Suppose is an elliptic curve. We will need some standard facts. First, if is a field automorphism, then will denote the elliptic curve formed by acting on the coefficients of a Weierstrass equation by . We have an isomorphism by . If has CM by the full ring of integers in a quadratic imaginary , then . This follows by exploiting the fact that to argue that lies in a finite extension of .

The natural map from isomorphism classes of curves over with CM by to isomorphism classes over is an isomorphism. This follows easily from the fact that . It turns out that if we use our normalized isomorphism , we get the nice relation for all and .

If is defined over and has CM by in , then every endomorphism of is defined over . It is useful to see why this is. Let fix . Then by definition , so using the normalized relation we get that for all . Thus if also fixes , then will be fixed by and hence by Galois descent every endomorphism descends to .

We can use this result to show that if has CM by the full ring of integers , then the class number of . As an example application, without knowing the Weierstrass equation for our curve we can argue that the unique curve with CM by must be defined over . This is because , so the degree of the extension is and hence . Of course, we already knew that because we said last time it is given by . Likewise, we could make the same argument for the curve with CM by where .

Now we get to our first real relation between CM and class field theory. Suppose has CM by . The field generated by the -invariant and the torsion points is an abelian extension of . The proof uses our field of definition result above about the endomorphisms. If , then we can reduce to showing that each is abelian over . Using the standard Galois representation, we see that injects into .

Since we have CM floating around we know every endomorphism is already defined over . Using the fact that the image of commutes with , we get that injects into an abelian group and hence is abelian. This result about being abelian I think is an exercise in Silverman’s first book.

Fix a quadratic imaginary field . Let’s return to the idea that we have a simply transitive action of on by . We also have a nice action of on given by . Fix some . This gives us a map by sending to the unique element with the property that . It turns out that is a group homomorphism that is independent of the choice of . This is actually not so easy to prove, and the underlying philosophy of why it is hard is that the action of is analytic in nature since it acts on the lattice , whereas the Galois action is purely algebraic. To prove the independence of requires translating between the two. Although it is sort of a fun proof, we’ll skip it in order to get to the meat of CM theory more quickly.