2

Suppose $\mathbb{Z}[X]/\Phi_m(x)$ where $\Phi_m(x)$ is a $m$-th cyclotomic polynomial .

And it can be factorized over modulo $p$ as follows

$\Phi_m(x)\equiv F_1(X)F_2(X)\ldots F_k(X) \mod p$

Is there any relationship between $F_i(X)$ ($i=1,\ldots,k$) and $\mathbb{Z}_m^*/\langle p \rangle$ ??

mallea
  • 829
  • Just checking: By “$\Bbb Z_m^*/\langle p\rangle$“ you mean the units of the finite ring modulo the multiplicative subgroup generated by $p$? – Lubin Sep 19 '16 at 19:55
  • yeah that's right – mallea Sep 20 '16 at 03:28
  • 1
    Assuming that $\gcd(m,p)=1$ the following can be said. Let's fix a primitive root $\zeta$. Then the exponents $j$ of zeros $\zeta^j$ of a factor $F_i(x)$ form a coset of $\langle p\rangle$ inside $\Bbb{Z}_m^$. Such cosets are often called cyclotomic cosets* for this very reason. – Jyrki Lahtonen Sep 20 '16 at 13:00
  • Related questions: 1,2, 3. More links to related questions collected at those. The question is at the core of much finite field Galois theory because over a finite field ALL irreducible polynomials are factors of some cyclotomic polynomial :-) – Jyrki Lahtonen Sep 20 '16 at 13:06
  • Thanks for replying me. Cyclotomic polynomial is a very beautiful entity :-) – mallea Sep 20 '16 at 13:33
  • @JyrkiLahtonen So those cosets are factors themselves??? – mallea Sep 20 '16 at 17:21
  • Not quite. You get an irreducible factor $F(X)=\prod(x-\zeta^j)$ with $j$ ranging over a coset. – Jyrki Lahtonen Sep 20 '16 at 19:32
  • I got it!!! But how about the case when $\Phi_m(X)$ cannot be factored into linear terms? In that case, order of each coset is more than 1. – mallea Sep 21 '16 at 08:54
  • Correct. This happens if and only if $m$ is a factor of $p-1$ (still assuming $\gcd(m,p)=1$). – Jyrki Lahtonen Sep 21 '16 at 11:56

0 Answers0