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Thread: Irreducibility in Z

  1. #1
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    Irreducibility in Z

    Prove that polynomial $\displaystyle \prod_{1\leq i\leq j\leq \phi(n)}(x-\omega_n^{q_i+q_j})$ is irreducible in $\displaystyle Z$ where $\displaystyle gcd(q_i, n)=1$, $\displaystyle \omega_n=e^{i\frac{2\pi}n}$ and $\displaystyle \phi$ Euler's totient function.
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  2. #2
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    This is wrong. Let $\displaystyle \zeta = e^{2\pi i/5}$.
    Then this polynomial is:
    $\displaystyle (x-\zeta^2)(x-\zeta^3)(x-\zeta^4)(x-\zeta^5)$
    $\displaystyle (x-\zeta^4)(x-\zeta^5)(x-\zeta^6)$
    $\displaystyle (x-\zeta^6)(x-\zeta^7)$
    $\displaystyle (x-\zeta^8)$

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

    Use the fact that $\displaystyle (x-\zeta)(x-\zeta^2)(x-\zeta^3)(x-\zeta^4) = x^4+x^3+x^2+x+1$
    This gives:
    $\displaystyle (x-1)(x^4+x^3+x^2+x+1)^2$
    This is not irreducible.
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  3. #3
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    You forgot one member $\displaystyle (x-1)$ but polynomial is still not irreducible.
    Thanks!
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  4. #4
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    Quote Originally Posted by rodeo View Post
    You forgot one member $\displaystyle (x-1)$ but polynomial is still not irreducible.
    Thanks!
    Your problem looks similar to this, hopefully you never seen it before. Let $\displaystyle \zeta = e^{2\pi i/n}$. Now define $\displaystyle f(x) = \prod_{1\leq k\leq n}^{\gcd(n,k)=1} (x - \zeta^k)$.
    It turns out that $\displaystyle f(x) \in \mathbb{Z}[x]$ and $\displaystyle f(x)$ is irreducible.
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  5. #5
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    I know for that so called cyclotomic polynomials and there are my primerily idea to creating the problem.

    Right question is :
    find $\displaystyle S$ such $\displaystyle \prod_{1\leq i<j\leq k}(x-\omega_n^{q_iq_j})\in Z[x]$, $\displaystyle q_i\in S$, $\displaystyle 1\leq i\leq k$.
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