%! TEX root = EC.tex % vim: tw=50 % 05/02/2025 11AM % Examples Class % Friday 14th February % 3:30PM MR9 % 2, 9, 10 \newpage \section{Isogenies} Let $E_1, E_2$ be \glspl{ellc}. \begin{fcdefnstar}[Isogeny] \glsnoundefn{isog}{isogeny}{isogenies}% \glsnoundefn{isogous}{isogenous}{\glspl{ellc}}% An \emph{isogeny} $\phi : E_1 \to E_2$ is a nonconstant morphism with $\phi(0_{E_1}) = 0_{E_2}$ (by \cref{thm:2.8}, a morphism is nonconstant if and only if surjective on $\ol{K}$-points). We say $E_1$ and $E_2$ are isogenous in this case. \end{fcdefnstar} \begin{fcdefnstar}[] $\Hom(E_1, E_2) = \{\text{\glspl{isog} $E_1 \to E_2$}\} \cup \{0\}$. \end{fcdefnstar} This is an abelian group under \[ (\phi + \psi)(P) = \phi(P) + \psi(P) .\] If $E_1 \stackrel{\phi}{\to} E_2 \stackrel{\psi}{\to} E_3$ are \glspl{isog} then $\psi \circ \phi$ is an \gls{isog}. Tower Law implies $\deg(\psi \phi) = \deg(\phi)\deg(\psi)$. \begin{fcprop}[] \label{prop:5.1} Assuming: - $0 \neq n \in \Zbb$ Then: $[n] : E \to E$ is an \gls{isog}. \end{fcprop} \begin{proof} $[n]$ is a morphism by \cref{thm:4.4}. We must show $[n] \neq [0]$. Assume $\char K \neq 2$. Case $n = 2$: \cref{lemma:4.5} implies $E\tors[2] \neq E$ implies $[2] \neq [0]$. Case $n$ odd: \cref{lemma:4.5} implies $\exists 0 \neq T \in E\tors[2]$. Then $nT = T \neq 0$ which gives $[n] \neq [0]$. Now use $[mn] = [m] \circ [n]$. If $\char K = 2$, then could rpelace \cref{lemma:4.5} with an explicit lemma about $3$-torsion points. \end{proof} \begin{fccoro}[] \label{coro:5.2} \cloze{$\Hom(E_1, E_2)$} is a torsion free $\Zbb$-module. \end{fccoro} \noproof \begin{fcthm}[] \label{thm:5.3} Assuming: - $\phi : E_1 \to E_2$ an \gls{isog} Then: $\phi(P + Q) = \phi(P) + \phi(Q)$ for all $P, Q \in E_1$. \end{fcthm} \begin{proof}[Proof (sketch)] $\phi$ incudes \begin{align*} \phi_* : \Div^0(E_1) &\to \Div^0(E_2) \\ \sum_{P \in E_1} n_P P &\mapsto \sum_{P \in E_1} n_P \phi(P) \end{align*} Recall $\phi^* : K(E_2) \hookrightarrow K(E_1)$. \begin{picmath} \begin{tikzcd} K(E_1) \ar[d, no head] \ar[d, bend left=50, "\text{norm map}"] \\ K(E_2) \end{tikzcd} \end{picmath} \textbf{Fact:} If $f \in \ol{K}(E_1)^*$ then \[ \div(N_{K(E_1) / K(E_2)} f) = \phi_*(\div f) .\] So $\phi_*$ sends principal divisors to principal divisors. Since $\phi(0_{E_1}) = 0_{E_2}$, the following diagram commutes: \begin{picmath} \begin{tikzcd} P \ar[d, mapsto] & E_1 \ar[r, "\phi"] \ar[d, "\cong"] & E_2 \ar[d, "\cong"] & Q \ar[d, mapsto] \\ {[(P) - (0_{E_1})]} & \Pic^0(E_1) \ar[r, "\phi_*"] & \Pic^0(E_2) & {[(Q) - (0_{E_2})]} \end{tikzcd} \end{picmath} $\phi_*$ a group homomorphism implies $\phi$ is a group homomorphism. \end{proof} \begin{fclemma}[] \label{lemma:5.4} Assuming: - $\phi : E_1 \to E_2$ an \gls{isog} Then: there exists a morphism $\xi$ making the following diagram commute: \begin{picmath} \begin{tikzcd} E_1 \ar[r, "\phi"] \ar[d, "x_1"] & E_2 \ar[d, "x_2"] \\ \Pbb^1 \ar[r, "\xi"] & \Pbb^1 \end{tikzcd} \end{picmath} ($x_i = $ $x$-coordinate on a Weierstrass equation for $E_i$). Moreover if $\xi(t) = \frac{r(t)}{s(t)}$, $r, s \in K[t]$ coprime, then \[ \deg \phi = \deg \xi = \max(\deg(r), \deg(s)) .\] \end{fclemma} \begin{proof} For $i = 1, 2$. $K(E_i) / K(x_i)$ is a degree $2$ Galois extension, with Galois group generated by $[-1]^*$. \cref{thm:5.3} implies that $\phi \circ [-1] = [-1] \circ \phi$. So if $f \in K(x_2)$ then \[ [-1]^* \phi^* f = \phi^* [-1]^* f = \phi^* f .\] Therefore $\phi^* f \in K(x_1)$. \begin{picmath} \begin{tikzcd} & K(E_1) = K(x_1, y_1) \ar[ld, no head, "2"] \ar[dd, no head, "\deg \phi"] \\ K(x_1) \ar[dd, no head, "\deg \xi"] \\ & K(E_2) = K(x_2, y_2) \ar[ld, no head, "2"] \\ K(x_2) \end{tikzcd} \end{picmath} In particular, $\phi^* = \xi(x_1)$ for some rational function $\xi$. Tower Law implies $\deg \phi = \deg \xi$. Now $K(x_2) \hookrightarrow K(x_1)$, $x_2 \mapsto \xi(x_1) = \frac{r(x_1)}{s(x_1)}$, $r, s \in K[t]$ coprime. We claim the minimal polynomial of $x_1$ over $K(x_2)$ is \[ F(t) = r(t) - s(t) x_2 \in K(x_2) [t] .\] Check: \begin{itemize} \item $F(x_1) = 0$: true by the definition of our embedding. \item $F$ is irreducible in $K[x_2, t]$ (since $r, s$ coprime), hence irreducible in $K(x_2)[t]$ by Gauss's lemma. \end{itemize} Therefore \[ \deg \xi = [K(x_1) : K(x_2)] = \deg F = \max(\deg r, \deg s) . \qedhere \] \end{proof} \begin{fclemma}[] $\deg[2] = 4$. \end{fclemma} \begin{proof} Assume $\char K \neq 2, 3$ (the result is true even in the case of $\char K \in \{2, 3\}$, but we will only prove the simpler case). $E$: $y^2 = x^3 + ax + b = f(x)$. If $P = (x, y) \in E$, then \begin{align*} x(2P) &= \left( \frac{3x^2 + a}{2y} \right)^2 - 2x \\ &= \frac{(3x^2 + a)^2 - 8x f(x)}{4f(x)} \\ &= \frac{x^4 + \cdots}{f(x)} \end{align*} The numerator and denominator are coprime. Indeed, otherwise there exists $\theta \in \ol{K}$ with $f(\theta) = f'(\theta) = 0$ and hence $f$ has a multiple root (contradiction). Now \cref{lemma:5.4} implies $\deg[2] = \max(4, 3) = 4$. \end{proof} \begin{fcdefnstar}[Quadratic form in an abelian group] \glsnoundefn{qf}{quadratic form}{quadratic forms}% Let $A$ be an abelian group. We say $q : A \to \Zbb$ is a \emph{quadratic form} if \begin{enumerate}[(i)] \item $q(nx) = n^2 q(x)$ for all $n \in \Zbb$, $x \in A$. \item $(x, y) \mapsto q(x + y) - q(x) - q(y)$ is $\Zbb$-bilinear. \end{enumerate} \end{fcdefnstar} \begin{fclemma}[] \label{lemma:5.6} \begin{iffc} \lhs $q : A \to \Zbb$ is a \gls{qf} \rhs it satisfies the parallelogram law: \[ q(x + y) + q(x - y) = 2q(x) = 2q(y) \qquad \forall x, y \in A .\] \end{iffc} \end{fclemma} \begin{iffproof} \rightimpl Let $\langle x, y \rangle = q(x + y) - q(x) - q(y)$. Then \[ \langle x, x \rangle = q(2x) - 2q(x) = 2q(x) \] (the equality in the last step is using property (i) with $n = 2$). But by (ii), \[ \langle x + y, x + y \rangle + \langle x - y, x - y \rangle = 2\langle x, x \rangle + 2\langle y, y \rangle .\] Hence (upon dividing by $2$), \[ q(x + y) + q(x - y) = 2q(x) + 2q(y) .\] \leftimpl See \es{2}. \end{iffproof}