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?


A cyclic group has a unique, normal subgroup of every possible order, and all these subgroups and their quotients are also cyclic.

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 $K = \mathbb{Q}(\sqrt{l^*})$.

  • $\begingroup$ We can assume $Z_{\mathfrak{p}}|\mathbb{Q}$ is Galois because $\text{Gal}(\mathbb{Q}(\zeta)|\mathbb{Q})$ is cyclic, right? $\endgroup$ – rmdmc89 May 17 at 16:45
  • 1
    $\begingroup$ That's correct. Every subgroup of a cyclic (or even just abelian) group is normal, so the corresponding subextension is Galois. $\endgroup$ – Ted May 18 at 3:14

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