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Number of elements in a factor ring



The 2019 Stack Overflow Developer Survey Results Are InFactor into a product of irreducible polynomialsNumber of elements in Quotient ring over $mathbb Z_5[x]$ and $mathbb Z_11[x]$Find element in factor ringHow many elements does this ring have?Find the number of elements in the factor ring $R/A$ and describe the cosets.Construct a field of 25 elements.Factor ring and prime elementsHow many elements does the factor ring $mathbbZ_5[x]/(x^2-2)$ have? Find the multiplicative inverses of $x, x+1$ and $3x+2$ in this ring.Constructing a non-commutative ring with $5^9$ elementsFactor Rings over Finite Fields










3












$begingroup$


I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbbZ_5[x]$ and need to explain why $F = fracmathbbZ_5[x]f(x)$ 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 $mathbbZ_5[x]$, then the factor ring $fracmathbbZ_5[x]f(x)$ 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










share|cite|improve this question









$endgroup$
















    3












    $begingroup$


    I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbbZ_5[x]$ and need to explain why $F = fracmathbbZ_5[x]f(x)$ 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 $mathbbZ_5[x]$, then the factor ring $fracmathbbZ_5[x]f(x)$ 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










    share|cite|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbbZ_5[x]$ and need to explain why $F = fracmathbbZ_5[x]f(x)$ 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 $mathbbZ_5[x]$, then the factor ring $fracmathbbZ_5[x]f(x)$ 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










      share|cite|improve this question









      $endgroup$




      I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbbZ_5[x]$ and need to explain why $F = fracmathbbZ_5[x]f(x)$ 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 $mathbbZ_5[x]$, then the factor ring $fracmathbbZ_5[x]f(x)$ 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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 8 at 10:29









      MathsRookieMathsRookie

      1237




      1237




















          4 Answers
          4






          active

          oldest

          votes


















          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 $$textnumber of such $f(x)+I$ =5^3=|Bbb Z_5[x]/I|$$






              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









                $endgroup$













                  Your Answer





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                  4 Answers
                  4






                  active

                  oldest

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                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  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 Apr 8 at 14:14

























                      answered Apr 8 at 10:35









                      WuestenfuxWuestenfux

                      5,5131513




                      5,5131513





















                          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 Apr 8 at 10:34









                              lhflhf

                              168k11172404




                              168k11172404





















                                  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 $$textnumber of such $f(x)+I$ =5^3=|Bbb Z_5[x]/I|$$






                                  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 $$textnumber of such $f(x)+I$ =5^3=|Bbb Z_5[x]/I|$$






                                    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 $$textnumber of such $f(x)+I$ =5^3=|Bbb Z_5[x]/I|$$






                                      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 $$textnumber of such $f(x)+I$ =5^3=|Bbb Z_5[x]/I|$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Apr 8 at 10:39









                                      Chinnapparaj RChinnapparaj R

                                      6,34521029




                                      6,34521029





















                                          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 Apr 8 at 10:47









                                              P VanchinathanP Vanchinathan

                                              15.6k12136




                                              15.6k12136



























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