%! TEX root = EC.tex % vim: tw=50 % 14/03/2025 11AM \textbf{Application:} Assume $\char K \nmid n$. There is a short exact sequence of $\Gal(\ol{K} / K)$-modules \begin{align*} 0 \to \mu_n \to \ol{K}^* &\to \ol{K}^* \to 0 \\ x &\mapsto x^n \end{align*} Long exact sequence: \begin{align*} K^* &\to K^* \to \Hone(\Gal(\ol{K} / K), \mu_n) \to \ob{\Hone(\Gal(\ol{K} / K), \ol{K}^*)}^{= 0 \text{ by \gls{hilb90}}} \\ x &\mapsto x^n \end{align*} Therefore $\Hone(\Gal(\ol{K} / K), \mu_n) \cong K^* / (K^*)^n$. If $\mu_n \subset K$ then \[ \Hom_{\text{cts}}(\Gal(\ol{K} / K), \mu_n) \cong K^* / (K^*)^n \] Finite subgroups of $\Hom_{\text{cts}}(\Gal(\ol{K} / K), \mu_n)$ are of the form $\Hom(\Gal(L / K), \mu_n)$ for $L / K$ a finite abelian extension of $K$ of exponent dividing $n$. This gives another proof of \cref{thm:11.2}. \begin{notation*} $\Hone(K, -)$ means $\Hone(\Gal(\ol{K} / K), -)$. \end{notation*} Let $\phi : E \to E'$ be an \gls{isog} of \glspl{ellc} over $K$. Short exact sequence of $\Gal(\ol{K} / K)$-modules \[ 0 \to E[\phi] \to E \stackrel{\phi}{\to} E' \to 0 \] has long exact seqeucne \[ E(K) \stackrel{\phi}{\to} E'(K) \stackrel{\delta}{\to} \Hone(K, E[\phi]) \to \Hone(K, E) \stackrel{\phi_*}{\to} \Hone(K, E') .\] We get a short exact sequence \begin{picmath} \begin{tikzcd} 0 \ar[r] & \frac{E'(K)}{\phi E(K)} \ar[r, "\delta"] \ar[d] & \Hone(K, E[\phi]) \ar[r] \ar[d, "\res_v"] \ar[rd] & \Hone(K, E)[\phi_*] \ar[r] \ar[d, "\res_v"] & 0 \\ 0 \ar[r] & \prod_v \frac{E'(K_v)}{\phi E(K_v)} \ar[r, "\delta_V"] & \prod_v \Hone(K_v, E[\phi]) \ar[r] & \prod_v \Hone(K_v, E)[\phi^*] \ar[r] & 0 \end{tikzcd} \end{picmath} Now take $K$ a number field. For each place $v$, fix an embedding $\ol{K} \subset \ol{K}_v$. Then $\Gal(\ol{K}_v / K_v) \subset \Gal(\ol{K} / K)$. \begin{fcdefnstar}[Selmer-group] \glssymboldefn{selmer}% We define the \emph{$\phi$-Selmer group} \begin{align*} S^{(\phi)}(E / K) &= \ker\left(\Hone(K, E[\phi]) \to \prod_v \Hone(K_v, E)\right) \\ &= \{\alpha \in \Hone(K, E[\phi]) \st \res_v(\alpha) \in \Im(\delta_v) ~\forall v\} \end{align*} (the map $\Hone(K, E[\phi]) \to \prod_v \Hone(K_v, E)$ is as in the commutative diagram above). \end{fcdefnstar} \begin{fcdefnstar}[Tate-Shafarevich group] \glssymboldefn{TS}% The \emph{Tate-Shafarevich group} is \[ \Sha(E / K) = \ker(\Hone(K, E) \to \prod_v \Hone(K_v, E)) .\] \end{fcdefnstar} We get a short exact sequence \[ 0 \to \frac{E'(K)}{\phi E(K)} \to \Selmer(E / K) \to \TS(E / K)[\phi_K] \to 0 .\] Taking $\phi = [n]$ gives \[ 0 \to \frac{E(K)}{n E(K)} \to \Selmer[n](E / K) \to \TS(E / K)\tors[n] \to 0 .\] Reorganising the proof of \gls{mwt} gives \begin{fcthm}[] \label{thm:15.3} $\Selmer[n](E / K)$ is \cloze{finite.} \end{fcthm} \begin{proof} For $L / K$ a finite Galois extension, \begin{picmath} \begin{tikzcd} 0 \ar[r] & \ob{\Hone(\Gal(L / K), E(L)\tors[n])}^{\text{finite}} \ar[r, "\inf"] & \Hone(K, E\tors[n]) \ar[r, "\res"] \ar[d, phantom, sloped, "\supset"] & \Hone(L, E\tors[n]) \ar[d, phantom, sloped, "\supset"] \\ & & \Selmer[n](E / K) \ar[r] & \Selmer[n](E / L) \end{tikzcd} \end{picmath} Therefore by extending our field, we may assume $E\tors[n] \subset E(K)$ and hence by the Weil pairing $\mu_n \subset K$. Therefore $E\tors[n] \cong \mu_n \times \mu_n$ as a $\Gal(\ol{K} / K)$-module. Then \begin{align*} \Hone(K, E\tors[n]) &\cong \Hone(K, \mu_n) \times \Hone(K, \mu_n) \\ &\cong K^* / (K^*)^n \times K^* / (K^*)^n \end{align*} Let \[ S = \{\text{primes of \gls{bred} for $E$}\} \cup \{v \mid n \infty\} \] (a finite set of places). Define the subgroup of $\Hone(K, A)$ unramified outside of $S$ as \[ \Hone(K, A; S) = \ker(\Hone(K, A) \to \prod_{v \notin S} \Hone(K_v^{\text{nr}}, A)) .\] There is a commutative diagram with exact rows: \begin{picmath} \begin{tikzcd} E(K_v) \ar[r, "\times n"] \ar[d] & E(K_v) \ar[r, "\delta_v"] \ar[d] & \Hone(K_v, E\tors[n]) \ar[d, "\res"] \\ E(K_v^\nr) \ar[r, "\times n"] & E(Kv^\nr) \ar[r, "0"] & \Hone(K_v^\nr, E\tors[n]) \end{tikzcd} \end{picmath} The bottom map $\times n$ is surjective $\forall v \notin S$ (see \cref{thm:9.8}). Therefore \begin{align*} \Selmer[n](E / K) &\subset \Hone(K, E\tors[n]; S) \\ &\cong \Hone(K, \mu_n; S) \times \Hone(K, \mu_n; S) \end{align*} But \begin{align*} \Hone(K, \mu_n; S) &= \ker \left( \frac{K^*}{(K^*)^n} \to \prod_{v \notin S} \frac{(K_v^\nr)^*}{(K_v^\nr)^{*n}} \right) \\ &\subset K(S, n) \end{align*} which is finite by \cref{lemma:11.4}. \end{proof} \begin{remark*} $\Selmer[n](E / K)$ is finite and effectively computable. It is conjectured that $|\TS(E / K)| < \infty$. This would imply that $\rank E(K)$ is effectively computable. \end{remark*} \newpage \section{Descent by Cyclic Isogeny} Let $E, E'$ be \glspl{ellc} over a number field $K$, and $\phi : E \to E'$ an \gls{isog} of degree $n$. Suppose $E'[\hat{\phi}] \cong \Zbb / n\Zbb$ generated by $T \in E'(K)$. Then $E[\phi] \cong \mu_n$ as a $\Gal(\ol{K} / K)$-module \[ S \mapsto \ephi(S, T) .\] Have a short exact sequence of $\Gal(\ol{K} / K)$-modules \[ 0 \to \mu_n \to E \stackrel{\phi}{\to} E' \to 0 .\] Get long exact sequence \begin{picmath} \begin{tikzcd} E(K) \ar[r, "\phi"] & E'(K) \ar[r, "\delta"] \ar[rd, "\alpha"] & \Hone(K, \mu_n) \ar[r] \ar[d, phantom, sloped, "\cong"] & \Hone(K, E) \\ & & K^* / (K^*)^n \end{tikzcd} \end{picmath} (the $\cong$ is by \gls{hilb90}). \begin{fcthm}[] \label{thm:16.1} Assuming: - $f \in K(E')$ and $g \in K(E)$ - $\div(f) = n(T) - n(0)$ - $\phi^* f = g^n$ Then: $\alpha(P) = f(P) \pmod{(K^*)^n}$ for all $P \in E;(K) \setminus \{0, T\}$. \end{fcthm}