When is $x^{2^n} + 1$ Reducible in $mathbf{F}_p$ For All Primes $p$












1












$begingroup$


Define $f_n = x^{2^n} + 1$.



Then we want to show that there is an integer $n$ such that $f_n$ is reducible in $mathbb{F}_p[x]$ for all primes, $p$.



However, I want to do this using the hint




The group of units modulo $2^r$, $(mathbb{Z}/2^rmathbb{Z})^*$, is not cyclic for $r ge 3 $.





Indeed, there are ways to show this without the hint - one such method forms the basis of this answer.



However, I just can't seem to figure out how to use the hint! Hopefully using the hint will also shine some light on the question of for which $n$ does the above hold?






galois-theory finite-fields cyclic-groups splitting-field







share|cite|improve this question









$endgroup$












  • $begingroup$
    I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 6:30












  • $begingroup$
    Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 6:41
















1












$begingroup$


Define $f_n = x^{2^n} + 1$.



Then we want to show that there is an integer $n$ such that $f_n$ is reducible in $mathbb{F}_p[x]$ for all primes, $p$.



However, I want to do this using the hint




The group of units modulo $2^r$, $(mathbb{Z}/2^rmathbb{Z})^*$, is not cyclic for $r ge 3 $.





Indeed, there are ways to show this without the hint - one such method forms the basis of this answer.



However, I just can't seem to figure out how to use the hint! Hopefully using the hint will also shine some light on the question of for which $n$ does the above hold?






galois-theory finite-fields cyclic-groups splitting-field







share|cite|improve this question









$endgroup$












  • $begingroup$
    I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 6:30












  • $begingroup$
    Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 6:41














1












1








1


2



$begingroup$


Define $f_n = x^{2^n} + 1$.



Then we want to show that there is an integer $n$ such that $f_n$ is reducible in $mathbb{F}_p[x]$ for all primes, $p$.



However, I want to do this using the hint




The group of units modulo $2^r$, $(mathbb{Z}/2^rmathbb{Z})^*$, is not cyclic for $r ge 3 $.





Indeed, there are ways to show this without the hint - one such method forms the basis of this answer.



However, I just can't seem to figure out how to use the hint! Hopefully using the hint will also shine some light on the question of for which $n$ does the above hold?






galois-theory finite-fields cyclic-groups splitting-field







share|cite|improve this question









$endgroup$




Define $f_n = x^{2^n} + 1$.



Then we want to show that there is an integer $n$ such that $f_n$ is reducible in $mathbb{F}_p[x]$ for all primes, $p$.



However, I want to do this using the hint




The group of units modulo $2^r$, $(mathbb{Z}/2^rmathbb{Z})^*$, is not cyclic for $r ge 3 $.





Indeed, there are ways to show this without the hint - one such method forms the basis of this answer.



However, I just can't seem to figure out how to use the hint! Hopefully using the hint will also shine some light on the question of for which $n$ does the above hold?






galois-theory finite-fields cyclic-groups splitting-field




galois-theory finite-fields cyclic-groups splitting-field






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 15:14









John DonJohn Don

337115




337115












  • $begingroup$
    I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 6:30












  • $begingroup$
    Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 6:41


















  • $begingroup$
    I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 6:30












  • $begingroup$
    Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 6:41
















$begingroup$
I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:30






$begingroup$
I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:30














$begingroup$
Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:41




$begingroup$
Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:41










2 Answers
2






active

oldest

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2












$begingroup$

It's always reducible modulo $2$.



Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity.
Let $alpha$ be any of them. Then $Bbb F_p(alpha)=Bbb F_{p^k}$
where $k$ is the least positive integer with $alpha^{p^k}=alpha$.
That's equivalent to $p^kequiv1pmod{2^{m+1}}$. If $k<2^m$, then the
degree of $Bbb F_p(alpha)$ over $Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$
cannot be irreducible over $Bbb F_p$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Assume $f_n$ is irreducible in some $K=mathbb{F}_p$, where $p$ is odd. Then let $omega in L=mathbb{F}_q$ a root of $f_n$ in some field extension.



    Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.



    When $p=2$, then $f_n=(X+1)^{2^n}$.






    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









      2












      $begingroup$

      It's always reducible modulo $2$.



      Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity.
      Let $alpha$ be any of them. Then $Bbb F_p(alpha)=Bbb F_{p^k}$
      where $k$ is the least positive integer with $alpha^{p^k}=alpha$.
      That's equivalent to $p^kequiv1pmod{2^{m+1}}$. If $k<2^m$, then the
      degree of $Bbb F_p(alpha)$ over $Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$
      cannot be irreducible over $Bbb F_p$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        It's always reducible modulo $2$.



        Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity.
        Let $alpha$ be any of them. Then $Bbb F_p(alpha)=Bbb F_{p^k}$
        where $k$ is the least positive integer with $alpha^{p^k}=alpha$.
        That's equivalent to $p^kequiv1pmod{2^{m+1}}$. If $k<2^m$, then the
        degree of $Bbb F_p(alpha)$ over $Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$
        cannot be irreducible over $Bbb F_p$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          It's always reducible modulo $2$.



          Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity.
          Let $alpha$ be any of them. Then $Bbb F_p(alpha)=Bbb F_{p^k}$
          where $k$ is the least positive integer with $alpha^{p^k}=alpha$.
          That's equivalent to $p^kequiv1pmod{2^{m+1}}$. If $k<2^m$, then the
          degree of $Bbb F_p(alpha)$ over $Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$
          cannot be irreducible over $Bbb F_p$.






          share|cite|improve this answer









          $endgroup$



          It's always reducible modulo $2$.



          Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity.
          Let $alpha$ be any of them. Then $Bbb F_p(alpha)=Bbb F_{p^k}$
          where $k$ is the least positive integer with $alpha^{p^k}=alpha$.
          That's equivalent to $p^kequiv1pmod{2^{m+1}}$. If $k<2^m$, then the
          degree of $Bbb F_p(alpha)$ over $Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$
          cannot be irreducible over $Bbb F_p$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 15:28









          Lord Shark the UnknownLord Shark the Unknown

          104k1160132




          104k1160132























              1












              $begingroup$

              Assume $f_n$ is irreducible in some $K=mathbb{F}_p$, where $p$ is odd. Then let $omega in L=mathbb{F}_q$ a root of $f_n$ in some field extension.



              Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.



              When $p=2$, then $f_n=(X+1)^{2^n}$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Assume $f_n$ is irreducible in some $K=mathbb{F}_p$, where $p$ is odd. Then let $omega in L=mathbb{F}_q$ a root of $f_n$ in some field extension.



                Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.



                When $p=2$, then $f_n=(X+1)^{2^n}$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Assume $f_n$ is irreducible in some $K=mathbb{F}_p$, where $p$ is odd. Then let $omega in L=mathbb{F}_q$ a root of $f_n$ in some field extension.



                  Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.



                  When $p=2$, then $f_n=(X+1)^{2^n}$.






                  share|cite|improve this answer









                  $endgroup$



                  Assume $f_n$ is irreducible in some $K=mathbb{F}_p$, where $p$ is odd. Then let $omega in L=mathbb{F}_q$ a root of $f_n$ in some field extension.



                  Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.



                  When $p=2$, then $f_n=(X+1)^{2^n}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 15:31









                  MindlackMindlack

                  3,86018




                  3,86018






























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