Let $z = e^ frac 2pi i7 $ and let $p = z+z^2+z^4 $.Then which of the following options are correct?cube root of 2 not in Q(primitive root)For which of the following fields $mathbb F$ the polynomial $x^3-312312x+123123$ is irreducible in $mathbb F[x]$?Perfect field of characteristic $p>0$ which is not an algebraic extension of the prime fieldLet $F$ be a field of 8 elements and $A$= $xin F$. Then the number of elements in A isWhich of these statements about the field extension $mathbbR/mathbbQ$ are true?Let $(F,+,cdot)$ is the finite field with $9$ elements. Then which of the following are true?Which of the following field properties are correct?Are the following options correct in case of a field?Are the extensions $mathbbQ(sqrt2,sqrt3)$ and $mathbbQ(sqrt[3]5)$ normal over $mathbbQ$A problem from Neukirch's algebraic number theory book.

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Let $z = e^ frac 2pi i7 $ and let $p = z+z^2+z^4 $.Then which of the following options are correct?


cube root of 2 not in Q(primitive root)For which of the following fields $mathbb F$ the polynomial $x^3-312312x+123123$ is irreducible in $mathbb F[x]$?Perfect field of characteristic $p>0$ which is not an algebraic extension of the prime fieldLet $F$ be a field of 8 elements and $A$= $xin F$. Then the number of elements in A isWhich of these statements about the field extension $mathbbR/mathbbQ$ are true?Let $(F,+,cdot)$ is the finite field with $9$ elements. Then which of the following are true?Which of the following field properties are correct?Are the following options correct in case of a field?Are the extensions $mathbbQ(sqrt2,sqrt3)$ and $mathbbQ(sqrt[3]5)$ normal over $mathbbQ$A problem from Neukirch's algebraic number theory book.













1












$begingroup$


Let $z= e^ frac 2pi i7 $ and let $p= z+z^2+z^4 $ then



  1. $p$ is in $ mathbb Q $


  2. $p$ is in $ mathbbQ (sqrt D) $ for some $D gt 0$


  3. $p$ is in $ mathbbQ(sqrt D) $ for some $D lt 0$


  4. $p$ is in $i mathbb R $


Option $1$ is clearly false. please give me some hints for other options.



Thanks in advance.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Look up Gauss sums. This local search should give you enough.
    $endgroup$
    – Jyrki Lahtonen
    yesterday















1












$begingroup$


Let $z= e^ frac 2pi i7 $ and let $p= z+z^2+z^4 $ then



  1. $p$ is in $ mathbb Q $


  2. $p$ is in $ mathbbQ (sqrt D) $ for some $D gt 0$


  3. $p$ is in $ mathbbQ(sqrt D) $ for some $D lt 0$


  4. $p$ is in $i mathbb R $


Option $1$ is clearly false. please give me some hints for other options.



Thanks in advance.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Look up Gauss sums. This local search should give you enough.
    $endgroup$
    – Jyrki Lahtonen
    yesterday













1












1








1


1



$begingroup$


Let $z= e^ frac 2pi i7 $ and let $p= z+z^2+z^4 $ then



  1. $p$ is in $ mathbb Q $


  2. $p$ is in $ mathbbQ (sqrt D) $ for some $D gt 0$


  3. $p$ is in $ mathbbQ(sqrt D) $ for some $D lt 0$


  4. $p$ is in $i mathbb R $


Option $1$ is clearly false. please give me some hints for other options.



Thanks in advance.










share|cite|improve this question











$endgroup$




Let $z= e^ frac 2pi i7 $ and let $p= z+z^2+z^4 $ then



  1. $p$ is in $ mathbb Q $


  2. $p$ is in $ mathbbQ (sqrt D) $ for some $D gt 0$


  3. $p$ is in $ mathbbQ(sqrt D) $ for some $D lt 0$


  4. $p$ is in $i mathbb R $


Option $1$ is clearly false. please give me some hints for other options.



Thanks in advance.







field-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









SNEHIL SANYAL

618110




618110










asked yesterday









suchanda adhikarisuchanda adhikari

807




807







  • 1




    $begingroup$
    Look up Gauss sums. This local search should give you enough.
    $endgroup$
    – Jyrki Lahtonen
    yesterday












  • 1




    $begingroup$
    Look up Gauss sums. This local search should give you enough.
    $endgroup$
    – Jyrki Lahtonen
    yesterday







1




1




$begingroup$
Look up Gauss sums. This local search should give you enough.
$endgroup$
– Jyrki Lahtonen
yesterday




$begingroup$
Look up Gauss sums. This local search should give you enough.
$endgroup$
– Jyrki Lahtonen
yesterday










2 Answers
2






active

oldest

votes


















4












$begingroup$

Note that
$$p^2=z^2+z^4+z+2z^3+2z^5+2z^6=-p-2+2(1+z+z^2+z^3+z^4+z^5+z^6)$$
and $1+z+cdots+z^6=0$.






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    Since $p^ast=z^6+z^5+z^3=z^2p$ and $z^2nepm1$, $p$ is neither real nor imaginary. eliminating $1$, $2$ and $4$. As Lord Shark the Unknown already noted, $p^2+p+2implies p=frac-1pmsqrt-72$ so option $3$ is correct.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      why $ p= frac -1+sqrt -72 $ @J.G. ,please explain, I can't understand how you delete the possibility that p can be $ frac -1-sqrt -72 $
      $endgroup$
      – suchanda adhikari
      yesterday






    • 1




      $begingroup$
      @suchandaadhikari Sorry, I meant to write $pm$ there, which I do as of my latest edit. We don't need to determine the sign to work out which options are applicable.
      $endgroup$
      – J.G.
      yesterday










    • $begingroup$
      yes but is it possible to determine the sign easily?
      $endgroup$
      – suchanda adhikari
      yesterday






    • 1




      $begingroup$
      @suchandaadhikari In general, no! Gauss himself spent a while figuring out the correct sign (for primes much larger than $7$), even though the solution is in many a book now. For $7$th roots of unity it is easy. Draw them on the complexplane. Surely you can figure which side of the real axis the sum is!
      $endgroup$
      – Jyrki Lahtonen
      yesterday











    • $begingroup$
      Thank you sir I will do it.
      $endgroup$
      – suchanda adhikari
      yesterday










    Your Answer





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    2 Answers
    2






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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Note that
    $$p^2=z^2+z^4+z+2z^3+2z^5+2z^6=-p-2+2(1+z+z^2+z^3+z^4+z^5+z^6)$$
    and $1+z+cdots+z^6=0$.






    share|cite|improve this answer









    $endgroup$

















      4












      $begingroup$

      Note that
      $$p^2=z^2+z^4+z+2z^3+2z^5+2z^6=-p-2+2(1+z+z^2+z^3+z^4+z^5+z^6)$$
      and $1+z+cdots+z^6=0$.






      share|cite|improve this answer









      $endgroup$















        4












        4








        4





        $begingroup$

        Note that
        $$p^2=z^2+z^4+z+2z^3+2z^5+2z^6=-p-2+2(1+z+z^2+z^3+z^4+z^5+z^6)$$
        and $1+z+cdots+z^6=0$.






        share|cite|improve this answer









        $endgroup$



        Note that
        $$p^2=z^2+z^4+z+2z^3+2z^5+2z^6=-p-2+2(1+z+z^2+z^3+z^4+z^5+z^6)$$
        and $1+z+cdots+z^6=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Lord Shark the UnknownLord Shark the Unknown

        106k1161133




        106k1161133





















            3












            $begingroup$

            Since $p^ast=z^6+z^5+z^3=z^2p$ and $z^2nepm1$, $p$ is neither real nor imaginary. eliminating $1$, $2$ and $4$. As Lord Shark the Unknown already noted, $p^2+p+2implies p=frac-1pmsqrt-72$ so option $3$ is correct.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              why $ p= frac -1+sqrt -72 $ @J.G. ,please explain, I can't understand how you delete the possibility that p can be $ frac -1-sqrt -72 $
              $endgroup$
              – suchanda adhikari
              yesterday






            • 1




              $begingroup$
              @suchandaadhikari Sorry, I meant to write $pm$ there, which I do as of my latest edit. We don't need to determine the sign to work out which options are applicable.
              $endgroup$
              – J.G.
              yesterday










            • $begingroup$
              yes but is it possible to determine the sign easily?
              $endgroup$
              – suchanda adhikari
              yesterday






            • 1




              $begingroup$
              @suchandaadhikari In general, no! Gauss himself spent a while figuring out the correct sign (for primes much larger than $7$), even though the solution is in many a book now. For $7$th roots of unity it is easy. Draw them on the complexplane. Surely you can figure which side of the real axis the sum is!
              $endgroup$
              – Jyrki Lahtonen
              yesterday











            • $begingroup$
              Thank you sir I will do it.
              $endgroup$
              – suchanda adhikari
              yesterday















            3












            $begingroup$

            Since $p^ast=z^6+z^5+z^3=z^2p$ and $z^2nepm1$, $p$ is neither real nor imaginary. eliminating $1$, $2$ and $4$. As Lord Shark the Unknown already noted, $p^2+p+2implies p=frac-1pmsqrt-72$ so option $3$ is correct.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              why $ p= frac -1+sqrt -72 $ @J.G. ,please explain, I can't understand how you delete the possibility that p can be $ frac -1-sqrt -72 $
              $endgroup$
              – suchanda adhikari
              yesterday






            • 1




              $begingroup$
              @suchandaadhikari Sorry, I meant to write $pm$ there, which I do as of my latest edit. We don't need to determine the sign to work out which options are applicable.
              $endgroup$
              – J.G.
              yesterday










            • $begingroup$
              yes but is it possible to determine the sign easily?
              $endgroup$
              – suchanda adhikari
              yesterday






            • 1




              $begingroup$
              @suchandaadhikari In general, no! Gauss himself spent a while figuring out the correct sign (for primes much larger than $7$), even though the solution is in many a book now. For $7$th roots of unity it is easy. Draw them on the complexplane. Surely you can figure which side of the real axis the sum is!
              $endgroup$
              – Jyrki Lahtonen
              yesterday











            • $begingroup$
              Thank you sir I will do it.
              $endgroup$
              – suchanda adhikari
              yesterday













            3












            3








            3





            $begingroup$

            Since $p^ast=z^6+z^5+z^3=z^2p$ and $z^2nepm1$, $p$ is neither real nor imaginary. eliminating $1$, $2$ and $4$. As Lord Shark the Unknown already noted, $p^2+p+2implies p=frac-1pmsqrt-72$ so option $3$ is correct.






            share|cite|improve this answer











            $endgroup$



            Since $p^ast=z^6+z^5+z^3=z^2p$ and $z^2nepm1$, $p$ is neither real nor imaginary. eliminating $1$, $2$ and $4$. As Lord Shark the Unknown already noted, $p^2+p+2implies p=frac-1pmsqrt-72$ so option $3$ is correct.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited yesterday

























            answered yesterday









            J.G.J.G.

            29.4k22846




            29.4k22846











            • $begingroup$
              why $ p= frac -1+sqrt -72 $ @J.G. ,please explain, I can't understand how you delete the possibility that p can be $ frac -1-sqrt -72 $
              $endgroup$
              – suchanda adhikari
              yesterday






            • 1




              $begingroup$
              @suchandaadhikari Sorry, I meant to write $pm$ there, which I do as of my latest edit. We don't need to determine the sign to work out which options are applicable.
              $endgroup$
              – J.G.
              yesterday










            • $begingroup$
              yes but is it possible to determine the sign easily?
              $endgroup$
              – suchanda adhikari
              yesterday






            • 1




              $begingroup$
              @suchandaadhikari In general, no! Gauss himself spent a while figuring out the correct sign (for primes much larger than $7$), even though the solution is in many a book now. For $7$th roots of unity it is easy. Draw them on the complexplane. Surely you can figure which side of the real axis the sum is!
              $endgroup$
              – Jyrki Lahtonen
              yesterday











            • $begingroup$
              Thank you sir I will do it.
              $endgroup$
              – suchanda adhikari
              yesterday
















            • $begingroup$
              why $ p= frac -1+sqrt -72 $ @J.G. ,please explain, I can't understand how you delete the possibility that p can be $ frac -1-sqrt -72 $
              $endgroup$
              – suchanda adhikari
              yesterday






            • 1




              $begingroup$
              @suchandaadhikari Sorry, I meant to write $pm$ there, which I do as of my latest edit. We don't need to determine the sign to work out which options are applicable.
              $endgroup$
              – J.G.
              yesterday










            • $begingroup$
              yes but is it possible to determine the sign easily?
              $endgroup$
              – suchanda adhikari
              yesterday






            • 1




              $begingroup$
              @suchandaadhikari In general, no! Gauss himself spent a while figuring out the correct sign (for primes much larger than $7$), even though the solution is in many a book now. For $7$th roots of unity it is easy. Draw them on the complexplane. Surely you can figure which side of the real axis the sum is!
              $endgroup$
              – Jyrki Lahtonen
              yesterday











            • $begingroup$
              Thank you sir I will do it.
              $endgroup$
              – suchanda adhikari
              yesterday















            $begingroup$
            why $ p= frac -1+sqrt -72 $ @J.G. ,please explain, I can't understand how you delete the possibility that p can be $ frac -1-sqrt -72 $
            $endgroup$
            – suchanda adhikari
            yesterday




            $begingroup$
            why $ p= frac -1+sqrt -72 $ @J.G. ,please explain, I can't understand how you delete the possibility that p can be $ frac -1-sqrt -72 $
            $endgroup$
            – suchanda adhikari
            yesterday




            1




            1




            $begingroup$
            @suchandaadhikari Sorry, I meant to write $pm$ there, which I do as of my latest edit. We don't need to determine the sign to work out which options are applicable.
            $endgroup$
            – J.G.
            yesterday




            $begingroup$
            @suchandaadhikari Sorry, I meant to write $pm$ there, which I do as of my latest edit. We don't need to determine the sign to work out which options are applicable.
            $endgroup$
            – J.G.
            yesterday












            $begingroup$
            yes but is it possible to determine the sign easily?
            $endgroup$
            – suchanda adhikari
            yesterday




            $begingroup$
            yes but is it possible to determine the sign easily?
            $endgroup$
            – suchanda adhikari
            yesterday




            1




            1




            $begingroup$
            @suchandaadhikari In general, no! Gauss himself spent a while figuring out the correct sign (for primes much larger than $7$), even though the solution is in many a book now. For $7$th roots of unity it is easy. Draw them on the complexplane. Surely you can figure which side of the real axis the sum is!
            $endgroup$
            – Jyrki Lahtonen
            yesterday





            $begingroup$
            @suchandaadhikari In general, no! Gauss himself spent a while figuring out the correct sign (for primes much larger than $7$), even though the solution is in many a book now. For $7$th roots of unity it is easy. Draw them on the complexplane. Surely you can figure which side of the real axis the sum is!
            $endgroup$
            – Jyrki Lahtonen
            yesterday













            $begingroup$
            Thank you sir I will do it.
            $endgroup$
            – suchanda adhikari
            yesterday




            $begingroup$
            Thank you sir I will do it.
            $endgroup$
            – suchanda adhikari
            yesterday

















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