# 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.

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I too got the same idea as the modular forms are said to have beautiful symmetry, but does this symmetry imply something ? @fretty –  Iyengar Feb 27 '12 at 15:54
So, why can't you post a nice answer, combining all comments together into a beautiful elegant account. Anyway thanks a lot, for helping me. But is there any measure on this case ? ( like haar measure ) on the corresponding groups. @fretty –  Iyengar Feb 27 '12 at 16:15
Actually, I didn't think I knew that much about CM! I will turn it into an answer. –  fretty Feb 27 '12 at 16:17
In fact I had a glance about your blog, your intention is awesome, but have you done something to improve it ? @fretty –  Iyengar Feb 27 '12 at 16:19
In a certain sense, elliptic curves with CM are "induced" from simpler objects. The $L$-function of a CM elliptic curve is a Hecke $L$-function (a "$GL(1)$" object on the automorphic side), whereas the $L$-function of a non-CM elliptic curve is the $L$-function of a cusp form (a "$GL(2)$" object). More-or-less. –  B R Feb 27 '12 at 16:52

## 1 Answer

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.

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Good try, let us wait and see what it turns out to be. But how can we say that CM is a special property, and what can we do if we can know about the representations of primes and their unique factorization ? –  Iyengar Feb 27 '12 at 16:28
There is a big theory about the representation numbers of binary quadratic forms (of negative discriminant)...but the beautiful thing is that $\text{SL}_2(\mathbb{Z})$-equivalence classes of these things are the same as homotheity classes of lattices in the complex plane...which in turn are the same as isomorphism classes of elliptic curves...which are also connected with modular forms. –  fretty Feb 27 '12 at 16:32