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
    13 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
    13 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 13 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
    13 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
    13 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
13 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
13 hours ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

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









$endgroup$





















    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
      13 hours ago










    • $begingroup$
      @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
      $endgroup$
      – Eureka
      13 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
      13 hours ago










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











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    3












    $begingroup$

    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









    $endgroup$


















      3












      $begingroup$

      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









      $endgroup$
















        3












        3








        3





        $begingroup$

        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 13 hours ago









        egregegreg

        185k1486206




        185k1486206























            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
              13 hours ago










            • $begingroup$
              @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
              $endgroup$
              – Eureka
              13 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
              13 hours ago










            • $begingroup$
              @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
              $endgroup$
              – Sil
              13 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
              13 hours ago










            • $begingroup$
              @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
              $endgroup$
              – Eureka
              13 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
              13 hours ago










            • $begingroup$
              @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
              $endgroup$
              – Sil
              13 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 13 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
              13 hours ago










            • $begingroup$
              @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
              $endgroup$
              – Eureka
              13 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
              13 hours ago










            • $begingroup$
              @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
              $endgroup$
              – Sil
              13 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
              13 hours ago










            • $begingroup$
              @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
              $endgroup$
              – Eureka
              13 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
              13 hours ago










            • $begingroup$
              @EthanMacBrough I understand, my point though was that things like that should be in answer itself.
              $endgroup$
              – Sil
              13 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
            13 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
            13 hours ago












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






            $begingroup$
            @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
            $endgroup$
            – Eureka
            13 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
            13 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
            13 hours ago












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




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


















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            Bunad

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            Tradisjonelle folkedrakter frå Aust-Telemark, truleg frå 1880-åra. Drakter som dette er grunnlaget for moderne bunadar. Foto: Axel Lindahl Bunad frå Eggedal i Buskerud fotografert i 1906. Bunad (frå kledebunad , der norrønt «búnaðr» har same rot som (føre)bu) er ei samnemning brukt om klesdrakter i tradisjonell nordisk stil — ofte utvikla frå folkedrakter, men likevel i ein eigen tradisjon. Ein bunad er typisk ei forseggjort og kostbar drakt, med mykje broderi, og med handsmidde smykke i gull eller sølv, særleg dei særeigne søljene, som obligatorisk prynad i tillegg. Bunad og bunadbruk er eit debattemne med mange ulike oppfatningar, til dømes om kva som er ein bunad, og om ein kan endra på utforminga av draktene eller ikkje. Den norske bunadrørsla går tilbake til folkedansrørsla og føregangsfolk som Hulda Garborg og Klara Semb. I dag blir bunad brukt som nasjonaldrakt. Festbunad er likestilt med galla-antrekk, og ofte nytta ved særskilde høve, som konfirmasjonar eller d
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            Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

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            FæreyjarHestar færeyskahjaltlandshestinumíslenski hesturinnBretlandskolanámumenskuWikipediafæreyskuWikipediaRossið - síða um færeyska hrossið á færeysku Færeyskur hestur Úr Wikipediu, frjálsa alfræðiritinu Jump to navigation Jump to search Rani og Grani. Grani í vetrarfeldi. Færeyski hesturinn (færeyska: Føroyska rossið ) er hestur sem hefur lifað í Færeyjum í hundruði ára. Hann er smágerður, 120-132 cm á hæð, og helst skyldur hjaltlandshestinum. Hlutverk hans var að draga vagna og plóg og bera klyfjar. Hann hefur fjórar gangtegundir eins og íslenski hesturinn (þar með talið tölt) og er með fjölda litaafbrigði. Nú til dags er hann notaður til frístunda. Í lok 19. aldar voru til um 800 hross. Mörg þeirra voru seld til Bretlands til að vinna í kolanámum og um 1960 voru aðeins nokkrir hestar eftir. Átak var gert í að varðveita hestinn og eru nú til 70-80 hross í Færeyjum. [1] Heimild |   Wikimedia Commons er með margmiðlunarefni sem te
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