Let $E$ be a CM elliptic curve defined over a quadratic imaginary field $K$ with maximal order, that is, $\mathrm{End}\_K(E)\cong \mathcal{O}\_K$. Suppose the class number of $K$ is equal to $1$. Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}\_K$ which satisfying $E(K)/\mathfrak{p}E(K)=0$. In this case, is it possible to derive finiteness of $E(K)$ using Mordell-Weil?



Mordell-Weil and finiteness of rational points
Let $Q\in E(K)$. Since $E(K)/\mathfrak{p}E(K)=0$, we know that $Q\in\mathfrak{p}E(K)$. You've assumed that $\mathfrak{p}$ is principal, say $\mathfrak{p}=(\pi)$, so $Q=u\_1\pi Q\_1$ for some unit $u\_1\in\mathcal{O}\_K^\*$ and some point $Q\_1\in E(K)$. Repeating $n$ times, we get $Q=u\_n\pi^nQ\_n$. (For simplicity, I'll assume that $p=N\mathfrak{p}$ is odd.) Then the easiest way to finish the proof at this point is probably to use canonical heights to conclude that 
$$ \hat h(Q)=\hat h(u\_n\pi^nQ\_n)=p^n\hat h(Q\_n). $$
Next we observe that
$$ h(Q\_n) = \hat h(Q\_n) + O(1) = p^{-n}\hat h(Q) + O(1), $$
so $(Q\_n)\_{n\ge1}$ is a set of bounded height in $E(K)$, hence is finite. Hence we can find $n>m$ with $Q\_n=Q\_m$, and hence
$$ Q = u\_n\pi^nQ\_n = u\_n\pi^nQ\_m = u\_nu\_m^{-1}\pi^{n-m}u\_m\pi^mQ\_m = u\_nu\_m^{-1}\pi^{n-m}Q. $$ Therefore
$$ (u\_nu\_m^{-1}\pi^{n-m}-1)Q = 0, $$
which proves that every point in $E(K)$ is a torsion point.


(There may well be easier ways to do this, but this is the first thing that occurred to me.)


