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 $\sigma \in \mathrm{Gal}(L∕K)$ we have

$$n(\sigma Q-Q)=\sigma P-P=0.$$

So $\sigma Q-Q\in E[n]$.

Since $\mathrm{Gal}(L∕K)$ and $E[n]$ are finite, there are only finitely many possibilities for the map

$$\begin{array}{llll}\hfill \mathrm{Gal}(L∕K)& \to E[n]& \hfill & \\ \hfill \sigma & \mapsto \sigma Q-Q& \hfill & \end{array}$$

(even without requiring it to be a group homomorphism!).

So we have a map

$$\begin{array}{llllll}\hfill \ker \left(\frac{E(K)}{nE(K)}\to \frac{E(L)}{nE(L)}\right)& \stackrel{(\ast )}{\to }\mathrm{Maps}(\mathrm{Gal}(L∕L),E[n])& \hfill & & \hfill & \\ \hfill P+nE(K)& \mapsto (\sigma \mapsto \sigma Q-Q)& \hfill & \text{where }nQ=P& \hfill & & \hfill & \end{array}$$

It remains to show ($\ast$) is injective.

So suppose $P_1,P_2\in E(K)$, $P_i=n{Q}_i$ for $i=1,2$, and suppose $\sigma Q_1-Q_1=\sigma Q_2-Q_2$ for all $\sigma \in \mathrm{Gal}(L∕K)$. Then $\sigma (Q_1-Q_2)=Q_1-Q_2$ for all $\sigma \in \mathrm{Gal}(L∕K)$, hence $Q_1-Q_2\in E(K)$, so $P_1-P_2\in nE(K)$. Hence $P_1+nE(K)=P_2+nE(K)$ as desired. □

Proof. Theorem 12.1 tells us that we may replace $K$ by a finite Galois extension.

So without loss of generality ${\mu }_n\subset K$ and $E[n]\subset E(K)$. Let

$$S=\{𝔭|n\}\cup \{\text{primes of }\text{bad reduction}\text{ for }E∕K\}.$$

For each $P\in E(K)$, the extension $K({[n]}^{-1}P)∕K$ is unramified outisde $S$ by Theorem 9.8. Since $\mathrm{Gal}(\overline{K}∕K)$ acts on ${[n]}^{-1}P$, it follows that $\mathrm{Gal}(\overline{K}∕K({[n]}^{-1}P))$ is a normal subgroup of $\mathrm{Gal}(\overline{K}∕K)$ and hence $K({[n]}^{-1}P)∕K$ is a Galois extension.

Let $Q\in {[n]}^{-1}P$. Since $E[n]\subset E(K)$, we have $K(Q)=K({[n]}^{-1}P)$. Consider

$$\begin{array}{llll}\hfill \mathrm{Gal}(K(Q)∕K)& \to E[n]\cong {(\mathbb{Z}∕n\mathbb{Z})}^2& \hfill & \\ \hfill \sigma & \mapsto \sigma Q-Q& \hfill & \end{array}$$

Group homomorphism: $\sigma \tau Q-Q=\sigma (\tau Q-Q)+(\sigma Q-Q)$.

Injective: If $\sigma Q=Q$ then $\sigma$ fixes $K(Q)$ pointwise, i.e. $\sigma =1$.

Therefore $K(Q)∕K$ is an abelian extension of exponent $n$, unramified outside $S$.

Proposition 11.3 shows that as we vary $P\in E(K)$ there are only finitely many possibilities for $K(Q)$.

Let $L$ be the composite of all such extensions of $K$. Then $L∕K$ is finite and Galois, and $\frac{E(K)}{nE(K)}\to \frac{E(L)}{nE(L)}$ is the zero map. Theorem 12.1 gives $\left|\frac{E(K)}{nE(K)}\right|<\infty$. □

Proof. Fix an integer $n\ge 2$.

Weak Mordell-Weil Theorem implies that $\left|\frac{E(K)}{nE(K)}\right|\infty$. Pick coset representatives $P_1,P_2,\dots ,P_m$. Let

$$\mathrm{\Sigma }=\{P\in E(K)|\widehat{(P)}\le \underset{1\le i\le m}{\max }ĥ(P_i)\}.$$

Claim: $\mathrm{\Sigma }$ generates $E(K)$.

If not, then there exists $P\in E(K)\setminus (\text{subgroup generated by }\mathrm{\Sigma })$ of minimal height (exists by ($\ast$)). Then $P=P_i+nQ$ for some $i$ and $Q\in E(K)$. Note that $Q\in E(K)\setminus (\text{subgroup generated by }\mathrm{\Sigma })$.

Minimal choice of $P$ gives

$$\begin{array}{llllll}\hfill 4ĥ(P)& \le 4ĥ(Q)& \hfill & & \hfill & \\ \hfill & \le n^2ĥ(Q)& \hfill & & \hfill & \\ \hfill & =ĥ(nQ)& \hfill & & \hfill & \\ \hfill & =ĥ(P-P_i)& \hfill & & \hfill & \\ \hfill & \le ĥ(P-P_i)+ĥ(P+P_i)& \hfill & & \hfill & \\ \hfill & =2ĥ(P)+2ĥ(P_i)& \hfill & \text{parallelogram law}& \hfill & & \hfill & \end{array}$$

Therefore $ĥ(P)\le ĥ(P_i)$. Hence $P\in \mathrm{\Sigma }$ (by definition of $\mathrm{\Sigma }$), which contradicts the choice of $P$.

This proves the claim.

By ($\ast$), $\mathrm{\Sigma }$ is finite. □