# Construction of genus class fields

Given a finite extension $$K/\mathbb{Q}$$, the genus class field $$L$$ is defined to be the maximal abelian extension of $$\mathbb{Q}$$ that is a subfield of the Hilbert class field $$H$$ of $$K$$. I am trying to understand the proof of an alternative construction of $$G$$ in the case when $$K$$ is a cyclic extension with prime degree $$l$$ as follows.

Let $$p_1, p_2, ..., p_n$$ be the primes in $$\mathbb{Q}$$ that ramify in $$K$$. Let $$L_1, L_2, ..., L_n$$ be cyclic extensions of degree $$l$$ over $$\mathbb{Q}$$ such that $$L_i$$ ramifies only at $$p_i$$. Then $$L = L_1 L_2...L_n$$.

I was reading the proof in "Construction of class Fields - Carl Herz". The basic strategy is to associate $$L$$ and the $$L_i$$'s to norm subgroups of $$K$$ using class field theory and relate them. But I find it hard to comprehend and justify certain crucial claims in the proof one of which I posted on MO earlier (Norm groups of number fields).

Can someone refer me to an alternate proof of this fact or provide some light into Carl's proof if possible?

• @YCor Just edited, replaced the $G$'s with L May 8, 2021 at 16:00
• Herz's first name is Carl. The claim is almost obvious: by Abhyankar's Lemma, the extension in question is unramified, and adjoining any other cyclic extensions ramified at a different prime gives you a ramified extension of $K$. May 9, 2021 at 6:14
• @FranzLemmermeyer Thank you for your comment. Edited Herz's name. But what if I adjoin another field ramified only at say $\mathfrak{p_1}$ and degree say $l^2$? May 9, 2021 at 14:45
• Then the ramification index is divisible by $\ell^2$, which survives and gives you a ramified extension of $K$. May 10, 2021 at 5:24
• It's a standard result from the theory of Hilbert ramification groups. If the ramification index is $\ell$, the unramified extension (inertia subfield) must be at the bottom. May 12, 2021 at 11:41