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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.

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  • $\begingroup$ We don't all have a copy of Marcus lying around. Please reproduce the statement of the proof you need help on. $\endgroup$
    – user147556
    Commented Jul 6, 2021 at 5:25
  • $\begingroup$ @MichaelBarz I am extremely sorry that I did not mention the statements clearly. I have now edited my question. $\endgroup$ Commented Jul 6, 2021 at 6:14

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Consider the cyclotomic field $\mathbb Q(\zeta_p)$, which has a unique quadratic subfield $\mathbb Q(\sqrt{p^*})$, where $p^*=(-1)^{\frac{p-1}2}p$. Then $q$ splits completely in $\mathbb Q(\sqrt{p^*})$ if and only if $q$ is a quadratic residue modulo $p$ by Theorem 30. On the other hand, $q$ splits completely in $\mathbb Q(\sqrt{p^*})$ if and only if $p^*$ is a quadratic residue modulo $q$ by Theorem 25. Thus $$\left(\frac qp\right)=\left(\frac{p^*}q\right)=(-1)^{\frac{(p-1)(q-1)}4}\left(\frac pq\right).$$

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  • $\begingroup$ Thank you very much! It is a simple yet beautiful explanation. @CHENRK $\endgroup$ Commented Jul 7, 2021 at 6:22

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