# Is it possible to derive the quadratic reciprocity law from the decomposition law in a quadratic extension?

How can I derive the quadratic reciprocity law from the decomposition law for a prime in a quadratic extension?

There is no need to read the following texts.

I know the necessary and sufficient conditions for a prime to be inert/ split/ ramified in a quadratic extension, also I know how to prove these conditions.

Let $$K=\mathbb{Q}(\sqrt{m})$$, and let's denote its discriminant by $$\Delta_K$$. And Let $$p$$ be an odd prime number. Then

• $$p$$ is ramified, $$p\mathcal{O}_K=\mathfrak{p}^2$$, if and only if $$p \mid \Delta_K$$.
• $$p$$ is split, $$p\mathcal{O}_K=\mathfrak{p}_1\mathfrak{p}_2$$, if and only if $$\left(\dfrac{\Delta_K}{p}\right)=\left(\dfrac{m}{p}\right)=1$$.
• $$p$$ is inert, $$p\mathcal{O}_K=\mathfrak{p}$$, if and only if $$\left(\dfrac{\Delta_K}{p}\right)=\left(\dfrac{m}{p}\right)=-1$$.

Also, we can list criteria for $$p=2$$.

• I imagine so: quadratic reciprocity is equivalent to the totality of binary quadratic forms theory, according to Kaplansky (verbal). A text that follows forms and ideals side by side is Lehman, Quadratic Number Theory: An Invitation to Algebraic Methods in the Higher Arithmetic. Oct 22, 2020 at 15:00
• @WillJagy I didn't aware of such a beautiful book by Lehman. Thank you so much for introducing that book. I read the first two chapters of the second edition (2012 or maybe 2013) of the book "Primes of the form $x^2+ny^2$" by D. Cox. I am aware of the correspondence between the binary quadratic forms of discriminant $D$ and the proper ideals of the order $\mathcal{O}$ of discriminant $D$, Cox, Chapter 2, Theorem 7.7, pages 123-124. Thank you so much for your fruitful comment and that beautiful book. Oct 22, 2020 at 16:09

Yes it is. Let $$K = \mathbb Q(\sqrt{p^*})$$ be the quadratic subfield of $$\mathbb Q(\zeta_p)$$ (so $$p^* = p$$ if $$p\equiv 1\pmod 4$$ and $$-p$$ if $$p\equiv -1\pmod 4$$. In particular, $$K$$ is unramified away from $$p$$.
On the one hand, by the Kummer-Dedekind theorem, $$q$$ splits in $$K$$ if and only if $$X^2-p^*$$ is reducible mod $$q$$ – i.e. if $$\left(\frac {p^*}q\right)=1$$.
On the other hand, $$K\subset \mathbb Q(\zeta_p)$$. Using the action of Frobenius in $$\mathbb Q(\zeta_p)$$ one can show that $$q$$ splits in $$K$$ if and only if $$\left(\frac {q}p\right)=1$$
It follows that $$\left(\frac qp\right)=\left(\frac {p^*}q\right)$$, which is exactly quadratic reciprocity.
• What I've written is exactly derived from the Kummer-Dedikind theorem. And I think the second Paragraph has a mistake: "For $q\ne p$ another (odd) prime, $q$ splits in $K = \mathbb Q(\sqrt{p^*})$ if and only if $\left(\frac p{q*}\right)=1$." But I think pointing out to the special quadratic extension $K = \mathbb Q(\sqrt{p^*})$ is a key point. Thank you so much for your fruitfull help. Oct 22, 2020 at 16:17
• You're right - I misread what you'd written. The other computation involves using that fact that $K\subset\mathbb Q(\zeta_p)$ and playing with Frobenius elements. Wikipedia has the full proof! Oct 22, 2020 at 16:42