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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 $\mathfrak{l} = (1 - \zeta)A$. $\mathfrak{l}$ is a prime ideal lying over $l$. Let $\mathfrak{l}_f = \mathfrak{l} \cap A_f$.

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


(1) $lA = \mathfrak{l}^{l-1}$.

(2) $lA_f = \mathfrak{l}_f^e$.

(3) $\mathfrak{l}_fA = \mathfrak{l}^f$

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1 Answer 1

up vote 3 down vote accepted

Note: I am going to use $p$ everywhere instead of $l$.

Yes, and it all follows from the fact that $p$ is totally ramified. To see this, show that $N_{K/\mathbb{Q}}(1-\zeta) = p = [A : \mathfrak{p}]$ and notice that $p = \prod_{k \in (\mathbb{Z}/p\mathbb{Z})^\times} (1 - \zeta^k) = \epsilon (1 - \zeta)^{p-1}$ for some cyclotomic unit $\epsilon$. Hence $pA = \mathfrak{p}^{p-1}$.

$(2)$ and $(3)$ are immediate consequences of $(1)$.

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