Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $K = \mathbb{Q}(\zeta)$. Let $A$ be the ring of algebraic integers in $K$. Let $G$ be the Galois group of $\mathbb{Q}(\zeta)/\mathbb{Q}$. $G$ is isomorphic to $(\mathbb{Z}/l\mathbb{Z})^*$. Hence $G$ is a cyclic group of order $l - 1$. Let $f = (l - 1)/2$. There exists a unique subgroup $G_f$ of $G$ whose order is $f$. Let $K_f$ be the fixed subfield of $K$ by $G_f$. $K_f$ is a unique quadratic subfield of $K$. Let $d$ be the discriminant of $K_f$.

My question: Is the following proposition true? If yes, how would you prove this?


(1) If $l \equiv 1$ (mod 4), then $d = l$.

(2) If $l \equiv -1$ (mod 4), then $d = -l$.

Remark I think, together with this, we can get the quadratic reciprocity law.


1 Answer 1


There are various ways to prove this. Write $l^* = \pm l$, the sign chosen so that $l^* \equiv 1 \pmod 4$. (This is what you call $d$.) First note that $\mathbb Q(\sqrt{l^*})$ is the unique quad. ext. of $\mathbb Q$ with discriminant equal to $l^*$ (by the explicit computation of discriminants of quad. number fields), and so one has to prove that $K_f = \mathbb Q(\sqrt{l^*})$.

  1. Theoretical approach: By ramification theory, $K_f$ is a quad. ext. of $\mathbb Q$ ramified only at $\ell$. Thus it equals $\mathbb Q(\sqrt{l^*})$, since this is the unique quad. number field ramified only at $l$.

  2. Explicit approach: Consider the Gauss sum $\sum_{n = 0}^{l-1} \bigl( \dfrac{n}{l} \bigr) e^{2\pi i n/l}$. This is obviously an element of $K$, and is checked to equal a square root of $l^*$. Thus $\mathbb Q(\sqrt{l^*})$ is a quadratic subfield of $K$. By the uniques of $K_f$, we find that $K_f = \mathbb Q(\sqrt{l^*})$.

By the way, as a cultural note, the proof of QR that you are aiming towards is the proof that is generalized by class field theory (and especially Artin's reciprocity law).

In its explicit form (i.e. using the above explicit approach), I think it goes back to Gauss, and a modern treatment can be found in Ireland and Rosen.

  • $\begingroup$ There's some other trick for calculating the discriminant by hand that I think I learned from Dave Savitt. Hopefully I'll remember it in the morning. Of course, Gauss sums are useful anyway, but [you might remember!] I had no appreciation for them in the beginning. $\endgroup$ Jul 25, 2012 at 5:15

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