DEGREES OF THE ELLIPTIC TEICHMÜLLER LIFT

LUÍS R. A. FINOTTI

Abstract. The purpose of this paper is to find upper bounds for the degrees, or equivalently, for the order of the poles at O, of the coordinate functions of the elliptic Teichmüller lift of an ordinary elliptic curve over a perfect field of characteristic p. We prove the following bounds:

ord_{0}(x_{n}) ≥ −(n + 2)p^{n} + np^{n−1}, ord_{0}(y_{n}) ≥ −(n + 3)p^{n} + np^{n−1}. Also, we prove that the bound for x_{n} is not the exact order if, and only if, p divides (n + 1), and the bound for y_{n} is not the exact order if, and only if, p divides (n + 1)(n + 2)/2. Finally, we give an algorithm to compute the reduction modulo p^{3} of the canonical lift for p ≠ 2, 3.

1. Introduction

Voloch and Walker in [5] applied the theory of canonical lifts of elliptic curves to construct error-correcting codes. In that paper, the degrees of some polynomials, that we shall make precise later, have some importance in estimating exponential sums. We here try to analyze those degrees, giving upper bounds and finding when the degrees are strictly less than those bounds. Also, we describe an algorithm to compute the reduction modulo p^{3} of canonical lifts explicitly for p ≠ 2, 3.

We will consider an ordinary elliptic curve over a perfect field k of characteristic p > 0. The curve can be given by a Weierstrass equation:

E/k : y_{0}^{2} + a_{0}x_{0}y_{0} + b_{0}y_{0} = x_{0}^{3} + c_{0}x_{0}^{2} + d_{0}x_{0} + e_{0}.

Such an elliptic curve has a canonical lifting to an elliptic curve over the ring of Witt vectors W (k),

E/W (k) : y^{2} + axy + by = x^{3} + cx^{2} + dx + e,

with a = (a_{0}, a_{1}, …), b = (b_{0}, b_{1}, …), …, e = (e_{0}, e_{1}, …) ∈ W (k), for which we can lift the p^{th} power Frobenius map. (See [2].)

We have an injective group homomorphism (given by a section of the reduction map) τ : E(k̄) → E(W (k̄)), called the elliptic Teichmüller lift of E:

(x_{0}, y_{0}) ↦^{τ} (x, y) = ((x_{0}, x_{1}, x_{2}, …), (y_{0}, y_{1}, y_{2}, …)).

1991 Mathematics Subject Classification. Primary 11G20; Secondary 11T71.

Key words and phrases. elliptic curves, canonical liftings, elliptic Teichmüller lift.

This work was funded by CAPES (an agency of the Brazilian government).

2 LUÍS R. A. FINOTTI

We notice that we can identify E/W (k) with its Greenberg transform G(E)/k, for which τ becomes simply

(x_{0}, y_{0}) ↦^{τ} (x_{0}, x_{1}, x_{2}, …, y_{0}, y_{1}, y_{2}, …).

By theorem 4.1 of [5], with G = O (the origin of E), the functions x_{n} and y_{n} are regular except at O (the origin of E), so are of the form R(x_{0}) + y_{0}S(x_{0}) for some polynomials R, S ∈ k[x_{0}]. For p ≠ 2, −(y_{0}, y_{1}, …) = (−y_{0}, −y_{1}, …), and using τ(−P) = −τ(P), one can deduce that x_{n} ∈ k[x_{0}] and y_{n} = y_{0} · F_{n}(x_{0}), with F_{n} ∈ k[x_{0}]. For p = 2, a similar argument also gives us that x_{n} ∈ k[x_{0}], but y_{n} does not have to be of the form y_{0} · F_{n}.

Our first goal is to get good bounds for the degrees of these polynomials, or equivalently, for the order of poles of x_{n} and y_{n} at O. We prove Theorem 1.1. Let v =^{def} ord_{O}, i.e., v is the valuation on the function field K of E given by the order of vanishing of functions at O. Then, v(x_{n}) ≥ −((n + 2)p^{n} − np^{n−1}) and v(y_{n}) ≥ −((n + 3)p^{n} − np^{n−1}), for all n ≥ 0.

The case n = 1 was proved by Voloch and Walker in [5]. We will get the theorem 1.1 as a special case of the theorem 3.1 below.

2. Witt Vectors and Valuations

Let p be a prime, and for any non-negative integer n consider

W_{n}(X_{0}, …, X_{n}) =^{def} X_{0}^{p^{n}} + pX_{1}^{p^{n−1}} + ⋯ + p^{n−1}X_{n−1}^{p} + p^{n}X_{n},

the corresponding Witt polynomial. Then, there exist polynomials S_{n}, P_{n} ∈ ℤ[X_{0}, …, X_{n}, Y_{0}, …, Y_{n}] satisfying:

W_{n}(S_{0}, …, S_{n}) = W_{n}(X_{0}, …, X_{n}) + W_{n}(Y_{0}, …, Y_{n})  (2.1)

and

W_{n}(P_{0}, …, P_{n}) = W_{n}(X_{0}, …, X_{n}) · W_{n}(Y_{0}, …, Y_{n}).  (2.2)

(See [4].)

Thus, if s = (s_{0}, s_{1}, …) and t = (t_{0}, t_{1}, …) are Witt vectors, we have by definition

s + t =^{def} (S_{0}(s_{0}, t_{0}), S_{1}(s_{0}, s_{1}, t_{0}, t_{1}), …)

and

s · t =^{def} (P_{0}(s_{0}, t_{0}), P_{1}(s_{0}, s_{1}, t_{0}, t_{1}), …).

We may write, to simplify the notation,

S_{n}(s, t) =^{def} S_{n}(s_{0}, …, s_{n}, t_{0}, …, t_{n})

and

P_{n}(s, t) =^{def} P_{n}(s_{0}, …, s_{n}, t_{0}, …, t_{n}).

DEGREES OF THE ELLIPTIC TEICHMÜLLER LIFT 3

Now, let K be a field of characteristic p > 0, and let us consider W (K). (Note that although soon we will consider K as the function field of E, as in the theorem 1.1, for now K is any field of characteristic p.) Since the entries of our Witt vectors are in characteristic p, we can use the polynomials S̄_{n}, P̄_{n} ∈ 𝔽_{p}[X_{0}, …, X_{n}, Y_{0}, …, Y_{n}], that are the reductions of S_{n}, P_{n} modulo p, to give us the sum and product of Witt vectors.

We now introduce three useful technical lemmas.. Lemma 2.1. The monomials ∏ X_{i}^{a_{i}} ∏ Y_{j}^{b_{j}} (disregarding the coefficient) occurring in P̄_{n} satisfy

∑ a_{i} p^{i} = ∑ b_{j} p^{j} = p^{n} and ∑ i a_{i} p^{i} + ∑ j b_{j} p^{j} ≤ n p^{n}.

Moreover,

P̄_{n} = ∑_{i=0}^{n} X_{i}^{p^{n−i}}Y_{n−i}^{p^{i}} + Q̄_{n},

where Q̄_{n} ∈ 𝔽_{p}[X_{0}, …, X_{n−1}, Y_{0}, …, Y_{n−1}] and has its monomials (as above) satisfying ∑ i a_{i} p^{i} + ∑ j b_{j} p^{j} < n p^{n}.

Proof. We prove it by induction. The case n = 0 is trivial, since P̄_{0} = X_{0}Y_{0}. Now assume the lemma true for all t ≤ n − 1. We have:

P_{n} = \frac{1}{p^{n}} [(X_{0}^{p^{n}} + ⋯ + p^{n}X_{n})(Y_{0}^{p^{n}} + ⋯ + p^{n}Y_{n})− ; (P_{0}^{p^{n}} + ⋯ + p^{n−1}P_{n−1}^{p})] ; = (X_{0}^{p^{n}}Y_{n} + X_{1}^{p^{n−1}}Y_{n−1}^{p} + ⋯ + X_{n}Y_{0}^{p^{n}}) ; + \frac{1}{p}(X_{0}^{p^{n}}Y_{n−1}^{p} + ⋯ + X_{n−1}^{p}Y_{0}^{p^{n}}) ; ⋮ ; + \frac{1}{p^{n}} (X_{0}^{p^{n}}Y_{0}^{p^{n}}) − \frac{1}{p^{n}} P_{0}^{p^{n}} − ⋯ − \frac{1}{p}P_{n−1}^{p} ; + p (X_{1}^{p^{n−1}}Y_{n} + X_{2}^{p^{n−2}}(Y_{n−1}^{p} + pY_{n}) + …) .  (2.3)

First we observe that the above polynomial has its coefficients in ℤ. Also, the part that is a multiple of p doesn’t contribute to P̄_{n}, and so we can disregard that last line of the equation above.

For t = 0, …, n − 1, write P_{t} = P̃_{t} + pR_{t}, where we collected all the monomials of P_{t} that have coefficients divisible by p in pR_{t}. By the induction hypothesis, P̃_{t} also satisfy the lemma. So, now we look at the contribution of \frac{1}{p^{n−t}} P_{t}^{p^{n−t}} to P̄_{n}: that is given by the monomials of P̃_{t}^{p^{n−t}}, which have the form

∏ X_{i}^{∑_{r=1}^{p^{n−t}} a_{ir}} ∏ Y_{j}^{∑_{r=1}^{p^{n−t}} b_{jr}} ,

where the ∏ X_{i}^{a_{ir}} ∏ Y_{j}^{b_{jr}} are monomials of P̃_{t} for r = 1, …, p^{n−t}. So,

∑_{i} [∑_{r=1}^{p^{n−t}} a_{i_{r}} ] p^{i} = ∑_{r=1}^{p^{n−t}} [∑_{i} a_{i_{r}}p^{i}] = ∑_{r=1}^{p^{n−t}} p^{t} = p^{n},

(and the analogous for the b_{j_{r}} also holds) and

∑_{i}i [∑_{r=1}^{p^{n−t}} a_{i_{r}} ] p^{i} + ∑_{j} j [∑_{r=1}^{p^{n−t}} b_{j_{r}} ] p^{j}

= ∑_{r=1}^{p^{n−t}} [∑_{i} ia_{i_{r}}p^{i} + ∑_{j} jb_{j_{r}}p^{j}] ≤ tp^{n} < np^{n}.

Observing that the last line of the equation (2.3) won’t contribute to P̄_{n}, all the remaining terms are of the form X_{i}^{p^{n−i}}Y_{j}^{p^{n−j}}. Excluding the ones of the form X_{i}^{p^{n−i}}Y_{n−i}^{p^{i}}, the remaining are such that i + j < n, and the lemma follows.

□

Now, let v : K → ℝ ∪ {∞} be a valuation of field K. (In the applications below, we will choose K to be the function field of E/k and v to be the order of vanishing at a point P ∈ E(k̄).) For e ≥ 0, define:

U(e) =^{def} {s = (s_{0}, s_{1}, …) ∈ W (K)^{×} | v(s_{n}) ≥ p^{n}(v(s_{0}) − ne), ∀n > 0} .

(Note that W (K)^{×} = {s = (s_{0}, s_{1}, …) ∈ W (K) | s_{0} ≠ 0}.) Lemma 2.2. The set U(e) is a subgroup of W (K)^{×}.

Proof. Let s, t ∈ U(e). The (n + 1)-th coordinate of s t is given by P̄_{n}(s, t). By lemma 2.1, for each monomial of P̄_{n}(s, t) we have:

\begin{aligned}v (∏ s_{i}^{a_{i}} ∏ t_{j}^{b_{j}}) & = ∑ a_{i}v(s_{i}) + ∑ b_{j}v(t_{j}) \\ & ≥ ∑ a_{i}p^{i}(v(s_{0}) − ie) + ∑ b_{j}p^{j}(v(t_{0}) − je) \\ & ≥ p^{n}(v(s_{0}) + v(t_{0}) − ne). \end{aligned}  (2.4)

Therefore, v(P̄_{n}(s, t)) ≥ p^{n}(v(s_{0}t_{0}) − ne) for all n, i.e., s t ∈ U(e). (Note that since all elements of 𝔽_{p}^{×} are roots of unity, v is zero on all its elements, and we don’t have to worry about the coefficients of the monomials in P̄_{n}.)

We prove that t =^{def} s^{−1} ∈ U(e) by induction on the coordinate: assume that for all i < n we have v(t_{i}) ≥ p^{i}(v(t_{0}) − ie). We observe that:

P̄_{n}(s, t) = t_{n} s_{0}^{p^{n}} + ⋯ = 0

where no omitted term involves t_{n}. So, v(t_{n} s_{0}^{p^{n}}) is equal to the valuation of the omitted terms. But for those, we can use (2.4), and so

v(t_{n} s_{0}^{p^{n}}) ≥ p^{n}(v(s_{0}) + v(t_{0}) − ne),

and this gives us v(t_{n}) ≥ p^{n}(v(t_{0}) − ne).

□

Lemma 2.3. If v(s_{0}) = 1 and v(s_{n}) ≥ 1 for all n, then s ∈ U((p − 1)/p).

Proof. Just note that v(s_{n}) ≥ 1 ≥ p^{n}[1 − n(p − 1)/p] = np^{n−1} − (n − 1)p^{n}. □

3. Upper Bounds

Now let K denote the function field of E/k̄ and K be the function field of E over the field of fractions k of W (k̄). An element g ∈ K can be written as a quotient g_{1}/g_{2}, where g_{1}, g_{2} ∈ W (k̄)[x, y]. Let R be ring of functions g = g_{1}/g_{2} ∈ K (as above), such that g_{2} ≢ 0 (mod p). (Then R is the valuation ring of K with respect to the valuation associated to p). We can identify R with a subring of W (K) (via τ^{∗}). We can then write for every g ∈ R, g = (g_{0}, g_{1}, …) ∈ W (K), and if g is regular at τ(P), for P ∈ E(k̄), then g_{i} is regular at P for every i ≥ 0 and g(τ(P)) = (g_{0}(P), g_{1}(P), …).

Define, for P ∈ E(k̄),

U(P) =^{def} {g ∈ R^{×} | ord_{τ(P)}(g) = ord_{P} (g_{0})} ,

and

U_{0}(P) =^{def} {g ∈ U(P) | ord_{P} (g_{0}) = 0} .

Observe that clearly U(P) is a subgroup of R^{×} and U_{0}(P) is a subgroup of U(P). Theorem 3.1. Let g = (g_{0}, g_{1}, …) ∈ U(P). Then

ord_{P} (g_{n}) ≥ p^{n}(ord_{P} (g_{0}) − n) + np^{n−1}, for all n ≥ 0.

Proof. Let π ∈ U(P) be such that ord_{τ(P)}(π) = 1. (Note we can choose π as either (x − x(τ(P))), y or x/y.) By lemma 2.3, π ∈ U((p − 1)/p), now with v =^{def} ord_{P}. In the same way, π^{1−v(g_{0})}g ∈ U((p − 1)/p). Since U((p − 1)/p) is a group, g ∈ U((p − 1)/p).

□

Theorem 1.1 then follows, applying the previous theorem with P = O and g = x, y.

We observe that if ord_{τ(P)}(g) < 0, then theorem 3.1 gives us upper bounds for the order of the poles of the g_{n}’s, for all n ≥ 0. If ord_{τ(P)}(g) > 0, the theorem still gives us some information: it gives lower bounds for the order of the zeros for n < p(ord_{P} (g_{0}))/(p − 1).

6 LUÍS R. A. FINOTTI

4. Leading Coefficients

Our main goal in this section is to verify when we don’t have the equality in the upper bounds of theorem 1.1. But since the same techniques give stronger results, we will obtain these results first, and then get our main goal as a corollary.

As observed in the proof of the theorem 3.1, we can always take a uniformizer π_{0} at P that is a reduction of a uniformizer π at τ(P). Let π_{0} be such a uniformizer at P and let g ∈ U(P). Also, let the expansion of g_{n} in terms of π_{0} be

g_{n} = b_{n}(g)π_{0}^{p^{n}(ord_{P} (g_{0})−n)+np^{n−1}} + …,  (4.1)

where the omitted terms have higher powers of π_{0} (by theorem 3.1). We call b_{n}(g) ∈ k the n-th leading coefficient of g at P, relative to π_{0}.

Finally, define

Φ(g) =^{def} ∑_{n=0}^{∞} b_{n}(g)^{p^{−n}} T^{n}.

Theorem 4.1. The function Φ : U(P) → (k[[T]])^{×} is a group homomorphism.

Proof. Let g, h ∈ U(P). We must prove that Φ(gh) = Φ(g)Φ(h), i.e.,

b_{n}(gh) = ∑_{i=0}^{n} b_{i}(g)^{p^{n−i}} b_{n−i}(h)^{p^{i}}.

This is another application of lemma 2.1: taking valuations v =^{def} ord_{P} on the terms of P̄_{n}(g, h) (the (n + 1)-th coordinate of gh), the part with valuation p^{n}(v(g_{0} h_{0}) − n) + np^{n−1} comes from

∑_{i=0}^{n} g_{i}^{p^{n−i}}h_{n−i}^{p^{i}},

and thus, the n-th leading coefficient of gh is obtained by multiplying the leading coefficients of the terms in the sum.

□

Theorem 4.2. For g ∈ U_{0}(P), Φ(g) = g_{0}(P).

Proof. Just observe that p^{n}(ord_{P} (g_{0}) − n) + np^{n−1} < 0 for n ≥ 1, and it is zero for n = 0. □

Theorem 4.3. If A is the Hasse invariant of E relative to the invariant differential λ such that (dπ_{0}/λ)(P) = 1, then

Φ(π) = (1 + A^{−p^{−1}}T).

Proof. Since ord_{τ(P)}(π) = ord_{P} (π_{0}) = 1, then p^{n}(ord_{P} (π_{0}) − n) + np^{n−1} is equal to 1 for n = 0, 1, and it is negative for n > 1. So

Φ(π) = 1 + α^{p^{−1}}T,

DEGREES OF THE ELLIPTIC TEICHMÜLLER LIFT 7

where π_{1} = α π_{0} + . . .. Hence we need to prove that

\frac{dπ_{1}}{dπ_{0}} (P) = A^{−1}.  (4.2)

So, let u ∈ U(P) such that u dπ is an invariant differential (i.e., holomorphic) on E, with u_{0}(P) = 1. Thus, λ = u_{0} dπ_{0}.

Now, let ϕ be the lift of the Frobenius to E. Then, ϕ^{∗}(u^{σ} dπ^{σ})/p, where v^{σ}, for v ∈ R, is obtained by applying the Frobenius σ for Witt vectors on the coefficients of v, is a well defined homomorphic differential on E, and its reduction modulo p, say ω, depends only on u_{0} dπ_{0}. (See [3].) Thus,

ω = cu_{0} dπ_{0} = c λ,  (4.3)

for some c ∈ k.

If we apply the Cartier operator, we get

C(ω) = C (c λ) = c^{1/p} A^{1/p} λ.  (4.4)

On the other hand, by [1], we know that, for v ∈ R, the p-derivation

δv =^{def} \frac{v^{σ} ∘ ϕ − v^{p}}{p} ,  (4.5)

is such that the reduction modulo p of δ^{i}v is equal to v_{i}+B_{i}, where B_{i} is a polynomial in v_{0}, … v_{i−1} that we can compute explicitly. (We observe that this polynomial is zero for i = 1.) Therefore,

\frac{1}{p}ϕ^{∗}(u^{σ} dπ^{σ}) = (pδu + u^{p}) (d(δπ) + π^{p−1}dπ)

and, reducing modulo p, we deduce that

ω = u_{0}^{p}(dπ_{1} + π_{0}^{p−1}dπ_{0})  (4.6)

Applying the Cartier operator in this new expression for ω we get C(ω) = u_{0} dπ_{0} = λ, and comparing with (4.4), we get c = A^{−1}. So, comparing equations (4.3) and (4.6), we obtain (4.2).

□

Corollary 4.4. If g = cπ^{v}, with c ∈ U_{0}(P), then

b_{n}(g) = c_{0}(P)^{p^{n}} \binom{v}{n}A^{−np^{n−1}}.

(With A as in the statement of the theorem 4.3.)

Proof. We have

Φ(g) = Φ(c) Φ(π)^{v} = c_{0}(P) (1 + A^{−p^{−1}}T )^{v} .

Hence

b_{n}(g)^{p^{−n}} = c_{0}(P)\binom{v}{n}A^{−np^{−1}}.

□

Corollary 4.5. The inequality in theorem 3.1 is an equality unless \binom{ord_{P} (g_{0})}{n} ≡ 0 (mod p), in which case it is a strict inequality.

Proof. This a simple consequence of the previous corollary (and the definition of b_{n}(g)). Note that A ≠ 0 since our elliptic curve is ordinary. □

Corollary 4.6. Let v =^{def} ord_{O}. Then, v(x_{n}) > −((n + 2)p^{n} − np^{n−1}) if, and only if, p divides (n + 1), and v(y_{n}) > −((n + 3)p^{n} − np^{n−1}), if, and only if, p divides (n + 1)(n + 2)/2.

Proof. The valuation of x_{n} (resp. y_{n}) is larger than −((n+2)p^{n}−np^{n−1}) (resp. −((n+3)p^{n}−np^{n−1})) if, and only if, b_{n}(x) = 0 (resp. b_{n}(y) = 0), relative to the uniformizer π = x/y.

We observe that

x = (\frac{x}{y})^{−2} + … and y = (\frac{x}{y})^{−3} + …,

and hence, by corollary 4.4,

b_{n}(x) = \binom{−2}{n}A^{−np^{n−1}} = (−1)^{n}(n + 1)A^{−np^{n−1}}

and

b_{n}(y) = \binom{−3}{n}A^{−np^{n−1}} = (−1)^{n} \frac{(n + 1)(n + 2)}{2}A^{−np^{n−1}},

what gives the result.

□

Remark. Theorem 1.1 tells us that the degree of x_{n} as a polynomial in x_{0} is less than or equal to

r =^{def} [(n + 2)p^{n} − np^{n−1}] /2. Since

x_{0} = (\frac{x_{0}}{y_{0}} )^{−2} + … and y_{0} = (\frac{x_{0}}{y_{0}} )^{−3} + …,  (4.7)

one can see that

b_{n}(x) = (−1)^{n}(n + 1)A^{−np^{n−1}}

is also the coefficient of x_{0}^{r} in x_{n}. Also, if p ≠ 2 and we write y_{n} = y_{0}F_{n}, where F_{n} is a polynomial in x_{0}, then the degree of F_{n} as a polynomial in x_{0} is less than or equal to s =^{def} [(n + 3)p^{n} − np^{n−1} − 3] /2, and its coefficient of x_{0}^{s} is

b_{n}(y) = (−1)^{n} \frac{(n + 1)(n + 2)}{2}A^{−np^{n−1}}

(again, using (4.7)).

DEGREES OF THE ELLIPTIC TEICHMÜLLER LIFT 9

5. Reduction Modulo p^{3}

In the next section we will describe an algorithm to compute the reduction modulo p^{3} of the canonical lift and the elliptic Teichmüller map explicitly for p ≠ 2, 3. To make sure that our computation gives us the right answer, we introduce the following sufficient condition (true for all primes):

Proposition 5.1. Let k be a perfect field of characteristic p > 0. If E/W_{n+1}(k) is an elliptic curve with reduction E, and if we have a section τ over E\{O} of the reduction from G(E) to E in the category of k-schemes, given by

(x_{0}, y_{0}) ↦ (x, y) = ((x_{0}, …, x_{n}), (y_{0}, …, y_{n})),

where x/y is regular at O with x/y(O) = 0, then E is the canonical lift of E and τ is the elliptic Teichmüller lift.

Proof. The proof is just the last paragraph of the proof of proposition 4.2 in [5]. □

Note that in the general case, in contrast to what happens for the second coordinate (see proposition 4.2 in [5]), it is not enough that deg(x_{i}) ≤ (n+2)p^{n} −np^{n−1} and deg(y_{i}) ≤ (n+3)p^{n} −np^{n−1} instead of x/y(O) = 0: e.g., in characteristic 5, considering just the first three coordinates, the elliptic curve

y^{2} = x^{3} + x

has reduction y_{0}^{2} = x_{0}^{3} + x_{0}, and the map

ν(x_{0}, y_{0}) =^{def} ((x_{0}, 4x_{0}^{7} + x_{0}^{3}, 4x_{0}^{5} + 3x_{0}^{13} + 2x_{0}^{15} + 2x_{0}^{17} + x_{0}^{19} + 4x_{0}^{23}+

3x_{0}^{25} + x_{0}^{27} + 4x_{0}^{31} + 3x_{0}^{33} + 2x_{0}^{37}),

(y_{0}, y_{0}(x_{0}^{8} + 2x_{0}^{6} + 2x_{0}^{4} + x_{0}^{2} + 3),
y_{0}(x_{0}^{56} + 2x_{0}^{54} + x_{0}^{52} + 3x_{0}^{48} + 3x_{0}^{44} + 2x_{0}^{42} + x_{0}^{40} + 2x_{0}^{38} + x_{0}^{36} + 2x_{0}^{34} + 3x_{0}^{32} + 4x_{0}^{30}

+ x_{0}^{26} + 3x_{0}^{24} + x_{0}^{16} + x_{0}^{14} + x_{0}^{10} + 4x_{0}^{8} + 3x_{0}^{6} + 3x_{0}^{2} + 4))),

is a section of the reduction, but this map is not such that

ν^{∗}(x/y) = (x ∘ ν) / (y ∘ ν)

is regular at O, and therefore, this is not the elliptic Teichmüller lift. Using the techniques introduced later, we can compute the correct map:

τ(x_{0}, y_{0}) =^{def} ((x_{0}, 4x_{0}^{7} + x_{0}^{3}, 4x_{0}^{5} + 3x_{0}^{13} + 4x_{0}^{15} + 2x_{0}^{17} + x_{0}^{19} + 4x_{0}^{23} + x_{0}^{27} + 4x_{0}^{31}+

3x_{0}^{33} + x_{0}^{35} + 2x_{0}^{37} + 2x_{0}^{45}),

(y_{0}, y_{0}(x_{0}^{8} + 2x_{0}^{6} + 2x_{0}^{4} + x_{0}^{2} + 3),
y_{0}(4x_{0}^{56} + 3x_{0}^{54} + 4x_{0}^{52} + 3x_{0}^{48} + 3x_{0}^{44} + 2x_{0}^{42} + x_{0}^{40} + 2x_{0}^{38} + 2x_{0}^{32} + 4x_{0}^{30} + 4x_{0}^{26} + 4x_{0}^{24}

+ 3x_{0}^{22} + 4x_{0}^{14} + 4x_{0}^{12} + x_{0}^{10} + 4x_{0}^{8} + 4x_{0}^{6} + 2x_{0}^{4} + 4x_{0}^{2} + 4))).

We now try to find properties that will allow us to compute explicitly coordinates of the coefficients of the canonical lift and the elliptic Teichmüller. We first observe that a method to compute the second coordinates can be derived from results in [5]. So, we try to obtain the analogues of those results to deduce a way to compute the third coordinates.

From the proof of proposition 4.2 in [5], one can deduce:

\frac{dx_{1}}{dx_{0}} = A^{−1}y_{0}^{p−1} − x_{0}^{p−1},  (5.1)

for p ≠ 2, where A is the Hasse invariant of the curve associated to the invariant differential dx_{0}/y_{0} (from this point on, A will always denote this particular Hasse invariant). Following the same idea: Proposition 5.2. For p ≠ 2, we have

\frac{dx_{2}}{dx_{0}} = A^{−(p+1)}y_{0}^{p^{2}−1} − x_{0}^{p^{2}−1} − x_{1}^{p−1} \frac{dx_{1}}{dx_{0}} .

Proof. We consider the differential

\frac{1}{p}ϕ^{∗} (\frac{1}{p}ϕ^{∗} (\frac{dx}{y})) ,

where ϕ is the lift of the Frobenius. Its reduction modulo p, say ω, is of the form c dx_{0}/y_{0}, for some c ∈ k.

If we apply the Cartier operator, we get

C(ω) = C (c \frac{dx_{0}}{y_{0}} ) = c^{1/p} A^{1/p} \frac{dx_{0}}{y_{0}} .  (5.2)

On the other hand,

\frac{1}{p}ϕ^{∗} (\frac{1}{p}ϕ^{∗} (\frac{dx}{y})) = \frac{1}{p}ϕ^{∗} (\frac{d(δx) + x^{p−1}dx}{pδy + y^{p}})

= \frac{d(δ^{2}x) + (δx)^{p−1}d(δx) + (pδx + x^{p})^{p−1}(d(δx) + x^{p−1}dx)}{p(p δ^{2}y + (δy)^{p}) + (pδy + y^{p})^{p}}.

Since the universal polynomial B_{2}(x_{0}, x_{1}) = −x_{0}^{p(p−1)}x_{1}, the reduction of the differential above modulo p, that is again ω, is

\frac{d(x_{2} − x_{0}^{p(p−1)}x_{1}) + x_{1}^{p−1}dx_{1} + x_{0}^{p(p−1)}(dx_{1} + x_{0}^{p−1}dx_{0})}{y_{0}^{p^{2}}}

= \frac{dx_{2} + x_{1}^{p−1}dx_{1} + x_{0}^{p^{2}−1}dx_{0}}{y_{0}^{p^{2}}} ,

and computing the Cartier operator using this form of ω and using (5.1), we get

C(ω) = \frac{1}{y_{0}^{p}} (dx_{1} + x_{0}^{p−1}dx_{0}) = A^{−1} \frac{dx_{0}}{y_{0}} .  (5.3)

Comparing equations (5.2) and (5.3), we get that c = A^{−(p+1)}, and comparing the two forms for ω, we have

\frac{dx_{2}}{dx_{0}} = A^{−(p+1)}y_{0}^{p^{2}−1} − x_{0}^{p^{2}−1} − x_{1}^{p−1}(A^{−1}y_{0}^{p−1} − x_{0}^{p−1}).

□

Remark. We note that for characteristic 2, similar computations would give

\frac{dx_{1}}{dx_{0}} = \frac{dx_{2}}{dx_{0}} = 0.

Hence, the proposition above allows us to find x_{2}, except for finitely many terms of the form d_{n}x_{0}^{np}. (We can find the number of missing terms from the bounds for the degree.)

Now, we take a closer look at the quotient x/y up to the third coordinate. In this case we have:

\frac{x}{y} = (\frac{x_{0}}{y_{0}} , \frac{x_{1}}{y_{0}^{p}} − \frac{y_{1}x_{0}^{p}}{y_{0}^{2p}} , −\frac{x_{1}^{p}y_{1}^{p}}{y_{0}^{2p^{2}}} + \frac{x_{2}}{y_{0}^{p^{2}}} + \frac{x_{0}^{p^{2}}y_{1}^{2p}}{y_{0}^{3p^{2}}} − \frac{x_{0}^{p^{2}}y_{2}}{y_{0}^{2p^{2}}}

+\frac{1}{p} (\frac{x_{1}^{p}}{y_{0}^{p^{2}}} − \frac{y_{1}^{p}x_{0}^{p^{2}}}{y_{0}^{2p^{2}}} − (\frac{x_{1}}{y_{0}^{p}} − \frac{y_{1}x_{0}^{p}}{y_{0}^{2p}} )^{p})) .

(We have here a small notation problem, since we cannot divide by p. But notice that the polynomial

\frac{1}{p} (X^{p} − Y^{p} − (X − Y)^{p})

has integer coefficients, and we can substitute

X = \frac{x_{1}}{y_{0}^{p}} , Y = \frac{y_{1}x_{0}^{p}}{y_{0}^{2p}}

in this polynomial to obtain what we write as

\frac{1}{p} (\frac{x_{1}^{p}}{y_{0}^{p^{2}}} − \frac{y_{1}^{p}x_{0}^{p^{2}}}{y_{0}^{2p^{2}}} − (\frac{x_{1}}{y_{0}^{p}} − \frac{y_{1}x_{0}^{p}}{y_{0}^{2p}} )^{p})

12 LUÍS R. A. FINOTTI

in characteristic p. This abuse of notation will appear again below, but we hope that no confusion will arise from it.)

Looking at the orders in the third coordinate, we see that

\frac{1}{p} (\frac{x_{1}^{p}}{y_{0}^{p^{2}}} − \frac{y_{1}^{p}x_{0}^{p^{2}}}{y_{0}^{2p^{2}}} − (\frac{x_{1}}{y_{0}^{p}} − \frac{y_{1}x_{0}^{p}}{y_{0}^{2p}} )^{p})

has already positive order at O, and that, for p ≠ 2, 3, all the summands in

−\frac{x_{1}^{p}y_{1}^{p}}{y_{0}^{2p^{2}}} + \frac{x_{2}}{y_{0}^{p^{2}}} + \frac{x_{0}^{p^{2}}y_{1}^{2p}}{y_{0}^{3p^{2}}} − \frac{x_{0}^{p^{2}}y_{2}}{y_{0}^{2p^{2}}}  (5.4)

have the same order, namely −p^{2}+2p. (Note that the orders of x_{2} and y_{2} are precisely −4p^{2}+2p and −5p^{2} + 2p, as we may see from our analysis of the leading coefficients.) But since τ^{∗}(x/y)(O) = 0, those terms have to add up to have positive order.

So now we restrict ourselves to p ≠ 2, 3, and then we may assume that E is given by an equation of the form

E/k : y_{0}^{2} = x_{0}^{3} + a_{0}x_{0} + b_{0},  (5.5)

and that the canonical lift is given by

E/W (k) : y^{2} = x^{3} + ax + b,  (5.6)

Looking at the third coordinates of the expression of (5.6), we see

\frac{x_{0}^{p^{2}}y_{2}}{y_{0}^{2p^{2}}} = \frac{x_{0}^{p^{2}}}{2y_{0}^{3p^{2}}} (2y_{0}^{p^{2}}y_{2})

= \frac{x_{0}^{p^{2}}}{2y_{0}^{3p^{2}}} (3x_{0}^{2p^{2}}x_{2} + 3x_{0}^{p^{2}}x_{1}^{2p} − y_{1}^{2p} + …) ,

where the terms not shown have order greater than −7p^{2}, and so when multiplied by x_{0}^{p^{2}}/2y_{0}^{3p^{2}}, they give terms of positive order.

So, the part of (5.4) that has to add up to have positive order is

−\frac{x_{1}^{p}y_{1}^{p}}{y_{0}^{2p^{2}}} + \frac{x_{2}}{y_{0}^{p^{2}}} + \frac{x_{0}^{p^{2}}y_{1}^{2p}}{y_{0}^{3p^{2}}} − \frac{x_{0}^{p^{2}}}{2y_{0}^{3p^{2}}} (3x_{0}^{2p^{2}}x_{2} + 3x_{0}^{p^{2}}x_{1}^{2p} − y_{1}^{2p})

= −\frac{x_{1}^{p}y_{1}^{p}}{y_{0}^{2p^{2}}} + \frac{x_{2}}{y_{0}^{p^{2}}} + \frac{2^{−1}3x_{0}^{p^{2}}y_{1}^{2p}}{y_{0}^{3p^{2}}} − \frac{2^{−1}3x_{0}^{3p^{2}}x_{2}}{y_{0}^{3p^{2}}} − \frac{2^{−1}3x_{0}^{2p^{2}}x_{1}^{2p}}{y_{0}^{3p^{2}}} .

Looking at the second coordinate of (5.6), we get

y_{1} = \frac{2^{−1}3x_{0}^{2p}x_{1} + …}{y_{0}^{p}},

where all the terms on the numerator omitted are of order greater than −6p. Then, the part of x_{1}^{p}y_{1}^{p}/y_{0}^{2p^{2}} that has negative order is

\frac{2^{−1}3x_{0}^{2p^{2}}x_{1}^{2p}}{y_{0}^{3p^{2}}} ,

and the part of 2^{−1}3x_{0}^{p^{2}}y_{1}^{2p}/y_{0}^{3p^{2}} that has negative order is

\frac{8^{−1}27x_{0}^{5p^{2}}x_{1}^{2p}}{y_{0}^{5p^{2}}} .

So, the part of (5.4) that has to add up to have positive order is

−\frac{2^{−1}3x_{0}^{2p^{2}}x_{1}^{2p}}{y_{0}^{3p^{2}}} + \frac{x_{2}}{y_{0}^{p^{2}}} + \frac{8^{−1}27x_{0}^{5p^{2}}x_{1}^{2p}}{y_{0}^{5p^{2}}} − \frac{2^{−1}3x_{0}^{3p^{2}}x_{2}}{y_{0}^{3p^{2}}} − \frac{2^{−1}3x_{0}^{2p^{2}}x_{1}^{2p}}{y_{0}^{3p^{2}}}

= y_{0}^{−5p^{2}} [−\frac{3}{2}x_{0}^{2p^{2}}x_{1}^{2p}y_{0}^{2p^{2}} + x_{2} y_{0}^{4p^{2}} + \frac{27}{8}x_{0}^{5p^{2}}x_{1}^{2p} − \frac{3}{2}x_{0}^{3p^{2}}x_{2} y_{0}^{2p^{2}} − \frac{3}{2}x_{0}^{2p^{2}}x_{1}^{2p}y_{0}^{2p^{2}}] .

Using (5.5) and noticing that the part of the above expression that has to add up to have positive order is the part inside the brackets that has order at most −15p^{2}, we get that

y_{0}^{−5p^{2}} [−\frac{3}{2}x_{0}^{5p^{2}}x_{1}^{2p} + x_{2} x_{0}^{6p^{2}} + \frac{27}{8}x_{0}^{5p^{2}}x_{1}^{2p} − \frac{3}{2}x_{0}^{6p^{2}}x_{2} − \frac{3}{2}x_{0}^{5p^{2}}x_{1}^{2p}]

= \frac{x_{0}^{5p^{2}}}{y_{0}^{5p^{2}}} [\frac{3}{8}x_{1}^{2p} − \frac{1}{2}x_{0}^{p^{2}}x_{2}]

has to add up to have positive order, i.e., the parts of order smaller or equal to −5p^{2} inside the brackets above have to cancel out. Since those terms are polynomials in x_{0}, we get that the coefficient of x_{0}^{np} in x_{2} is 3/4 times the p^{th} power of the coefficient of x_{0}^{n+p} in x_{1}^{2}, for all n ≥ (3p+1)/2. (Note that by proposition 5.2, we knew that all the terms of degree, as a polynomial in x_{0}, higher than (3p^{2} − 1)/2 in x_{2} have to come from terms of the form d_{n}x_{0}^{np}.) Therefore, in the computation of the elliptic Teichmüller, some of the missing coefficients of x_{2} can be obtained from coefficients of x_{1}.

Thus, this analysis, along with proposition 5.1, allows us to deduce the following theorem: Theorem 5.3. If p ≠ 2, 3 and E/W_{3}(k) is an elliptic curve with reduction E, and if we have a section τ over E\{O} of the reduction from G(E) to E, in the category of k-schemes, given by

(x_{0}, y_{0}) ↦ (x, y) = ((x_{0}, x_{1}, x_{2}), (y_{0}, y_{1}, y_{2})),

such that E and τ are the canonical lift and the elliptic Teichmüller modulo p^{2}, then the same is true modulo p^{3} if, and only if, the degree of the polynomial (in x_{0}) [x_{0}^{p^{2}}x_{2} − 3/4 x_{1}^{2p}] is less than or equal to (5p^{2} − 1)/2. In fact, if this inequality holds, we must have the equality.

14 LUÍS R. A. FINOTTI

Proof. The only part not discussed before is the last statement. For it, it just suffices to observe that proposition 5.2 implies that the coefficient of x_{0}^{(3p^{2}−1)/2} in x_{2} is not zero (it is −2A^{−(p+1)}) and that x_{1}^{2p} just has powers of x_{0} multiples of p. □

6. The Algorithm

So now we see how to compute the canonical lifting and the elliptic Teichmüller explicitly, up to the third coordinate. In this whole section, we assume p ≠ 2, 3, and that E and E are given by equations (5.5) and (5.6).

First we compute x_{1} by integrating formally the formula (5.1), and we leave the constant term, say c_{0}, and the coefficient of the term in x_{0}^{p}, say c_{1}, as indeterminates. (Note that in this case, the Hasse invariant A is the coefficient of x_{0}^{p−1} of (x_{0}^{3} + a_{0}x_{0} + b_{0})^{(p−1)/2}.)

The second coordinate of the equation of E/W_{2}(k), namely

(y_{0}, y_{1})^{2} = (x_{0}, x_{1})^{3} + (a_{0}, a_{1})(x_{0}, x_{1}) + (b_{0}, b_{1}),

is given by:

\begin{aligned}2y_{0}^{p}y_{1} & = 3x_{0}^{2p}x_{1} + a_{0}^{p}x_{1} + a_{1}x_{0}^{p} + b_{1} \\ & + \frac{1}{p} (x_{0}^{3p} + a_{0}^{p}x_{0}^{p} + b_{0}^{p} − (x_{0}^{3} + a_{0}x_{0} + b_{0})^{p}) . \end{aligned}  (6.1)

Since y_{1} is y_{0} times a polynomial in x_{0}, equation (6.1) (keeping a_{1} and b_{1} as indeterminates) tells us that the division of polynomials (in x_{0})

\frac{3x_{0}^{2p}x_{1} + a_{0}^{p}x_{1} + a_{1}x_{0}^{p} + b_{1} + \frac{1}{p} (x_{0}^{3p} + a_{0}^{p}x_{0}^{p} + b_{0}^{p} − (x_{0}^{3} + a_{0}x_{0} + b_{0})^{p})}{2(x_{0}^{3} + a_{0}x_{0} + b_{0})^{(p+1)/2}}

must be exact. So we compute its remainder, which is a polynomial that has coefficients that depend on a_{1}, b_{1}, c_{0} and c_{1}. Forcing that remainder to be zero gives us a linear system on those indeterminates. Solving that system gives us the canonical lift (i.e., a_{1} and b_{1}) and x_{1} (i.e., c_{0} and c_{1}). And y_{1} is just y_{0} times the quotient of that exact division above.

We observe that the converse of the proposition 4.2 in [5] guarantees that the elliptic curve and map found are the right ones. Also, note that the solution of the system above does not have to be unique, since the canonical lift is only unique up to isomorphism.

The way to compute the third coordinate is analogous: we integrate formally the formula in proposition 5.2, and add the terms of degree in x_{0} greater than 3p^{2} from x_{1}^{2} as explained in the end of the previous section, and consider the coefficients in x_{0}^{np}, say d_{n}, for n from 0 to [(3p^{2} − 1)/2p], as indeterminates.

Then, we just look at the third coordinate of the expression of the elliptic curve, use the fact that y_{2} is also y_{0} times a polynomial in x_{0}, and force the corresponding remainder of the analogous division of polynomials to be zero. We get another system, that we solve to get the desired values

DEGREES OF THE ELLIPTIC TEICHMÜLLER LIFT 15

for the indeterminates, i.e., a_{2}, b_{2} and the d_{i}’s. Theorem 5.3 then guarantees that this gives the canonical lift and the elliptic Teichmüller. We used this method to compute the canonical lift (the first three coordinates) of

y_{0}^{2} = x_{0}^{3} + x_{0}

in characteristic p = 5 shown in section 5. In fact, we were able to compute, using that algorithm, the canonical lift for a generic ordinary elliptic curve in characteristic 5: if

y_{0}^{2} = x_{0}^{3} + a_{0}x_{0} + b_{0}

is such curve (a_{0} ≠ 0, since the curve is ordinary), then its canonical lift has

a_{1} = a_{0}^{2}b_{0}^{2} + \frac{b_{0}^{4}}{a_{0}} ,
a_{2} = 2a_{0}^{25} + a_{0}^{22}b_{0}^{2} + a_{0}^{19}b_{0}^{4} + 3a_{0}^{16}b_{0}^{6} + 2a_{0}^{13}b_{0}^{8} + a_{0}^{7}b_{0}^{12} + 4a_{0}b_{0}^{16}

+ \frac{3b_{0}^{18}}{a_{0}^{2}} + \frac{4b_{0}^{20}}{a_{0}^{5}} + \frac{4b_{0}^{22}}{a_{0}^{8}} + \frac{4b_{0}^{24}}{a_{0}^{11}} ,

b_{1} = 4a_{0}^{6}b_{0} + a_{0}^{3}b_{0}^{3} + b_{0}^{5},
b_{2} = a_{0}^{36}b_{0} + 4a_{0}^{33}b_{0}^{3} + 3a_{0}^{27}b_{0}^{7} + 4a_{0}^{21}b_{0}^{11} + 4a_{0}^{15}b_{0}^{15} + a_{0}^{12}b_{0}^{17} + 3a_{0}^{6}b_{0}^{21} + b_{0}^{25}.

(The polynomials for the the elliptic Teichmüller map are too long to be put in here.) We also were able to compute the generic cases for p = 7, 11, 13. A not too long particular case for p = 7 would be:

y_{0}^{2} = x_{0}^{3} + 1,

for which we have

a_{1} = 0,  a_{2} = 0,
b_{1} = 4,  b_{2} = 0,

and

x_{1} = 5x_{0} + 2x_{0}^{4} + 4x_{0}^{10},
x_{2} = 4x_{0} + 3x_{0}^{4} + 5x_{0}^{7} + 4x_{0}^{10} + 6x_{0}^{13} + 6x_{0}^{19} + 2x_{0}^{22} + 3x_{0}^{25} + x_{0}^{28}+
2x_{0}^{31} + 5x_{0}^{34} + 6x_{0}^{37} + 2x_{0}^{43} + 2x_{0}^{46} + 2x_{0}^{52} + 6x_{0}^{55} + 4x_{0}^{58} + 2x_{0}^{61}+
2x_{0}^{64} + 3x_{0}^{67} + 3x_{0}^{70} + 6x_{0}^{73} + 5x_{0}^{91},
y_{1} = y_{0}(2x_{0}^{3} + 3x_{0}^{6} + 4x_{0}^{9} + 6x_{0}^{12}),
y_{2} = y_{0}(2 + 6x_{0}^{3} + 3x_{0}^{6} + 6x_{0}^{9} + x_{0}^{12} + x_{0}^{15} + 2x_{0}^{18} + 5x_{0}^{21} + x_{0}^{24} + 3x_{0}^{30}+
x_{0}^{39} + 6x_{0}^{42} + x_{0}^{51} + x_{0}^{54} + 5x_{0}^{57} + 6x_{0}^{60} + 4x_{0}^{66} + 3x_{0}^{72} + 6x_{0}^{75}+
6x_{0}^{81} + 4x_{0}^{84} + 5x_{0}^{87} + 6x_{0}^{93} + 2x_{0}^{96} + 3x_{0}^{105} + 2x_{0}^{108} + 2x_{0}^{111} + 3x_{0}^{114}).

16 LUÍS R. A. FINOTTI

We first had implemented the algorithm using the software Mathematica and then, for convenience and speed, we switched to Magma, and the files are available at

http://www.ma.utexas.edu/users/finotti/can_lifts.html

where we also put the generic formulas for characteristic 2, 3, 5, 7, 11 and 13 and some more examples.

We also observe that the algorithm described also seems to “work” if you don’t introduce the terms of x_{2} from x_{1}^{2}, i.e., you use for x_{2} just the formal integral of the derivative in proposition 5.2, and the terms of the form d_{i}x_{0}^{ip} for i < (3p^{2} − 1)/2p. The algorithm will give you back a_{1}, a_{2}, x_{1}, x_{2}, y_{1} and y_{2}, where ν = ((x_{0}, x_{1}, x_{2}), (y_{0}, y_{1}, y_{2})) is a section of the reduction. But since ν^{∗}(x/y) is not regular at O, the curve obtained is in principle not necessarily the canonical lift, and the map is certainly not the elliptic Teichmüller. (This was how we obtained the “wrong lifting” ν in section 5.) But it seems that this lift may be used for some applications in coding theory, and it would be nicer than the canonical lift itself, since it has smaller degrees.

Acknowledgments: The author would like to thank J. F. Voloch for many of the ideas used in this paper, the referee, whose suggestions made this paper simpler and with stronger results, and CAPES (an agency of the Brazilian government) for financial support.

References

[1] A. Buium. Geometry of p-jets. Duke Math. Journal, 82:349–367, 1996.

[2] J. Lubin, J.-P. Serre, and J. Tate. Elliptic curves and formal groups. Proc. of Woods Hole summer institute in algebriac geometry, 1964. Unpublished. Available at http://www.ma.utexas.edu/users/voloch/lst.html.

[3] B. Mazur. Frobenius and the Hodge filtration, estimates. Ann. Math., 98:58–95, 1973.

[4] J.-P. Serre. Local Fields, volume 67 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1979.

[5] J. F. Voloch and J. L. Walker. Euclidean weights of codes from elliptic curves over rings. Trans. Amer. Math. Soc., 352(11):5063–5076 (electronic), 2000.

University of California, Department of Mathematics, Santa Barbara, CA – 93106

   E-mail address: finotti@math.ucsb.edu