Proof. A sheaf $\mathcal {G}_A'$ satisfying all of the above properties but (4) is given by the $\ell $ -adic Fourier transform of the Artin–Schreier sheaf $\mathcal {L}_{e(f(X)/p)}$ , where $f(X)=A/X^2\in \mathbb {F}_p(X)$ . The resulting trace function (the left-hand side of (15)) is indeed the discrete Fourier transform of $e(f(x)/p)$ , namely, $K_2$ . The statements of the lemma are all contained in [Reference Katz12, Section 7.3, Section 7.12 ( $\text{SL}$ -Example(3)), Theorem 7.12.3.1], with $h=f$ , $\alpha =0$ and $n_0=\operatorname {ord}_0(f)=2$ . In particular, the rank is $1+n_0=3$ .

Regarding the last property, since $f(x)+f(-x)$ is nonconstant, the geometric monodromy group is isomorphic to $\operatorname {SL}_3(\mathbb {C})$ for $p> 7$ by the theorem [Reference Katz12, Theorem 7.12.3.1]. By [Reference Katz12, Section 7.12 ( $\text{SL}$ -Example(3))], $\det \mathcal {G}^{\prime }_A$ is geometrically trivial, whence arithmetically isomorphic to $\beta \otimes \overline {\mathbb {Q}}_\ell $ , where $\beta $ is a p-Weil number of weight $0$ . Therefore, setting $\mathcal {G}_A=\beta ^{-1/3}\otimes \mathcal {G}^{\prime }_A$ yields

$$ \begin{align*} \operatorname{SL}_3(\mathbb{C})=G_{\mathrm{geom}}(\mathcal{G}_A)=G_{\mathrm{geom}}(\mathcal{G}^{\prime}_A)\le G_{\mathrm{arith}}(\mathcal{G}_A)\le\operatorname{SL}_3(\mathbb{C}), \end{align*} $$

with $\mathcal {G}_A$ still verifying the previous properties. The theorem then holds with the twist $\gamma =\iota \beta ^{-1/3}\in \mathbb {C}$ .

Proof. Let us first consider the case where $\psi $ is trivial. By the Goursat–Kolchin–Ribet criterion [Reference Katz12, Section 1.8], since the pair $(\operatorname {SL}_3(\mathbb {C}),\operatorname {Std})_{i={1,2}}$ is Goursat-adapted (in the language given there), it suffices to show that there exists no geometric isomorphism of the form

$$ \begin{align*}[+h]^*\mathcal{G}_A\cong [+h']^*\mathcal{G}_A\otimes\mathcal{L},\quad h\neq h',\end{align*} $$

for a one-dimensional sheaf $\mathcal {L}$ . Without loss of generality, $h'=0$ . Let us assume by contradiction that $h\neq 0$ . Since the left-hand side is ramified exactly at $-h$ and $\infty $ , $\mathcal {L}$ must be ramified at $0$ and $-h$ (and possibly at $\infty $ ). It follows that

$$ \begin{align*}\mathcal{G}_A^{I_0}\cong([+h]^*\mathcal{G}_A)^{I_{-h}}\cong (\mathcal{G}_A\otimes\mathcal{L})^{I_{-h}}\cong\mathcal{G}_A\otimes\mathcal{L}^{I_{-h}}=0.\end{align*} $$

However, this contradicts Lemma 3.3(3), which states that the stalk at $0$ of $\mathcal {G}_A$ is a rank 2 pseudoreflection.

If $\psi $ is nontrivial, let us write $\mathcal {H}^+_{A,\boldsymbol {h}, \psi }=\mathcal {L}_\psi \oplus \mathcal {F}$ , where we may assume by the above that the arithmetic and geometric monodromy groups of $\mathcal {F}$ coincide and are isomorphic to $\operatorname {SL}_3(\mathbb {C})^H$ . We have

$$ \begin{align*}G_{\mathrm{geom}} \left(\mathcal{H}^+_{A,\boldsymbol{h}, \psi}\right)\le G_{\mathrm{arith}} \left(\mathcal{H}^+_{A,\boldsymbol{h}, \psi}\right)\le \mathbb{F}_p\times\operatorname{SL}_3(\mathbb{C})^H,\end{align*} $$

and $G_{\mathrm {geom}}$ surjects onto both $\mathbb {F}_p$ and $\operatorname {SL}_3(\mathbb {C})^H$ . The surjection onto $\mathbb {F}_p$ implies that $G_{\mathrm {geom}}$ has at least p connected components, of which the component at the identity contains $\operatorname {SL}_3(\mathbb {C})^H$ . It follows that $G_{\mathrm {geom}} = \mathbb {F}_p \times \operatorname {SL}_3(\mathbb {C})^H$ , as claimed.

Proof. Set $r := \left \lceil \frac {\log (20D^3)}{\log p}\right \rceil \leq n/2$ . Assume the claim is false, so that $p^r|(\alpha _1+\alpha _2u_0 + \alpha _3 u_0^2)$ and $\alpha _1,\alpha _2,\alpha _3$ are not all the same. As $p \equiv 1 \pmod {3}$ and $u_0$ is a primitive cube root, we must have $u_0 \not \equiv 1 \pmod {p^{\left \lfloor n/2\right \rfloor }}$ . By assumption, at least one of $\alpha _1,\alpha _2,\alpha _3$ does not vanish and up to multiplication by $u_0^j$ for $j \in \{0,1,2\}$ (which does not change the p-adic valuation), we may suppose $\alpha _3$ does not. As $u_0^3 \equiv 1 \pmod {p^r}$ , we have

$$ \begin{align*} \alpha_1 + \alpha_2 u_0 + \alpha_3 u_0^2 &= (\alpha_1-\alpha_3) + (\alpha_2-\alpha_3)u_0 + \alpha_3(1+u_0+u_0^2) \\ &\equiv (\alpha_1-\alpha_3) + (\alpha_2-\alpha_3)u_0 \pmod{p^r}. \end{align*} $$

Now, on one hand this implies that

(20) $$ \begin{align} \alpha_1-\alpha_3 \equiv -u_0(\alpha_2-\alpha_3) \pmod{p^r}. \end{align} $$

On the other, taking cubes, we obtain that

$$ \begin{align*}(\alpha_1-\alpha_3)^3 + (\alpha_2-\alpha_3)^3 \equiv 0 \pmod{p^r}. \end{align*} $$

However, as

$$ \begin{align*}|(\alpha_1-\alpha_3)^3 + (\alpha_2-\alpha_3)^3| \leq 2\cdot (2D)^3 < 20D^3 \leq p^r, \end{align*} $$

it follows that $(\alpha _1-\alpha _3)^3 = -(\alpha _2-\alpha _3)^3$ and therefore that $\alpha _1-\alpha _3 = -(\alpha _2-\alpha _3)$ . Plugging this into the earlier congruence implies that either $\alpha _1-\alpha _3 = \alpha _2-\alpha _3 = 0$ , or else $u_0 \equiv 1 \pmod {p^r}$ . Since the $\alpha _j$ are not all the same by assumption, the first of these is impossible. To see that the second fails as well, note that if $u_0 \not \equiv 1 \pmod {p^{\left \lfloor n/2 \right \rfloor }}$ but $u_0 = 1 +mp^r$ for some $m \in \mathbb {Z}$ , then of course $p^{\left \lfloor n/2 \right \rfloor -r} \nmid m$ . On the other hand, since $u_0$ is a cube root of unity modulo $p^{\left \lfloor n/2\right \rfloor }$ ,

$$ \begin{align*}u_0^3 = (1+mp^r)^3 = 1+mp^r(3+3p^rm+m^2p^{2r}) \equiv 1 \pmod{p^{\left\lfloor n/2\right\rfloor}}, \end{align*} $$

which is only possible for $p \neq 3$ if $\nu _p(m) \geq \left \lfloor n/2 \right \rfloor - r$ , a contradiction. The claim follows.

Proof. In each case, $\phi (\boldsymbol {0}) \neq 0$ if and only if there exist tuples $\boldsymbol {j} \in \mathcal {U}$ , $\boldsymbol {j}' \in \mathcal {V}$ such that

$$ \begin{align*}0 = \sum_{1 \leq i \leq \mu(\tau)} u_0^{j_i(\tau)} - \sum_{1 \leq i' \leq \nu(\tau)} u_0^{j^{\prime}_{i'}(\tau)} \text{ for all } \tau \in T. \end{align*} $$

a) When $p \equiv 2 \pmod {3}$ , we may (trivially) take $u_0 = 1$ and $j_i(\tau ) = j^{\prime }_{i'}(\tau ) = 0$ for all $i,i'$ and $\tau $ , which leads immediately to the conclusion $\mu (\tau ) = \nu (\tau )$ for all $\tau \in T$ .

b) When $p \equiv 1 \pmod {3}$ , we write the above expression as

$$ \begin{align*}0 = \left(m_{\tau} + n_{\tau} u_0 + p_{\tau}u_0^2\right) - \left(m_{\tau}' + n_{\tau}'u_0 + p_{\tau}'u_0^2\right) = (m_{\tau} - m_{\tau}') + (n_{\tau}-n_{\tau}')u_0 + (p_{\tau} - p_{\tau}')u_0^2, \end{align*} $$

where $m_{\tau },n_{\tau },p_{\tau }$ and $m_{\tau }',n_{\tau }',p_{\tau }'$ are nonnegative integers satisying

$$ \begin{align*} &m_{\tau} + n_{\tau}+p_{\tau} = \mu(\tau) &m_{\tau}' + n_{\tau}'+p_{\tau}' = \nu(\tau), \end{align*} $$

for all $\tau \in T$ . By Lemma 4.4, it follows that $m_{\tau } - m_{\tau }' = n_{\tau } - n_{\tau }' = p_{\tau } - p_{\tau }' = \ell $ , say. We conclude that

$$ \begin{align*}\mu(\tau)-\nu(\tau) = \left(m_{\tau} + n_{\tau} + p_{\tau}\right) - \left(m_{\tau}' + n_{\tau}' + p_{\tau}'\right) = 3\ell, \end{align*} $$

which implies the claim.

Proof. a) When $p \equiv 2 \pmod {3}$ every residue class modulo p is a cube. Thus, if d satisfies the condition in the sum on the left-hand side, then as $p> 3$ this is equivalent to $\prod _{\tau \in \tilde {T}} (d+\tau ) \in (\mathbb {Z}/p\mathbb {Z})^{\times }$ , which is satisfied for all but $O(|\tilde {T}|)$ residue classes d modulo p. Thus, the sum in question is simply

$$ \begin{align*} \sum_{d \pmod{p}} e_p(dC) + O(|\tilde{T}|) = p1_{C \equiv 0 \pmod{p}} + O(|\tilde{T}|), \end{align*} $$

as required.

b) Let $\Xi _3(p) := \{\chi \pmod {p} : \chi ^3 = \chi _0\}$ , where $\chi _0$ denotes the trivial multiplicative character modulo p. Since the set of multiplicative characters modulo p is a cyclic group of order $p-1$ and $3|(p-1)$ , we can write

$$ \begin{align*}\Xi_3(p) = \{\xi^j : j \in \{-1,0,1\}\}, \text{ where } \xi := \chi_1^{(p-1)/3} \end{align*} $$

for some fixed generator $\chi _1$ for the group of characters mod p. We note that for any $b \in \mathbb {Z}/p\mathbb {Z}$ ,

$$ \begin{align*}\sum_{-1 \leq j \leq 1} \xi^j(b) = \begin{cases} 3 &\text{ if } b \in (\mathbb{Z}/p\mathbb{Z})^{\times 3} \\ 0 &\text{ otherwise}\end{cases}, \end{align*} $$

and so the exponential sum in question is

(21) $$ \begin{align} &\sum_{d \pmod{p}} e_p(dC) \prod_{\tau \in \tilde{T}} 1_{4A^2(d+\tau) \in (\mathbb{Z}/p\mathbb{Z})^{\times 3}} \nonumber \\ &= 3^{-|\tilde{T}|} \sum_{\boldsymbol{j} \in \{-1,0,1\}^{|\tilde{T}|}} \xi(4A^2)^{t(\boldsymbol{j})} \sum_{d \pmod{p}} \xi\left(\prod_{\tau \in \tilde{T}} (d+\tau)^{j_{\tau}}\right)e_p(dC), \end{align} $$

where we have written $t(\boldsymbol {j}) := \sum _{\tau \in \tilde {T}} j_{\tau }$ .

When $\boldsymbol {j} = \boldsymbol {0}$ we get

(22) $$ \begin{align} &3^{-|\tilde{T}|} \sum_{d \pmod{p}} \chi_0\left(\prod_{\tau \in \tilde{T}} (d+\tau)\right) e_p(dC) \nonumber\\ &= 3^{-|\tilde{T}|} \sum_{d \pmod{p}} e_p(dC) + O(|\tilde{T}|3^{-|\tilde{T}|}) = 3^{-|\tilde{T}|}\left(p1_{C \equiv 0 \pmod{p}} + O(|\tilde{T}|)\right). \end{align} $$

For $\boldsymbol {j} \neq \boldsymbol {0}$ , we define

$$ \begin{align*} g_{\boldsymbol{j}}(d) := \prod_{\tau \in \tilde{T} \atop j_{\tau} = 1} (d+\tau), \ \ \ \ h_{\boldsymbol{j}}(d) &:= \prod_{\tau \in \tilde{T} \atop j_{\tau} = -1} (d+\tau), \end{align*} $$

and what remains to be estimated is the expression

$$ \begin{align*}3^{-|\tilde{T}|} \sum_{\substack{\boldsymbol{j} \in \{-1,0,1\}^{|\tilde{T}|} \\ \boldsymbol{j} \neq \boldsymbol{0}}} \sum_{\substack{d \pmod{p} \\ p\nmid (d+\tau) \forall \tau \in T}} \xi(g_{\boldsymbol{j}}(d) \overline{h}_{\boldsymbol{j}}(d)) e_p(dC) \ll \max_{\boldsymbol{j} \neq \boldsymbol{0}} \left|\sum_{\substack{d \pmod{p} \\ p\nmid (d+\tau) \forall \tau \in T}} \xi(g_{\boldsymbol{j}}(d) \overline{h}_{\boldsymbol{j}}(d)) e_p(dC)\right|. \end{align*} $$

We interpret the latter sum as a sum over $\mathbb {F}_p$ and estimate the sum using the language of $\ell $ -adic trace functions, for an auxiliary prime $\ell \neq p$ . For each $\boldsymbol {j} \neq \boldsymbol {0}$ let $\mathcal {K}_{\boldsymbol {j}}$ denote the Kummer sheaf attached to $\xi \circ (g_{\boldsymbol {j}}/h_{\boldsymbol {j}})$ and let $\mathcal {L}_C$ be the Artin–Schreier sheaf attached $d \mapsto e_p(dC)$ . Then the sum we must estimate is

$$ \begin{align*} \sum_{d \in U_{\boldsymbol{j}}(\mathbb{F}_p)} t_{\mathcal{K}_{\boldsymbol{j}}}(d) t_{\mathcal{L}_C}(d) = \sum_{d \in U_{\boldsymbol{j}}(\mathbb{F}_p)} t_{\mathcal{F}}(d), \end{align*} $$

where $\mathcal {F} := \mathcal {K}_{\boldsymbol {j}} \otimes \mathcal {L}_C$ and $U_{\boldsymbol {j}}(\mathbb {F}_p) := \{d \in \mathbb {F}_p : d \neq -\tau \ \forall \tau \in T\}$ . By Corollary 2.31 of [Reference Perret-Gentil17], this is

$$ \begin{align*}p1_{\mathcal{F} \text{ geom. trivial}} + O(\text{cond}(\mathcal{F})^2p^{1/2}). \end{align*} $$

When $\boldsymbol {j} \neq \boldsymbol {0}$ , we claim that $\mathcal {F}$ is not geometrically trivial; that is, that $\mathcal {K}_{\boldsymbol {j}}$ and $D(\mathcal {L}_{C})$ are not geometrically isomorphic. Indeed, on one hand, if $C \neq 0$ , then $\mathcal {L}_C$ (and thus $D(\mathcal {L}_C)$ ) has a lone wild ramification point at $\infty $ (with $\text {Swan}_{\infty }(\mathcal {L}_C) = 1$ ), while in contrast $\mathcal {K}_{\boldsymbol {j}}$ has only tame ramification points at the zeros of $g_{\boldsymbol {j}}h_{\boldsymbol {j}}$ (and, in particular, $\text {Swan}_{\infty }(\mathcal {K}_{\boldsymbol {j}}) = 0$ ). If $C = 0$ , then because $g_{\boldsymbol {j}}(d)h_{\boldsymbol {j}}(d)$ has only distinct roots and is nonconstant, it is not the cube of a polynomial. Thus, $\mathcal {K}_{\boldsymbol {j}}$ is ramified in at least one point and hence not geometrically trivial in this case as well. Finally,

$$ \begin{align*}\text{cond}(\mathcal{F}) = \text{cond}(\mathcal{K}_{\boldsymbol{j}}) \text{cond}(\mathcal{L}_C) \leq 3(\deg{g_{\boldsymbol{j}}} + \deg{h_{\boldsymbol{j}}}+1) \ll |\tilde{T}|, \end{align*} $$

so that indeed

$$ \begin{align*}\sum_{\substack{d \pmod{p} \\ p \nmid g_{\boldsymbol{j}}(d)h_{\boldsymbol{j}}(d) \\ p\nmid (d+\tau) \forall \tau \in T}} \xi(g_{\boldsymbol{j}}(d) \overline{h}_{\boldsymbol{j}}(d)) e_p(dC) \ll |\tilde{T}|^2p^{1/2}, \end{align*} $$

for all $\boldsymbol {j} \neq \boldsymbol {0}$ . Inserting this and (22) into (21) and summing over $\boldsymbol {j}$ , we reach the claim.

Proof of Proposition 4.3

Suppose $\boldsymbol {\epsilon } = \boldsymbol {0}$ . Since $f_{T,\boldsymbol {0}}(b) = cb$ , we have

$$ \begin{align*} \sideset{}{^{\Diamond}}\sum_{d \pmod{p}} \mathcal{S}^{\ast, \boldsymbol{0}}_{p^n}(c,\mu,\nu;a,d) &= \phi(\boldsymbol{0}) p^{(N+M)n/2} \sideset{}{^{\Diamond}}\sum_{d \pmod{p}} \sum_{t \pmod{p^{n-1}}} e_{p^{n}}(c(pt + d)) \\ &= \phi(\boldsymbol{0})p^{(N+M+2)n/2-1} 1_{p^{n-1}|c} \sideset{}{^{\Diamond}}\sum_{d \pmod{p}} e_p(dc'), \end{align*} $$

where $c' = 0$ if $p^{n-1} \nmid c$ and otherwise $c \equiv c'p^{n-1} \pmod {p^{n}}$ . By Lemma 4.6, we have

$$ \begin{align*}\sideset{}{^{\Diamond}}\sum_{d \pmod{p}} e_p(dc') = 3^{-|\tilde{T}|1_{p\equiv 1 \pmod{3}}} p1_{c' \equiv 0 \pmod{p}} + O_{|\tilde{T}|}(p^{1/2}1_{p \equiv 1 \pmod{3}} + 1_{p \equiv 2 \pmod{3}}), \end{align*} $$

where $\tilde {T} := \{\tau \pmod {p} : \tau \in T\}$ . As $|\tilde {T}| \leq |T| \leq N+M$ , the above expression is thus

$$ \begin{align*}\ll_{N,M} p^{(N+M+2)n/2} \left(1_{c \equiv 0 \pmod{p^{n}}} + 1_{p^{n-1}|c}\left(p^{-1/2} 1_{p \equiv 1 \pmod{3}} + p^{-1}1_{c = 0}1_{p \equiv 2 \pmod{3}}\right)\right). \end{align*} $$

By Lemma 4.5, this gives

$$ \begin{align*} &\sideset{}{^{\Diamond}}\sum_{d \pmod{p}} \mathcal{S}^{\ast, \boldsymbol{0}}_{p^n}(c,\mu,\nu;a,d)\\ &\quad\ll_{N,M} p^{(N+M+2)n/2} 1_{p \equiv 1 \pmod{3}} \left(1_{c \equiv 0 \pmod{p^{n}}} + p^{-1/2}1_{p^{n-1}|c}\right) \prod_{\tau \in T} 1_{3|(\mu(\tau)-\nu(\tau))} \\ &\qquad+ p^{(N+M+2)n/2} 1_{p \equiv 2 \pmod{3}}\left( 1_{c \equiv 0 \pmod{p^{n}}} + p^{-1}1_{p^{n-1}|c}\right) \prod_{\tau \in T} 1_{\mu(\tau) = \nu(\tau)}, \end{align*} $$

which implies the claim.

Proof. Notice that

$$ \begin{align*} \binom{2/3}{j} = (j!)^{-1}\prod_{0 \leq l \leq j-1}(2/3-l) = \frac{(-1)^{j-1}2}{3^{j}j!} \prod_{1 \leq l \leq j-1}(3l-2), \end{align*} $$

and so as $p> 3$ ,

$$ \begin{align*} \nu_p\left(\binom{2/3}{j}\right) = \nu_p\left(\prod_{1 \leq l \leq j-1} (3l-2)\right) - \nu_p(j!). \end{align*} $$

Note that

$$ \begin{align*} \max_{0 \leq j \leq k} \nu_p\left(\binom{2/3}{j}\right) \geq \nu_p\left(\binom{2/3}{0}\right) \geq 0. \end{align*} $$

By Legendre’s formula for p-adic valuations of factorials, it suffices to show that

$$ \begin{align*} \nu_p\left(\prod_{1 \leq l \leq j-1} (3l-2)\right) \leq \sum_{r \geq 1} \left\lfloor \frac{j}{p^r} \right\rfloor + 1_{p \leq 3j/2-1}, \end{align*} $$

for each $1 \leq j \leq k$ . To check this inequality, we observe first that if $1 \leq l_0 \leq j$ is minimal such that $p|(3l_0-2)$ and $l> l_0$ also has this property then $p|(l-l_0)$ . Clearly, by minimality we must have $\nu _p(3l_0-2) = 1$ : if $p \equiv 1 \pmod {3}$ then $l_0$ satisfies $p = 3l_0-2$ ; if $p \equiv 2 \pmod {3}$ then $2p = 3l_0-2$ . In particular, if $p> 3j/2-1$ , then $l_0$ does not exist.

Consider next when $p \leq 3j/2-1$ . Then we have

$$ \begin{align*} \nu_p\left(\prod_{1 \leq l \leq j-1}(3l-2)\right) = \sum_{r \geq 1} |\{1 \leq l \leq j-1 : l \equiv 2\overline{3} \pmod{p^r}\}| \leq \sum_{r \geq 1} \left\lfloor \frac{j-a_r}{p^r}\right\rfloor, \end{align*} $$

where $0 \leq a_r \leq p^r-1$ is a minimal representative of the residue class $2\overline {3} \pmod {p^r}$ . We clearly have $\left \lfloor (j-a_r)/p^r\right \rfloor \leq \left \lfloor j/p^r \right \rfloor $ , so that, summing over r, we obtain the desired bound.

Proof of Proposition 4.8

We begin by noting that if $|\{\tau \in T : \epsilon _{\tau } \not \equiv 0 \pmod {p^{n}}\}| \leq 1$ , then since $\boldsymbol {\epsilon }$ is nonzero the p-adic stationary phase method (see, e.g., Lemma 1 (1) of [Reference Milićević and Zhang15]) implies that for some $\tau _0 \in T$ ,

$$ \begin{align*} \sum_{b \in \mathbb{Z}/p^{n}\mathbb{Z} \atop b \equiv d \pmod{d}} e_{p^{n}}(f_{T,\boldsymbol{\epsilon}}(b)) &= \sum_{b \in \mathbb{Z}/p^{n}\mathbb{Z} \atop b \equiv d \pmod{d}} e_{p^{n}}(bc + \epsilon_{\tau_0} s(\overline{2a}(b+\tau_0))^2) \\ &= p^{n/2} \sum_{\substack{b \in \mathbb{Z}/p^{n}\mathbb{Z} \\ b \equiv d\pmod{p} \\ \epsilon_{\tau_0}s(\overline{2a}(b+\tau_0))^{2} \equiv - c \pmod{p^{\left\lfloor n/2\right\rfloor}}}} e_{p^{n}}(bc + \epsilon_{\tau_0}s(\overline{2a}(b+\tau_0))^2) \\ &\ll p^{n/2}, \end{align*} $$

since the inner sum has at most a single summand. As $N+M \geq 1$ this satisfies the first claim, so henceforth we may assume that $|\{\tau \in T : \epsilon _{\tau } \not \equiv 0 \pmod {p^{n}}\}| \geq 2$ .

The second condition is now vacuously satisfied if $0 < \delta _2n < 1$ , so we may assume that $\delta _2 n \geq 1$ . We put $X := \{b \pmod {p^{n}} : b \equiv d \pmod {p}\}$ , noting that this is $p^{\left \lfloor \delta _2n \right \rfloor } \mathbb {Z}/p^{n}\mathbb {Z}$ -invariant. Putting

$$ \begin{align*} \mathcal{E}_{|T|} := \left\{b \in \mathbb{Z}/p^{n}\mathbb{Z} : f^{(j)}(b) \equiv 0 \pmod{p^{\left\lfloor \delta_2 n\right\rfloor}} \text{ for all } 1 \leq j \leq |T|\right\}, \end{align*} $$

the proof of Proposition 8 of [Reference Milićević and Zhang15] shows that

(25) $$ \begin{align} \sum_{b \in X} e_{p^{n}}(f_{T,\boldsymbol{\epsilon}}(b)) = \sum_{b \in X \cap \mathcal{E}_{|T|}} e_{p^{n}}(f_{T,\boldsymbol{\epsilon}}(b)) + O_{|T|}\left(p^{(1-\delta_1)n}\right), \end{align} $$

as long as (writing $f_{T,\boldsymbol {\epsilon }}^{(0)} = f_{T,\boldsymbol {\epsilon }}$ ) the relations

$$ \begin{align*} f^{(j)}_{T,\boldsymbol{\epsilon}}(b+p^{\kappa}t) \equiv f^{(j)}_{T,\boldsymbol{\epsilon}}(b) + p^{\kappa} t f^{(j+1)}_{T,\boldsymbol{\epsilon}}(b) \pmod{p^{\min\{2\kappa,n\}}} \end{align*} $$

hold for $0 \leq j \leq |T|-1$ ; this is guaranteed by a calculation analogous to (24). The proof of Proposition 9 there shows that if

$$ \begin{align*} \rho_0 := \max_{1 \leq j \leq |T|} \nu_p \left( \binom{2/3}{j} \right) + \min_{\tau \in T} \nu_p(\epsilon_{\tau}), \end{align*} $$

then $\mathcal {E}_T = \emptyset $ whenever

$$ \begin{align*} \min\{|\tau-\tau|_p : \tau, \tau' \in T, \tau \neq \tau', \epsilon_{\tau},\epsilon_{\tau'} \not \equiv 0 \pmod{p^n}\} \geq p^{-\delta_3(\left\lfloor \delta_2n\right\rfloor - \rho_0)} \end{align*} $$

or, equivalently, whenever

$$ \begin{align*} \tau \not \equiv \tau' \pmod{p^{\left\lfloor \delta_3(\left\lfloor \delta_2 n\right\rfloor-\rho)\right\rfloor}} \text{ for all distinct } \tau,\tau' \text{ with nonzero } \epsilon_{\tau},\epsilon_{\tau'}. \end{align*} $$

Let us assume the latter condition. Then (25) yields the estimate

$$ \begin{align*} \sum_{b \in X} e_{p^{n}}(f_{T,\boldsymbol{\epsilon}}(b)) \ll_{|T|} p^{(1-\delta_1)n}, \end{align*} $$

and it remains to check that the required constraints on the constants hold. As noted above the proof of Proposition 8 in [Reference Milićević and Zhang15], we may take $\delta _1 = \delta _2 = 2^{-|T|}$ and $\delta _3 = \binom {|T|}{2}^{-1}$ . Now, if $p \equiv 2 \pmod {3}$ , then $\epsilon _{\tau }$ can be identified with an integer in $[-N,M]$ and we deduce that $\nu _p(\epsilon _{\tau }) \leq \left \lceil \log (2(N+M))/\log p \right \rceil $ , which is acceptable. Thus, consider when $p \equiv 1 \pmod {3}$ . We may then write

$$ \begin{align*} \epsilon_{\tau} = \alpha_{\tau} + \beta_{\tau} u_0 + \gamma_{\tau}u_0^2, \end{align*} $$

and as $\phi (\boldsymbol {\epsilon }) \neq 0$ such a representation exists with $|\alpha _{\tau }|,|\beta _{\tau }|,|\gamma _{\tau }| \leq N+M$ for all $\tau \in T$ . Since $\boldsymbol {\epsilon }$ is nonzero we may find $\tau ' \in T$ such that $\epsilon _{\tau '} \not \equiv 0 \pmod {p^n}$ and therefore $\alpha _{\tau '},\beta _{\tau '}$ and $\gamma _{\tau '}$ are not the same. By Lemma 4.4 we get

$$ \begin{align*} \min_{\tau \in T} \nu_p(\epsilon_{\tau}) \leq \nu_p(\epsilon_{\tau'}) \leq \left\lceil \frac{\log(20(N+M)^3)}{\log p}\right\rceil. \end{align*} $$

Finally, by Lemma 4.9, we have

$$ \begin{align*} \max_{1 \leq j \leq |T|} \nu_p\left(\binom{2/3}{j}\right) \leq 1_{p \leq 3|T|/2-1}. \end{align*} $$

We thus obtain

$$ \begin{align*} \rho_0 \leq 1_{p \leq 3|T|/2-1} + \left\lceil \frac{\log(20(N+M)^3)}{\log p} \right\rceil =: \rho. \end{align*} $$

The claim then follows by replacing $\rho _0$ with its upper bound $\rho $ in the second alternative of the proposition, which relaxes the condition there.

Proof of Proposition 4.7

Note that by the assumed lower bound for $n$ , we must have $p^{n=2} > 20(N +M)^3$ . We may partition the nonzero $\boldsymbol {\epsilon }$ into the sets

$$ \begin{align*} \mathcal{A}_T &:= \left\{\boldsymbol{\epsilon} \in (\mathbb{Z}/p^{n}\mathbb{Z})^{|T|} \backslash \{\boldsymbol{0}\} : \min_{\substack{\tau \neq \tau' \\ \epsilon_{\tau},\epsilon_{\tau'} \not\equiv 0 \pmod{p^{n}}}} |\tau-\tau'|_p \geq p^{-\delta_3(\left\lfloor \delta_2 n \right\rfloor - \rho)}\right\}\\ \mathcal{B}_T &:= \left((\mathbb{Z}/p^{n}\mathbb{Z})^{|T|} \backslash \{\boldsymbol{0}\}\right) \backslash \mathcal{A}_T. \end{align*} $$

Noting that $|T| \leq N+M$ , we apply Proposition 4.8 to bound

$$ \begin{align*} p^{(N+M)n/2} \sideset{}{^{\Diamond}} \sum_{d \pmod{p}} \sum_{\boldsymbol{\epsilon} \in \mathcal{A}_T} \phi(\boldsymbol{\epsilon}) \sum_{b \pmod{p^{n}} \atop b \equiv d \pmod{p}} e_{p^{n}}\left(f_{T,\boldsymbol{\epsilon}}(b)\right) &\ll_{N,M} p^{(N+M)n/2 + 1} \cdot p^{(1-\delta_1)n}\\ &\ll p^{(N+M+2-2\delta_1)n/2 + 1}. \end{align*} $$

On the other hand, if $\mathcal {B}_T \neq \emptyset $ , then the second alternative of the proposition holds. This implies the claim.

Proof. The proof when $j = 2$ is completely similar to that of $j = 1$ , so we focus only on the latter case.

As in the proof of Lemma 4.6, we translate the problem to $\mathbb {F}_p$ , so that, for example, $\tilde {T}$ is identified with a subset of $\mathbb {F}_p$ . Put $\mathbb {F} := \mathbb {F}_p$ if $p \equiv 1 \pmod {3}$ and $\mathbb {F} := \mathbb {F}_p[X]/(X^2+X+1)\mathbb {F}_p[X]$ when $p \equiv 2 \pmod {3}$ . Define also

$$ \begin{align*} N_{\tilde{T}}(w) := |\{d \in \mathbb{F} : 4a^2(d+\tau) \in \mathbb{F}^{\times 3} \forall \ \tau \in \tilde{T} \text{ and } \sum_{\tau \in \tilde{T}} \tilde{\epsilon}_{\tau} s((2a)^{-1}(d+\tau))^{-1} = w\}|. \end{align*} $$

When $p \equiv 1 \pmod {3}$ , $N_{\tilde {T}}(w)$ is precisely the count on the left-hand side in the statement of the lemma; when $p \equiv 2 \pmod {3}$ , $N_{\tilde {T}}(w)$ is an upper bound for the desired quantity, since $\mathbb {F}_p \subseteq \mathbb {F}$ . In what follows we let $U_0$ denote a primitive cube root of unity in $\mathbb {F}$ ; naturally, this satisfies $1+U_0 + U_0^2 = 0$ .

Let $\boldsymbol {a} \in \mathbb {F}_p^{|\tilde {T}|}$ . Consider the product

$$ \begin{align*} Q_w(\boldsymbol{a}) := \prod_{\boldsymbol{j} \in \{-1,0,1\}^{|\tilde{T}|}} \left(w - \sum_{\tau \in \tilde{T}} U_0^{j_{\tau}} a_{\tau}\right), \end{align*} $$

which is a polynomial of total degree $\leq 3^{|\tilde {T}|}$ in the variables $(a_{\tau })_{\tau \in \tilde {T}} \in \mathbb {F}_p^{|\tilde {T}|}$ . Note that $Q_w(\boldsymbol {a}) = Q_w((a_{\tau }U_0^{j_{\tau }})_{\tau \in \tilde {T}})$ for all $\boldsymbol {j} \in \{-1,0,1\}^{|\tilde {T}|}$ . This implies that we can find a polynomial $\tilde {Q}_w$ , defined over $\mathbb {F}$ , such thatFootnote ⁹ $Q_w(\boldsymbol {a}) = \tilde {Q}_w((a_{\tau }^3)_{\tau \in \tilde {T}})$ . In particular, we can write

$$ \begin{align*} Q_w(\boldsymbol{a}) = \sum_{r_1,\ldots,r_{|\tilde{T}|} \geq 0 \atop r_1 + \cdots + r_{|\tilde{T}|} \leq 3^{|\tilde{T}|-1}} b_{r_1,\ldots,r_{|\tilde{T}|}}(w) a_{\tau_1}^{3r_1} \cdots a_{\tau_{|\tilde{T}|}}^{3r_{|\tilde{T}|}}, \end{align*} $$

where $\{\tau _1,\ldots ,\tau _{|\tilde {T}|}\}$ is an enumeration of $\tilde {T}$ . Now, define

$$ \begin{align*} \tilde{R}_w(Y) &:= \left(\prod_{\tau \in \tilde{T}}(Y+\tau)^{3^{|\tilde{T}|-1}}\right) \cdot Q_w((\tilde{\epsilon}_{\tau}s((2a)^{-1}(Y+\tau))^{-1})_{\tau \in \tilde{T}}) \\ &= \sum_{r_1,\ldots,r_{|\tilde{T}|} \geq 0 \atop r_1 + \cdots + r_{|\tilde{T}|} \leq 3^{|\tilde{T}|-1}} b_{r_1,\ldots,r_{|\tilde{T}|}}(w) \prod_{1 \leq j \leq |\tilde{T}|} (2a \tilde{\epsilon}_{\tau_j}^3)^{r_j} \cdot \prod_{1 \leq j \leq |\tilde{T}|} (Y+\tau_j)^{3^{|\tilde{T}|-1}-r_j}, \end{align*} $$

which is a polynomial in Y of degree $\leq |\tilde {T}|3^{|\tilde {T}|-1} \ll _{|\tilde {T}|} 1$ over $\mathbb {F}$ . Every $d \in \mathbb {F}$ counted by $N_{\tilde {T}}(w)$ is a root of $\tilde {R}_w(Y)$ over $\mathbb {F}$ , since it is a root of $Q_w((\epsilon _{\tau }s((2a)^{-1}(Y+\tau ))^{-1})_{\tau \in \tilde {T}})$ . It follows that, provided $\tilde {R}_w(Y)$ is a nonzero polynomial, we get $N_{\tilde {T}}(w) \ll _{|\tilde {T}|} 1$ as required. It therefore suffices to show that $\tilde {R}_w(Y)$ is nonzero.

Assume first that $w \neq 0$ . The leading coefficient (in Y) of $R_w(Y)$ is $b_{0,\ldots ,0}(w) = w^{3^{|\tilde {T}|}} \neq 0$ , so that $\tilde {R}_w(Y)$ is necessarily nonzero in this case.

Suppose next that $w = 0$ . Note that by hypothesis there is a $\tau _1 \in \tilde {T}$ such that $\tilde {\epsilon }_{\tau _1} \neq 0$ . Setting $Y = -\tau _1$ and expanding the product we see that only the term with $r_1 = 3^{|\tilde {T}|-1}$ (and $r_j = 0$ for $j \neq 1$ ) survives, arising as the monomial in $a_{\tau _1}^{3^{|\tilde {T}|}}$ in $Q_0(\boldsymbol {a})$ . Explicitly, this term has coefficient

$$ \begin{align*} [a_{\tau_1}^{3^{|\tilde{T}|}}] Q_0(\boldsymbol{a}) = \prod_{\boldsymbol{j} \in \{-1,0,1\}} (-U_0^{j_{\tau_1}}) = -1. \end{align*} $$

It follows that $b_{3^{|\tilde {T}|-1},0,\ldots ,0} = -(2a\tilde {\epsilon }_{\tau _1}^3)^{3^{|\tilde {T}|-1}}$ and thus

$$ \begin{align*} \tilde{R}_0(-\tau_1) = -((2a)\tilde{\epsilon}_{\tau_1}^3)^{3^{|\tilde{T}|-1}} \in \mathbb{F}^{\times}. \end{align*} $$

The claim thus follows when $w \neq 0$ as well.

Proof of Proposition 4.10

Recall that

$$ \begin{align*} P_{\boldsymbol{\epsilon},d}(t) = \sum_{0 \leq j \leq n-1} a_j(\boldsymbol{\epsilon},d)t^j \text{ for } 0 \leq t \leq p^{n-1}-1. \end{align*} $$

We wish to estimate

$$ \begin{align*} \sideset{}{^{\Diamond}}\sum_{d\pmod{p}} \sum_{t \pmod{p^{n-1}}} e_{p^n}\left(P_{\boldsymbol{\epsilon},d}(t)\right). \end{align*} $$

Let $R \geq 1$ , $N:= \frac {1}{2}p^{n-1}$ . Also, put

$$ \begin{align*} P_{\boldsymbol{\epsilon},d,t}(z) := P_{\boldsymbol{\epsilon},d}(t+z) = \sum_{0 \leq i \leq n-1} z^i\sum_{i \leq j \leq n-1} \binom{j}{i} t^{j-i} a_j(\boldsymbol{\epsilon},d) = \sum_{0 \leq i \leq n-1} A_i(\boldsymbol{\epsilon},d,t)z^i, \end{align*} $$

where we have defined

$$ \begin{align*} A_i(\boldsymbol{\epsilon},d,t) := p^i \sum_{i \leq j \leq n-1} (pt)^{j-i} \binom{j}{i}\left(\binom{2/3}{i} (2a)^j\sum_{\tau \in T} \epsilon_{\tau}s(\overline{2a}(d+\tau))^{2-3j} + cd^{1-j}1_{j \in \{0,1\}}\right). \end{align*} $$

We define

$$ \begin{align*} F_{\boldsymbol{\epsilon}}(d) := R^{-2} \sum_{t \pmod{p^n}} S_{\boldsymbol{\epsilon},d,t}(R) := R^{-2}\sum_{|t| \leq N} \sum_{1 \leq y,z \leq R} e_{p^n}(P_{\boldsymbol{\epsilon},d,t}(yz)) + O(R^2). \end{align*} $$

Set $\ell := n(n-1)/2$ . We invoke Vinogradov’s method, as exposed in Subsection 8.5 of [Reference Iwaniec and Kowalski10]. Given $\alpha \in \mathbb {R}$ and $Y \geq 1$ , define

$$ \begin{align*} D(\alpha,Y) := Y^{-2} \sum_{|m| \leq Y} \left|\sum_{|n| \leq Y} e(\alpha mn)\right|. \end{align*} $$

Now, for each $|t| \leq \frac {1}{2}p^{n-1}$ , we obtain (cf. [Reference Iwaniec and Kowalski10, equation (8.76)])

(26) $$ \begin{align} |S_{\boldsymbol{\epsilon},d,t}(R)| \leq \left(\ell^{2(n-1)}R^{4\ell(\ell-1)+n(n-1)} J_{\ell,n-1}^2(R)\Delta(t)\right)^{\frac{1}{2\ell^2}}, \end{align} $$

where we have written

$$ \begin{align*} \Delta(t) := \prod_{1 \leq h \leq n-1} D(A_h(\boldsymbol{\epsilon},d,t)/p^n,\ell R^h). \end{align*} $$

Our goal is to extract savings over the trivial bound $\Delta (t) \ll 1$ . To this end, we consider $D(\alpha ,Y)$ for $\alpha $ a p-adic rational. Put $\alpha = a/p^s$ , for $(a,p) = 1$ and $s \geq 1$ . We obtain

$$ \begin{align*} D(a/p^s,Y) &\ll Y^{-2} \sum_{|m| \leq Y} \min\{Y,\|am/p^s\|^{-1}\} \\ &= Y^{-2}\sum_{0 \leq r \leq s} \sum_{1 \leq |u| \leq \max\{1,\frac{1}{2}p^{s-r}\} \atop p \nmid u} \sum_{|m| \leq Y \atop am \equiv up^r \pmod{p^s}} \min\{Y,p^{s-r}/|u|\} \\ &= Y^{-2}\sum_{0 \leq r \leq s-1} p^{s-r} \sum_{1 \leq |u| \leq \frac{1}{2}p^{s-r} \atop p \nmid u} \frac{1}{|u|} \sum_{|m| \leq Y \atop am \equiv up^r \pmod{p^s}} 1 + Y^{-1} \sum_{|m| \leq Y \atop p^s|m} 1 \\ & \ll Y^{-1}(p^s/Y+1)\log(p^s) + p^{-s} \end{align*} $$

in this case. We specialise $Y = \ell R^h$ and

$$ \begin{align*} \alpha = A_h(\boldsymbol{\epsilon},d,t)/p^n = \tilde{A}_h/p^{n-\theta_h}, \end{align*} $$

where $\theta _h = \theta _h(\boldsymbol {\epsilon },d,t) := \nu _p(A_h(\boldsymbol {\epsilon },d,t))$ and $p \nmid \tilde {A}_h$ , for each $1 \leq h \leq n-1$ . Note that $\theta _h(\boldsymbol {\epsilon },d,t) \geq h$ , with equality if and only if we have

$$ \begin{align*} (2a)^h\binom{2/3}{h} \sum_{\tau \in T} \epsilon_{\tau} s(\overline{2a}(d+\tau))^{2-3h} \not \equiv -c1_{h = 1} \pmod{p}, \end{align*} $$

this condition being independent of t.

For $\boldsymbol {\epsilon },d$ and t fixed, let

$$ \begin{align*} \mathfrak{D} = \mathfrak{D}(\boldsymbol{\epsilon},d,t) := \max\{1 \leq j \leq n-1 : \theta_j(\boldsymbol{\epsilon},d,t) < n\}, \end{align*} $$

if this maximum is defined, and let $\mathfrak {D}(\boldsymbol {\epsilon },d,t) = 0$ otherwise. If $\mathfrak {D} \geq 1$ , then we obtain

$$ \begin{align*} \Delta(t) \ll_{n} R^{-\mathfrak{D}}(p^{n-\mathfrak{D}}R^{-\mathfrak{D}} + 1)\log p + p^{\theta_{\mathfrak{D}}-n}. \end{align*} $$

Suppose now that $\mathfrak {D} \geq 2$ . Applying Theorem 4.13, we obtain from (26) that

$$ \begin{align*} |S_{\boldsymbol{\epsilon},d,t}(R)| &\ll_{\varepsilon,n} \left(R^{4\ell(\ell-1)+n(n-1)+\varepsilon}\left(R^{2\ell} + R^{4\ell - n(n-1)}\right)\left(R^{-\mathfrak{D}}(p^{n-\mathfrak{D}}R^{-\mathfrak{D}}+1) + p^{\theta_{\mathfrak{D}}-n}\right)\right)^{\frac{1}{2\ell^2}} \\ &\ll R^{2+\varepsilon}\left(R^{-\mathfrak{D}/2\ell^2}((p^{n-\mathfrak{D}}/R^{\mathfrak{D}})^{1/2\ell^2}+1) + p^{-(n-\theta_{\mathfrak{D}})/2\ell^2}\right), \end{align*} $$

recalling that $2\ell = n(n-1)$ . Taking $R = (N/\sqrt {p})^{1/2} \asymp p^{n/2-3/4}$ , we deduce that

$$ \begin{align*} F_{\boldsymbol{\epsilon}}(d) &\ll_{\varepsilon,d,t} |\{t \pmod{p^{n-1}} : \exists d : 4a^2(d+\tau) \in (\mathbb{Z}/p\mathbb{Z})^{\times 3} \forall \tau \in T \text{ and }\mathcal{D}(\boldsymbol{\epsilon},d,t) \in \{0,1\}\}| \\ &+p^{\varepsilon} \sum_{|t| \leq N \atop \mathfrak{D} = \mathfrak{D}(\boldsymbol{\epsilon},d,t) \geq 2} \left((N/\sqrt{p})^{-\mathfrak{D}/4\ell^2}+(p^{n-\mathfrak{D}}(N/\sqrt{p})^{-\mathfrak{D}})^{1/2\ell^2} + p^{-(n-\theta_{\mathfrak{D}})/2\ell^2}\right) + (N/\sqrt{p})^2 \\ &\ll_{\varepsilon} |\{t \pmod{p^{n-1}} : \exists d : 4a^2(d+\tau) \in (\mathbb{Z}/p\mathbb{Z})^{\times 3} \forall \tau \in T \text{ and }\mathcal{D}(\boldsymbol{\epsilon},d,t) \in \{0,1\}\}| \\ &+p^{n-1-\frac{1}{n^2(n-1)^2}+\varepsilon}. \end{align*} $$

In this way, we obtain

$$ \begin{align*} &\sideset{}{^{\Diamond}} \sum_{d \pmod{p}} F_{\boldsymbol{\epsilon}}(d)\\ &\ll_{\varepsilon,n} \sum_{t \pmod{p^{n-1}}} |\{d \pmod{p} : 4a^2(d+\tau) \in (\mathbb{Z}/p\mathbb{Z})^{\times 3} \forall \tau \in T \text{ and } \mathcal{D}(\boldsymbol{\epsilon},d,t) \in \{0,1\}\}| \\ &+p^{n-\frac{1}{n^2(n-1)^2} + \varepsilon}. \end{align*} $$

It remains to treat the contribution from pairs $(t,d)$ for which $\mathcal {D}(\boldsymbol {\epsilon },d,t) \in \{0,1\}$ . By construction, if $\mathcal {D}(\boldsymbol {\epsilon },d,t) = 0$ , then $\theta _1(\boldsymbol {\epsilon },d,t) \geq n> 1$ . Similarly, if $\mathcal {D}(\boldsymbol {\epsilon },d,t) = 1$ and $n \geq 3$ , then $\theta _2(d,t) \geq n> 2$ . In either of these cases, provided p is large enough in terms of n, we get

$$ \begin{align*} \sum_{\tau \in T} \epsilon_{\tau} s(\overline{2a}(d+\tau))^{2-3(\mathcal{D}+1)} \equiv 0 \pmod{p}. \end{align*} $$

Applying Lemma 4.12, we find that for each $t \pmod {p^{n-1}}$ the number of d satisfying $(\Diamond )$ such that this latter congruence holds is $\ll _{N,M} 1$ . Since $n \ll _{N,M} 1$ , we deduce that

$$ \begin{align*} \sideset{}{^{\Diamond}} \sum_{d \pmod{p}} F_{\boldsymbol{\epsilon}}(d) \ll_{\varepsilon,N,M} p^{n-\frac{1}{n^2(n-1)^2}+\varepsilon} + p^{n-1} \ll p^{n-\frac{1}{n^2(n-1)^2}+\varepsilon}. \end{align*} $$

It remains to consider those d for which $\mathcal {D}(\boldsymbol {\epsilon },d,t) = 1$ and $n = 2$ . In this case, we know that $p^2 \nmid A_1(\boldsymbol {\epsilon },d,t)$ and, moreover, that $A_1(\boldsymbol {\epsilon },d,t) = a_1(\boldsymbol {\epsilon },d)$ ; that is, $A_1(\boldsymbol {\epsilon },d,t)$ is independent of t. The sum we wish to bound is therefore

$$ \begin{align*} \sideset{}{^{\Diamond}} \sum_{d \pmod{p} \atop p^2\nmid a_1(\boldsymbol{\epsilon},d)} e_{p^2}(a_0(\boldsymbol{\epsilon},d)) \sum_{t \pmod{p}} e_p\left(t(a_1(\boldsymbol{\epsilon},d)/p)\right) = 0, \end{align*} $$

so the bound required is satisfied in this case as well.