MAN) and subsequently were confused by the alarm or thought it was faulty.
...that the following factors could have contributed to the accident: omission of returning the cabin pressurization mode selector to the AUTO position after non-scheduled maintenance on the aircraft;
This is a procedural failure on the implementation end. The report would cite bad procedures for this
...the report also identifies a number of additional safety deficiencies pertaining to: maintenance procedures; pilot training, normal and emergency procedures; organizational issues of the Operator; organizational issues related to safety oversight of maintenance and flight operations by Cyprus DCA, EASA/JAA and ICAO; issues related to the aircraft manufacturer's documentation for maintenance and flight operations;
There are some assumptions here of course first and foremost that the plane is still maneuverable. There are any number of things that could cause a depressurization and some may also cause flight control issues as we saw Aloha Airlines Flight 243 a situation where the crew did not immediately know what was going on. However generally speaking, once the issue has been identified the plane should start a decent if possible.
This of course brings up another debate. In almost all cases you should alert the crew to anything you feel is strange our out of place. Of course in this instance it may be hard to move about the cabin, and potentially hard to shout or get your message to the crew. In some cases the crew may be working the issue and getting in the way of that can be problematic. Again generally alerting them is a good thing but its a situation by situation basis for sure.
In regards to the comments on other answers you will not die, at least one flight attendant (who was also a pilot) survived the ordeal and made his way into the cockpit as was heard on the CVR
At 08:54:18 h, the following distress was recorded by the CVR "MAYDAY, MAYDAY, MAYDAY, Helios Airways Flight 522 Athens … (unintelligible word)". A few seconds later, another "MAYDAY, MAYDAY" with a very weak voice was recorded.When the Boeing 737 was at about 7 000 ft, the person in the Captain's seat for the first time appeared to acknowledge the presence of the F-16s and he made a hand motion.
Unlikely as these days its hard to get in the cockpit, or even get a message to the pilots. But you should, in any event at least try. This saved some lives on Air Florida 90 when a passenger realized icing on the wings and assumed the crash position and informed the other around him.
DaveDave
$\begingroup$ The pilot did not regain conciousness, it was a flight attendant named Andreas Prodromou with a commercial pilots license (not rated for that aircraft) that was sitting in the pilots seat. $\endgroup$ – Ron Beyer Mar 28 '17 at 19:05
$\begingroup$ interesting, ill remove that section. $\endgroup$ – Dave Mar 28 '17 at 19:06
It seems from the Wikipedia page that there was indeed an knowledgeable passenger. Not fully qualified, but ...
At 11:49, flight attendant Andreas Prodromou entered the cockpit and sat down in the captain's seat, having remained conscious by using a portable oxygen supply. Prodromou held a UK Commercial Pilot License, but was not qualified to fly the Boeing 737. Crash investigators concluded that Prodromou's experience was insufficient for him to gain control of the aircraft under the circumstances. Prodromou waved at the F16s very briefly, but almost as soon as he entered the cockpit, the left engine flamed out due to fuel exhaustion and the plane left the holding pattern and started to descend. Ten minutes after the loss of power from the left engine, the right engine also flamed out, and just before 12:04 the aircraft crashed into hills near Grammatiko.
So depending on how knowledgeable you're looking for, this seems pretty clear-cut.
https://en.wikipedia.org/wiki/Helios_Airways_Flight_522
David AldridgeDavid Aldridge
$\begingroup$ I agree that he was a pilot, but why did he wait until the fuel was exhausted before entering the cockpit? Maybe he didn't know that he should act quickly? $\endgroup$ – David Glickman Mar 28 '17 at 16:02
$\begingroup$ @David Glickman The CVR only recorded the last 30 min of the flight and the F16 intercept did not arrive until 2 hours after the last contact with the pilots. So we don't know what the cabin crew may have done during that period of time. The FDR does not register the operation of the intercom or the cockpit door. It's quite possible the FA's may have tried to call the pilots or entered the cockpit earlier. They may have tried calling ATC for help. The report says that when the FA tried to call mayday it did not go out over VHF, so he was probably not familiar with how to operate the radio. $\endgroup$ – TomMcW Mar 29 '17 at 5:43
A medic could have had chances to recover the pilot with the portable source of the oxygen available. May make more sense than to attempt the landing.
h22h22
Not the answer you're looking for? Browse other questions tagged emergency passenger or ask your own question.
Why don't pilots parachute from small planes that are in distress?
How long will one remain conscious in the event of a sudden cabin depressurization?
Why do pilots' masks have more oxygen?
Why do airlines have intermediate stopovers with no passenger boarding?
Would a 19,000 ft descent in 6 minutes feel noticably different to passengers?
How fast does cabin air pressure leak after a total engine failure?
How often are passenger oxygen masks deployed?
Have any commercial airline passengers ever been injured from hypoxia due to a depressurization event?
What do flight attendants do if a passenger asks what is going on in case of a potential emergency?
Has a passenger or cabin crew member ever flown an airliner?


text:
What the Bruins were screwing up while everyone was thinking about the Patriots
Most of New England was expected to be distracted by the anticipation for the Patriots' wild card game against Tennessee.
But that wasn't supposed to include the Bruins.
Oh well, at least their latest dud of a performance will fly in the radar regardless of how things go down in Foxboro.
At least this time they didn't blow a third-period lead and lose in a shootout or overtime the way they have most of their recent defeats. This time they lost the lead in the second period and lost in regulation, 4-1 to the Edmonton Oilers at TD Garden.
Now they're 0-1-2 in their past three games – with losses to New Jersey, Columbus and Edmonton (not exactly Murderer's Row) – and just 4-5-6 (one game belowNHL .500, but seven below .500 in the real world) in their past 15 games. The Toronto Maple Leafs and Tampa Bay Lightning are getting within smelling distance in the Atlantic Division.
The Bruins' leadership isn't panicking.
"Well this time I don't think you can get frustrated," center Patrice Bergeron said. "I think that's the last thing you want to do. You know it's a tough league, there's going to be times where things like that happen. It's about rolling up your sleeves and getting to the net. I think there's a lot of fancy plays that we can get rid of in our game and go back to maybe funneling the puck and knowing that we're going there, and there's a lot of loose pucks that were there even tonight that we can find and maybe find some urgency to get them and bear down on those goals.
"It's not always going to be pretty and that's what we need to rely on right now."
There are two glaring issues right now that stick out more than Jake DeBrusk's giveaway that led to Gaetan Haas' game-tying goal in the second period.
A brutal turnover from Jake DeBrusk leads to a goal for Gaetan Haas.
1-1 game. pic.twitter.com/uK5i78JTwi
— Conor Ryan (@ConorRyan_93) January 4, 2020
One, the forward depth has been pushed to its limit and needs to be supplemented. The Bruins are fully healthy up front, yet they've scored one even-strength goal among their four goals in the past three games. And that goal was a net-mouth scramble by Joakim Nordstrom.
If general manager Don Sweeney isn't ready to make a trade, or the players that could make a difference here aren't available yet, then the Bruins have to dip into the Providence pool. Zach Senyshyn's healthy, Paul Carey is scoring at a strong clip and Jack Studnicka couldn't do worse as a third- or fourth-line center right now than Par Lindholm. It'd be easy enough to clear a roster spot by waiving Lindholm or Brett Ritchie, and the Bruins have just enough breathing room under the salary-cap ceiling to make at least a minor change to try to spark the offense.
The second problem is the lack of devotion from the veteran players that are expected to produce at the offensive end. No one's paying the price. Sean Kuraly was Boston's best forward Saturday. That just can't happen. Bergeron's right about funneling the puck and following it to the net. But there has to be more determination along the boards and in the slot. It's time to bang some bodies on the forecheck. And it's time for the defenseman to hit the net – we're looking at you Charlie McAvoy – with their shots. It's time to start taking better care of the puck instead of giving up odd-man rushes and expecting Jaroslav Halak and Tuukka Rask to stop everything.
Coach Bruce Cassidy juggled his forward lines again in the third period against Edmonton, even breaking up Bergeron's line and replacing David Pastrnak with Charlie Coyle. Pastrnak skated with David Krejci and Danton Heinen. There's just so much a coach can do with this group of players that we know are talented and should be able to keep the Bruins afloat until Sweeney supplements the lineup, but seem to have just hit a wall.
The Bruins right now remind one of the Lightning from last April, the team that cruised through the last three months of the regular season and then got punked in four straight by Columbus. Maybe things came too easy to the Bruins as they built their divisional lead and looked like they found a Stanley Cup runner-up hangover cure.
Luckily for the Bruins, they have more time to get back on track than the Lightning did eight months ago. For the long haul, they're going to need Sweeney to swing a deal or two. In the short term, though, we saw in May and June, and again in November and early December, that these guys can play like one of the elite teams in the NHL. Whether they get a spark from a newcomer or not, the Bruins have to rededicate themselves to playing a harder-nosed brand of hockey and a smarter brand of hockey to grind out some victories and not let that divisional lead disappear.
Bruce Cassidy to represent Bruins at NHL All-Star Game
The Skate Podcast: Looking back at the Bruins of the 2010s, looking ahead to 2020s
Patrice Bergeron
David Krejci
David Pastrnak
Charlie McAvoy


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David Wain is back with new episodes of the fifth season of "Wainy Days," in which he bags all of the hot chicks you've dreamed of. Only bagged doesn't mean what you think it means. You know what I mean. Roll the clips!
Future episodes will show up on Mondays, with the season finale on Tuesday, Jan. 3, 2012. And that Fiat car that Jennifer Lopez touts is now sponsoring the show. Even though none of it is shot in the Bronx. Shocking! But true!
Wainy Days follows David Wain's (slightly) fictionalized adventures looking for love in New York City. In season 5, Wain is joined by an all-star cast of characters, many of which serve as targets of his pursuit and/or get in the way as he gives it his best try. These include Ken Marino (Childrens Hospital, Party Down), Steven Weber (Falling Skies, Wings), Lizzy Caplan (Party Down, True Blood), Erinn Hayes (Childrens Hospital, Parenthood), Jorma Taccone (Saturday Night Live, The Lonely Island), Zandy Hartig (Childrens Hospital) and Thomas Lennon (The State, Reno 911). Wain & Marino have collaborated often. The two wrote and produced the new comedy feature Wanderlust, opening February 24, also directed by Wain and starring Paul Rudd and Jennifer Aniston. Wain's many comedy credits include The State, Stella, Wet Hot American Summer, The Ten, Role Models and Childrens Hospital.


The lifetime of human beings is being influenced on a regular basis by the fast-paced actions made in science. Spending time reading science stories is good preparation for writing one's own science news articles. The articles are rewritten at a primary reading stage and arranged into classes akin to Cash & Work, Health & Schooling, and so forth. Tales about immigrant issues and immigration coverage get extensive coverage.
I happen to imagine that the civilisations which were on this planet long ago at the time of Mu and Atlantis have been wiped out but that does not mean that humans had been doomed as a race or that every one the opposite life kinds died. Whereas science doesn't usually handle religious issues, I consider that the seek for the placement of memory will assist us to know the character of the human spirit and greater than that.
Another curious facet of the idea of evolution is that everybody thinks he understand it. I imply philosophers, social scientists, and so on. While in truth only a few folks understand it, truly because it stands, even as it stood when Darwin expressed it, and even less as we now might be able to understand it in biology.
Physics – and science usually – is the examine of what appears to be the target content of reality. This isn't good news for a lot of people. Many information outlets, together with USA At the moment and The Boston Globe, not have a chosen science" web page. A wonderful opportunity to showcase the scientific and analysis work being undertaken in Queensland and to current the fascinating world of science to the Queensland public – to all walks of life.
Another road taken, this time by the Kilcullen GAA U14 girls who turned All Eire Champions on the latest Féile na nÓg competitors, is detailed in a diffusion with a lot of footage of their journey. On the same time in a bunch of people talking about the news you won't wish to be the only one who is not conscious of what's going on as this could make you feel very isolated and out of contact.