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Cyclotomic polynomial identity proof with primes.

Cyclotomic Polynomial of a PrimeProperties of cyclotomic polynomialIdentity Involving Cyclotomic PolynomialsA cyclotomic polynomial whose index has a large prime divisor cannot be too sparsecyclotomic polynomial $Phi_2n(x)$Finding the $18$th cyclotomic polynomial $phi_18(X))$.Product of cyclotomic polynomialsCoefficients of $pq$-th cyclotomic polynomial for distinct primes $pq$Several identities regarding cyclotomic polynomials.

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If p is prime, show that $Phi_p(x^p^k-1)=Phi_p^k(x)$.

Here is my attempt:

$x^p^k-1=prod_p^kPhi_d(x)=prod_i=1^kPhi_p^i(x)=Phi_p^k(x)(Phi_pPhi_p^2cdots Phi_p^k-1)=(x^p^k-1)^p-1=(x^p^k-1-1)Phi_p(x^p^k-1)$

But how do I show that $x^p^k-1-1=Phi_pPhi_p^2cdots Phi_p^k-1$?

prime-numbers proof-explanation cyclotomic-polynomials

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asked Mar 21 at 20:09

numericalorangenumericalorange

1,879313

$endgroup$

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  $begingroup$
  In this product you omitted a factor $Phi_1(x)$. In fact, $x^p^k-1=Phi_1(x)Phi_p(x)Phi_p^2(x)cdotsPhi_p^k(x)$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 21 at 20:52
  
  

  
  
  
  


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  @LordSharktheUnknown Oops, that is totally my bad. Then it is just by definition (?) that they are equal (of course when I include the factor $Phi_1$...!)
  $endgroup$
  – numericalorange
  Mar 21 at 20:56
  
  

  
  
  
  


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  What is your definition of $Phi_n(x)$? My definition is that it is the minimal polynomial of a primitive $n$th root of unity. By standard Galois theory, one can show the roots of $Phi_n(x)$ are precisely all of the primitive $n$th roots, and therefore $x^n-1$ is $prod_dmid nPhi_d(x)$ since both are monic and have the exact same set of (distinct) roots.
  $endgroup$
  – arctic tern
  Mar 22 at 16:21
  
  

  
  
  
  

add a comment | 

1

1

$begingroup$

If p is prime, show that $Phi_p(x^p^k-1)=Phi_p^k(x)$.

Here is my attempt:

$x^p^k-1=prod_p^kPhi_d(x)=prod_i=1^kPhi_p^i(x)=Phi_p^k(x)(Phi_pPhi_p^2cdots Phi_p^k-1)=(x^p^k-1)^p-1=(x^p^k-1-1)Phi_p(x^p^k-1)$

But how do I show that $x^p^k-1-1=Phi_pPhi_p^2cdots Phi_p^k-1$?

prime-numbers proof-explanation cyclotomic-polynomials

share|cite|improve this question

asked Mar 21 at 20:09

numericalorangenumericalorange

1,879313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In this product you omitted a factor $Phi_1(x)$. In fact, $x^p^k-1=Phi_1(x)Phi_p(x)Phi_p^2(x)cdotsPhi_p^k(x)$.
  $endgroup$
  – Lord Shark the Unknown
  Mar 21 at 20:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown Oops, that is totally my bad. Then it is just by definition (?) that they are equal (of course when I include the factor $Phi_1$...!)
  $endgroup$
  – numericalorange
  Mar 21 at 20:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is your definition of $Phi_n(x)$? My definition is that it is the minimal polynomial of a primitive $n$th root of unity. By standard Galois theory, one can show the roots of $Phi_n(x)$ are precisely all of the primitive $n$th roots, and therefore $x^n-1$ is $prod_dmid nPhi_d(x)$ since both are monic and have the exact same set of (distinct) roots.
  $endgroup$
  – arctic tern
  Mar 22 at 16:21
  
  

  
  
  
  

add a comment | 

$begingroup$

If p is prime, show that $Phi_p(x^p^k-1)=Phi_p^k(x)$.

Here is my attempt:

$x^p^k-1=prod_p^kPhi_d(x)=prod_i=1^kPhi_p^i(x)=Phi_p^k(x)(Phi_pPhi_p^2cdots Phi_p^k-1)=(x^p^k-1)^p-1=(x^p^k-1-1)Phi_p(x^p^k-1)$

But how do I show that $x^p^k-1-1=Phi_pPhi_p^2cdots Phi_p^k-1$?

prime-numbers proof-explanation cyclotomic-polynomials

share|cite|improve this question

asked Mar 21 at 20:09

numericalorangenumericalorange

1,879313

$endgroup$

share|cite|improve this question

asked Mar 21 at 20:09

numericalorangenumericalorange

1,879313

share|cite|improve this question

asked Mar 21 at 20:09

numericalorangenumericalorange

1,879313

asked Mar 21 at 20:09

numericalorangenumericalorange

1,879313

asked Mar 21 at 20:09

numericalorangenumericalorange

1,879313