# Is there an excplicit number field of definition for an Abelian Variety $A/\mathbb{C}$ with CM?

Consider a simple abelian variety $A/\mathbb{C}$ with sufficiently many CMs by $\mathcal{O}$, where $\mathcal{O}$ is an order in a CM field $K$. Specifically, $K$ is a CM field of degree $2g$, where $g = \dim A$. Let $(K,\, \Phi)$ be the CM type. It can be shown that there is a field $F$ such that $A$ is defined and has smCM over $F$.

Is there an explicit field $F_0(\mathcal{O},\, \Phi)$ such that $A$ is defined over $F_0$, with CM over $F_0$?

As motivation: If $A$ is an elliptic curve, $K$ an imaginary quadratic field, and $\mathcal{O}$ is the ring of integers in $K$, then the answer is yes: $A$ is defined with CM over the Hilbert class field $H(K)$, and moreover there is no better field of definition, so $F_0$ can be taken to be $H(K)$. It is moreover my understanding that if $\mathcal{O}$ is a different order, then $F_0$ can be taken to be the Hilbert class field of the order $\mathcal{O}$, though this is something I heard and might be misinformed.

My question, then, is whether this extends to higher-dimensional abelian varieties. It's known that such an abelian variety can be defined over some number field, but the arguments I've seen in this direction are of the form "if $A$ is defined over $\mathbb{C}$, then it can be defined over a finite-type ring over $\mathbb{Q}$, so reduce modulo a maximal ideal and check that it works." There is, on the other hand, an old result of Taniyama-Shimura that says: if $A$ has CM type $(K,\, \Phi)$ and $A$ is defined over $F$, then $F$ contains the reflex field $E(K,\, \Phi)$.

So, can the result: An elliptic curve $A/\mathbb{C}$ with CM by $\mathcal{O}$ can be defined over the Hilbert class field of $\mathcal{O}$ be extended, in some form, to higher-dimensional abelian varieties? And if so, in what form?

• @VesselinDimitrov: Chapter IV in that book is quite long, consisting of many subsections. At least in the more re-issued version of the book by Shimura, there is no "Theorem 1" in Chapter IV, but there is a "Main Theorem 1" in section 15.3 which says nothing about fields of definition and imposes severe restrictions (CM type must be primitive, and CM order must be maximal). Moreover, that result involves a specification of a $\mathbf{Q}_{>0}^{\times}$-homothety class of polarizations (rather than a specific polarization), so it is not determined by the abelian variety alone. Please clarify. – user27920 Aug 6 '14 at 16:56
• Although it is weaker than what you want, within the $K$-linear isogeny class of $A$ there is always a member which (together with its CM-structure) is defined over the field of moduli $F$ of the reflex norm for the CM type $\Phi$ of $A$ (equipped with its CM-structure), and such a descent is unique up to $K$-linear $F$-isogeny. This amounts to constructing on the ideles of that field of moduli an algebraic Hecke character with the reflex norm as its algebraic part; see Cor. A.4.6.5 and Theorem 2.5.2 in the book "Complex multiplication and lifting problems". – user27920 Aug 6 '14 at 17:12
• @VesselinDimitrov: No, fields of moduli are not fields of definition. Very far from it. This is related to one reason that coarse moduli spaces (rather than moduli stacks) generally suck: their rational points have very limited meaning. If you don't allow isogeny change over $\mathbf{C}$ then it is false that a field of definition can be predicted by the CM type which is isogeny-invariant. It is a very subtle problem beyond dimension 1 to nail down a specific field of definition for a specific CM abelian variety (equipped with its CM-structure) without moving around in the isogeny class. – user27920 Aug 7 '14 at 0:22