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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 $f = (l - 1)/2$. There exists a unique subgroup $G_f$ of $G$ whose order is $f$. Let $K_f$ be the fixed subfield of $K$ by $G_f$. $K_f$ is a unique quadratic subfield of $K$. Let $A_f$ be the ring of algebraic integers in $K_f$.

Let $p$ be a prime number such that $p \neq l$. Let $pA_f = P_1\cdots P_r$, where $P_1, \dots, P_r$ are distinct prime ideals of $A_f$.

Since $p^{l - 1} \equiv 1$ (mod $l$), $p^f = p^{(l - 1)/2} \equiv \pm$1 (mod $l$).

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

Proposition

(1) If $p^{(l - 1)/2} \equiv 1$ (mod $l$), then $r = 2$.

(2) If $p^{(l - 1)/2} \equiv -1$ (mod $l$), then $r = 1$.

This is a related question.

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    @MattE Dear Matt, If someone asks a question which can be answered by using the above result, I think it's easier and perhaps better to show him the link(under the condition that it has a good answer) than to write the result and say this can be proved by the notion of decomposition groups and refer to Neukirch, etc.. Regards,2012-07-26

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This is most easily established by decomposition group calculations; it is a special case of the more general result proved here (which is an answer to OP's linked question).