Splitting primes and quadratic reciprocity.

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I'm reading through Marcus's wonderful book Number Fields, and I had two questions on his proof of quadratic reciprocity in chapter $4$.

Marcus states first that if $p$ is an odd prime, and $q$ is any prime, that in the field $\mathbb{Q}[\omega]$ where $\omega = e^{2\pi i /p}$ that $q$ will split into $r$ primes, where $r = (p-1)/f$ and $f$ is the multiplicative order of $q \mod p$. He says this like it's obvious, so maybe I'm missing something simple, but I do not understand this line.

My second question is that Marcus also claims that the following are equivalent

$1)$ $q$ is a $d^{th}$ power $\mod p$

$2)$ $f|(p-1)/d$

$3)$ $d|r$

$4)$ $F_d \subset F_r$

Where $F_d$ is the unique subfield of $\mathbb{Q}[\omega]$ of degree $d$ over $\mathbb{Q}$ guaranteed by the fact that the Galois group of $\mathbb{Q}[\omega]/\mathbb{Q}$ is cyclic.

I think $1\implies 2\implies 3 \implies 4$ are pretty trivial, as well as $4\implies 3 \implies 2$, but I can't quite see the equivalence of $1$ with the rest of these statements.

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Your first question has its answer in the Corollary of Theorem 26. Observe that as $q$ does not divide $p$, so it should split into exactly $\frac{p-1}{f}$ factors, which he has denoted by $r$, and where $f$ is the multiplicative order of $q$ modulo $p$. So, $r = \frac{p-1}{f}$. In your second question, understanding the equivalence between two statements just as you have already mentioned will suffice to make sure that all are equivalent.