Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 125 / 4

The natural maphas finite image by hypothesis. It remains to bound its kernel. If $P\in E(K)$ becomes $nQ$ for $Q\in E(L)$, thenis a one-cocycle for $G=\mathrm{Gal}(L/K)$. Changing $Q$ by an $n$-torsion point changes this cocycle by a coboundary, producing a well-defined map from the kernel toIf its cohomology class is zero, subtracting the corresponding torsion point from $Q$ makes $Q$ Galois fixed, so $P\in nE(K)$. The map is therefore injective. Both $G$ and $E(L)[n]$ are finite, so this group cohomology set is finite. A finite kernel and finite image give

If $A$ is finitely generated, the structure theorem for finitely generated modules over a principal ideal domain immediately makes $A/nA$ finite.

Conversely, first replace the given height by a quadratic one. SetCondition (ii) makes this limit converge and givesThus $\hat{h}$ still has finite bounded subsets. Condition (i) also gives a global lower bound for $h$, so $\hat{h}\ge 0$. Applying condition (iii) to $2^{r}P,2^{r}Q$, dividing by $4^r$ and passing to the limit gives one direction of the parallelogram identity. Applying the same inequality to $P+Q$ and $P-Q$, and using $\hat{h}(2P)=4\hat{h}(P)$, gives the reverse direction. Henceand induction yields $\hat{h}(mP)=m^2\hat{h}(P)$ for every integer $m$.

Now suppose $A/nA$ is finite and choose representatives $R_1,\dots ,R_s$. Put $R={\max }_i\hat{h}(R_i)$. For any $P$, write $P=nQ+R_i$. Nonnegativity and the parallelogram identity giveso, since $n\ge 2$,Repeated division modulo $nA$ therefore reaches the finite set $B=\{P:\hat{h}(P)\le R+1\}$. Reversing the recursion expresses every element of $A$ using $B$ and the finitely many $R_i$. This is the height descent lemma, and proves

For a reduced rational number $x=a/b$ with $b>0$, the height of a rational number isCondition (i) holds because only finitely many coprime integer pairs have bounded maximum.

Condition (iii) also holds. The standard height inequalitygivesCondition (ii) fails: for positive integers $m$,whose absolute value is unbounded. Thus precisely conditions $\boxed{\text{(i) and (iii)}}$ hold. This is consistent with $\mathbb{Q}/n\mathbb{Q}=0$ although the additive group $\mathbb{Q}$ is not finitely generated.