Special privilege enjoyed by Elliptic Curves with Complex Multiplication I think after reading the title one may understand the intention of me, this question is concerned about the Elliptic curves having a Complex Multiplication. 

I have been reading many theorems, ( celebrated papers of Zagier and Kolyvagin ). In majority, much of the proofs I came across, considers Elliptic Curves with Complex Multiplication. And many results have been discovered in that direction. I know what is meant by Complex Multiplication and endomorphism rings. But if some one asks for the reason behind such privilege enjoyed by the elliptic curves with CM ( Complex Multiplication ), what are the precise things one can tell ?

To put in other way, how come the proofs are discovered about Elliptic curves with CM are discovered so easily and why not for the other case ? ( Is there some bird's eye view ? )
Thank you.
 A: I have never really known the answer to this question either. There has to have been a motivation for this. All I can think is that in having CM, the lattice associated the the curve (if working over $\mathbb{C}$) has more symmetry, so that the geometry dictates much more towards the arithmetic.
I am only just studying this stuff myself so I am no expert. The main use of CM that I know of (but have never read much into) is that given a number field $K$, you can use an elliptic curve with CM by $\mathfrak{O}_K$ to get the Hilbert class field of $K$, by simply adjoining the $j$-invariant of the curve to $K$. Knowing the Hilbert class field is a very powerful thing...for example it tells you great things about representations of primes by binary quadratic forms.
For example for certain $n$ (namely those that are square-free and not $3$ mod $4$) we may study when a prime $p$ can be written as $x^2+ny^2$ for integers $x,y$. Now algebraically this is the same as $p$ splitting, into principal prime ideals, in the ring of integers $\mathbb{Z}[\sqrt{−n}]$  of the number field $\mathbb{Q}(\sqrt{−n})$. Now the principal part is important here because when we involve the Hilbert class field, this is the same as saying that the ideal splits completely in this field. So $p$ can be written in this form iff $p$ splits completely in the Hilbert class field of $Q(\sqrt{n})$
But as I just said, you can get this Hilbert class field by finding an elliptic curve with CM by $\mathbb{Z}[\sqrt{−n}]$ and adjoining its $j$-invariant to $\mathbb{Q}(\sqrt{-n})$. So really, once this has been accomplished, the minimal polynomial $f(x)$ of the $j$-invariant over $\mathbb{Q}$ gives a criterion for the possible $p$'s that can be written as $x^2+ny^2$ (for those $n$'s we restricted to).
Basically (for $p$ not dividing $n$ or the discriminant of $f(x)$): 
$p = x^2 + ny^2$ if and only if $\left(\frac{-n}{p}\right) = 1$ and $f(x) \equiv 0$ mod $p$ has a solution.
