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.






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







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$endgroup$








  • 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
    3 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
    3 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






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share|cite|improve this question













share|cite|improve this question




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edited 3 hours ago









Servaes

27.8k34098




27.8k34098










asked 3 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
    3 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
    3 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
3 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
3 hours ago










2 Answers
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oldest

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$begingroup$

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









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      2 Answers
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      2 Answers
      2






      active

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      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






























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