Algebraic equation (find $b$ and $c$)












3












$begingroup$


The goal is, given the field extension $,mathbb{Q}subsetmathbb{C}$, to find the minimal polynomial for the element
$$eta=cosleft(frac{2pi}{5}right)$$




I define the element
$$xi=cosleft(frac{2pi}{5}right)+isinleft(frac{2pi}{5}right)$$
and due to $,left|xiright|=1$, I rewrite $,eta,$ as
$$eta=frac{xi+overline{xi}}{2}=frac{1}{2}left(xi+frac{1}{xi}right)$$
The next step is to find $,b,cinmathbb{Q},$ such that $,fleft(etaright)=0$, where $,f(x)=x^2+bx+cinmathbb{Q}left[Xright].$ So, I am trying to solve (unsuccessfully) the following equation
$$frac{1}{4}left(xi+frac{1}{xi}right)^2+frac{b}{2}left(xi+frac{1}{xi}right)+c=0$$



I know the solution is $,b=frac{1}{2},,c=-frac{1}{4},$ but when it comes to obtaining those values myself, I fail again and again. HELP!






abstract-algebra cryptography







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$endgroup$












  • $begingroup$
    No finite fields in site. Removing the tag.
    $endgroup$
    – Jyrki Lahtonen
    Dec 31 '18 at 12:09
















3












$begingroup$


The goal is, given the field extension $,mathbb{Q}subsetmathbb{C}$, to find the minimal polynomial for the element
$$eta=cosleft(frac{2pi}{5}right)$$




I define the element
$$xi=cosleft(frac{2pi}{5}right)+isinleft(frac{2pi}{5}right)$$
and due to $,left|xiright|=1$, I rewrite $,eta,$ as
$$eta=frac{xi+overline{xi}}{2}=frac{1}{2}left(xi+frac{1}{xi}right)$$
The next step is to find $,b,cinmathbb{Q},$ such that $,fleft(etaright)=0$, where $,f(x)=x^2+bx+cinmathbb{Q}left[Xright].$ So, I am trying to solve (unsuccessfully) the following equation
$$frac{1}{4}left(xi+frac{1}{xi}right)^2+frac{b}{2}left(xi+frac{1}{xi}right)+c=0$$



I know the solution is $,b=frac{1}{2},,c=-frac{1}{4},$ but when it comes to obtaining those values myself, I fail again and again. HELP!






abstract-algebra cryptography







share|cite|improve this question











$endgroup$












  • $begingroup$
    No finite fields in site. Removing the tag.
    $endgroup$
    – Jyrki Lahtonen
    Dec 31 '18 at 12:09














3












3








3





$begingroup$


The goal is, given the field extension $,mathbb{Q}subsetmathbb{C}$, to find the minimal polynomial for the element
$$eta=cosleft(frac{2pi}{5}right)$$




I define the element
$$xi=cosleft(frac{2pi}{5}right)+isinleft(frac{2pi}{5}right)$$
and due to $,left|xiright|=1$, I rewrite $,eta,$ as
$$eta=frac{xi+overline{xi}}{2}=frac{1}{2}left(xi+frac{1}{xi}right)$$
The next step is to find $,b,cinmathbb{Q},$ such that $,fleft(etaright)=0$, where $,f(x)=x^2+bx+cinmathbb{Q}left[Xright].$ So, I am trying to solve (unsuccessfully) the following equation
$$frac{1}{4}left(xi+frac{1}{xi}right)^2+frac{b}{2}left(xi+frac{1}{xi}right)+c=0$$



I know the solution is $,b=frac{1}{2},,c=-frac{1}{4},$ but when it comes to obtaining those values myself, I fail again and again. HELP!






abstract-algebra cryptography







share|cite|improve this question











$endgroup$




The goal is, given the field extension $,mathbb{Q}subsetmathbb{C}$, to find the minimal polynomial for the element
$$eta=cosleft(frac{2pi}{5}right)$$




I define the element
$$xi=cosleft(frac{2pi}{5}right)+isinleft(frac{2pi}{5}right)$$
and due to $,left|xiright|=1$, I rewrite $,eta,$ as
$$eta=frac{xi+overline{xi}}{2}=frac{1}{2}left(xi+frac{1}{xi}right)$$
The next step is to find $,b,cinmathbb{Q},$ such that $,fleft(etaright)=0$, where $,f(x)=x^2+bx+cinmathbb{Q}left[Xright].$ So, I am trying to solve (unsuccessfully) the following equation
$$frac{1}{4}left(xi+frac{1}{xi}right)^2+frac{b}{2}left(xi+frac{1}{xi}right)+c=0$$



I know the solution is $,b=frac{1}{2},,c=-frac{1}{4},$ but when it comes to obtaining those values myself, I fail again and again. HELP!






abstract-algebra cryptography




abstract-algebra cryptography






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 31 '18 at 12:09









Jyrki Lahtonen

110k13171378




110k13171378










asked Dec 31 '18 at 10:52









CarlIOCarlIO

957




957












  • $begingroup$
    No finite fields in site. Removing the tag.
    $endgroup$
    – Jyrki Lahtonen
    Dec 31 '18 at 12:09


















  • $begingroup$
    No finite fields in site. Removing the tag.
    $endgroup$
    – Jyrki Lahtonen
    Dec 31 '18 at 12:09
















$begingroup$
No finite fields in site. Removing the tag.
$endgroup$
– Jyrki Lahtonen
Dec 31 '18 at 12:09




$begingroup$
No finite fields in site. Removing the tag.
$endgroup$
– Jyrki Lahtonen
Dec 31 '18 at 12:09










2 Answers
2






active

oldest

votes


















4












$begingroup$

Note $xi = e^{2ipi/5}$. Since $xi$ is a complex fifth root of unity, we know $xi^4 + xi^3 + xi^2 + xi + 1 = 0$. Now divide through by $xi^2$ to get $$xi^2 + xi + 1 + frac{1}{xi} + frac{1}{xi^2} = 0$$ Hence $$left(xi+ frac{1}{xi}right)^2 - 2 + left(xi + frac{1}{xi}right) + 1 = 0$$ and so $$4eta^2 + 2eta - 1 = 0$$ and since $etanotin mathbb{Q}$, the minimal polynomial of $eta$ is $4x^2 +2x - 1$, or making it monic $x^2 + frac{1}{2} x - frac{1}{4}$, as per your solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much
    $endgroup$
    – CarlIO
    Dec 31 '18 at 11:20





















-1












$begingroup$

Steps with CAS (Maple, Mathematica, Maxima, ...):




  • $x=cosfrac{2pi}{5}$

  • $arccos(x)=frac{2pi}{5}$

  • $5arccos(x)=2pi$

  • $cos(5arccos(x))=1$

  • $5 x, {{left( 1-{{x}^{2}}right) }^{2}}+{{x}^{5}}-10 {{x}^{3}}, left( 1-{{x}^{2}}right) =1$

  • $left( x-1right) , {{left( 4 {{x}^{2}}+2 x-1right) }^{2}}=0$


Then $x=cosfrac{2pi}{5}$ is solution of $4x^2+2x-1=0$.



This method can apply for other elements. For example, $x=cosfrac{2pi}{15}$ is solution of



$$16x^4-8x^3-16x^2+8x+1=0$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






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    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

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    4












    $begingroup$

    Note $xi = e^{2ipi/5}$. Since $xi$ is a complex fifth root of unity, we know $xi^4 + xi^3 + xi^2 + xi + 1 = 0$. Now divide through by $xi^2$ to get $$xi^2 + xi + 1 + frac{1}{xi} + frac{1}{xi^2} = 0$$ Hence $$left(xi+ frac{1}{xi}right)^2 - 2 + left(xi + frac{1}{xi}right) + 1 = 0$$ and so $$4eta^2 + 2eta - 1 = 0$$ and since $etanotin mathbb{Q}$, the minimal polynomial of $eta$ is $4x^2 +2x - 1$, or making it monic $x^2 + frac{1}{2} x - frac{1}{4}$, as per your solution.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you very much
      $endgroup$
      – CarlIO
      Dec 31 '18 at 11:20


















    4












    $begingroup$

    Note $xi = e^{2ipi/5}$. Since $xi$ is a complex fifth root of unity, we know $xi^4 + xi^3 + xi^2 + xi + 1 = 0$. Now divide through by $xi^2$ to get $$xi^2 + xi + 1 + frac{1}{xi} + frac{1}{xi^2} = 0$$ Hence $$left(xi+ frac{1}{xi}right)^2 - 2 + left(xi + frac{1}{xi}right) + 1 = 0$$ and so $$4eta^2 + 2eta - 1 = 0$$ and since $etanotin mathbb{Q}$, the minimal polynomial of $eta$ is $4x^2 +2x - 1$, or making it monic $x^2 + frac{1}{2} x - frac{1}{4}$, as per your solution.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you very much
      $endgroup$
      – CarlIO
      Dec 31 '18 at 11:20
















    4












    4








    4





    $begingroup$

    Note $xi = e^{2ipi/5}$. Since $xi$ is a complex fifth root of unity, we know $xi^4 + xi^3 + xi^2 + xi + 1 = 0$. Now divide through by $xi^2$ to get $$xi^2 + xi + 1 + frac{1}{xi} + frac{1}{xi^2} = 0$$ Hence $$left(xi+ frac{1}{xi}right)^2 - 2 + left(xi + frac{1}{xi}right) + 1 = 0$$ and so $$4eta^2 + 2eta - 1 = 0$$ and since $etanotin mathbb{Q}$, the minimal polynomial of $eta$ is $4x^2 +2x - 1$, or making it monic $x^2 + frac{1}{2} x - frac{1}{4}$, as per your solution.






    share|cite|improve this answer









    $endgroup$



    Note $xi = e^{2ipi/5}$. Since $xi$ is a complex fifth root of unity, we know $xi^4 + xi^3 + xi^2 + xi + 1 = 0$. Now divide through by $xi^2$ to get $$xi^2 + xi + 1 + frac{1}{xi} + frac{1}{xi^2} = 0$$ Hence $$left(xi+ frac{1}{xi}right)^2 - 2 + left(xi + frac{1}{xi}right) + 1 = 0$$ and so $$4eta^2 + 2eta - 1 = 0$$ and since $etanotin mathbb{Q}$, the minimal polynomial of $eta$ is $4x^2 +2x - 1$, or making it monic $x^2 + frac{1}{2} x - frac{1}{4}$, as per your solution.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 31 '18 at 11:08









    ODFODF

    1,486510




    1,486510












    • $begingroup$
      Thank you very much
      $endgroup$
      – CarlIO
      Dec 31 '18 at 11:20




















    • $begingroup$
      Thank you very much
      $endgroup$
      – CarlIO
      Dec 31 '18 at 11:20


















    $begingroup$
    Thank you very much
    $endgroup$
    – CarlIO
    Dec 31 '18 at 11:20






    $begingroup$
    Thank you very much
    $endgroup$
    – CarlIO
    Dec 31 '18 at 11:20













    -1












    $begingroup$

    Steps with CAS (Maple, Mathematica, Maxima, ...):




    • $x=cosfrac{2pi}{5}$

    • $arccos(x)=frac{2pi}{5}$

    • $5arccos(x)=2pi$

    • $cos(5arccos(x))=1$

    • $5 x, {{left( 1-{{x}^{2}}right) }^{2}}+{{x}^{5}}-10 {{x}^{3}}, left( 1-{{x}^{2}}right) =1$

    • $left( x-1right) , {{left( 4 {{x}^{2}}+2 x-1right) }^{2}}=0$


    Then $x=cosfrac{2pi}{5}$ is solution of $4x^2+2x-1=0$.



    This method can apply for other elements. For example, $x=cosfrac{2pi}{15}$ is solution of



    $$16x^4-8x^3-16x^2+8x+1=0$$






    share|cite|improve this answer









    $endgroup$


















      -1












      $begingroup$

      Steps with CAS (Maple, Mathematica, Maxima, ...):




      • $x=cosfrac{2pi}{5}$

      • $arccos(x)=frac{2pi}{5}$

      • $5arccos(x)=2pi$

      • $cos(5arccos(x))=1$

      • $5 x, {{left( 1-{{x}^{2}}right) }^{2}}+{{x}^{5}}-10 {{x}^{3}}, left( 1-{{x}^{2}}right) =1$

      • $left( x-1right) , {{left( 4 {{x}^{2}}+2 x-1right) }^{2}}=0$


      Then $x=cosfrac{2pi}{5}$ is solution of $4x^2+2x-1=0$.



      This method can apply for other elements. For example, $x=cosfrac{2pi}{15}$ is solution of



      $$16x^4-8x^3-16x^2+8x+1=0$$






      share|cite|improve this answer









      $endgroup$
















        -1












        -1








        -1





        $begingroup$

        Steps with CAS (Maple, Mathematica, Maxima, ...):




        • $x=cosfrac{2pi}{5}$

        • $arccos(x)=frac{2pi}{5}$

        • $5arccos(x)=2pi$

        • $cos(5arccos(x))=1$

        • $5 x, {{left( 1-{{x}^{2}}right) }^{2}}+{{x}^{5}}-10 {{x}^{3}}, left( 1-{{x}^{2}}right) =1$

        • $left( x-1right) , {{left( 4 {{x}^{2}}+2 x-1right) }^{2}}=0$


        Then $x=cosfrac{2pi}{5}$ is solution of $4x^2+2x-1=0$.



        This method can apply for other elements. For example, $x=cosfrac{2pi}{15}$ is solution of



        $$16x^4-8x^3-16x^2+8x+1=0$$






        share|cite|improve this answer









        $endgroup$



        Steps with CAS (Maple, Mathematica, Maxima, ...):




        • $x=cosfrac{2pi}{5}$

        • $arccos(x)=frac{2pi}{5}$

        • $5arccos(x)=2pi$

        • $cos(5arccos(x))=1$

        • $5 x, {{left( 1-{{x}^{2}}right) }^{2}}+{{x}^{5}}-10 {{x}^{3}}, left( 1-{{x}^{2}}right) =1$

        • $left( x-1right) , {{left( 4 {{x}^{2}}+2 x-1right) }^{2}}=0$


        Then $x=cosfrac{2pi}{5}$ is solution of $4x^2+2x-1=0$.



        This method can apply for other elements. For example, $x=cosfrac{2pi}{15}$ is solution of



        $$16x^4-8x^3-16x^2+8x+1=0$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 12:03









        Aleksas DomarkasAleksas Domarkas

        1,37216




        1,37216






























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