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









$endgroup$













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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

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    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






























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