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Let \(k\) be a number field isomorphic to \(\mathbb {Q}(\zeta _m)\). The Stickelberger element \(\Theta \) annihilates the ideal class group \(\mathrm{Cl}(k)\): for every ideal class \(c\),
For each prime \(\mathfrak {q}\) of \(\mathcal{O}_k\) above \(p\), choose \(a_{\mathfrak {q}} \in (\mathbb {Z}/m\mathbb {Z})^\times \) with \(\sigma _{a_{\mathfrak {q}}}^{-1}(\mathfrak {p}) = \mathfrak {q}\). The map
is a bijection, and the image \(a\) of \((\mathfrak {q}, j)\) satisfies \(\sigma _a^{-1}(\mathfrak {p}) = \mathfrak {q}\).
An element \(\tau \in I(Q)\) acts trivially on \(S/Q\), so it acts on \(Q/Q^2\) by an \(S/Q\)-linear automorphism. The tame character is the determinant of this action,
It is a group homomorphism because the determinant is multiplicative.
Assume that \(P\) has Teichmüller lifts of order \(q - 1\) (2). The Teichmüller character \(\omega : \mathcal{O}_K/P \to \mathcal{O}_K\) is the multiplicative character whose restriction to \((\mathcal{O}_K/P)^\times \) is the inverse of the reduction map \(\mu _{q-1}(\mathcal{O}_K) \xrightarrow {\sim } (\mathcal{O}_K/P)^\times \), extended by \(\omega (0) = 0\).
Assume moreover that \(G\) is abelian and acts faithfully on \(S\), and that \(Q\) is tame. Then \(\# I(Q)\) divides \(N\mathfrak {p}- 1\).
Restriction to \(K\) induces an isomorphism \(\mathrm{Gal}(L/F) \cong \mathrm{Gal}(K/\mathbb {Q}) \cong (\mathbb {Z}/(q-1)\mathbb {Z})^\times \). For \(\sigma \in \mathrm{Gal}(L/F)\), we denote by \(\mathfrak {n}(\sigma ) \in \{ 0, \ldots , q-2\} \) the representative of its image, so that \(\sigma (\zeta _{q-1}) = \zeta _{q-1}^{\mathfrak {n}(\sigma )}\).
For all \(a \in \mathbb {Z}\), we have \(g_a \neq 0\). If \((q-1) \nmid (a+b)\), then \(J(a, b) \neq 0\).
Let \(L/K\) be a finite Galois extension with group \(G\), let \(E\) be an intermediate field, and let \(I\) be a normal subgroup of \(G\) that is a complement of \(\mathrm{Gal}(L/E)\). Then the fixed field \(F = L^{I}\) satisfies \(FE = L\), and \(G/I \cong \mathrm{Gal}(L/E)\).
Let \(L/\mathbb {Q}\) be a finite abelian extension, and let \(K, E \subseteq L\) be subfields with \(KE = L\), \([K:\mathbb {Q}] = p^m\) and \(E \cong \mathbb {Q}(\zeta _q)\), where \(q \neq p\) are primes and \(q \nmid [L:\mathbb {Q}]\). Let \(\mathfrak {Q}\) be a prime of \(L\) above \(q\). Then the inertia group \(I = I(\mathfrak {Q}) \subseteq \mathrm{Gal}(L/\mathbb {Q})\) is a complement of \(\mathrm{Gal}(L/E)\): one has \(I \cap \mathrm{Gal}(L/E) = 1\) and \(\# I \cdot \# \mathrm{Gal}(L/E) = [L:\mathbb {Q}]\).
Let \(C \subseteq A\) be an abelian extension of \(\mathbb {Q}\), unramified outside \(p\), and let \(H\) be a subgroup of \(\mathrm{Gal}(C/\mathbb {Q})\) of index \(p^m\) with cyclic quotient. Then the fixed field \(C^{H}\) is a cyclic extension of \(\mathbb {Q}\) of degree \(p^m\), unramified outside \(p\).
Let \(K \subseteq A\) be a cyclic extension of \(\mathbb {Q}\) of degree \(p^m\), and let \(q \neq p\) be a prime that ramifies in \(K\). Then there is a cyclic extension \(F \subseteq A\) of \(\mathbb {Q}\) of \(p\)-power degree such that
the prime \(q\) is unramified in \(F\);
every prime that is unramified in \(K\) is unramified in \(F\);
one has \(K\mathbb {Q}(\zeta _q) = F\mathbb {Q}(\zeta _q)\).
Let \(K \subseteq A\) be a cyclic extension of \(\mathbb {Q}\) of degree \(p^m\). Then there exist a cyclic extension \(F \subseteq A\) of \(\mathbb {Q}\) of \(p\)-power degree, unramified outside \(p\), and an integer \(n \geq 1\) such that \(K\mathbb {Q}(\zeta _n) = F\mathbb {Q}(\zeta _n)\).
Let \(R\) be a ring containing \(\mathcal{O}_K\) and let \(\sigma \) be a ring endomorphism of \(R\) such that \(\sigma (\zeta _{q-1}) = \zeta _{q-1}^m\) for some \(m \in \mathbb {N}\). Then \(\sigma (\omega (x)^a) = \omega (x)^{am}\) for all \(x \in \mathcal{O}_K/P\) and \(a \in \mathbb {Z}\).
Let \(K/\mathbb {Q}\) be a finite Galois extension, \(E, F\) two Galois subextensions with \(EF = K\), and \(\mathfrak {P}\) a prime of \(K\) above a prime number \(p\) with \(p \nmid [K:\mathbb {Q}]\). Then
Assume that \(G\) acts faithfully on \(S\) and that \(Q\) is tame. Then \(\theta \) is injective, and the inertia group \(I(Q)\) is cyclic.
Let \(\mathfrak {q}\) be a prime of \(\mathcal{O}_k\) above \(p\) and \(\mathfrak {Q}\) a prime of \(\mathcal{O}_L\) above \(\mathfrak {q}\). Then
where \(b_j \in \{ 0, \ldots , m-1\} \) represents the image of \((\mathfrak {q}, j)\) under the bijection of 53.
Let \(\mathfrak {q}\) be a prime of \(\mathcal{O}_k\) above \(p\) and \(a \in (\mathbb {Z}/m\mathbb {Z})^\times \) with \(\sigma _a^{-1}(\mathfrak {p}) = \mathfrak {q}\). Then the sum of the (representatives of the) images of \((\mathfrak {q}, j)\), \(0 \leq j {\lt} f\), under the bijection of 53 is \(\sum _{j=0}^{f-1} (ap^j \bmod m)\). In particular, \(v_{\mathfrak {q}}(\mathfrak {p}^{\Theta }) = \sum _{j=0}^{f-1} (ap^j \bmod m)\).