Proof Let $K = \mathbb {Q}(\sqrt {N})$ with $N$ a square-free integer. Recall that one can describe the ring of integers as

$$\begin{align*}\mathcal{O}_K = \begin{cases} \mathbb{Z}[\sqrt{N}], & \text{if } N \not\equiv 1 \bmod 4, \\[1em] \mathbb{Z}\!\left[\dfrac{1 + \sqrt{N}}{2}\right], & \text{if } N \equiv 1 \bmod 4. \end{cases} \end{align*}$$

Let $\{1, \omega \}$ be a $\mathbb {Z}$ -basis of $\mathcal {O}$ . If $N \not \equiv 1 \pmod {4}$ , then Lemma 2.1 implies that $\omega = f\sqrt {N}$ , which has minimal polynomial $m_1(x) = x^2 - f^2 N$ . If $N \equiv 1 \pmod {4}$ , then Lemma 2.1 implies that $\omega = \left (\dfrac {1 + \sqrt {N}}{2}\right )f$ , which has minimal polynomial

$$\begin{align*}m_2(x) = x^2 - f x + \dfrac{f^2(1 - N)}{4}.\\[-37pt] \end{align*}$$

Proof We just need to show that $F(\boldsymbol {v})$ also satisfies the recurrence relation of elliptic nets. To do so, first write (2.5) and (2.6) in multiplicative form and multiply them together. We then utilize the symmetry in quadratic forms. We omit the computation here; interested readers may refer to [Reference Stange14, Proposition 6.1].

Proof We first assume that $ w $ and $ \nu $ are finite places not associated with $ 2 $ . Note that by performing the linear transformation $(x, y)\mapsto \left (x, \frac {1}{2}(y-a_{1}x-a_3)\right )$ , we can always transform the Weierstrass equation of $E/K$ into the form

(2.12) $$ \begin{align} E\colon y^2=4x^3+b_2^2+2b_4x+b_6. \end{align} $$

Since the finite places are assumed not to be associated with $2$ , this does not affect the type of reduction $ {\mathrm{mod}} \ \nu $ . Therefore, without loss of generality, we will assume that $E/K$ has a Weierstrass equation of the form (2.12).

Following the proof of [Reference Silverman13, Chapter III.1, Proposition 1.7(a), p. 49], we know that if we take the substitution $(x', y')=(x, 2y),$ then we can factorize the cubic equation over $\bar {\mathbb {Q}}$ to get the equation

$$\begin{align*}y^{\prime2}=(x'-e_1)(x'-e_2)(x'-e_3), \end{align*}$$

with $e_{i}$ ’s are seen to be distinct. Define $L=K\left (e_1, (e_2-e_1)^{\frac {3}{2}}\right )$ , and make the substitution

$$\begin{align*}x_{K}'=(e_2-e_1)x_{L}+e_{1}; \quad y_{K}'=(e_{2}-e_{1})^{\frac{3}{2}}y_{L} \end{align*}$$

to give the elliptic curve $E/L$ in Legendre form

$$\begin{align*}E\colon y^2=x(x-1)(x-\lambda), \lambda =\frac{e_{3}-e_{1}}{e_{2}-e_{1}} \in L\setminus \{0, 1\}. \end{align*}$$

We claim that this last equation has coefficients which are $w$ -adic integers, and that it has good reduction modulo $w$ , so this is indeed a $w$ -minimal Weierstrass equation for $E/L$ . Indeed, as split multiplicative reduction is invariant under field extensions, we see from the proof of [Reference Silverman13, Chapter VII.5, Proposition 5.4(c), p. 197] that we must have $w(\lambda ) \geq 0$ and good reduction. By assumption, $P$ is a cusp point $ {\mathrm{mod}} \ \nu $ and therefore $e_1\equiv e_2 \equiv e_3 \mod w$ , and in particular $e_2-e_1 \equiv 0 \mod w$ . From (2.12), we then have $(x_K', y_K') \equiv (e_1, 0) \ \mod w$ .

In the case where $w$ and $v$ are associated with $2$ , we use the Deuring normal form and the proof is an elaboration of [Reference Silverman13, Appendix A, Propositions 1.2(b) and 1.3, pp. 410–412]. Given a general Weierstrass equation (2.10), in characteristic 2, the $j$ -invariant is given by $j=\frac {a_{1}}{\Delta }$ . Since we are assuming that $E/K$ is singular under reduction modulo $\nu $ and $j$ is integral (so $j$ is not infinity modulo $\nu $ ), we have $\Delta \equiv 0 \ {\mathrm{mod}} \ \nu $ and this implies that we must have $a_{1} \equiv 0 \ {\mathrm{mod}} \ \nu $ . According to [Reference Silverman13, Appendix A, Proposition 1.1(c)], we use the transformation

$$\begin{align*}x=x'+a_{2}; \quad y=y' \end{align*}$$

to give

(2.13) $$ \begin{align} E'/K\colon y^{\prime2}+a_{1}x'y'+(a_1a_2+a_3)y'=x^{\prime3}+4a_2x^{\prime2}+(5a_{2}^2+a_4)x'+(2a_{2}^3{+}a_{4}a_{2}{+}a_{6}), \end{align} $$

which has the form

$$\begin{align*}y^2+b_3y=x^3+b_4x+b_6 \end{align*}$$

under reduction modulo $\nu $ , with $\Delta = b_3^4\equiv (a_1a_2+a_3)^4 \ {\mathrm{mod}} \ \nu $ . This implies

$$\begin{align*}a_1a_2+a_3 \equiv a_3^4 \equiv b_3^4 \equiv 0\ {\mathrm{mod}} \ \nu. \end{align*}$$

Since $E'/K$ has potential good reduction, we can transform the reduction of $E' \bmod \nu $ into the Deuring normal form

(2.14) $$ \begin{align} E_{\alpha}/\mathbb{F}_{w}\colon y^2+\alpha xy +y =x^3, \mathbb{F}_{w}\,\, \text{ an extension of the finite field} \,\, \mathbb{F}_{\nu}, \end{align} $$

which has good reduction everywhere, modulo $\nu $ in particular. To do so, we will need to find an extension $L/K$ such that $w$ is a prime of $L$ , and a transformation

$$\begin{align*}x^{\prime}_{K}=x_{L}u^2+\gamma; \quad y^{\prime}_{K}=y_{L}u^3+u^2sx_{L}+t \end{align*}$$

such that $E_{\alpha }$ is the reduction of $E/L \bmod w$ . Apply the above transformation to (2.13), we get

$$ \begin{align*} E/L\colon y_{L}^2 &+ u^{-1}(2s+a_1)x_{L}y_{L} + u^{-3}(a_1a_2+a_{1}\gamma+a_3+2t)y_{L} \\ &= x_{L}^3 + u^{-2}(3\gamma-4a_{2}-a_1s-s^2)x_{L}^{2} \\ &\quad + u^{-4}(3\gamma^2-a_1a_2s-a_1\gamma s-5a_2^2-8a_2\gamma-a_3s-a_1t-2st-a_4)x_{L} \\ &\quad + (\gamma^3-2a_2^3-5a_2^2\gamma-4a_2\gamma^2-a_1a_2t-a_1\gamma t-a_2a_4-a_4\gamma-a_3t-t^2-a_6). \end{align*} $$

Reduction modulo $w$ will give the Deuring normal form. Recall that $\nu , w$ are valuations associated with $2$ and $a_1\equiv 0 \bmod \nu $ , hence $a_1\equiv 0 \bmod w$ . Under reduction modulo $w$ , $E/L$ has the form

(2.15) $$ \begin{align} \begin{aligned} y_{L}^2+u^{-3}a_3y_L &\equiv x^3+u^{-2}(\gamma-s^2)x_L^{2}+u^{-4}(\gamma^2-a_2^2-a_3s-a_4)x_L \\ &\quad+(\gamma^3-a_2^2\gamma-a_2a_4-a_4\gamma-a_3t-t^2-a_6). \end{aligned} \end{align} $$

After reduction modulo $w$ , (2.15) is in Deuring form, by comparing coefficients of (2.14) and (2.15), we found the following relationships over $L$ modulo $w$ :

$$\begin{align*}a_3u^{-3}\equiv 0, \alpha \equiv 0, \gamma\equiv s^2, \gamma^2\equiv a_4+a_2^2+a_3s, \gamma^3-a_2^2\gamma-a_4\gamma-a_4a_2-a_3t-a_6-t^2\equiv 0. \end{align*}$$

Note that we always have $u\equiv 0 \bmod w$ , so the first relation implies $a_3 \equiv 0 \bmod w$ necessarily and hence we have

$$\begin{align*}\gamma^2\equiv a_4+a_2^2, t^2\equiv a_4a_2+a_6 \ \pmod w. \end{align*}$$

On the other hand, if we compute the partial derivatives of (2.13) to find the coordinates of the singular point mod $\nu $ , we require

$$\begin{align*}\frac{\partial}{\partial y} \equiv a_3\equiv 0, \frac{\partial}{\partial x}\equiv -3x^{\prime2}+a_4+a_2^2\equiv 0 \ \pmod \nu. \end{align*}$$

This implies that the $x$ -coordinate of the singular point satisfies $(x^{\prime }_{K})^2 \equiv a_{4}+a_{2}^2 \equiv \gamma ^{2} \bmod w$ . This implies $(x^{\prime }_{K})^{2}-\gamma ^{2} \equiv (x^{\prime }_{K}-\gamma )^2 \equiv 0 \bmod w$ , hence we have $x^{\prime }_{K}\equiv \gamma \bmod w$ .

Now reduce $E'/L$ reduction modulo $w$ , using the identities above, we obtain

$$ \begin{align*} (y^{\prime}_K)^2 &\equiv (x')_K^3+(a_2^2+a_4)x^{\prime}_{K}+(a_4a_2+a_6) \\ &\equiv 2(x')_K^2+t^2 \\ &\equiv t^2 \pmod w. \end{align*} $$

This implies $(y^{\prime }_{K})^{2}-t^{2} \equiv (y^{\prime }_{K}-t)^2 \equiv 0 \bmod w$ , hence we have $y^{\prime }_{K}\equiv t \bmod w$ , as we required.

Proof Let $P = (x_K, y_K) \in E_0(K)$ . Using the notations in (2.11), we know that $x_K \not \equiv \gamma \ {\mathrm{mod}} \ {w}$ by Lemma 2.12, so we must have $w\left (x_L\right ) \leq -2w\left (u\right )$ . This implies that there is an injection map $E_0(K) \hookrightarrow {} E_{i_0}(L)$ , where $i_0 = w(u) \geq 1$ and for $n\geq 1$ , we also have a filtration

(2.16) $$ \begin{align} \cdots E_{3}(L_{w}) \subset E_{2}(L_{w}) \subset E_{1}(L_{w}) \subset E_0(L_{w}), \end{align} $$

where $E_{n}(L_{w})=\{P\in E(L_{w})\colon w(x_L(P))\leq -2nw(\pi _w)\} \cup \{O\}$ and $\pi _w$ is a uniformizer for $w$ . Also note that if $P=(x_{L}, y_L) \in E_{i_0}(L)$ , then by definition $w(x_L) \leq -2w(u) = -2i_0$ . From (2.11), this tells us that $w(x_K-\gamma )\leq 0$ , which is equivalent to $x_K \not \equiv \gamma \bmod w$ . Overall, we have $E_0(K) = E(K) \cap E_{i_0}(L)$ .

From Theorem 2.9, we know that both sets on the right-hand side are closed under the $[\omega ]$ map. Therefore, $E_0(K)$ is also closed under the $[\omega ]$ map, which proves that it is a $\mathbb {Z}[\omega ]$ -submodule of $E(K_\nu )$ .

Proof We would first induct on $a \in \mathbb {Z}_{\geq 0}$ , where our base cases are $a=0$ and $a=1$ . For $a=1$ , we will do subproofs of induction on $b\geq -1$ and $b\leq 0$ . Next, we will apply the inductive hypothesis to a and once it is done, we use the symmetry argument $\widetilde {\lambda }(P) = \widetilde {\lambda }(-P)$ to establish the statement for $(a, b) \in \mathbb {Z}^2$ .

We first dispose of $(a, 0), (0, b),$ and $(a, b)=(1, -1)$ . Note that if one of $a$ or $b$ is 0, then we can apply Corollary 2.7 to the point $[a]P$ or $[b]Q$ directly and the result holds. While for $(a, b)=(1, -1)$ , this is just the quasi-parallelogram law: if $(a, b)=(1, -1)$ , then

$$ \begin{align*} \text{LHS}&=\widetilde{\lambda}(P-Q)\\ \text{RHS} &=2\widetilde{\lambda}(P)+2 \widetilde{\lambda}(Q)-\widetilde{\lambda}(P+Q)+\nu\left(\Psi_{(1, -1)}(\boldsymbol{P})\right)\\ &=2\widetilde{\lambda}(P)+2 \widetilde{\lambda}(Q)-\widetilde{\lambda}(P+Q)+\nu\left(x(Q)-x(P)\right), \end{align*} $$

where $\Psi _{(1, -1)}(\boldsymbol {P})=x(Q)-x(P)$ was given in [Reference Stange14, Proposition 3.8]. Since $\nu \left (x(Q)-x(P)\right )=\nu \left (x(P)-x(Q)\right )$ , rearrange the terms to resemble (2.7).

For the inductive step, we first induct on $a=1$ and $b\geq -1$ . Fix $b$ and suppose the statement is true for $a=1$ and each $b \in [-1, k]$ for some $k \in \mathbb {N}$ , consider $(a, b)=(1, k+1)$ . Apply the quasi-parallelogram law with $P = P+[k]Q$ and $Q = Q$ and so we get

$$ \begin{align*} \widetilde{\lambda}(P+[k+1]Q) &= \widetilde{\lambda}\left((P+[k]Q) + Q\right) \\ &= 2\widetilde{\lambda}\left(P+[k]Q\right) + 2\widetilde{\lambda}(Q) - \widetilde{\lambda}(P + [k-1]Q)\\ & \quad + \nu\left(x\left(P+[k]Q\right) - x(Q)\right). \end{align*} $$

Now apply the inductive hypothesis to $\widetilde {\lambda }(P + [k]Q)$ and $\widetilde {\lambda }(P + [k-1]Q)$ , and then apply property (d) in Proposition 2.4 to rewrite the expression for the valuation of the difference between the two $x$ -coordinates, so that we have

$$ \begin{align*} \widetilde{\lambda}(P+[k+1]Q) &= 2\widetilde{\lambda}(P) + 2k^{2}\widetilde{\lambda}(Q) + 2k\left(\widetilde{\lambda}(P+Q) - \widetilde{\lambda}(P) - \widetilde{\lambda}(Q)\right) \\ &\quad + 2\nu\left(\Psi_{(1, k)}(\boldsymbol{P})\right)+ 2\widetilde{\lambda}(Q)- \widetilde{\lambda}(P) \\ &\quad - (k-1)^{2}\widetilde{\lambda}(Q) - (k-1)\left(\widetilde{\lambda}(P+Q) - \widetilde{\lambda}(P) - \widetilde{\lambda}(Q)\right) \\ &\quad - \nu\left(\Psi_{(1, k-1)}(\boldsymbol{P})\right) + \nu\left(-\frac{\Psi_{(1, k+1)}(\boldsymbol{P})\Psi_{(1, k-1)}(\boldsymbol{P})}{\Psi_{(1, k)}^{2}(\boldsymbol{P})}\right) \\ &= \widetilde{\lambda}(P) + (k+1)^2\widetilde{\lambda}(Q) + (k+1)\left(\widetilde{\lambda}(P+Q) - \widetilde{\lambda}(P) - \widetilde{\lambda}(Q)\right) \\ &\quad + \nu\left(\Psi_{(1, k+1)}(\boldsymbol{P})\right). \end{align*} $$

This shows that the statement holds for any $b \geq -1$ when $a=1$ . Next, we induct on $a=1$ and $b\leq 0$ . Again, fix $b$ and suppose the statement is true for $a=1$ and each $b \in [-k, -1]$ for some $k \in \mathbb {N}$ , consider $(a, b)=(1, -k-1)$ . The proof goes similarly as above: first apply the quasi-parallelogram law with $P = P+[-k]Q$ and $Q = Q$ , then the inductive hypothesis to $\widetilde {\lambda }\left (P+[-k]Q\right )$ and $\widetilde {\lambda }(P + [-k+1]Q)$ and finally apply property (d) in Proposition 2.4. We get

$$ \begin{align*} \widetilde{\lambda}(P+[-k-1]Q) &= \widetilde{\lambda}\left((P+[-k]Q) - Q\right) \\ &= 2\widetilde{\lambda}\left(P+[-k]Q\right) + 2\widetilde{\lambda}(Q) - \widetilde{\lambda}(P + [-k+1]Q) \\ &\quad + \nu\left(x\left(P+[-k]Q\right) - x(Q)\right) \\ &= 2\widetilde{\lambda}(P) + 2k^{2}\widetilde{\lambda}(Q) - 2k\left(\widetilde{\lambda}(P+Q) - \widetilde{\lambda}(P) - \widetilde{\lambda}(Q)\right) \\ &\quad + 2\nu\left(\Psi_{(1, -k)}(\boldsymbol{P})\right) + 2\widetilde{\lambda}(Q) - \widetilde{\lambda}(P) - (-k+1)^{2}\widetilde{\lambda}(Q) \\ &\quad - (-k+1)\left(\widetilde{\lambda}(P+Q) - \widetilde{\lambda}(P) - \widetilde{\lambda}(Q)\right) \\ &\quad - \nu\left(\Psi_{(1, -k+1)}(\boldsymbol{P})\right) + \nu\left(-\frac{\Psi_{(1, -k+1)}(\boldsymbol{P})\Psi_{(1, -k-1)}(\boldsymbol{P})}{\Psi_{(1, -k)}^{2}(\boldsymbol{P})}\right) \\ &= \widetilde{\lambda}(P) + (-k-1)^2\widetilde{\lambda}(Q) \\ &\quad - (k+1)\left(\widetilde{\lambda}(P+Q) - \widetilde{\lambda}(P) - \widetilde{\lambda}(Q)\right) + \nu\left(\Psi_{(1, -k-1)}(\boldsymbol{P})\right). \end{align*} $$

Overall, we have shown that the statement holds for $a=1$ and $b\in \mathbb {Z}$ . Now we need to induct on $a\geq 0$ . Suppose the statement holds for each $a \in [-1, k]$ for some $k\in \mathbb {N}$ , consider $(a, b)=(k+1, b)$ . As before, we apply the quasi-parallelogram law with $P = [k]P+[b]Q$ and $Q = P$ , then the inductive hypothesis to $\widetilde {\lambda }([k]P+[b]Q)$ and $\widetilde {\lambda }([k-1]P + [b]Q)$ , and property (d) in Proposition 2.4. We get

$$ \begin{align*} \widetilde{\lambda}([k+1]P+[b]Q) &= \widetilde{\lambda}\left(([k]P+[b]Q) + P\right) \\ &= 2\widetilde{\lambda}([k]P + [b]Q) + 2\widetilde{\lambda}(P) - \widetilde{\lambda}([k-1]P + [b]Q) \\ &\quad + \nu\left(x([k]P + [b]Q) - x(P)\right) \\ &= 2k^{2}\widetilde{\lambda}(P) + 2b^{2}\widetilde{\lambda}(Q) + 2kb\left(\widetilde{\lambda}(P+Q) - \widetilde{\lambda}(P) - \widetilde{\lambda}(Q)\right) \\ &\quad + 2\nu\left(\Psi_{(k, b)}(\boldsymbol{P})\right) + 2\widetilde{\lambda}(P)- (k-1)^{2}\widetilde{\lambda}(P) - b^{2}\widetilde{\lambda}(Q) \\ &\quad - (k-1)b\left(\widetilde{\lambda}(P+Q) - \widetilde{\lambda}(P) - \widetilde{\lambda}(Q)\right) \\ &\quad - \nu\left(\Psi_{(k-1, b)}(\boldsymbol{P})\right) + \nu\left(-\frac{\Psi_{(k+1, b)}(\boldsymbol{P})\Psi_{(k-1, b)}(\boldsymbol{P})}{\Psi_{(k, b)}^{2}(\boldsymbol{P})}\right) \\ &= (k + 1)^{2}\widetilde{\lambda}(P) + b^{2}\widetilde{\lambda}(Q) + \nu\left(\Psi_{(k+1, b)}(\boldsymbol{P})\right)\\ &\quad +(k+1)b\left(\widetilde{\lambda}(P+Q) - \widetilde{\lambda}(P) - \widetilde{\lambda}(Q)\right). \end{align*} $$

This shows that if the statement holds for $(a, b)$ , then it holds for $(a+1, b)$ for $a\in \mathbb {Z}_{\geq 0}, b\in \mathbb {Z}$ . By the symmetry argument, this also implies that the statement holds for $a\in \mathbb {Z}_{\leq 0}$ and $b \in \mathbb {Z}$ . Hence, the statement holds for any $(a, b) \in \mathbb {Z}^2$ by induction.

Proof First notice that $[\alpha r]P$ is non-singular $ {\mathrm{mod}} \ \mathfrak {p}$ , so we always have

$$\begin{align*}&-\min\left(2\nu\left(\Psi_{(X, Y)}(\boldsymbol{P})\right), \nu\left(\Phi_{(X, Y)}(\boldsymbol{P})\right)\right)\\&\quad = \begin{cases} -2v(\Psi_{(X, Y)}(\boldsymbol{P})) & \text{if } [\alpha r]P \in E_0(K) \setminus E_1(K); \\ -v(\Phi_{(X, Y)}(\boldsymbol{P})) & \text{if } [\alpha r]P \in E_1(K). \end{cases} \end{align*}$$

By Corollary 3.2, we have

$$\begin{align*}\widetilde{\lambda}([\alpha r]P) &= X^{2}\widetilde{\lambda}(P) +Y^{2}\widetilde{\lambda}(\omega P) + XY\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right)\\&\quad + \nu(\Psi_{(X, Y)}(\boldsymbol{P})). \end{align*}$$

Recall that for all $P \in E_{0}(K)$ , we have $\widetilde {\lambda }(P)=\frac {1}{2}\max \left (\nu \left (x(P)^{-1}\right ), 0\right )$ and we always have $\nu \left (x([\alpha r]P)\right )=\nu \left (\Phi _{(X, Y)}(\boldsymbol {P})\right )-2\nu \left (\Psi _{(X, Y)}(\boldsymbol {P})\right )$ .

If $[\alpha r]P \in E_{1}(K)$ , then

$$\begin{align*}\widetilde{\lambda}([\alpha r]P)=-\frac{1}{2}\nu\left(x([\alpha r]P)\right), \,\,\text{and} \end{align*}$$

$$\begin{align*}2X\widetilde{\lambda}(P)+2Y\widetilde{\lambda}(\omega P)+2XY\left(\widetilde{\lambda}([1+\omega]P)-\widetilde{\lambda}(P)-\widetilde{\lambda}(\omega P)\right)=-\nu\left(\Phi_{(X, Y)}(\boldsymbol{P})\right). \end{align*}$$

For $[\alpha r]P \in E_{0}(K) \setminus E_{1}(K)$ , note that we must have $\widetilde {\lambda }([\alpha r]P) = 0$ . Therefore,

$$\begin{align*}2X\widetilde{\lambda}(P)+2Y\widetilde{\lambda}(\omega P)+2XY\left(\widetilde{\lambda}([1+\omega]P)-\widetilde{\lambda}(P)-\widetilde{\lambda}(\omega P)\right)= -2\nu\left(\Psi_{(X, Y)}(\boldsymbol{P})\right). \end{align*}$$

Proof By the semi-stable reduction theorem (see [Reference Silverman13, Chapter VII.5, Proposition 5.4, p. 197]), there exists a field extension $L/K$ such that $E$ has good reduction and a finite place $w$ in $L$ that extends $\nu $ . From (2.11), we know that the new $x$ -coordinate is $x_{L}=\frac {x_{K}-\gamma }{u^2}$ . Now the point $[r]P$ is non-singular $ {\mathrm{mod}} \ \nu $ , so, by Lemma 2.12, we have $x_{K}([r]P)\not \equiv \gamma \ {\mathrm{mod}} \ w$ and $w\left (x_{L}([r]P)\right )=-2w(u)<0$ . However, since $[\alpha ]P$ is singular $ {\mathrm{mod}} \ \nu $ , the lemma tells us that $w\left (x_{L}([\alpha ]P)\right )>-2w(u)=w\left (x_{L}([r]P)\right )$ . We know that there is a $\mathbb {Z}[\omega ]$ -module filtration as in (2.16), which tells us that if $[r]P \in E_{i}(L_{w})$ and $[\alpha ]P \in E_{j}(L_{w})$ , then we have $i>j$ . Therefore, $[\alpha +r]P \in E_{j}(L_{w})$ . Recall that the points $[r]P$ and $[r\pm \alpha ]P$ have good reduction in $E(L)$ , so the formula in Theorem 2.8 applies and tell us that

$$\begin{align*}\lambda_{w}\left([r\pm \alpha]P\right)=\lambda_{w}\left([\pm\alpha]P\right)=\lambda_{w}\left([\alpha]P\right). \end{align*}$$

Since the local height function is invariant under field extension, we also get

$$\begin{align*}\lambda_{\nu}\left([r\pm \alpha]P\right)=\lambda_{\nu}\left([\alpha]P\right).\\[-34pt] \end{align*}$$

Proof The proof mostly follows from [Reference Akbary, Bleaney and Yazdani1, Proposition 1.7]. One can verify, or refer back to [Reference Akbary, Bleaney and Yazdani1, Lemma 2.11], that the function $\varepsilon : \mathbb {Z}^{r} \rightarrow \mathbb {Z}$ , defined by

(4.3) $$ \begin{align} \varepsilon(\boldsymbol{v})= \begin{cases} \lambda(\boldsymbol{v} \cdot \boldsymbol{P})-\frac{1}{12}\nu(\Delta)-d_{\mathfrak{p}}(\Psi_{\boldsymbol{v}}(\boldsymbol{P})\mathcal{O}_{K}), & \text{if } \boldsymbol{v}\neq 0, \\ 0, & \text{otherwise}. \end{cases} \end{align} $$

is a quadratic form, where $\lambda $ is the Néron height function associated with the prime $\mathfrak {p}$ . Since $\nu $ is normalized, we know $\nu (\alpha )=d_{\mathfrak {p}}(\alpha \mathcal {O}_{K})$ for $\alpha \in K$ . Recall from (2.8) that if $P \ {\mathrm{mod}} \ \mathfrak {p}$ is non-singular, then we have

$$\begin{align*}\lambda(P)=\max\left(-\frac{1}{2}d_{\mathfrak{p}}(x(P)\mathcal{O}_{K}), 0\right)+\frac{1}{12}\nu(\Delta). \end{align*}$$

Observe that in this case we have $d_{\mathfrak {p}}(D_{P}) = \max \left (-\tfrac {1}{2} d_{\mathfrak {p}}(x(P)\mathcal {O}_K), 0\right )$ , so for $\boldsymbol {v} \in \mathbb {Z}^{r} \setminus \{\boldsymbol {0}\}$ , we can rewrite the quadratic form $\varepsilon (\boldsymbol {v})$ as

$$ \begin{align*} \varepsilon(\boldsymbol{v}) &=\lambda(\boldsymbol{v} \cdot \boldsymbol{P})-\frac{1}{12}\nu(\Delta)-d_{\mathfrak{p}}(\Psi_{\boldsymbol{v}}(\boldsymbol{P})\mathcal{O}_{K}) \\ &=d_{\mathfrak{p}}(D_{\boldsymbol{v} \cdot \boldsymbol{P}})-d_{\mathfrak{p}}\left(\Psi_{\boldsymbol{v}}(\boldsymbol{P})\mathcal{O}_{K}\right). \end{align*} $$

Now we further define another quadratic form

$$ \begin{align*} \widehat{\varepsilon}: \mathbb{Z}^{r} \rightarrow \mathbb{Z}, \quad \widehat{\varepsilon}(\boldsymbol{v}) &= \varepsilon(\boldsymbol{v}) - d_{\mathfrak{p}}(F_{\boldsymbol{v}}(\boldsymbol{P})) \\ &= d_{\mathfrak{p}}(D_{\boldsymbol{v} \cdot \boldsymbol{P}}) - d_{\mathfrak{p}}(\widehat{\Psi}_{\boldsymbol{v}}(\boldsymbol{P})), \end{align*} $$

where we remind readers that $F_{\boldsymbol {v}}(\boldsymbol {P})$ is as defined in (4.1).

The difference between two quadratic forms is also a quadratic form, hence $\hat {\varepsilon }$ is a quadratic form. Note that for any $1 \leq i < j \leq n$ , $\widehat {\varepsilon }$ satisfies

$$ \begin{align*} \widehat{\varepsilon}(\boldsymbol{e_{i}}) &= d_{\mathfrak{p}}(D_{P_{i}}) - d_{\mathfrak{p}}(\widehat{\Psi}_{\boldsymbol{e_{i}}}(\boldsymbol{P}))=0 \\ \widehat{\varepsilon}(\boldsymbol{e_{i}+e_{j}}) &= d_{\mathfrak{p}}(D_{P_{i}+P_{j}}) - d_{\mathfrak{p}}(\widehat{\Psi}_{\boldsymbol{e_{i}+e_{j}}}(\boldsymbol{P}))=0. \end{align*} $$

Note that if a quadratic form $G \ \colon \ \mathbb {Z}^r \rightarrow I_{K}$ satisfies $G(\boldsymbol {v})=0$ for all $\boldsymbol {v}=\boldsymbol {e_{i}}$ and $\boldsymbol {v}=\boldsymbol {e_{i}+e_{j}}$ , then G( $\boldsymbol {v})=0$ for all $\boldsymbol {v}$ (see [Reference Stange14, Lemma 4.5]), so $\widehat {\varepsilon }(\boldsymbol {v})=0$ for all $\boldsymbol {v} \in \mathbb {Z}^{n}$ , which we conclude using the definition of $\widehat {\varepsilon }$ that for all $\boldsymbol {v} \in \mathbb {Z}^{r}$ ,

$$\begin{align*}d_{\mathfrak{p}}(D_{\boldsymbol{v} \cdot \boldsymbol{P}})=d_{\mathfrak{p}}(\widehat{\Psi}_{\boldsymbol{v}}(\boldsymbol{P})). \end{align*}$$

Proof The main idea of the proof is as follows: we first use the local height formula from Corollary 3.2 to express $\nu \left (\psi _{\alpha }(rP)\right )$ in terms of $\widetilde {\lambda }\left ([\alpha ]rP\right ), \widetilde {\lambda }\left (rP\right ), \widetilde {\lambda }\left ([\omega ]rP\right )$ and $\widetilde {\lambda }\left ([1+\omega ]rP\right )$ . Since $rP$ is a non-singular point $\bmod \nu$ , we can use the local height formula for non-singular point in (2.8), which says

$$\begin{align*}\widetilde{\lambda}(rP) = \frac{1}{2} \max \left(\nu(x(rP)^{-1}), 0 \right), \text{ for all } rP \in E_0(K), \end{align*}$$

to replace all the local height terms above so we can express $\nu \left (\psi _{\alpha }(rP)\right )$ in terms of the valuations of the division polynomials $\phi _{z}$ and $\psi _{z}$ instead. Finally, recall from (4.4) that we have

(4.5) $$ \begin{align} \psi_{\alpha r}(P)=\psi_{\alpha}(rP)\psi_{r}(P)^{a^{2}-ab}\psi_{r\omega}(P)^{b^{2}-ab}\psi_{r(1+\omega)}(P)^{ab}. \end{align} $$

By substituting the expression we obtained for $\nu \left (\psi _{\alpha }(rP)\right )$ into (4.5) and carefully handle the cases of whether the points reduce to the identity modulo $\mathfrak {p}$ or not, we yield the result. Overall, we have five different cases to examine.

Case I: $rP \equiv O \bmod \mathfrak {p}$ , so $r\omega P \equiv O \bmod \mathfrak {p}$ as well.

First, Corollary 3.2 tells us that

$$ \begin{align*} \nu\left(\psi_{\alpha}(rP)\right) &= \widetilde{\lambda}\left([\alpha]rP\right) - a^{2}\widetilde{\lambda}(rP) - b^{2}\widetilde{\lambda}([\omega]rP)\\& \quad - ab\left(\widetilde{\lambda}([1+\omega]rP) - \widetilde{\lambda}(rP) - \widetilde{\lambda}([\omega]rP)\right)\\ &= \nu\left(\psi_{\alpha r}\right) - \frac{1}{2}\nu\left(\phi_{\alpha r}\right) - a^{2}{\left(\nu\left(\psi_{r}\right) - \frac{1}{2}\nu\left(\phi_{r}\right)\right)} - b^{2}{\left(\nu\left(\psi_{\omega r}\right) - \frac{1}{2}\nu\left(\phi_{\omega r}\right)\right)} \\ & \quad - ab\left(\nu\left(\psi_{(1+\omega)r}\right) - \frac{1}{2}\nu\left(\phi_{(1+\omega)r}\right) - {\left(\nu\left(\psi_{r}\right) - \frac{1}{2}\nu\left(\phi_{r}\right)\right)}\right.\\ &\quad\left. - {\left(\nu\left(\psi_{\omega r}\right) - \frac{1}{2}\nu\left(\phi_{\omega r}\right)\right)} \right) \\ &= \nu\left(\psi_{\alpha r}\right) - \frac{1}{2}\nu\left(\phi_{\alpha r}\right) - (a^{2} - ab){\left(\nu\left(\psi_{r}\right) - \frac{1}{2}\nu\left(\phi_{r}\right)\right)} \\ & \quad - (b^{2} - ab){\left(\nu\left(\psi_{\omega r}\right) - \frac{1}{2}\nu\left(\phi_{\omega r}\right)\right)} - ab{\left(\nu\left(\psi_{(1+\omega)r}\right) - \frac{1}{2}\nu\left(\phi_{(1+\omega)r}\right)\right)}. \end{align*} $$

Now substitute the expression of $\nu \left (\psi _{\alpha }(rP)\right )$ into (4.5), we have

$$\begin{align*}\nu\left(\psi_{\alpha r}\right)=\nu\left(\psi_{\alpha r}\right)-\frac{1}{2}\nu\left(\phi_{\alpha r}\right)+\frac{a^{2}-ab}{2}\nu\left(\phi_{r}\right)+\frac{b^{2}-ab}{2}\nu\left(\phi_{r\omega}\right)+\frac{ab}{2}\nu\left({\phi}_{r+r\omega}\right). \end{align*}$$

Rearrange and multiply both sides by $2$ , then we have

$$\begin{align*}\nu\left(\phi_{\alpha r}\right)=(a^{2}-ab)\nu\left(\phi_{r}\right)+(b^{2}-ab)\nu\left(\phi_{r\omega}\right)+(ab)\nu\left({\phi}_{r+r\omega}\right). \end{align*}$$

Since $rP \equiv r\omega P \equiv r(1+\omega )P \equiv O \bmod \mathfrak {p}$ , we also have $\alpha rP\equiv O \bmod \mathfrak {p}$ and therefore the following holds:

$$\begin{align*}2\nu(\psi_{r})\geq \nu(\phi_{r}), 2\nu(\psi_{r\omega})\geq \nu(\phi_{r\omega}), 2\nu(\psi_{r+r\omega})\geq \nu(\phi_{r+r\omega})\,\, \text{and} \,\,2\nu(\psi_{\alpha r})\geq \nu(\phi_{\alpha r}). \end{align*}$$

Overall, when $rP \equiv O \mod \mathfrak {p}$ ,

$$ \begin{align*} &{\mathrm{min}}\left(2\nu\left(\psi_{\alpha r}(P)\right), \nu\left(\phi_{\alpha r}(P)\right)\right)\\ &=\nu(\phi_{\alpha r}) \\ &=(a^{2}-ab)\mu(r)+(b^{2}-ab)\mu_{r\omega}(r)+(ab)\mu_{r+r\omega}(r). \end{align*} $$

Case II: $rP, r\omega P, r(1+\omega )P,$ and $\alpha rP \not \equiv O \bmod \mathfrak {p.}$

We start by considering the term $\nu (\psi _{\alpha }(rP))$ . Recall that if a point $P \in E_{0}(K)$ satisfies $P \not \equiv O \ \mod \mathfrak {p}$ , then $\widetilde {\lambda }(P) = 0$ . Hence, using (2.8) and noting that we assumed $\alpha rP \not \equiv O \ \mod \mathfrak {p}$ , we have

$$ \begin{align*} \nu(\psi_{\alpha}(rP)) &= \widetilde{\lambda}([\alpha r]P) - a^{2}(0) - b^{2}(0) - ab\left(0-0-0\right) \\ &=\widetilde{\lambda}([\alpha r]P) \\ &=\frac{1}{2}\max\left(\nu\left(x(\alpha rP)^{-1}\right), 0\right)\\ &=0. \end{align*} $$

Then (4.5) tells us that

$$ \begin{align*} &{\mathrm{min}}\left(2\nu\left(\psi_{\alpha r}(P)\right), \nu\left(\phi_{\alpha r}(P)\right)\right)\\ &=2\nu\left(\psi_{\alpha r}\right)\\ &=0+2(a^{2}-ab)\nu\left(\psi_{r}\right)+2(b^2-ab)\nu\left(\psi_{r\omega}\right)+(2ab)\nu\left(\psi_{r(1+\omega)}\right) \\ &=(a^{2}-ab)\mu(r)+(b^2-ab)\mu_{\omega}(r)+(ab)\mu_{1+\omega}(r), \end{align*} $$

where the last equality follows from the fact that

$$\begin{align*}2\nu(\psi_{r})\leq \nu(\phi_{r}), 2\nu(\psi_{r\omega})\leq \nu(\phi_{r\omega}) \text{ and } 2\nu(\psi_{r+r\omega})\leq \nu(\phi_{r+r\omega}) \end{align*}$$

under the assumption that none of the points reduces to the identity modulo $\mathfrak {p}$ .

Case III: $rP, r\omega P, r(1+\omega )P \not \equiv O \bmod \mathfrak {p}$ but $\alpha rP \equiv O \bmod \mathfrak {p.}$

As in Case II, (2.8) tells us that

$$\begin{align*}\nu(\psi_{\alpha}(rP))=\widetilde{\lambda}([\alpha r]P)=\frac{1}{2}\max\left(\nu\left(x(\alpha rP)^{-1}\right), 0\right)=\nu(\psi_{\alpha r})-\frac{1}{2}\nu(\phi_{\alpha r}). \end{align*}$$

Substitute this into (4.5) to give

$$\begin{align*}\nu\left(\psi_{\alpha r}\right)=\nu\left(\psi_{\alpha r}\right)-\frac{1}{2}\nu\left(\phi_{\alpha r}\right)+(a^{2}-ab)\nu\left(\psi_{r}\right)+(b^{2}-ab)\nu\left(\psi_{r\omega}\right)+ab\nu\left(\psi_{r+r\omega}\right). \end{align*}$$

Rearrange and multiply both sides by 2, we then get

$$\begin{align*}\nu\left(\phi_{\alpha r}\right)=2(a^{2}-ab)\nu\left(\psi_{r}\right)+2(b^{2}-ab)\nu\left(\psi_{r\omega}\right)+(2ab)\nu\left(\psi_{r+r\omega}\right). \end{align*}$$

This shows that

$$ \begin{align*} &{\mathrm{min}}\left(2\nu\left(\psi_{\alpha r}(P)\right), \nu\left(\phi_{\alpha r}(P)\right)\right) \\ &=\nu\left(\phi_{\alpha r}\right) \\ &=2(a^{2}-ab)\nu\left(\psi_{r}\right)+2(b^{2}-ab)\nu\left(\psi_{r\omega}\right)+(2ab)\nu\left(\psi_{r+r\omega}\right) \\ &=(a^{2}-ab)\mu(r)+(b^2-ab)\mu_{\omega}(r)+(ab)\mu_{1+\omega}(r), \end{align*} $$

where the last equality follows from the assumption that $2\nu (\psi _r)\leq \nu (\phi _r), 2\nu (\psi _{r\omega })\leq \nu (\phi _{r\omega })$ , and $2\nu (\psi _{r+r\omega })\leq \nu (\phi _{r+r\omega }).$

Case IV: $rP \not \equiv O \bmod \mathfrak {p}$ , but $r\omega P \equiv O \bmod \mathfrak {p}$ and so $r(1+\omega )P \equiv rP \not \equiv O \bmod \mathfrak {p.}$ Again, we start by considering $\nu (\psi _{\alpha }(rP))$ . Using (2.8), we have

$$ \begin{align*} \nu(\psi_{\alpha}(rP)) &=\widetilde{\lambda}([\alpha r]P)-0-b^{2}\widetilde{\lambda}(\omega rP)-ab\left(0-0-\widetilde{\lambda}(\omega rP)\right) \\ &=\widetilde{\lambda}([\alpha r]P)-(b^{2}-ab)\widetilde{\lambda}(\omega rP) \\ &=\widetilde{\lambda}([\alpha r]P)-(b^{2}-ab)\left(\nu(\psi_{\omega r})-\frac{1}{2}\nu(\phi_{\omega r})\right), \end{align*} $$

where the last equality follows from the fact that $r\omega P \equiv O \bmod \mathfrak {p}$ . Now we want to substitute this back into (4.5), which depends on the behavior of $\alpha rP$ reduction modulo $\mathfrak {p}$ .

Case IV.1: $\alpha rP \equiv O \bmod \mathfrak {p.}$

Since $\alpha rP$ is a non-singular point $\bmod \mathfrak {p}$ , we have

$$\begin{align*}\nu(\psi_{\alpha}(rP)) =\left(\nu(\psi_{\alpha r})-\frac{1}{2}\nu(\phi_{\alpha r})\right)-(b^{2}-ab)\left(\nu(\psi_{\omega r})-\frac{1}{2}\nu(\phi_{\omega r})\right). \end{align*}$$

In this case, we have

$$\begin{align*}2\nu(\psi_{\alpha r}) \geq \nu(\phi_{\alpha r}), 2\nu(\psi_{r})\leq \nu(\phi_{r}), 2\nu(\psi_{r\omega})\geq \nu(\phi_{r\omega}),\text{ and } 2\nu(\psi_{r+r\omega})\leq \nu(\phi_{r+r\omega}), \end{align*}$$

so if we substitute the expression for $\nu (\psi _{\alpha }(rP))$ back into (4.5) and rearrange as in Case I, we can conclude that

$$ \begin{align*} &{\mathrm{min}}\left(2\nu\left(\psi_{\alpha r}(P)\right), \nu\left(\phi_{\alpha r}(P)\right)\right) \\ &=\nu(\phi_{\alpha r}) \\ &=2(a^{2}-ab)\nu(\phi_{r})+(b^{2}-ab)\nu(\phi_{\omega r})+2ab\nu(\phi_{r+r\omega}) \\ &=(a^{2}-ab)\mu(r)+(b^{2}-ab)\mu_{\omega}(r)+ab\mu_{1+\omega}(r). \end{align*} $$

Case IV.2: $\alpha rP \not \equiv O \bmod \mathfrak {p.}$

Then as shown in Case III, we now have $\widetilde {\lambda }([\alpha r]P)=0$ and the following inequalities hold:

$$\begin{align*}2\nu(\psi_{\alpha r}) \leq \nu(\phi_{\alpha r}), 2\nu(\psi_{r})\leq \nu(\phi_{r}), 2\nu(\psi_{r\omega})\geq \nu(\phi_{r\omega}), \text{ and } 2\nu(\psi_{r+r\omega})\leq \nu(\phi_{r+r\omega}). \end{align*}$$

Therefore, substitute the expression for $\nu (\psi _{\alpha }(rP))$ back into (4.5) and conclude that

$$ \begin{align*} &{\mathrm{min}}\left(2\nu\left(\psi_{\alpha r}(P)\right), \nu\left(\phi_{\alpha r}(P)\right)\right) \\ &=2\nu(\psi_{\alpha r}) \\ &=(a^{2}-ab)\nu(\phi_{r})+(b^{2}-ab)\nu(\phi_{\omega r})+2ab\nu(\phi_{r+r\omega}) \\ &=(a^{2}-ab)\mu(r)+(b^{2}-ab)\mu_{\omega}(r)+ab\mu_{1+\omega}(r). \end{align*} $$

$\textbf{Case V} \ : \ rP, r\omega P \not \equiv O \bmod \mathfrak {p}$ , but $r(1+\omega )P \equiv O \bmod \mathfrak {p}$ .

By the assumption, we have $\widetilde {\lambda }(rP) = \widetilde {\lambda }(r\omega P) = 0$ . Together with (2.8), this gives

$$ \begin{align*} \nu(\psi_{\alpha}(rP)) & =\widetilde{\lambda}([\alpha r]P)-ab\widetilde{\lambda}([r(1+\omega)]P)\\&=\widetilde{\lambda}([\alpha r]P)-ab\left[\nu(\psi_{r+r\omega})-\frac{1}{2}\nu(\phi_{r+r\omega})\right]. \end{align*} $$

As in Case IV, the final result depends on the behavior of $\alpha rP$ reduction modulo $\mathfrak {p}$ .

$\textbf {Case V.1} \ : \ \alpha rP \equiv O \bmod \mathfrak {p}$ .

Then we have

$$\begin{align*}\nu(\psi_{\alpha}(rP)) =\left(\nu(\psi_{\alpha r})-\frac{1}{2}\nu(\phi_{\alpha r})\right)-ab\left(\nu(\psi_{r+r\omega})-\frac{1}{2}\nu(\phi_{r+r\omega})\right). \end{align*}$$

Substitute this back into (4.5), rearrange and simplify to get

$$\begin{align*}\nu\left(\phi_{\alpha r}\right)=2(a^{2}-ab)\nu\left(\psi_{r}\right)+2(b^{2}-ab)\nu\left(\psi_{r\omega}\right)+(ab)\nu\left(\phi_{r+r\omega}\right). \end{align*}$$

Finally, recall from the assumption that we have the following inequalities:

$$\begin{align*}2\nu(\psi_{\alpha r}) \geq \nu(\psi_{\alpha r}), 2\nu(\psi_{r})\leq \nu(\phi_{r}), 2\nu(\psi_{r\omega})\leq \nu(\phi_{r\omega}), \text{ and } 2\nu(\psi_{r+r\omega})\geq \nu(\phi_{r+r\omega}). \end{align*}$$

Therefore, we conclude that

$$ \begin{align*} &{\mathrm{min}}\left(2\nu\left(\psi_{\alpha r}(P)\right), \nu\left(\phi_{\alpha r}(P)\right)\right)\\&=\nu(\phi_{\alpha r}) \\&=2(a^{2}-ab)\nu\left(\psi_{r}\right)+2(b^{2}-ab)\nu\left(\psi_{r\omega}\right)+(ab)\nu\left(\phi_{r+r\omega}\right)\\&=(a^{2}-ab)\mu(r)+(b^{2}-ab)\mu_{\omega}(r)+ab\mu_{1+\omega}(r). \end{align*} $$

$\textbf {Case V.2} \ : \ \alpha rP \not \equiv O \bmod \mathfrak {p}$ .

As in Case IV.2, we now have $\widetilde {\lambda }([\alpha r]P)=0$ and

$$\begin{align*}2\nu(\psi_{\alpha r}) \geq \nu(\phi_{\alpha r}), 2\nu(\psi_{r})\leq \nu(\phi_{r}), 2\nu(\psi_{r\omega})\geq \nu(\phi_{r\omega}), \text{ and } 2\nu(\psi_{r+r\omega})\leq \nu(\phi_{r+r\omega}). \end{align*}$$

Therefore, substitute the expression for $\nu (\psi _{\alpha }(rP))$ back into (4.5) and conclude that

$$ \begin{align*} &{\mathrm{min}}\left(2\nu\left(\psi_{\alpha r}(P)\right), \nu\left(\phi_{\alpha r}(P)\right)\right)\\&=2\nu(\psi_{\alpha r}) \\&=2(a^{2}-ab)\nu(\psi_{r})+2(b^{2}-ab)\nu(\psi_{\omega r})+ab\nu(\phi_{r+r\omega})\\&=(a^{2}-ab)\mu(r)+(b^{2}-ab)\mu_{\omega}(r)+ab\mu_{1+\omega}(r). \end{align*} $$

Proof Recall that we have two cases to consider: for orders in quadratic imaginary fields, their generators could have two different behaviors: either $\omega ^2=-D$ , or $\omega ^2=\omega -D$ for some square-free integer $D$ .

Case I: $\omega ^{2}=-D$ for some positive integer $D$ . First take note of the following:

• • $\alpha r=(ar_1-br_2D)+(ar_2+br_1)\omega $ ;

• • $ r\omega = -r_2D+r_1\omega $ ;

• • $r(1+\omega )=(r_1-r_2D)+(r_1+r_2)\omega $ .

Apply Corollary 3.2 to the points $[\alpha r +1]P$ and $[\alpha r-1]P$ to get

$$ \begin{align*} &\nu\left(\psi_{\alpha r+1}\psi_{\alpha r-1}\right) \\ &=2\widetilde{\lambda}(P)-2\left((ar_1-br_2D)^2+1\right)\widetilde{\lambda}(P)-2(ar_2+br_1)^2\widetilde{\lambda}(\omega P) \\ &\quad-2(ar_1-br_2D)(ar_2+br_1)\left(\widetilde{\lambda}([1+\omega]P)-\widetilde{\lambda}(P)-\widetilde{\lambda}(\omega P)\right)\\ &=-2(a^2r_{1}^{2}-2abr_1r_2D+b^2r_2^2D^2)\widetilde{\lambda}(P)-2(a^2r_{2}^{2}+2abr_1r_2+b^2r_1^2)\widetilde{\lambda}(\omega P)\\ &\quad-2(a^2r_{1}r_{2}+abr_{1}^{2}-abr_{1}^{2}-abr_{2}^{2}D-b^{2}r_1r_2D)\left(\widetilde{\lambda}([1+\omega]P)-\widetilde{\lambda}(P)-\widetilde{\lambda}(\omega P)\right). \end{align*} $$

Now rearrange the terms and we will see some familiar expressions:

$$ \begin{align*} &=-2a^2r_1^2\widetilde{\lambda}(P)-2a^2r_2^2\widetilde{\lambda}(\omega P)-2ra^2_1r_2\left(\widetilde{\lambda}([1+\omega]P)-\widetilde{\lambda}(P)-\widetilde{\lambda}(\omega P)\right) \\ &\quad-2b^2r_2^2D^2\widetilde{\lambda}(P)-2b^2r_1^2\widetilde{\lambda}(\omega P)+2b^2r_1r_2D\left(\widetilde{\lambda}([1+\omega]P)-\widetilde{\lambda}(P)-\widetilde{\lambda}(\omega P)\right) \\ &\quad-2ab\left(-2r_1r_2D\widetilde{\lambda}(P)+2r_1r_2\widetilde{\lambda}(\omega P)+(r_1^2-r_2^2D)\left(\widetilde{\lambda}([1{+}\omega]P){-}\widetilde{\lambda}(P){-}\widetilde{\lambda}(\omega P)\right)\right). \end{align*} $$

Note that applying Proposition 3.3 to the points $[r]P$ and $[r \omega ]P$ produces the terms in the first two lines of the equality sign above, which correspond to $\mu (P)$ and $\mu _{\omega }(P)$ , respectively. Thus, we have

$$ \begin{align*} &\nu\left(\psi_{\alpha r+1}\psi_{\alpha r-1}\right) \\ &=a^{2}\cdot\mu(r)+b^{2}\cdot \mu_{\omega}(r) \\ &\quad-2ab\left(-2r_1r_2D\widetilde{\lambda}(r)+2r_1r_2\widetilde{\lambda}(\omega P)+(r_1^2-r_2^2D)\left(\widetilde{\lambda}([1+\omega]P){-}\widetilde{\lambda}(P){-}\widetilde{\lambda}(\omega P)\right)\right). \end{align*} $$

For the remaining terms, we once again use Corollary 3.2 to find an expression for $\widetilde {\lambda }\left ([r(1+\omega )]P\right )-\nu \left (\psi _{r(1+\omega )}\right )$ :

$$ \begin{align*} & \widetilde{\lambda}\left([r(1+\omega)]P\right) - \nu\left(\psi_{r(1+\omega)}\right) \\ &= (r_1 - r_2D)^2\widetilde{\lambda}(P) + (r_1 + r_2)^2\widetilde{\lambda}(\omega P) \\ &\quad + (r_1^2 + r_1r_2 - r_1r_2D - r_2^2D)\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right) \\ &= (r_1^2 - 2r_1r_2D + r_2^2D^2)\widetilde{\lambda}(P) + (r_1^2 + 2r_1r_2 + r_2^2)\widetilde{\lambda}(\omega P) \\ &\quad + (r_1^2 + r_1r_2 - r_1r_2D - r_2^2D)\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right). \end{align*} $$

As before, by rearranging the terms and applying Proposition 3.3 to the points $ [r]P $ and $ [r\omega ]P $ , we get, in order, $ -\frac {1}{2}\mu (P) $ , $ \frac {1}{2}\mu _{\omega }(P) $ , and the remaining terms:

$$ \begin{align*} &= r_1^2\widetilde{\lambda}(P) + r_2^2\widetilde{\lambda}(\omega P) + r_1r_2\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right) \\ &\quad + r_2^2D^2\widetilde{\lambda}(P) + r_1^2\widetilde{\lambda}(\omega P) - r_1r_2D\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right) \\ &\quad - 2r_1r_2D\widetilde{\lambda}(P) + 2r_1r_2\widetilde{\lambda}(\omega P) + (r_1^2 - r_2^2D)\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right) \\ &= -\frac{1}{2}\mu(P) - \frac{1}{2}\mu_{\omega}(P) \\ &\quad - 2r_1r_2D\widetilde{\lambda}(P) + 2r_1r_2\widetilde{\lambda}(\omega P) + (r_1^2 - r_2^2D)\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right). \end{align*} $$

Hence, an alternative expression for the remaining terms is $\frac {1}{2}(\mu (r) + \mu _{\omega }(r) - \mu _{1+\omega }(r))$ . Substituting this back into $ \nu \left (\psi _{\alpha r+1}\psi _{\alpha r-1}\right ) $ , we obtain

$$ \begin{align*} & \nu\left(\psi_{\alpha r+1}\psi_{\alpha r-1}\right) \\ &= a^2\cdot\mu(P) + b^2\cdot \mu_{\omega}(P) - 2ab\left(\frac{1}{2}\left(\mu(P) + \mu_{\omega}(P) - \mu_{1+\omega}(P)\right)\right) \\ &= (a^2 - ab)\mu(r) + (b^2 - ab)\mu_{\omega}(r) + (ab)\mu_{1+\omega}(r). \end{align*} $$

Case II: $\omega ^2=f\omega -D$ for some positive integer $D.$

Again, we need to first take note of some expressions:

• • $\alpha r=(ar_1-br_2D)+(ar_2+br_1+br_2f)\omega $ ,

• • $ r\omega = -r_2D+(r_1+r_2f)\omega $ ,

• • $r(1+\omega )=(r_1-r_2D)+(r_1+r_2+r_2f)\omega $ .

The proof proceeds exactly as in Case I: first, use Corollary 3.2 to expand $\nu \left (\psi _{\alpha r+1}\psi _{\alpha r-1}\right )$ , then apply Proposition 3.3 to identify $\mu (P)$ and $\mu _{\omega }(P)$ . Finally, clear up the remaining terms by using Corollary 3.2 on $\widetilde {\lambda }\left ([r(1+\omega )]P\right )-\nu \left (\psi _{r(1+\omega )}\right )$ . Since the strategy is the same, we will omit intermediate explanations for conciseness:

$$ \begin{align*} & \widetilde{\lambda}\left([r(1+\omega)]P\right)-\nu\left(\psi_{r(1+\omega)}\right) \\ &=(r_1^2 - 2r_1r_2D + r_2^2D^2)\widetilde{\lambda}(P) + (f^2 r_2^2 + 2 f r_1 r_2 + 2 f r_2^2 + r_1^2 + 2 r_1 r_2 + r_2^2)\widetilde{\lambda}(\omega P) \\ &\quad + (-D f r_2^2 - D r_1 r_2 - D r_2^2 + f r_1 r_2 + r_1^2 + r_1 r_2)\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right) \\ &= -\frac{1}{2}\mu(r) -\frac{1}{2}\mu_{\omega}(r)-2r_1r_2D\widetilde{\lambda}(P) + (2r_1r_2 + 2fr_2^2)\widetilde{\lambda}(\omega P) \\ &\quad + (r_1^2 +fr_1r_2 - r_2^2D)\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right). \end{align*} $$

Therefore, we get

$$ \begin{align*} & \nu\left(\psi_{\alpha r+1}\psi_{\alpha r-1}\right) \\ &=-2(a^2r_{1}^{2}-2abr_1r_2D+b^2r_2^2D^2)\widetilde{\lambda}(P) \\ &\quad - 2(a^2 r_2^2 + 2 a b f r_2^2 + 2 a b r_1 r_2 + b^2 f^2 r_2^2 + 2 b^2 f r_1 r_2 + b^2 r_1^2)\widetilde{\lambda}(\omega P) \\ &\quad -2(a^2 r_1 r_2 - a b D r_2^2 + a b f r_1 r_2 + a b r_1^2 - b^2 D f r_2^2 - b^2 D r_1 r_2)\\&\qquad\left(\widetilde{\lambda}([1+\omega]P)-\widetilde{\lambda}(P)-\widetilde{\lambda}(\omega P)\right) \\ &=a^2\cdot \mu(r) + b^2 \cdot \mu_{\omega}(r) \\ &\quad - 2ab\left(-2r_1r_2D\widetilde{\lambda}(P) + (2r_1r_2+2fr_2^2)\widetilde{\lambda}(\omega P) + (r_1^2+fr_1r_2-r_2^2D)\right.\\&\qquad\left.\left(\widetilde{\lambda}([1+\omega]P)-\widetilde{\lambda}(P)-\widetilde{\lambda}(\omega P)\right)\right) \\ &=a^2\cdot \mu(r) + b^2 \cdot \mu_{\omega}(r) - 2ab\left(\frac{1}{2}\left(\mu(r)+\mu_{\omega}(r)-\mu_{1+\omega}(r)\right)\right) \\ &=(a^{2}-ab)\mu(r) + (b^{2}-ab)\mu_{\omega}(r) + (ab)\mu_{1+\omega}(r), \end{align*} $$

as required, which completes the proof.

Proof Since the point $[\alpha r +\beta ]P$ is singular $ {\mathrm{mod}} \ \mathfrak {p}$ , we know $[\alpha r +\beta ]P \not \equiv O \ {\mathrm{mod}} \ \mathfrak {p}$ and therefore

$$\begin{align*}{\mathrm{min}}\left(2\nu\left(\psi_{\alpha r+\beta}(P)\right), \nu\left(\phi_{\alpha r+\beta}(P)\right)\right)=2\nu\left(\psi_{\alpha r+\beta}(P)\right). \end{align*}$$

As in Lemma 4.4, we have two cases, although the strategies are identically the same.

Case I: $\omega ^{2}=-D$ for some positive integer $D$ .

For convenience, take note of the following expansions:

• • $\alpha r=(ar_1-br_2D)+(ar_2+br_1)\omega $

• • $r-\beta =(r_1-c)+(r_2-d)\omega $

• • $(r_1-c)(r_2-d)=r_1r_2-dr_1-cr_2+d$

• • $r\omega =-r_2D+r_1\omega $

• • $r\omega - \beta =(-r_2D-c)+(r_1-d)\omega $

• • $(-r_2D-c)(r_1-d)=-r_1r_2D+r_2dD-cr_1+cd.$

We first use Corollary 3.2 to evaluate $a\left (\nu \left (\frac {\psi _{\beta }}{\psi _{r - \beta }}\right )\right )$ and $b\left (\nu \left (\frac {\psi _{\beta }}{\psi _{r\omega - \beta }}\right )\right )$ :

$$ \begin{align*} & a\left(\nu\left(\psi_{\beta}\right)-\nu\left(\psi_{r - \beta}\right)\right) \\ &= a\left( \tilde{\lambda}\left([\beta]P\right) - c^2\tilde{\lambda}(P) - d^2\tilde{\lambda}(\omega P) - cd\left[\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right] \right) \\ &\quad - a\left( \widetilde{\lambda}\left([r-\beta]P\right) - (r_1^2 + 2r_1c + c^2)\widetilde{\lambda}\left(P\right) - (r_2^2 - 2r_2d + d^2)\tilde{\lambda}\left(\omega P\right) \right) \\ &\quad - a\left((-r_1r_2 + dr_1 - cr_2 + cd)\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right) \right). \end{align*} $$

Note that terms in $c$ and $d$ only cancel out each other, while $\widetilde {\lambda }\left ([\beta ]P\right )$ cancels with $\widetilde {\lambda }\left ([r\omega -\beta ]P\right )$ by Proposition 3.5. By noticing the terms in $r_1$ and $r_2$ resemble $-\frac {1}{2}\mu (r)$ , we obtain

$$ \begin{align*} & a\left(\nu\left(\psi_{\beta}\right)-\nu\left(\psi_{r - \beta}\right)\right) \\ &= a\left(-\frac{\mu(r)}{2}\right) + 2car_1\widetilde{\lambda}(P) - 2dar_2\widetilde{\lambda}(\omega P)\\ &\quad - (dar_1 + acr_2) \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right). \end{align*} $$

Next, we evaluate $b\left [\nu \left (\psi _{\beta }\right )-\nu \left (\psi _{r\omega - \beta }\right )\right ] $ :

$$ \begin{align*} & b\left(\nu\left(\psi_{\beta}\right)-\nu\left(\psi_{r\omega - \beta}\right)\right) \\ &= b\left( \widetilde{\lambda}\left([\beta]P\right) - c^2\widetilde{\lambda}(P) - d^2\widetilde{\lambda}(\omega P) - cd\left[\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right] \right) \\ &\quad - b\left(\widetilde{\lambda}\left([r\omega-\beta]P\right) - (r_2^2D^2 + 2r_2cD + c^2)\widetilde{\lambda}\left(P\right) - (r_1^2 - 2r_1d + d^2)\widetilde{\lambda}\left(\omega P\right) \right) \\ &\quad - b\left(-( -r_1r_2D + r_2dD - cr_1 + cd)\left[\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right] \right). \end{align*} $$

Similarly, the terms in $c$ and $d$ cancel out each other, while $\widetilde {\lambda }\left ([\beta ]P\right )$ cancels with $\widetilde {\lambda }\left ([r\omega -\beta ]P\right )$ by Proposition 3.5. By noticing the terms in $r_1$ and $r_2D$ resemble $-\frac {1}{2}\mu _{\omega }(r)$ , we get

$$ \begin{align*} & b\left(\nu\left(\psi_{\beta}\right)-\nu\left(\psi_{r\omega - \beta}\right)\right) \\ &= b\left(-\frac{\mu_{\omega}(r)}{2}\right) + 2cbr_2D\widetilde{\lambda}(P) - 2dbr_1\widetilde{\lambda}(\omega P)\\&\quad + (dbr_2D - bcr_1) \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right). \end{align*} $$

Now, if we use Corollary 3.2 to expand $\nu \left (\psi _{\alpha r + \beta }\right )$ and substitute these expressions, we obtain

$$ \begin{align*} & \nu\left(\psi_{\alpha r + \beta}\right) \\ &= \widetilde{\lambda}\left([\beta]P\right) - \left((ar_1 - br_2D) + c\right)^2 \widetilde{\lambda}(P) - \left((ar_2 + br_1)^2 + d\right)^2 \widetilde{\lambda}(\omega P) \\ &\quad - (ar_1 - br_2D + c)(ar_2 + br_1 + d) \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right) \\ &= \widetilde{\lambda}\left([\beta]P\right) - \left( (ar_1 - br_2D)^2 + 2c(ar_1 - br_2D) + c^2 \right) \widetilde{\lambda}(P) \\ &\quad - \left( (ar_2 + br_1)^2 + 2d(ar_2 + br_1) + d^2 \right) \widetilde{\lambda}(\omega P) \\ &\quad - \left( (ar_1 - br_2D)(ar_2 + br_1) + d(ar_1 - br_2D) + c(ar_2 + br_1) + cd \right) \\ &\quad \cdot \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right). \end{align*} $$

Using Corollary 3.2, one can see that the terms in $ c $ and $ d $ resemble $ \nu \left (\psi _{\beta }\right ) $ . On the other hand, by recalling that $ \alpha r = (ar_1-br_2D)+(ar_2+br_1) $ and comparing with the proof of Lemma 4.4 (Case I), we obtain $ \frac {1}{2}\nu \left (\psi _{\alpha r + 1}\psi _{\alpha r - 1}\right ) $ , so we get

$$ \begin{align*} & \nu\left(\psi_{\alpha r + \beta}\right) \\ &= \nu\left(\psi_{\beta}\right) + \frac{1}{2} \nu\left(\psi_{\alpha r + 1}\psi_{\alpha r - 1}\right) \\ &\quad + 2car_1 \widetilde{\lambda}(P) - 2dar_2 \widetilde{\lambda}(\omega P) - (dar_1 + acr_2) \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right) \\ &\quad + 2cbr_2D \tilde{\lambda}(P) - 2dbr_1 \widetilde{\lambda}(\omega P)\\ &\quad + (dbr_2D - bcr_1) \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right). \end{align*} $$

Finally, assembling everything we obtained before and using Lemma 4.4 (Case I) for $ \nu \left (\psi _{\alpha r + 1}\psi _{\alpha r - 1}\right ) $ , we obtain

$$ \begin{align*} & \nu\left(\psi_{\alpha r + \beta}\right) \\ &= \nu\left(\psi_{\beta}\right) + a\left(\nu\left(\frac{\psi_{\beta}}{\psi_{r - \beta}}\right) + \frac{\mu(r)}{2}\right) + b\left(\nu\left(\frac{\psi_{\beta}}{\psi_{r \omega - \beta}}\right) + \frac{\mu_{\omega}(r)}{2}\right) \\ &\quad + \frac{a^{2} - ab}{2} \mu(r) + \frac{b^{2} - ab}{2} \mu_{\omega}(r) + \frac{ab}{2} \mu_{1+\omega}(r). \end{align*} $$

Case II: $\omega ^{2}=\omega -D$ for some square-free integer $D$ .

For convenience, take note of the following expansions:

• • $\alpha r=(ar_1-br_2D)+(ar_2+br_1+br_2f)\omega $

• • $r-\beta =(r_1-c)+(r_2-d)\omega $

• • $(r_1-c)(r_2-d)=r_1r_2-dr_1-cr_2+d$

• • $r\omega =-r_2D+(r_1+fr_2)\omega $

• • $r\omega - \beta =(-r_2D-c)+(r_1+fr_2-d)\omega $

• • $(-r_2D-c)(r_1+fr_2-d)=c d - c f r_2 - c r_1 + d D r_2 - f r_2^2D - r_1 r_2D.$

As mentioned, the proof proceeds exactly as in Case I. Note that the behavior of $\omega $ does not affect $r-\beta $ , so $a\left (\nu \left (\frac {\psi _{\beta }}{\psi _{r - \beta }}\right )\right )$ is the same as in Case I, which is

$$\begin{align*}&a\left(-\frac{\mu(r)}{2}\right) + 2car_1\widetilde{\lambda}(P) - 2dar_2\widetilde{\lambda}(\omega P)\\&\quad - (dar_1 + acr_2) \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right). \end{align*}$$

Next, use Corollary 3.2 to evaluate and $b\left (\nu \left (\frac {\psi _{\beta }}{\psi _{r\omega - \beta }}\right )\right )$ , then apply Corollary 3.2 to $\nu \left (\psi _{\alpha r + \beta }\right )$ and assemble the terms to get the result:

$$ \begin{align*} & b\left(\nu\left(\psi_{\beta}\right)-\nu\left(\psi_{r\omega - \beta}\right)\right) \\ &= b\left( \widetilde{\lambda}\left([\beta]P\right) - c^2\widetilde{\lambda}(P) - d^2\widetilde{\lambda}(\omega P) - cd\left(\widetilde{\lambda}([1+\omega]P) - \tilde{\lambda}(P) - \widetilde{\lambda}(\omega P)\right) \right) \\ &\quad - b\left( \widetilde{\lambda}\left([r\omega-\beta]P\right) - (r_2^2D^2 + 2r_2cD + c^2)\widetilde{\lambda}\left(P\right)\right.\\&\qquad \left. - (r_1^2 + 2 f r_1 r_2 + f^2 r_2^2 + d^2 - 2 d f r_2 - 2 d r_1 )\widetilde{\lambda}\left(\omega P\right) \right) \\ &\quad -b\left(-(c d - c f r_2 - c r_1 + d D r_2 - D f r_2^2 - D r_1 r_2)\left(\widetilde{\lambda}([1+\omega]P) - \widetilde{\lambda}(P) - \tilde{\lambda}(\omega P)\right) \right) \\ &= b\left(-\frac{\mu_{\omega}(r)}{2}\right)+ 2bcr_2D\widetilde{\lambda}(P) -(2bdr_1+2br_2df)\widetilde{\lambda}(\omega P) \\ &\quad + (bdr_2D - bcr_1-bcfr_2) \left(\widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right). \end{align*} $$

Therefore, we have

$$ \begin{align*} & \nu\left(\psi_{\alpha r + \beta}\right) \\ &= \widetilde{\lambda}\left([\beta]P\right) - \left((ar_1 - br_2D) + c\right)^2 \widetilde{\lambda}(P) - \left((ar_2+br_1+br_2f)^2 + d\right)^2 \widetilde{\lambda}(\omega P) \\ &\quad - (ar_1 - br_2D + c)(ar_2+br_1+br_2f + d)\left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right) \\ &= \widetilde{\lambda}\left([\beta]P\right) - \left((ar_1 - br_2D)^2 + 2c(ar_1 - br_2D)+ c^2 \right) \widetilde{\lambda}(P) \\ &\quad - \left((ar_2+br_1+br_2f)^2 + 2d(ar_2+br_1+br_2f)+ d^2 \right) \widetilde{\lambda}(\omega P) \\ &\quad - \left((ar_1 - br_2D)(ar_2+br_1+br_2f) + d(ar_1 - br_2D) + c(ar_2+br_1+br_2f)+ cd \right) \\ &\quad \cdot \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right) \\ &= \nu\left(\psi_{\beta}\right) + \frac{1}{2}\nu\left(\psi_{\alpha r + 1}\psi_{\alpha r - 1}\right) \\ &\quad + 2car_1 \tilde{\lambda}(P) - 2dar_2 \tilde{\lambda}(\omega P) - (dar_1 + acr_2) \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right) \\ &\quad +2bcr_2D\tilde{\lambda}(P) -(2bdr_1+2bdr_2f)\tilde{\lambda}(\omega P) \\ &\quad + (bdr_2D - bcr_1-bcr_2f) \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right) \\ &= \nu\left(\psi_{\beta}\right) + a\left(\nu\left(\frac{\psi_{\beta}}{\psi_{r - \beta}}\right) + \frac{\mu(r)}{2}\right]+ b\left(\nu\left(\frac{\psi_{\beta}}{\psi_{r \omega - \beta}}\right)+\frac{\mu_{\omega}(r)}{2}\right) \\ &\quad + \frac{a^{2} - ab}{2} \mu(r) + \frac{b^{2} - ab}{2} \mu_{\omega}(r) + \frac{ab}{2} \mu_{1+\omega}(r). \end{align*} $$

For $\nu \left (\psi _{\alpha r - \beta }\right )$ , the proof and the computation follow in the exact same way. We will do the case for $\omega ^2=-D$ for some square-free integer $D$ as an example:

$$ \begin{align*} & \nu\left(\psi_{\alpha r - \beta}\right) \\ &= \widetilde{\lambda}\left([\beta]P\right) - \left((ar_1 - br_2D) - c\right)^2 \widetilde{\lambda}(P) - \left((ar_2 + br_1)^2 - d\right)^2 \widetilde{\lambda}(\omega P) \\ &\quad - (ar_1 - br_2D - c)(ar_2 + br_1 - d)\left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right) \\ &= \widetilde{\lambda}\left([\beta]P\right) - \left((ar_1 - br_2D)^2 - 2c(ar_1 - br_2D)+ c^2 \right) \widetilde{\lambda}(P) \\ &\quad - \left((ar_2 + br_1)^2 - 2d(ar_2 + br_1)+ d^2 \right) \widetilde{\lambda}(\omega P) \\ &\quad - \left((ar_1 - br_2D)(ar_2 + br_1) - d(ar_1 - br_2D) - c(ar_2 + br_1) + cd \right) \\ &\quad \cdot \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right) \\ &= \nu\left(\psi_{\beta}\right)+ \frac{1}{2}\nu\left(\psi_{\alpha r + 1}\psi_{\alpha r - 1}\right) \\ &\quad - 2car_1 \widetilde{\lambda}(P) + 2dar_2 \widetilde{\lambda}(\omega P) + (dar_1 + acr_2) \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right) \\ &\quad - 2cbr_2D \widetilde{\lambda}(P) + 2dbr_1 \widetilde{\lambda}(\omega P) - (dbr_2D - bcr_1) \left( \widetilde{\lambda}([1 + \omega]P) - \widetilde{\lambda}(P) - \widetilde{\lambda}(\omega P) \right) \\ &= \nu\left(\psi_{\beta}\right) - a\left(\nu\left(\frac{\psi_{\beta}}{\psi_{r - \beta}}\right) + \frac{\mu(r)}{2}\right)- b\left(\nu\left(\frac{\psi_{\beta}}{\psi_{r \omega - \beta}}\right)+\frac{\mu_{\omega}(r)}{2}\right) \\ &\quad + \frac{a^{2} - ab}{2} \mu(r) + \frac{b^{2} - ab}{2} \mu_{\omega}(r) + \frac{ab}{2} \mu_{1+\omega}(r). \end{align*} $$