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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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
$endgroup$
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
$endgroup$
$begingroup$
In this product you omitted a factor $Phi_1(x)$. In fact, $x^p^k-1=Phi_1(x)Phi_p(x)Phi_p^2(x)cdotsPhi_p^k(x)$.
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– 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 |
$begingroup$
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
$endgroup$
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
prime-numbers proof-explanation cyclotomic-polynomials
asked Mar 21 at 20:09
numericalorangenumericalorange
1,879313
1,879313
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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 |
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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
$begingroup$
@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