I have been studying from the book Number Fields by Marcus and I come across a proof for the Law of Quadratic Reciprocity using the splitting of prime numbers in quadratic extensions. To be precise, it is stated as a corollary to Theorem 30. Although I know independent proofs of the Law, using Gauss' Lemma, I could not figure out how this one can be solved.
Theorem 30: Let $p$ be an odd prime and $q$ be another prime not same as $q$. Then for a fixed divisor $d$ of $p-1$, $q$ is a $d^{th}$ power modulo $p$, if and only if $q$ splits completely in $\mathbb{F}_d$, where $\mathbb{F}_d$ is the fixed field of the unique subgroup of $Gal(\mathbb{Q}[e^{\frac{2 \pi i}{p}}]/\mathbb{Q})$ of order $\frac{p-1}{d}$.
The corollary to this theorem is to prove the Law of Quadratic Reciprocity. I understand that a prime splits in a quadratic extension $\mathbb{Q}[\sqrt{q}]$ if it is a quadratic residue modulo $q$. But I do not understand the following part where it says the result follows from Theorem 25, which is the explicit formula for the splitting of primes in such extensions. I would really appreciate any kind of help to write down the proof. Thanks in Advance!
I am mentioning the Theorem 25 below.
Theorem 25: Let $R = \mathbb{Z}[\sqrt{m}]$, where $m$ is square-free.
1. If $p\;|\;m$, then
\begin{eqnarray}
pR = (p,\sqrt{m})^2
\end{eqnarray}
2. If $m$ is odd, then
\begin{eqnarray}
2R = \begin{cases}
(2,1+\sqrt{m})^2,\;\text{when}\; m\equiv 3\;(mod\;4) \\
\left(2,\frac{1+\sqrt{m}}{2}\right)\left(2,\frac{1-\sqrt{m}}{2}\right),\;\text{when}\; m\equiv 1\;(mod\;8) \\
prime,\;\text{when}\;m \equiv 5\;(mod\;8)
\end{cases}
\end{eqnarray}
3. If $p$ is odd and $p \not |\;m$, then
\begin{eqnarray}
pR = \begin{cases}
(p,n+\sqrt{m})(p,n-\sqrt{m}),\;\text{when $m \equiv n^2 \;(mod\;p)$} \\
prime,\;\text{otherwise}
\end{cases}
\end{eqnarray}
Also, the factors involved in 2nd case of (2) and 1st case of (3), are distinct.
