Is $frac{R[x]}{(f(x))}$ finite for $R$ finite and $f$ not monic polynomial?












0












$begingroup$



If $R$ is a finite commutative ring, then is $frac{R[x]}{(f(x))}$ finite with $f$ not monic polynomial?




I can prove above claim if f(x) is monic polynomial using division algorithm? But I am not possible further as if f(x) is not monic polynomials then I can not use division algorithm.



Any Help will be appreciated










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    0












    $begingroup$



    If $R$ is a finite commutative ring, then is $frac{R[x]}{(f(x))}$ finite with $f$ not monic polynomial?




    I can prove above claim if f(x) is monic polynomial using division algorithm? But I am not possible further as if f(x) is not monic polynomials then I can not use division algorithm.



    Any Help will be appreciated










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      If $R$ is a finite commutative ring, then is $frac{R[x]}{(f(x))}$ finite with $f$ not monic polynomial?




      I can prove above claim if f(x) is monic polynomial using division algorithm? But I am not possible further as if f(x) is not monic polynomials then I can not use division algorithm.



      Any Help will be appreciated










      share|cite|improve this question











      $endgroup$





      If $R$ is a finite commutative ring, then is $frac{R[x]}{(f(x))}$ finite with $f$ not monic polynomial?




      I can prove above claim if f(x) is monic polynomial using division algorithm? But I am not possible further as if f(x) is not monic polynomials then I can not use division algorithm.



      Any Help will be appreciated







      abstract-algebra polynomials ring-theory finite-rings






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      edited Jan 19 at 16:44









      user26857

      39.3k124183




      39.3k124183










      asked Jan 18 at 11:09









      SRJSRJ

      1,7831620




      1,7831620






















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          $begingroup$

          In $Bbb Z_4[x]/(2x)$, the elements $1, x, x^2, x^3, ldots$ are all distinct.






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            3












            $begingroup$

            In $Bbb Z_4[x]/(2x)$, the elements $1, x, x^2, x^3, ldots$ are all distinct.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              In $Bbb Z_4[x]/(2x)$, the elements $1, x, x^2, x^3, ldots$ are all distinct.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                In $Bbb Z_4[x]/(2x)$, the elements $1, x, x^2, x^3, ldots$ are all distinct.






                share|cite|improve this answer









                $endgroup$



                In $Bbb Z_4[x]/(2x)$, the elements $1, x, x^2, x^3, ldots$ are all distinct.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 18 at 11:13









                ArthurArthur

                115k7116198




                115k7116198






























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