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Is it possible to determine how many irreducible factors $X^p-1$ in the polynomial ring $(\mathbb{Z}/q \mathbb{Z})[X]$ has and maybe even the degrees of the irreducible factors? ($p,q$ are a primes with $\gcd(p,q)=1$).

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$x^p-1=(x-1)^p$ –  user8268 Jul 6 '12 at 13:41
    
sorry, I wrote $p$ instead of $q$ –  Narayan Jul 6 '12 at 13:45

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up vote 4 down vote accepted

It has one factor of degree $1$, namely $x-1$. All the remaining factors have the same degree, namely the order of $q$ in the multiplicative group $(\mathbb{Z}/p \mathbb{Z})^*$. To see it: this is the length of every orbit of the action of the Frobenius $a\mapsto a^q$ on the set of the roots of $(x^p-1)/(x-1)$ in the algebraic closure.

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ahh, ok, thank I got it. Thank you –  Narayan Jul 6 '12 at 14:04

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