For $K$ a number field, denote by $\mathcal{O}_K$ its ring of integers and by $H_K$ its Hilbert class field.

For which imaginary quadratic field $K$ does there exist an elliptic curve $E$, defined over $H$, with complex multiplication by $\mathcal{O}_K$ having everywhere good reduction (on $H$)?

By Fontaine's Il n'y a pas de variété abélienne sur Z, corollary of Théorème B, there do not exist such curves for $K=\mathbb{Q}(\sqrt{-1})$ or $\mathbb{Q}(\sqrt{-3})$.

  • $\begingroup$ I am nothing close to a specialist but there is a paper Elliptic curves with everywhere good reduction $\endgroup$ May 26 at 17:17
  • 2
    $\begingroup$ There are clearly no examples with $H_K = K$, since in that case the Groessencharacter attached to $E$ would be a Groessencharacter of conductor 1 and infinity-type $(1, 0)$, which is impossible. $\endgroup$ May 26 at 19:40
  • 1
    $\begingroup$ I have a feeling there might be one for $K = \mathbf{Q}(\sqrt{-21})$. For this field, $H_K$ has class number 1 and all units of $H_K$ get sent to 1 by the norm map to $K$. So there is a conductor 1 Groessencharacter of $H_K$ which sends $\mathfrak{a}$ to $\sigma_1(\operatorname{N}_{H_K / K} \alpha)$ where $\sigma_1$ is your favourite embedding of $K$ into $\mathbf{C}$ and $\alpha$ is any generator of $\mathfrak{a}$. This has the correct infinity-type to be the GC of a CM elliptic curve. $\endgroup$ May 27 at 10:28

1 Answer 1


Here is an example.

Let $K = \mathbf{Q}(\sqrt{-21})$. Then the class group of $K$ is $C_2 \times C_2$ and its Hilbert class field is $H_K = \mathbf{Q}(\sqrt{-1}, \sqrt{3}, \sqrt{7})$. In particular, $H_K$ is a CM-field and its maximal totally real subfield $H_K^+$ is $\mathbf{Q}(\sqrt{3}, \sqrt{7})$.

The LMFDB database has an elliptic curve 4.4.7056.1-1.1.a1 over $H_K^+$ which has everywhere good reduction and has CM by $\mathcal{O}_K$. Base-extending this from $H_K^+$ to $H_K$ gives the example you seek.

Similar examples should exist whenever $H_K$ has class number 1 and all units of $H_K$ are in the kernel of the norm map to $K$. (The class number condition may well not be needed, but the condition on the units certainly is). I have no idea if there are infinitely many such fields $K$, but there definitely some! This happens for $\mathbf{Q}(\sqrt{-d})$ for $d = 21, 33, 42, 57, 66, 77, 93$ (and no others for $d < 100$).

  • $\begingroup$ This is enlightening, thank you very much for your answer! And your computations, as well $\endgroup$
    – Stabilo
    May 27 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.