SKW

6 Auxiliary results

Lemma 74
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Any abelian extension of \(\mathbb {Q}(\zeta _m)\) contained in \(\mathbb {Q}(\zeta _m, \zeta _p)\) and unramified at all primes is trivial.

Proof
Lemma 75
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Let \(F\) be a number field containing \(\mu _p\), let \(\mu \in F^\times \), and let \(L = F(\sqrt[p]{\mu })\). Let \(G_0 \leq \mathrm{Gal}(F/\mathbb {Q})\) be a subgroup. Then \(L/\mathbb {Q}\) is Galois with \(\mathrm{Gal}(L/\mathbb {Q})\) extending \(G_0\) if and only if for every \(\sigma \in G_0\) there exists \(\xi \in F^\times \) such that \(\sigma (\mu ) = \xi ^p \mu ^{a(\sigma )}\) for some \(a(\sigma ) \in \mathbb {Z}\). In particular, \(L/\mathbb {Q}\) is abelian if and only if for every \(\sigma _a \in \mathrm{Gal}(F/\mathbb {Q})\) there exists \(\xi \in F^\times \) such that \(\sigma _a(\mu ) = \xi ^p \mu ^a\).

Proof
Lemma 76
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The ramification index of \(\mathfrak {P}\) above \(p\) is \(p - 1\), i.e.,

\[ e(\mathfrak {P}\mid p) = p - 1. \]
Proof

This follows from the general formula for the ramification index in cyclotomic extensions.

Lemma 77
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We have \(\zeta _p - 1 \notin \mathfrak {P}^2\).

Proof

By 76, \(e(\mathfrak {P}\mid p) = p - 1\). Since \(\zeta _p\) is a primitive \(p\)-th root of unity, \((\zeta _p - 1)\) generates the unique prime above \(p\) in \(\mathbb {Q}(\zeta _p)\), which has ramification index \(p - 1\). Hence \(v_{\mathfrak {P}}(\zeta _p - 1) = 1\), so \(\zeta _p - 1 \notin \mathfrak {P}^2\).

Lemma 78
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For any additive character \(\psi \) of \(\mathcal{O}_K/P\),

\[ g(1, \psi ) = \begin{cases} q - 1 & \text{if } \psi = 1 \\ -1 & \text{otherwise,} \end{cases} \]

where \(\psi = 1\) denotes the trivial character (identically equal to \(1\)).

Proof