I want to prove the assertion:
The unique quadratic subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}(\sqrt{p})$ when $p \equiv 1 \pmod{4}$, respectively $\mathbb{Q}(\sqrt{-p})$ when $p \equiv 3 \pmod{4}$.
My first attempt is this. In $\mathbb{Z}[\zeta_p]$, $1-\zeta_p$ is prime and
$$
p = \epsilon^{-1} (1-\zeta_p)^{p-1}
$$
where $\epsilon$ is a unit. Since $p$ is an odd prime, $(p-1)/2$ is an integer and
$$
\sqrt{\epsilon p } = (1-\zeta_p)^{(p-1)/2}
$$
makes sense and belongs to $\mathbb{Z}[\zeta_p]$.
How do I deal with the $\epsilon$ under the square root? I guess the condition on the congruence class of $p$ comes from that. Is this even the right way to proceed?
Uniqueness is not clear to me either. I thought about looking at the valuation $v_p$ on $\mathbb{Q}$, extending it to two possible quadratic extensions beneath $\mathbb{Q}(\zeta_p)$, then seeing how those have to extend to common valuations on $\mathbb{Q}(\zeta_p)$, but I didn't see how to make it work.
I would appreciate some help.
Best Answer
My favorite way to prove this is to explicitly write down the quadratic Gauss sum: $$g_p = \sum_{a \in \mathbb F_p} \left( \frac{a}{p} \right) \zeta_p^{a}$$ Then you can show $g_p^2 = (-1)^{\frac{p-1}{2}} p$ by direct manipulation. This gives the result very explicitly!