Since \((q-1) \mid a\), \(\omega ^{-a} = 1\) by 2, so \(g_a = \sum _{x} \psi (x)\), which equals \(0\) by 78 since \(\psi \) is nontrivial by 6. Hence \(s(a) = v_{\mathfrak {P}}(0) = 0\).

For any prime ideal \(\mathfrak {Q}\) of \(\mathcal{O}_L\) and \(\alpha \in \mathcal{O}_L\), \(v_{\sigma (\mathfrak {Q})}(\alpha ) = v_{\mathfrak {Q}}(\sigma ^{-1}(\alpha ))\) by definition of the pushforward valuation. Applying this with \(\mathfrak {Q}= \mathfrak {P}\) and \(\alpha = g_1\), \(v_{\sigma (\mathfrak {P})}(g_1) = v_{\mathfrak {P}}(\sigma ^{-1}(g_1))\). By 19 applied to \(\sigma ^{-1}\), \(\sigma ^{-1}(g_1) = g_{\mathfrak {n}(\sigma ^{-1})}\). Hence \(v_{\sigma (\mathfrak {P})}(g_1) = v_{\mathfrak {P}}(g_{\mathfrak {n}(\sigma ^{-1})}) = s(\mathfrak {n}(\sigma ^{-1}))\).

If \((q-1) \mid (a+b)\) then \(s(a+b) = 0\) by 26 and the result is trivial. Otherwise, by 21,

Taking valuations and using the fact that \(J(\omega ^{-a}, \omega ^{-b}) \in \mathcal{O}_L\), hence \(v_{\mathfrak {P}}(J) \geq 0\), we obtain \(s(a) + s(b) \geq s(a+b)\).

Since \((q-1) \mid k\), we have \(s(k-a) = s(-a)\) by periodicity of \(g\): by 12, \(g_{a+(q-1)} = g_a\), so \(s\) has period \(q-1\).

If \((q-1) \mid a\), then \((q-1) \mid (-a)\), so \(s(a) = s(-a) = 0\) by 26 and the result is clear.

If \((q-1) \nmid a\), by 14, \(g_a \cdot g_{-a} = \omega ^{a}(-1) \cdot p^f\). Since \(\omega ^{a}(-1)\) is a root of unity, it is a unit in \(\mathcal{O}_L\), hence \(v_{\mathfrak {P}}(\omega ^{a}(-1)) = 0\). By 76, \(v_{\mathfrak {P}}(p^f) = f \cdot e(\mathfrak {P}\mid p) = f(p-1)\). Therefore

Case \((q-1) \nmid (a+b)\): We distinguish subcases.

If \((q-1) \mid a\), then \(s(a) = 0\) by 26, and \(g_{a+b} = g_b\) by 12, so \(s(a+b) = s(b)\). Hence \(s(a) + s(b) = s(a+b) + 0 \cdot (p-1)\). The case \((q-1) \mid b\) is symmetric.

If \((q-1) \nmid a\) and \((q-1) \nmid b\), then by 21, \(g_a \cdot g_b = g_{a+b} \cdot J(\omega ^{-a}, \omega ^{-b})\), and \(J \neq 0\) by 23. Taking valuations,

By 24, \(v_{\mathfrak {P}}(J) = (p-1) \cdot v_P(J)\), so \(k = v_P(J) \geq 0\).

Case \((q-1) \mid (a+b)\): Then \(s(a+b) = 0\) by 26. By 29 with \(k = a+b\): if \((q-1) \mid a\) then \(s(a) = s(b) = 0\) by 26, so \(k = 0\); otherwise \(s(a) + s(b) = (p-1) \cdot f\), so \(k = f \geq 1\).

By 20, \(g_{pa} = g_a\), hence \(s(pa) = v_{\mathfrak {P}}(g_{pa}) = v_{\mathfrak {P}}(g_a) = s(a)\).

Since \((q-1) \nmid a\), we have \(g_a \neq 0\) by 16 and \(g_a \in \mathfrak {P}\) by 13. Hence \(v_{\mathfrak {P}}(g_a) \geq 1 {\gt} 0\).

By 13, \(g_1 \in \mathfrak {P}\), so \(s(1) \geq 1\). By 17, \(g_1 \equiv -(\zeta _p - 1) \pmod{\mathfrak {P}^2}\). Since \(\zeta _p - 1 \notin \mathfrak {P}^2\) by 77, we have \(g_1 \notin \mathfrak {P}^2\), so \(s(1) \leq 1\).

By induction on \(a\). The case \(a = 0\) follows from 26 (since \((q-1) \mid 0\)). For \(a \geq 1\), by 28 and the induction hypothesis,

By induction on \(a\). The case \(a = 0\) gives \(s(0) = 0\) by 26, so \(k = 0\). For \(a \geq 1\), by 30 applied to \((1, a-1)\), there exists \(k' \geq 0\) such that \(s(1) + s(a-1) = s(a) + k'(p-1)\). By induction, there exists \(k'' \geq 0\) such that \(a - 1 = s(a-1) + k''(p-1)\). Using \(s(1) = 1\) by 33,

For \(a = 0\) the result is immediate. For \(1 \leq a \leq p-2\), by 35 there exists \(k \geq 0\) such that \(a = s(a) + k(p-1)\). Since \(s(a) {\gt} 0\) by 32, we have \(k(p-1) = a - s(a) \leq a - 1 \leq p-3 {\lt} p-1\), hence \(k = 0\) and \(s(a) = a\).

Since \(f \geq 2\), we have \(q = p^f \geq p^2 {\gt} p\), so \(q - 1 {\gt} p - 1\) and \((q-1) \nmid (p-1)\). By 35, there exists \(k \geq 0\) such that \(p - 1 = s(p-1) + k(p-1)\). Since \((q-1) \nmid (p-1)\), we have \(s(p-1) {\gt} 0\) by 32, hence \(k {\lt} 1\), so \(k = 0\) and \(s(p-1) = p-1\).

By induction on the list of digits \((a_0, \ldots , a_{t-1})\). The base case \(a = 0\) is immediate. For the inductive step, write \(a = a_0 + p \cdot a'\) where \(a' = a_1 + a_2 p + \cdots + a_{t-1} p^{t-2}\). By 28,

By 34, \(s(a_0) \leq a_0\). By 31, \(s(p \cdot a') = s(a')\). By the induction hypothesis, \(s(a') \leq a_1 + \cdots + a_{t-1}\). Hence \(s(a) \leq a_0 + a_1 + \cdots + a_{t-1}\).

We pair each \(a \in \{ 0, \ldots , q-2\} \) with \(q-1-a\). By 29 with \(k = q-1\), for any \(a\) with \((q-1) \nmid a\),

Note that \(s(0) = 0\) and \(s(q-1) = 0\) by 26.

Since \(p\) is odd, \(q - 1 = p^f - 1\) is even. The set \(\{ 1, \ldots , q-2\} \) consists of \((q-3)/2\) pairs \(\{ a, q-1-a\} \) with \(a \neq q-1-a\), plus the central term \(a = (q-1)/2\) which pairs with itself. Since \(q \geq 3\), we have \(1 \leq (q-1)/2 \leq q-2\), so \((q-1) \nmid (q-1)/2\), and 29 gives \(2s((q-1)/2) = (p-1)f\), hence \(s((q-1)/2) = (p-1)f/2\). Therefore

By 38, \(s(a) \leq a_0 + \cdots + a_{f-1}\) for all \(a\). Summing over \(0 \leq a \leq q - 2\), we get

By Nat.sum_sum_digits_eq, the right-hand side equals \(\frac{(q-2) \cdot f \cdot (p-1)}{2}\), which equals the left-hand side by 39. Since each term satisfies \(s(a) \leq a_0 + \cdots + a_{f-1}\) and the sums are equal, we must have \(s(a) = a_0 + \cdots + a_{f-1}\) for all \(a\).