Number of elements in a factor ring












3












$begingroup$


I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbb{Z_5}[x]$ and need to explain why $F = frac{mathbb{Z_5}[x]}{f(x)}$ is a field and also find how many elements are in F.



So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,




If $f(x)$ is an irreducible polynomial in $mathbb{Z_5}[x]$, then the factor ring $frac{mathbb{Z_5}[x]}{f(x)}$ is also a field.




Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.



Thanks in advance






abstract-algebra







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbb{Z_5}[x]$ and need to explain why $F = frac{mathbb{Z_5}[x]}{f(x)}$ is a field and also find how many elements are in F.



    So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,




    If $f(x)$ is an irreducible polynomial in $mathbb{Z_5}[x]$, then the factor ring $frac{mathbb{Z_5}[x]}{f(x)}$ is also a field.




    Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.



    Thanks in advance






    abstract-algebra







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbb{Z_5}[x]$ and need to explain why $F = frac{mathbb{Z_5}[x]}{f(x)}$ is a field and also find how many elements are in F.



      So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,




      If $f(x)$ is an irreducible polynomial in $mathbb{Z_5}[x]$, then the factor ring $frac{mathbb{Z_5}[x]}{f(x)}$ is also a field.




      Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.



      Thanks in advance






      abstract-algebra







      share|cite|improve this question









      $endgroup$




      I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbb{Z_5}[x]$ and need to explain why $F = frac{mathbb{Z_5}[x]}{f(x)}$ is a field and also find how many elements are in F.



      So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,




      If $f(x)$ is an irreducible polynomial in $mathbb{Z_5}[x]$, then the factor ring $frac{mathbb{Z_5}[x]}{f(x)}$ is also a field.




      Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.



      Thanks in advance






      abstract-algebra




      abstract-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked yesterday









      MathsRookieMathsRookie

      1237




      1237






















          4 Answers
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          active

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          3












          $begingroup$

          Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.



          If $f$ is an irreducible polynomial of degree $n$ over, say, ${Bbb Z}_p$, then
          the quotient field ${Bbb Z}_p[x]/langle frangle$ has $p^n$ elements and is a vector space over ${Bbb Z}_p$ of dimension $n$.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Outline:



              You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb {Z_5} (Bbb Z_5[x]/I)=3=deg f$$



              Call ${v_1,v_2,v_3}$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$



              Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$



              There are $5 times 5 times 5$ choices to choose the coefficients, so $$text{number of such $f(x)+I$ }=5^3=|Bbb Z_5[x]/I|$$






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
                $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.



                Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
                such that $h(x)=q(x)f(x) + r(x)$



                Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.






                share|cite|improve this answer









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                  4 Answers
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                  4 Answers
                  4






                  active

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                  active

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                  active

                  oldest

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                  3












                  $begingroup$

                  Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.



                  If $f$ is an irreducible polynomial of degree $n$ over, say, ${Bbb Z}_p$, then
                  the quotient field ${Bbb Z}_p[x]/langle frangle$ has $p^n$ elements and is a vector space over ${Bbb Z}_p$ of dimension $n$.






                  share|cite|improve this answer











                  $endgroup$


















                    3












                    $begingroup$

                    Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.



                    If $f$ is an irreducible polynomial of degree $n$ over, say, ${Bbb Z}_p$, then
                    the quotient field ${Bbb Z}_p[x]/langle frangle$ has $p^n$ elements and is a vector space over ${Bbb Z}_p$ of dimension $n$.






                    share|cite|improve this answer











                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.



                      If $f$ is an irreducible polynomial of degree $n$ over, say, ${Bbb Z}_p$, then
                      the quotient field ${Bbb Z}_p[x]/langle frangle$ has $p^n$ elements and is a vector space over ${Bbb Z}_p$ of dimension $n$.






                      share|cite|improve this answer











                      $endgroup$



                      Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.



                      If $f$ is an irreducible polynomial of degree $n$ over, say, ${Bbb Z}_p$, then
                      the quotient field ${Bbb Z}_p[x]/langle frangle$ has $p^n$ elements and is a vector space over ${Bbb Z}_p$ of dimension $n$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited yesterday

























                      answered yesterday









                      WuestenfuxWuestenfux

                      5,4841513




                      5,4841513























                          2












                          $begingroup$

                          Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.






                              share|cite|improve this answer









                              $endgroup$



                              Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered yesterday









                              lhflhf

                              167k11172404




                              167k11172404























                                  0












                                  $begingroup$

                                  Outline:



                                  You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb {Z_5} (Bbb Z_5[x]/I)=3=deg f$$



                                  Call ${v_1,v_2,v_3}$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$



                                  Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$



                                  There are $5 times 5 times 5$ choices to choose the coefficients, so $$text{number of such $f(x)+I$ }=5^3=|Bbb Z_5[x]/I|$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Outline:



                                    You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb {Z_5} (Bbb Z_5[x]/I)=3=deg f$$



                                    Call ${v_1,v_2,v_3}$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$



                                    Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$



                                    There are $5 times 5 times 5$ choices to choose the coefficients, so $$text{number of such $f(x)+I$ }=5^3=|Bbb Z_5[x]/I|$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Outline:



                                      You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb {Z_5} (Bbb Z_5[x]/I)=3=deg f$$



                                      Call ${v_1,v_2,v_3}$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$



                                      Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$



                                      There are $5 times 5 times 5$ choices to choose the coefficients, so $$text{number of such $f(x)+I$ }=5^3=|Bbb Z_5[x]/I|$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      Outline:



                                      You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb {Z_5} (Bbb Z_5[x]/I)=3=deg f$$



                                      Call ${v_1,v_2,v_3}$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$



                                      Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$



                                      There are $5 times 5 times 5$ choices to choose the coefficients, so $$text{number of such $f(x)+I$ }=5^3=|Bbb Z_5[x]/I|$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered yesterday









                                      Chinnapparaj RChinnapparaj R

                                      6,1612929




                                      6,1612929























                                          0












                                          $begingroup$

                                          Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
                                          $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.



                                          Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
                                          such that $h(x)=q(x)f(x) + r(x)$



                                          Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
                                            $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.



                                            Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
                                            such that $h(x)=q(x)f(x) + r(x)$



                                            Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
                                              $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.



                                              Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
                                              such that $h(x)=q(x)f(x) + r(x)$



                                              Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.






                                              share|cite|improve this answer









                                              $endgroup$



                                              Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
                                              $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.



                                              Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
                                              such that $h(x)=q(x)f(x) + r(x)$



                                              Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered yesterday









                                              P VanchinathanP Vanchinathan

                                              15.5k12136




                                              15.5k12136






























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