2

Let $r\geq1$ and $p\not\mid r $ prime, where $p\in(\mathbb{Z}/r\mathbb{Z})^*$ has order $a$.

How do I prove that $\Phi_r\in\mathbb{F}_p[x]$ is the product of $\varphi(r)/a$ irreducible factors of degree $a$?

To be honest, I currently would not know where to approach this problem, except for thinking to use Galois theory.
However I got the hint:

What is $[\mathbb{F}_p(b):\mathbb{F}_p]$ for a root $b\in\overline{\mathbb{F}}_p$ of $\Phi_r$?

I was thinking it is $a$, but I'm not sure why or how to show it.

Edit: $\Phi_r=\prod_{k\in(\mathbb{Z}/r\mathbb{Z})^*}(x-\zeta^k_n)$ is the $r$-th cyclotomic polynomial, where $\zeta_r$ is the $r$-th primitve unit root.

  • @NobleMushtak Can you help me with the problem? –  Apr 17 '16 at 12:38
  • I'm currently answering another question, but I will get back to this one and see if I can solve it. – Noble Mushtak Apr 17 '16 at 12:39
  • @NobleMushtak Thankyou, much appreciated –  Apr 17 '16 at 12:39
  • 1
    Would probably be a good idea to edit the question to include what you mean by the notation $\Phi_r$. – Gerry Myerson Apr 17 '16 at 12:46
  • Assuming that $\Phi_r(x)$ stands for the cyclotomic polynomial, then I have answered this very question many times on our site (in various guises). You are right. Galois theory is all you need. – Jyrki Lahtonen Apr 17 '16 at 12:50
  • @GerryMyerson Thanks for the headsup! –  Apr 17 '16 at 12:54
  • @JyrkiLahtonen Could you maybe point out where you answered it? because I could not find an answer for this case. –  Apr 17 '16 at 12:55
  • Related posts I can find from my own history are 1,2,3, 4. Not sure that any of them is an exact match. – Jyrki Lahtonen Apr 17 '16 at 12:56
  • 1
    Anyway, the idea is that if $g$ is a primitive root of unity of order $r$, then by Galois theory the zeros of the minimal polynomial of $g$ are exactly $g,g^p, g^{p^2},\ldots,g^{p^{a-1}}$. The list stops here, because $g^{p^a}=g$ as $p^a\equiv1\pmod r$. So irrespective of the choice of $g$, its minimal polynomial has degree $a$. – Jyrki Lahtonen Apr 17 '16 at 13:05
  • Your definition of $\Phi_r$ involves the undefined symbol $n$. – Gerry Myerson Apr 17 '16 at 22:32

1 Answers1

1

Let's say we have a root $b$ of $\Phi_r$. We wonder what the degree of the minimal polynomial of $b$ in $\Bbb{F}_p$ so we can find a factor of $\Phi_r$. We can do this using Galois theory by finding the number of automorphisms of $\Bbb{F}_p(b)$ that fix $\Bbb{F}_p$.

It is well known that all automorphisms of $\Bbb{F}_p$ must be in the form of $\sigma(x)=x^{p^n}$ for some integer $n$. If we have a primitive $r$-root of unity $b$, then for any integer $n$, $b^{p^n}$ must also be a primitive $r$-root of unity because $p^n$ is coprime with $r$. Thus, all of the automorphisms $\sigma(x)=x^{p^n}$ are valid because they all map $b$ to another root of $\Phi_r$. However, once we get to $\sigma(x)=x^{p^a}$, since $p^a \equiv 1 \pmod 4$, this is just the identity automorphism again, so we are repeating ourselves after $a$ automorphisms. Thus, there are $a$ automorphisms in the Galois group of $\Bbb{F}_p(b)$ over $\Bbb{F}_p$.

This means for all of the roots of $\Phi_r$, their minimal polynomial has degree $a$. All of these minimal polynomials must divide $\Phi_r$ because otherwise, there would be no way $\Phi_r$ would have them as roots. Therefore, since all of the irreducible polynomials dividing $\Phi_r$ have degree $a$ and $\Phi_r$ has degree $\phi(r)$, $\Phi_r$ is a product of $\phi(r)/a$ irreducible factors with degree $a$.

Noble Mushtak
  • 18,402
  • 28
  • 44