THE l-ADIC GALOIS REPRESENTATION ASSOCIATED TO AN ELLIPTIC CURVE

DINA ROUMIANTSEVA

1. Some Basic Definitions

We will begin with some basic definitions in order to develop the notion of an elliptic curve. We will show that there is a natural addition law on the points of an elliptic curve and then we will discuss n-torsion points and use them to construct a representation associated to an elliptic curve.

Definition 1. Let k be a field, then projective n-space over k

kℙ^{n} = (k^{n+1} − {0})/(v ∼ λv, ∀λ ∈ k^{∗})

Lemma 1. E = {(x, y, z) | y^{2}z = x^{3} + axz^{2} + bz^{3}, a, b ∈ k} is well-defined as a subspace of kℙ^{2}

(Note: Any homogeneous cubic in x, y and z can be reduced to this form by an invertible change of coordinates as long as the characteristic of k is not 2 or 3, see for example [2, pp. 46–48, 63–64].)

Proof. It suffices to show that (x, y, z) ∈ E implies (λx, λy, λz) ∈ E, ∀λ ∈ k^{∗} because E is defined as a subspace of k^{3} − {0} and (λx, λy, λz) ∈ E implies (\frac{1}{λ}λx, \frac{1}{λ}λy, \frac{1}{λ}λz) = (x, y, z) ∈ E as well. But this statement is clear because y^{2}z = x^{3} + axz^{2} + bz^{3} ⟹ λ^{3}y^{2}z = λ^{3}(x^{3} + axz^{2} + bz^{3}) ⟹ (λy)^{2}λz = (λx)^{3} + aλx(λz)^{2} + b(λz)^{3}.

Definition 2. The discriminant of E as above is D = −4a^{3} − 27b^{2}.

Definition 3. An elliptic curve is the set E as above such that x^{3} + ax + b = 0 has distinct roots over \overline{k}, the algebraic closure of k.

Lemma 2. E is an elliptic curve ⟺ D ≠ 0

Proof. Write x^{3} + ax + b as (x − α_{1})(x − α_{2})(x − α_{3}) and observe that a = α_{1}α_{2} + α_{1}α_{3} + α_{2}α_{3}, b = −α_{1}α_{2}α_{3}, and α_{1} + α_{2} + α_{3} = 0. Then it is simply a direct calculation to verify that D = (α_{1} − α_{2})^{2}(α_{1} − α_{3})^{2}(α_{2} − α_{3})^{2}. But then it follows directly that D is nonzero if and only if the roots are distinct.

Now let (x, y, z) be a representative of a class in E and suppose z ≠ 0 then (\frac{x}{z}, \frac{y}{z} , 1) is the unique representative of [(x, y, z)], the class of (x, y, z), in the plane z = 1. So we can think of projective points on E with nonzero z-coordinate as corresponding bijectively to solutions (x, y) of y^{2} = x^{3} + ax + b. Now suppose z = 0, then [(x, y, z)] ∈ E implies that x^{3} = 0, so there is exactly one such class, it is [(0, 1, 0)]. When looking at the curve in affine coordinates, we will call this class O, the point at infinity.

Definition 4. Let +: E ×E → E such that O is the identity and A+B = A^{∗}B^{∗}O, where X^{∗}Y is the third point on the line \overline{XY}. We also define \overline{XX} to be the line tangent to the curve at X, O^{∗}O = O, and say that a line intersects O ⟺ it is vertical.

Given the notion of intersection multiplicity, this definition makes sense for all cases, but we will make it more precise without using algebraic geometry shortly. First, if \overline{AO} goes through no other points on the curve besides those two, we define A^{∗}O = A. Next, suppose A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), and \overline{AB} is not a vertical line, then we write its equation as y = mx + d. To get the third point, we make the substitution y^{2} = (mx+d)^{2} = x^{3}+ax+b, so we have x^{3}−m^{2}x^{2}−(2md−a)x+b−d^{2} = (x − x_{1})(x − x_{2})(x − x_{3}) ⟹ x_{3} = m^{2} − x_{1} − x_{2}. So we can define A^{∗}B to be the unique point on \overline{AB} with x_{3} as its x-coordinate.

Now we can compute explicit formulas for addition of two points: if A + B ≠ O,

A + B = (m^{2} − x_{1} − x_{2}, m(m^{2} − 2x_{1} − x_{2}) + y_{1}) (1)

where m = \frac{y_{2}−y_{1}}{x_{2}−x_{1}} if A ≠ B, and m = \frac{dy}{dx} = \frac{3x^{2}+a}{2y} if A = B. The fact that m is well defined in the second case means that if y = 0 then we need A + A = O, therefore we need to show that if y = 0, then 3x^{2} + a ≠ 0 since then \frac{dx}{dy} = 0, so A^{∗}A = O and A + A = O. However, a polynomial and its derivative have a root in common if and only if it is a double root of the polynomial, but we specified that it have all distinct roots. Now we can look at the points on the elliptic curve as a group under this operation, where O is the identity, A^{−1} = (x_{1}, −y_{1}), and the associativity follows quite directly, albeit tediously, from the formulas (there is also a nice geometric argument for it on p. 21 of [3]). This group is abelian because the ∗ operation is clearly commutative.

2. n-torsion points on E

The main goal of this section is to show that the group of n-torsion points over \overline{k} is isomorphic to ℤ/nℤ × ℤ/nℤ. We present the outline of an elementary proof suggested in [3], we then give the description of a more standard but less elementary proof which uses complex analysis, however this proof has the disadvantage that it only works for subfields of ℂ.

Definition 5. An n-torsion point of an elliptic curve is a point such that nA =

\overbrace{A + ⋯ + A}^{n terms} = O

From here on we assume that the characteristic of k is 0.

Proposition 1. ∃P_{n} ∈ ℤ[x] of degree n^{2} − 1 such that (x, y) ∈ E is an n-torsion point ⟺ P_{n}(x) = 0. Additionally, P_{n} has a double root at x if y ≠ 0 and a simple root otherwise.

Proof. See [3, exercise 6.4, (a)–(f)]. Our P_{n} corresponds to ψ_{n}^{2} in the book. Note also that the statement of part (e) is incorrect and should be changed to the statement given above.

Corollary 1. E[n], the group of n-torsion points over \overline{k} is isomorphic to ℤ/nℤ × ℤ/nℤ.

Proof. First, there are exactly n^{2} n-torsion points because if x is a root of P_{n}, then there are two points on the curve with x as their x-coordinate unless their y-coordinate is 0, in which case there is one. Hence the number of points with x as their x-coordinate is the same as the order of the zero of P_{n} at x. Therefore, by Proposition 1, there are n^{2} − 1 points (x, y) such that P_{n}(x) = 0 and hence there are n^{2} − 1 nontrivial points in E[n]. Together with O we get n^{2} points.

Now to show that E[n]is actually ℤ/nℤ × ℤ/nℤ, let n = p_{1}^{r_{1}}p_{2}^{r_{2}} … p_{m}^{r_{m}} be the prime factorization of n. First, it is enough to show that no more than two ℤ/p_{i}^{r}ℤ terms can appear in the abelian decomposition of E[n]for any i. This is because if s such terms appear with exponents t_{1}, …, t_{s}, we must have Σ_{j}t_{j} = 2r_{i} but each t_{j} ≤ r_{i} because otherwise there would be a point of order not dividing n. But if s ≤ 2, we immediately have that s = 2 and t_{1} = t_{2} = r_{i}. To finish the proof, choose generators a_{1}, …, a_{s} of each copy of ℤ/p_{i}^{t_{j}}ℤ and note that p_{i}^{t_{j}−1}a_{j} has order exactly p_{i}. Thus {p_{i}^{t_{j}−1}a_{j}} generate a subgroup of p_{i}-torsion points of order p_{i}^{s}. But the number of p_{i}-torsion points is p_{i}^{2} so s ≤ 2 as desired.

We will now outline the complex-analytic proof of the goal stated earlier.

Theorem 1. Every elliptic curve over ℂ is isomorphic as an abelian group to ℂ/L where L is some lattice in the complex plane.

Proof. The proof that ℂ/L is isomorphic to an elliptic curve requires no more than basic complex analysis and a particularly elegant approach is given in [1, pp. 14– 35]. The fact that every elliptic curve is of this form is more difficult and there are several common approaches. One approach using j-invariants is given in [1, pp. 119–121].

Corollary 2. Let k be a subfield of ℂ and E defined over k. Then E[n] is ℤ/nℤ × ℤ/nℤ over \overline{k}.

Proof. It is clear from Theorem 1 that E[n] is isomorphic to ℤ/nℤ × ℤ/nℤ over ℂ since if ω_{1}, ω_{2} ∈ ℂ generate L such that ℂ/L is isomorphic to E, then it is clear that \{c_{1} \frac{ω_{1}}{n} + c_{2} \frac{ω_{2}}{n} \} is exactly the set of n-torsion points. However, it follows from an argument identical to the one in Proposition 2 that if σ is an automorphism of ℂ then the map (x, y) → (σx, σy) permutes the n-torsion points of E and since there are only finitely many n-torsion points, it follow that their coordinates generate an extension of finite degree over k. Therefore, all n^{2} n-torsion points lie inside \overline{k}.

3. The Galois Representation

Definition 6. Given σ ∈ Gal(\overline{k}/k), we define an action of σ on \overline{k}^{2} by σ(x, y) = (σx, σy).

Proposition 2. f_{σ}: (x, y) → σ(x, y) induces an automorphism of E, where we define f_{σ}(O) = O.

Proof. First, we will show that f_{σ} sends E to E, so say y^{2} = x^{3} + ax + b, a, b ∈ k, then (σ(y))^{2} = σ(y^{2}) = σ(x^{3}+ax+b) = σ(x^{3})+σ(a)σ(x)+σ(b) = (σ(x))^{3}+aσ(x)+b because σ fixes k, so (σx, σy) ∈ E. By a similar argument, f_{σ} is a homomorphism simply because the addition map is given by rational functions with coefficients in k and is therefore preserved by f_{σ}. Finally, it is actually an automorphism since σ is invertible.

(Note that here we mean automorphisms of E as an abelian group. Aut(E) often refers to group automorphisms given by an algebraic map, which f_{σ} isn’t in general.)

This actually gives a homomorphism from Gal(\overline{k}/k) → Aut(E) since it is trivial to see that (στ)(x, y) = σ(τ(x, y)).

Since f_{σ} is an automorphism of E it sends E[n] to E[n]. In order to take advantage of this to develop a representation of Gal(\overline{k}/k), we will construct a convenient module, called the Tate module.

Definition 7. Let G_{1}, G_{2}, … be any finite groups and φ_{i}: G_{i+1} → G_{i} any homomorphisms. The inverse limit, G = \underleftarrow{lim} G_{i} is the set of all sequences g = \{g_{i}\} such that g_{i} ∈ G_{i} and φ_{i}(g_{i+1}) = g_{i}.

Definition 8. ℤ_{l}, the l-adic integers, are \underleftarrow{lim} ℤ/l^{i}ℤ with the φ_{i} just taking the residue modulo l^{i}.

Definition 9. The Tate module, T_{l}, is \underleftarrow{lim} E[l^{i}] with the φ_{i} being the multiplication by l maps.

Lemma 3. T_{l} is a free ℤ_{l}-module of rank 2.

Proof. First, T_{l} is a ℤ_{l}-module since if P = (P_{1}, P_{2}, …) ∈ T_{l} and m = (m_{1}, m_{2}, …) ∈ ℤ_{l}, we can define mP = (m_{1}P_{1}, m_{2}P_{2}, …). This is well-defined because the m_{i} are defined mod l^{i} and the P_{i} are l^{i}-torsion points. Furthermore, it is in T_{l} because lm_{i}P_{i} = m_{i}P_{i−1} = m_{i−1}P_{i−1} because m_{i} ≡ m_{i−1} mod l^{i−1}. T_{l} is clearly a free module because if P ≠ 0 is in the Tate module, and m ≠ 0 in ℤ_{l} then mP ≠ 0 directly from the definition. To show it is of rank 2, start by choosing any point P ∈ T_{l} such that P_{i} ≠ 0, then∀i, for each i, choose a P′_{i} outside of {jP_{i}} such that lP′_{i} = P′_{i−1}. Then it is clear that P and P′ are linearly independent and span T_{l}.

Now we claim that given σ ∈ Gal(\overline{k}/k), f_{σ} defined above induces an invertible ℤ_{l}- linear map from T_{l} to itself, by sending (P_{1}, P_{2}, …) to ρ(σ)(P) = (f_{σ}(P_{1}), f_{σ}(P_{2}), …). First, we want to show that ρ(σ)(P) ∈ T_{l}. But since f_{σ} is an automorphism, it takes E[n]to E[n], so we only need to show that l(f_{σ}(P_{i+1})) = f_{σ}(P_{i}), which is clear because l commutes with f_{σ} since f_{σ} is a homomorphism. Similarly, ρ(σ) is ℤ_{l}-linear because f_{σ} is a homomorphism and addition and scalar multiplication can be evaluated term by term. The above discussion gives us the following:

Proposition 3. ρ is a homomorphism from Gal(\overline{k}/k) to GL(T_{l}).

If we choose a basis {e_{1}, e_{2}}, we get an isomorphism between GL(T_{l}) and GL_{2}(ℤ_{l}), the set of invertible 2 × 2 matrices with coefficients in ℤ_{l}. Since ℤ_{l} ⊂ ℚ_{l}, we can think of the 2 × 2 matrices as invertible linear maps from ℚ_{l}^{2} to itself. This gives us:

Theorem 2. Given a choice of basis of T_{l}, ρ is a 2-dimensional representation of Gal(\overline{k}/k) over ℚ_{l}.

4. Cool Facts

The remarkable thing about this representation is that one can recover much of the information about the elliptic curve from it alone. First we show that Proposition 4. The field extension of k generated by the coordinates of E[l^{i}] can be determined solely from the Galois representation of E.

Proof. First, let k′ be an extension of k, and suppose that E[l^{i}] has coefficients in k′, then take σ ∈ Gal(\overline{k}/k) such that σ fixes k′. So σ acts as the identity E[l^{i}] which means that the first i terms of the coefficients of the matrix for ρ(σ) are 1 on the diagonal and 0 otherwise. But this is actually if and only if. To determine what field is generated by coefficients of E[l^{i}], this is simply the fixed field of {σ such that the first i terms of the coefficients of the matrix for ρ(σ) are the same as the identity matrix }.

We will state two further examples, justification of which is far beyond the scope of this paper, which demonstrate the power of this representation. First, given an elliptic curve over ℚ we can look at the set of solutions to the same equation over ℤ/pℤ for any p. This will define an elliptic curve for all but finitely many p, since the discriminant vanishes mod p exactly when p divides the discriminant of E over ℚ. Therefore in all the other cases, the discriminant mod p will be nonzero which by Lemma 2 means that we get an elliptic curve over ℤ/pℤ. It is a theorem that one can actually recover the number of points of E over ℤ/pℤ for any p simply by examining the l-adic Galois representation of E over ℚ for any l.

Finally, one can define an l-adic Galois representation attached to a modular form of weight 2 in an analogous but substantially more technical manner. It has long been known that given an elliptic curve E and a modular form of weight 2, E is parametrized by a modular form if and only if they have the same l-adic Galois representation for some l. Therefore, when Andrew Wiles proved that every semistable elliptic curve can be parametrized by a modular form of weight 2, what he did was to show that given any semistable elliptic curve he could produce a modular form of weight 2 whose 3-adic Galois representation was the same as that of the elliptic curve.

References

1. Neal Koblitz, Introduction to elliptic curves and modular forms, 2nd ed., Springer-Verlag, New York, US, 1993.

2. Joseph Silverman, The arithmetic of elliptic curves, Springer-Verlag, New York, US, 1986.

3. Joseph Silverman and John Tate, Rational points on elliptic curves, Springer-Verlag, New York, US, 1992.