I am attempting to solve Artin 16.10.9, part (b). I have already solved (a).
Let $f(x)=(x-α_1) \cdots (x-α_n)$.
(a) Prove that the discriminant of $f$ is $\pm f'(α_1) \cdots f'(α_n)$, where $f'$ is the derivative of $f$, and determine the sign.
(b) Use the formula to compute the discriminant of the polynomial $x^p-1$, and use it to give another proof of Theorem 16.10.12.
Here is Theorem 16.10.2, which we are asked to prove.
Theorem: Let $p$ be a prime different from $2$, and let $L$ be the unique quadratic extension of $\mathbb{Q}$ contained in the cyclotomic field $\mathbb{Q}(ζ_p)$. If $p \equiv 1$ (mod 4), then $L=\mathbb{Q}(\sqrt p)$, and if $p \equiv 3$ (mod 4), then $L=\mathbb{Q}(\sqrt{-p})$.
My progress so far
I have shown, for part (a), that the discriminant of $f$ is $(-1)^{n \choose 2}f'(α_1) \cdots f'(α_n)$. Then, using this result, I have shown that the discriminant of $x^p-1$ is $(-1)^{p \choose 2}p^p$. If I have made a mistake, please correct me.
Now, I want to use this formula to prove Theorem 16.10.2. So let $p\neq2$ and let $L$ be the unique quadratic extension contained in $\mathbb{Q}(ζ_p)$. Write $L=\mathbb{Q}(\sqrt k)$. For now, assume $p \equiv 1$ (mod 4).
I see that $\mathbb{Q}(ζ_p)/ \mathbb{Q}$ is a Galois extension, with its Galois group isomorphic to $C_{p-1}$. Additionally, I see that $\mathbb{Q}(ζ_p)/\mathbb{Q}(\sqrt k)$ is a Galois extension as well.
I'm not sure how to show that $L=\mathbb{Q}(\sqrt p)$ though. How do it do it? Please help!
Since both cases are similar, here we'll just prove the theorem for the $p \equiv 1$ (mod 4) case.
To show $L=\mathbb{Q}(\sqrt p)$, it is enough to show that $L=\mathbb{Q}(\sqrt D)$ since these two fields are the same since $D=p^p$.
To show $L=\mathbb{Q}(\sqrt D)$, we just need to show $\sqrt D \in \mathbb{Q}(ζ_p$), but this is obvious since $\sqrt D$ will be a product of differences of roots of $x^p-1$, hence a product of $(ζ_p^j-ζ_p^k)$ terms.