%! TEX root = EC.tex % vim: tw=50 % 28/02/2025 10AM \begin{example*} $E_D$: $y^2 = x^3 - D^2 x = f(x)$. $D \in \Zbb$ square-free, $\Delta = 2^6 D^6$. If $p \nmid 2D$, then \[ \#\reduction{E}_D(\Fbb_p) = 1 + \sum_{x \in \Fbb_p} \left( \left( \frac{f(x)}{p} \right) + 1 \right) .\] If $p \equiv 3 \pmod 4$ then since $f$ is odd, \[ \left( \frac{f(-x)}{p} \right) = \left( \frac{-f(x)}{p} \right) = \left( \frac{-1}{p} \right) \left( \frac{f(x)}{p} \right) = - \left( \frac{f(x)}{p} \right) .\] Hence \[ \#\reduction{E}_D(\Fbb_p) = p + 1 .\] \[ E_D(\Qbb)_{\text{tors}} \supset \{0, (0, 0), (\pm D, 0)\} \cong (\Zbb / 2\Zbb)^2 .\] Let $m = \#E_D(\Qbb)_{\text{tors}}$. We have $4 \mid m \mid p + 1$ forr all sufficiently large ($p \nmid 2mD$) primes $p$ with $p \equiv 3 \pmod 4$. Hence $m = 4$ (otherwise get contradiction to Dirichlet's theorem on primes on arithmetic progressions). So \begin{align*} \rank E_D(\Qbb) \ge 1 &\iff \exists x, y \in \Qbb, y \neq 0 \wedge y^2 = x^3 - D^2 x \\ &\stackrel{\text{Lecture 1}}{\iff} \text{$D$ is a congruent number} \end{align*} \end{example*} \begin{fclemma}[] \label{lemma:10.4} Assuming: - $E / \Qbb$ is given by a \gls{weq} with $a_1, \ldots, a_6 \in \Zbb$ - $0 \neq T = (x, y) \in E(\Qbb)_{\text{tors}}$ Thens:[(i)] - $4x, 8y \in \Zbb$ - If $2 \mid a_1$ or $2T \neq 0$ then $x, y \in \Zbb$ \end{fclemma} \begin{proof} \phantom{} \begin{enumerate}[(i)] \item The \gls{weq} defines a formal group $\hat{E}$ over $\Zbb$. For $r \ge 1$, \[ \hat{E}(p^r \Zbb_p) = \{(x, y) \in E(\Qbb_p) \st v_p(x) \le -2r, v_p(y) \le -3r\} \cup \{0\} .\] Then \cref{thm:9.2} gives $\hat{E}(p^r \Zbb_p) \cong (\Zbb_p, +)$ if $r > \frac{1}{p - 1}$. Hence $\hat{E}(4 \Zbb_2)$ and $\hat{E}(p\Zbb_p)$ for $p$ are odd are torsion free. So if $0 \neq T = (x, y) \in E(\Qbb)_{\text{tors}}$ then \begin{align*} v_2(x) &\ge -2 \\ v_2(y) &\ge -3 \\ v_p(x) &\ge 0 \\ v_p(y) &\ge 0 \end{align*} for all odd primes $p$. \item Suppose $T \in \hat{E}(2\Zbb_2)$, i.e. $v_2(x) = -2$, $v_2(y) = -3$. Since \[ \frac{\hat{E}(2\Zbb_2)}{\hat{E}(4\Zbb_2)} \cong (\Fbb_2, +) \] and $\hat{E}(4\Zbb_2)$ is torsion free, we get $2T = 0$. Also $(x, y) = T = -T = (x, -y - a_1 x - a_3)$. So \begin{align*} 2y + a_1 x + a_3 &= 0 \\ \ub{8y}_{\text{odd}} + a_1 \ub{(4x)}_{\text{odd}} + \ub{4a_3}_{\text{even}} &= 0 \\ \end{align*} Hence $a_1$ is odd. So if $2T \neq 0$ or $a_1$ is even then $T \notin \hat{E}(2\Zbb_2)$, so $x, y \in \Zbb$. \qedhere \end{enumerate} \end{proof} \begin{example*} $y^2 + xy = x^3 + 4x + 1$, $\left( -\frac{1}{4}, \frac{1}{8} \right) \in E(\Qbb)\tors[2]$. \end{example*} \begin{fcthm}[Lutz Nagell] \label{thm:10.5} Assuming: - $E / \Qbb$: $y^2 = x^3 + ax + b$, $a, b \in \Zbb$ - $0 \neq T = (x, y) \in E(\Qbb)_{\text{tors}}$ Then: $x, y \in \Zbb$ and either $y = 0$ or $y^2 \mid (4a^3 + 27b^2)$. \end{fcthm} \begin{proof} \cref{lemma:10.4} gives that $x, y \in \Zbb$. If $2T = 0$ then $y = 0$. Otherwise $0 \neq 2T = (x_2, y_2) \in E(\Qbb)_{\text{tors}}$. \cref{lemma:10.4} gives $x_2, y_2 \in \Zbb$. But \[ x_2 = \left( \frac{f'(x)}{2y} \right)^2 - 2x \] hence $y \mid f'(x)$. $E$ non-singular gives that $f(X)$ adn $f'(X)$ are coprime, so $f(X)$ and $f'(X)^2$ are coprime. Therefore there exists $g, h \in \Qbb[X]$ satisfying \[ g(X) f(X) + h(X) f'(X)^2 = 1 .\] Doing this and clearing denominators gives \[ (3X^2 + 4a) f'(X)^2 - 27(X^3 + aX - b) f(X) = 4a^3 + 27 b^2 .\] Since $y \mid f'(x)$ and $y^2 = f(x)$, we get $y^2 \mid (4a^3 + 27b^2)$. \end{proof} \begin{remark*} Mazur showed that if $E / \Qbb$ is an \gls{ellc} then \[ E(\Qbb)_{\text{tors}} \cong \begin{cases} \Zbb / n \Zbb & 1 \le n \le 12, n \neq 11 \\ \Zbb / 2\Zbb \times \Zbb / 2n \Zbb & 1 \le n \le 4 \end{cases} \] Moreover, all 15 possibilities occur. \end{remark*} \newpage \section{Kummer Theory} Let $K$ be a field and $\char K \nmid n$. Assume $\mu_n \subset K$. \begin{fclemma}[] \label{lemma:11.1} Assuming: - $\Delta \subset K^* / (K^*)^n$ a finite subgroup - $L = K(\sqrt[n]{\Delta})$ Then: $L / K$ is Galois and $\Gal(L / K) \cong \Hom(\Delta, \mu_n)$. \end{fclemma} \begin{proof} $\mu_n \subset K$ gives $L / K$ normal, and $\char K \nmid n$ gives that $L / K$ is separable. So $L / K$ is Galois. Define the Kummer pairing \begin{align*} \langle , \rangle : \Gal(L / K) \times \Delta &\to \mu_n \\ (\sigma, x) &\mapsto \frac{\sigma(\sqrt[n]{x})}{\sqrt[n]{x}} \end{align*} \textbf{Well-defined:} Suppose $\alpha, \beta \in L$ with $\alpha^n = \beta^n = x$. Then $\left( \frac{\alpha}{\beta} \right)^n = 1$, so $\frac{\alpha}{\beta} \in \mu_n \subset K$. Then $\sigma \left( \frac{\alpha}{\beta} \right) = \frac{\alpha}{\beta}$ for all $\sigma \in \Gal(L / K)$, hence $\frac{\sigma(\alpha)}{\alpha} = \frac{\sigma(\beta)}{\beta}$ for all $\sigma \in \Gal(L / K)$. \textbf{Bilinear:} \begin{align*} \langle \sigma\tau, x \rangle &= \frac{\sigma(\tau \sqrt[n]{x})}{\tau \sqrt[n]{x}} \frac{\tau \sqrt[n]{x}}{\sqrt[n]{x}} &= \langle \sigma, x \rangle \langle \tau, x \rangle \\ \langle \sigma, xy \rangle &= \frac{\sigma \sqrt[n]{xy}}{\sqrt[n]{xy}} \\ &= \frac{\sigma \sqrt[n]{x}}{\sqrt[n]{x}} \frac{\sigma \sqrt[n]{y}}{\sqrt[n]{y}} \\ &= \langle \sigma, x \rangle \langle \sigma, y \rangle \end{align*} \textbf{Non-degenerate:} Let $\sigma \in \Gal(L / K)$. If $\langle \sigma, x \rangle = 1$ for all $x \in \Delta$, then \[ \sigma \sqrt[n]{x} = \sqrt[n]{x} \qquad \forall x \in \Delta \] adn hence $\sigma$ fixes $L$ pointwise, i.e. $\sigma = 1$. Let $x (K^*)^n \in \Delta$. If $\langle \sigma, x \rangle = 1$ for all $\sigma \in \Gal(L / K)$, then \[ \sigma \sqrt[n]{x} = \sqrt[n]{x} \qquad \forall \sigma \in \Gal(L / K) .\] So $\sqrt[n]{x} \in K$, so $x \in (K^*)^n$, i.e. $x(K^*)^n \in \Delta$ is the identity element. We get injective group homomorphisms \begin{enumerate}[(i)] \item $\Gal(L / K) \injto \Hom(\Delta, \mu_n)$. \item $\Delta \injto \Hom(\Gal(L / K), \mu_n)$. \end{enumerate}