%! TEX root = EC.tex % vim: tw=50 % 03/03/2025 11AM % (i) implies $\Gal(L / K)$ is abelian and of exponent dividing $n$. \textbf{Fact:} If $G$ is a finite abelian group of exponent dividing $n$ then $\Hom(G, \mu_n) \cong G$ (non canonically). So \[ |\Gal(L / K)| \stackrel{\text{(i)}}{\le} |\Delta| \stackrel{\text{(ii)}}{\le} |\Gal(L / K)| .\] Therefore (i) and (ii) are isomorphisms. \end{proof} \begin{example*} $\Gal(\Qbb(\sqrt{2}, \sqrt{3}, \sqrt{5}) / \Qbb) \cong (\Zbb / 2\Zbb)^3$. \end{example*} \begin{fcdefnstar}[Abelian extension] We say $L / K$ is \emph{abelian} if it is Galois, and has abelian Galois group. Similarly for other group terminology (e.g. we can say that $L / K$ has exponent dividing $n$ to mean that it is Galois, with Galois group having exponent dividing $n$). \end{fcdefnstar} \begin{fcthm}[] \label{thm:11.2} There is a bijection \begin{align*} \mathcloze[1]{ \left\{\substack{ \text{finite subgroups} \\ \Delta \subset K^* / (K^*)^n }\right\} } &\leftrightarrow \mathcloze[2]{ \left\{\substack{ \text{finite abelian extension $L / K$} \\ \text{of exponent dividing $n$} }\right\} } \\ \mathcloze[1]{\Delta} &\mapsto \mathcloze[2]{K(\sqrt[n]{\Delta})} \\ \mathcloze[1]{\frac{(L^*)^n \cap K^*}{(K^*)^n}} &\mapsfrom \mathcloze[2]{L} \end{align*} \end{fcthm} \begin{proof} \phantom{} \begin{enumerate}[(i)] \item Let $\Delta \subset K^* / (K^*)^n$ be a finite subgroup. Let $L = K(\sqrt[n]{\Delta})$ and $\Delta' = \frac{(L^*)^n \cap K^*}{(K^*)^n}$. We must show $\Delta = \Delta'$. Clearly $\Delta \subset \Delta'$. So \[ L = K(\sqrt[n]{\Delta}) \subset K(\sqrt[n]{\Delta'}) \subset L .\] So $K(\sqrt[n]{\Delta}) = K(\sqrt[n]{\Delta'})$. So \cref{lemma:11.1} gives $|\Delta| = |\Delta'|$. Since $\Delta \subset \Delta'$, it follows that $\Delta = \Delta'$. \item Let $L / K$ be a finite abelian extension of exponent dividing $n$. Let $\Delta = \frac{(L^*)^n \cap K^*}{(K^*)^n}$. Then $K(\sqrt[n]{\Delta}) \subset L$ and we aim to prove this inclusion is an equality. Let $G = \Gal(L / K)$. Thu Kummer pairing gives an injection $\Delta \injto \Hom(G, \mu_n)$. \textbf{Claim:} This map is surjective. Granted the claim, \begin{align*} [K(\sqrt[n]{\Delta}) : K] &\stackrel{\text{\cref{lemma:11.1}}}{=} |\Delta| \\ &\stackrel{\text{by claim}}{=} |G| \\ &= [L : K] \end{align*} Since $K(\sqrt[n]{\Delta}) \subset L$ it follows that $L = K(\sqrt[n]{\Delta})$. \textbf{Proof of claim:} Let $\chi : G \to \mu_n$ be a group homomorphism. Distinct automorphisms are linearly independent. So there exists $a \in L$ such that \[ \ub{\sum_{\tau \in G} \chi(\tau)^{-1} \tau(a)}_{= y} \neq 0 .\] Let $\sigma \in G$. Then \begin{align*} \sigma(y) &= \sum_{\tau \in G} \chi(\tau)^{-1} \sigma \tau(a) \\ &= \sum_{\tau \in G} \chi(\sigma^{-1} \tau)^{-1} \tau(a) \\ &= \chi(\sigma) \sum_{\tau \in G} \chi(\tau)^{-1} \tau(a) \\ &= \chi(\sigma) \cdot y \tag{$*$} \label{lec17eq1} \end{align*} Therefore $\sigma(y^n) = y^n$ for all $\sigma \in G$. Hence $y^n \in K$. Let $x = y^n$. Then $x \in K^* \cap (L^*)^n$. Then $x(K^*)^n \in \Delta$. Also, by \eqref{lec17eq1}, $\chi : \sigma \mapsto \frac{\sigma(y)}{y} = \frac{\sigma \sqrt[n]{x}}{\sqrt[n]{x}}$. So the map $\Delta \injto \Hom(G, \mu_n)$ sends $x \mapsto \chi$. This proves the claim. \qedhere \end{enumerate} \end{proof} \begin{fcprop}[] \label{prop:11.3} Assuming: - $K$ a number field - $\mu_n \subset K$ - $S$ a finite set of promes of $K$ Then there are only finitely many extensions $L / K$ such that \begin{enumerate}[(i)] \item $L / K$ is a finite abelian extension of exponent dividing $n$ \item $L / K$ is unramified at all $\mathfrak{p} \notin S$ \end{enumerate} \end{fcprop} \begin{proof} \cref{thm:11.2} gives $L = K(\sqrt[n]{\Delta})$ for some $\Delta \subset K^* / (K^*)^n$ a finite subgroup. Let $\mathfrak{p}$ be a prime of $K$. \[ \mathfrak{p} \mathcal{O}_L = \mathcal{P}_1^{e_1} \cdots \mathcal{P}_r^{e_r} \] for $\mathcal{P}_1, \ldots, \mathcal{P}_r$ some distinct primes in $\mathcal{O}_L$. If $x \in K^*$ represents an element of $\Delta$ then \[ nv_{\mathcal{P}_i}(\sqrt[n]{x}) = v_{\mathcal{P}_i}(x) = e_i v_{\mathfrak{p}}(x) .\] If $\mathfrak{p} \notin S$ then all $e_i = 1$, so $v_{\mathfrak{p}}(x) \equiv 0 \pmod n$. Therefore $\Delta \subset K(S, n)$ where \[ K(S, n) = \{x \in K^* / (K^*)^n \st v_{\mathfrak{p}}(x) \equiv 0 \pmod n ~\forall \mathfrak{p} \notin S\} .\] The proof is completed by the next lemma. \end{proof} \begin{fclemma}[] \label{lemma:11.4} $K(S, n)$ is \cloze{finite.} \end{fclemma} \begin{proof} The map \begin{align*} K(S, n) &\to (\Zbb / n\Zbb)^{|S|} \\ x &\mapsto (v_{\mathfrak{p}}(x) \pmod n)_{\mathfrak{p} \in S} \end{align*} is a group homomorphism with kernel $K(\emptyset, n)$. Since $|S| < \infty$, it suffices to prove the lemma with $S = \emptyset$. If $x \in K^*$ represents an element of $K(\emptyset, n)$ then $(x) = \mathfrak{a}^n$ for some fractional ideal $\mathfrak{a}$. There is a short exact sequence \begin{align*} 0 \to \mathcal{O}_K^* / (\mathcal{O}_K^*)^n \to K(\emptyset, n) &\to \Cl_K \to 0 \\ x(K^*)^n &\mapsto [\mathfrak{a}] \end{align*} $|\Cl_K| < \infty$ and $\mathcal{O}_K^*$ being a finitely generated abelian group (Dirichlet's unit theorem) gives us that $K(\phi, n)$ is finite. \end{proof} \newpage \section{Elliptic Curves over Number Fields: The weak Mordell-Weil Theorem} \begin{fcthm}[] \label{lemma:12.1} Assuming: - $E / K$ an \gls{ellc} - $L / K$ a finite Galois extension Then: the natural map $\frac{E(K)}{n E(K)} \to \frac{E(L)}{n E(L)}$ has finite kernel. \end{fcthm} \begin{proof} For each element in the kernel we pick a coset representative $P \in E(K)$ and then $Q \in E(L)$ such that $nQ = P$. For any $\sigma \in \Gal(L / K)$ we have \[ n(\sigma Q - Q) = \sigma P - P = 0 .\] So $\sigma Q - Q \in E\tors[n]$. Since $\Gal(L / K)$ and $E\tors[n]$ are finite, there are only finitely many possibilities for the map \begin{align*} \Gal(L / K) &\to E\tors[n] \\ \sigma &\mapsto \sigma Q - Q \end{align*} (even without requiring it to be a group homomorphism!).