Let $A$ be an abelian variety. Let $K$ be a CM field. ${\rm End}^0(A):={\rm End}(A)\otimes_{\Bbb{Z}}\Bbb{Q} $. $A$ is said to have CM by $K$ if only if ${\rm End}^0(A)$ contains $K$.
Then, why $[K:\Bbb{Q}]\leq2\dim A$ holds ?
This maybe famous fact as though I couldn't find the statement exactly. To be honestly,I don't understand why $K$ should be CM field except for the case of CM elliptic curves. Reference only is also appreciated.
Edit: I realize that we can show it with $A[n] \cong \Bbb{Z}/(n)^{2d}$ and some algebraic number theory result. If $End^0(A)$ contains a field $K$ then there is a primitive element $b\in End(A)$ such that $K=\Bbb{Q}(b)$. The minimal polynomial $f\in \Bbb{Z}[x]$ of $b$ is irreducible $\bmod n$ for some $n$ (coprime with the leading coefficient), with which I mean that $f\not \equiv gh\bmod n$ with $g,h\in \Bbb{Z}[x]$ non-constant whose leading coefficient is a unit $\bmod n$. Clearly the minimal polynomial of $b|_{A[n]}$ has degree $\le 2d$. So $[K:\Bbb{Q}]=\deg(f)\le 2d$.
There is certainly an abstract algebra argument, but it is worth knowing that it becomes evident on the complex torus side.
Let $A=\Bbb{C}^d/ \Lambda$ be a complex torus.
$End(A)$ is the space of $\Bbb{C}$-linear maps $h:\Bbb{C}^d\to \Bbb{C}^d$ such that $h(\Lambda)\subset \Lambda$.
If $End(A)\otimes_\Bbb{Z}\Bbb{Q}$ contains a field $K$ then by the primitive element theorem $K = \Bbb{Q}(b)$ for some $b\in End(A)$.
Let $\tilde{b} = b|_\Lambda \in End_\Bbb{Z}(\Lambda)$.
$\Lambda \cong \Bbb{Z}^{2d}$ so the minimal polynomial $f\in \Bbb{Z}[x]$ of $\tilde{b}$ has degree $\le 2d$.
$\Bbb{C}^d = \Lambda \otimes_\Bbb{Z}\Bbb{R}$ so $f(b)=0\in End(A)$ and $[K:\Bbb{Q}]\le \deg(f)\le 2d$.