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Factoring of polynomial over the rationals (quick question)



The 2019 Stack Overflow Developer Survey Results Are InFinding an irreducible polynomial over the integers.What are the prime ideals of $mathbbR[x_1,x_2,x_3,…]$How to prove how many ireducible polynomials are in a polynomial ring over a finite field.$X^3+2$ is irreducible in $mathbbF_7[X]$Confused on notions of maximal ideal and some notationFind all the maximal ideals in the ring $mathbbR[x]$.Show that the polynomial $x^2+y^2-1$ is irreducible over $mathbbQ[x,y]$Extending a principal prime of $mathbbZ[X]$ to a maximal oneFactor a polynomial over a finite fieldMaximal ideals of $Bbb F_2[x]$










3












$begingroup$


Okay this might be a stupid question, but couldn't find any clear explanation on the internet, so here goes.



If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?



For clarification, here is the problem im working on:



Let $J$ be the ideal of $mathbbQ[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textitand$ $g(x)=(x+1)^2 (x+2)^2.$ Is the ideal $J$ prime? Is it maximal? Is it principal?



Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbbQ$?



If this is the right approach, I think I can handle the rest of the problem myself.



Thanks in advance :)










share|cite|improve this question









New contributor




aaalg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$
















    3












    $begingroup$


    Okay this might be a stupid question, but couldn't find any clear explanation on the internet, so here goes.



    If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?



    For clarification, here is the problem im working on:



    Let $J$ be the ideal of $mathbbQ[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textitand$ $g(x)=(x+1)^2 (x+2)^2.$ Is the ideal $J$ prime? Is it maximal? Is it principal?



    Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbbQ$?



    If this is the right approach, I think I can handle the rest of the problem myself.



    Thanks in advance :)










    share|cite|improve this question









    New contributor




    aaalg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      3












      3








      3





      $begingroup$


      Okay this might be a stupid question, but couldn't find any clear explanation on the internet, so here goes.



      If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?



      For clarification, here is the problem im working on:



      Let $J$ be the ideal of $mathbbQ[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textitand$ $g(x)=(x+1)^2 (x+2)^2.$ Is the ideal $J$ prime? Is it maximal? Is it principal?



      Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbbQ$?



      If this is the right approach, I think I can handle the rest of the problem myself.



      Thanks in advance :)










      share|cite|improve this question









      New contributor




      aaalg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Okay this might be a stupid question, but couldn't find any clear explanation on the internet, so here goes.



      If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?



      For clarification, here is the problem im working on:



      Let $J$ be the ideal of $mathbbQ[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $textitand$ $g(x)=(x+1)^2 (x+2)^2.$ Is the ideal $J$ prime? Is it maximal? Is it principal?



      Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbbQ$?



      If this is the right approach, I think I can handle the rest of the problem myself.



      Thanks in advance :)







      abstract-algebra polynomials irreducible-polynomials






      share|cite|improve this question









      New contributor




      aaalg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      aaalg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited Apr 5 at 12:23









      Martin Sleziak

      45k10122277




      45k10122277






      New contributor




      aaalg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked Apr 5 at 10:09









      aaalgaaalg

      232




      232




      New contributor




      aaalg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      New contributor





      aaalg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$


          If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?




          No, quite the opposite in fact. In fact, that's basically what irreducible means: $f$ is irreducible if $f=gh$ implies one of $g,h$ are constant, i.e. there's no nontrivial factorization. So if you take irreducible $g,h$ (which requires them to be have degree at least one) and multiply them together, you get a polynomial $f$ whose factorization is $gh$ .



          Think of it like prime numbers - irreducible polynomials are like primes. Multiply two primes, you get a composite number, never, ever a prime number.




          Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbbQ$?




          It would not be irreducible by the previous discussion as a result.






          share|cite|improve this answer











          $endgroup$




















            2












            $begingroup$

            Simply $p(x)$ is not irreducible, because you have exhibited a factorisation. The factorisation you have given is a factorisation of $p(x)$ into irreducible factors.






            share|cite|improve this answer









            $endgroup$








            • 1




              $begingroup$
              Aha thank you! So i should think of it as $p(x)=x^2 +3x +2$ being reducible and then argue from there, right?
              $endgroup$
              – aaalg
              Apr 5 at 10:58











            • $begingroup$
              @aaalg That's the way to go.
              $endgroup$
              – Mark Bennet
              Apr 5 at 11:09











            Your Answer





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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            active

            oldest

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            2












            $begingroup$


            If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?




            No, quite the opposite in fact. In fact, that's basically what irreducible means: $f$ is irreducible if $f=gh$ implies one of $g,h$ are constant, i.e. there's no nontrivial factorization. So if you take irreducible $g,h$ (which requires them to be have degree at least one) and multiply them together, you get a polynomial $f$ whose factorization is $gh$ .



            Think of it like prime numbers - irreducible polynomials are like primes. Multiply two primes, you get a composite number, never, ever a prime number.




            Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbbQ$?




            It would not be irreducible by the previous discussion as a result.






            share|cite|improve this answer











            $endgroup$

















              2












              $begingroup$


              If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?




              No, quite the opposite in fact. In fact, that's basically what irreducible means: $f$ is irreducible if $f=gh$ implies one of $g,h$ are constant, i.e. there's no nontrivial factorization. So if you take irreducible $g,h$ (which requires them to be have degree at least one) and multiply them together, you get a polynomial $f$ whose factorization is $gh$ .



              Think of it like prime numbers - irreducible polynomials are like primes. Multiply two primes, you get a composite number, never, ever a prime number.




              Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbbQ$?




              It would not be irreducible by the previous discussion as a result.






              share|cite|improve this answer











              $endgroup$















                2












                2








                2





                $begingroup$


                If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?




                No, quite the opposite in fact. In fact, that's basically what irreducible means: $f$ is irreducible if $f=gh$ implies one of $g,h$ are constant, i.e. there's no nontrivial factorization. So if you take irreducible $g,h$ (which requires them to be have degree at least one) and multiply them together, you get a polynomial $f$ whose factorization is $gh$ .



                Think of it like prime numbers - irreducible polynomials are like primes. Multiply two primes, you get a composite number, never, ever a prime number.




                Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbbQ$?




                It would not be irreducible by the previous discussion as a result.






                share|cite|improve this answer











                $endgroup$




                If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?




                No, quite the opposite in fact. In fact, that's basically what irreducible means: $f$ is irreducible if $f=gh$ implies one of $g,h$ are constant, i.e. there's no nontrivial factorization. So if you take irreducible $g,h$ (which requires them to be have degree at least one) and multiply them together, you get a polynomial $f$ whose factorization is $gh$ .



                Think of it like prime numbers - irreducible polynomials are like primes. Multiply two primes, you get a composite number, never, ever a prime number.




                Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $mathbbQ$?




                It would not be irreducible by the previous discussion as a result.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Apr 5 at 12:25









                Martin Sleziak

                45k10122277




                45k10122277










                answered Apr 5 at 10:14









                Eevee TrainerEevee Trainer

                10.4k31742




                10.4k31742





















                    2












                    $begingroup$

                    Simply $p(x)$ is not irreducible, because you have exhibited a factorisation. The factorisation you have given is a factorisation of $p(x)$ into irreducible factors.






                    share|cite|improve this answer









                    $endgroup$








                    • 1




                      $begingroup$
                      Aha thank you! So i should think of it as $p(x)=x^2 +3x +2$ being reducible and then argue from there, right?
                      $endgroup$
                      – aaalg
                      Apr 5 at 10:58











                    • $begingroup$
                      @aaalg That's the way to go.
                      $endgroup$
                      – Mark Bennet
                      Apr 5 at 11:09















                    2












                    $begingroup$

                    Simply $p(x)$ is not irreducible, because you have exhibited a factorisation. The factorisation you have given is a factorisation of $p(x)$ into irreducible factors.






                    share|cite|improve this answer









                    $endgroup$








                    • 1




                      $begingroup$
                      Aha thank you! So i should think of it as $p(x)=x^2 +3x +2$ being reducible and then argue from there, right?
                      $endgroup$
                      – aaalg
                      Apr 5 at 10:58











                    • $begingroup$
                      @aaalg That's the way to go.
                      $endgroup$
                      – Mark Bennet
                      Apr 5 at 11:09













                    2












                    2








                    2





                    $begingroup$

                    Simply $p(x)$ is not irreducible, because you have exhibited a factorisation. The factorisation you have given is a factorisation of $p(x)$ into irreducible factors.






                    share|cite|improve this answer









                    $endgroup$



                    Simply $p(x)$ is not irreducible, because you have exhibited a factorisation. The factorisation you have given is a factorisation of $p(x)$ into irreducible factors.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Apr 5 at 10:13









                    Mark BennetMark Bennet

                    82k984183




                    82k984183







                    • 1




                      $begingroup$
                      Aha thank you! So i should think of it as $p(x)=x^2 +3x +2$ being reducible and then argue from there, right?
                      $endgroup$
                      – aaalg
                      Apr 5 at 10:58











                    • $begingroup$
                      @aaalg That's the way to go.
                      $endgroup$
                      – Mark Bennet
                      Apr 5 at 11:09












                    • 1




                      $begingroup$
                      Aha thank you! So i should think of it as $p(x)=x^2 +3x +2$ being reducible and then argue from there, right?
                      $endgroup$
                      – aaalg
                      Apr 5 at 10:58











                    • $begingroup$
                      @aaalg That's the way to go.
                      $endgroup$
                      – Mark Bennet
                      Apr 5 at 11:09







                    1




                    1




                    $begingroup$
                    Aha thank you! So i should think of it as $p(x)=x^2 +3x +2$ being reducible and then argue from there, right?
                    $endgroup$
                    – aaalg
                    Apr 5 at 10:58





                    $begingroup$
                    Aha thank you! So i should think of it as $p(x)=x^2 +3x +2$ being reducible and then argue from there, right?
                    $endgroup$
                    – aaalg
                    Apr 5 at 10:58













                    $begingroup$
                    @aaalg That's the way to go.
                    $endgroup$
                    – Mark Bennet
                    Apr 5 at 11:09




                    $begingroup$
                    @aaalg That's the way to go.
                    $endgroup$
                    – Mark Bennet
                    Apr 5 at 11:09










                    aaalg is a new contributor. Be nice, and check out our Code of Conduct.









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