Subfield of cyclotomic extension of degree 2 is either $\Bbb Q(\sqrt{p})$ or $\Bbb Q(\sqrt{-p})$. From my previous post, I conlcude that 'If $\zeta$ is a primitive $p$th root of 1 ($p\geq 3$), then there is a unique subfield $K\subset\Bbb Q(\zeta)$ such that $[K:\Bbb Q] = 2$'. I want to find the field $K$ explicitly. I hear that such $K = \Bbb Q(\sqrt{p})$ or $\Bbb Q(\sqrt{-p})$. Why is that?
I also heard that this is the special case of 'Kroneker-Weber' theorem.
 A: The cyclotomic extension $E := \mathbb{Q}(\zeta)$ of $\mathbb{Q}$ is Galois with Galois group isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\times} \cong \mathbb{Z}/(p-1)\mathbb{Z}$. Since $p \geq 3$, $p-1$ is odd, and so $\mathbb{Z}/(p-1)\mathbb{Z}$ contains a (unique) subgroup of index $2$. If we call this subgroup $H$, then the fixed field $K := \mathbb{Q}(\zeta)^{H}$ is of degree $2$ over $\mathbb{Q}$ by Galois theory. We want to show that $K = \mathbb{Q}(\sqrt{p})$ or $K = \mathbb{Q}(\sqrt{-p})$.
Suppose that $K = \mathbb{Q}(\sqrt{D})$. Recall that the discriminant of $E$ over $\mathbb{Q}$ is (up to some sign) $p^{p-1}$. Since the discriminant of $K$ divides the discriminant of $E$, we conclude that the only prime divisor of the discriminant of $K$ is $p$. But the discriminant of $K$ is known: it is $D$ if $D \equiv 1 \pmod 4$, and $4D$ otherwise. Hence, since the only divisor of the discriminant of $K$ is $p$, we must have that $D$ is a power of $p$ or $-p$, depending on which of the two is congruent to $1$ mod $4$. We can ignore any even powers of $p$ in $D$, whence we conclude that $K = \mathbb{Q}(\sqrt{p})$ or $K = \mathbb{Q}(\sqrt{-p})$.
A: It's not really Kronecker-Weber, which asserts (as a special case) that every quadratic extension of $\mathbb Q$ is a subfield of a cyclotomic field. This is a much easier question.
A non-intuitive, but profoundly illuminating, argument is via Gauss sums. Namely, letting $\chi(a)$ be the quadratic character mod $p$, and $\psi(a)=e^{2\pi ia/p}$, the Gauss sum is $g(\chi)=\sum_{a\mod p} \chi(a)\psi(a)$.
A standard computation gives $|g(\chi)|=\sqrt{p}$, and $g(\chi)^2=\varepsilon\cdot p$, where $\varepsilon=\pm 1$ depending whether $-1$ is a square mod $p$ or not.
(This computation is very standard, but/and is also eminently do-able, once one knows that it is possible.)
