Factor Rings over Finite 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.










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.










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.










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






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 4 hours ago

























          answered 4 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 4 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

                      flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror

                      Mangá

                      Eduardo VII do Reino Unido