If $zeta_n$ is the $nth$ root of unity, find $dim_Bbb QBbb Q[zeta_n]$

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For $n=1,2,3,4$ I found it to be $1,1,2,2$ respectively. So I hypothesize that $dim_Bbb QBbb Q[zeta_2n]=n$ and $dim_Bbb QBbb Q[zeta_2n+1]=n+1$.
But how to prove it?
When we have $2n$, the roots lie on the vertices of a regular $2n-$gon centred at $0$. The roots occur in pairs; placed on the extremities of a diagonal through $0$. So the dimension is no more than $n$, since half of them are simply scalar multiples of the other half.



But this is all I got. Any help is appreciated.




vector-spaces roots-of-unity




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  • "The $n$th root of unity" does not make sense. $1$ is a $n$th root of unity for any $n$. You probably meant "a primitive $n$th root of unity".
    – fkraiem
    Aug 16 at 5:09










  • Try $n=5$......
    – lhf
    Aug 16 at 10:26














up vote
0
down vote

favorite
1












For $n=1,2,3,4$ I found it to be $1,1,2,2$ respectively. So I hypothesize that $dim_Bbb QBbb Q[zeta_2n]=n$ and $dim_Bbb QBbb Q[zeta_2n+1]=n+1$.
But how to prove it?
When we have $2n$, the roots lie on the vertices of a regular $2n-$gon centred at $0$. The roots occur in pairs; placed on the extremities of a diagonal through $0$. So the dimension is no more than $n$, since half of them are simply scalar multiples of the other half.



But this is all I got. Any help is appreciated.




vector-spaces roots-of-unity




share|cite|improve this question






















  • "The $n$th root of unity" does not make sense. $1$ is a $n$th root of unity for any $n$. You probably meant "a primitive $n$th root of unity".
    – fkraiem
    Aug 16 at 5:09










  • Try $n=5$......
    – lhf
    Aug 16 at 10:26












up vote
0
down vote

favorite
1









up vote
0
down vote

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For $n=1,2,3,4$ I found it to be $1,1,2,2$ respectively. So I hypothesize that $dim_Bbb QBbb Q[zeta_2n]=n$ and $dim_Bbb QBbb Q[zeta_2n+1]=n+1$.
But how to prove it?
When we have $2n$, the roots lie on the vertices of a regular $2n-$gon centred at $0$. The roots occur in pairs; placed on the extremities of a diagonal through $0$. So the dimension is no more than $n$, since half of them are simply scalar multiples of the other half.



But this is all I got. Any help is appreciated.




vector-spaces roots-of-unity




share|cite|improve this question














For $n=1,2,3,4$ I found it to be $1,1,2,2$ respectively. So I hypothesize that $dim_Bbb QBbb Q[zeta_2n]=n$ and $dim_Bbb QBbb Q[zeta_2n+1]=n+1$.
But how to prove it?
When we have $2n$, the roots lie on the vertices of a regular $2n-$gon centred at $0$. The roots occur in pairs; placed on the extremities of a diagonal through $0$. So the dimension is no more than $n$, since half of them are simply scalar multiples of the other half.



But this is all I got. Any help is appreciated.





vector-spaces roots-of-unity





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share|cite|improve this question




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edited Aug 16 at 8:46









Bernard

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asked Aug 16 at 4:53









Hrit Roy

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  • "The $n$th root of unity" does not make sense. $1$ is a $n$th root of unity for any $n$. You probably meant "a primitive $n$th root of unity".
    – fkraiem
    Aug 16 at 5:09










  • Try $n=5$......
    – lhf
    Aug 16 at 10:26
















  • "The $n$th root of unity" does not make sense. $1$ is a $n$th root of unity for any $n$. You probably meant "a primitive $n$th root of unity".
    – fkraiem
    Aug 16 at 5:09










  • Try $n=5$......
    – lhf
    Aug 16 at 10:26















"The $n$th root of unity" does not make sense. $1$ is a $n$th root of unity for any $n$. You probably meant "a primitive $n$th root of unity".
– fkraiem
Aug 16 at 5:09




"The $n$th root of unity" does not make sense. $1$ is a $n$th root of unity for any $n$. You probably meant "a primitive $n$th root of unity".
– fkraiem
Aug 16 at 5:09












Try $n=5$......
– lhf
Aug 16 at 10:26




Try $n=5$......
– lhf
Aug 16 at 10:26










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When $n=6$, the sixth roots of unity are the cube roots of
unity and their negatives, so that $Bbb Z(zeta_6)=Bbb Z(zeta_3)$ which has degree $2$ over $Bbb Q$. That rather puts the kibosh on your conjecture.



The minimum polynomial of $zeta_n$ is called the $n$-th
cyclotomic polynomial.






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  • 1




    ... and the dimension the author is looking for is $varphi(n)$, where $varphi$ is Euler's totient function, see en.wikipedia.org/wiki/Euler%27s_totient_function
    – A. Pongrácz
    Aug 16 at 5:08











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When $n=6$, the sixth roots of unity are the cube roots of
unity and their negatives, so that $Bbb Z(zeta_6)=Bbb Z(zeta_3)$ which has degree $2$ over $Bbb Q$. That rather puts the kibosh on your conjecture.



The minimum polynomial of $zeta_n$ is called the $n$-th
cyclotomic polynomial.






share|cite|improve this answer


















  • 1




    ... and the dimension the author is looking for is $varphi(n)$, where $varphi$ is Euler's totient function, see en.wikipedia.org/wiki/Euler%27s_totient_function
    – A. Pongrácz
    Aug 16 at 5:08















up vote
0
down vote













When $n=6$, the sixth roots of unity are the cube roots of
unity and their negatives, so that $Bbb Z(zeta_6)=Bbb Z(zeta_3)$ which has degree $2$ over $Bbb Q$. That rather puts the kibosh on your conjecture.



The minimum polynomial of $zeta_n$ is called the $n$-th
cyclotomic polynomial.






share|cite|improve this answer


















  • 1




    ... and the dimension the author is looking for is $varphi(n)$, where $varphi$ is Euler's totient function, see en.wikipedia.org/wiki/Euler%27s_totient_function
    – A. Pongrácz
    Aug 16 at 5:08













up vote
0
down vote










up vote
0
down vote









When $n=6$, the sixth roots of unity are the cube roots of
unity and their negatives, so that $Bbb Z(zeta_6)=Bbb Z(zeta_3)$ which has degree $2$ over $Bbb Q$. That rather puts the kibosh on your conjecture.



The minimum polynomial of $zeta_n$ is called the $n$-th
cyclotomic polynomial.






share|cite|improve this answer














When $n=6$, the sixth roots of unity are the cube roots of
unity and their negatives, so that $Bbb Z(zeta_6)=Bbb Z(zeta_3)$ which has degree $2$ over $Bbb Q$. That rather puts the kibosh on your conjecture.



The minimum polynomial of $zeta_n$ is called the $n$-th
cyclotomic polynomial.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 16 at 13:40









Did

242k23208443




242k23208443










answered Aug 16 at 5:01









Lord Shark the Unknown

87.3k952113




87.3k952113







  • 1




    ... and the dimension the author is looking for is $varphi(n)$, where $varphi$ is Euler's totient function, see en.wikipedia.org/wiki/Euler%27s_totient_function
    – A. Pongrácz
    Aug 16 at 5:08













  • 1




    ... and the dimension the author is looking for is $varphi(n)$, where $varphi$ is Euler's totient function, see en.wikipedia.org/wiki/Euler%27s_totient_function
    – A. Pongrácz
    Aug 16 at 5:08








1




1




... and the dimension the author is looking for is $varphi(n)$, where $varphi$ is Euler's totient function, see en.wikipedia.org/wiki/Euler%27s_totient_function
– A. Pongrácz
Aug 16 at 5:08





... and the dimension the author is looking for is $varphi(n)$, where $varphi$ is Euler's totient function, see en.wikipedia.org/wiki/Euler%27s_totient_function
– A. Pongrácz
Aug 16 at 5:08













 

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