Factor Rings over Finite FieldsQuotient Rings of Polynomials Over Finite FieldsPolynomial rings — Inherited...

Graphic representation of a triangle using ArrayPlot

How do we create new idioms and use them in a novel?

Professor forcing me to attend a conference, I can't afford even with 50% funding

School performs periodic password audits. Is my password compromised?

Can I take the the bonus-action attack from Two-Weapon Fighting without taking the Attack action?

Rationale to prefer local variables over instance variables?

"If + would" conditional in present perfect tense

Factor Rings over Finite Fields

(Codewars) Linked Lists-Sorted Insert

The (Easy) Road to Code

Idiom for feeling after taking risk and someone else being rewarded

Trocar background-image com delay via jQuery

Can't make sense of a paragraph from Lovecraft

Is there a logarithm base for which the logarithm becomes an identity function?

What will happen if my luggage gets delayed?

How do spaceships determine each other's mass in space?

Is it possible to clone a polymorphic object without manually adding overridden clone method into each derived class in C++?

How do you make a gun that shoots melee weapons and/or swords?

What is Tony Stark injecting into himself in Iron Man 3?

Why aren't there more Gauls like Obelix?

Is there a way to make cleveref distinguish two environments with the same counter?

Short scifi story where reproductive organs are converted to produce "materials", pregnant protagonist is "found fit" to be a mother

What is better: yes / no radio, or simple checkbox?

Use Mercury as quenching liquid for swords?



Factor Rings over Finite Fields


Quotient Rings of Polynomials Over Finite FieldsPolynomial rings — Inherited properties from coefficient ringAddition in finite fieldsFactor Ring fundamental questionIs division allowed in rings and fields?Extensions Fields and Zeros of PolynomialsAre these rings fields?Finitely generated modules over non-commutative principal ideal ringsIrreducible polynomials of degree greater than 4 over finite fieldsSplitting Fields over arbitrary fields













2












$begingroup$


Given a polynomial ring over a field $F[x]$, I can factor, for example, the ideal generated by an irreducible polynomial $ax^2 + bx + c$: $F[x]/left<ax^2 + bx + cright>$, and guarantee that this factor ring is also a field.



My question concerns the structure of this factor ring. For example, if I consider the factor ring $Z_p[x] / left<ax^2 + bx + cright>$ for some irreducible polynomial $ax^2 + bx + c$, I can guarantee, for example, that this field has $p^2$ elements.



I am unsure why this is the case. My understanding is that the coset representitives of this factor ring are possible remainders by division by $ax^2 + bx + c$. Is this the right idea, and how would I know that two different remainders aren't in the same coset? Thanks.






abstract-algebra ring-theory field-theory finite-fields quotient-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Roughly speaking, elements in your said ring are linear polynomials $Ax+B$. There are $p$ many choices for each of the coefficients $A$ and $B$.
    $endgroup$
    – thedilated
    4 hours ago
















2












$begingroup$


Given a polynomial ring over a field $F[x]$, I can factor, for example, the ideal generated by an irreducible polynomial $ax^2 + bx + c$: $F[x]/left<ax^2 + bx + cright>$, and guarantee that this factor ring is also a field.



My question concerns the structure of this factor ring. For example, if I consider the factor ring $Z_p[x] / left<ax^2 + bx + cright>$ for some irreducible polynomial $ax^2 + bx + c$, I can guarantee, for example, that this field has $p^2$ elements.



I am unsure why this is the case. My understanding is that the coset representitives of this factor ring are possible remainders by division by $ax^2 + bx + c$. Is this the right idea, and how would I know that two different remainders aren't in the same coset? Thanks.






abstract-algebra ring-theory field-theory finite-fields quotient-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Roughly speaking, elements in your said ring are linear polynomials $Ax+B$. There are $p$ many choices for each of the coefficients $A$ and $B$.
    $endgroup$
    – thedilated
    4 hours ago














2












2








2





$begingroup$


Given a polynomial ring over a field $F[x]$, I can factor, for example, the ideal generated by an irreducible polynomial $ax^2 + bx + c$: $F[x]/left<ax^2 + bx + cright>$, and guarantee that this factor ring is also a field.



My question concerns the structure of this factor ring. For example, if I consider the factor ring $Z_p[x] / left<ax^2 + bx + cright>$ for some irreducible polynomial $ax^2 + bx + c$, I can guarantee, for example, that this field has $p^2$ elements.



I am unsure why this is the case. My understanding is that the coset representitives of this factor ring are possible remainders by division by $ax^2 + bx + c$. Is this the right idea, and how would I know that two different remainders aren't in the same coset? Thanks.






abstract-algebra ring-theory field-theory finite-fields quotient-spaces







share|cite|improve this question











$endgroup$




Given a polynomial ring over a field $F[x]$, I can factor, for example, the ideal generated by an irreducible polynomial $ax^2 + bx + c$: $F[x]/left<ax^2 + bx + cright>$, and guarantee that this factor ring is also a field.



My question concerns the structure of this factor ring. For example, if I consider the factor ring $Z_p[x] / left<ax^2 + bx + cright>$ for some irreducible polynomial $ax^2 + bx + c$, I can guarantee, for example, that this field has $p^2$ elements.



I am unsure why this is the case. My understanding is that the coset representitives of this factor ring are possible remainders by division by $ax^2 + bx + c$. Is this the right idea, and how would I know that two different remainders aren't in the same coset? Thanks.






abstract-algebra ring-theory field-theory finite-fields quotient-spaces




abstract-algebra ring-theory field-theory finite-fields quotient-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Servaes

27.8k34098




27.8k34098










asked 4 hours ago









Solarflare0Solarflare0

773




773








  • 1




    $begingroup$
    Roughly speaking, elements in your said ring are linear polynomials $Ax+B$. There are $p$ many choices for each of the coefficients $A$ and $B$.
    $endgroup$
    – thedilated
    4 hours ago














  • 1




    $begingroup$
    Roughly speaking, elements in your said ring are linear polynomials $Ax+B$. There are $p$ many choices for each of the coefficients $A$ and $B$.
    $endgroup$
    – thedilated
    4 hours ago








1




1




$begingroup$
Roughly speaking, elements in your said ring are linear polynomials $Ax+B$. There are $p$ many choices for each of the coefficients $A$ and $B$.
$endgroup$
– thedilated
4 hours ago




$begingroup$
Roughly speaking, elements in your said ring are linear polynomials $Ax+B$. There are $p$ many choices for each of the coefficients $A$ and $B$.
$endgroup$
– thedilated
4 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Indeed the elements of the factor ring $Bbb{F}_p[x]/langle ax^2+bx^2+crangle$ can be represented by the remainders by division by $ax^2+bx+c$. This is true because we can divide polynomials in $Bbb{F}_p[x]$ by $ax^2+bx+c$ with remainder. What this means is that




For every polynomial $finBbb{F}_p[x]$ there exist unique $q,rinBbb{F}_p[x]$ with $deg r<2$ such that
$$f=q(ax^2+bx+c)+r.tag{1}$$




This equality shows that $f$ and $r$ are in the same coset of $langle ax^2+bx+crangle$, and hence they are mapped to the same element of the factor ring $Bbb{F}_p[x]/langle ax^2+bx+crangle$. Hence the image of
$f$ in the factor ring is represented by $r$, and so every element of the factor ring is represented by a linear polynomial.



To see that no two linear polynomials represent the same element of $Bbb{F}_p[x]/langle ax^2+bx+crangle$, it suffices to note that the remainder $r$ in $(1)$ is unique for every $finBbb{F}_p[x]$, meaning in particular that every linear polynomial is represented only by itself.



Alternatively, if two linear polynomials $r$ and $r'$ represent the same coset of $langle ax^2+bx+crangle$ in the factor ring, then $r-r'$ is a multiple of $ax^2+bx+c$. Because $deg r-r'<deg(ax^2+bx+c)$ it follows that $r-r'=0$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    The quotient you get is a two-dimensional vector space over your field. Can you show that any such vector space must have $p^2$ elements?






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3141833%2ffactor-rings-over-finite-fields%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Indeed the elements of the factor ring $Bbb{F}_p[x]/langle ax^2+bx^2+crangle$ can be represented by the remainders by division by $ax^2+bx+c$. This is true because we can divide polynomials in $Bbb{F}_p[x]$ by $ax^2+bx+c$ with remainder. What this means is that




      For every polynomial $finBbb{F}_p[x]$ there exist unique $q,rinBbb{F}_p[x]$ with $deg r<2$ such that
      $$f=q(ax^2+bx+c)+r.tag{1}$$




      This equality shows that $f$ and $r$ are in the same coset of $langle ax^2+bx+crangle$, and hence they are mapped to the same element of the factor ring $Bbb{F}_p[x]/langle ax^2+bx+crangle$. Hence the image of
      $f$ in the factor ring is represented by $r$, and so every element of the factor ring is represented by a linear polynomial.



      To see that no two linear polynomials represent the same element of $Bbb{F}_p[x]/langle ax^2+bx+crangle$, it suffices to note that the remainder $r$ in $(1)$ is unique for every $finBbb{F}_p[x]$, meaning in particular that every linear polynomial is represented only by itself.



      Alternatively, if two linear polynomials $r$ and $r'$ represent the same coset of $langle ax^2+bx+crangle$ in the factor ring, then $r-r'$ is a multiple of $ax^2+bx+c$. Because $deg r-r'<deg(ax^2+bx+c)$ it follows that $r-r'=0$.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Indeed the elements of the factor ring $Bbb{F}_p[x]/langle ax^2+bx^2+crangle$ can be represented by the remainders by division by $ax^2+bx+c$. This is true because we can divide polynomials in $Bbb{F}_p[x]$ by $ax^2+bx+c$ with remainder. What this means is that




        For every polynomial $finBbb{F}_p[x]$ there exist unique $q,rinBbb{F}_p[x]$ with $deg r<2$ such that
        $$f=q(ax^2+bx+c)+r.tag{1}$$




        This equality shows that $f$ and $r$ are in the same coset of $langle ax^2+bx+crangle$, and hence they are mapped to the same element of the factor ring $Bbb{F}_p[x]/langle ax^2+bx+crangle$. Hence the image of
        $f$ in the factor ring is represented by $r$, and so every element of the factor ring is represented by a linear polynomial.



        To see that no two linear polynomials represent the same element of $Bbb{F}_p[x]/langle ax^2+bx+crangle$, it suffices to note that the remainder $r$ in $(1)$ is unique for every $finBbb{F}_p[x]$, meaning in particular that every linear polynomial is represented only by itself.



        Alternatively, if two linear polynomials $r$ and $r'$ represent the same coset of $langle ax^2+bx+crangle$ in the factor ring, then $r-r'$ is a multiple of $ax^2+bx+c$. Because $deg r-r'<deg(ax^2+bx+c)$ it follows that $r-r'=0$.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Indeed the elements of the factor ring $Bbb{F}_p[x]/langle ax^2+bx^2+crangle$ can be represented by the remainders by division by $ax^2+bx+c$. This is true because we can divide polynomials in $Bbb{F}_p[x]$ by $ax^2+bx+c$ with remainder. What this means is that




          For every polynomial $finBbb{F}_p[x]$ there exist unique $q,rinBbb{F}_p[x]$ with $deg r<2$ such that
          $$f=q(ax^2+bx+c)+r.tag{1}$$




          This equality shows that $f$ and $r$ are in the same coset of $langle ax^2+bx+crangle$, and hence they are mapped to the same element of the factor ring $Bbb{F}_p[x]/langle ax^2+bx+crangle$. Hence the image of
          $f$ in the factor ring is represented by $r$, and so every element of the factor ring is represented by a linear polynomial.



          To see that no two linear polynomials represent the same element of $Bbb{F}_p[x]/langle ax^2+bx+crangle$, it suffices to note that the remainder $r$ in $(1)$ is unique for every $finBbb{F}_p[x]$, meaning in particular that every linear polynomial is represented only by itself.



          Alternatively, if two linear polynomials $r$ and $r'$ represent the same coset of $langle ax^2+bx+crangle$ in the factor ring, then $r-r'$ is a multiple of $ax^2+bx+c$. Because $deg r-r'<deg(ax^2+bx+c)$ it follows that $r-r'=0$.






          share|cite|improve this answer











          $endgroup$



          Indeed the elements of the factor ring $Bbb{F}_p[x]/langle ax^2+bx^2+crangle$ can be represented by the remainders by division by $ax^2+bx+c$. This is true because we can divide polynomials in $Bbb{F}_p[x]$ by $ax^2+bx+c$ with remainder. What this means is that




          For every polynomial $finBbb{F}_p[x]$ there exist unique $q,rinBbb{F}_p[x]$ with $deg r<2$ such that
          $$f=q(ax^2+bx+c)+r.tag{1}$$




          This equality shows that $f$ and $r$ are in the same coset of $langle ax^2+bx+crangle$, and hence they are mapped to the same element of the factor ring $Bbb{F}_p[x]/langle ax^2+bx+crangle$. Hence the image of
          $f$ in the factor ring is represented by $r$, and so every element of the factor ring is represented by a linear polynomial.



          To see that no two linear polynomials represent the same element of $Bbb{F}_p[x]/langle ax^2+bx+crangle$, it suffices to note that the remainder $r$ in $(1)$ is unique for every $finBbb{F}_p[x]$, meaning in particular that every linear polynomial is represented only by itself.



          Alternatively, if two linear polynomials $r$ and $r'$ represent the same coset of $langle ax^2+bx+crangle$ in the factor ring, then $r-r'$ is a multiple of $ax^2+bx+c$. Because $deg r-r'<deg(ax^2+bx+c)$ it follows that $r-r'=0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 3 hours ago









          ServaesServaes

          27.8k34098




          27.8k34098























              3












              $begingroup$

              The quotient you get is a two-dimensional vector space over your field. Can you show that any such vector space must have $p^2$ elements?






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                The quotient you get is a two-dimensional vector space over your field. Can you show that any such vector space must have $p^2$ elements?






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The quotient you get is a two-dimensional vector space over your field. Can you show that any such vector space must have $p^2$ elements?






                  share|cite|improve this answer









                  $endgroup$



                  The quotient you get is a two-dimensional vector space over your field. Can you show that any such vector space must have $p^2$ elements?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Santana AftonSantana Afton

                  2,9132629




                  2,9132629






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3141833%2ffactor-rings-over-finite-fields%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

                      Image
                      Feminismo en EspañaAborto en EspañaManifestaciones y protestas en EspañaManifestaciones y protestas en 2014España en 2014Historia de las mujeres en España movimiento socialfeministaderechos sexuales y reproductivosmujeresLey del aborto en España1 de febrero2014MadridAlberto Ruíz Gallardónla ley del aborto vigente del 2010AsturiasTertulia Feminista Les ComadresCongreso de DiputadosAlicia MiyaresAna MatoAlberto Ruiz GallardónEdimburgoRomaParísLuxemburgoBuenos AiresGracia QuerejetaÁngeles González-SindeMabel LozanoIcíar BollaínIsabel CoixetGeorgina CisquellaTeresa FontVirginia Yagüe2010José Luis Rodríguez ZapateroCongreso de Diputados19 de diciembre20095 de julio2010Partido PopularTribunal Constitucional201120 de diciembre2013Partido PopularMariano RajoyGijónTertulia Feminista Les ComadresBegoña PiñeroRenfeAlicia MiyaresTertulia Feminista Les Comadres1 de febrero2014Congreso de DiputadosMariano RajoyAlberto Ruiz GallardónAna MatoAsturiasAndalucíaCanariasCataluñaComunidad ValencianaCasti...
                      Read more

                      Castillo d'Acher Características Menú de navegación

                      Image
                      Montañas de los Pirineos de AragónMontañas de la provincia de Huesca msnmSelva de OzaPirineo aragonésEspañabosque Este artículo o sección necesita referencias que aparezcan en una publicación acreditada. Este aviso fue puesto el 11 de agosto de 2011. El pico desde Selva de Oza. El Castillo d' Acher es una cima de 2.384msnm en la zona de la Selva de Oza en la parte occidental del Pirineo aragonés, en España. Debe su nombre a su aspecto de fortaleza que presenta hacia todas sus vertientes. Características En orden ascendente, la cima pasa de un bosque frondoso de tonalidades verde-amarillenta a laderas rojizas. Su cima, que suele presentarse mucha nieve, es un pequeño valle rectangular rodeado en sus cuatro caras por una muralla que le da un aspecto de fortaleza. This page is only for reference, If you need detailed information, please check here
                      Read more

                      Connecting two nodes from the same mother node horizontallyTikZ: What EXACTLY does the the |- notation for...

                      Image
                      Pigeonhole Principle Problem Was Hulk present at this event? Attending a conference where my ex-supervisor and his collaborator are present, should I attend? How did Arya manage to disguise herself? What is the word which sounds like "shtrass"? Does higher resolution in an image imply more bits per pixel? What was the state of the German rail system in 1944? Packet sniffer for MacOS Mojave and above Pressure to defend the relevance of one's area of mathematics Why are there synthetic chemicals in our bodies? Where do they come from? Does the Darkness spell dispel the Color Spray ...
                      Read more