Ramification in cyclotomic field I was trying to prove that:
$p$ is a odd prime and $\zeta$ is a primitive $p$-th root of unity and let $p^{*} = (-1)^{\frac{p-1}{2}}p$. Show that $\mathbb{Q}(\sqrt{p^{*}})$ is a quadratic subfield of $\mathbb{Q}(\zeta)$.
This is what I got so far:
The only prime ramified in $\mathbb{Q}(\zeta)$ is $p$, and the prime factorization in $\mathbb{Q}(\zeta)$ is: $(p) = (\mathcal{P_{1}}...\mathcal{P}_{n})^{\phi(p)=p-1}$ (Since it is Galois, so all primes have the same ramification degree). $p-1$is even, so I looked at $(\mathcal{P_{1}}...\mathcal{P}_{n})^{\frac{p-1}{2}}$. But I have trouble to determine what is this ideal exactly? Is this $(\sqrt{p^{*}})$? If it is, where does $(-1)^{\frac{p-1}{2}}$ comes from?
Any help is appreciated!

First, an informal answer to your question is, since you know that $2$ cannot ramify in $\mathbb Q(\zeta)$, it also cannot ramify in a quadratic subfield of $\mathbb Q(\zeta)$. But $2$ does ramify in $\mathbb Q(\sqrt{-p^*})$ but not in $\mathbb Q(\sqrt{p^*})$ (why?). This is why $(-1)^{\frac{p-1}{2}}$ comes from.
Let's try to find $\sqrt{p^*}$ in the field $\mathbb Q(\zeta)$.
A frequently used trick is that the value $(\zeta-1)(\zeta^2-1)\cdots(\zeta^{p-1}-1) = p$. To see this, note that $\zeta, \cdots, \zeta^{p-1}$ is the complete set of roots of $\frac{x^p-1}{x-1}$, so $\zeta-1, \cdots, \zeta^{p-1}-1$ is the complete set of roots of $\frac{(x+1)^p-1}{x}$, and $(\zeta-1)(\zeta^2-1)\cdots(\zeta^{p-1}-1)$ is the constant term of $\frac{(x+1)^p-1}{x}$, which is $p$.
Now, consider $A = (\zeta - 1)(\zeta^2 - 1)\cdots(\zeta^{\frac{p-1}{2}}-1)$ and $B = (\zeta^{\frac{p+1}{2}} - 1)\cdots(\zeta^{p-1}-1)$. Note that $AB = p$ by the previous paragraph.
We want one more relation between $A$ and $B$. Note that $(\zeta^l - 1) = -\zeta^l(\zeta^{p-l}-1)$, we have $A = (-1)^{\frac{p-1}2}\zeta^KB$ for some integer $K$. So $A^2\zeta^{-K} = p^*$ and hence $\sqrt{p^*} \in \mathbb Q(\zeta)$ (why?).
