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# 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

$j(\tau)=\frac{1}{q}+744+196884q+21493760q^2+...$

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

$\Delta(\tau)=(2\pi)^{12}q\prod_{n\geq1}(1-q^n)^{24}$

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

$\frac{1}{q}+744+196884q+21493760q^2+...\in\mathbb{Z}$

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.