Let $f(x)$ be irreducible in $F[x]$, where $F$ is a field. If $f(x) | p(x)q(x)$, prove that either $f(x) |...












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Let $f(x)$ be irreducible in $F[x]$, where $F$ is a field. If $f(x) | p(x)q(x)$, prove that either $f(x) | p(x)$ or $f(x) | q(x)$.




Is this true because every irreducible element in a field is prime?






abstract-algebra field-theory irreducible-polynomials







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    0












    $begingroup$



    Let $f(x)$ be irreducible in $F[x]$, where $F$ is a field. If $f(x) | p(x)q(x)$, prove that either $f(x) | p(x)$ or $f(x) | q(x)$.




    Is this true because every irreducible element in a field is prime?






    abstract-algebra field-theory irreducible-polynomials







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$



      Let $f(x)$ be irreducible in $F[x]$, where $F$ is a field. If $f(x) | p(x)q(x)$, prove that either $f(x) | p(x)$ or $f(x) | q(x)$.




      Is this true because every irreducible element in a field is prime?






      abstract-algebra field-theory irreducible-polynomials







      share|cite|improve this question









      $endgroup$





      Let $f(x)$ be irreducible in $F[x]$, where $F$ is a field. If $f(x) | p(x)q(x)$, prove that either $f(x) | p(x)$ or $f(x) | q(x)$.




      Is this true because every irreducible element in a field is prime?






      abstract-algebra field-theory irreducible-polynomials




      abstract-algebra field-theory irreducible-polynomials






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 6 '18 at 23:40









      numericalorangenumericalorange

      1,728311




      1,728311






















          3 Answers
          3






          active

          oldest

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          1












          $begingroup$

          Prove it by contradiction(using factorisation of $p(x) and q(x)$),i.e.,



          Case 1: Both of p(x) and q(x) are irreducible



          This is not possible because then $p(x)$ and $q(x)$ are the only factors of $p(x)q(x)$.



          Case 2: Exactly one of them is reducible



          In this case, $f(x)$ must be a factor of the reducible one($because$ only factors of $p(x)q(x)$ are the irreducible polynomial and the factors of the reducible one)



          Case 3: Both of them are reducible



          Say $$p(x) = prod_{i=1}^mg_i^{k_i}(x)$$ $$q(x) = prod_{i=1}^nh_i^{l_i}(x)$$
          Then, $f(x)$ being a factor of $p(x)q(x)$, must be factored in terms of $g_i(x)$ and $h_i(x)$,which is a contradiction to its irreducbility.



          Hope it is helpful






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This is a very great answer!
            $endgroup$
            – numericalorange
            Dec 7 '18 at 3:49



















          2












          $begingroup$

          Yes, every irreducible element in $F[x]$ geberates a prime ideal, since $F[x]$ (polynomial ring in one indeterminate over a field) is a P.I.D.



          The same would be true for several indeterminates since $F[x_1,dots ,x_n]$ is a U.F.D.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            $F[X]$ with $F$ a field is a unique factorisation domain, so the irreducible elements are prime and the conclusion is straightforward.






            share|cite|improve this answer









            $endgroup$













              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Prove it by contradiction(using factorisation of $p(x) and q(x)$),i.e.,



              Case 1: Both of p(x) and q(x) are irreducible



              This is not possible because then $p(x)$ and $q(x)$ are the only factors of $p(x)q(x)$.



              Case 2: Exactly one of them is reducible



              In this case, $f(x)$ must be a factor of the reducible one($because$ only factors of $p(x)q(x)$ are the irreducible polynomial and the factors of the reducible one)



              Case 3: Both of them are reducible



              Say $$p(x) = prod_{i=1}^mg_i^{k_i}(x)$$ $$q(x) = prod_{i=1}^nh_i^{l_i}(x)$$
              Then, $f(x)$ being a factor of $p(x)q(x)$, must be factored in terms of $g_i(x)$ and $h_i(x)$,which is a contradiction to its irreducbility.



              Hope it is helpful






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                This is a very great answer!
                $endgroup$
                – numericalorange
                Dec 7 '18 at 3:49
















              1












              $begingroup$

              Prove it by contradiction(using factorisation of $p(x) and q(x)$),i.e.,



              Case 1: Both of p(x) and q(x) are irreducible



              This is not possible because then $p(x)$ and $q(x)$ are the only factors of $p(x)q(x)$.



              Case 2: Exactly one of them is reducible



              In this case, $f(x)$ must be a factor of the reducible one($because$ only factors of $p(x)q(x)$ are the irreducible polynomial and the factors of the reducible one)



              Case 3: Both of them are reducible



              Say $$p(x) = prod_{i=1}^mg_i^{k_i}(x)$$ $$q(x) = prod_{i=1}^nh_i^{l_i}(x)$$
              Then, $f(x)$ being a factor of $p(x)q(x)$, must be factored in terms of $g_i(x)$ and $h_i(x)$,which is a contradiction to its irreducbility.



              Hope it is helpful






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                This is a very great answer!
                $endgroup$
                – numericalorange
                Dec 7 '18 at 3:49














              1












              1








              1





              $begingroup$

              Prove it by contradiction(using factorisation of $p(x) and q(x)$),i.e.,



              Case 1: Both of p(x) and q(x) are irreducible



              This is not possible because then $p(x)$ and $q(x)$ are the only factors of $p(x)q(x)$.



              Case 2: Exactly one of them is reducible



              In this case, $f(x)$ must be a factor of the reducible one($because$ only factors of $p(x)q(x)$ are the irreducible polynomial and the factors of the reducible one)



              Case 3: Both of them are reducible



              Say $$p(x) = prod_{i=1}^mg_i^{k_i}(x)$$ $$q(x) = prod_{i=1}^nh_i^{l_i}(x)$$
              Then, $f(x)$ being a factor of $p(x)q(x)$, must be factored in terms of $g_i(x)$ and $h_i(x)$,which is a contradiction to its irreducbility.



              Hope it is helpful






              share|cite|improve this answer











              $endgroup$



              Prove it by contradiction(using factorisation of $p(x) and q(x)$),i.e.,



              Case 1: Both of p(x) and q(x) are irreducible



              This is not possible because then $p(x)$ and $q(x)$ are the only factors of $p(x)q(x)$.



              Case 2: Exactly one of them is reducible



              In this case, $f(x)$ must be a factor of the reducible one($because$ only factors of $p(x)q(x)$ are the irreducible polynomial and the factors of the reducible one)



              Case 3: Both of them are reducible



              Say $$p(x) = prod_{i=1}^mg_i^{k_i}(x)$$ $$q(x) = prod_{i=1}^nh_i^{l_i}(x)$$
              Then, $f(x)$ being a factor of $p(x)q(x)$, must be factored in terms of $g_i(x)$ and $h_i(x)$,which is a contradiction to its irreducbility.



              Hope it is helpful







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 7 '18 at 0:07

























              answered Dec 6 '18 at 23:44









              MartundMartund

              1,413212




              1,413212












              • $begingroup$
                This is a very great answer!
                $endgroup$
                – numericalorange
                Dec 7 '18 at 3:49


















              • $begingroup$
                This is a very great answer!
                $endgroup$
                – numericalorange
                Dec 7 '18 at 3:49
















              $begingroup$
              This is a very great answer!
              $endgroup$
              – numericalorange
              Dec 7 '18 at 3:49




              $begingroup$
              This is a very great answer!
              $endgroup$
              – numericalorange
              Dec 7 '18 at 3:49











              2












              $begingroup$

              Yes, every irreducible element in $F[x]$ geberates a prime ideal, since $F[x]$ (polynomial ring in one indeterminate over a field) is a P.I.D.



              The same would be true for several indeterminates since $F[x_1,dots ,x_n]$ is a U.F.D.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Yes, every irreducible element in $F[x]$ geberates a prime ideal, since $F[x]$ (polynomial ring in one indeterminate over a field) is a P.I.D.



                The same would be true for several indeterminates since $F[x_1,dots ,x_n]$ is a U.F.D.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Yes, every irreducible element in $F[x]$ geberates a prime ideal, since $F[x]$ (polynomial ring in one indeterminate over a field) is a P.I.D.



                  The same would be true for several indeterminates since $F[x_1,dots ,x_n]$ is a U.F.D.






                  share|cite|improve this answer









                  $endgroup$



                  Yes, every irreducible element in $F[x]$ geberates a prime ideal, since $F[x]$ (polynomial ring in one indeterminate over a field) is a P.I.D.



                  The same would be true for several indeterminates since $F[x_1,dots ,x_n]$ is a U.F.D.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 23:47









                  BernardBernard

                  119k639112




                  119k639112























                      1












                      $begingroup$

                      $F[X]$ with $F$ a field is a unique factorisation domain, so the irreducible elements are prime and the conclusion is straightforward.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        $F[X]$ with $F$ a field is a unique factorisation domain, so the irreducible elements are prime and the conclusion is straightforward.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          $F[X]$ with $F$ a field is a unique factorisation domain, so the irreducible elements are prime and the conclusion is straightforward.






                          share|cite|improve this answer









                          $endgroup$



                          $F[X]$ with $F$ a field is a unique factorisation domain, so the irreducible elements are prime and the conclusion is straightforward.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 7 '18 at 12:38









                          Math_QEDMath_QED

                          7,33631450




                          7,33631450






























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