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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 $\mathbb{Q}(\sqrt{-3})$.

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  • $\begingroup$ I am nothing close to a specialist but there is a paper Elliptic curves with everywhere good reduction $\endgroup$ May 26 at 17:17
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    $\begingroup$ 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
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    $\begingroup$ 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

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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 $d < 100$).

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

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