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 $l - 1$. 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 $p \neq l$ and $f$ is the order of $p$ mod $l$. Let $P$ be a prime ideal of $A_f$ lying over $p$.

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

Proposition For every $\alpha \in A_f$, $\alpha^p \equiv \alpha$ (mod $P$).

share|improve this question

1 Answer 1

up vote 2 down vote accepted

This follows from the theory of decomosition groups. By assumption the decomposition group at $p$ in $G$ equals $G_f$, and so the decomposition group at $p$ in $G/G_f$ is trivial, which says that $p$ splits completely in $K_f$, which says that the residue field of $P$ is just $\mathbb F_p$, which gives the proposition.

share|improve this answer

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.