# Is the ring of integers of any imaginary quadratic field equal to $\mathrm{End}(E)$ for some $E / \Bbb Q$?

Let $$K / \Bbb Q$$ be an imaginary quadratic field, and let $$O_K$$ be its ring of integers. Is there an elliptic curve $$E / \Bbb Q$$ such that its ring of integers $$\mathrm{End}_{\overline{ \Bbb{Q}}}(E)$$ is isomorphic to $$O_K$$ ?

This is clearly true over $$\Bbb C$$ (I think one can take the torus $$C / \Bbb O_K$$), but it is not clear when it can be defined as an algebraic curve with rational coefficients.

(The analoguous result over finite fields holds: this is Deuring correspondence).

• See also doc.sagemath.org/html/en/reference/arithmetic_curves/sage/…, there are indeed finitely many j-invariants of CM elliptic curves over Q. – Alphonse May 7 at 14:52
• Over finite fields F_q, we know that either for supersingular curves, one has Deuring result, and for ordinary we know that End(E) is an order (not nec. maximal) in an imaginary quadratic field K, which is generated by Frobenius. We have $K = Q(\sqrt{a_q(E)^2 - 4q})$, where $a_q(E) = q+1-|E(F_q)|$. Hasse bound implies that $K$ is imaginary (at least if $q=p$ is prime). – Alphonse May 9 at 12:19
By complex multiplication, the set of $$j$$-invariants of isomorphism classes of elliptic curves $$E/\mathbb{C}$$ with $$\operatorname{End}(E) = \mathcal{O}_K$$ is exactly $$\{j(\mathfrak{a}) : \mathfrak{a} \in \mathcal{Cl}(\mathcal{O}_K)\}$$ ($$\mathcal{Cl}$$ denoting the ideal class group). Every element of this set has the same minimal polynomial. That minimal polynomial is the Hilbert class polynomial of $$\mathcal{O}_K$$, and the degree of the Hilbert class polynomial of $$\mathcal{O}_K$$ is equal to the ideal class number of $$\mathcal{O}_K$$. Hence the answer to your question is: There exists $$E/\mathbb{Q}$$ with $$\operatorname{End}(E) = \mathcal{O}_K$$ if and only if $$\mathcal{O}_K$$ has class number $$1$$.
There is an extensive body of literature on the class number $$1$$ problem which this post is too small to contain.