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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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marked as duplicate by joriki, Gerry Myerson, Brandon Carter, J. M., t.b. Aug 11 '12 at 9:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

this question was answered here:… – user8268 Jul 23 '12 at 8:51
@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? – Makoto Kato Jul 23 '12 at 10:13
The question is (in a sense) also answered here and here. 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 $f$. – Jyrki Lahtonen Jul 23 '12 at 10:37

We denote by $|S|$ the number of elements of a finite set $S$.

Let $F = \mathbb{Z}/p\mathbb{Z}$. Let $\Omega$ be the algebraic closure of $F$. Let $\omega \neq 1$ be a root of $X^l - 1$ in $\Omega$. Let $K$ be the unique subfield of $\Omega$ such that $|K| = p^f$. $K$ is the set of all the roots of $X^{p^f} - X$ in $\Omega$. $K$ is a finite extension of $F$ and $[K : F] = f$. Let $K^* = K -$ {$0$} be the multiplicative group of $K$. It is well known that $K^*$ is a cyclic group. Since $|K^*| = p^f - 1$ and $l|p^f - 1$, $K^*$ has a unique cyclic subgroup of order $l$. Hence $\omega \in K^*$.

Let $L$ be a proper subfiled of $K$. Let $[L : F] = p^r$. Suppose $\omega \in L$. Then $l|p^r - 1$, i.e. $p^r \equiv 1$ (mod $l$). Since $r < f$, this is a contradiction. Hence $K = F(\omega)$. Hence the minimal polynomial of $\omega$ over $F$ has degree $f$. This completes the proof.

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