Perhaps the most basic fact about abelian varieties with CM is they have an everywhere potential good reduction (Serre-Tate). On the face of it it might appear that there isn't much more to be added to this classical theorem, for already the elliptic curve $y^2 = x^3+x$ with endomorphism ring $\mathbb{Z}[i]$ has a bad reduction at the prime $2$, while its quadratic twist $fy^2 = x^3+x$ has bad reductions at all the prime factors of $f$.

The following argument seems to indicate to the contrary that a lot more can be said about the possible bad reduction characteristics, once we fix a dimension $g$, require the CM order to be maximal, and take care of the twist feature.

Theorem. (Weber; Granville-Stark). Let $E/K$ be an elliptic curve with $\mathrm{End}(E)$ equal to the full ring of integers of a complex quadratic number field. Then $E/K$ is $\bar{K}$-isomorphic to an elliptic curve $E'/K$ (a $K$-form of $E/K$), such that $E'/K$ has a good reduction in all primes of $K$ having residual characteristic $\notin \{2,3\}$.

Proof. This is implicit in the key Lemma 1 of Granville and Stark's paper $ABC$ implies no Siegel zeros$\ldots$, that is the substance of their implication that the uniform $abc$-conjecture precludes a Siegel zero for the odd quadratic Dirichlet characters. Let $j(\tau)$ be the $j$-invariant of $E$ under some complex embedding $K \hookrightarrow \mathbb{C}$.

We lose no generality in assuming that $K = H$, the Hilbert class field of the quadratic field $k := \mathrm{End}(E) \otimes \mathbb{Q}$. With $\gamma_2$ and $\gamma_3$ the Weber functions used in loc. cit. (they are modular functions for the group $\Gamma(6)$ fulfilling $j = \gamma_2^3 = \gamma_3^2 + 1728$), the Weierstrass equation $$ \quad Y^2 = X^3 - 27\gamma_2(\tau)^3X - 54 \gamma_3(\tau)^2 $$ has $j$-invariant $\gamma_2(\tau)^3 = j(\tau)$, and hence defines a complex elliptic curve isomorphic to $E_{\mathbb{C}}$. By the Shimura reciprocity law (cf. Thm. 5.1.2 in Schertz's book Complex Multiplication), which Granville and Stark cite in their Lemma 1 with the comment that this case had already been known to Weber, the special values $\gamma_2(\tau)$ and $\gamma_3(\tau)$ are algebraic [of course, and] belong to the ring of integers of the mod $6$ ray class field $F$ of $k$. But the functional relation $\gamma_2(\tau)^3 - \gamma_3(\tau)^2 = 1728$ means that our Weierstrass equation has a unit discriminant, and since its coefficients lie in $O_F$, this explicit Weierstrass model constitutes an everywhere good reduction model over $F$. We thus have an elliptic curve $E'/H$ whose base-change to $F/H$ has an everywhere good reduction. But this extension $F/H$ ramifies only at $\{2,3\}$, and since the formation of Neron's model commutes with unramified extensions of the base, we conclude that $E'/K$ had an everywhere good extension outside of $\{2,3\}$ to begin with. QED

Obviously the role of the special primes $2$ and $3$ here has to do with the level $6$ of the modularity group $\Gamma(6)$ of the Weber special functions $\gamma_2(\tau)$ and $\gamma_3(\tau)$. This is sharp upon considering the elliptic curves with invariants $1728$ and $0$ (indeed there are no abelian schemes whatsoever over $\mathbb{Z}[i]$ or $\mathbb{Z}[\omega]$, as Fontaine proved).

My question is how all of this is supposed to extend to $g$-dimensional abelian varieties $A/K$ with $\mathrm{End}(A)$ equal to the full integer ring $O_E$ of a CM field $E$ of degree $2g$.

Is there an abelian variety $A'/K$ over $K$ that is $\bar{K}$-isomorphic to $A/K$ and has an everywhere good reduction outside some fixed finite set $S_g \subset \mathrm{Spec}\, \mathbb{Z}$ of residual characteristics that depends on $g$ alone?

Is there a reference that I am missing that takes care of this? It does look surprising to me that such a basic question may have been left untreated by the literature.

  • $\begingroup$ Doesn't the quadratic twist have CM by all of $\mathbb Z[i]$, because negation commutes with every endomorphism? Moreover, an elliptic curve defined over $F$ that has CM by $K$ over $\mathbb C$ necessarily has CM by $K$ over $FK$, because each endomorphism is determined by its action on the tangent space and the identity. $\endgroup$ – Will Sawin Jul 2 '18 at 19:37
  • $\begingroup$ @WillSawin: I suppose what this modular functions argument is saying is that $E/K$ is $\bar{K}$-isomorphic to some other elliptic curve $E'/K$ (a $K$-form of $E$) that has a good reduction apart from the characteristics $2$ and $3$. The point being that the field $K$ is unchanged; while we merely ask for the existence of a form over that same field $K$. That remains a sharp statement, of course, in the sense that $\{2,3\}$ must be definitely excluded. Let me edit my question to this effect. (And thank you for these comments!) $\endgroup$ – Vesselin Dimitrov Jul 2 '18 at 20:05
  • $\begingroup$ @WillSawin: After another revision, I believe this should now be the correct statement and question. (I apologize if it is still not tight.) In the $abc$ connection what mattered was precisely having the form $A'/K$ over the same field, that acquires an everywhere good reduction over an almost unramified field $F$ (so that the root discriminant term is essentially unchanged on passing to $F$). For the stable Faltings heights in the elliptic case satisfy $2h^{\mathrm{st}}(E) = 2h^{\mathrm{st}}(E') = \log{\sqrt{|\mathrm{disc}(k)|}} + \frac{L'}{L}(1,\chi_k) + \mathrm{const}$. $\endgroup$ – Vesselin Dimitrov Jul 2 '18 at 21:08

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