Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 28 / 4 / b / Solution

Let $K$ be a field, $K_s$ a separable closure and $G_K$ its absolute Galois group. A discrete Galois module $A$ is an abelian group with a continuous $G_K$-action, so each element has an open stabilizer. Galois cohomology uses continuous cochains. Its zeroth group is $H^0(K,A)=A^{G_K}$, and its first group iswhereThus cocycles measure a failure of descent, and coboundaries account for changing a chosen lift. A short exact sequence of Galois modules gives a long exact sequence in group cohomology. Its connecting map sends an invariant element of a quotient to the cocycle $\sigma b-b$ of a chosen lift $b$; changing the lift changes this by a coboundary. These constructions provide the bridge between arithmetic divisibility and cohomology.

For Kummer theory, assume the characteristic does not divide $n$. The multiplicative exact sequence isThe final arrow is surjective because a root of $T^n-a$ is separable for $a\ne 0$. Hilbert's theorem 90 says $H^1(K,K_s^{\times })=0$. Here is its usual finite-extension proof. For a multiplicative cocycle $c_{\sigma }$ on a finite Galois extension $L/K$, choose $a\in L$ so thatSuch an $a$ exists by the Artin independence theorem. The cocycle identity gives $\tau A=c_{\tau }^{-1}A$. With $b=A^{-1}$ this becomes $c_{\tau }=\tau b/b$, a coboundary. Every continuous cocycle with values in $K_s^{\times }$ is defined over a sufficiently large finite Galois extension, so the same result holds for the absolute group.

The long exact sequence in group cohomology now givesThe class of $a$ maps to $c_{\sigma }=\sigma (\alpha )/\alpha$, where ${\alpha }^n=a$. This cohomological statement does not require ${\mu }_n\subset K$; it retains the actual Galois action on ${\mu }_n$. If all $n$th roots of unity do lie in $K$, the action is trivial, and after choosing a primitive root of unity the cocycles are characters into $\mathbb{Z}/n\mathbb{Z}$. Thus adjoining an $n$th root of $a$ describes a cyclic extension of degree dividing $n$, and Kummer classes classify the corresponding characters. For $n=2$, this gives the familiar quadratic extensions and the square-class group of a field.

Now let $K$ be a number field and $E/K$ an elliptic curve. The multiplication-by-n morphism is a surjective isogeny on $E(\overline{K})$, with kernel $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$. The Kummer exact sequence of an elliptic curveyieldsExplicitly, choose $Q$ with $nQ=P$ and set $\delta (P{)}_{\sigma }=\sigma Q-Q$. This lies in $E[n]$, satisfies the cocycle identity and is unchanged as a class if $Q$ is replaced by another division point. If it is a coboundary $\sigma T-T$ for $T\in E[n]$, then $Q-T$ is rational over $K$ and $P=n(Q-T)$. Conversely such a rational division point makes the cocycle zero. This proves the injectivity of the Kummer map of an elliptic curve on $E(K)/nE(K)$.

To prove the Weak Mordell-Weil theorem, it remains to prove that this particular cohomological image is finite. The full group $H^1(K,E[n])$ need not be finite, so the injection alone is not enough. Choose a finite set $S$ of finite places containing the bad-reduction places and those dividing $n$. At any other place, the construction in Question 3 works over the local field: reduce $P$, find an $n$-division point over a finite residue extension, lift it by the Hensel lemma, and correct its error by the inverse formal multiplication. Thus $P$ has an $n$-division point over the maximal unramified extension. Choosing that lift locally makes the cocycle zero on inertia. For an arbitrary lift its restriction is a coboundary, so the cohomology class is unramified. Therefore the image lies in first Galois cohomology unramified outside a finite set:We prove the required finiteness of this group. Take a finite Galois extension $L/K$ containing all of $E[n]$ and ${\mu }_n$, and enlarge $S$ to include any remaining ramification of $L/K$. Let $S_L$ be the places above it. The inflation-restriction exact sequence bounds the kernel of restriction to $L$ bya finite group, since both its group of arguments and its group of values are finite. Over $L$ the torsion module is constant and can be identified with ${\mu }_n^2$, so Kummer theory givesRestrictions of our unramified classes remain unramified outside $S_L$. If a Kummer class $[a]$ is unramified at such a place $v$, its $n$th root can be taken in an unramified local extension. The extended valuation still has integer values, and $nv(\sqrt[n]{a})=v(a)$. Hence $v(a)$ is divisible by $n$. Each coordinate therefore lies in the S-unramified power class groupThis group is finite. Indeed, the ideal of $a$ away from $S_L$ is ${\mathfrak{a}}^n$. Sending $[a]$ to $[\mathfrak{a}]$ yields the exact sequenceFor its kernel, a principal $\mathfrak{a}$ lets us divide $a$ by an $n$th power to obtain an $S_L$-unit. For surjectivity, an ideal class killed by $n$ has an $n$th power which is principal away from $S_L$, giving such an $a$. The S-unit group is finitely generated and the ideal class group is finite; hence both end groups are finite, proving finiteness of S-unramified Kummer classes. Restriction has finite image in $L(S_L,n{)}^2$ and finite kernel, so $H_S^1(K,E[n])$ is finite. We concludewhich is the Weak Mordell-Weil theorem and its Kummer-theoretic proof of the weak Mordell-Weil theorem.

This framework also defines the Selmer group of an elliptic curve by imposing that a class in $H^1(K,E[n])$ lie in the local Kummer image at every completion. Its classes are unramified outside a fixed finite set, so the same finiteness proof applies. Its relation to rational points isThe Tate–Shafarevich group here measures the difference between locally soluble torsors and globally soluble ones. Finally, the weak theorem gives finite divisibility quotients; adding the height descent lemma and bounded-height finiteness proves the stronger Mordell-Weil theorem. Those height inputs are additional to the cohomological finiteness just established.