%! TEX root = EC.tex % vim: tw=50 % 07/03/2025 11AM \[ |\kappa a_i^{2d - 1}| \le \max_{j = 1, 2} \ub{|f_j(a_1, a_2)|}_{\le \kappa \height(F(P))} \ub{\sum_{j = 1}^2 |g_{ij}(a_1, a_2)|}_{\le \gamma_i \height(D)^{d - 1}} \] where \[ \gamma_i = \sum_{j = 1}^2 \text{(sum of absolute values of coefficients of $g_{ij}$)} .\] Therefore \[ \kappa|a_i|^{2d - 1} \le \kappa \height(F(P)) \gamma_i \height(P)^{d - 1} \] so \[ \height(P)^{2d - 1} \le \max(\gamma_1, \gamma_2) \height(F(P)) \height(P)^{d - 1} \] so \[ \ub{\frac{1}{\max(\gamma_1, \gamma_2)}}_{= c_1} \height(P)^d \le \height(F(P)) . \qedhere \] \end{proof} \begin{notation*} For $x \in \Qbb$, $\height(X) = \height((x : 1)) = \max(|u|, |v|)$, where $x = \frac{u}{v}$, $u, v \in \Zbb$ coprime. \end{notation*} \begin{fcdefnstar}[Height of a point] \glssymboldefn{ellh}% Let $E / \Qbb$ be an \gls{ellc}, $y^2 = x^3 + ax + b$. Define the \emph{height} \begin{align*} H : E(\Qbb) &\to \Rbb_{\ge 1} \\ P &\mapsto \begin{cases} H(x) & \text{if $P = (x, y)$} \\ 1 & \text{if $P = 0_E$} \end{cases} \end{align*} Alsdefine \emph{logarithmic height} \begin{align*} h : E(\Qbb) &\to \Rbb_{\ge 0} \\ P &\mapsto \log H(P) \end{align*} \end{fcdefnstar} \begin{fclemma} \label{lemma:13.2} Assuming: - $E$, $E'$ are \glspl{ellc} over $\Qbb$ - $\phi : E \to E'$ an \gls{isog} defined over $\Qbb$ Then: there exists $c > 0$ such that \[ |\ellh(\phi(P)) - (\deg\phi) \ellh(P)| \le c \qquad \forall P \in E(\Qbb) .\] \end{fclemma} \begin{note*} $c$ depends on $E$ and $E'$, but \emph{not} $P$. \end{note*} \begin{proof} Recall (\cref{lemma:5.4}) \begin{picmath} \begin{tikzcd} E \ar[r, "\phi"] \ar[d, "x"] & E' \ar[d, "x"] \\ \Pbb^1 \ar[r, "\xi"] & \Pbb^1 \end{tikzcd} \end{picmath} $\deg \phi = \deg \xi$ ($= d$ say). \cref{lemma:13.1} tells us that there exists $c_1, c_2 > 0$ such that \[ c_1 \ellH(P)^d \le \ellH(\phi(P)) \le c_2 \ellH(P)^d \qquad \forall P \in E(\Qbb) .\] Taking logs gives \[ |\ellh(\phi(P)) - d\ellh(P)| \le \ub{\max(\log c_2 - \log c_1)}_{=c} \qedhere \] \end{proof} \begin{example*} $\phi = [2] : E \to E$. Then there exists $c > 0$ such that \[ |\ellh(2P) - 4\ellh(P)| \le c \qquad \forall P \in E(\Qbb) .\] \end{example*} \begin{fcdefnstar}[Canonical height of a point] \glssymboldefn{canh}% The \emph{canonical height} is \[ \hat{h}(P) = \lim_{n \to \infty} \frac{1}{4^n} \ellh(2^n P) .\] \end{fcdefnstar} We check convergence: Let $m \ge n$. Then \begin{align*} \left| \frac{1}{4^m} \ellh(2^m P) - \frac{1}{4^n} \ellh(2^n P) \right| &\le \sum_{r = n}^{m - 1} \left| \frac{1}{4^{r + 1}} \ellh(2^{r + 1} P) - \frac{1}{4^r} \ellh(2^r P) \right| \\ &= \sum_{r = n}^{m - 1} \frac{1}{4^{r + 1}} |\ellh(2(2^r P)) - 4\ellh(2^r P)| \\ &< c \sum_{r = n}^\infty \frac{1}{4^{r + 1}} \\ &= \frac{c}{4^{n + 1}} \frac{1}{1 - \quarter} \\ &= \frac{c}{3 \times 4^n} \\ &\to 0 \end{align*} as $n \to \infty$. So the sequence is Cauchy, $\canh(P)$ exists. \begin{fclemma}[] \label{lemma:13.3} $|\ellh(P) - \canh(P)|$ is \cloze{bounded for $P \in E(\Qbb)$.} \end{fclemma} \begin{proof} Put $n = 0$ in above calculation to get \[ \left| \frac{1}{4^m} \ellh(2^m P) - \ellh(P) \right| < \frac{c}{3} .\] Take limit $m \to \infty$. \end{proof} \begin{fclemma}[] \label{lemma:13.4} Assuming: - $B > 0$ Then: \[ \#\{P \in E(\Qbb) \st \canh(P) \le B\} < \infty .\] \end{fclemma} \begin{proof} $\canh(P)$ bounded means we have a bound on $\ellh(P)$ (by \cref{lemma:13.3}). So only finitely many possibilities for $x$. Each $x$ gives $\le 2$ choices for $y$. \end{proof} \begin{fclemma}[] \label{lemma:13.5} Assuming: - $\phi : E \to E'$ an \gls{isog} defined over $\Qbb$ Then: \[ \canh(\phi P) = (\deg \phi) \canh(P) \qquad \forall P \in E(\Qbb) .\] \end{fclemma} \begin{proof} By \cref{lemma:13.2}, there exists $c > 0$ such that \[ |\ellh(\phi P) - (\deg \phi) \ellh(P)| < c \qquad \forall P \in E(\Qbb) .\] Replace $P$ by $2^n P$, divide by $4^n$ and take limit $n \to \infty$. \end{proof} \begin{remark*} \phantom{} \begin{enumerate}[(i)] \item Case $\deg \phi = 1$ shows that $\canh$ (unlike $\ellh$) is independent of the choice of \gls{weq}. \item Taking $\phi = [n] : E \to E$ shows \[ \canh(nP) = n^2 \canh(P) \qquad \forall n \in \Zbb, P \in E(\Qbb) .\] \end{enumerate} \end{remark*} \begin{fclemma}[] \label{lemma:13.6} Assuming: - $E / \Qbb$ an \gls{ellc} Then: there exists $c > 0$ such that for all $P, Q \in E(\Qbb)$ with $P, Q, P + Q, P - Q \neq 0$, we have \[ \ellH(P + Q) \ellH(P - Q) \le c \ellH(P)^2 \ellH(Q)^2 .\] \end{fclemma} \begin{proof} Let $E$ have \gls{weq} $y^2 = x^3 + ax + b$, $a, b \in \Zbb$. Let $P, Q, P + Q, P - Q$ have $x$ coordinates $x_1, \ldots, x_4$. By \cref{lemma:5.8}, there exists $W_0, W_1, W_2 \in \Zbb[x_1, x_2]$ of degree $\le 2$ in $x_2$ such that \[ (1 : x_3 + x_4 : x_3 x_4) = (\ub{W_0}_{= (x_1 - x_2)^2} : W_1 : W_2) .\] Write $x_i = \frac{r_i}{s_i}$ with $r_i, s_i \in \Zbb$ coprime. \begin{align*} \ub{(s_3 s_4 : r_3 s_4 + r_4 s_3 : r_3 r_4)}_{\gcd = 1} &= ((r_1 s_2 - r_2 s_1)^2 : \cdots) \\ \ellH(P + Q) \ellH(P - Q) &= \max(|r_3|, |s_3|) \max(|r_4|, |s_4|) \\ &\le 2\max(|s_3 s_4|, |r_3 s_4 + r_4 s_3|, |r_3 r_4|) \\ &\le 2 \max(|r_1 s_2 - r_2 s_1|^2, \ldots) \\ &\le c \max(|r_1|^2, |s_1|)^2 \max(|r_2|, |s_2|)^2 \\ &= c\ellH(P)^2 \ellH(Q)^2 \end{align*} where $c$ depends on $E$, but not on $P$ and $Q$. \end{proof} \begin{fcthm}[] \label{thm:13.7} $\canh : E(\Qbb) \to \Rbb_{\ge 0}$ is a \gls{qf}. \end{fcthm} \begin{proof} \cref{lemma:13.6} and $|\ellh(2P) - 4\ellh(P)|$ bounded gives that there exists $c \in \Rbb$ such that \[ \ellh(P + Q) + \ellh(P - Q) \le 2\ellh(P) + 2\ellh(Q) + c \qquad \forall P, Q \in E(\Qbb) .\] Replacing $P$, $Q$ by $2^n P$, $2^n Q$, dividing by $4^n$ and taking the limit $n \to \infty$ gives \[ \canh(P + Q) + \canh(P - Q) \le 2 \canh(P) + 2\canh(Q) .\] Replacing $P, Q$ by $P + Q$, $P - Q$ and $\canh(2P) = 4\canh(P)$ gives the reverse inequality. Therefore $\canh$ satisfies the parallelogram law, and hence $\canh$ is a \gls{qf}. \end{proof}