Irreducibility of a simple polynomialShow $x^6 + 1.5x^5 + 3x - 4.5$ is irreducible in $mathbb Q[x]$.Determine...

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Irreducibility of a simple polynomial


Show $x^6 + 1.5x^5 + 3x - 4.5$ is irreducible in $mathbb Q[x]$.Determine whether the polynomial $x^2-12$ in $mathbb Z[x]$ satisfies an Eisenstein criterion for irreducibility over $mathbb Q$Proving Irreducibility of $x^4-16x^3+20x^2+12$ in $mathbb Q[x]$Constructibility of roots of a polynomialEisenstein's criterion for polynomials in Z mod pProving irreducibility; What is this method and what is the logic behind it?Irreducibility of special cyclotomic polynomial.Irreducibility of a Polynomial after a substitutionIrreducibility of Non-monic Quartic Polynomials in Q[x]Irreducible monic polynomial in $mathbb{Q}[x]$













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
    1 hour 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
    1 hour 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 1 hour 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
    1 hour 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
    1 hour 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
1 hour 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
1 hour 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
    1 hour ago










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










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









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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
      1 hour ago










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










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










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










    • $begingroup$
      @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
      $endgroup$
      – Sil
      1 hour 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 1 hour ago









    EurekaEureka

    22611




    22611




    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
      1 hour ago










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










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










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










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












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






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












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




    $begingroup$
    @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
    $endgroup$
    – Sil
    1 hour 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 1 hour ago









        egregegreg

        185k1486206




        185k1486206






























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            How to remove border from elements in the last row?Targeting flex items on the last rowHow to vertically wrap...

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            Does the Idaho Potato Commission associate potato skins with healthy eating? Finitely generated matrix groups whose eigenvalues are all algebraic How dangerous is XSS How could indestructible materials be used in power generation? Getting extremely large arrows with tikzcd How to aggregate categorical data in R? What is this scratchy sound on the acoustic guitar called? How obscure is the use of 令 in 令和? What exactly is ineptocracy? How to travel to Japan while expressing milk? Can someone clarify Hamming's notion of important problems in relation to modern academia? ...
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            Tecnologías entrañables Índice Antecedentes Desarrollo Tecnologías Entrañables en la...

            Tecnología en la sociedad desarrollo tecnológicoalienación tecnológicaMiguel Ángel QuintanillaMarxEllulOrtega y GassetMiguel Angel QuintanillaMartín ParselisDiego LawlerDarío SandroneJosé Antonio López Cerezo Las tecnologías entrañables son un conjunto de criterios para el desarrollo tecnológico orientados a evitar la alienación tecnológica. Estos criterios pueden ser aplicados en las instancias de diseño tecnológico, y también pueden ser utilizados en la evaluación de proyectos tecnológicos. Índice 1 Antecedentes 2 Desarrollo 3 Tecnologías Entrañables en la actualidad 4 Referencias Antecedentes La idea original de las Tecnologías Entrañables fue propuesta por Miguel Ángel Quintanilla, identificando el problema como sigue: Ya hemos aludido a la paradoja que supone el hecho de que, en general, cuanto más fácil de usar es una tecnología, más incomprensible resulta para el usuario: las tecnologías transparentes exigen ciudadanos ciegos. Hay que recon...
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