2 The Gauss sum
Throughout this chapter and the next two, we assume that \(p\) is an odd prime.
2.1 Gauss and Jacobi sums
For \(a \in \mathbb {Z}\), the Gauss sum is
For \(a, b \in \mathbb {Z}\), the Jacobi sum is
If \((q-1) \mid k\), then for all \(a \in \mathbb {Z}\), \(g_{a+k} = g_a\).
Since \(\omega ^{q-1} = 1\) by 5, we have \(\omega ^{-k} = 1\), so \(\omega ^{-(a+k)} = \omega ^{-a}\).
If \((q-1) \mid a\), then \(g_a = -1\).
For all \(a \in \mathbb {Z}\), \(g_{pa} = g_a\).
Substitute \(x \mapsto x^p\), a bijection of \(\mathcal{O}_K/P\), in the sum defining \(g_a\). We have \(\omega (x^p)^{-a} = \omega (x)^{-pa}\) and \(\psi (x^p) = \psi (x)\) by 11.
If \((q-1) \nmid (a+b)\), then
The character \(\omega ^{-(a+b)}\) is nontrivial by 6, so this is jacobiSum_mul_nontrivial applied to \(\omega ^{-a}\) and \(\omega ^{-b}\).
If \((q-1) \nmid a\), then
The character \(\omega ^{-a}\) is nontrivial by 6 and \(\psi \) is primitive by 10. By gaussSum_mul_gaussSum_eq_card, \(g(\omega ^{-a}, \psi ) \cdot g(\omega ^{a}, \psi ^{-1}) = |\mathcal{O}_K/P| = p^f\), and the change of variable \(x \mapsto -x\) gives \(g(\omega ^{a}, \psi ^{-1}) = \omega ^{a}(-1) \cdot g_{-a}\).
2.2 Congruences
If \((q-1) \nmid a\), then \(g_a \in (\zeta _p - 1)\mathcal{O}_L\).
If \((q-1) \nmid a\), then \(g_a \in \mathfrak {P}\).
If \((q-1) \nmid a\), then \(g_a\) is not a unit of \(\mathcal{O}_L\).
The element \(\zeta _p - 1\) is not a unit of \(\mathcal{O}_L\), so this follows from 23.
If \((q-1) \nmid a\), then there exists \(k \geq 1\) such that \(|N_{L/\mathbb {Q}}(g_a)| = p^k\).
For all \(a \in \mathbb {Z}\), we have \(g_a \neq 0\). If \((q-1) \nmid (a+b)\), then \(J(a, b) \neq 0\).
We have
The extension \(K/\mathbb {Q}\) is unramified at \(p\), so \(e(\mathfrak {P}\mid P) = p - 1 \geq 2\) by 105, hence \(P \subseteq \mathfrak {P}^2\). Since \(\zeta _p - 1 \in \mathfrak {P}\) by 14, 12 gives
The first sum vanishes by 6 since \(q - 1 \geq 2\). For the second, we have \(\omega (x) \equiv x \pmod{P}\) by 4 and \(\mathrm{tr}(x) = \sum _{i{\lt}f} x^{p^i}\), and \(\sum _{x \in \mathcal{O}_K/P} x^{-1} x^{p^i}\) is \(-1\) for \(i = 0\) and \(0\) for \(0 {\lt} i {\lt} f\). Hence the second sum is \(-1\) modulo \(\mathfrak {P}^2\).
2.3 Galois action
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 \(\sigma (\zeta _{q-1}) = \zeta _{q-1}^{\mathfrak {n}(\sigma )}\).
Since \(\gcd (q-1, p) = 1\), the fields \(K\) and \(F\) are linearly disjoint over \(\mathbb {Q}\) and \(L = KF\).
For all \(\sigma \in \mathrm{Gal}(L/F)\) and \(a \in \mathbb {Z}\),
The automorphism \(\sigma \) fixes \(\zeta _p\), hence the values of \(\psi \), and \(\sigma (\omega (x)^{-a}) = \omega (x)^{-a \mathfrak {n}(\sigma )}\) by 7.
Let \(\tau \in \mathrm{Gal}(L/K)\) and \(e \in (\mathbb {Z}/p\mathbb {Z})^\times \) with \(\tau (\zeta _p) = \zeta _p^e\). Then for all \(a \in \mathbb {Z}\),
The automorphism \(\tau \) fixes the values of \(\omega \) and \(\tau (\psi (x)) = \psi (ex)\). The substitution \(y = ex\) in the sum gives \(\tau (g_a) = \sum _y \omega (y/e)^{-a}\psi (y) = \omega (e)^{a} g_a\).