Since \(p\) is totally ramified in \(F = \mathbb {Q}(\zeta _p)\), there is a unique prime above \(p\) in \(\mathcal{O}_F\).

Let \(\mathfrak {Q}\) be a prime of \(\mathcal{O}_L\) above \(p\). By 41, \(\mathfrak {Q}\cap \mathcal{O}_F = \mathfrak {P}\cap \mathcal{O}_F\), so \(\mathfrak {Q}\) lies above \(\mathfrak {P}\cap \mathcal{O}_F\). Since \(L/F\) is Galois by 18, \(\mathrm{Gal}(L/F)\) acts transitively on the primes of \(\mathcal{O}_L\) above \(\mathfrak {P}\cap \mathcal{O}_F\), so \(\mathfrak {Q}= \sigma (\mathfrak {P})\) for some \(\sigma \in \mathrm{Gal}(L/F)\). The second statement follows from 42.

Since \((q-1) \nmid a\), we have \(g_a \neq 0\) by 15. The only primes dividing \((g_a)\) are those above \(p\), since by 15 the only rational prime dividing \(N_{L/\mathbb {Q}}(g_a)\) is \(p\). By 43, these are exactly the \(\sigma (\mathfrak {P})\) for \(\sigma \in \mathrm{Gal}(L/F)\).

For each prime \(\mathfrak {Q}= \sigma _0(\mathfrak {P})\) above \(p\), the exponent of \(\mathfrak {Q}\) in \((g_a)\) is \(v_{\mathfrak {Q}}(g_a) = s(\mathfrak {n}(\sigma _0^{-1}))\) by 27. By 43, \(\mathfrak {Q}\) appears exactly \(f\) times in the family \(\{ \sigma (\mathfrak {P}) \mid \sigma \in \mathrm{Gal}(L/F)\} \), namely for \(\sigma = \sigma _0 \cdot \tau \) with \(\tau \in \langle p \bmod (q-1) \rangle \). By 31, \(s(\mathfrak {n}(\sigma _0^{-1} \cdot \tau ^{-1})) = s(\mathfrak {n}(\sigma _0^{-1}))\) for all such \(\tau \). Therefore the exponent of \(\mathfrak {Q}\) in the right-hand side is \(f \cdot s(\mathfrak {n}(\sigma _0^{-1})) = f \cdot v_{\mathfrak {Q}}(g_a)\), which equals the exponent of \(\mathfrak {Q}\) in \((g_a)^f\).

Let \(h = a_0 + a_1 p + \cdots + a_{f-1} p^{f-1}\) be the \(p\)-adic expansion of \(h\). By 40, \(s(h) = a_0 + \cdots + a_{f-1}\). On the other hand, \(p^i h \equiv a_0 p^i + \cdots + a_{f-1} p^{i+f-1} \pmod{q-1}\), so

where the indices are taken mod \(f\). Summing over \(i = 0, \ldots , f-1\), each digit \(a_j\) appears multiplied by \((p^0 + p^1 + \cdots + p^{f-1})/(q-1) = 1/(p-1)\). Therefore

Since \(\tau \) fixes \(\zeta _p\), it fixes \(\psi \). Let \(b = \mathfrak {n}(\tau ) \in (\mathbb {Z}/(q-1)\mathbb {Z})^\times \). By 3, \(\tau (\omega (x)) = \omega (x)^b\) for all \(x\), so \(\tau (\chi (x)) = \chi ^b(x)\). Therefore

Since \(\tau \) fixes \(\zeta _m\), we have \(\mathfrak {n}(\tau ) \equiv 1 \pmod{m}\), hence \(\chi ^{\mathfrak {n}(\tau )} = \chi \) (as \(\chi ^m = 1\)). By 46, \(\tau (g(\chi )) = \chi (\mathfrak {n}(\tau ))^{-1} g(\chi )\). Therefore \(\tau (g(\chi )^m) = \chi (\mathfrak {n}(\tau ))^{-m} g(\chi )^m = g(\chi )^m\).

Since \(\tau \in \mathrm{Gal}(L/K)\), it fixes \(K = \mathbb {Q}(\zeta _{q-1})\), hence fixes all \((q-1)\)-th roots of unity, hence fixes \(\omega (x)\) for all \(x\). Since \(\tau (\zeta _p) = \zeta _p^e\), we have \(\tau (\psi (x)) = \tau (\zeta _p^{\mathrm{tr}(x)}) = \zeta _p^{e \cdot \mathrm{tr}(x)} = \psi (ex)\). Therefore

By the substitution \(y = ex\) (which is a bijection on \(\mathcal{O}_K/P\) since \(e \in (\mathbb {Z}/p\mathbb {Z})^\times \subset \mathbb {F}_q^\times \)),

By 48, \(\tau (g(\chi )) = \chi ^{-1}(e) g(\chi )\) where \(e\) is such that \(\tau (\zeta _p) = \zeta _p^e\). Since \(\chi ^m = 1\), we have \(\chi ^{-1}(e)^m = 1\), hence \(\tau (g(\chi )^m) = \chi ^{-1}(e)^m g(\chi )^m = g(\chi )^m\).

By 47, \(g(\chi )^m\) is fixed by \(\mathrm{Gal}(L/\mathbb {Q}(\zeta _m, \zeta _p))\), so \(g(\chi )^m \in \mathbb {Q}(\zeta _m, \zeta _p)\). By 49, \(g(\chi )^m\) is fixed by \(\mathrm{Gal}(L/K)\). Since \(\mathbb {Q}(\zeta _m, \zeta _p) = KF\) and \(\mathrm{Gal}(KF/\mathbb {Q}(\zeta _m))\) is generated by the restrictions of \(\mathrm{Gal}(L/\mathbb {Q}(\zeta _m, \zeta _p))\) and \(\mathrm{Gal}(L/K)\), we conclude \(g(\chi )^m \in \mathbb {Q}(\zeta _m)\). Since \(g(\chi ) \in \mathcal{O}_L\), we have \(g(\chi )^m \in \mathcal{O}_{\mathbb {Q}(\zeta _m)}\).

This follows immediately from 74.

By 44 applied with \(a = d = (q-1)/m\),

Raising to the \(m\)-th power,

By 45, \(s(\mathfrak {n}(\sigma ^{-1})) = (p-1)\sum _{i=0}^{f-1} \{ p^i \mathfrak {n}(\sigma ^{-1})/(q-1)\} \). Summing over \(\sigma \in \mathrm{Gal}(L/F)\) and regrouping by prime of \(\mathbb {Q}(\zeta _m)\), the right-hand side becomes \(\mathfrak {p}_0^{mf\theta }\). The descent from \(\mathcal{O}_L\) to \(\mathcal{O}_{\mathbb {Q}(\zeta _m)}\) is justified by 50.

By 52, \((g(\chi )^{mf}) = \mathfrak {p}_0^{mf\theta }\), i.e., \(((g(\chi )^m)^f) = (\mathfrak {p}_0^{m\theta })^f\). Since the monoid of fractional ideals of a Dedekind domain is torsion-free, we conclude \((g(\chi )^m) = \mathfrak {p}_0^{m\theta }\).

By 53, \((g(\chi _{\mathfrak {p}_0})^m) = \mathfrak {p}_0^{m\theta }\). Since \(g(\chi _{\mathfrak {p}_0})^m \in \mathbb {Q}(\zeta _m)\) by 50, \(\mathfrak {p}_0^{m\theta }\) is principal in \(\mathbb {Q}(\zeta _m)\). Since the monoid of fractional ideals is torsion-free, \(\mathfrak {p}_0^{\theta }\) is principal.