We’ve now talked about quaternion algebras, and today I’ll talk about the surprisingly close connection between quaternion algebras and elliptic curves.

We first recall a fundamental fact about isogenies:

Proposition:The set of isogenies $E_1 \to E_2$ forms a torsion-free $\mathbf{Z}$-module.

That we have a module structure stems from the fact that we have the structure of an abelian group coming from the group structure of $E$ ( $(\psi + \phi )(P) = \psi(P) + \phi(P)$) and that isogenies commute with addition on the curve ($\psi(P +P) = \psi(P) +\psi(P)$). Now suppose there were torsion, i.e. there were a  nonzero integer $m$ and a nonconstant isogeny $\psi$ such that $[m] \circ \psi = [0]$. By taking degrees, we see that $m^2\deg(\psi) = \deg([m])\deg(\psi) = \deg([0]) = 0$. Thus either $m = 0$ or $\psi$ is constant.

When we take $E_1 = E_2 = E$, we can do more than just add isogenies, we can also compose them, and the above gives that the group of endomorphisms form a $\mathbf{Z}$-algebra. Note also that this implies we have a possibly noncommutative ring of characteristic zero.

Moreover, note that $End(E)$ must have no zero divisors by the same multiplicativity of degrees. That is to say, if $\phi,\psi$ are endomorphisms such that $\phi \circ \psi = [0]$ then $\deg(\phi)\deg(\psi) = 0$ so either $\phi = [0]$ or $\psi = [0]$.

We wish to show that we get a subalgebra of a quaternion order as described in the last post. To do this we introduce the Tate module. Let $\ell$ be a prime and $E[\ell^n] = \ker([\ell^n])$. We have natural maps $E[\ell^{n+1}] \stackrel{[\ell]}{\to} E[\ell^n]$ which form a compatible system and allow us to build an inverse limit. We denote this $\ell$-th inverse limit by $T_\ell E$.

Consider now the structure of $E[\ell^n]$. If $\ell$ is coprime to the characteristic of the field of definition for $E$, we know that $[\ell^n]$ is separable of degree $\ell^{2n}$. Since the size of the group is equal to the separable degree of the map and $E[\ell^n]$ consists of cyclic subgroups of order $\ell^n$, the group structure will be isomorphic to $\mathbf{Z}/\ell^n\mathbf{Z} \times \mathbf{Z}/\ell^n\mathbf{Z}$. The Tate module is then $\mathbf{Z}_\ell^2$. What then if $\ell = char(k)$?

In that case, the map $[\ell^n]$ will not be a separable map.  We then consider the Absolute Frobenius Map : the map of schemes whose  topological map is identity and whose pullback on all local rings is $x \mapsto x^{\ell^n}$.  We use this to build a curve $E^{\ell^n}$ and a map called the Relative Frobenius which is inseparable of degree $\ell^n$. To do this, consider the diagram

$\begin{array}{ccc} E & \to & E \\ \downarrow & & \downarrow \\ Spec(k) & \to & Spec(k) \end{array}$

where the horizontal arrows are the Absolute Frobenius $F$. The fiber product of the lower right corner of this diagram $Spec(k) \times_{Spec(k)} E$ is defined to be $E^{\ell^n}$ and the morphism $\phi: E \to E^{\ell^n}$ is given by the universal mapping property of fiber products. A number of useful facts about this construction(which easily extends to larger schemes) can be found in Qing Liu’s book in section 3.2.4 or in Silverman’s book section II.2 . We use only the fact that the map is purely inseparable degree $\ell^n$ and so by the earlier construction of the Dual Isogeny, $[\ell^n] = \phi\circ \hat{\phi}$. As was mentioned earlier, $E[\ell^n]$ has size equal to the separable degree of $[\ell^n]$ and is made up of cyclic subgroups of size $\ell^n$. If $\hat{\phi}$ is also purely inseparable, then $E[\ell^n] = O$ and we say that $E$ is supersingular. If $\hat{\phi}$ is separable, then $E[\ell^n] = \mathbf{Z}/\ell^n\mathbf{Z}$.

Thus upshot is then that the $\ell$-th Tate module of an elliptic curve is a free $\mathbf{Z}_\ell$-module of rank at most 2. Thus the endomorphism ring of the Tate module is free of rank at most 4. How can we then connect this to standard homomorphisms?

Let $\phi:E_1 \to E_2$ be an isogeny. Then since $\phi([\ell^n]P) = [\ell^n]\phi(P)$, we have induced maps $\hspace{.1mm} E_1[\ell^n] \to E_2[\ell^n]$ which induce a map $T_\ell E_1 \to T_\ell E_2$. We thus have a homomorphism $Hom(E_1,E_2) \to Hom(T_\ell E_1 ,T_\ell E_2)$ or even after tensoring with $\mathbf{Z}_\ell$, $Hom(E_1, E_2) \otimes \mathbf{Z}_\ell \to Hom(T_\ell E_1, T_\ell E_2)$. We then have the following:

Theorem: The above map is injective. That is to say, if $\phi:E_1 \to E_2$ is a nonzero isogeny and $\alpha \in \mathbf{Z}_\ell -\{0\}$, the induced map $(\phi\otimes\alpha)_\ell: T_\ell E_1 \to T_\ell E_2$ is a nonzero isogeny.

A very nice proof of this fact is in Silverman III.7.4. The upshot of this theorem is that for any 2 elliptic curves, $Hom(E_1,E_2)$ is a free $\mathbf{Z}$-module of rank at most 4. Finally we have the object of all this buildup:

Theorem: If $E$ is an elliptic curve, $End(E)$ is either $\mathbf{Z}$, an order(edit:thanks, Pete!) of an imaginary quadratic field or an order in a definite quaternion algebra(and this last case occurs only over a field of characteristic $p$.

We do this by rank. If the rank of $End(E)$ is 1, we are done. This is the generic case over a field of characteristic zero. If the rank is greater than one, pick some endomorphism $\phi$ which is not of the form $[m]$ where $m\in \mathbf{Z}$.

We develop a version of trace and norm analogous to the quaternion algebra case, $n(\phi) = \phi\hat{\phi}$ and $t(\phi) = \phi + \hat{\phi}$. Note that by definition $n(\phi)$ is a non-negative integer. Thus if we alter $\phi$ so that the trace is zero, we get $\mathbf{Z} + \phi\mathbf{Z}$ as an order in an imaginary quadratic field. If the rank of our endomorphism ring is 2, we are again done.

If not, consider that $\psi \to \phi\psi \phi^{-1}$ is a linear map of order 2 and thus decomposes into $+1,-1$ eigenspaces. As in the writeup I posted in addition to my last post, if we take $\psi$ in the -1 eigenspace of that linear map, $1, \phi,\psi, \phi\psi$ forms a basis for our rank 4 $\mathbf{Z}$-module.

Now to show something a bit more specialized:

If we work over a field of characteristic zero, the rank of our endomorphism ring can be at most 2. This follows from the Lefschetz principle if we can show this for the complex numbers. This is easy however, as the theory of modular forms allows us to view elliptic curves over the complex numbers to be in bijection with a fundamental domain of numbers in the upper half-plane($\tau \in \mathcal{F}$ in bijection with $\mathbf{C}/\Lambda$ where $\Lambda \cong \mathbf{Z} + \tau\mathbf{Z}$).

Over the complex numbers,  the endomorphism ring is isomorphic to the set of complex numbers such that $z\Lambda \subset \Lambda$, so if $z$  is an endomorphism of $E$,  $z = a + b\tau$ and $z = c + d\tau$ so $z$ satisfies the characteristic polynomial of $\left(\begin{array}{cc} a & b \\ c & d\end{array}\right)$, which is a monic polynomial of degree 2 with integral coefficients, so $End(E)$ is an integral extension of $\mathbf{Z}$. If $z\not\in \mathbf{Z}$, then $\tau = (z-a)/b$ so $End(E) \subset \mathbf{Q}(\tau)$, which we can see is an imaginary quadratic field by substituting $z = a + b\tau$ into the characteristic polynomial for $z$.

Thus the quaternion order case can only occur over characteristic $p$. We can say even more, explicitly what quaternion algebra this order must lie in. Note that for $\ell \ne p$, the endomorphism ring of the tate module is $M_2(\mathbf{Z}_\ell)$. Since this is already a maximal order in the split quaternion algebra $M_2(\mathbf{Q}_\ell)$, we know $End(E)$ must be split at every prime ideal of $\mathbf{Z}$ besides $p$. Since quaternion algebras over $\mathbf{Q}$ are determined, up to isomorphism by an even number of either prime ideals or $\infty$, the quaternion algebra in question must be either the one which is nonsplit exactly at $p,\infty$ or the matrix algebra $M_2(\mathbf{Q})$. However, note that $M_2(\mathbf{Q})$ has plentiful zero divisors, while $End(E)$ has none, so it must be the quaternion algebra which is nonsplit precisely at $p,\infty$.(edit 2: Thanks to Pete again)

This shall conclude for some time the discussion on elliptic curves. Already at the end of this post we’ve touched on a pair of topics, complex multiplication and class field theory, which require some heavy algebraic number theory to confront. Developing that will be the aim for the next several posts.