# Proof of Quadratic Reciprocity using the splitting of primes

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.

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

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 $$\left(\frac qp\right)=\left(\frac{p^*}q\right)=(-1)^{\frac{(p-1)(q-1)}4}\left(\frac pq\right).$$