What is the inverse of X modulo $1 + X + X^2 + X^3 + X^4$? [duplicate]












-1












$begingroup$



This question already has an answer here:




  • Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$

    2 answers




What is the inverse of X modulo $1 + X + X^2 + X^3 + X^4$?
Is there any open softwares to calculate such things easily?






ring-theory inverse irreducible-polynomials







share|cite|improve this question











$endgroup$



marked as duplicate by Saad, GNUSupporter 8964民主女神 地下教會, jgon, Bill Dubuque ring-theory
Users with the  ring-theory badge can single-handedly close ring-theory questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 Dec 4 '18 at 15:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    Hint $bmod, 1+xf!:, (-f)xequiv 1 $
    $endgroup$
    – Bill Dubuque
    Dec 4 '18 at 14:40


















-1












$begingroup$



This question already has an answer here:




  • Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$

    2 answers




What is the inverse of X modulo $1 + X + X^2 + X^3 + X^4$?
Is there any open softwares to calculate such things easily?






ring-theory inverse irreducible-polynomials







share|cite|improve this question











$endgroup$



marked as duplicate by Saad, GNUSupporter 8964民主女神 地下教會, jgon, Bill Dubuque ring-theory
Users with the  ring-theory badge can single-handedly close ring-theory questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 Dec 4 '18 at 15:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    Hint $bmod, 1+xf!:, (-f)xequiv 1 $
    $endgroup$
    – Bill Dubuque
    Dec 4 '18 at 14:40
















-1












-1








-1





$begingroup$



This question already has an answer here:




  • Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$

    2 answers




What is the inverse of X modulo $1 + X + X^2 + X^3 + X^4$?
Is there any open softwares to calculate such things easily?






ring-theory inverse irreducible-polynomials







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$

    2 answers




What is the inverse of X modulo $1 + X + X^2 + X^3 + X^4$?
Is there any open softwares to calculate such things easily?





This question already has an answer here:




  • Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$

    2 answers







ring-theory inverse irreducible-polynomials




ring-theory inverse irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 9:26









José Carlos Santos

157k22126227




157k22126227










asked Dec 4 '18 at 9:23









malleamallea

29919




29919




marked as duplicate by Saad, GNUSupporter 8964民主女神 地下教會, jgon, Bill Dubuque ring-theory
Users with the  ring-theory badge can single-handedly close ring-theory questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 Dec 4 '18 at 15:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Saad, GNUSupporter 8964民主女神 地下教會, jgon, Bill Dubuque ring-theory
Users with the  ring-theory badge can single-handedly close ring-theory questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 Dec 4 '18 at 15:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    Hint $bmod, 1+xf!:, (-f)xequiv 1 $
    $endgroup$
    – Bill Dubuque
    Dec 4 '18 at 14:40
















  • 1




    $begingroup$
    Hint $bmod, 1+xf!:, (-f)xequiv 1 $
    $endgroup$
    – Bill Dubuque
    Dec 4 '18 at 14:40










1




1




$begingroup$
Hint $bmod, 1+xf!:, (-f)xequiv 1 $
$endgroup$
– Bill Dubuque
Dec 4 '18 at 14:40






$begingroup$
Hint $bmod, 1+xf!:, (-f)xequiv 1 $
$endgroup$
– Bill Dubuque
Dec 4 '18 at 14:40












1 Answer
1






active

oldest

votes


















1












$begingroup$

Hint: $1+X+X^2+X^3+X^4=1+X(1+X+X^2+X^3)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Got it! Thank you very much :)
    $endgroup$
    – mallea
    Dec 4 '18 at 13:33


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Hint: $1+X+X^2+X^3+X^4=1+X(1+X+X^2+X^3)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Got it! Thank you very much :)
    $endgroup$
    – mallea
    Dec 4 '18 at 13:33
















1












$begingroup$

Hint: $1+X+X^2+X^3+X^4=1+X(1+X+X^2+X^3)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Got it! Thank you very much :)
    $endgroup$
    – mallea
    Dec 4 '18 at 13:33














1












1








1





$begingroup$

Hint: $1+X+X^2+X^3+X^4=1+X(1+X+X^2+X^3)$.






share|cite|improve this answer









$endgroup$



Hint: $1+X+X^2+X^3+X^4=1+X(1+X+X^2+X^3)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 9:25









José Carlos SantosJosé Carlos Santos

157k22126227




157k22126227












  • $begingroup$
    Got it! Thank you very much :)
    $endgroup$
    – mallea
    Dec 4 '18 at 13:33


















  • $begingroup$
    Got it! Thank you very much :)
    $endgroup$
    – mallea
    Dec 4 '18 at 13:33
















$begingroup$
Got it! Thank you very much :)
$endgroup$
– mallea
Dec 4 '18 at 13:33




$begingroup$
Got it! Thank you very much :)
$endgroup$
– mallea
Dec 4 '18 at 13:33



-

Popular posts from this blog

Keirin

Image
Keirin in Kanada, Juli 2006. Keirin (von jap. 競輪 , Keirin , dt. „Radrennen“) ist eine Disziplin des Bahnradsports. Es handelt sich um eine aus Japan stammende Variante des Sprints; sie wird auch als „Kampfsprint“ bezeichnet. Inhaltsverzeichnis 1 Geschichte 2 Regeln 3 Keirin in Japan 4 Ergebnisse bei UCI-Bahn-Weltmeisterschaften 4.1 Männer (Profis/Elite) 4.2 Frauen 5 Literatur 6 Weblinks 7 Einzelnachweise Geschichte | Keirin-Lauf bei den UEC-Bahn-Europameisterschaften 2011 in Apeldoorn Keirin wurde 1948 in Japan als Wettsport eingeführt; die Einnahmen aus den Wetten waren damals für den Wiederaufbau nach dem Zweiten Weltkrieg gedacht. 1980 wurde Keirin in das Programm von UCI-Bahn-Weltmeisterschaften für Männer aufgenommen, seit 2002 auch für Frauen. Seit dem Jahr 2000 ist Keirin für Männer ein olympischer Radsportwettbewerb. Bei den Olympischen Spielen 2012 wurden erstmals Medaillen in einem Keirinwettbewerb für Fraue...
Read more

how to set this table in IEEE paper

Image
1 begin{table}[h] begin{center} caption{ Comparison of Maximum Accuracy of Different Methods on BCI comp III dataset IIIa }label{tab:2} begin{tabular}{c c c c c c} hline % after \: hline or cline{col1-col2} cline{col3-col4} ... $Feature Extraction^a$ & $Classifier^b$ & & Accuracy(%) & & \ hline & & K3 & K6 & L1 & $mean^c$ \ AAR & MDA[9] & 66.6 & 38.5 & 49.5 & 51.5 \ CAR+CSP & NN[10] & 81.7 & 76.7 & 73.3 & 77.2 \ Barlow method & SVM[11] & 63.3 & 45.0 & 60.0 & 56.1 \ WPD & ME[12] & 90.4 & 66.0 & 79.6 & 78.7\ WPD+CSP & SVM+NN[13] & 83.1 & 84.4 & 85.6 & 84.4 \ PLV & SVM+Quicksort[14] ...
Read more

Santa Maria sopra Minerva (Rom)

Image
Santa Maria sopra Minerva [1] Konfession: römisch-katholisch Patrozinium: Hl. Maria Weihejahr: 1370 Rang: Basilica minor Orden: Dominikaner (OP) Kardinalpriester: António Marto Anschrift: Piazza della Minerva, 42 00187 Roma 41.898029 12.478194 Koordinaten: 41° 53′ 52,9″  N , 12° 28′ 41,5″  O Santa Maria sopra Minerva ( lateinisch Sanctae Mariae supra Minervam ), vollständig Basilica di Santa Maria sopra Minerva , ist der einzig bedeutende Kirchenbau Roms aus der Zeit der Gotik und eine der Hauptkirchen des Dominikanerordens in Rom. Seit 1566 ist die Basilica minor eine Titelkirche der römisch-katholischen Kirche. Sie befindet sich an der Piazza della Minerva im Rione Pigna im historischen Zentrum Roms, dem Marsfeld südöstlich des Pantheon. Südlich grenzt die Straße Via di S. Caterina da Siena an, benannt nach der Heiligen Katharina von Siena, deren Grab sich unter dem Hauptaltar der Kirche befindet. Die Kirche wurde über den Ruin...
Read more