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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 $l - 1$. 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 $P$).

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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.

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