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 $p=2$.

  • $\begingroup$ 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. $\endgroup$
    – Will Jagy
    Oct 22, 2020 at 15:00
  • $\begingroup$ @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. $\endgroup$ Oct 22, 2020 at 16:09

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Oct 22, 2020 at 16:17
  • 1
    $\begingroup$ 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! $\endgroup$
    – Mathmo123
    Oct 22, 2020 at 16:42
  • $\begingroup$ Thank you so much for your explanations and for a very fruitful wiki link. $\endgroup$ Oct 22, 2020 at 16:55

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