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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?

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  • $\begingroup$ @YCor Just edited, replaced the $G$'s with L $\endgroup$
    – Melanka
    Commented May 8, 2021 at 16:00
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    $\begingroup$ 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 $K$. $\endgroup$
    – Franz Lemmermeyer
    Commented May 9, 2021 at 6:14
  • $\begingroup$ @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 $l^2$? $\endgroup$
    – Melanka
    Commented May 9, 2021 at 14:45
  • $\begingroup$ Then the ramification index is divisible by $\ell^2$, which survives and gives you a ramified extension of $K$. $\endgroup$
    – Franz Lemmermeyer
    Commented May 10, 2021 at 5:24
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    $\begingroup$ 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. $\endgroup$
    – Franz Lemmermeyer
    Commented May 12, 2021 at 11:41

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