Endomorphisms of Elliptic Curves and the Tate module

Dear Readers,

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.

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10 Responses to Endomorphisms of Elliptic Curves and the Tate module

  1. In essence the first proposition says that any one point may have a period in the group structure on the elliptic curve, but that the points in the image of an isogeny can’t all have the same period, right?

    • Jim Stankewicz says:

      I would say this is absolutely correct although we can get far more(as we see) because we are working with irreducible curves, so any morphism between them has image either a point or the whole curve.

      Your point also reminds me that I should clarify that we are talking about these curves geometrically. Of course if we consider an elliptic curve over a finite field we have a group of finite size m so [m] is rationally the constant map, but that’s a different question dealing with the rationality of m-torsion points.

      • Dr. Armstrong is correct. For an algebraically closed field k, it is possible for E(k) to be a torsion abelian group. This occurs iff k is the algebraic closure of a finite field. (One direction is pretty clear. The other is more challenging and is a good exercise for those with some background in elliptic curves.) What Jim has established is that, in every case, there does not exist a single positive integer n such that E(k) = E(k)[n].

        In the statement of the second theorem, the adjective “maximal” should apply to the third case (the quaternion order) and not the second case (CM order).

      • Anon says:

        Can you give me a hint how to prove the direction “E(k) not torsion => k not an algebraic closure of a finite field”? I thought about using division polynomials. This works for uncountable k, but then the result can also be seen just by using the known structure of E(k)_tors.

        I also thought about making a case differentiation: If char(k) = 0, use the Lefschetz principle (does this give points of infinite order?—I don’t know the exact formulation as I don’t know model theory), if char(k) = p > 0 use an element transcendental over the prime field to somehow construct a point of infinite order.

      • Jim Stankewicz says:

        This is a fairly standard contradiction argument. If k is the algebraic closure of a finite field F then if P= (x,y) is a point in E(k) then P is also in E(F(x,y)). What can you say about the size of the field F(x,y)? What can you say about 2P, 3P, etc?

        That being said, I wonder if given the approaches you suggest, what you are trying to prove is “k algebraically closed+ k not the algebraic closure of a finite field => E(k) not torsion”

      • P.L. Clark says:

        Yes, probably he means the converse.

        Let me sketch some approaches.

        First, let K be a field which is a finite extension either of Q, of Q(t), or of F_p(t). It’s enough to show that for every elliptic curve defined over K, there exists an algebraic extension L of K and an L-rational point of E which is nontorsion.

        [You need to know something about j-invariants and twists to see that every elliptic curve defined over an algebraically closed field of transcendence degree at least 2 is also defined over an algebraically closed subfield of transcendence degree at most 1.)

        1) Galois theory. The field extension generated by a torsion point P has very particular properties: in particular its Galois closure is a subfield of a GL_2(Z/nZ)-Galois extension, so has Galois group a quotient of GL_2(Z/nZ). The Jordan-Holder factors of such groups are very restricted: the only nonabelian simple ones are PSL_2(Z/pZ). On the other hand, there are much more complicated Galois extensions of any field K as above: for instance it is easy to show that there are S_n-extensions for every n.

        2) Local methods

        Write

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

        (Small modifications to be made in characteristic 2…)

        Then there are infinitely many points P on E which are quadratic over K, i.e., such that K(P)/K has degree at most 2: choose x to be any element of K. On the other hand, I claim that the torsion subgroup of E over the maximal multiquadratic extension of K is finite.

        This is even true locally:

        One can embed K into a locally compact field K’ (yes, even if the transcendence degree of K is positive), and then the maximal multiquadratic extension L’ of K’ has finite degree over K’ so is itself a
        local field.

        But the torsion subgroup of any elliptic curve over any local field is finite, as follows for instance from p-adic Lie theory.

        Another way to go is to embed in a local field and use the fact that there are standard integrality properties that torsion points must satisfy, so that e.g. in characteristic 0 a point whose x-coordinate is 1/2^n cannot be a torsion point if n is sufficiently large.

        There must be many other proofs as well; I would be interested to hear what others can come up with.

  2. Another comment: in your determination of the isomorphism class of the quaternion algebra, you showed that the algebra is not ramified at any finite place except at the prime p. Therefore by the classification of quaternion algebras, there are two choices:

    (i) It could be B_{p,oo}, the definite
    quaternion algebra ramified at p and no other finite place,

    or

    (ii) It could be the matrix algebra M_2(Q).

    Although you mentioned earlier that the quaternion algebra was definite, I couldn’t find where you actually showed that the answer was necessarily (i) and not (ii). Can you do this?

    (One relatively high tech method: apply Poincare’s complete irreducibility theorem on the endomorphism algebra of any abelian variety.)

    • Jim Stankewicz says:

      Thanks for your comments. Those points you brought up really needed to be fixed.

      As you can see from my recent edit, the issue you brought up about the ramification of End(E) can be dealt with in a much more low-tech way than Poincare’s irreducibility theorem.

  3. Cederash says:

    Мне кажется очень полезная штука

  4. Avertedd says:

    Спасибо. То, что нужно ))

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