In this post I want to sketch the proof that the j-invariant of an elliptic curve over with complex multiplication is an algebraic number. With more work, one can show that it is in fact an algebraic integer which underpins the famous observation that is remarkably close to an integer ( to 40 significant digits ).
We start with some background about elliptic curves over .
A lattice is a rank 2 discrete subgroup for -linearly independent. The quotient carries a natural group structure and the Weierstrass – function
gives a group isomorphism sending to where denotes the -points of the elliptic curve
for the Eisenstein series of weight . To eliminate unwanted constants later we define and .
The uniformisation theorem shows that every elliptic curve is isomorphic to for some .
For an elliptic curve given by the Weierstrass equation
the discriminant of E is defined to be the quantity (more or less the discriminant of the RHS of the equation defining E) and the j-invariant of E to be the quantity . As one might hope from the name, if E’ is isomorphic to E and is itself given by a Weierstrass equation of the form above then we have (as any elliptic curve over can be put in Weierstrass form we will refer to the j-invariant of E without first specifying an equation). In light of the correspondence between elliptic curves over and lattices, we can think of j and as functions on lattices. We find and .
Now suppose and are lattices and for some . Then multiplication by induces a homomorphism and it turns out that this gives rise to an isogeny (morphism of algebraic varieties which is also a group homomorphism) and that every isogeny arises in this way. In particular, elliptic curves and are isomorphic if and only if and are homothetic, i.e. for some . Moreover, the endomorphism ring of (isogenies with addition and composition) is isomorphic to the ring
Note that any lattice is homothetic to one of the form for some . In this case, it is a simple check to see that either or is an imaginary quadratic extension of and R is an order in K (a subring S of K, finitely generated as a -module such that , for example the ring of integers ). In the latter case we say that has complex multiplication (by K or R).
If we start with an imaginary quadratic field K then it is easy to construct an elliptic curve with complex multiplication by K. Indeed, let be any non-zero fractional ideal of K. After embedding K in , becomes a lattice and we may consider the elliptic curve corresponding to . By definition, is an -module so . Conversely, any order of K is a subring of so actually . Moreover, if two fractional ideals and have the same image in the class group of K then for some and so and are isomorphic. This lets us define a map
sending to which is easily seen to be injective. Moreover, if has complex multiplication by for some then whence is a fractional ideal of . We have therefore established a bijection between and and so, crucially, the latter set is finite.
We are now in a position to complete the proof that the j-invariant of an elliptic curve with complex multiplication is an algebraic number but let’s first pause to consider the implications of this result. Since the j-invariant is a well defined function on isomorphism classes of elliptic curves it may also be viewed as a function on lattices up to homothety. That is, we have for all . If we define the function on the upper half plane by then is a modular function of weight for . Its q-expansion begins
where and so we are asserting that this function, when evaluated at any element of that lies in an imaginary quadratic field, yields an algebraic number. This seemed truly remarkable to me when I first saw it. It is also worth remarking that the function defined in the obvious way is, in fact, a weight 12 modular form and famously has the product expansion
Continuing with the proof, fix E an elliptic curve with complex multiplication by (the general case where is replaced by an arbitrary order in K is more complicated and I will omit it for simplicity) and j-invariant . It suffices to show that the set is finite. Indeed, if is transcendental over then we may find with isomorphic to the function field over in n indeterminates. Note that has (at least) n distinct images under automorphisms of L (given by permuting the generators) and since is algebraically closed these extend to automorphisms of . As we may do this for every n we see that the set is infinite. Note conversely that if is algebraic over then any automorphism of must send to some root of the minimal polynomial of over and so is finite and has order equal to the degree of over .
Fix now and define to be the elliptic curve obtained from E by acting on the coefficients of E. That is, if E is the elliptic curve for some lattice then is the elliptic curve . In particular, we have . If is an endomorphism of E then , obtained from by applying to the equations defining (recall that is a morphism of algebraic varieties), is an endomorphism of . So and in particular, also has complex multiplication by . The upshot of this is that for some and we have already seen that the set is finite, thus is an algebraic number.
In fact, the proof tells us more, namely that the degree of over is at most the class number of K. In particular, if lies in an imaginary quadratic field K of class number one and the corresponding lattice is a fractional ideal of K, then will actually be rational. As mentioned at the beginning, one may show that the j-invariant of a CM elliptic curve is in fact an algebraic integer so will actually be in in this case. Choosing to be , a generator of the ring of integers of the class number one field and substituting into the q-expansion for j, we have
where . Since q is very small () this predicts that should be very close to an integer, as indeed is the case.