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Jul 18, 2016 at 15:56 comment added nfdc23 The endomorphism rings are orders $O$ in imaginary quadratic fields. For $E$ over $\mathbf{C}$ with ${\rm{End}}(E) = O$, ${\rm{H}}_1(E(\mathbf{C}),\mathbf{Z})$ is an invertible $O$-module and $K(j(E))/K$ is a ring class field. For a CM field $L$ and CM type $\Phi$ on $L$, $L$-linear isogeny classes of abelian varieties with CM type $\Phi$ over a finite extension $K$ of the reflex field correspond to certain algebraic Hecke characters $\mathbf{A}^{\times}_K\to L^{\times}$. See 2.5.1-2.5.2 and A.4.6 of amazon.com/…
Jul 18, 2016 at 10:56 comment added Jeff Yelton It would be better to say that, loosely, CM for abelian varieties corresponds to the case when the endomorphism ring is "as big as possible". If an abelian variety $X$ of dimension $g$ is simple with $\mathrm{End}^{0}(X) = \mathrm{End}(X) \otimes \mathbb{Q}$, then $de$ divides $g$, where $d^{2}$ is the degree of the division algebra $\mathrm{End}^{0}(X)$ over its center, and $e$ is the degree of its center over $\mathbb{Q}$. The CM case is where $de = g$.
Jul 18, 2016 at 8:15 answer added user19475 timeline score: 2
Jul 18, 2016 at 6:16 history asked dorebell CC BY-SA 3.0