\nu)+h\bar p
\lambda\lambda'}{a}\right).\end{aligned}$$ Here $\bar p$ denotes a multiplicative inverse of $p$ modulo $a$. We first observe that the sum vanishes unless $r=(h,a)$ divides $\nu$. This is easily seen by replacing $\lambda\mapsto \lambda+\frac{a}{r} \tau$ for $\tau\in
{\mathcal{O}}_F$ in the sum. Therefore, put $r=(h,a)$ and write $h=h_1 r$, $a= a_1 r$, and $\nu=\nu_1 r$. Then the sum becomes $$\begin{aligned}
&\frac{\epsilon_p(a)}{ a }\sum_{\substack{\lambda\in {\mathcal{O}}_F/a{\mathcal{O}}_F}} e\left(\frac{{\operatorname{tr}}(\lambda \nu_1)+h_1\bar p \lambda\lambda'}{a_1}\right)\\
&=\frac{\epsilon_p(a)}{ a }r^2 e\left(-\frac{p \bar h_1 \nu_1\nu_1'}{a_1}\right)
\sum_{\substack{\lambda\in {\mathcal{O}}_F/a_1{\mathcal{O}}_F}} e\left(\frac{h_1\bar p(\lambda +p\bar h_1 \nu_1')(\lambda' +p\bar h_1 \nu_1)}{a_1}\right)\\
&=\frac{\epsilon_p(a)}{ a_1 }r e\left(-\frac{p \bar h_1 \nu_1\nu_1'}{a_1}\right)\sum_{\substack{\lambda\in
{\mathcal{O}}_F/a_1{\mathcal{O}}_F}} e\left(\frac{h_1\bar p\lambda \lambda' }{a_1}\right).\end{aligned}$$ The latter sum is computed in [@Za] §4 Lemma 2. Inserting its value $$\begin{aligned}
\frac{1}{ a_1 }\sum_{\substack{\lambda\in {\mathcal{O}}_F/a_1{\mathcal{O}}_F}} e\left(\frac{h_1\bar p\lambda \lambda'
}{a_1}\right)= \epsilon_p( a_1)\end{aligned}$$ we finally find that the left hand side of is equal to $$\begin{aligned}
\epsilon_p(r) r \cdot e\left(-\frac{p \bar h_1 \nu\nu'/r^2}{a/r}\right).\end{aligned}$$

Let us now consider the left hand side of in the case that $p\mid a$. Then it is given by $$\begin{aligned}
&\frac{1}{ a }\sum_{m\,(a)}\sum_{\substack{\lambda\in {\mathcal{O}}_F/a{\mathcal{O}}_F\\ \lambda\lambda'\equiv-mp\,(a)}}
\epsilon_p\left(\frac{\lambda\lambda'+mp}{a}\right)  e\left(\frac{{\operatorname{tr}}(\nu\lambda)-mh}{ a}\right).\end{aligned}$$ We substitute $m=m_1+m_2a/p$, where $m_1$ runs modulo $a/p$ and $m_2$ modulo $p$. Moreover, we notice that in the sum over $\lambda$, we actually only sum over $\lambda\in \partial_F/a{\mathcal{O}}_F$, since $p\mid a$. Substituting 