Since \(\omega ^{q-1} = 1\) by 2, \((q-1) \mid k\) implies \(\omega ^{-k} = 1\), so \(\omega ^{-(a+k)} = \omega ^{-a}\), hence \(g_{a+k} = g_a\).

By 10, \(\psi (x) \equiv 1 \pmod{\mathfrak {P}}\) for all \(x \in \mathcal{O}_K/P\). Therefore

Since \((q-1) \nmid a\), we have \(\omega ^{-a} \neq 1\) by 2. Hence \(\sum _{x \in \mathcal{O}_K/P} \omega (x)^{-a} = 0\) by 4. Therefore \(g_a \equiv 0 \pmod{\mathfrak {P}}\), i.e. \(g_a \in \mathfrak {P}\).

Since \((q-1) \nmid a\), \(\omega ^{-a}\) is nontrivial and \(\psi \) is primitive by 7. By gaussSum_mul_gaussSum_eq_card, \(g(\omega ^{-a}, \psi ) \cdot g(\omega ^{a}, \psi ^{-1}) = p^f\). By the change of variable \(x \mapsto -x\), \(g(\omega ^{a}, \psi ^{-1}) = \omega ^{a}(-1) \cdot g_{-a}\). Hence \(g_a \cdot g_{-a} = \omega ^a(-1) \cdot p^f\).

By 14, \(g_a \cdot g_{-a} = \omega ^a(-1) \cdot p^f = \pm p^f\). Taking norms, \(N_{L/\mathbb {Q}}(g_a) \cdot N_{L/\mathbb {Q}}(g_{-a}) = \pm p^{f \cdot [L:\mathbb {Q}]}\). Since both norms are integers and their product is \(\pm p^{f \cdot [L:\mathbb {Q}]}\), each is of the form \(\pm p^k\) for some \(k \geq 0\). By 13, \(g_a \in \mathfrak {P}\), so \(g_a\) is not a unit in \(\mathcal{O}_L\), hence \(N_{L/\mathbb {Q}}(g_a) \neq \pm 1\), thus \(k \geq 1\).

By 15, \(N_{L/\mathbb {Q}}(g_a) = \pm p^k \neq 0\), hence \(g_a \neq 0\).

By 9, \(\psi (x) \equiv 1 + \mathrm{tr}(x) \cdot (\zeta _p - 1) \pmod{\mathfrak {P}^2}\) for all \(x \in \mathcal{O}_K/P\). Therefore

Since \(p\) is odd, \(q - 1 \geq 2\), so \(\omega ^{-1} \neq 1\) by 2, and \(\sum _x \omega (x)^{-1} = 0\) by orthogonality of characters. Moreover, Computing the trace via powers of Frobenius, \(\sum _{x \in \mathcal{O}_K/P} \omega (x)^{-1} \cdot \mathrm{tr}(x) = -1\). Hence \(g_1 \equiv -(\zeta _p - 1) \pmod{\mathfrak {P}^2}\).

Since \(\gcd (q-1, p) = 1\), the fields \(K = \mathbb {Q}(\zeta _{q-1})\) and \(F = \mathbb {Q}(\zeta _p)\) are linearly disjoint over \(\mathbb {Q}\). Therefore \(L = KF\) is Galois over \(F\) with \(\mathrm{Gal}(L/F) \cong \mathrm{Gal}(K/\mathbb {Q}) \cong (\mathbb {Z}/(q-1)\mathbb {Z})^\times \).

Since \(\sigma \in \mathrm{Gal}(L/F)\) and \(F = \mathbb {Q}(\zeta _p)\), \(\sigma \) fixes \(\zeta _p\), hence \(\sigma (\psi (x)) = \psi (x)\) for all \(x\). By definition of \(\mathfrak {n}(\sigma )\), \(\sigma (\zeta _{q-1}) = \zeta _{q-1}^{\mathfrak {n}(\sigma )}\), so by 3, \(\sigma (\omega (x)) = \omega (x)^{\mathfrak {n}(\sigma )}\) for all \(x\). Therefore

The Frobenius map \(\varphi : x \mapsto x^p\) is a bijection on \(\mathcal{O}_K/P\). Using the substitution \(y = \varphi (x)\),

By multiplicativity of \(\omega \), \(\omega (\varphi ^{-1}(y))^p = \omega (\varphi ^{-1}(y)^p) = \omega (y)\), so \(\omega (\varphi ^{-1}(y))^{-pa} = \omega (y)^{-a}\). By 8, \(\psi (\varphi ^{-1}(y)) = \psi (y)\). Therefore \(g_{pa} = g_a\).

This follows from jacobiSum_mul_nontrivial applied to \(\chi = \omega ^{-a}\) and \(\phi = \omega ^{-b}\).

By 16, \(g_a, g_b, g_{a+b} \neq 0\). By 21, \(g_a \cdot g_b = g_{a+b} \cdot J(\omega ^{-a}, \omega ^{-b})\). Since \(g_{a+b} \neq 0\), we conclude \(J(\omega ^{-a}, \omega ^{-b}) \neq 0\).

By 22, \(J(\omega ^{-a}, \omega ^{-b}) \in \mathcal{O}_K\). For any \(\alpha \in \mathcal{O}_K\), the relation between the valuations \(v_{\mathfrak {P}}\) and \(v_P\) is given by the ramification index:

By 76, \(e(\mathfrak {P}\mid P) = p - 1\), hence