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Possible Duplicate:
Irreducible factors of $X^p-1$ in $(\mathbb{Z}/q \mathbb{Z})\[X\]$

Let $p$ be a prime number. Let $l$ be an odd prime number such that $l \neq p$.

Let $X^l - 1 \in \mathbb{Z}[X]$. Since $(X^l - 1)' = lX^{l-1}$, $X^l - 1$ has no multiple irreducible factor mod $p$. Since $X^l - 1 = (X - 1)(1 + X + ... + X^{l-1})$, $1 + X + ... + X^{l-1}$ has no multiple irreducible factor mod $p$, either.

Let $1 + X + ... + X^{l-1} \equiv f_1(X)...f_r(X)$ (mod $p$), where $f_i(X)$ is a monic irreducible polynomial mod $p$.

Let $f$ be the smallest positive integer such that $p^f \equiv 1$ (mod $l$).

My question: Can we prove that the degree of each $f_i(X)$ is $f$?

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    this question was answered here: http://math.stackexchange.com/questions/167486/irreducible-factors-of-xp-1-in-mathbbz-q-mathbbzx2012-07-23
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    @user8268 I suppose you think about the splitting field $K$ of $1 + X + ... + X^{l-1}$ over $F = \mathbb{Z}/p\mathbb{Z}$ and the Galois group $G$ of $K/F$, right?2012-07-23
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    The question is (in a sense) also answered [here](http://math.stackexchange.com/q/153429/11619) and [here](http://math.stackexchange.com/q/172468/11619). In your case of prime $\ell$ it is easy to see that the ($p$-)cyclotomic cosets modulo $\ell$, other than the coset of $0$, all have the same size $f$.2012-07-23

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