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 $p=2$.
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.