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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

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.

field-theory

share|cite|improve this question

edited yesterday

SNEHIL SANYAL

618110

asked yesterday

suchanda adhikarisuchanda adhikari

807

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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.

field-theory

share|cite|improve this question

edited yesterday

SNEHIL SANYAL

618110

asked yesterday

suchanda adhikarisuchanda adhikari

807

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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.

field-theory

share|cite|improve this question

edited yesterday

SNEHIL SANYAL

618110

asked yesterday

suchanda adhikarisuchanda adhikari

807

$endgroup$

share|cite|improve this question

edited yesterday

SNEHIL SANYAL

618110

asked yesterday

suchanda adhikarisuchanda adhikari

807

share|cite|improve this question

edited yesterday

SNEHIL SANYAL

618110

asked yesterday

suchanda adhikarisuchanda adhikari

807

edited yesterday

SNEHIL SANYAL

618110

edited yesterday

SNEHIL SANYAL

618110

asked yesterday

suchanda adhikarisuchanda adhikari

807

asked yesterday

suchanda adhikarisuchanda adhikari

807

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

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

add a comment | 

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

edited yesterday

answered yesterday

J.G.J.G.

29.4k22846

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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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

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

add a comment | 

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

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

add a comment | 

$begingroup$

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

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

share|cite|improve this answer

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

106k1161133

answered yesterday

Lord Shark the UnknownLord Shark the Unknown

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

edited yesterday

answered yesterday

J.G.J.G.

29.4k22846

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited yesterday

answered yesterday

J.G.J.G.

29.4k22846

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited yesterday

answered yesterday

J.G.J.G.

29.4k22846

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

J.G.J.G.

29.4k22846

answered yesterday

J.G.J.G.

29.4k22846

answered yesterday

J.G.J.G.

29.4k22846