= 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. Set
$$
\widehat h(P)=\lim_{r\to\infty}4^{-r}h(2^rP).
$$
Condition (ii) makes this limit converge and gives
$$
|\widehat h(P)-h(P)|\leq\sum_{r\geq0}c_1,4^{-r-1}=\frac{c_1}{3}.
$$
Thus $\widehat h$ still has finite bounded subsets. Condition (i) also gives a global lower bound for $h$, so $\widehat h\geq0$. Applying condition (iii) to $2^rP,2^rQ$, 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 $\widehat h(2P)=4\widehat h(P)$, gives the reverse direction. Hence
$$
\widehat h(P+Q)+\widehat h(P-Q)
=2\widehat h(P)+2\widehat h(Q),
$$
and induction yields $\widehat h(mP)=m^2\widehat h(P)$ for every integer $m$.
Now suppose $A/nA$ is finite and choose representatives $R_1,\ldots,R_s$. Put $R=\max_i\widehat h(R_i)$. For any $P$, write $P=nQ+R_i$. Nonnegativity and the parallelogram identity give
$$
n^2\widehat h(Q)=\widehat h(P-R_i)
\leq2\widehat h(P)+2\widehat h(R_i),
$$
so, since $n\geq2$,
$$
\widehat h(Q)-R\leq\frac12(\widehat h(P)-R).
$$
Repeated division modulo $nA$ therefore reaches the finite set $B=\{P:\widehat h(P)\leq R+1\}$. Reversing the recursion expresses every element of $A$ using $B$ and the finitely many $R_i$. This is the <height descent lemma>, and proves
$$
\boxed{A\text{ is finitely generated}\iff|A/nA|<\infty.}
$$
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