Decomposition of a prime number $p \neq l$ in the quadratic subfield of a cyclotomic number field of an odd prime order $l$ 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.

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).
