4 Ideal factorization of Gauss sums
4.1 Setup for Stickelberger’s theorem
Let \(m\) be a positive integer with \(p \nmid m\), and assume that \(f\) is the multiplicative order of \(p\) modulo \(m\). Then \(m \mid q - 1\), and we set \(d = (q-1)/m\). Let \(k = \mathbb {Q}(\zeta _m)\), a subfield of \(L\), and let \(\mathfrak {p}\) be a maximal ideal of \(\mathcal{O}_k\) lying above \(p\). We assume that \(\mathfrak {P}\) lies above \(\mathfrak {p}\).
Let \(G = \mathrm{Gal}(k/\mathbb {Q}) \cong (\mathbb {Z}/m\mathbb {Z})^\times \). For \(a \in (\mathbb {Z}/m\mathbb {Z})^\times \), let \(\sigma _a \in G\) denote the element with \(\sigma _a(\zeta _m) = \zeta _m^a\). The Stickelberger element is
where \(a\) is identified with its representative in \(\{ 0, \ldots , m-1\} \). For an ideal \(\mathfrak {a}\) of \(\mathcal{O}_k\), we write \(\mathfrak {a}^{\Theta } = \prod _{a \in (\mathbb {Z}/m\mathbb {Z})^\times } \sigma _a^{-1}(\mathfrak {a})^{a}\).
With the notation of Washington, \(\Theta = m\theta \) where \(\theta = \sum _{(a,m)=1} \{ a/m\} \sigma _a^{-1} \in \mathbb {Q}[G]\).
4.2 Descent of \(g_d^m\) to \(k\)
For any \(\tau \in \mathrm{Gal}(L/K)\), we have \(\tau (g_d^m) = g_d^m\).
Let \(E = \mathbb {Q}(\zeta _{mp}) \subseteq L\). For any \(\tau \in \mathrm{Gal}(L/E)\), we have \(\tau (g_d) = g_d\).
There exists \(\Gamma \in \mathcal{O}_k\) with \(g_d^m = \Gamma \).
4.3 Factorization in \(L\)
For all \(a \in \mathbb {Z}\), we have the equality of ideals of \(\mathcal{O}_L\)
where the product runs over the primes of \(\mathcal{O}_L\) above \(p\).
4.4 Primes of \(k\) above \(p\)
The stabilizer of \(\mathfrak {p}\) in \(G\) is the subgroup \(\langle \sigma _p \rangle \), which has order \(f\).
The stabilizer of \(\mathfrak {p}\) is its decomposition group, which acts on the residue field through powers of the Frobenius, and \(\sigma _p\) induces the Frobenius. The two groups have order \(f\).
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 \(\sigma _a^{-1}(\mathfrak {p}) = \mathfrak {q}\).
The group \(G\) acts transitively on the primes above \(p\), so \(a_{\mathfrak {q}}\) exists. The bijectivity and the last claim follow from 52.
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, \(v_{\mathfrak {q}}(\mathfrak {p}^{\Theta }) = \sum _{j=0}^{f-1} (ap^j \bmod m)\).
The images of the \((\mathfrak {q}, j)\) are the \(a p^j\) up to a cyclic shift of \(j\) modulo \(f\). The second claim follows since the exponents of \(\mathfrak {q}\) in \(\mathfrak {p}^{\Theta }\) are the \(b\) with \(\sigma _b^{-1}(\mathfrak {p}) = \mathfrak {q}\).
4.5 Valuations and the factorization
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 \(e(\mathfrak {Q}\mid \mathfrak {q}) = p - 1\).
We have \(e(\mathfrak {Q}\mid p) = p - 1\) by 105, and \(k/\mathbb {Q}\) is unramified at \(p\) since \(p \nmid m\).
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.
The group \(\mathrm{Gal}(L/F)\) acts transitively on the primes of \(\mathcal{O}_L\) above \(p\), since \(F\) has a single prime above \(p\). Choose \(\tau \in \mathrm{Gal}(L/F)\) with \(\tau (\mathfrak {P}) = \mathfrak {Q}\) and let \(a \in (\mathbb {Z}/m\mathbb {Z})^\times \) be the image of \(\mathfrak {n}(\tau ^{-1})\) (29). The restriction of \(\tau \) to \(k\) is \(\sigma _a^{-1}\), so \(\sigma _a^{-1}(\mathfrak {p}) = \mathfrak {q}\). By 41 and 33, \(s_{\mathfrak {Q}}(d) = s_{\mathfrak {P}}(da)\), and we conclude with 47 and 54.
Let \(\Gamma \in \mathcal{O}_k\) with \(g_d^m = \Gamma \) (50). Then
as ideals of \(\mathcal{O}_k\).
Since \(\mathcal{O}_L\) is faithfully flat over \(\mathcal{O}_k\), it suffices to compare the extensions of both sides to \(\mathcal{O}_L\). By 51, the ideal \((\Gamma )\mathcal{O}_L = (g_d)^m\) is \(\prod _{\mathfrak {Q}\mid p} \mathfrak {Q}^{m s_{\mathfrak {Q}}(d)}\). Regrouping the factors of \(\mathfrak {p}^{\Theta }\) with 53 and using \(e(\mathfrak {Q}\mid \mathfrak {q}) = p - 1\) (55), the extension of \(\mathfrak {p}^{\Theta }\) is \(\prod _{\mathfrak {q}\mid p} \prod _{\mathfrak {Q}\mid \mathfrak {q}} \mathfrak {Q}^{(p-1) \sum _j b_j}\), with the notation of 56. The exponents agree by 56.
4.6 Stickelberger’s theorem
In the setting of this chapter, the ideal \(\mathfrak {p}^{\Theta }\) of \(\mathcal{O}_k\) is principal.
Let \(k\) be a number field isomorphic to \(\mathbb {Q}(\zeta _m)\), let \(p\) be an odd prime with \(p \nmid m\) and let \(\mathfrak {p}\) be a maximal ideal of \(\mathcal{O}_k\) above \(p\). Then the ideal \(\mathfrak {p}^{\Theta }\) is principal.
Let \(f\) be the order of \(p\) modulo \(m\) and \(q = p^f\). The field \(L = \mathbb {Q}(\zeta _{p(q-1)})\) contains a subfield \(k_0 = \mathbb {Q}(\zeta _m)\), and an isomorphism \(k_0 \cong k\) commutes with the action of the \(\sigma _a\). Apply 58 to the preimage of \(\mathfrak {p}\) in \(\mathcal{O}_{k_0}\) and transport the result to \(k\).
Let \(R\) be a Dedekind domain and \(J\) a nonzero ideal of \(R\). Every ideal class of \(R\) contains an integral ideal coprime to \(J\).
Let \(I_0\) be an integral ideal in the inverse class. By the Chinese remainder theorem, there is \(x \in I_0\) such that \((x) I_0^{-1}\) is coprime to \(J\), and this ideal lies in the given class.
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 \(c\),
By 60, choose an integral ideal \(\mathfrak {b} \in c\) coprime to \(2m\). Every prime factor of \(\mathfrak {b}\) lies above an odd prime not dividing \(m\), so 59 applied to each prime factor shows that \(\mathfrak {b}^{\Theta }\) is principal. Hence \(c^{\Theta } = [\mathfrak {b}^{\Theta }] = 1\).