Irreducibility of a simple polynomial












3












$begingroup$


For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.



What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.



For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.



A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.






abstract-algebra field-theory irreducible-polynomials







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
    $endgroup$
    – Sil
    6 hours ago


















3












$begingroup$


For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.



What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.



For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.



A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.






abstract-algebra field-theory irreducible-polynomials







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
    $endgroup$
    – Sil
    6 hours ago
















3












3








3


1



$begingroup$


For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.



What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.



For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.



A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.






abstract-algebra field-theory irreducible-polynomials







share|cite|improve this question









$endgroup$




For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.



What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.



For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.



A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.






abstract-algebra field-theory irreducible-polynomials




abstract-algebra field-theory irreducible-polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 6 hours ago









JonHalesJonHales

520311




520311








  • 2




    $begingroup$
    If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
    $endgroup$
    – Sil
    6 hours ago
















  • 2




    $begingroup$
    If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
    $endgroup$
    – Sil
    6 hours ago










2




2




$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
6 hours ago






$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
6 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

Notice that we are trying to reduce that polynomial by this way:



$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$



We need:



$$2a-b^2=0$$
$$b=sqrt{2a}$$



But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:



$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$



Which is also known as Sophie Germain Identity.






share|cite|improve this answer








New contributor




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






$endgroup$













  • $begingroup$
    I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
    $endgroup$
    – Sil
    5 hours ago










  • $begingroup$
    @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
    $endgroup$
    – Eureka
    5 hours ago








  • 1




    $begingroup$
    @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
    $endgroup$
    – Ethan MacBrough
    5 hours ago










  • $begingroup$
    @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
    $endgroup$
    – Sil
    5 hours ago



















2












$begingroup$

Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    4












    $begingroup$

    Notice that we are trying to reduce that polynomial by this way:



    $$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$



    We need:



    $$2a-b^2=0$$
    $$b=sqrt{2a}$$



    But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:



    $$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$



    Which is also known as Sophie Germain Identity.






    share|cite|improve this answer








    New contributor




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






    $endgroup$













    • $begingroup$
      I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
      $endgroup$
      – Sil
      5 hours ago










    • $begingroup$
      @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
      $endgroup$
      – Eureka
      5 hours ago








    • 1




      $begingroup$
      @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
      $endgroup$
      – Ethan MacBrough
      5 hours ago










    • $begingroup$
      @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
      $endgroup$
      – Sil
      5 hours ago
















    4












    $begingroup$

    Notice that we are trying to reduce that polynomial by this way:



    $$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$



    We need:



    $$2a-b^2=0$$
    $$b=sqrt{2a}$$



    But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:



    $$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$



    Which is also known as Sophie Germain Identity.






    share|cite|improve this answer








    New contributor




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






    $endgroup$













    • $begingroup$
      I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
      $endgroup$
      – Sil
      5 hours ago










    • $begingroup$
      @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
      $endgroup$
      – Eureka
      5 hours ago








    • 1




      $begingroup$
      @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
      $endgroup$
      – Ethan MacBrough
      5 hours ago










    • $begingroup$
      @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
      $endgroup$
      – Sil
      5 hours ago














    4












    4








    4





    $begingroup$

    Notice that we are trying to reduce that polynomial by this way:



    $$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$



    We need:



    $$2a-b^2=0$$
    $$b=sqrt{2a}$$



    But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:



    $$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$



    Which is also known as Sophie Germain Identity.






    share|cite|improve this answer








    New contributor




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






    $endgroup$



    Notice that we are trying to reduce that polynomial by this way:



    $$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$



    We need:



    $$2a-b^2=0$$
    $$b=sqrt{2a}$$



    But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:



    $$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$



    Which is also known as Sophie Germain Identity.







    share|cite|improve this answer








    New contributor




    Eureka 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 answer



    share|cite|improve this answer






    New contributor




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









    answered 5 hours ago









    EurekaEureka

    25611




    25611




    New contributor




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





    New contributor





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






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












    • $begingroup$
      I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
      $endgroup$
      – Sil
      5 hours ago










    • $begingroup$
      @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
      $endgroup$
      – Eureka
      5 hours ago








    • 1




      $begingroup$
      @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
      $endgroup$
      – Ethan MacBrough
      5 hours ago










    • $begingroup$
      @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
      $endgroup$
      – Sil
      5 hours ago


















    • $begingroup$
      I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
      $endgroup$
      – Sil
      5 hours ago










    • $begingroup$
      @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
      $endgroup$
      – Eureka
      5 hours ago








    • 1




      $begingroup$
      @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
      $endgroup$
      – Ethan MacBrough
      5 hours ago










    • $begingroup$
      @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
      $endgroup$
      – Sil
      5 hours ago
















    $begingroup$
    I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
    $endgroup$
    – Sil
    5 hours ago




    $begingroup$
    I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
    $endgroup$
    – Sil
    5 hours ago












    $begingroup$
    @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
    $endgroup$
    – Eureka
    5 hours ago






    $begingroup$
    @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
    $endgroup$
    – Eureka
    5 hours ago






    1




    1




    $begingroup$
    @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
    $endgroup$
    – Ethan MacBrough
    5 hours ago




    $begingroup$
    @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
    $endgroup$
    – Ethan MacBrough
    5 hours ago












    $begingroup$
    @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
    $endgroup$
    – Sil
    5 hours ago




    $begingroup$
    @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
    $endgroup$
    – Sil
    5 hours ago











    2












    $begingroup$

    Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
    $$
    x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
    (x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
    $$
    and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
      $$
      x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
      (x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
      $$
      and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
        $$
        x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
        (x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
        $$
        and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.






        share|cite|improve this answer









        $endgroup$



        Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
        $$
        x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
        (x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
        $$
        and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 5 hours ago









        egregegreg

        185k1486206




        185k1486206






























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