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Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $K = \mathbb{Q}(\zeta)$. Let $A$ be the ring of algebraic integers in $K$. Let $G$ be the Galois group of $\mathbb{Q}(\zeta)/\mathbb{Q}$. $G$ is isomorphic to $(\mathbb{Z}/l\mathbb{Z})^*$. Hence $G$ is a cyclic group of order $l - 1$. Let $f$ be a positive divisor of $l - 1$. Let $e = (l - 1)/f$. There exists a unique subgroup $G_f$ of $G$ whose order is $f$. Let $K_f$ be the fixed subfield of $K$ by $G_f$. Let $A_f$ be the ring of algebraic integers in $K_f$.

Let $p$ be a prime number such that $p \neq l$. Let $f'$ be the order of $p$ mod $l$. Let $pA_f = P_1\cdots P_r$, where $P_1, \dots, P_r$ are distinct prime ideals of $A_f$. Let $P_1A = Q_1\cdots Q_s$, where $Q_1, \dots, Q_s$ are distinct prime ideals of $A$.

My question: Is the following proposition true? If yes, how would you prove this?

Proposition

(1) $r = (l - 1)/\operatorname{lcm}(f, f')$.

(2) $s = f/\gcd(f, f')$.

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1 Answer

The decomposition group at $p$ in $G$ is generated by the residue class of $p$. It a cyclic subgroup of the cylic group $G$ of order $f'$.

We have $Gal(K/K_f) = G_f$, and $Gal(K_f/\mathbb Q) = G/G_f$.

Then the general theory of how primes split shows that $r$ is the index in $G/G_f$ of the decomposition group at $p$, i.e. the index in $G/G_f$ of the subgroup generated by the image of $p$ in $G/G_f$, which by elementary cyclic group theory equals $(l-1)/lcm(f,f')$.

Similary, $s$ equals the index in $G_f$ of the subgroup generated by the decomposition group of $P_1$, which is the intersection with $G_f$ of the subgroup of $G$ generated by the image of $p$. Again, elementary cyclic group theory shows then that $s = f/gcd(f,f')$.

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