# Heegner Points on $X_0(N)$ when some primes dividing $N$ are inert in the imaginary quadratic field

If $K = \mathbb{Q}(\sqrt{-D})$ is a imaginary quadratic field with discriminant $-D$, then we get Heegner points on $X_0(N)$ as long as there exists $\mathfrak{n} \subset \mathcal{O}_K$ such that $\mathcal{O}_K/\mathfrak{n} \simeq \mathbb{Z}/N$. For this to happen, we need all of the primes $p \mid N$ to either split or ramify in $\mathcal{O}_K$. Gross-Zagier was originally proved when the "Heegner hypothesis" is satisfied, i.e. all of the primes $p$ split. However, in applications, this hypothesis is not always satisfied.

However, Heegner's original paper used $D \equiv 3 \pmod{8}$ and $N = 32$, so $2 \mid N$ is inert. How do you get points on $X_0(N)$ in this situation? I couldn't find an English translation or summary of Heegner's paper, and am confused as to how this works.

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I think it historically known that Heegner only used "mock" Heegner points, see papers of Monsky (see page 46:

"It should be noted that the modular points that Heegner uses to show that $2p_3$ is a congruent number aren't "Heegner points" either; they arise from the field $Q(\sqrt{ip_3})$ in which 2 is inert."

In the paragraph below that, Monsky gives an outline of the method (he is unifying various cases, including Heegner's). He works on $2Y^2=X^4+1$ rather than $E$ directly. He does not use the language of modular curves per se, but I suspect it can be made equivalent. Perhaps if one does so, there could also be "partial traces" from a covering modular curve, corresponding to genus theory when summing over quadratic forms.

The idea is that the Hilbert class field of $Q(\sqrt{-dl})$ contains that of $Q(\sqrt{-d})$, and if we construct a point in the former, then tracing by the genus character $\chi_l$ gives a point in $Q(\sqrt{-d})$. An explicit example is then given, where $-d=-3$ for 76A, and this is not square. You twist by $l=5$, to leverage.
I've looked at the paper of Monsky, and the sentence you quoted was what led me to ask this question. But Monsky uses an imaginary quadratic field in which 2 ramifies and doesn't (I think) say anything more about the fields $\mathbb{Q}(\sqrt{i p_3})$ which Heegner used. – stl Apr 19 '13 at 6:24