If $zeta_n$ is a primitive $n$th root of unity, why is $textdim_Bbb QBbb Q[zeta_n]=phi(n)$? [duplicate]

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  • showing that $n$th cyclotomic polynomial $Phi_n(x)$ is irreducible over $mathbbQ$

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I have no idea what cyclotomic polynomials are and how we can get the result using that. Is there another way to prove it? Any hint is appreciated.




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marked as duplicate by Dietrich Burde, Lord Shark the Unknown, John Ma, José Carlos Santos, Xander Henderson Aug 17 at 0:01


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  • @DietrichBurde as I said; we don't know what cyclotomic polynomials are. Our professor asked us this knowing that fact. There must be some other way to prove it.
    – Hrit Roy
    Aug 16 at 12:05











  • I see. Still I think it's worth to look up what a cyclotomic polynomial is, and you asked anyway how we can get the result using that polynomial.
    – Dietrich Burde
    Aug 16 at 12:09










  • @DietrichBurde could you tell me how cyclotomic polynomials help us prove it?
    – Hrit Roy
    Aug 16 at 15:40










  • Hrit, the $BbbQ$-dimension of $BbbQ[zeta_n]$ is by definition the degree of the minimal polynomial $Phi_n(x)$ of $zeta_n$, which obviously is $phi(n)$; see the duplicate.
    – Dietrich Burde
    Aug 16 at 18:15















up vote
0
down vote

favorite













This question already has an answer here:



  • showing that $n$th cyclotomic polynomial $Phi_n(x)$ is irreducible over $mathbbQ$

    6 answers



I have no idea what cyclotomic polynomials are and how we can get the result using that. Is there another way to prove it? Any hint is appreciated.




vector-spaces roots-of-unity




share|cite|improve this question












marked as duplicate by Dietrich Burde, Lord Shark the Unknown, John Ma, José Carlos Santos, Xander Henderson Aug 17 at 0:01


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • @DietrichBurde as I said; we don't know what cyclotomic polynomials are. Our professor asked us this knowing that fact. There must be some other way to prove it.
    – Hrit Roy
    Aug 16 at 12:05











  • I see. Still I think it's worth to look up what a cyclotomic polynomial is, and you asked anyway how we can get the result using that polynomial.
    – Dietrich Burde
    Aug 16 at 12:09










  • @DietrichBurde could you tell me how cyclotomic polynomials help us prove it?
    – Hrit Roy
    Aug 16 at 15:40










  • Hrit, the $BbbQ$-dimension of $BbbQ[zeta_n]$ is by definition the degree of the minimal polynomial $Phi_n(x)$ of $zeta_n$, which obviously is $phi(n)$; see the duplicate.
    – Dietrich Burde
    Aug 16 at 18:15













up vote
0
down vote

favorite









up vote
0
down vote

favorite












This question already has an answer here:



  • showing that $n$th cyclotomic polynomial $Phi_n(x)$ is irreducible over $mathbbQ$

    6 answers



I have no idea what cyclotomic polynomials are and how we can get the result using that. Is there another way to prove it? Any hint is appreciated.




vector-spaces roots-of-unity




share|cite|improve this question













This question already has an answer here:



  • showing that $n$th cyclotomic polynomial $Phi_n(x)$ is irreducible over $mathbbQ$

    6 answers



I have no idea what cyclotomic polynomials are and how we can get the result using that. Is there another way to prove it? Any hint is appreciated.





This question already has an answer here:



  • showing that $n$th cyclotomic polynomial $Phi_n(x)$ is irreducible over $mathbbQ$

    6 answers





vector-spaces roots-of-unity





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 16 at 11:29









Hrit Roy

837113




837113




marked as duplicate by Dietrich Burde, Lord Shark the Unknown, John Ma, José Carlos Santos, Xander Henderson Aug 17 at 0:01


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Dietrich Burde, Lord Shark the Unknown, John Ma, José Carlos Santos, Xander Henderson Aug 17 at 0:01


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • @DietrichBurde as I said; we don't know what cyclotomic polynomials are. Our professor asked us this knowing that fact. There must be some other way to prove it.
    – Hrit Roy
    Aug 16 at 12:05











  • I see. Still I think it's worth to look up what a cyclotomic polynomial is, and you asked anyway how we can get the result using that polynomial.
    – Dietrich Burde
    Aug 16 at 12:09










  • @DietrichBurde could you tell me how cyclotomic polynomials help us prove it?
    – Hrit Roy
    Aug 16 at 15:40










  • Hrit, the $BbbQ$-dimension of $BbbQ[zeta_n]$ is by definition the degree of the minimal polynomial $Phi_n(x)$ of $zeta_n$, which obviously is $phi(n)$; see the duplicate.
    – Dietrich Burde
    Aug 16 at 18:15

















  • @DietrichBurde as I said; we don't know what cyclotomic polynomials are. Our professor asked us this knowing that fact. There must be some other way to prove it.
    – Hrit Roy
    Aug 16 at 12:05











  • I see. Still I think it's worth to look up what a cyclotomic polynomial is, and you asked anyway how we can get the result using that polynomial.
    – Dietrich Burde
    Aug 16 at 12:09










  • @DietrichBurde could you tell me how cyclotomic polynomials help us prove it?
    – Hrit Roy
    Aug 16 at 15:40










  • Hrit, the $BbbQ$-dimension of $BbbQ[zeta_n]$ is by definition the degree of the minimal polynomial $Phi_n(x)$ of $zeta_n$, which obviously is $phi(n)$; see the duplicate.
    – Dietrich Burde
    Aug 16 at 18:15
















@DietrichBurde as I said; we don't know what cyclotomic polynomials are. Our professor asked us this knowing that fact. There must be some other way to prove it.
– Hrit Roy
Aug 16 at 12:05





@DietrichBurde as I said; we don't know what cyclotomic polynomials are. Our professor asked us this knowing that fact. There must be some other way to prove it.
– Hrit Roy
Aug 16 at 12:05













I see. Still I think it's worth to look up what a cyclotomic polynomial is, and you asked anyway how we can get the result using that polynomial.
– Dietrich Burde
Aug 16 at 12:09




I see. Still I think it's worth to look up what a cyclotomic polynomial is, and you asked anyway how we can get the result using that polynomial.
– Dietrich Burde
Aug 16 at 12:09












@DietrichBurde could you tell me how cyclotomic polynomials help us prove it?
– Hrit Roy
Aug 16 at 15:40




@DietrichBurde could you tell me how cyclotomic polynomials help us prove it?
– Hrit Roy
Aug 16 at 15:40












Hrit, the $BbbQ$-dimension of $BbbQ[zeta_n]$ is by definition the degree of the minimal polynomial $Phi_n(x)$ of $zeta_n$, which obviously is $phi(n)$; see the duplicate.
– Dietrich Burde
Aug 16 at 18:15





Hrit, the $BbbQ$-dimension of $BbbQ[zeta_n]$ is by definition the degree of the minimal polynomial $Phi_n(x)$ of $zeta_n$, which obviously is $phi(n)$; see the duplicate.
– Dietrich Burde
Aug 16 at 18:15











1 Answer
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If $zeta_n$ is primitive, then $zeta_n^jneq zeta_n^k$ whenever $n nmid (k-j)$. Knowing this you could write $mathbb Q [zeta_n]$ concretely. The major point is $(zeta_n^j)_gcd(j,n)=1$ is a $mathbb Q$-basis of $mathbb Q [zeta_n]$.






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  • @ArnaudD. Yeah, i found that. Thanks.
    – xbh
    Aug 16 at 11:37

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













If $zeta_n$ is primitive, then $zeta_n^jneq zeta_n^k$ whenever $n nmid (k-j)$. Knowing this you could write $mathbb Q [zeta_n]$ concretely. The major point is $(zeta_n^j)_gcd(j,n)=1$ is a $mathbb Q$-basis of $mathbb Q [zeta_n]$.






share|cite|improve this answer






















  • @ArnaudD. Yeah, i found that. Thanks.
    – xbh
    Aug 16 at 11:37














up vote
0
down vote













If $zeta_n$ is primitive, then $zeta_n^jneq zeta_n^k$ whenever $n nmid (k-j)$. Knowing this you could write $mathbb Q [zeta_n]$ concretely. The major point is $(zeta_n^j)_gcd(j,n)=1$ is a $mathbb Q$-basis of $mathbb Q [zeta_n]$.






share|cite|improve this answer






















  • @ArnaudD. Yeah, i found that. Thanks.
    – xbh
    Aug 16 at 11:37












up vote
0
down vote










up vote
0
down vote









If $zeta_n$ is primitive, then $zeta_n^jneq zeta_n^k$ whenever $n nmid (k-j)$. Knowing this you could write $mathbb Q [zeta_n]$ concretely. The major point is $(zeta_n^j)_gcd(j,n)=1$ is a $mathbb Q$-basis of $mathbb Q [zeta_n]$.






share|cite|improve this answer














If $zeta_n$ is primitive, then $zeta_n^jneq zeta_n^k$ whenever $n nmid (k-j)$. Knowing this you could write $mathbb Q [zeta_n]$ concretely. The major point is $(zeta_n^j)_gcd(j,n)=1$ is a $mathbb Q$-basis of $mathbb Q [zeta_n]$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 16 at 11:36

























answered Aug 16 at 11:34









xbh

2,232113




2,232113











  • @ArnaudD. Yeah, i found that. Thanks.
    – xbh
    Aug 16 at 11:37
















  • @ArnaudD. Yeah, i found that. Thanks.
    – xbh
    Aug 16 at 11:37















@ArnaudD. Yeah, i found that. Thanks.
– xbh
Aug 16 at 11:37




@ArnaudD. Yeah, i found that. Thanks.
– xbh
Aug 16 at 11:37


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