Irreducibility of $x^4 + x^3 + 1$ over finite field $mathbb{F}_{2^{a}}$, $1 leq a leq 6$












0














I have to discuss the irreducibility of $P(X) = X^4 + X^3 + 1$ over finite field $mathbb{F}_{2^{a}}$, $1 leq a leq 6$.



So, for $a = 1$, we have that $P$ is irreducible since is has no roots in $mathbb{F}_2$ and not the square of the only irreducible quadratic polynomial over $mathbb{F}_2$ which is $X^2 + X + 1$



And if $P$ were reducible over $mathbb{F}_{2^b}$, where $b$ is odd $< 6$, then each of its roots will generate $mathbb{F}_{2^4}$, but this is a contradiction as $mathbb{F}_{p^k}$ is a subfield of $mathbb{F}_{p^k}$ if and only if $k | n$; this is clearly impossible, since $b$ is odd.



Therefore, for $a = 3,5$ i.e. over $mathbb{F}_{8}$ and $mathbb{F}_{32} P$ is irreducible; and it follows from the above discussion that $P$ is reducible over $mathbb{F}_{16}$, since a root of this polynomial generates the field.



Now, if we let $mathbb{F}_4 = {0, 1, a, a+1 vert a^2 + a + 1=0}$ then we see that $P(X) = (X^2 + aX +a)(X^2 + (a+1)X + (a+1))$, hence $P$ is reducible in $mathbb{F}_4$



Which leaves irreducibility of $P$ over $mathbb{F}_{64}$. It is here that I'm drawing a blank. Any help is appreciated!






finite-fields irreducible-polynomials







share|cite|improve this question
























  • "...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
    – DonAntonio
    Nov 19 at 0:53










  • Ah, a silly mistake on my part. Corrected
    – Naweed G. Seldon
    Nov 19 at 0:55
















0














I have to discuss the irreducibility of $P(X) = X^4 + X^3 + 1$ over finite field $mathbb{F}_{2^{a}}$, $1 leq a leq 6$.



So, for $a = 1$, we have that $P$ is irreducible since is has no roots in $mathbb{F}_2$ and not the square of the only irreducible quadratic polynomial over $mathbb{F}_2$ which is $X^2 + X + 1$



And if $P$ were reducible over $mathbb{F}_{2^b}$, where $b$ is odd $< 6$, then each of its roots will generate $mathbb{F}_{2^4}$, but this is a contradiction as $mathbb{F}_{p^k}$ is a subfield of $mathbb{F}_{p^k}$ if and only if $k | n$; this is clearly impossible, since $b$ is odd.



Therefore, for $a = 3,5$ i.e. over $mathbb{F}_{8}$ and $mathbb{F}_{32} P$ is irreducible; and it follows from the above discussion that $P$ is reducible over $mathbb{F}_{16}$, since a root of this polynomial generates the field.



Now, if we let $mathbb{F}_4 = {0, 1, a, a+1 vert a^2 + a + 1=0}$ then we see that $P(X) = (X^2 + aX +a)(X^2 + (a+1)X + (a+1))$, hence $P$ is reducible in $mathbb{F}_4$



Which leaves irreducibility of $P$ over $mathbb{F}_{64}$. It is here that I'm drawing a blank. Any help is appreciated!






finite-fields irreducible-polynomials







share|cite|improve this question
























  • "...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
    – DonAntonio
    Nov 19 at 0:53










  • Ah, a silly mistake on my part. Corrected
    – Naweed G. Seldon
    Nov 19 at 0:55














0












0








0







I have to discuss the irreducibility of $P(X) = X^4 + X^3 + 1$ over finite field $mathbb{F}_{2^{a}}$, $1 leq a leq 6$.



So, for $a = 1$, we have that $P$ is irreducible since is has no roots in $mathbb{F}_2$ and not the square of the only irreducible quadratic polynomial over $mathbb{F}_2$ which is $X^2 + X + 1$



And if $P$ were reducible over $mathbb{F}_{2^b}$, where $b$ is odd $< 6$, then each of its roots will generate $mathbb{F}_{2^4}$, but this is a contradiction as $mathbb{F}_{p^k}$ is a subfield of $mathbb{F}_{p^k}$ if and only if $k | n$; this is clearly impossible, since $b$ is odd.



Therefore, for $a = 3,5$ i.e. over $mathbb{F}_{8}$ and $mathbb{F}_{32} P$ is irreducible; and it follows from the above discussion that $P$ is reducible over $mathbb{F}_{16}$, since a root of this polynomial generates the field.



Now, if we let $mathbb{F}_4 = {0, 1, a, a+1 vert a^2 + a + 1=0}$ then we see that $P(X) = (X^2 + aX +a)(X^2 + (a+1)X + (a+1))$, hence $P$ is reducible in $mathbb{F}_4$



Which leaves irreducibility of $P$ over $mathbb{F}_{64}$. It is here that I'm drawing a blank. Any help is appreciated!






finite-fields irreducible-polynomials







share|cite|improve this question















I have to discuss the irreducibility of $P(X) = X^4 + X^3 + 1$ over finite field $mathbb{F}_{2^{a}}$, $1 leq a leq 6$.



So, for $a = 1$, we have that $P$ is irreducible since is has no roots in $mathbb{F}_2$ and not the square of the only irreducible quadratic polynomial over $mathbb{F}_2$ which is $X^2 + X + 1$



And if $P$ were reducible over $mathbb{F}_{2^b}$, where $b$ is odd $< 6$, then each of its roots will generate $mathbb{F}_{2^4}$, but this is a contradiction as $mathbb{F}_{p^k}$ is a subfield of $mathbb{F}_{p^k}$ if and only if $k | n$; this is clearly impossible, since $b$ is odd.



Therefore, for $a = 3,5$ i.e. over $mathbb{F}_{8}$ and $mathbb{F}_{32} P$ is irreducible; and it follows from the above discussion that $P$ is reducible over $mathbb{F}_{16}$, since a root of this polynomial generates the field.



Now, if we let $mathbb{F}_4 = {0, 1, a, a+1 vert a^2 + a + 1=0}$ then we see that $P(X) = (X^2 + aX +a)(X^2 + (a+1)X + (a+1))$, hence $P$ is reducible in $mathbb{F}_4$



Which leaves irreducibility of $P$ over $mathbb{F}_{64}$. It is here that I'm drawing a blank. Any help is appreciated!






finite-fields irreducible-polynomials




finite-fields irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 at 1:21

























asked Nov 19 at 0:50









Naweed G. Seldon

1,299419




1,299419












  • "...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
    – DonAntonio
    Nov 19 at 0:53










  • Ah, a silly mistake on my part. Corrected
    – Naweed G. Seldon
    Nov 19 at 0:55


















  • "...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
    – DonAntonio
    Nov 19 at 0:53










  • Ah, a silly mistake on my part. Corrected
    – Naweed G. Seldon
    Nov 19 at 0:55
















"...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
– DonAntonio
Nov 19 at 0:53




"...the only quadratic irreducible polynomial..." ...and then you wrote down a cubic ...
– DonAntonio
Nov 19 at 0:53












Ah, a silly mistake on my part. Corrected
– Naweed G. Seldon
Nov 19 at 0:55




Ah, a silly mistake on my part. Corrected
– Naweed G. Seldon
Nov 19 at 0:55










2 Answers
2






active

oldest

votes


















1














$P$ is reducible over $mathbb F_{64}$, because reducible over the subfield $mathbb F_4subset mathbb F_{64}$ ($2mid6$).






share|cite|improve this answer





















  • Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
    – Naweed G. Seldon
    Nov 19 at 2:05






  • 1




    Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
    – Chris Custer
    Nov 19 at 2:28



















0














$Bbb{F}_{64}not supsetBbb{F}_{16}$ which you know is the splitting field of $P$, but what about $Bbb{F}_{4096}$? It's Galois over $Bbb{F}_{64}$ and contains $Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $Bbb{F}_{64}$? What does that tell you about $P$ over $Bbb{F}_{64}$?






share|cite|improve this answer























    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%2f3004358%2firreducibility-of-x4-x3-1-over-finite-field-mathbbf-2a-1-le%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









    1














    $P$ is reducible over $mathbb F_{64}$, because reducible over the subfield $mathbb F_4subset mathbb F_{64}$ ($2mid6$).






    share|cite|improve this answer





















    • Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
      – Naweed G. Seldon
      Nov 19 at 2:05






    • 1




      Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
      – Chris Custer
      Nov 19 at 2:28
















    1














    $P$ is reducible over $mathbb F_{64}$, because reducible over the subfield $mathbb F_4subset mathbb F_{64}$ ($2mid6$).






    share|cite|improve this answer





















    • Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
      – Naweed G. Seldon
      Nov 19 at 2:05






    • 1




      Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
      – Chris Custer
      Nov 19 at 2:28














    1












    1








    1






    $P$ is reducible over $mathbb F_{64}$, because reducible over the subfield $mathbb F_4subset mathbb F_{64}$ ($2mid6$).






    share|cite|improve this answer












    $P$ is reducible over $mathbb F_{64}$, because reducible over the subfield $mathbb F_4subset mathbb F_{64}$ ($2mid6$).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 19 at 1:54









    Chris Custer

    10.8k3724




    10.8k3724












    • Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
      – Naweed G. Seldon
      Nov 19 at 2:05






    • 1




      Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
      – Chris Custer
      Nov 19 at 2:28


















    • Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
      – Naweed G. Seldon
      Nov 19 at 2:05






    • 1




      Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
      – Chris Custer
      Nov 19 at 2:28
















    Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
    – Naweed G. Seldon
    Nov 19 at 2:05




    Does it also follow that $P$ has no roots over $mathbb{F}_{64}$
    – Naweed G. Seldon
    Nov 19 at 2:05




    1




    1




    Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
    – Chris Custer
    Nov 19 at 2:28




    Such a root would generate $mathbb F_{16}$, right? But $4notmid6$.
    – Chris Custer
    Nov 19 at 2:28











    0














    $Bbb{F}_{64}not supsetBbb{F}_{16}$ which you know is the splitting field of $P$, but what about $Bbb{F}_{4096}$? It's Galois over $Bbb{F}_{64}$ and contains $Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $Bbb{F}_{64}$? What does that tell you about $P$ over $Bbb{F}_{64}$?






    share|cite|improve this answer




























      0














      $Bbb{F}_{64}not supsetBbb{F}_{16}$ which you know is the splitting field of $P$, but what about $Bbb{F}_{4096}$? It's Galois over $Bbb{F}_{64}$ and contains $Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $Bbb{F}_{64}$? What does that tell you about $P$ over $Bbb{F}_{64}$?






      share|cite|improve this answer


























        0












        0








        0






        $Bbb{F}_{64}not supsetBbb{F}_{16}$ which you know is the splitting field of $P$, but what about $Bbb{F}_{4096}$? It's Galois over $Bbb{F}_{64}$ and contains $Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $Bbb{F}_{64}$? What does that tell you about $P$ over $Bbb{F}_{64}$?






        share|cite|improve this answer














        $Bbb{F}_{64}not supsetBbb{F}_{16}$ which you know is the splitting field of $P$, but what about $Bbb{F}_{4096}$? It's Galois over $Bbb{F}_{64}$ and contains $Bbb{F}_{16}$, the splitting field of $P$. What's it's degree over $Bbb{F}_{64}$? What does that tell you about $P$ over $Bbb{F}_{64}$?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 19 at 2:49

























        answered Nov 19 at 1:53









        sharding4

        4,0961823




        4,0961823






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3004358%2firreducibility-of-x4-x3-1-over-finite-field-mathbbf-2a-1-le%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

            AnyDesk - Fatal Program Failure

            Image
            up vote 0 down vote favorite Good day all. I have a window that pops up saying Fatal Program Failure (AnyDesk). This is what I have done so far: Uninstalled AnyDesk and reinstalled it. ( The window still appeared ) I disabled AnyDesk start up in Msconfig and the window seems to have disapeared. Lastly I made sure that AnyDesk worked by remoting into it, and it worked. Today the same window appeared. How do I get rid of this window? remote-desktop share | improve this question edited 10 hours ago Tomasz Jakub Rup 552 3 13 ...
            Read more

            How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

            Image
            up vote 0 down vote favorite I need help with an issue we're facing in a project where we are moving from a Windows-based POS environment to one running on CentOS7. The problem consists of a number of factors, please find the details below: The supplier of our POS application will only deliver an application that is hard-coded to run in a 1024:768 4:3 format. The hardware we are using is a POS unit from HP that has a built-in 16:9 touch-screen. The employees using this solution is not supposed to see nor have access to anything but the POS application. Our sales organization demands that touch is an available feature when going live with the new POS environment. Going further into this, allow me to explain what we have encountered on the way trying to achieve the goal of having a touch-screen: We've simply change...
            Read more

            QoS: MAC-Priority for clients behind a repeater

            Image
            1 1 I've setup a wireless network using a Linksys-Router (DD-WRT) as an AP and a TP-Link repeater for range extension. To manage bandwith for specific users, I've setup QoS-Rules on the AP (DD-WRT) using MAC-Priority: However the AP shows only the Repeater's MAC and the MACs of users inside the AP-Range. The MAC addresses of users behind the Repeater are not listed : The purple MAC is not listed in the AP-Clientlist. Its related to a device connected via the TP-Link repeater. Q: How can I setup MAC-Priority for repeater-clients? Do I simply add the purple MAC in AP-MAC priority list although it's not listed as a client? I want to setup specific priority instead of giving one priority for all clients behind the repeater by adding the repeater's MAC in the MAC priority list in the AP as ...
            Read more