Let $A$ be an abelian variety over $\overline{\mathbb{Q}}$ and assume that $A$ has complex multiplication by the ring of integers of a CM field $K$. Then $A$ has potentially good reduction, that is: there exists a number field $F$ and an abelian scheme $\mathcal{A} \to \mathrm{Spec}(\mathcal{O}_F)$ whose generic fiber is isomorphic $A$ after extending the scalars from $F$ to $\overline{\mathbb{Q}}$.

The question is: how to determine $F$ in terms of $K$?

[EDIT: The following is false: I'm more familiar with the case of elliptic curve, in which case one can choose $F$ to be the Hilbert class field of $K$, that is, the maximal abelian unramified extension]

Does something similar hold for higher dimensional abelian varieties?

  • $\begingroup$ See Serre-Tate, "Good reduction of abelian varieties", (Annals of Math. 88, no. 3 (1968)), Th. 7, p. 505. $\endgroup$ – Damian Rössler Jan 25 '14 at 20:25
  • $\begingroup$ What you say in the elliptic curve case is false, since the CM order might not be maximal (so the $j$-invariant might generate over $K$ a much bigger class field). $\endgroup$ – user76758 Jan 25 '14 at 20:41
  • $\begingroup$ Actually, the statement for elliptic curves isn't even true for maximal orders. Consider the case that $\text{End}(E)=\mathbb{Z}[i]$, so in the OP's notation, $F=K=\mathbb{Q}(i)$. But $E:y^2=x^3+x$ doesn't acquire good reduction over $\mathbb{Q}(i)$. As Damian has said, you can get an answer from Serre-Tate; roughly speaking, if you take a minimal field of definition for $A$ and then adjoin some torsion, you'll get $F$. $\endgroup$ – Joe Silverman Jan 25 '14 at 22:31
  • $\begingroup$ @Joe: There is a hidden subtlety in the formulation of the question which "rules out" twisting counterexamples (which is why I focused on non-maximal orders): the OP is beginning with an abelian variety over $\overline{\mathbf{Q}}$ and seems to allow the option to choose whatever descent to a number field we like best. So your (natural) viewpoint of first fixing a descent and then going up from there is sort of "opposite" to how the question is posed. Though I really have no idea if the OP recognizes the subtlety of this aspect of how the question was posed, and whether it is intended. $\endgroup$ – user76758 Jan 25 '14 at 22:54
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    $\begingroup$ @conduc Even for elliptic curves with CM by the maximal order, you generally can't take $F$ to be the Hilbert class field of $K$ and acquire everywhere good reduction. Also, you ask "how to determine $F$ in terms of $K$"? You do realize that there may not be a unique minimal $F$. Indeed, abelian varieties of dim $\ge2$ often have many different minimal fields of definition, since the field of moduli may not itself be a field of definition. $\endgroup$ – Joe Silverman Jan 25 '14 at 23:16

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