Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 125 / 4

Let $G=\mathrm{Gal}(L/K)$. If $a\in K^{\times }$ becomes an $n$th power in $L$, choose $\alpha \in L$ with ${\alpha }^n=a$. Thenis a cocycle with values in the finite group ${\mu }_n(L)$. Changing $\alpha$ changes it by a coboundary, and its class is trivial exactly when $a$ was already an $n$th power in $K$. Thus the kernel injects into the finite group $H^1(G,{\mu }_n(L))$ and is finite. This is the multiplicative case of the finite-extension kernel of a Kummer map.

Assume ${\mu }_n\subset K$ and put $\alpha =\sqrt[n]{a}$. All roots of $X^n-a$ are $\zeta \alpha$ with $\zeta \in {\mu }_n\subset K$, so $K(\alpha )$ is its splitting field and is a Finite Galois extension. The mapis an injective group homomorphism.

For a finite Galois extension $L/K$, the kernel ofis finite. Indeed, if $P=nQ$ for $Q\in E(L)$, then $\sigma \mapsto \sigma Q-Q$ is a cocycle in the finite $G$-module $E[n](L)$, and the resulting map from the kernel to $H^1(G,E[n](L))$ is injective.

The analogue of part b assumes $E[n]\subset E(K)$. Given $P\in E(K)$ and $Q\in E(\overline{K})$ with $nQ=P$, every conjugate of $Q$ is $Q+T$ for some $T\in E[n]$. Hence $K(Q)/K$ is Galois andis an injective homomorphism. These are the two elliptic forms of the Kummer pairing.

Pass to the finite extension $L=K(E[n])$. Part c shows that the kernel of $E(K)/nE(K)\to E(L)/nE(L)$ is finite, so it is enough to control the image after $n$-torsion becomes rational. The Kummer map of an elliptic curveassociates to $P$ the finite extension generated by one $n$-division point of $P$.

Let $S$ contain the places over $n$, all archimedean places, and all places of bad reduction. The local theory of elliptic curves shows that these Kummer classes are unramified outside $S$. Because $E[n]$ is finite and constant over $L$, such classes are controlled by finitely many $S$-unramified power classes. Finiteness of the class group and finite generation of the unit group make that power-class group finite. The Kummer image is therefore finite, proving the Weak Mordell-Weil theorem that $E(K)/nE(K)$ is finite; this is the Kummer-theoretic proof of the weak Mordell-Weil theorem.