# Discriminant of the quadratic subfield of the cyclotomic number field of an odd prime order $l$

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?

Proposition

(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.

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 $K_f = \mathbb Q(\sqrt{l^*})$.