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Degrees of factors of pth cyclotomic polynomial in GF(q^n)

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Given p, q distinct primes, what are possible degrees of factors of $Phi_p(x)$ in $mathbbF_q^n[x]$? In the case of n=1, every factor has degree the order of q in $mathbbZ/pmathbbZ$ so if q is primitive mod p then this should be irreducible. But what about the case when n > 1?

abstract-algebra number-theory

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asked Mar 31 at 19:11

JouxJoux

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  Do you know the proof of the assertion in your second sentence? Does it work when $Bbb F_q$ is replaced by $Bbb F_q^n$? If not, where does the proof fail?
  $endgroup$
  – Greg Martin
  Mar 31 at 19:29
  

  
  
  
  


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  The proof I worked out (not 100% sure if it's right), is that if you have a root $a$ in the algebraic closure of $F_q$ then using the frobenius automorphism gives d distinct roots (distinct mod p), where d is the order of q mod p, so any irreducible factor of $Phi_p(x)$ has degree d in $F_q$. Extending to $q^n$ doesn't seem to change this argument, so would the factors still have degree d? Not the order of $q^n$ mod p? I feel like I'm missing something here.
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  – Joux
  Mar 31 at 20:00
  

  
  
  
  


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  In your notation, $Phi_p$ splits in $Bbb F_q^d$, so there certainly is some dependence on $n$. Have you worked out a specific example, say $q=2$ and $p=13$ (or $p=17$)?
  $endgroup$
  – Greg Martin
  Mar 31 at 21:49
  

  
  
  
  


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  You're right. I just went through a few examples in sage. It seems like $Phi_p$ either splits or all its factors have degree d (order of q mod p). From some experimentation it seems like $mathbbF_q^d$ might always be the splitting field, but I'm not sure how to go about proving that.
  $endgroup$
  – Joux
  Mar 31 at 23:33
  

  
  
  
  

add a comment | 

0

$begingroup$

Given p, q distinct primes, what are possible degrees of factors of $Phi_p(x)$ in $mathbbF_q^n[x]$? In the case of n=1, every factor has degree the order of q in $mathbbZ/pmathbbZ$ so if q is primitive mod p then this should be irreducible. But what about the case when n > 1?

abstract-algebra number-theory

share|cite|improve this question

asked Mar 31 at 19:11

JouxJoux

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know the proof of the assertion in your second sentence? Does it work when $Bbb F_q$ is replaced by $Bbb F_q^n$? If not, where does the proof fail?
  $endgroup$
  – Greg Martin
  Mar 31 at 19:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The proof I worked out (not 100% sure if it's right), is that if you have a root $a$ in the algebraic closure of $F_q$ then using the frobenius automorphism gives d distinct roots (distinct mod p), where d is the order of q mod p, so any irreducible factor of $Phi_p(x)$ has degree d in $F_q$. Extending to $q^n$ doesn't seem to change this argument, so would the factors still have degree d? Not the order of $q^n$ mod p? I feel like I'm missing something here.
  $endgroup$
  – Joux
  Mar 31 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In your notation, $Phi_p$ splits in $Bbb F_q^d$, so there certainly is some dependence on $n$. Have you worked out a specific example, say $q=2$ and $p=13$ (or $p=17$)?
  $endgroup$
  – Greg Martin
  Mar 31 at 21:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. I just went through a few examples in sage. It seems like $Phi_p$ either splits or all its factors have degree d (order of q mod p). From some experimentation it seems like $mathbbF_q^d$ might always be the splitting field, but I'm not sure how to go about proving that.
  $endgroup$
  – Joux
  Mar 31 at 23:33
  

  
  
  
  

add a comment | 

$begingroup$

Given p, q distinct primes, what are possible degrees of factors of $Phi_p(x)$ in $mathbbF_q^n[x]$? In the case of n=1, every factor has degree the order of q in $mathbbZ/pmathbbZ$ so if q is primitive mod p then this should be irreducible. But what about the case when n > 1?

abstract-algebra number-theory

share|cite|improve this question

asked Mar 31 at 19:11

JouxJoux

11

$endgroup$

share|cite|improve this question

asked Mar 31 at 19:11

JouxJoux

11

share|cite|improve this question

asked Mar 31 at 19:11

JouxJoux

11

asked Mar 31 at 19:11

JouxJoux

11

asked Mar 31 at 19:11

JouxJoux

11