Quadratic reciprocity and decomposition of primes in cyclotomic fields

In Neukirch's Algebraic Number Theory, there is a proof of the quadratic reciprocity which makes use of proposition $$10.5$$: $$p\text{ is totally split in }\mathbb{Q}(\sqrt{\ell^*})\Leftrightarrow p\text{ splits in }\mathbb{Q}(\zeta)\text{ into an even number of primes}$$ [here $$p,\ell\in\mathbb{Z}$$ are odd primes, $$\zeta$$ is a primitive $$\ell$$-th root of unity and $$\ell^*:=(-1)^{\frac{\ell-1}{2}}\ell$$]

I'm trying to understand the proof of $$(\Leftarrow)$$.

He observes that, if $$\mathfrak{p}$$ is a prime above $$p$$, then the even number of primes is equivalent to $$[Z_{\mathfrak{p}}:\mathbb{Q}]$$ being even (where $$Z_{\mathfrak{p}}$$ is the decomposition field). I'm ok with that.

The part I don't understand is when he says "since $$\text{Gal}(\mathbb{Q}(\zeta)| \mathbb{Q})$$ is cyclic, it follows that $$\mathbb{Q}(\sqrt{\ell^*})\subset Z_{\mathfrak{p}}$$". How does one follow from the other?

Therefore by the correspondence between Galois groups and extension fields, if $$Z_{\frak{p}}$$ has even degree over $$\mathbb{Q}$$, then there is a unique extension $$K$$ of $$\mathbb{Q}$$ of degree 2 contained $$Z_\frak{p}$$. But by the same reasoning, $$K$$ is also the unique extension of $$\mathbb{Q}$$ of degree 2 contained in $$\mathbb{Q}(\zeta)$$. Since we know (I'm assuming we already know this) that $$\mathbb{Q}(\sqrt{l^*})$$ is a quadratic extension of $$\mathbb{Q}$$ contained in $$\mathbb{Q}(\zeta)$$, it follows that $$K = \mathbb{Q}(\sqrt{l^*})$$.
• We can assume $Z_{\mathfrak{p}}|\mathbb{Q}$ is Galois because $\text{Gal}(\mathbb{Q}(\zeta)|\mathbb{Q})$ is cyclic, right? – rmdmc89 May 17 at 16:45