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?

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    $\begingroup$ "Using Mordell-Weil" ? If you know that $E(K)$ is a finitely generated $\mathcal{O}_K$-module, then it is quite clear that $E(K)$ won't contain a copy of $\mathcal{O}_K$ using the decomposition theorem for finitely generated $\mathcal{O}_K$-modules because $\mathcal{O}_K/\mathfrak{p}\neq 0$ . $\endgroup$ – Chris Wuthrich Dec 21 '13 at 19:42
  • $\begingroup$ @ChrisWuthrich I am sorry, could you explain more details? In general, for a fixed ring $R$, a finitely generated $R$-module contains a copy of $R$, doesn't it? $\endgroup$ – Kevin.lijh Dec 22 '13 at 1:40
  • $\begingroup$ No. $\mathcal{O}_K/\mathfrak p$ is a counter-example for $R=\mathcal{O}_K$ and $\mathbb{Z}/2\mathbb{Z}$ a counter-example for $R=\mathbb{Z}$. $\endgroup$ – Chris Wuthrich Dec 22 '13 at 15:09

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.)

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  • $\begingroup$ Maybe, I cannot get the result just from the statement of Mordell-Weil. $\endgroup$ – Kevin.lijh Dec 21 '13 at 3:27
  • $\begingroup$ The proof of MW has two parts, first show $E(K)/mE(K)$ is finite, then use heights to deduce finite generation of $E(K)$. So you just want to use the fact that $E(K)$ is finitely generated, without using heights. Doesn't the fact that $Q$ is infinitely $\pi$-divisible contradict the finite generation of $E(K)$ (assuming that $Q$ is non-torsion, of course)? $\endgroup$ – Joe Silverman Dec 21 '13 at 16:45

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