# Thread: Irreducibility in Z

1. ## Irreducibility in Z

Prove that polynomial $\prod_{1\leq i\leq j\leq \phi(n)}(x-\omega_n^{q_i+q_j})$ is irreducible in $Z$ where $gcd(q_i, n)=1$, $\omega_n=e^{i\frac{2\pi}n}$ and $\phi$ Euler's totient function.

2. This is wrong. Let $\zeta = e^{2\pi i/5}$.
Then this polynomial is:
$(x-\zeta^2)(x-\zeta^3)(x-\zeta^4)(x-\zeta^5)$
$(x-\zeta^4)(x-\zeta^5)(x-\zeta^6)$
$(x-\zeta^6)(x-\zeta^7)$
$(x-\zeta^8)$

Using the fact that $\zeta^5 = 1$ we get:
$(x-1)(x-\zeta)^2(x-\zeta^2)^2(x-\zeta^3)^2(x-\zeta^4)^2$.

Use the fact that $(x-\zeta)(x-\zeta^2)(x-\zeta^3)(x-\zeta^4) = x^4+x^3+x^2+x+1$
This gives:
$(x-1)(x^4+x^3+x^2+x+1)^2$
This is not irreducible.

3. You forgot one member $(x-1)$ but polynomial is still not irreducible.
Thanks!

4. Originally Posted by rodeo
You forgot one member $(x-1)$ but polynomial is still not irreducible.
Thanks!
Your problem looks similar to this, hopefully you never seen it before. Let $\zeta = e^{2\pi i/n}$. Now define $f(x) = \prod_{1\leq k\leq n}^{\gcd(n,k)=1} (x - \zeta^k)$.
It turns out that $f(x) \in \mathbb{Z}[x]$ and $f(x)$ is irreducible.

5. I know for that so called cyclotomic polynomials and there are my primerily idea to creating the problem.

Right question is :
find $S$ such $\prod_{1\leq i, $q_i\in S$, $1\leq i\leq k$.