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Let $m \in \mathbb{N}$, $K$ a field of characteristic $k$, $k \nmid m$. Let $\alpha \in K$. Then, if $\phi_m(x)$ is the cyclotomic polynomial. $\phi_m(\alpha)=0$ iff $\alpha$ is a primitive $m^{th}$ root of unity.

Proof:

$\Rightarrow )$ $\phi_m(\alpha)= \prod_i (\alpha - w_i)=0$, where $w_i$ are primitive $m^{th}$ root of unity. Then, $\alpha$ must be a primitive $m^{th}$ root of unity.

$\Leftarrow )$ ?

How i go?

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What is your definition of cyclotomic polynomial? With the one in Wikipedia, it's easy. –  lhf Nov 27 '12 at 5:18
    
@GerryMyerson, is not a number field, my mistake! Is only a field. Thanks! –  P. M. O. Nov 27 '12 at 16:06
    
Maybe the question you are asking is whether the polynomial that works over the rationals also works over fields in general. I think you'll find that follows from facts about the number of zeros of a polynomial over a field, facts about reduction modulo $k$ being a homomorphism, and suchlike. But the question could do with some editing/clarification. –  Gerry Myerson Nov 27 '12 at 21:59
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