Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Proposition

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

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

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

share|improve this question
add comment

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 $(1)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.