Proving polynomial over Q is irreducible












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Prove that if $a$ and $b$ are odd then the polynomial $$x^3+ax+b$$is irreducible over $mathbb{Q}$




I would be very much thankful if someone could help me with this one.






abstract-algebra elementary-number-theory polynomials







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  • $begingroup$
    Eisenstein criterion. Its not ''linear algebra''.
    $endgroup$
    – Wuestenfux
    Dec 9 '18 at 9:42








  • 4




    $begingroup$
    reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
    $endgroup$
    – Peter
    Dec 9 '18 at 9:53


















2












$begingroup$



Prove that if $a$ and $b$ are odd then the polynomial $$x^3+ax+b$$is irreducible over $mathbb{Q}$




I would be very much thankful if someone could help me with this one.






abstract-algebra elementary-number-theory polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    Eisenstein criterion. Its not ''linear algebra''.
    $endgroup$
    – Wuestenfux
    Dec 9 '18 at 9:42








  • 4




    $begingroup$
    reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
    $endgroup$
    – Peter
    Dec 9 '18 at 9:53
















2












2








2


1



$begingroup$



Prove that if $a$ and $b$ are odd then the polynomial $$x^3+ax+b$$is irreducible over $mathbb{Q}$




I would be very much thankful if someone could help me with this one.






abstract-algebra elementary-number-theory polynomials







share|cite|improve this question











$endgroup$





Prove that if $a$ and $b$ are odd then the polynomial $$x^3+ax+b$$is irreducible over $mathbb{Q}$




I would be very much thankful if someone could help me with this one.






abstract-algebra elementary-number-theory polynomials




abstract-algebra elementary-number-theory polynomials






share|cite|improve this question















share|cite|improve this question













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edited Dec 9 '18 at 10:11









greedoid

39.2k114797




39.2k114797










asked Dec 9 '18 at 9:38









Torbjörn OlssonTorbjörn Olsson

134




134












  • $begingroup$
    Eisenstein criterion. Its not ''linear algebra''.
    $endgroup$
    – Wuestenfux
    Dec 9 '18 at 9:42








  • 4




    $begingroup$
    reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
    $endgroup$
    – Peter
    Dec 9 '18 at 9:53




















  • $begingroup$
    Eisenstein criterion. Its not ''linear algebra''.
    $endgroup$
    – Wuestenfux
    Dec 9 '18 at 9:42








  • 4




    $begingroup$
    reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
    $endgroup$
    – Peter
    Dec 9 '18 at 9:53


















$begingroup$
Eisenstein criterion. Its not ''linear algebra''.
$endgroup$
– Wuestenfux
Dec 9 '18 at 9:42






$begingroup$
Eisenstein criterion. Its not ''linear algebra''.
$endgroup$
– Wuestenfux
Dec 9 '18 at 9:42






4




4




$begingroup$
reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
$endgroup$
– Peter
Dec 9 '18 at 9:53






$begingroup$
reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $mathbb Z_2$
$endgroup$
– Peter
Dec 9 '18 at 9:53












2 Answers
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$begingroup$

Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$



where $c,d,rin mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.



Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.






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    3












    $begingroup$

    A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $mathbb Z$ and hence $mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $mathbb Z_2$,
    $$x^3+ax+b=x^3+x+1.$$
    Suppose this is reducible, then it has a root over $mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1neq0$ and $1^3+1+1=1neq0$, so neither of the only two elements in $mathbb Z_2$ are a root. We conclude that it is irreducible over $mathbb Z_2$, hence over $mathbb Q$ as well.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      2












      $begingroup$

      Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$



      where $c,d,rin mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.



      Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.






      share|cite|improve this answer









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        2












        $begingroup$

        Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$



        where $c,d,rin mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.



        Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$



          where $c,d,rin mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.



          Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.






          share|cite|improve this answer









          $endgroup$



          Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$



          where $c,d,rin mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.



          Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 10:00









          greedoidgreedoid

          39.2k114797




          39.2k114797























              3












              $begingroup$

              A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $mathbb Z$ and hence $mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $mathbb Z_2$,
              $$x^3+ax+b=x^3+x+1.$$
              Suppose this is reducible, then it has a root over $mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1neq0$ and $1^3+1+1=1neq0$, so neither of the only two elements in $mathbb Z_2$ are a root. We conclude that it is irreducible over $mathbb Z_2$, hence over $mathbb Q$ as well.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $mathbb Z$ and hence $mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $mathbb Z_2$,
                $$x^3+ax+b=x^3+x+1.$$
                Suppose this is reducible, then it has a root over $mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1neq0$ and $1^3+1+1=1neq0$, so neither of the only two elements in $mathbb Z_2$ are a root. We conclude that it is irreducible over $mathbb Z_2$, hence over $mathbb Q$ as well.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $mathbb Z$ and hence $mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $mathbb Z_2$,
                  $$x^3+ax+b=x^3+x+1.$$
                  Suppose this is reducible, then it has a root over $mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1neq0$ and $1^3+1+1=1neq0$, so neither of the only two elements in $mathbb Z_2$ are a root. We conclude that it is irreducible over $mathbb Z_2$, hence over $mathbb Q$ as well.






                  share|cite|improve this answer









                  $endgroup$



                  A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $mathbb Z$ and hence $mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $mathbb Z_2$,
                  $$x^3+ax+b=x^3+x+1.$$
                  Suppose this is reducible, then it has a root over $mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1neq0$ and $1^3+1+1=1neq0$, so neither of the only two elements in $mathbb Z_2$ are a root. We conclude that it is irreducible over $mathbb Z_2$, hence over $mathbb Q$ as well.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 12:02









                  YiFanYiFan

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