# Decomposition of a prime number in a cyclotomic field

Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $A$ be the ring of algebraic integers in $\mathbb{Q}(\zeta)$. Let $p$ be a prime number such that $p \neq l$. Let $f$ be the smallest positive integer such that $p^f \equiv 1$ (mod $l$). Then $pA = P_1...P_r$, where $P_i's$ are distinct prime ideals of $A$ and each $P_i$ has the degree $f$ and $r = (l - 1)/f$.

My question: How would you prove this?

This is a related question.

Once you know what the ring of integers in $\mathbf{Q}(\zeta)$ is, you know that the factorization of a rational prime $p$ therein is determined by the factorization of the minimal polynomial of $\zeta$ over $\mathbf{Q}$, which is the cyclotomic polynomial $\Phi_\ell$, mod $p$. So you basically just need to determine the degree of a splitting field over $\mathbf{F}_p[X]$ of the image of $\Phi_\ell$ in $\mathbf{F}_p$. The degree is the $f$ in your question. This can be determined using the Galois theory of finite fields, mainly the fact that the Galois group is cyclic with a canonical generator. The details are carried out in many books, e.g., Neukrich's book on algebraic number theory.
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 $p$. By the answer to this question, the degree of each $f_i(X)$ is $f$. By this question, $P_i = (p, f_i(\zeta))$ is a prime ideal of $\mathbb{Z}[\zeta]$ lying over $p\mathbb{Z}$. It is easy to see that $\mathbb{Z}[\zeta]/P_i$ is a finite extension of $\mathbb{Z}/p\mathbb{Z}$ of degree $f$. It is well known that $\mathbb{Z}[\zeta]$ is the ring of algebraic integers in $\mathbb{Q}[\zeta]$. It is also well known that each $P_i$ has the same ramification index $e$ and $l - 1 = efg$, where $g$ is the number of prime ideals of $\mathbb{Z}[\zeta]$ lying over $p\mathbb{Z}$. Since $l - 1 = fr$, $e = 1$ and $r = g$. Hence $p\mathbb{Z} = P_1\cdots P_r$ in $\mathbb{Z}[\zeta]$.