The minimal polynomial satisfied by the primitive generator












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I am trying to do the following problem that appears in Dummit and Foote book. But I have no idea how to start the problem. Can anyone please give me a hint? Thank you.



Section 14.5, Problem #1(Page#603): Determine the minimal polynomials satisfied by the primitive generators given in the text for the subfields of $mathbb{Q(zeta_{13})}$.



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    $begingroup$


    I am trying to do the following problem that appears in Dummit and Foote book. But I have no idea how to start the problem. Can anyone please give me a hint? Thank you.



    Section 14.5, Problem #1(Page#603): Determine the minimal polynomials satisfied by the primitive generators given in the text for the subfields of $mathbb{Q(zeta_{13})}$.



    enter image description here






    abstract-algebra field-theory galois-theory







    share|cite|improve this question









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      0












      0








      0





      $begingroup$


      I am trying to do the following problem that appears in Dummit and Foote book. But I have no idea how to start the problem. Can anyone please give me a hint? Thank you.



      Section 14.5, Problem #1(Page#603): Determine the minimal polynomials satisfied by the primitive generators given in the text for the subfields of $mathbb{Q(zeta_{13})}$.



      enter image description here






      abstract-algebra field-theory galois-theory







      share|cite|improve this question









      $endgroup$




      I am trying to do the following problem that appears in Dummit and Foote book. But I have no idea how to start the problem. Can anyone please give me a hint? Thank you.



      Section 14.5, Problem #1(Page#603): Determine the minimal polynomials satisfied by the primitive generators given in the text for the subfields of $mathbb{Q(zeta_{13})}$.



      enter image description here






      abstract-algebra field-theory galois-theory




      abstract-algebra field-theory galois-theory






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 27 at 1:45









      usmndjusmndj

      524




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          1 Answer
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          $begingroup$

          I’ll show you a trick that I use, really just a variation of what’s obvious, but organized so that you can trap careless errors. I promise you that I did this all by hand, no machine aid.



          I’ll attack the “simplest” one, finding the minimal polynomial (it has to be quadratic) for $xi=zeta+zeta^3+zeta^4+zeta^9+zeta^{10}+zeta^{12}$. I’m going to be consistent in rewriting this and all quantities so as to use the fact that $zeta^{13}=1$. In particular, we have
          $$
          0=zeta^6+zeta^5+zeta^4+zeta^3+zeta^2+zeta+1+zeta^{-1}+zeta^{-2}+zeta^{-3}+zeta^{-4}+zeta^{-5}+zeta^{-6},.
          $$
          And then, calculating,
          $$
          begin{matrix}
          xi=&zeta^4&+zeta^3&+zeta&+zeta^{-1}&+zeta^{-3}&+zeta^{-4}\
          xi^2=&zeta^8&+2zeta^7&+2zeta^5&+2zeta^3&+2zeta&+2\
          &&+zeta^6&+2zeta^4&+2zeta^2&+2&+2zeta^{-2}\
          &&&+zeta^2&+2&+2zeta^{-2}&+2zeta^{-3}\
          &&&&+zeta^{-2}&+2zeta^{-4}&+2zeta^{-5}\
          &&&&&+zeta^{-6}&+2zeta^{-7}\
          &&&&&&+zeta^{-8},,
          end{matrix}
          $$
          in which you collect like terms, using $zeta^8=zeta^{-5}$ etc., to get
          $$
          begin{align}
          xi^2&=3zeta^6+3zeta^5+2zeta^4+2zeta^3+3zeta^2+2zeta+6+2zeta^{-1}+3zeta^{-2}+2zeta^{-3}+2zeta^{-4}+3zeta^{-5}+3zeta^{-6}\
          &=-zeta^4-zeta^3-zeta+3-zeta^{-1}-zeta^{-3}-zeta^{-4}\
          &=-xi+3,,
          end{align}
          $$
          in which I subtracted $3$ times zero to go from the first line to the second. And so you see that $xi^2+xi-3=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, Lubin.
            $endgroup$
            – usmndj
            Jan 27 at 22:03











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

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          0












          $begingroup$

          I’ll show you a trick that I use, really just a variation of what’s obvious, but organized so that you can trap careless errors. I promise you that I did this all by hand, no machine aid.



          I’ll attack the “simplest” one, finding the minimal polynomial (it has to be quadratic) for $xi=zeta+zeta^3+zeta^4+zeta^9+zeta^{10}+zeta^{12}$. I’m going to be consistent in rewriting this and all quantities so as to use the fact that $zeta^{13}=1$. In particular, we have
          $$
          0=zeta^6+zeta^5+zeta^4+zeta^3+zeta^2+zeta+1+zeta^{-1}+zeta^{-2}+zeta^{-3}+zeta^{-4}+zeta^{-5}+zeta^{-6},.
          $$
          And then, calculating,
          $$
          begin{matrix}
          xi=&zeta^4&+zeta^3&+zeta&+zeta^{-1}&+zeta^{-3}&+zeta^{-4}\
          xi^2=&zeta^8&+2zeta^7&+2zeta^5&+2zeta^3&+2zeta&+2\
          &&+zeta^6&+2zeta^4&+2zeta^2&+2&+2zeta^{-2}\
          &&&+zeta^2&+2&+2zeta^{-2}&+2zeta^{-3}\
          &&&&+zeta^{-2}&+2zeta^{-4}&+2zeta^{-5}\
          &&&&&+zeta^{-6}&+2zeta^{-7}\
          &&&&&&+zeta^{-8},,
          end{matrix}
          $$
          in which you collect like terms, using $zeta^8=zeta^{-5}$ etc., to get
          $$
          begin{align}
          xi^2&=3zeta^6+3zeta^5+2zeta^4+2zeta^3+3zeta^2+2zeta+6+2zeta^{-1}+3zeta^{-2}+2zeta^{-3}+2zeta^{-4}+3zeta^{-5}+3zeta^{-6}\
          &=-zeta^4-zeta^3-zeta+3-zeta^{-1}-zeta^{-3}-zeta^{-4}\
          &=-xi+3,,
          end{align}
          $$
          in which I subtracted $3$ times zero to go from the first line to the second. And so you see that $xi^2+xi-3=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, Lubin.
            $endgroup$
            – usmndj
            Jan 27 at 22:03
















          0












          $begingroup$

          I’ll show you a trick that I use, really just a variation of what’s obvious, but organized so that you can trap careless errors. I promise you that I did this all by hand, no machine aid.



          I’ll attack the “simplest” one, finding the minimal polynomial (it has to be quadratic) for $xi=zeta+zeta^3+zeta^4+zeta^9+zeta^{10}+zeta^{12}$. I’m going to be consistent in rewriting this and all quantities so as to use the fact that $zeta^{13}=1$. In particular, we have
          $$
          0=zeta^6+zeta^5+zeta^4+zeta^3+zeta^2+zeta+1+zeta^{-1}+zeta^{-2}+zeta^{-3}+zeta^{-4}+zeta^{-5}+zeta^{-6},.
          $$
          And then, calculating,
          $$
          begin{matrix}
          xi=&zeta^4&+zeta^3&+zeta&+zeta^{-1}&+zeta^{-3}&+zeta^{-4}\
          xi^2=&zeta^8&+2zeta^7&+2zeta^5&+2zeta^3&+2zeta&+2\
          &&+zeta^6&+2zeta^4&+2zeta^2&+2&+2zeta^{-2}\
          &&&+zeta^2&+2&+2zeta^{-2}&+2zeta^{-3}\
          &&&&+zeta^{-2}&+2zeta^{-4}&+2zeta^{-5}\
          &&&&&+zeta^{-6}&+2zeta^{-7}\
          &&&&&&+zeta^{-8},,
          end{matrix}
          $$
          in which you collect like terms, using $zeta^8=zeta^{-5}$ etc., to get
          $$
          begin{align}
          xi^2&=3zeta^6+3zeta^5+2zeta^4+2zeta^3+3zeta^2+2zeta+6+2zeta^{-1}+3zeta^{-2}+2zeta^{-3}+2zeta^{-4}+3zeta^{-5}+3zeta^{-6}\
          &=-zeta^4-zeta^3-zeta+3-zeta^{-1}-zeta^{-3}-zeta^{-4}\
          &=-xi+3,,
          end{align}
          $$
          in which I subtracted $3$ times zero to go from the first line to the second. And so you see that $xi^2+xi-3=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, Lubin.
            $endgroup$
            – usmndj
            Jan 27 at 22:03














          0












          0








          0





          $begingroup$

          I’ll show you a trick that I use, really just a variation of what’s obvious, but organized so that you can trap careless errors. I promise you that I did this all by hand, no machine aid.



          I’ll attack the “simplest” one, finding the minimal polynomial (it has to be quadratic) for $xi=zeta+zeta^3+zeta^4+zeta^9+zeta^{10}+zeta^{12}$. I’m going to be consistent in rewriting this and all quantities so as to use the fact that $zeta^{13}=1$. In particular, we have
          $$
          0=zeta^6+zeta^5+zeta^4+zeta^3+zeta^2+zeta+1+zeta^{-1}+zeta^{-2}+zeta^{-3}+zeta^{-4}+zeta^{-5}+zeta^{-6},.
          $$
          And then, calculating,
          $$
          begin{matrix}
          xi=&zeta^4&+zeta^3&+zeta&+zeta^{-1}&+zeta^{-3}&+zeta^{-4}\
          xi^2=&zeta^8&+2zeta^7&+2zeta^5&+2zeta^3&+2zeta&+2\
          &&+zeta^6&+2zeta^4&+2zeta^2&+2&+2zeta^{-2}\
          &&&+zeta^2&+2&+2zeta^{-2}&+2zeta^{-3}\
          &&&&+zeta^{-2}&+2zeta^{-4}&+2zeta^{-5}\
          &&&&&+zeta^{-6}&+2zeta^{-7}\
          &&&&&&+zeta^{-8},,
          end{matrix}
          $$
          in which you collect like terms, using $zeta^8=zeta^{-5}$ etc., to get
          $$
          begin{align}
          xi^2&=3zeta^6+3zeta^5+2zeta^4+2zeta^3+3zeta^2+2zeta+6+2zeta^{-1}+3zeta^{-2}+2zeta^{-3}+2zeta^{-4}+3zeta^{-5}+3zeta^{-6}\
          &=-zeta^4-zeta^3-zeta+3-zeta^{-1}-zeta^{-3}-zeta^{-4}\
          &=-xi+3,,
          end{align}
          $$
          in which I subtracted $3$ times zero to go from the first line to the second. And so you see that $xi^2+xi-3=0$.






          share|cite|improve this answer









          $endgroup$



          I’ll show you a trick that I use, really just a variation of what’s obvious, but organized so that you can trap careless errors. I promise you that I did this all by hand, no machine aid.



          I’ll attack the “simplest” one, finding the minimal polynomial (it has to be quadratic) for $xi=zeta+zeta^3+zeta^4+zeta^9+zeta^{10}+zeta^{12}$. I’m going to be consistent in rewriting this and all quantities so as to use the fact that $zeta^{13}=1$. In particular, we have
          $$
          0=zeta^6+zeta^5+zeta^4+zeta^3+zeta^2+zeta+1+zeta^{-1}+zeta^{-2}+zeta^{-3}+zeta^{-4}+zeta^{-5}+zeta^{-6},.
          $$
          And then, calculating,
          $$
          begin{matrix}
          xi=&zeta^4&+zeta^3&+zeta&+zeta^{-1}&+zeta^{-3}&+zeta^{-4}\
          xi^2=&zeta^8&+2zeta^7&+2zeta^5&+2zeta^3&+2zeta&+2\
          &&+zeta^6&+2zeta^4&+2zeta^2&+2&+2zeta^{-2}\
          &&&+zeta^2&+2&+2zeta^{-2}&+2zeta^{-3}\
          &&&&+zeta^{-2}&+2zeta^{-4}&+2zeta^{-5}\
          &&&&&+zeta^{-6}&+2zeta^{-7}\
          &&&&&&+zeta^{-8},,
          end{matrix}
          $$
          in which you collect like terms, using $zeta^8=zeta^{-5}$ etc., to get
          $$
          begin{align}
          xi^2&=3zeta^6+3zeta^5+2zeta^4+2zeta^3+3zeta^2+2zeta+6+2zeta^{-1}+3zeta^{-2}+2zeta^{-3}+2zeta^{-4}+3zeta^{-5}+3zeta^{-6}\
          &=-zeta^4-zeta^3-zeta+3-zeta^{-1}-zeta^{-3}-zeta^{-4}\
          &=-xi+3,,
          end{align}
          $$
          in which I subtracted $3$ times zero to go from the first line to the second. And so you see that $xi^2+xi-3=0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 27 at 4:57









          LubinLubin

          45.3k44688




          45.3k44688












          • $begingroup$
            Thank you, Lubin.
            $endgroup$
            – usmndj
            Jan 27 at 22:03


















          • $begingroup$
            Thank you, Lubin.
            $endgroup$
            – usmndj
            Jan 27 at 22:03
















          $begingroup$
          Thank you, Lubin.
          $endgroup$
          – usmndj
          Jan 27 at 22:03




          $begingroup$
          Thank you, Lubin.
          $endgroup$
          – usmndj
          Jan 27 at 22:03


















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