Heegner point

Remarks

Heegner points for $K$ exist exactly when $\mathcal{O}_K$ has an ideal $\mathfrak{n}$ with $\mathcal{O}_K/\mathfrak{n} \cong \Z/N\Z$; then $(\C/\mathcal{O}_K, \mathfrak{n}^{-1}/\mathcal{O}_K)$ is one. This holds, for example, under the Heegner hypothesis that every prime dividing $N$ splits in $K$. By the theory of complex multiplication, Heegner points are defined over the Hilbert class field $H$ of $K$, and the trace $y_K = \sum_{\sigma \in \Gal(H/K)} \varphi(x)^\sigma$ lies in $E(K)$.

Example: for $N = 1$, $X_0(1)$ is the $j$-line and the condition is only that $E$ has CM by $\mathcal{O}_K$. For $K = \Q(i)$, the curve $\C/\Z[i]$ has endomorphism ring $\Z[i] = \mathcal{O}_K$, so the point $j = 1728$ is a Heegner point for $K$.

Non-example: the curve $\C/(\Z + 2i\Z)$ has endomorphism ring $\Z[2i]$, an order of conductor $2$ in $\Q(i)$. This is not the maximal order $\Z[i]$, so its point $j(2i)$ of $X_0(1)$ is not a Heegner point for $\Q(i)$ in this sense.

Some authors also allow pairs with CM by a non-maximal order of $K$, and speak of Heegner points of conductor $c$.