The j-invariant of CM elliptic curves

In this post I want to sketch the proof that the j-invariant of an elliptic curve over \mathbb{C} 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 e^{\pi\sqrt{163}} is remarkably close to an integer ( to 40 significant digits e^{\pi\sqrt{163}}=262537412640768743.999999999999250072597).

We start with some background about elliptic curves over \mathbb{C}.

A lattice \Lambda \subseteq \mathbb{C} is a rank 2 discrete subgroup \Lambda =\mathbb{Z}w_1 \oplus \mathbb{Z} w_2 for w_1, w_2 \mathbb{R}-linearly independent. The quotient \mathbb{C}/\Lambda carries a natural group structure and the Weierstrass \wp– function

\wp(z;\Lambda)=\frac{1}{z^2} +\sum_{w\in\Lambda-{0}}\frac{1}{(z-w)^2}-\frac{1}{w^2}

 gives a group isomorphism \mathbb{C}/\Lambda\longrightarrow E_{\Lambda}(\mathbb{C}) sending z\in\mathbb{C} to (\wp(z;\Lambda),\wp'(z;\Lambda)) where E(\mathbb{C}) denotes the \mathbb{C}-points of the elliptic curve

E_{\Lambda} : y^2=4x^3-60G_{4}(\Lambda)x-140G_{6}(\Lambda)

for G_{2k}(\Lambda)=\sum_{w\in\Lambda-{0}}w^{-2k} the Eisenstein series of weight 2k. To eliminate unwanted constants later we define g_{2}(\Lambda)=60G_4(\Lambda) and g_3(\Lambda)=140G_6(\Lambda).

The uniformisation theorem  shows that every elliptic curve E/\mathbb{C} is isomorphic to E_{\Lambda} for some \Lambda\subseteq\mathbb{C}.

For an elliptic curve given by the Weierstrass equation

E: y^2=4x^3+Ax+B

the discriminant of E is defined to be the quantity \Delta(E)=-(A^3+27B^2) (more or less the discriminant of the RHS of the equation defining E) and the j-invariant of E to be the quantity j(E)=-1728\frac{A^3}{\Delta}. 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 j(E')=j(E) (as any elliptic curve E over \mathbb{C} 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 \mathbb{C} and lattices, we can think of j and \Delta as functions on lattices. We find \Delta(\Lambda)=g_2(\Lambda)^3-27g_3(\Lambda)^2 and j(\Lambda)=1728\frac{g_2(\Lambda)^3}{\Delta(\Lambda)}.

Now suppose \Lambda_{1} and \Lambda_{2} are lattices and \alpha\Lambda_{1}\subseteq\Lambda_{2} for some \alpha\in\mathbb{C}. Then multiplication by \alpha induces a homomorphism \mathbb{C}/\Lambda_{1}\longrightarrow\mathbb{C}/\Lambda_{2} and it turns out that this gives rise to an isogeny (morphism of algebraic varieties which is also a group homomorphism) \phi_{\alpha}:E_{\Lambda_{1}}\longrightarrow E_{\Lambda_{2}} and that every isogeny arises in this way. In particular, elliptic curves E_{\Lambda_{1}} and E_{\Lambda_{2}} are isomorphic if and only if \Lambda_{1} and \Lambda_{2} are homothetic, i.e. \Lambda_{1}=\alpha\Lambda_{2} for some \alpha\in\mathbb{C}^{\times}. Moreover, the endomorphism ring of  E_{\Lambda} (isogenies E_{\Lambda}\rightarrow E_{\Lambda} with addition and composition) is isomorphic to the ring

 R=\{\alpha\in\mathbb{C} : \alpha\Lambda\subseteq\Lambda\}

Note that any lattice is homothetic to one of the form \mathbb{Z}\oplus\tau\mathbb{Z} for some \tau\in\mathbb{C}. In this case, it is a simple check to see that either R=\mathbb{Z} or K=\mathbb{Q}(\tau) is an imaginary quadratic extension of \mathbb{Q} and R is an order in K (a subring S of K, finitely generated as a \mathbb{Z}-module such that S\otimes\mathbb{Q}=K, for example the ring of integers \mathcal{O}_{K}). In the latter case we say that E_{\Lambda} 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 \mathfrak{a} be any non-zero fractional ideal of K. After embedding K in \mathbb{C}\mathfrak{a} becomes a lattice and we may consider the elliptic curve E_{\mathfrak{a}} corresponding to \mathbb{C}/\mathfrak{a}. By definition, \mathfrak{a} is an \mathcal{O}_{K}-module so \mathcal{O}_{K}\subseteq \text{End}(E). Conversely, any order of K is a subring of \mathcal{O}_{K} so actually \text{End}(E)=\mathcal{O}_{K}. Moreover, if two fractional ideals \mathfrak{a} and \mathfrak{b} have the same image in the class group \text{cl}(\mathcal{O}_{K}) of K then \mathfrak{a}=\alpha\mathfrak{b} for some \alpha\in K^{\times} and so E_\mathfrak{a} and E_{\mathfrak{b}} are isomorphic. This lets us define a map

\text{cl}(\mathcal{O}_{K})\longrightarrow\mathcal{EL}(\mathcal{O}_{K})=\{\text{elliptic curves with CM by } \mathcal{O}_{K}\}/\text{iso}

sending \mathfrak{a} to E_{\mathfrak{a}} which is easily seen to be injective. Moreover, if E_{\Lambda} has complex multiplication by \mathcal{O}_{K} for some \Lambda=\mathbb{Z}\oplus\tau\mathbb{Z} then \tau\in\mathcal{O}_{K} whence \Lambda is a fractional ideal of \mathcal{O}_{K}. We have therefore established a bijection between \text{cl}(\mathcal{O}_{K}) and \mathcal{EL}(\mathcal{O}_{K}) 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 j(\alpha\Lambda)=j(\Lambda) for all \alpha\in\mathbb{C}^{\times}. If we define the function j(\tau) on the upper half plane \mathbb{H} by j(\tau)=j(\mathbb{Z}\oplus \tau\mathbb{Z}) then j(\tau) is a modular function of weight 0 for SL(2,\mathbb{Z}). Its q-expansion begins


where q=e^{2\pi i\tau} and so we are asserting that this function, when evaluated at any element of \mathbb{H} 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 \Delta(\tau) 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 \mathcal{O}_{K} (the general case where \mathcal{O}_K  is replaced by an arbitrary order in K is more complicated and I will omit it for simplicity) and j-invariant j(E). It suffices to show that the set \{\sigma(j(E)) : \sigma \in \text{Aut}(\mathbb{C}) \} is finite. Indeed, if x_{1}\in\mathbb{C} is transcendental over \mathbb{Q} then we may find x_{2},...,x_{n}\in\mathbb{C} with L=\mathbb{Q}(x_{1},...,x_{n}) isomorphic to the function field over \mathbb{Q} in n indeterminates. Note that x_{1}  has (at least) n distinct images under automorphisms of L (given by permuting the generators) and since \mathbb{C} is algebraically closed these extend to automorphisms of \mathbb{C}. As we may do this for every n we see that the set \{\sigma(x_{1}) : \sigma \in \text{Aut}(\mathbb{C}) \} is infinite. Note conversely that if x_{1} is algebraic over \mathbb{Q} then any automorphism of \mathbb{C} must send x_{1} to some root of the minimal polynomial of x_{1} over \mathbb{Q} and so \{\sigma(x_{1}) : \sigma \in \text{Aut}(\mathbb{C}) \} is finite and has order equal to the degree of \mathbb{Q}(x_{1}) over \mathbb{Q}.

Fix now \sigma\in\text{Aut}(\mathbb{C}) and define E^{\sigma} to be the elliptic curve obtained from E by acting on the coefficients of E. That is, if E is the elliptic curve y^2=4x^3-60G_{4}(\Lambda)x-140G_{6}(\Lambda) for some lattice \Lambda then E^{\sigma} is the elliptic curve y^2=4x^3-60\sigma(G_{4}(\Lambda))x-140\sigma(G_{6}(\Lambda)) . In particular, we have j(E^{\sigma})=\sigma(j(E)). If \phi is an endomorphism of E then \phi^{\sigma}, obtained from \phi by applying \sigma to the equations defining \phi (recall that \phi is a morphism of algebraic varieties), is an endomorphism of E^{\sigma}. So \text{End}(E)\cong\text{End}(E^{\sigma}) and in particular, E^{\sigma} also has complex multiplication by \mathcal{O}_{K}. The upshot of this is that \sigma(j(E))=j(E') for some E'\in\mathcal{EL}(\mathcal{O}_{K}) and we have already seen that the set \mathcal{EL}(\mathcal{O}_{K}) is finite, thus j(E) is an algebraic number.

In fact, the proof tells us more, namely that the degree of \mathbb{Q}(j(E)) over \mathbb{Q} is at most the class number of K. In particular, if \tau\in\mathbb{H} lies in an imaginary quadratic field K of class number one and the corresponding lattice is a fractional ideal of K, then j(\tau) 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 j(\tau) will actually be in \mathbb{Z} in this case. Choosing \tau to be \frac{1+\sqrt{-163}}{2}, a generator of the ring of integers of the class number one field \mathbb{Q}(\sqrt{-163}) and substituting into the q-expansion for j, we have


where q=-e^{-\pi\sqrt{163}}. Since q is very small (\approx 10^{-18}) this predicts that \frac{-1}{q}=e^{\pi\sqrt{163}} should be very close to an integer, as indeed is the case.