 Solution

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