This exercise gives an elementary (highly computational) proof that the multiplication-by- $m$ map has degree $m^2$. We will assume $\operatorname{char}(K) \neq 2,3$, and take an elliptic curve
$$
E: y^2=x^3+A x+B .
$$
Define division polynomials $\psi_m \in \mathbb{Z}[A, B, x, y]$ inductively as follows:
$$
\begin{aligned}
\psi_1 & =1, \quad \psi_2=2 y \\
\psi_3 & =3 x^4+6 A x^2+12 B x-A^2 \\
\psi_4 & =4 y\left(x^6+5 A x^4+20 B x^3-5 A^2 x^2-4 A B x-8 B^2-A^3\right) \\
\psi_{2 m+1} & =\psi_{m+2} \psi_m^3-\psi_{m-1} \psi_{m+1}^3 \quad(m \geqslant 2) \\
2 y \psi_{2 m} & =\psi_m\left(\psi_{m+2} \psi_{m-1}^2-\psi_{m-2} \psi_{m+1}^2\right) \quad(m \geqslant 2) .
\end{aligned}
$$
(One easily checks that the $\psi_{2 m}$ 's are polynomials.) Further define polynomials $\phi_m$ and $\omega_m$ by
$$
\begin{aligned}
\phi_m & =x \psi_m^2-\psi_{m+1} \psi_{m-1} \\
4 y \omega_m & =\psi_{m+2} \psi_{m-1}^2-\psi_{m-2} \psi_{m+1}^2
\end{aligned}
$$
(a) Prove that $\psi_m, \phi_m, y^{-1} \omega_m$ (for $m$ odd) and $(2 y)^{-1} \psi_m, \phi_m, \omega_m$ (for $m$ even) are polynomials in $\mathbb{Z}\left[A, B, x, y^2\right]$. Hence replacing $y^2$ by $x^3+A x+B$, we will treat them as polynomials in $\mathbb{Z}[A, B, x]$.
(b) As polynomials in $x$, show that
$$
\begin{gathered}
\phi_m(x)=x^{m^2}+\text { lower order terms } \\
\psi_m(x)^2=m^2 x^{m^2-1}+\text { lower order terms }
\end{gathered}
$$
(c) If $\Delta=-16\left(4 A^3+27 B^2\right) \neq 0$, then $\phi_m(x)$ and $\psi_m(x)^2$ are relatively prime polynomials (in $K[x]$.)
(d) Again assume $\Delta \neq 0$, so $E$ is an elliptic curve. Let $P=\left(x_0, y_0\right) \in E$. Then
$$
[m] P=\left(\frac{\phi_m(P)}{\psi_m(P)^2}, \frac{\omega_m(P)}{\psi_m(P)^3}\right)
$$
(e) The map $[m]: E \rightarrow E$ has degree $m^2$.