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

346115




346115












  • $begingroup$
    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

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









$endgroup$





















    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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041591%2fwhen-is-x2n-1-reducible-in-mathbff-p-for-all-primes-p%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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

                  4,585210




                  4,585210






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041591%2fwhen-is-x2n-1-reducible-in-mathbff-p-for-all-primes-p%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Image
                      0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                      Read more

                      Aardman Animations

                      Image
                      Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                      Read more

                      Are they similar matrix

                      Image
                      0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                      Read more