The minimal polynomial is the determinant of $xI-L_{alpha}$.












1












$begingroup$


Let $K=F(a)$ a finite field extension of $F$. For $alpha in K$, let $L_{alpha} : K to K$ be the transformation $L_{alpha} (x)=alpha x$.
Show that $L_{alpha} $ is an $F$-linear transformation and show that $det(xI-L_a) =min(a,F)$. For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$?



Here, $min(alpha, F)$ denotes the minimal polynomial of $alpha$ in $F$, that is, the polynomial with minimal degree with coefficients in $F$ that has $alpha$ as a root.



It's clear that $L_{alpha}$ is a linear function. Now, I don't know how to manage the rest of the problem. I know that the basis of $K$ as an $F$-linear space is ${1,a,...,a^{n-1} }$, where $n$ is the degree of $min(a,F)$. But then I don't know what to do.



Any help will be very appreciated. Thank you so much!






field-theory galois-theory extension-field minimal-polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    A more general and interesting question is when the extension K is not of type F(a). See e.g. math.stackexchange.com/a/2710883/300700
    $endgroup$
    – nguyen quang do
    Dec 21 '18 at 17:00
















1












$begingroup$


Let $K=F(a)$ a finite field extension of $F$. For $alpha in K$, let $L_{alpha} : K to K$ be the transformation $L_{alpha} (x)=alpha x$.
Show that $L_{alpha} $ is an $F$-linear transformation and show that $det(xI-L_a) =min(a,F)$. For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$?



Here, $min(alpha, F)$ denotes the minimal polynomial of $alpha$ in $F$, that is, the polynomial with minimal degree with coefficients in $F$ that has $alpha$ as a root.



It's clear that $L_{alpha}$ is a linear function. Now, I don't know how to manage the rest of the problem. I know that the basis of $K$ as an $F$-linear space is ${1,a,...,a^{n-1} }$, where $n$ is the degree of $min(a,F)$. But then I don't know what to do.



Any help will be very appreciated. Thank you so much!






field-theory galois-theory extension-field minimal-polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    A more general and interesting question is when the extension K is not of type F(a). See e.g. math.stackexchange.com/a/2710883/300700
    $endgroup$
    – nguyen quang do
    Dec 21 '18 at 17:00














1












1








1


1



$begingroup$


Let $K=F(a)$ a finite field extension of $F$. For $alpha in K$, let $L_{alpha} : K to K$ be the transformation $L_{alpha} (x)=alpha x$.
Show that $L_{alpha} $ is an $F$-linear transformation and show that $det(xI-L_a) =min(a,F)$. For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$?



Here, $min(alpha, F)$ denotes the minimal polynomial of $alpha$ in $F$, that is, the polynomial with minimal degree with coefficients in $F$ that has $alpha$ as a root.



It's clear that $L_{alpha}$ is a linear function. Now, I don't know how to manage the rest of the problem. I know that the basis of $K$ as an $F$-linear space is ${1,a,...,a^{n-1} }$, where $n$ is the degree of $min(a,F)$. But then I don't know what to do.



Any help will be very appreciated. Thank you so much!






field-theory galois-theory extension-field minimal-polynomials







share|cite|improve this question









$endgroup$




Let $K=F(a)$ a finite field extension of $F$. For $alpha in K$, let $L_{alpha} : K to K$ be the transformation $L_{alpha} (x)=alpha x$.
Show that $L_{alpha} $ is an $F$-linear transformation and show that $det(xI-L_a) =min(a,F)$. For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$?



Here, $min(alpha, F)$ denotes the minimal polynomial of $alpha$ in $F$, that is, the polynomial with minimal degree with coefficients in $F$ that has $alpha$ as a root.



It's clear that $L_{alpha}$ is a linear function. Now, I don't know how to manage the rest of the problem. I know that the basis of $K$ as an $F$-linear space is ${1,a,...,a^{n-1} }$, where $n$ is the degree of $min(a,F)$. But then I don't know what to do.



Any help will be very appreciated. Thank you so much!






field-theory galois-theory extension-field minimal-polynomials




field-theory galois-theory extension-field minimal-polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 21 '18 at 1:17









user392559user392559

38118




38118












  • $begingroup$
    A more general and interesting question is when the extension K is not of type F(a). See e.g. math.stackexchange.com/a/2710883/300700
    $endgroup$
    – nguyen quang do
    Dec 21 '18 at 17:00


















  • $begingroup$
    A more general and interesting question is when the extension K is not of type F(a). See e.g. math.stackexchange.com/a/2710883/300700
    $endgroup$
    – nguyen quang do
    Dec 21 '18 at 17:00
















$begingroup$
A more general and interesting question is when the extension K is not of type F(a). See e.g. math.stackexchange.com/a/2710883/300700
$endgroup$
– nguyen quang do
Dec 21 '18 at 17:00




$begingroup$
A more general and interesting question is when the extension K is not of type F(a). See e.g. math.stackexchange.com/a/2710883/300700
$endgroup$
– nguyen quang do
Dec 21 '18 at 17:00










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $P=det(xI-L_alpha)$ Cayley Hamilton implies that $P(L_alpha)=0$ this implies that $P(L_alpha)(1)=P(alpha)=0$. Since $deg P=[F(alpha):F]$ we deduce that $P$ is the minimal polynomial of $alpha$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! This shows that $det(xI-L_a) =min(a,F)$.Now I'm having trouble with the general case: "For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$? " It would be great if you could help me with this. Thanks again!
    $endgroup$
    – user392559
    Dec 21 '18 at 21:28





















0












$begingroup$

Let $p(x) = det(xI - L_alpha)$. $p(x)$ is a monic polynomial of degree $n$, and by the Cayley-Hamilton theorem we have $p(L_alpha) = 0$. However, if there were a polynomial $q$ of degree $d$ strictly smaller than $n$ such that $q(alpha) = 0$, then we would conclude that the elements ${1, alpha, dots, alpha^{d}}$ fail to be linearly independent.






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%2f3048104%2fthe-minimal-polynomial-is-the-determinant-of-xi-l-alpha%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












    $begingroup$

    Let $P=det(xI-L_alpha)$ Cayley Hamilton implies that $P(L_alpha)=0$ this implies that $P(L_alpha)(1)=P(alpha)=0$. Since $deg P=[F(alpha):F]$ we deduce that $P$ is the minimal polynomial of $alpha$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you! This shows that $det(xI-L_a) =min(a,F)$.Now I'm having trouble with the general case: "For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$? " It would be great if you could help me with this. Thanks again!
      $endgroup$
      – user392559
      Dec 21 '18 at 21:28


















    1












    $begingroup$

    Let $P=det(xI-L_alpha)$ Cayley Hamilton implies that $P(L_alpha)=0$ this implies that $P(L_alpha)(1)=P(alpha)=0$. Since $deg P=[F(alpha):F]$ we deduce that $P$ is the minimal polynomial of $alpha$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you! This shows that $det(xI-L_a) =min(a,F)$.Now I'm having trouble with the general case: "For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$? " It would be great if you could help me with this. Thanks again!
      $endgroup$
      – user392559
      Dec 21 '18 at 21:28
















    1












    1








    1





    $begingroup$

    Let $P=det(xI-L_alpha)$ Cayley Hamilton implies that $P(L_alpha)=0$ this implies that $P(L_alpha)(1)=P(alpha)=0$. Since $deg P=[F(alpha):F]$ we deduce that $P$ is the minimal polynomial of $alpha$.






    share|cite|improve this answer









    $endgroup$



    Let $P=det(xI-L_alpha)$ Cayley Hamilton implies that $P(L_alpha)=0$ this implies that $P(L_alpha)(1)=P(alpha)=0$. Since $deg P=[F(alpha):F]$ we deduce that $P$ is the minimal polynomial of $alpha$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 21 '18 at 1:25









    Tsemo AristideTsemo Aristide

    59.9k11446




    59.9k11446












    • $begingroup$
      Thank you! This shows that $det(xI-L_a) =min(a,F)$.Now I'm having trouble with the general case: "For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$? " It would be great if you could help me with this. Thanks again!
      $endgroup$
      – user392559
      Dec 21 '18 at 21:28




















    • $begingroup$
      Thank you! This shows that $det(xI-L_a) =min(a,F)$.Now I'm having trouble with the general case: "For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$? " It would be great if you could help me with this. Thanks again!
      $endgroup$
      – user392559
      Dec 21 '18 at 21:28


















    $begingroup$
    Thank you! This shows that $det(xI-L_a) =min(a,F)$.Now I'm having trouble with the general case: "For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$? " It would be great if you could help me with this. Thanks again!
    $endgroup$
    – user392559
    Dec 21 '18 at 21:28






    $begingroup$
    Thank you! This shows that $det(xI-L_a) =min(a,F)$.Now I'm having trouble with the general case: "For which $alpha in K$ do we have that $det(xI-L_{alpha})= min(alpha, F)$? " It would be great if you could help me with this. Thanks again!
    $endgroup$
    – user392559
    Dec 21 '18 at 21:28













    0












    $begingroup$

    Let $p(x) = det(xI - L_alpha)$. $p(x)$ is a monic polynomial of degree $n$, and by the Cayley-Hamilton theorem we have $p(L_alpha) = 0$. However, if there were a polynomial $q$ of degree $d$ strictly smaller than $n$ such that $q(alpha) = 0$, then we would conclude that the elements ${1, alpha, dots, alpha^{d}}$ fail to be linearly independent.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let $p(x) = det(xI - L_alpha)$. $p(x)$ is a monic polynomial of degree $n$, and by the Cayley-Hamilton theorem we have $p(L_alpha) = 0$. However, if there were a polynomial $q$ of degree $d$ strictly smaller than $n$ such that $q(alpha) = 0$, then we would conclude that the elements ${1, alpha, dots, alpha^{d}}$ fail to be linearly independent.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $p(x) = det(xI - L_alpha)$. $p(x)$ is a monic polynomial of degree $n$, and by the Cayley-Hamilton theorem we have $p(L_alpha) = 0$. However, if there were a polynomial $q$ of degree $d$ strictly smaller than $n$ such that $q(alpha) = 0$, then we would conclude that the elements ${1, alpha, dots, alpha^{d}}$ fail to be linearly independent.






        share|cite|improve this answer









        $endgroup$



        Let $p(x) = det(xI - L_alpha)$. $p(x)$ is a monic polynomial of degree $n$, and by the Cayley-Hamilton theorem we have $p(L_alpha) = 0$. However, if there were a polynomial $q$ of degree $d$ strictly smaller than $n$ such that $q(alpha) = 0$, then we would conclude that the elements ${1, alpha, dots, alpha^{d}}$ fail to be linearly independent.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 1:28









        OmnomnomnomOmnomnomnom

        129k792185




        129k792185






























            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%2f3048104%2fthe-minimal-polynomial-is-the-determinant-of-xi-l-alpha%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

            Le Mesnil-Réaume

            Image
            Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more

            web3.py web3.isConnected() returns false always

            Image
            1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
            Read more