Copyright

by

Luís Renato Abib Finotti

CANONICAL AND MINIMAL DEGREE LIFTINGS OF CURVES

by

LUíS RENATO ABIB FINOTTI, B.S., M.S.

DISSERTATION

Presented to the Faculty of the Graduate School of

The University of Texas at Austin in Partial Fulfillment of the Requirements for the Degree of

DOCTOR OF PHILOSOPHY

THE UNIVERSITY OF TEXAS AT AUSTIN August 2001

The Dissertation Committee for Luís Renato Abib Finotti Certifies that this is the approved version of the following dissertation:

CANONICAL AND MINIMAL DEGREE LIFTINGS OF CURVES

Committee: José Felipe Voloch, Supervisor Fernando Rodriguez-Villegas Judy Walker John Tate Jeffrey Vaaler I dedicate this work to the people that guided me through the different paths

of life and knowledge: my parents, my godparents and my family; my advisors, Felipe Voloch and Paulo A. Martin; the professors Daniel Levcovitz, John Tate, Fernando Rodriguez-Villegas and Jeff Vaaler; my dearest friend,

Heather Lehr.

Acknowledgments

The author would like to thank: Felipe Voloch, for his many suggestions and ideas and most valuable discussions, the referee from the Journal of Number Theory, who is currently analyzing a paper consisting of the basic ideas of chapter 2, whose comments gave stronger results with easier proofs, and CAPES, an institution of the Brazilian government, for financial support. Also, the author acknowledges the use of the softwares Magma and Mathematica for the computations mentioned in the text.

CANONICAL AND MINIMAL DEGREE LIFTINGS OF CURVES

Publication No.

Luís Renato Abib Finotti, Ph.D. The University of Texas at Austin, 2001

Supervisor: José Felipe Voloch

We first find upper bounds for the degrees of the coordinate functions of the elliptic Teichmüller lift of an ordinary elliptic curve over a perfect field of characteristic p > 0, giving exact conditions for the bound to be achieved. Also, we give an algorithm to compute the reduction modulo p^{3} of the canonical lift and the elliptic Teichmüller.

Next, motivated by coding theory, we look for lifts with degrees smaller than the degrees of the elliptic Teichmüller, finding again some upper bounds. We obtain some more precise information about those lifts modulo p^{3}, such as precise degrees and verify that we can lift of the Frobenius on the affine parts. We again show how to compute those lifts.

We then compute lifts of hyperelliptic curves with “small” degrees, give a lower bound for these degrees and conditions to achieve this bound. Finally, we give an example of a lift that is possibly a Mochizuki lift.

Table of Contents

Acknowledgments v

Abstract vi

Chapter 1. Introduction 1

Chapter 2. Degrees of the Elliptic Teichmüller Lift 5 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2.2 Witt Vectors and Valuations . . . . . . . . . . . . . . . . . . . . . 8 2.3 Upper Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.4 Leading Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.5 Reduction Modulo p^{3} . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.6 The Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Chapter 3. Minimal Degree Liftings of Elliptic Curves 33 3.1 Minimal Degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.2 Minimal Degrees Modulo p^{3} . . . . . . . . . . . . . . . . . . . . . 38 3.3 Minimal Lifts and the Frobenius . . . . . . . . . . . . . . . . . . . 43 3.4 Characteristic 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 3.5 Characteristic 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Chapter 4. Hyperelliptic Curves 63 4.1 Minimal Degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 4.2 Mochizuki Lifts . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

Bibliography 72

Vita 74

Chapter 1

Introduction

Let k be a perfect field of characteristic p > 0. An elliptic curve E over k is ordinary if E[p^{r}] = ℤ/p^{r}ℤ, for all r ≥ 1. If p ≠ 2 and E is given by an equation

E/k : y_{0}^{2} = f(x_{0}),

then E is ordinary if, and only if, the coefficient of x_{0}^{p−1} of f(x_{0})^{(p−1)/2}, say A, is non zero. (Theorem V.4.1 of [7].) This A is called the Hasse invariant of E. (Note that this implies that if σ is the Frobenius of k and E is ordinary, then E^{σ} is also ordinary. Also, our definition is different from the one given in [7]: if A ≠ 0, Silverman defines the Hasse invariant to be one.)

Ordinary elliptic curves have associated to them canonical lifts over the ring of Witt vectors W (k) (see [2]), meaning that for every ordinary E/k we have a unique (up to isomorphism) elliptic curve E/W (k), whose reduction modulo p is E, and a lift of points τ : E(k̄) → E(W (k̄)), called the elliptic Teichmüller lift, that is an injective group homomorphism. Moreover, if we also denote by σ the Frobenius of W (k), i.e.,

(a_{0}, a_{1}, a_{2}, …)^{σ} = (a_{0}^{σ}, a_{1}^{σ}, a_{2}^{σ}, …) = (a_{0}^{p}, a_{1}^{p}, a_{2}^{p}, …),

the canonical lift of E^{σ} is E^{σ} and if ϕ : E → E^{σ} is the Frobenius (for curves over k), there exists a lift of the Frobenius associated to τ, i.e., a map, that we will also denote by ϕ, from E to E^{σ} that makes the following diagram commute:

E(W (k̄)) ⟶^{ϕ} E^{σ}(W (k̄))
τ↑ ↑^{τ}
E(k̄) ⟶^{ϕ} E^{σ}(k̄)

(In fact, one has that ϕ(τ(P)) = τ(P)^{σ}, for all P ∈ E(k̄).)

Thus

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

We notice that we can always identify E/W (k) with its Greenberg transform G(E)/k, that is the infinite dimensional scheme given by the equations (over k) that appear in the coordinates of the equation of E. With this identification, τ becomes simply

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

(In this perspective, the map τ is defined over k.) As we shall soon verify (proposition 2.2), the x_{n}’s and y_{n}’s are polynomials in x_{0} and y_{0}.

Voloch and Walker in [9] applied this theory of canonical lifts of elliptic curves to construct error-correcting codes. In that paper, the degrees of x_{n} and y_{n} have some importance in estimating exponential sums, that tells us how “good” the obtained codes possibly are.

In chapter 2 we analyze the degrees of the elliptic Teichmüller lift, 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.

While trying to compute canonical lifts, I found some lifts modulo p^{3} that I thought to be the canonical, but as it turned out, those lifts were defined only on the affine part of the curve. On the other hand, those lifts had degrees smaller than the elliptic Teichmüller itself, and for the purposes of constructing codes, those might be even better, since smaller degrees would probably give better codes. In chapter 3, we discuss the existence of such lifts, analyze the possible degrees and check when the Frobenius will lift for the affine parts of the curves.

Voloch and Walker in [8] also used non elliptic curves to construct codes. On the other hand, if the genus of the curve is greater than 1, there is no lift of the Frobenius. (See [5].) We can construct lifts though, by purely algebraic techniques, which is done also in chapter 3 for hyperelliptic curves, since the techniques are the same as the ones we would use for elliptic curves. But in chapter 4 we look at the minimal possible degrees with a more theoretical approach, finding a lower bound for the degrees and a necessary condition to achieve that bound. (The condition also seems to be sufficient!)

Finally, on the last part of chapter 4, we try to find a Mochizuki lift of a curve of genus 2 in characteristic 3. In [4], Mochizuki proves that if we take off an appropriate set of (p − 1)(g − 2) points of an “ordinary” curve of genus g over a perfect field of characteristic p, we have the existence of a lift with “small” degrees for which the Frobenius lifts. Not many examples of such lifts are known, and in this last part of the chapter, we follow an idea given to the author by Voloch, to try to find an example. We computed the lift modulo p^{3} to verify that we can indeed lift the Frobenius, at least modulo p^{3}.

Degrees of the Elliptic Teichmüller Lift

2.1 Introduction

Let E/k be an ordinary elliptic curve over a perfect field k of characteristic p > 0. If p ≠ 2, 3, we will assume that such a curve is given by the equation:

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

Let

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

be its canonical lift, where a = (a_{0}, a_{1}, …) and b = (b_{0}, b_{1}, …).

We will consider only ordinary elliptic curves, and thus, for p = 2 and p = 3, we will always assume, by extending k if necessary, that E has the forms y_{0}^{2} + x_{0}y_{0} = x_{0}^{3} + a_{0} and y_{0}^{2} = x_{0}^{3} + x_{0}^{2} + a_{0} respectively, and we can assume that their canonical liftings have similar forms.

Let τ : E(k̄) → E(W (k̄)), denote 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}, …)),

or, identifying E/W (k) with its Greenberg transform G(E)/k, we can see τ as

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

For n > 1, theorem 4.1 in [9] tells us that deg x_{n} ∘ τ ≤ 2^{n+1}p^{n} and deg y_{n} ∘ τ ≤ 3 (2p)^{n}. In that same paper, better bounds were proved for n = 1 in proposition 4.2, and we use similar ideas to improve the bounds for n > 1. The goal is to prove:

Theorem 2.1. An upper bound for the degrees of the composition of τ with coordinate functions x_{n}’s is (n + 2)p^{n} − np^{n−1}, and with the y_{n}’s is (n + 3)p^{n} − np^{n−1}.

We may sometimes write, for simplicity, x_{n} meaning x_{n} ∘ τ, and the same for y_{n}, x and y.

Proposition 2.2. The function x_{n} is a polynomial in x_{0} and the function y_{n} is a polynomial in x_{0} and y_{0}. Moreover, if p ≠ 2, then y_{n} is a product of y_{0} with a polynomial in x_{0}.

Proof. We have that τ is a morphism from E to G(E). Thus, the maps have to be given by rational functions on x_{0} and y_{0}, over k. But using the equation defining E, we may write

x_{n} = \frac{F_{1}(x_{0}) + y_{0}G_{1}(x_{0})}{F_{2}(x_{0}) + y_{0}G_{2}(x_{0})},

where the F_{i}’s and G_{i}’s are polynomials.

Since τ is injective, the only point mapped to O, the origin of E, is O, the origin of E, i.e., τ maps the affine part of E into the affine part of E. So x_{n} cannot have poles in the affine part of E, and therefore the denominator of x_{n} can be taken to be constant. Thus, we can write

x_{n} = F (x_{0}) + y_{0}G(x_{0}),

with F and G polynomials. A similar argument shows that y_{n} is also a polynomial in x_{0} and y_{0}.

If (x_{0}, y_{0}) ∈ E(k̄), then its inverse with respect to the group law is (x_{0}, −y_{0}), for p ≠ 2, or (x_{0}, x_{0} + y_{0}), for p = 2. Since τ is a homomorphism, for p ≠ 2, we have that τ(x_{0}, −y_{0}) is the inverse of τ(x_{0}, y_{0}). But if

τ(x_{0}, y_{0}) = (x_{0}, x_{1}, …, x_{i}, …, y_{0}, y_{1}, …, y_{i}, …) = (x, y),

since the inverse of (x, y) is (x, −y), we get

τ(x_{0}, −y_{0}) = (x, −y) = (x_{0}, x_{1}, …, x_{i}, …, −y_{0}, −y_{1}, …, −y_{i}, …),

observing that for p ≠ 2,

−s = −(s_{0}, s_{1}, …) = (−s_{0}, −s_{1}, …).

Thus, x_{n} ∘ τ(x_{0}, y_{0}) = x_{n} ∘ τ(x_{0}, −y_{0}), i.e., F (x_{0}) + y_{0}G(x_{0}) = F (x_{0}) − y_{0}G(x_{0}), what gives us G = 0.

A similar argument with inverses gives the result for y_{n}.

If p = 2, we just need to observe that x_{n}(x_{0}, y_{0}) = x_{n}(x_{0}, x_{0} + y_{0}), i.e., x_{0}G(x_{0}) = 0, for any x_{0}. Thus G = 0 and x_{n} is a polynomial in x_{0}.

Thus, deg x_{n} = − ord_{O} x_{n} and deg y_{n} = − ord_{O} y_{n}. (Unless mentioned otherwise, we will use the word “degree” for degree as functions on E, not as polynomials.) Then, we may look at ord_{O} instead of deg to prove the theorem, i.e., it suffices to prove that ord_{O} x_{n} ≥ −(n + 2)p^{n} + np^{n−1} and ord_{O} y_{n} ≥ −(n + 3)p^{n} + np^{n−1}.

2.2 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 [6].)

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}).

We start with an elementary lemma that we shall use soon:

Lemma 2.3. Let v_{p} denote the p-adic valuation on ℤ. Then

v_{p} (\binom{p^{t}}{i}p^{i}) = t + i − v_{p}(i), for i = 1, …, p^{t}.

In particular,

\binom{p^{t}}{i}p^{i} ≡ 0 (mod p^{t+1}).

Proof. We prove it by induction on i. The case i = 1 is trivial. Now suppose true for some i < p^{t} and we prove for i + 1. We have

v_{p} (\binom{p^{t}}{i + 1}p^{i+1}) = v_{p} (\binom{p^{t}}{i}p^{i} \frac{p^{t} − i}{i + 1} p)
= t + i − v_{p}(i) + v_{p}(p^{t} − i) − v_{p}(i + 1) + 1
= t + (i + 1) − v_{p}(i + 1) + (v_{p}(p^{t} − i) − v_{p}(i))
= t + (i + 1) − v_{p}(i + 1),

and the last equality holds since i < p^{t}.

The last part is trivially true, since i > v_{p}(i).

Now, let K be a field of characteristic p > 0, and let us consider W (K). 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 four technical lemmas that will be useful in estimating degrees.

Lemma 2.4. 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^{t}} P_{n−t}^{p^{t}} to P̄_{n}: that is given by the monomials of P̃_{n−t}^{p^{t}} (by lemma 2.3), 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 K. For e ≥ 0, define:

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

(Note that W (K)^{×} = {s = (s_{0}, s_{1}, …) ∈ W (K) | s_{0} ≠ 0}.) Lemma 2.5. The set U(e) is a subgroup of W (K)^{×}.

Proof. Let s, t ∈ U(e). We have that the (n + 1)-th coordinate of s t is given by P̄_{n}(s, t). By lemma 2.4, 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 2.6. Let s, t ∈ U(e), and assume v(t_{0}) > v(s_{0}). Then v(S̄_{n}(s, t)) ≥ p^{n}(v(s_{0}) − ne). Moreover, if e > 0, the equality holds if, and only if, v(s_{n}) = p^{n}(v(s_{0}) − ne).

Proof. We prove by induction on n. For n = 0, it is trivial, since S̄_{0}(s, t) = s_{0} + t_{0}. We have

S_{n} =(X_{n} + Y_{n}) + \frac{1}{p}(X_{n−1}^{p} + Y_{n−1}^{p} − S_{n−1}^{p}) + ⋯ + \frac{1}{p^{n}} (X_{0}^{p^{n}} + Y_{0}^{p^{n}} − S_{0}^{p^{n}}),

and again we observe that this polynomial has integer coefficients. Notice that, when taking the valuation in S̄_{n}, we can again disregard the coefficients of the monomials ∏ X_{i}^{a_{i}} ∏ Y_{j}^{b_{j}} that appear in S_{n}. So, we assume (inductively) that every monomial of that form in S_{t}, for t = 0, …, (n − 1), is such that v(∏ s_{i}^{a_{i}} ∏ t_{j}^{b_{j}}) ≥ p^{t}(v(s_{0}) − te). So, the monomials of S_{t}^{p^{n−t}} are products of p^{n−t} monomials of S_{t}, and therefore they have valuation, when computed at s and t, greater than or equal to p^{n−t}p^{t}(v(s_{0}) − te) ≥ p^{n}(v(s_{0}) − ne), and equality never holds if e > 0. The remaining monomials of S_{n} are of the forms X_{i}^{p^{n−i}} and Y_{j}^{p^{n−j}}. The former gives valuations greater than or equal to p^{n}(v(s_{0}) − ie) ≥ p^{n}(v(s_{0}) − ne), and if e > 0, we have the equality only for i = n and v(s_{n}) = p^{n}(v(s_{0}) − ne). The latter gives valuation greater than or equal to p^{n}(v(t_{0}) − je) > p^{n}(v(s_{0}) − ne), which finishes the proof.

Lemma 2.7. 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}.

2.3 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 of 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) (i.e., R is the valuation ring of K with respect to the valuation associated to p). Then, for every g ∈ R, we have that g = (g_{0}, g_{1}, …) ∈ W (K), and if g is regular at τ(P), for P ∈ E(k̄), then g(τ(P)) = (g_{0}(P), g_{1}(P), …).

Observe that since ordO x/y = ordO y/x^{2} = 1, both x/y and y/x^{2} satisfy the conditions of lemma 2.7, and thus they are in the subgroup U((p − 1)/p) of R^{×}. This implies that x and y are also in that group, which proves theorem 2.1.

In fact the same idea can be used to prove a more general statement:

Theorem 2.8. Let g = (g_{0}, g_{1}, …) ∈ R^{×} such that ord_{P} g_{0} = ord_{τ(P)} g for some P ∈ E(k̄). Then

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

Proof. Let π ∈ R^{×} be such that ord_{τ(P)} π = 1, i.e., a uniformizer at τ(P). (Note we can choose π as either (x−x(τ(P))), y or x/y, and so ord_{P} π_{0} = 1.) By lemma 2.7, 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).

We observe that if ord_{τ(P)} g < 0, then the theorem above 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).

2.4 Leading Coefficients

Our main goal in this section is to verify when we don’t have the equality in the upper bounds of theorem 2.1. But since the same techniques give stronger results, we will obtain these results first, and then get our main goal as a corollary.

Let g be a function as in the theorem 2.8 and let π_{0} be a uniformizer at P. 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}} + …,  (2.5)

where the omitted terms have higher powers of π_{0} (by theorem 2.8). We call b_{n}(g) ∈ k the n-th leading coefficient of g at P, relative to π_{0}.

Let P ∈ E(k̄) and, for p ≠ 2, define

π =^{def} \begin{cases}x/y, & if P = O; \\ (x − x(τ(P))), & if 2P ≠ O; \\ y, & otherwise. \end{cases}  (2.6)

For p = 2, we define

π =^{def} \begin{cases}x/y, & if P = O; \\ (x − x(τ(P))), & if x_{0}(P) ≠ 0; \\ y, & if x_{0}(P) = 0; \end{cases}  (2.7)

So π is a uniformizer at τ(P). (Remember that τ is an injective homomorphism.) Also, if we write π = (π_{0}, π_{1}, …), we have that π_{0} is a uniformizer at P. Moreover, ord_{P} π_{n} ≥ 0 for all n ≥ 0. Then, write

π_{1} = α π_{0} + …

where the omitted terms have larger powers of π_{0}. We have then:

Theorem 2.9. Let π be as above and g as in theorem 2.8. Let v_{0} =^{def} ord_{P} g_{0} and write

g = cπ^{v_{0}} + … (c = (c_{0}, c_{1}, …) ∈ W (k))

where the omitted terms have larger powers of π. We then have

\frac{b_{n}(g)}{c_{0}^{p^{n}}} = \binom{v_{0}}{n}α^{np^{n−1}}.

Proof. Note that the theorem is trivially true for π itself: for n > 1, p^{n}(1 − n) + np^{n−1} < 0, and since ord_{P} π_{n} > 0, b_{n}(π) = 0 for n > 1.

Now, by induction, we prove that the theorem is true for π^{r}, with r > 0: suppose the theorem is true for π^{r−1}. Write

π^{r−1} = (u_{0}, u_{1}, …) and π^{r} = (v_{0}, v_{1}, …).

Then

v_{n} = ∑_{i=0}^{n} π_{i}^{p^{n−i}}u_{n−i}^{p^{i}} + …

where all the omitted terms have order larger than p^{n}(r−n)+np^{n−1}, by lemma 2.4. Thus,

b_{n}(π^{r}) = ∑_{i=0}^{n} b_{i}(π)^{p^{n−i}}b_{n−i}(π^{r−1})^{p^{i}} = [∑_{i=0}^{n} \binom{1}{i}\binom{r − 1}{n − i}] α^{np^{n−1}}

= [∑_{i=0}^{1} \binom{1}{i}\binom{r − 1}{n − i}] α^{np^{n−1}} = \binom{r}{n}α^{np^{n−1}}.

For r < 0, we prove by induction on n. Let

π^{−r} = (u_{0}, u_{1}, …) and π^{r} = (v_{0}, v_{1}, …).

Suppose true for i = 0, …, n − 1. Then,

∑_{i=0}^{n} u_{i}^{p^{n−i}}v_{n−i}^{p^{i}} + ⋯ = 0

where all the omitted terms have order larger than −np^{n} + np^{n−1}. So, the terms of order −np^{n} + np^{n−1} have to cancel, i.e.,

b_{0}(π^{−r})^{p^{n}}b_{n}(π^{r}) = − ∑_{i=1}^{n} b_{i}(π^{−r})^{p^{n−i}}b_{n−i}(π^{r})^{p^{i}}

= − ∑_{i=1}^{n} \binom{−r}{i}\binom{r}{n − i}α^{np^{n−1}} = \binom{−r}{n}α^{np^{n−1}},

what gives the result, since b_{0}(π^{−r})^{p^{n}} = 1.

For cπ^{r}, by looking at the expression (2.3) and using lemma 2.4, one can check that the leading term will just be multiplied by c_{0}^{p^{n}}: write π^{r} = (u_{0}, u_{1}, …) and consider the monomials in P̄_{n} of the form ∏ X_{i}^{a_{i}} ∏ Y_{j}^{b_{j}}. Then, defining v =^{def} ord_{P}, in the (n + 1)-th coordinate of cπ^{r} we have valuations of the form

v (∏ u_{i}^{a_{i}} ∏ c_{j}^{b_{j}}) = ∑ a_{i} v(u_{i})

≥ r [∑ a_{i} p^{i}] − \frac{p − 1}{p} [∑ a_{i} i p^{i}]

≥ rp^{n} − \frac{p − 1}{p}np^{n},

and it is clear that the equality can occur only if the monomial is X_{n}Y_{0}^{p^{n}}. Thus, the formula is true for cπ^{r}.

Finally we observe that if we write g as in the statement of the theorem, the higher powers of π will contribute to the (n + 1)-th coordinate with terms with powers of π_{0} higher than p^{n}(v_{0} − n) + np^{n−1} (as one can see from the proof of lemma 2.6), and so b_{n}(g) = b_{n}(cπ^{v_{0}}), and our formula holds for g.

Then, using the same notation as before, the above theorem tells us that the map

Φ : g ↦ \frac{1}{c_{0}} [∑_{n=0}^{∞} b_{n}(g)^{p^{−n}}T^{n}] ,

is such that Φ(g) = (1 + α^{p^{−1}}T)^{v_{0}}.

Now, assuming p ≠ 2, we compute the α’s for the three distinct choices of π. Assume that E is given by

E/k : y_{0}^{2} = f(x_{0}).

Let P = (r_{0}, s_{0}), such that 2P ≠ O. Then, we take the uniformizer at τ(P) = (r, s) given by (x − r). So,

π_{1} = x_{1} − r_{1} + \frac{x_{0}^{p} − r_{0}^{p} − (x_{0} − r_{0})^{p}}{p} (r_{1} =^{def} x_{1}(r_{0})).

(We here face a small notation problem: the above expression is in characteristic p, and so it would make no sense dividing by p. One should read an expression as the one above in the following way: first to consider a corresponding polynomial in characteristic zero, in this case t(x, r) =^{def} ((x^{p}−r^{p})−(x−r)^{p})/p ∈ ℤ[x, r], and then read:

\frac{x_{0}^{p} − r_{0}^{p} − (x_{0} − r_{0})^{p}}{p} =^{def} t(x_{0}, r_{0}).

Unfortunately this abuse of notation will appear several times, but we hope that no confusion will arise from it.)

Note that π_{1} is a polynomial in x_{0}, and therefore, α = dπ_{1}/dx_{0}|_{x_{0}=r_{0}}. But

\frac{dπ_{1}}{dx_{0}} = \frac{dx_{1}}{dx_{0}} + x_{0}^{p−1} − (x_{0} − r_{0})^{p−1}.

We observe that from the proof of proposition 4.2 in [9], one can deduce:

\frac{dx_{1}}{dx_{0}} = A^{−1}y_{0}^{p−1} − x_{0}^{p−1} = A^{−1}f(x_{0})^{(p−1)/2} − x_{0}^{p−1},  (2.8)

for p ≠ 2, where A is the Hasse invariant of the curve. (Note that A ≠ 0, since our elliptic curve is ordinary.)

Therefore,

\frac{dπ_{1}}{dx_{0}} = A^{−1}f(x_{0})^{(p−1)/2} − (x_{0} − r_{0})^{p−1},

and hence,

α = A^{−1}f(r_{0})^{(p−1)/2}.

(Note that since 2P ≠ O, α ≠ 0.)

If P = (r_{0}, s_{0}) is finite and 2P = O, then f(r_{0}) = 0, or s_{0} = 0. So, we take π = y, and if we write y_{1} = y_{0}F_{1}(x_{0}), where F_{1} is a polynomial, then we have α = F_{1}(r_{0}).

To try to find a more explicit expression for F_{1}(r_{0}), we look at the reduction modulo p^{2} of the equation of E. We assume here p ≠ 3. The case p = 3 can be easily calculated, and gives similar results. So for p ≠ 2, 3, we take f(x_{0}) = x_{0}^{3} + a_{0}x_{0} + b_{0}. Then, 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}  (2.9)

Writing y_{0}^{2} = f(x_{0}) and y_{1} = y_{0}F_{1}(x_{0}), we get:

\begin{aligned}2f(x_{0})^{(p+1)/2}F_{1}(x_{0}) & = 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}

Taking derivatives with respect to x_{0} of both sides, one gets

f′(x_{0}) F_{1}(x_{0}) + 2f(x_{0})\frac{dF_{1}}{dx_{0}} (x_{0}) = f′(x_{0}) [A^{−1}f′(x_{0})^{p−1} − f(x_{0})^{(p−1)/2}] .

(We observe that the formula above is also true for p = 3, as one can easily check.) So, α = F_{1}(r_{0}) = A^{−1}f′(r_{0})^{p−1}. Note that since the curve is nonsingular, we have that f′(r_{0}) ≠ 0.

Finally let P = O. Then, we take π = x/y, and we have

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

We observe that

x_{0} = (\frac{x_{0}}{y_{0}} )^{−2} + … and y_{0} = (\frac{x_{0}}{y_{0}} )^{−3} + ….  (2.10)

So, if x_{1} = βx_{0}^{(3p−1)/2} + … and y_{1} = y_{0}(γx_{0}^{2p−2} + …), then we have

π_{1} = (β − γ)\frac{x_{0}}{y_{0}} + ….

Since dx_{1}/dx_{0} = A^{−1}f(x_{0})^{(p−1)/2} − x_{0}^{p−1}, we have that β = −2A^{−1}. For p ≠ 3, by looking at the terms of highest degrees in the equation (2.9), one gets that γ = 2/3 β = −3A^{−1}. So, in this case, α = A^{−1}. For p = 3, we refer to [9], where x_{1} and y_{1} where computed. We get then α = A^{−1} = 1. (Remember that, by our choice of the form of f(x_{0}) for p = 3, we always have A = 1.)

For P = O and p = 2, we can follow the same idea above, and we also have formulas for x_{1} and y_{1} in [9]. Those give us α = 1.

Hence:

Corollary 2.10. We have deg x_{n} < (n+2)p^{n} −np^{n−1} if, and only if, p divides (n+1), and deg y_{n} < (n+3)p^{n}−np^{n−1} if, and only if, p divides (n+1)(n+2)/2.

Proof. First, we remember that for x_{n} and y_{n} we have deg = − ord_{O}. So the order of x_{n} (resp. y_{n}) is larger than (−2−n)p^{n}+np^{n−1} (resp. (−3−n)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.

But by theorem 2.9,

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

and

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

what gives the result, since α ≠ 0.

Remark. Note that since x_{n} is a polynomial, we have 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, and b_{n}(x) is also the coefficient of x_{0}^{r} in x_{n} (using equation (2.10)). Also, if 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) (again, using (2.10)).

2.5 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: Proposition 2.11. 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 of the reduction from G(E) to E, in the category of k-schemes over E\{O}, 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 [9].

Note that in the general case, in contrast to what happens for the second coordinate (see proposition 4.2 in [9]), 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, we have that 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))).

Now, we proceed trying 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 [9]. So, we try to obtain the analogues of those results to deduce a way to compute the third coordinates.

As we observed before,

\frac{dx_{1}}{dx_{0}} = A^{−1}y_{0}^{p−1} − x_{0}^{p−1}

for p ≠ 2. Following the same idea:

Proposition 2.12. 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}(A^{−1}y_{0}^{p−1} − x_{0}^{p−1}).

Proof. We consider the differential

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

where ϕ is the lift of the Frobenius. This is a well-defined holomorphic differential on E, and its reduction modulo p, say ω, depends only on dx/y. (See [3].) On one hand, since this differential is holomorphic, it has the form αdx/y, where α ∈ W (k), and so its reduction modulo p has the form α_{0}dx_{0}/y_{0}, where α_{0} is just the reduction of α.

If we apply the Cartier operator, we get

C(ω) = C (α_{0} \frac{dx_{0}}{y_{0}} ) = α_{0}^{1/p}A^{1/p} \frac{dx_{0}}{y_{0}} .  (2.11)

On the other hand, by [1], we know that the p-derivation

δu =^{def} \frac{u^{σ} ∘ ϕ − u^{p}}{p} ,  (2.12)

where u^{σ} is obtained by applying the Frobenius σ for Witt vectors on the coefficients of u, is such that the reduction modulo p of δ^{i}x is equal to x_{i} + P_{i}, where P_{i} is a polynomial in x_{0}, … x_{i−1} that we can compute explicitly. (Note that P_{i} is not the polynomial that defines the Witt product. From now on, we won’t use those anymore, and the P_{i}’s will always denote these universal polynomials arising from the reduction modulo p of δ^{i}x.) For i = 1 such a polynomial is zero, and for i = 2 (and p ≠ 2),

P_{2}(x_{0}, x_{1}) =^{def} −x_{0}^{p(p−1)}x_{1}.

Since u ∘ ϕ = pδu + u^{p}, we get

\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}}.

The reduction of such differential 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 (2.8), we get

C(ω) = \frac{1}{y_{0}^{p}} (dx_{1} + x_{0}^{p−1}dx_{0}) = A^{−1} \frac{dx_{0}}{y_{0}} .  (2.13)

Comparing equations (2.11) and (2.13), we get that α_{0} = 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})) .

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 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}}}  (2.14)

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.

Now for p ≠ 2, 3, looking at the third coordinates of the expression of our elliptic curve, we have that

\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 (2.14) 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 the equation of the elliptic curve, 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 (2.14) 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

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

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 2.12, 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 2.11, allows us to deduce the following theorem:

Theorem 2.13. If p ≠ 2, 3 and E/W_{3}(k) is an elliptic curve with reduction E, and if we have a section τ of the reduction from G(E) to E, in the category of k-schemes over E\{O}, 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, deg[x_{0}^{p^{2}}x_{2} − 3/4 x_{1}^{2p}] ≤ 5p^{2} − 1. In fact, if this inequality holds, we must have the equality.

Proof. The only part not discussed before is the last statement. For it, it just suffices to observe that proposition 2.12 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 p powers of x_{0}.

2.6 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.

First we compute x_{1} by integrating formally the formula (2.8), and we leave the constant term, say c_{0}, and the coefficient of the term in x_{0}^{p}, say c_{1}, as indeterminates.

Since y_{1} is y_{0} times a polynomial in x_{0}, equation (2.9) (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 [9] 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 2.12, 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 for the indeterminates, i.e., a_{2}, b_{2} and the d_{i}’s. Theorem 2.13 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 2.5. 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}).

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 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 2.12, 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 2.5.) 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. In the next chapter we will verify the existence of those lifts and estimate the degrees obtained.

Minimal Degree Liftings of Elliptic Curves

3.1 Minimal Degrees

In section 2.5, we exhibited a map ν between the affine part of an elliptic curve (over a field k of characteristic 5) and its canonical lift with degree of x_{2} smaller than the degree of the corresponding x_{2} of the elliptic Teichmüller. We now study the existence of such maps with smaller degrees in more detail. We will show that the affine part of the canonical lifting always has a unique section of the reduction modulo p^{n+1}, for p ≠ 2, with deg x_{i} ≤ (3p^{i} + 1) and deg y_{i} ≤ (i + 3)p^{i} − ip^{i−1}, for i = 1, …, n.

In this section we will be a little more generic and consider hyperelliptic curves for p ≠ 2. For simplicity, we shall write

C/k : y_{0}^{2} = f(x_{0}),

where f is a monic polynomial of degree d ≥ 3 with simple roots, and

C/W (k) : y^{2} = f(x),

where f is a monic polynomial that reduces to f modulo p. Looking at the (n + 1)-th coordinate of the equation of C we have:

2y_{0}^{p^{n}}y_{n} + ⋯ = f′(x_{0})^{p^{n}}x_{n} + …  (3.1)

(where f′(x_{0}) represents the formal derivative of f(x_{0})) where neither x_{n} nor y_{n} appear in any of the omitted terms.

We shall use “deg_{x_{0}}” to denote the degree as polynomial in x_{0}, to not be confused with “deg” that denotes degrees as functions on the curve. To make our exposition clearer, we introduce the following useful lemma:

Lemma 3.1. Let a, b, c ∈ k[x_{0}], with deg_{x_{0}}(a) = n, deg_{x_{0}}(b) = m, deg_{x_{0}}(c) = r. Also, let s =^{def} max{r, n + m − 1} and assume (a, b) = 1. Then, there exists a unique pair of polynomials u, v ∈ k[x_{0}] with deg_{x_{0}}(u) ≤ m − 1 and deg_{x_{0}}(v) ≤ s − m such that au + bv = c.

Proof. We follow the basic idea of lemma IV.1 in [8]. Let L(i) denote the vector space of polynomials in k[x_{0}] with degrees less than or equal to i. Consider the linear map

ψ : L(m − 1) ⊕ L(s − m) → L(s),

given by ψ(u, v) =^{def} au + bv. Since (a, b) = 1, ψ(u, v) = 0 if, and only if, u = bz and v = −az, for some polynomial z ∈ k[x_{0}]. But deg_{x_{0}}(u) ≤ m−1 < deg_{x_{0}}(b), what implies u = z = 0. Thus ker ψ = {0}. Now, since dim L(i) = i + 1, comparing dimensions, we have that ψ is an isomorphism, and since c ∈ L(s), there exist a unique pair u, v as in the statement.

We need to introduce some new notation here: for r ≥ 0, let

U_{r}(e) =^{def} {s = (s_{0}, s_{1}, …) ∈ W (K)^{×} | v(s_{n}) ≥ p^{n}(v(s_{0}) − ne), for 0 ≤ n ≤ r} .

Note that the proof of lemma 2.5 (that tells us that U(e) is a group) can be used to prove that U_{r}(e) is also a group.

Proposition 3.2. Given any curve

C/W (k) : y^{2} = f(x)

where f(x) is a monic polynomial of degree d with reduction f(x_{0}) modulo p such that (f(x_{0}), f′(x_{0})) = 1, then there exists a unique lift of the affine part of the curve

C/k : y_{0}^{2} = f(x_{0}),

to the affine part of C, say

ν = ((x_{0}, x̃_{1}, x̃_{2}, …), (y_{0}, ỹ_{1}, ỹ_{2}, …)),

where x̃_{n} is a polynomial in x_{0} and ỹ_{n} is y_{0} times a polynomial in x_{0}, with deg x̃_{n} ≤ d(p^{n} +1)−2 and deg ỹ_{n} ≤ [n(d−2)+d]p^{n} +[n(d−2)]p^{n−1}. Moreover, the degrees of the x̃_{n} are minimal.

Proof. We will consider groups of the form U_{r}((d − 2)(p + 1)/p) (with an appropriate r), where the valuation is defined by v =^{def} − deg. (Note that will be dealing only with polynomials in x_{0} and y_{0}.)

We prove the theorem by induction on n. The case n = 0 is trivial. Now suppose we have ν up to the n-th coordinate. We construct x̃_{n} and ỹ_{n} in the following way: observe that (x_{0}, x̃_{1}, …, x̃_{n−1}) and (y_{0}, ỹ_{1}, …, ỹ_{n−1}) are both in the group U_{n−1}((d − 2)(p + 1)/p), by the induction hypothesis. We now “find” x̃_{n} and ỹ_{n}. Write down the (n + 1)-th coordinate of the equation of C/W_{n+1}(k), regarding x̃_{n} and ỹ_{n} as unknowns to be found. Since we want ỹ_{n} to be of the form y_{0} F̃_{n}(x_{0}), with F̃_{n}(x_{0}) ∈ k[x_{0}], we need in fact to find F̃_{n}.

We then have

−f′(x_{0})^{p^{n}} x̃_{n} + 2f(x_{0})^{(p^{n}+1)/2}F̃_{n} = …,  (3.2)

where no omitted term has x̃_{n} nor ỹ_{n}.

Looking carefully at the proof of 2.5, we notice that we actually bound each summand that appears in the (n + 1)-th coordinate of product s · t, with s, t ∈ U(e), by p^{n}(v(s_{0}) + v(t_{0}) − ne). So, in particular, if s, t ∈ U_{n−1}(e), all the terms in the (n+1)-th coordinate of s·t that do not have s_{n} and t_{n} have to satisfy this bound, since they depend only on (s_{0}, …, s_{n−1}) and (t_{0}, …, t_{n−1}) (even though we would not have control over the terms with s_{n} and t_{n}).

Hence all the terms in (3.2) that do not involve x̃_{n} or F̃_{n} have degrees less than or equal to [n(d−2)+2d]p^{n}+[n(d−2)]p^{n−1} (since the highest degrees should come from x^{d} and y^{2}, by lemma 2.6).

Now let c denote the omitted terms of (3.2). (Note that c is a polynomial in x_{0}.) Let a =^{def} −f′(x_{0})^{p^{n}} and b =^{def} 2f(x_{0})^{(p^{n}+1)/2}, and by lemma 3.1, there are u and v polynomials in x_{0} such that au + bv = c, and deg u ≤ d(p^{n} + 1) − 2 and deg v ≤ [n(d − 2) + d]p^{n} + [n(d − 2)]p^{n−1} − d. So, we take x̃_{n} =^{def} u, and ỹ_{n} =^{def} y_{0} v.

The minimality comes from the uniqueness in the lemma. We cannot have a ˜ x̃_{n} with degree less than the degree of x̃_{n}, unless we allow deg ˜ ỹ_{n} > [n(d − 2) + d]p^{n} + [n(d − 2)]p^{n−1}. But in this case the degree of the left hand side of the equation

−f′(x_{0})^{p^{n}} ˜x̃_{n} + 2y_{0}^{p^{n}} ˜ỹ_{n} = …,

would have degree larger than the upper bound for the degree of the right hand side. Therefore, there can be no such pair ˜ x̃_{n}, ˜ ỹ_{n}.

We call the above ν the minimal lift of C to C (with respect to x). Therefore, although we may neglect to mention it, whenever we talk about minimal lifts, we will be considering the affine parts of the curves only.

One can also use the same approach to minimize the degrees of the y_{n} instead. In such case we get:

Proposition 3.3. With the same hypothesis and notation as proposition 3.2, and assuming p does not divide d−1, there exists a unique lift ν with deg ỹ_{n} ≤ 2(d − 1)p^{n} + (d − 2) and deg x̃_{n} ≤ [n(d − 2) + 2]p^{n} + n(d − 2)p^{n−1}. In this case, deg ỹ_{n} is minimal.

Proof. The proof follows the exact same idea as the proof of proposition 3.2: again we will work in U((d − 2)(p + 1)/p) and we just apply lemma 3.1 with a =^{def} 2f(x_{0})^{(p^{n}+1)/2}, b =^{def} −f′(x_{0})^{p^{n}}, and c as before.

The above propositions have obvious applications to elliptic curves, by taking d = 3. But by theorem 2.1, we can see that taking E ordinary and E its canonical lift, we can have deg x_{1} ≤ 3p − 1, deg y_{1} ≤ 4p − 1. This gives the motivation for the following proposition, with better bounds to ỹ_{n}:

Proposition 3.4. Assume again the same hypothesis and notation from proposition 3.2 and suppose further that we have a lift modulo p^{2}, say

ν = ((x_{0}, x̃_{1}), (y_{0}, ỹ_{1})),

such that deg x̃_{1} ≤ dp − (d − 2) and deg ỹ_{1} ≤ (2d − 2)p − (d − 2). Then, we can complete ν to

ν(x_{0}, y_{0}) = ((x_{0}, x̃_{1}, x̃_{2}, …), (y_{0}, ỹ_{1}, ỹ_{2}, …)),

with deg x̃_{n} ≤ d(p^{n} + 1) − 2 and deg ỹ_{n} ≤ [n(d − 2) + d]p^{n} − n(d − 2)p^{n−1} in a unique way.

Proof. The idea is that the restrictions on x̃_{1} and ỹ_{1} allows us to work on U((d − 2)(p − 1)/p) instead of U((d − 2)(p + 1)/p). Inductively, the term c (as in the proof of proposition 3.2) will have degree less than or equal to [n(d − 2) + 2d]p^{n} − n(d − 2)p^{n−1}, and we just apply lemma 3.1 again.

3.2 Minimal Degrees Modulo p^{3}

In this section we will consider elliptic curves only and p ≠ 2, 3.

Proposition 3.2 gives us upper bounds for the minimal degrees. We notice that our choices for E (here denoting any curve with reduction E) give different x̃_{2}’s, and one can ask which choice would give us the minimal possible degree for x̃_{2}, which we shall call the absolute minimal degree lift of E, and exactly what degree would x̃_{2} have. (Note that the absolute minimal lift is not necessarily unique.) In this section we try to answer these questions for the reduction modulo p^{3}, at least for an ordinary E.

First of all, we observe that modulo p^{2}, those answers are given by proposition 4.2 of [9]: the choice of curve that gives the minimal possible degree for x̃_{1} is the canonical lift itself, the minimal degree map is the elliptic Teichmüller and the degree of x̃_{1} (or x_{1} in this case) is exactly 3p − 1.

We remember here that we mentioned in section 2.6 that we could change the algorithm presented to compute the canonical lift to give a x̃_{2} with degree 3p^{2} − 1, with the same derivative as x_{2}, at least in all cases we tried. By proposition 3.2, this map has minimal degree, meaning that no other lift to the canonical lift E could have smaller degrees.

We start our analysis with a conjecture:

Conjecture 3.5. The division of polynomials

\frac{3}{4}x_{1}^{2} = [x_{0}^{p} f(x_{0})^{(p+1)/2}]q(x_{0}) + r(x_{0})

(where x_{1} comes from the elliptic Teichmüller and E is given by y_{0}^{2} = f(x_{0})) is such that deg_{x_{0}} r(x_{0}) ≤ (5p − 1)/2.

First, observe that the division algorithm tells us that deg_{x_{0}} r(x_{0}) ≤ (5p + 1)/2, i.e., the bound is one more than the one stated in the conjecture. To verify the statement of conjecture 3.5 we wrote a routine in Magma that checked that it is true for any ordinary E in characteristic p from 5 to 877. We checked by formally integrating dx_{1}/dx_{0} and adding c_{1}x_{0}^{p} + c_{0}, leaving c_{1} and c_{0} as independent variables, so that we did not have to compute them precisely. Unfortunately, we don’t have yet a proof.

We now show the implications of such conjecture in answering the question modulo p^{3}:

Theorem 3.6. Assume conjecture 3.5 is true. Then, the absolute minimal degree lift

ν = ((x_{0}, x_{1}, x̃_{2}), (y_{0}, y_{1}, ỹ_{2}))

is such that deg x̃_{2} = 3p^{2}−1 and E is the canonical lift (modulo p^{3}). Moreover, we have that

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

Proof. First we observe, as we mentioned above, that E modulo p^{2} has to be the canonical lift, and x_{1} and y_{1} come from the elliptic Teichmüller.

We now prove that we always have such a lift to the canonical lift. We will actually give a way to construct the absolute minimal degree lift: if

τ = ((x_{0}, x_{1}, x_{2}), (y_{0}, y_{1}, y_{2}))

is the elliptic Teichmüller lift, we compute, using the division algorithm,

x_{2} = f(x_{0})^{(p^{2}+1)/2} q_{1}(x_{0}) + r_{1}(x_{0}) (deg_{x_{0}} r_{1} ≤ (3p^{2} + 1)/2).

Now define x̃_{2} =^{def} r_{1}(x_{0}) and ỹ_{2} =^{def} y_{2} − y_{0}[f′(x_{0})^{p^{2}}q_{1}(x_{0})]/2. Then, we have

2y_{0}^{p^{2}} ỹ_{2} − f′(x_{0})^{p^{2}} x̃_{2} = 2y_{0}^{p^{2}}y_{2} − f′(x_{0})^{p^{2}}x_{2},

and thus, by equation (3.1),

ν =^{def} ((x_{0}, x_{1}, x̃_{2}), (y_{0}, y_{1}, ỹ_{2}))

is another lift from E to its canonical lift, and since deg x̃_{2} ≤ (3p^{2} + 1), by proposition 3.2, it is the minimal lift. We now have to prove that deg x̃_{2} = 3p^{2}−1: let d(x_{0}) =^{def} x_{0}^{p^{2}}x_{2}−3/4x_{1}^{2p}. So, theorem 2.13 tells us that deg_{x_{0}} d(x_{0}) = (5p^{2} − 1)/2. We can then write

\begin{aligned}x_{0}^{p^{2}}x_{2} & = x_{0}^{p^{2}}f(x_{0})^{(p^{2}+p)/2}q(x_{0})^{p} + [r(x_{0})^{p} + d(x_{0})] \\ & = x_{0}^{p^{2}} f(x_{0})^{(p^{2}+1)/2} [f(x_{0})^{(p−1)/2}q(x_{0})^{p}] + [r(x_{0})^{p} + d(x_{0})], \end{aligned}  (3.3)

with deg_{x_{0}}[r(x_{0})^{p} + d(x_{0})] = (5p^{2} − 1)/2 (since we are assuming the conjecture to be true), and thus it is the remainder of the division of x_{0}^{p^{2}}x_{2} by x_{0}^{p^{2}}f(x_{0})^{(p^{2}+1)/2}. We see then that x_{0}^{p^{2}} divides this remainder and x̃_{2} = r_{1}(x_{0}) = [r(x_{0})^{p} + d(x_{0})]/x_{0}^{p^{2}}, which implies that deg_{x_{0}} x̃_{2} = (3p^{2} − 1)/2.

After we prove that this is indeed the absolute minimal lift of E, the last part of the statement of the theorem is just a consequence of equation (3.3): we have

\begin{aligned}x_{2} & = f(x_{0})^{(p^{2}+p)/2}q(x_{0})^{p} + \frac{r(x_{0})^{p} + d(x_{0})}{x_{0}^{p^{2}}} \\ & = f(x_{0})^{(p^{2}+p)/2}q(x_{0})^{p} + x̃_{2}, \end{aligned}

and taking derivatives gives us the result.

Now, we prove that this indeed gives us the absolute minimal degree lift: assume we have some lift of E to some curve E (not necessarily the canonical lift) with deg x̃_{2} ≤ 3p^{2} − 1. Let

˜x̃_{2} =^{def} x̃_{2} + [q(x_{0})f(x_{0})^{(p+1)/2}]^{p}

(with the q(x_{0}) from the conjecture) and

˜ỹ_{2} =^{def} ỹ_{2} + \frac{y_{0}}{2} [f′(x_{0})^{p^{2}}f(x_{0})^{(p−1)/2}q(x_{0})^{p}].

Then,

ν̃ : (x_{0}, y_{0}) ↦ ((x_{0}, x_{1}, ˜ x̃_{2}), (y_{0}, y_{1}, ˜ ỹ_{2})),

is another lift, since

2y_{0}^{p^{2}} ˜ỹ_{2} − f′(x_{0})^{p^{2}} ˜x̃_{2} = 2y_{0}^{p^{2}} ỹ_{2} − f′(x_{0})^{p^{2}} x̃_{2}.

But then, by hypothesis,

x_{0}^{p^{2}} ˜x̃_{2} − \frac{3}{4}x_{1}^{2p} = x_{0}^{p^{2}} x̃_{2} + [x_{0}^{p} q(x_{0}) f(x_{0})^{(p+1)/2}]^{p} − \frac{3}{4}x_{1}^{2p} = x_{0}^{p^{2}} x̃_{2} − r(x_{0})^{p}

has degree, as a polynomial in x_{0}, less than or equal to (5p^{2} − 1)/2. Theorem 2.13 then tells us that E is the canonical lift modulo p^{3}, ˜ x̃_{2} and ˜ ỹ_{2} are x_{2} and y_{2} from the elliptic Teichmüller, and deg_{x_{0}}[x_{0}^{p^{2}} x̃_{2} − r(x_{0})] = (5p^{2} − 1)/2, what implies deg x̃_{2} = 3p^{2} − 1.

Thus, if the conjecture is true, the minimal lift among all choices of curves E occurs for E equal to the canonical lift, and with this minimal degree of x̃_{2} exactly 3p^{2} − 1.

Theorem 3.6 justifies why our modified algorithm (when we did not introduce the terms from x_{1}^{2p} in the computation of x_{2}) described in the end of the section 2.6 seems to work. That modified algorithm computes the absolute minimal degree lift and the canonical lift. Also, note that we can verify the existence of such lift with deg x̃_{2} = 3p^{2} − 1 before we really have to compute x_{2}, since its existence depends only on the conjecture, that deals only with x_{1} (which can be computed fairly fast). Using the modified algorithm, we can then compute x̃_{2} and ỹ_{2} without computing x_{2} and y_{2}. Our function written for Magma gives you an option to compute this absolute minimal degrees lift instead of the Teichmüller is this exact way. Note that the curve obtained by this algorithm is then necessarily the canonical lift.

3.3 Minimal Lifts and the Frobenius

Having a lift from E to E/W_{2}(k) is equivalent to having a lift of the Frobenius (see [1]) in the affine part of E. In fact, if the lift is given by

(x_{0}, y_{0}) ↦ ((x_{0}, x_{1}), (y_{0}, y_{1})),

one can define ϕ(x, y) =^{def} (x^{p} + px_{1}, y^{p} + py_{1}), where x_{1} is any polynomial in W_{2}(k)[x] with reduction x_{1} and y_{1} is any polynomial in W_{2}(k)[x, y] with reduction y_{1}.

On the other hand, having a lift from E to E/W_{3}(k) just guarantees a lift of ϕ^{2}:

ϕ^{2}(x, y) =^{def} (x^{p^{2}} + px_{1}^{p} + p^{2}(x_{2} − x^{p(p−1)}x_{1}), (y^{p^{2}} + py_{1}^{p} + p^{2}(y_{2} − y^{p(p−1)}y_{1}))).

Of course, the canonical lift always has a lift of ϕ associated to τ. So, one could ask if the Frobenius also lifts (at least for the the affine parts) for these minimal lifts, making the diagram

\begin{matrix}E(W_{3}(k̄)) & ⟶^{ϕ} & E^{σ}(W_{3}(k̄)) \\ ^{ν}↑ & & ↑^{ν} \\ E(k̄) & ⟶^{ϕ} & E^{σ}(k̄) \end{matrix}  (3.4)

commute. (Here σ represents the Frobenius in k and W (k).) Note that in the case where E is the canonical lift of E (and therefore E^{σ} is the canonical lift of E^{σ}), we cannot use the lift of the Frobenius associated to the elliptic Teichmüller, since the diagram would not commute.

Lemma 3.7. Let P (X, Y) be a polynomial over a field of characteristic zero.

Then

\begin{matrix}P (X_{0} + p & X_{1}, Y_{0} + pY_{1}) \\ ≡ P (X_{0}, Y_{0}) + p (\frac{∂P}{∂X} (X_{0}, Y_{0})X_{1} + \frac{∂P}{∂Y} (X_{0}, Y_{0})Y_{1}) (mod p^{2}). \end{matrix}

Proof. This is an easy application of Taylor’s formula for P.

Proposition 3.8. Assume conjecture 3.5 is true. Let ν be the minimal lift from affine part of E (ordinary) to the affine part of E/W_{3}(k), such that, modulo p^{2}, ν gives us the Teichmüller. We have a lift of the Frobenius (for the affine parts) that make the diagram (3.4) commute if, and only if, deg x̃_{2} = 3p^{2} − 1 (and then ν is the absolute minimal degree lift and E is the canonical lift).

Proof. Note that the minimality of ν implies that E is the canonical lift modulo p^{2}. Now assume that we have a lift of ϕ associated to ν:

ϕ(x, y) = (x^{p} + px_{1} + p^{2}P , y^{p} + py_{1} + p^{2}Q).

Let δ be the p-derivation associated to ϕ (as in equation (2.12)). We have

δx = x_{1} + pP

and, using lemma 3.7,

\begin{aligned}δ^{2}x & = \frac{(x_{1} + pP)^{σ} ∘ ϕ − (x_{1} + pP)^{p}}{p} \\ & = \frac{x_{1}^{σ}(x^{p}) − x_{1}^{p}}{p} + \frac{dx_{1}}{dx}^{σ}(x^{p}) · x_{1} + P^{σ}(x^{p}) + p · (…). \end{aligned}  (3.5)

But, by [1], we have that the reduction modulo p of δ^{2}x must be equal to x̃_{2} − x_{0}^{p(p−1)}x_{1}. Taking derivatives, the reduction modulo p of P^{σ}(x^{p}) vanishes (it is a p-power). So all that is left after taking derivative does not depend on P, it depends only on x_{1}, and then, should give the same result as if we had used the lift of the Frobenius given by the elliptic Teichmüller (that maybe takes E to an E′ ≠ E, but necessarily E′ ≡ E (mod p^{2})). Therefore

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

Since deg x̃_{2} ≤ 3p^{2} + 1 and it has the same derivative as x_{2}, we have deg x̃_{2} = 3p^{2} − 1.

Conversely, assume ν is such that deg x̃_{2} = 3p^{2}−1. By theorem 3.6, the curve E is the canonical lift of E and dx_{2}/dx_{0} = dx̃_{2}/dx_{0}. Hence, Δx =^{def} x_{2}−x̃_{2} is a p-power. This implies that Δy =^{def} y_{2} − ỹ_{2} can also be written as a p-power, since

2y_{0}^{p^{2}}Δy = f′(x_{0})^{p^{2}}Δx.

Let

ϕ_{1}(x, y) = (x^{p} + px_{1} + p^{2}P_{1}, y^{p} + py_{1} + p^{2}Q_{1})

be the Frobenius associated to τ. Define now P_{2} as P_{1} minus a lift of (Δx)^{1/p}, that we shall call ΔP, and Q_{2} as Q_{1} minus a lift of (Δy)^{1/p}, that we shall call ΔQ. We claim that in this case, the ϕ_{2} as defined as

ϕ_{2}(x, y) = (x^{p} + px_{1} + p^{2}P_{2}, y^{p} + py_{1} + p^{2}Q_{2}),

is a lift of the Frobenius associated to ν. First, we check that it is well defined:

let 𝒳 =^{def} x^{p} + px_{1} + p^{2}P_{1} and 𝒴 =^{def} y^{p} + py_{1} + p^{2}Q_{1}. Then

𝒴^{2} ≡ f^{σ}(𝒳) (mod p^{3})

and we need to prove

(𝒴 − p^{2}ΔQ)^{2} ≡ f^{σ}(𝒳 − p^{2}ΔP) (mod p^{3}).

But, applying lemma 3.7, that is equivalent to

2𝒴ΔQ ≡ (f′)^{σ}(𝒳) · ΔP (mod p),

or

2 y_{0}^{p} (Δy)^{1/p} = f′(x_{0})^{p} (Δx)^{1/p},

that we can see is true, by raising both sides of the equation to the p-th power.

Clearly ϕ_{2} is a lift of the Frobenius, and the fact that it makes the diagram (3.4) commute is equivalent to the fact that δ_{2}x = x̃_{2} − x_{0}^{p(p−1)}x_{1} and δ_{2}y = ỹ_{2} − y_{0}^{p(p−1)}y_{1}, where δ_{2} is the p-derivation associated to ϕ_{2}. This can be easily checked by noticing that the reduction of

\begin{aligned}δ_{i}^{2}x & = \frac{(x_{1} + pP_{i})^{σ} ∘ ϕ_{i} − (x_{1} + pP_{i})^{p}}{p} \\ & = \frac{x_{1}^{σ}(x^{p}) − x_{1}^{p}}{p} + \frac{dx_{1}}{dx}^{σ}(x^{p}) · x_{1} + P _{i}^{σ}(x^{p}) + p · (…). \end{aligned}  (3.6)

for i = 2 differs from the reduction for i = 1 by Δx (that is, the p-power of the reduction of ΔP), and the analogue will hold for δ_{2}y.

The above proposition then says that conjecture 3.5 implies that the minimal degree lift from the affine part of E to the affine part of its canonical lift has a lift of the Frobenius associated to it.

Also, the proof the proposition also gives us:

Proposition 3.9. Let C be the hyperelliptic curve

C/k : y_{0}^{2} = f(x_{0}),

where f is a monic polynomial of degree d ≥ 3 with simple roots, and suppose that we have a lift of C

C/W_{3}(k) : y^{2} = f(x).

for which the Frobenius lift in the affine part of C. Let

ν =^{def} ((x_{0}, x_{1}, x_{2}), (y_{0}, y_{1}, y_{2}))

be the lift of points, and also assume that x_{i} is a polynomial in x_{0}. Then,

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

In particular, if dx_{1}/dx_{0} = λ y_{0}^{p−1} − x_{0}^{p−1} for some λ ∈ k^{∗}, then

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

Proof. Let x_{1} and y_{1} be lifts of x_{1} and y_{1} to W_{3}(k)[x] and W_{3}(k)[x, y] respectively. Then the lift of the Frobenius has the form

ϕ(x, y) = (x^{p} + px_{1} + p^{2}P , y^{p} + py_{1} + p^{2}Q),

for some polynomials P and Q. As in the proof of proposition 3.8, formula (3.5) also holds in this case. The reduction of this formula modulo p is equal to x_{2} − x_{0}^{p(p−1)}, and taking derivatives of both sides of this equation gives the result.

3.4 Characteristic 2

We now compute the canonical lifting and the elliptic Teichmüller modulo higher powers of 2 of an elliptic curve in characteristic 2, where our previous analysis does not apply. So, throughout this section we are going to fix p = 2.

An ordinary elliptic curve in characteristic 2 over an algebraic closed field (or over some finite extension of our original ground field k) can always be put in the form

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

and so, we consider the canonical lift having the same form:

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

Voloch and Walker in [9] computed the reduction modulo 4, obtaining

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

(Notice that for p = 2, y_{n} is not necessarily y_{0} times a polynomial in x_{0}.)

To compute the reduction modulo higher powers of 2, we first computed the derivative of x_{n}.

Theorem 3.10. For p = 2 and n ≥ 1, we have

\frac{dx_{n}}{dx_{0}} = 0.

We are going to break the proof in small lemmas. But first, we introduce the notation

ℱ =^{def} \frac{1}{p}ϕ^{∗},

where ϕ is the lift of the Frobenius. Also, we denote by δ the 2-derivation associated to ϕ.

Lemma 3.11. We have that

ℱ^{n}(dx) = d(δ^{n}x) + x^{2^{n}−1}dx + …,

where every omitted term that contains dx is a multiple of 2 and no other term contains d(δ^{n}x).

Proof. We prove it by induction on n. For n = 1, we have:

ℱ(dx) = \frac{d(x ∘ ϕ)}{2} = \frac{d(x^{2} + 2δx)}{2} = x dx + d(δx).

Suppose now the lemma true for n. We prove if for n + 1:

ℱ^{n+1}(dx) = ℱ(ℱ^{n}(dx)) = ℱ(d(δ^{n}x) + x^{2^{n}−1}dx + …)
= (δ^{n}x) d(δ^{n}x) + d(δ^{n+1}x)
+ (x^{2} + 2δx)^{2^{n}−1} (x dx + d(δx)) + …
= d(δ^{n+1}x) + x^{2^{n+1}−2}(x dx + d(δx)) + …
= d(δ^{n+1}x) + x^{2^{n+1}−1}dx + …,

and the omitted terms do not have a term with dx that is not a multiple of 2 or a term with d(δ^{n+1}x), for ℱ does not introduce denominators, and thus takes multiples of 2 to multiples of 2, and ℱ(d(δ^{i}x)) gives just terms with d(δ^{i}x) and d(δ^{i+1}x). Lemma 3.12. The reduction modulo 2 of d(δ^{n}x) can be written as

dx_{n} + ∑_{i=1}^{n−1} \frac{∂P_{n}}{∂x_{i}} dx_{i}.

where P_{n} ∈ 𝔽_{2}[x_{0}, …, x_{n−1}].

Proof. We recall that the reduction modulo 2 of δ^{n}x is x_{n} + P_{n}(x_{0}, …, x_{n−1}), for some polynomial P_{n}, that the proof of lemma 2.6 in [1] shows us how to compute. Therefore, the reduction modulo 2 of d(δ^{n}x) can be written as

dx_{n} + \frac{∂P_{n}}{∂x_{0}} dx_{0} + ∑_{i=1}^{n−1} \frac{∂P_{n}}{∂x_{i}} dx_{i}.

So, it suffices to prove that ∂P_{n}/∂x_{0} = 0 for all n ≥ 1, that is the same as saying that P_{n} just have even powers of x_{0}, what we prove by induction on n. For n = 1 it is trivial, since P_{1} = 0. Then, assume it is true for all i ≤ n. We recall how to compute P_{n}: let

a =^{def} ∑_{k≥0} x_{k} 2^{k},

considering a in characteristic zero and x_{k}’s as independent variables. Let δ^{0}a =^{def} a and ϕ(x_{i}) =^{def} x_{i}^{2} and extend ϕ linearly to power series and polynomials. Define

δ^{i+1}(a) =^{def} \frac{1}{2} (ϕ(δ^{i}a) − (δ^{i}a)^{2}) .

If we write

δ^{n}a = ∑_{k≥0} Θ_{k} 2^{k},

where Θ_{k} ∈ ℤ[x_{0}, x_{1}, …] with coefficients 0 or 1, then

P_{n}(x_{0}, x_{1}^{2}, …, x_{n−1}^{2^{n−1}}) = Θ_{0}(x_{0}, x_{1}, …, x_{n}) − x_{n}^{2^{n}}.  (3.7)

(Here we are identifying P_{n} with the corresponding polynomial with integers coefficients.) Now,

δ^{n+1}a = \frac{1}{2} [(ϕ(Θ_{0}) + 2ϕ(Θ_{1}) + …)

− (Θ_{0}^{2} + 4Θ_{0}(Θ_{1} + 2Θ_{2} + …) + 4(Θ_{1} + 2Θ_{2} + …)^{2})]

= \frac{ϕ(Θ_{0}) − Θ_{0}^{2}}{2} + ϕ(Θ_{1}) + 2(…).

By induction hypothesis and equation 3.7, we have that Θ_{0} just has even powers of x_{0}, and therefore, so does [ϕ(Θ_{0}) − Θ_{0}^{2}]/2. Clearly ϕ(Θ_{1}) just has even powers of x_{0}, which finishes the proof, since

P_{n+1}(x_{0}, x_{1}^{2}, …, x_{n}^{2^{n}}) ≡ \frac{ϕ(Θ_{0}) − Θ_{0}^{2}}{2} + ϕ(Θ_{1}) − x_{n+1}^{2^{n+1}} (mod 2).

Lemma 3.13. Let

ω_{n} =^{def} reduction modulo 2 of ℱ^{n} (\frac{dx}{2y + x}) .

Then, ω_{n} = dx_{0}/x_{0}.

Proof. Let C be the Cartier operator. Then, since ℱ is the “inverse” of C (see [3]), we have that C^{n}(ω_{n}) = dx_{0}/x_{0}. On the other hand, since ω_{n} is holomorphic, ω_{n} = α dx_{0}/x_{0}, and so C^{n}(ω_{n}) = α^{1/2^{n}}dx_{0}/x_{0}. Thus α = 1, and ω_{n} = dx_{0}/x_{0}.

With these lemmas, we prove the theorem:

Proof of Theorem 3.10. The theorem is equivalent to say that dx_{n} = 0 for n ≥ 1. We prove it by induction on n. With the same notation as in the the previous lemma, for n = 1 we have:

ω_{1} = \frac{x_{0} dx_{0} + dx_{1}}{x_{0}^{2}} = \frac{dx_{0}}{x_{0}}

where the last equality comes from lemma 3.13. This equation gives us dx_{1} = 0.

Now, assume dx_{i} = 0 for i = 1, …, n − 1. We prove that dx_{n} = 0: we have that ω_{n} is given by

ω_{n} = \frac{dx_{n} + x_{0}^{2^{n}−1}dx_{0} + …}{x_{0}^{2^{n}}},

where all omitted term can be written only with dx_{1}, …, dx_{n−1}, by lemmas 3.11 and 3.12. By induction hypothesis and lemma 3.13, we have

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

which proves that dx_{n} = 0.

We can now compute the canonical lift modulo 2^{3}: the algorithm described in section 2.6 does not work, but we can still compute it. We know that dx_{2}/dx_{0} = 0, and so x_{2} just has even powers of x_{0}, and from chapter 2, we know that deg x_{2} = 12, with leading coefficient 1, and that deg y_{2} < 16. The third coordinate of the equation is not too complicated, and we simply write

x_{2} = x_{0}^{6} + ∑_{i=0}^{2} A_{2i} x_{0}^{2i}

and

y_{2} = ∑_{i=0}^{7} B_{i} x_{0}^{i} + y_{0} ∑_{i=0}^{6} C_{i} x_{0}^{i},

leaving the A_{2i}’s, B_{i}’s, C_{i}’s and a_{2} as unknowns in the third coordinate of the equation of the curve, and force the equality by solving the linear system given by this equation.

Solving that we find an unique solution:

a_{2} = a_{0}^{4}
x_{2} = a_{0} x_{0}^{2} + x_{0}^{6}
y_{2} = (a_{0}^{2} + a_{0}^{4}) + (a_{0} + a_{0}^{3}) x_{0}^{2} + a_{0}^{4} x_{0}^{4} + (1 + a_{0}^{2}) x_{0}^{5} + (1 + a_{0} + a_{0}^{2}) x_{0}^{6}

+ [(1 + a_{0}^{2}) x_{0}^{3} + a_{0} x_{0}^{4} + x_{0}^{6}] y_{0}.

The uniqueness implies that this has to give us the canonical lift and the Teichmüller map, as one can also check by analyzing x/y.

To check solutions with smaller degrees, we can try to write

x_{2} = ∑_{i=0}^{6} A_{i} x_{0}^{i} + y_{0} ∑_{i=0}^{4} A′_{i} x_{0}^{i},

and y_{2} as before and try to solve it in the same way, but we will still find exactly one single solution to the corresponding system, again having the canonical lift and the Teichmüller map as the result.

But, in this case, we do have a lift

ν = ((x_{0}, x_{1}, x̃_{2}), (y_{0}, y_{1}, ỹ_{2}))

with deg x̃_{2} < deg x_{2}, but in this case, in contrast to what happens to characteristic different from 2, we have deg ỹ_{2} > deg y_{2}. Such a lift is:

a_{2} = a_{0}^{4}
x̃_{2} = a_{0} x_{0}^{2}
ỹ_{2} = (a_{0}^{2} + a_{0}^{4}) + (a_{0} + a_{0}^{2} + a_{0}^{3}) x_{0}^{2} + (a_{0} + a_{0}^{4}) x_{0}^{4}
+ (1 + a_{0}^{2}) x_{0}^{5} + (1 + a_{0} + a_{0}^{2}) x_{0}^{6} + x_{0}^{7} + x_{0}^{8} + x_{0}^{10}
+ [(1 + a_{0}^{2}) x_{0}^{3} + a_{0} x_{0}^{4} + x_{0}^{5} + x_{0}^{6}] y_{0}

If we compute another coordinate, we have deg x_{3} < 28 and deg y_{3} < 36, by our analysis of leading coefficients. Proceeding as described above, we can try to find the canonical lift and possibly some lift with smaller degrees, we can set

x_{3} = ∑_{i=0}^{13} A_{i} x_{0}^{i} + y_{0} ∑_{i=0}^{11} B_{i} x_{0}^{i}

and

y_{3} = ∑_{i=0}^{17} C_{i} x_{0}^{i} + y_{0} ∑_{i=0}^{16} D_{i} x_{0}^{i},

and solve the corresponding system. We get, as solutions a_{3} = a_{0}^{8},

x_{3} = a_{0}^{8} + a_{0} x_{0}^{6} + A_{8} x_{0}^{8} + A_{9} x_{0}^{9} + x_{0}^{10} + x_{0}^{12} + y_{0} (D_{16} x_{0}^{8}),

and

y_{3} = a_{0}^{4} + a_{0}^{5} + a_{0}^{4} A_{8} + a_{0}^{4} A_{9} x_{0} + a_{0}^{5} x_{0}^{2} + (a_{0}^{2} + a_{0}^{6} + a_{0}^{8} + a_{0}^{10} + a_{0}^{2} A_{8}) x_{0}^{4}

+ (a_{0}^{4} + a_{0}^{2} A_{9}) x_{0}^{5} + (a_{0} + a_{0}^{5} + a_{0}^{7} + a_{0}^{9} + a_{0} A_{8}) x_{0}^{6}

+ (a_{0}^{2} + a_{0}^{3} + a_{0}^{4} + a_{0} A_{9} + a_{0} D_{16}) x_{0}^{7} + (a_{0} + a_{0}^{2} + a_{0}^{3} + a_{0}^{4} + a_{0}^{6}) x_{0}^{8}
+ (1 + a_{0}^{6} + a_{0}^{8} + A_{8}) x_{0}^{9} + (1 + a_{0}^{6} + a_{0}^{8} + A_{8} + A_{9} + D_{16}) x_{0}^{10}
+ (1 + a_{0}^{2} + a_{0}^{3} + A_{9}) x_{0}^{11} + (1 + a_{0} + a_{0}^{3} + a_{0}^{8} + A_{8}) x_{0}^{12}
+ (1 + a_{0}^{2} + a_{0}^{4} + A_{9}) x_{0}^{13} + (1 + a_{0} + a_{0}^{2} + a_{0}^{4}) x_{0}^{14}
+ (1 + a_{0} + a_{0}^{2} + a_{0}^{4} + A_{8}) x_{0}^{16} + A_{9} x_{0}^{17} + y_{0} [a_{0}^{4} D_{16} + a_{0}^{4} x_{0}^{3}
+ (a_{0}^{5} + a_{0}^{2} D_{16}) x_{0}^{4} + a_{0}^{2} x_{0}^{5} + (a_{0}^{2} + a_{0}^{3} + a_{0} D_{16}) x_{0}^{6}

+ (1 + a_{0} + a_{0}^{2} + a_{0}^{3} + a_{0}^{6} + a_{0}^{8} + A_{8}) x_{0}^{7} + (a_{0} + a_{0}^{3} + a_{0}^{5} + A_{9} + D_{16}) x_{0}^{8}

+ (1 + a_{0} + D_{16}) x_{0}^{9} + (1 + a_{0}^{3} + a_{0}^{4} + D_{16}) x_{0}^{10} + (a_{0} + a_{0}^{4}) x_{0}^{11}

+ (a_{0} + a_{0}^{2} + D_{16}) x_{0}^{12} + x_{0}^{13} + x_{0}^{14} + D_{16} x_{0}^{16}].

Thus, clearly, to get the minimal and the Teichmüller lifts, we need to choose D_{16} = 0. This leaves us the choices of A_{8} and A_{9}. Again, to get minimal degrees and the Teichmüller (since dx_{3}/dx_{0} = 0), we choose A_{9} = 0. So, we have

x_{3} = a_{0}^{8} + a_{0} x_{0}^{6} + A_{8} x_{0}^{8} + x_{0}^{10} + x_{0}^{12},

and

y_{3} = a_{0}^{4} + a_{0}^{5} + a_{0}^{4} A_{8} + a_{0}^{5} x_{0}^{2} + (a_{0}^{2} + a_{0}^{6} + a_{0}^{8} + a_{0}^{10} + a_{0}^{2} A_{8}) x_{0}^{4} + a_{0}^{4} x_{0}^{5} + (a_{0} + a_{0}^{5} + a_{0}^{7} + a_{0}^{9} + a_{0} A_{8}) x_{0}^{6} + (a_{0}^{2} + a_{0}^{3} + a_{0}^{4}) x_{0}^{7}

+ (a_{0} + a_{0}^{2} + a_{0}^{3} + a_{0}^{4} + a_{0}^{6}) x_{0}^{8} + (1 + a_{0}^{6} + a_{0}^{8} + A_{8}) x_{0}^{9}
+ (1 + a_{0}^{6} + a_{0}^{8} + A_{8}) x_{0}^{10} + (1 + a_{0}^{2} + a_{0}^{3}) x_{0}^{11}
+ (1 + a_{0} + a_{0}^{3} + a_{0}^{8} + A_{8}) x_{0}^{12} + (1 + a_{0}^{2} + a_{0}^{4}) x_{0}^{13}
+ (1 + a_{0} + a_{0}^{2} + a_{0}^{4}) x_{0}^{14} + (1 + a_{0} + a_{0}^{2} + a_{0}^{4} + A_{8}) x_{0}^{16}
+ y_{0} [a_{0}^{4} x_{0}^{3} + a_{0}^{5} x_{0}^{4} + a_{0}^{2} x_{0}^{5} + (a_{0}^{2} + a_{0}^{3}) x_{0}^{6}

+ (1 + a_{0} + a_{0}^{2} + a_{0}^{3} + a_{0}^{6} + a_{0}^{8} + A_{8}) x_{0}^{7} + (a_{0} + a_{0}^{3} + a_{0}^{5}) x_{0}^{8} + (1 + a_{0}) x_{0}^{9}

+ (1 + a_{0}^{3} + a_{0}^{4}) x_{0}^{10} + (a_{0} + a_{0}^{4}) x_{0}^{11} + (a_{0} + a_{0}^{2}) x_{0}^{12} + x_{0}^{13} + x_{0}^{14}].

The choice of A_{8} = 1 + a_{0} + a_{0}^{2} + a_{0}^{3} + a_{0}^{6} + a_{0}^{8}, clearly gives us the smaller degrees, but in fact, it does not gives us the canonical lift. One can check that by looking at x/y. The fourth coordinate of x/y is given by,

\frac{x_{3}}{y_{0}^{8}} + \frac{x_{0}^{2} x_{1}^{3} y_{1}}{y_{0}^{10}} + \frac{x_{0}^{2} x_{1} x_{2} y_{1}}{y_{0}^{10}} + \frac{x_{0}^{4} x_{2} y_{1}^{2}}{y_{0}^{12}} + \frac{x_{1}^{2} x_{2} y_{1}^{2}}{y_{0}^{12}} + \frac{x_{0}^{2} x_{1}^{3} y_{1}^{3}}{y_{0}^{14}}

+ \frac{x_{0}^{8} y_{1}^{4}}{y_{0}^{16}} + \frac{x_{0}^{4} x_{1}^{2} y_{1}^{4}}{y_{0}^{16}} + \frac{x_{1}^{4} y_{1}^{4}}{y_{0}^{16}} + \frac{x_{0}^{4} x_{2} y_{1}^{4}}{y_{0}^{16}} + \frac{x_{2}^{2} y_{1}^{4}}{y_{0}^{16}} + \frac{x_{0}^{6} x_{1} y_{1}^{5}}{y_{0}^{18}} + \frac{x_{0}^{8} y_{1}^{6}}{y_{0}^{20}}

+ \frac{x_{0}^{4} x_{1}^{2} y_{1}^{6}}{y_{0}^{20}} + \frac{x_{1}^{4} y_{1}^{8}}{y_{0}^{24}} + \frac{x_{0}^{8} y_{1}^{12}}{y_{0}^{32}} + \frac{x_{0}^{4} x_{2} y_{2}}{y_{0}^{12}} + \frac{x_{0}^{6} x_{1} y_{1} y_{2}}{y_{0}^{14}} + \frac{x_{0}^{8} y_{1}^{2} y_{2}}{y_{0}^{16}}

+ \frac{x_{0}^{4} x_{1}^{2} y_{1}^{2} y_{2}}{y_{0}^{16}} + \frac{x_{0}^{8} y_{1}^{4} y_{2}}{y_{0}^{20}} + \frac{x_{0}^{8} y_{2}^{2}}{y_{0}^{16}} + \frac{x_{1}^{4} y_{2}^{2}}{y_{0}^{16}} + \frac{x_{0}^{8} y_{3}}{y_{0}^{16}}

The part that has non-positive order at infinity and the part in x_{3} and y_{3} is

\frac{x_{3}}{y_{0}^{8}} + \frac{x_{0}^{4} x_{2} y_{1}^{4}}{y_{0}^{16}} + \frac{x_{2}^{2} y_{1}^{4}}{y_{0}^{16}} + \frac{x_{0}^{8} y_{1}^{12}}{y_{0}^{32}} + \frac{x_{0}^{8} y_{3}}{y_{0}^{16}} .

Using the expressions of x_{1}, y_{1}, x_{2}, y_{2}, we get that

\frac{x_{0}^{4} x_{2} y_{1}^{4}}{y_{0}^{16}} + \frac{x_{2}^{2} y_{1}^{4}}{y_{0}^{16}} + \frac{x_{0}^{8} y_{1}^{12}}{y_{0}^{32}}

has actually positive order at infinity, and since x_{3}/y_{0}^{8} has order 0 (for any choice of A_{8}), we have that the Teichmüller lift had deg y_{3} = 32, and A_{8} = a_{0} + a_{0}^{2} + a_{0}^{4}. In this case, we have a lift

ν = ((x_{0}, x_{1}, x_{2}, x̃_{3}), (y_{0}, y_{1}, y_{2}, ỹ_{3})),

with deg x̃_{3} = deg x_{3} and deg ỹ_{3} < deg y_{3}.

On the other hand if we use x̃_{2} and ỹ_{2} as before, we can get a lift ν̃ = ((x_{0}, x_{1}, x̃_{2}, ˜ x̃_{3}), (y_{0}, y_{1}, ỹ_{2}, ˜ ỹ_{3}))

with

˜x̃_{3} = a_{0}^{8} + a_{0}^{6} x_{0}^{6}

and

˜ỹ_{3} = a_{0}^{4} + a_{0}^{6} + (a_{0}^{3} + a_{0}^{6} + a_{0}^{7}) x_{0}^{2} + (a_{0}^{3} + a_{0}^{4} + a_{0}^{5} + a_{0}^{8} + a_{0}^{10}) x_{0}^{4}

+ (a_{0}^{2} + a_{0}^{6}) x_{0}^{5} + (a_{0} + a_{0}^{2} + a_{0}^{3} + a_{0}^{4} + a_{0}^{5} + a_{0}^{6} + a_{0}^{7} + a_{0}^{9}) x_{0}^{6}

+ (a_{0}^{2} + a_{0}^{3}) x_{0}^{7} + (a_{0}^{2} + a_{0}^{6}) x_{0}^{8} + (1 + a_{0} + a_{0}^{6} + a_{0}^{8}) x_{0}^{9}
+ (a_{0} + a_{0}^{2} + a_{0}^{3} + a_{0}^{5} + a_{0}^{6} + a_{0}^{8}) x_{0}^{10} + (a_{0} + a_{0}^{2} + a_{0}^{3}) x_{0}^{11}
+ (1 + a_{0} + a_{0}^{3} + a_{0}^{8}) x_{0}^{12} + (1 + a_{0}) x_{0}^{13} + (a_{0}^{2} + a_{0}^{4}) x_{0}^{16}
+ (1 + a_{0}) x_{0}^{18} + x_{0}^{19} + x_{0}^{21} + x_{0}^{22} + x_{0}^{24}
+ y_{0} [(a_{0}^{2} + a_{0}^{6}) x_{0}^{3} + a_{0}^{5} x_{0}^{4} + (a_{0}^{2} + a_{0}^{4}) x_{0}^{5} + a_{0}^{2} x_{0}^{6}
+ (1 + a_{0}^{3} + a_{0}^{6} + a_{0}^{8}) x_{0}^{7} + (a_{0} + a_{0}^{3} + a_{0}^{5}) x_{0}^{8} + a_{0} x_{0}^{9}

+ (1 + a_{0}^{3} + a_{0}^{4}) x_{0}^{10} + (a_{0} + a_{0}^{2}) x_{0}^{11} + (1 + a_{0}^{2}) x_{0}^{12} + a_{0} x_{0}^{14} + x_{0}^{16} + x_{0}^{17} + x_{0}^{19}] And so, deg ˜ x̃_{3} = 12 and deg ˜ ỹ_{3} = 48.

3.5 Characteristic 3

In characteristic 3, an ordinary elliptic curve over an algebraic closed field (or over a finite extension of our original base field k) can be put in the form:

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

(and then, we have the Hasse invariant A = 1) and we consider the canonical lifting having the same form:

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

We notice that the formulas for the derivatives of x_{1} and x_{2} remain the same as for characteristic p ≥ 5 we used before. An analogous analysis to the one done to prove theorem 2.13 tells us, though, that deg x_{2} = 3p^{2} − 1, and in fact, we can use the modified algorithm described to produce a lifting modulo 3^{3}. What we obtain is:

a_{1} = 0, a_{2} = a_{0}^{9} + a_{0}^{12},

and

x_{1} =2 a_{0}^{2} + a_{0} x_{0} + (1 + 2 a_{0}) x_{0}^{3} + x_{0}^{4}
y_{1} =x_{0}^{2} y_{0}
x_{2} =2 a_{0}^{5} + 2 a_{0}^{9} + (a_{0}^{4} + 2 a_{0}^{5}) x_{0} + a_{0}^{4} x_{0}^{2} + (a_{0}^{3} + a_{0}^{4} + a_{0}^{5} + 2 a_{0}^{6}) x_{0}^{3} + 2 a_{0}^{2} x_{0}^{5}

+ (2 a_{0}^{2} + a_{0}^{3} + a_{0}^{4}) x_{0}^{6} + a_{0}^{2} x_{0}^{7} + a_{0} x_{0}^{8} + (2 a_{0} + a_{0}^{2} + a_{0}^{3}) x_{0}^{9} + 2 a_{0}^{2} x_{0}^{10} + (2 + a_{0}) x_{0}^{11} + (2 a_{0} + 2 a_{0}^{3}) x_{0}^{12} + x_{0}^{13}

y_{2} =y_{0} [2 a_{0}^{4} + a_{0}^{7} + a_{0}^{6} x_{0}^{2} + (2 a_{0}^{4} + a_{0}^{6}) x_{0}^{3} + 2 a_{0}^{2} x_{0}^{4} + (a_{0}^{2} + 2 a_{0}^{3}) x_{0}^{5}

+ (2 a_{0}^{2} + 2 a_{0}^{3} + a_{0}^{4}) x_{0}^{6} + a_{0} x_{0}^{7} + (a_{0} + a_{0}^{3}) x_{0}^{8} + (2 a_{0} + a_{0}^{3} + a_{0}^{4}) x_{0}^{9} +2 x_{0}^{10} + (2 + a_{0}^{3}) x_{0}^{11} + (1 + 2 a_{0} + a_{0}^{3}) x_{0}^{12} + 2 x_{0}^{14} + 2 x_{0}^{15}]

So, for p = 3, the Teichmüller lift is also the absolute minimal degree lift.

If we look modulo 3^{4}, though, we have that deg x_{3} = 108, with leading coefficient 2. So, in this case, proposition 3.4 tells us that we have a lift with deg x̃_{3} ≤ 82. We can try to look at such lift to see if similar properties happen to this minimal lift as it happened to minimal lifts modulo p^{3} for p ≥ 5, e.g. is dx̃_{3}/dx_{0} = dx_{3}/dx_{0}? What is its degree?

To compute the Teichmüller in this case, it would be useful to know again the derivative. We wrote a program in Magma to help us, and some numerical evidence allows us to conjecture:

Conjecture 3.14. For p ≠ 2 and n ≥ 1, we have that the reduction modulo p of ℱ^{n}(dx/y) is given by

\frac{1}{y_{0}^{p^{n}}} ∑_{i=0}^{n} x_{i}^{(p^{n−i}−1)} dx_{i},

which implies that

\frac{dx_{n}}{dx_{0}} = A^{−(p^{n}−1)/(p−1)} y_{0}^{p^{n}−1} − x_{0}^{p^{n}−1} − ∑_{i=1}^{n−1} x_{i}^{(p^{n−i}−1)} \frac{dx_{i}}{dx_{0}} .

Voloch was the first to “guess” the formula above, and we did some calculations to confirm it in some cases. We checked it to be true for: p = 3 and n ≤ 6, p = 5, 7, 11, 13 and n ≤ 4, p = 17 and n ≤ 3.

The main difficulty of proving this conjecture is maybe dealing with Buium’s polynomials P_{i}’s (as in lemma 3.12 – or check lemma 2.6 in [1]). Even with Magma running on fast computers, the computations of such polynomials take very long for large p’s or large i’s, and there does not seem to be a straightforward way to deal with them theoretically to work out a proof.

We now proceed to compute x_{3}. Since we know its derivative (the conjecture was verified for p = 3 and n = 3), we follow the same idea of the algorithm described in section 2.6, leaving the coefficients of x_{3} that cannot be obtained by the derivative as indeterminates. We can then compute the canonical lift and the elliptic Teichmüller map. The expressions are too long to be put in here, but as an example, taking a_{0} = 1, we would have:

a_{3} =1
x_{3} =2 + 2 x_{0}^{3} + x_{0}^{4} + 2 x_{0}^{6} + x_{0}^{7} + x_{0}^{8} + 2 x_{0}^{9} + x_{0}^{10} + x_{0}^{11} + 2 x_{0}^{12} + 2 x_{0}^{16}

+ 2 x_{0}^{17} + 2 x_{0}^{18} + 2 x_{0}^{20} + x_{0}^{25} + 2 x_{0}^{27} + 2 x_{0}^{28} + 2 x_{0}^{29} + x_{0}^{30} + x_{0}^{32} + 2 x_{0}^{34} + x_{0}^{35} + 2 x_{0}^{36} + x_{0}^{37} + 2 x_{0}^{38} + x_{0}^{40} + x_{0}^{45} + 2 x_{0}^{54}

y_{3} =y_{0} [2 + 2 x_{0}^{3} + 2 x_{0}^{4} + 2 x_{0}^{6} + x_{0}^{7} + x_{0}^{9} + 2 x_{0}^{10} + x_{0}^{12} + x_{0}^{13} + 2 x_{0}^{15}

+ 2 x_{0}^{17} + 2 x_{0}^{18} + 2 x_{0}^{19} + x_{0}^{20} + x_{0}^{21} + x_{0}^{22} + 2 x_{0}^{23} + x_{0}^{27} + 2 x_{0}^{28} + 2 x_{0}^{29} + x_{0}^{30} + 2 x_{0}^{31} + 2 x_{0}^{34} + 2 x_{0}^{35} + x_{0}^{36} + 2 x_{0}^{37} + 2 x_{0}^{38} + 2 x_{0}^{39} + x_{0}^{43} + 2 x_{0}^{44} + 2 x_{0}^{48} + 2 x_{0}^{50} + x_{0}^{53} + x_{0}^{54} + 2 x_{0}^{60} + 2 x_{0}^{62} + x_{0}^{63} + 2 x_{0}^{65} + 2 x_{0}^{66}]

In the generic case, we get always the same curve, having

a_{3} = 2 a_{0}^{27} + 2 a_{0}^{33} + a_{0}^{36} + 2 a_{0}^{45}.

The general formula can be found in the web address cited in chapter 2. To find the minimal degree lift to the canonical lift, we proceed in an analogous way as in theorem 3.6, we make the division of polynomials

x_{3} = f(x_{0})^{14} q(x_{0}) + x̃_{3}, (f(x_{0}) = x_{0}^{3} + x_{0}^{2} + a_{0})

and define

ỹ_{3} = y_{3} − \frac{y_{0}}{2} [f′(x_{0})^{27}q(x_{0})].

Again, (in the general case) we get that deg x̃_{3} = 80, not 82 as it could be. For a_{0} = 1 again, we would have:

x̃_{3} =1 + 2 x_{0}^{3} + x_{0}^{4} + 2 x_{0}^{6} + x_{0}^{7} + x_{0}^{8} + 2 x_{0}^{9} + x_{0}^{10} + x_{0}^{11} + 2 x_{0}^{12} + 2 x_{0}^{16} + 2 x_{0}^{17}

+ x_{0}^{18} + 2 x_{0}^{20} + x_{0}^{25} + x_{0}^{27} + 2 x_{0}^{28} + 2 x_{0}^{29} + x_{0}^{30} + x_{0}^{32} + 2 x_{0}^{34} + x_{0}^{35} + x_{0}^{37}

+ 2 x_{0}^{38} + x_{0}^{40}
ỹ_{3} =y_{0} [2 + 2 x_{0}^{3} + 2 x_{0}^{4} + 2 x_{0}^{6} + x_{0}^{7} + x_{0}^{9} + 2 x_{0}^{10} + x_{0}^{12} + x_{0}^{13} + 2 x_{0}^{15} + 2 x_{0}^{17}

+ 2 x_{0}^{18} + 2 x_{0}^{19} + x_{0}^{20} + x_{0}^{21} + x_{0}^{22} + 2 x_{0}^{23} + 2 x_{0}^{27} + 2 x_{0}^{28} + 2 x_{0}^{30} + 2 x_{0}^{31}

+ x_{0}^{33} + 2 x_{0}^{34} + 2 x_{0}^{37} + x_{0}^{43} + 2 x_{0}^{44} + 2 x_{0}^{48} + 2 x_{0}^{50} + x_{0}^{53} + x_{0}^{54} + 2 x_{0}^{60}
+ 2 x_{0}^{62} + x_{0}^{63} + 2 x_{0}^{65} + 2 x_{0}^{66}]

(Again, one can see the general formula at the web address.)

We also observe that x_{3} and x̃_{3} have the same derivative (even in the generic case), and many of the properties observed in section 3.3 for x̃_{2} will hold here for x̃_{3} in a very similar way: the Frobenius will also lift and x̃_{3} will give us the absolute minimal lift, and thus the minimal degree occurs for the canonical lift, and lifting to every other curve will give a map ˜ x̃_{3} with degree larger or equal to 82. For the first claim, just observe that x_{3} − x̃_{3} and y_{3} − ỹ_{3} are 3-powers. The second claim follows from analyzing the part of x/y that can be singular, namely

\frac{x_{3}}{y_{0}^{27}} + \frac{2 x_{1}^{9} y_{2}^{3}}{y_{0}^{54}} + \frac{2 x_{0}^{27} y_{3}}{y_{0}^{54}} .

Rewriting

y_{3} = \frac{1}{y_{0}^{27}} [y_{1}^{9} y_{2}^{3} + 2x_{0}^{27}x_{3} + …] ,

we get an expression for the terms of higher power in x_{0} of x_{3} that makes it the one from Teichmüller, and if deg ˜ x̃_{3} ≤ 80, we can add this terms and modify ˜ỹ_{3} to get the Teichmüller (again, as in section 3.3), which forces the curve to be the canonical lift and the degree to be 80.

Hyperelliptic Curves

4.1 Minimal Degrees

Again, let k be a perfect field of characteristic p ≠ 2, and consider the (nonsingular) hyperelliptic curve given by

C/k : y_{0}^{2} = f(x_{0}),

with deg_{x_{0}} f = d ≥ 3. Let ϵ be the number of points at infinity of C, and let U denote the affine part of C.

Theorem 4.1. Suppose that we have a curve

C/W_{n+1}(k) : y^{2} = f(x)

with reduction C modulo p such that there is a lift of the Frobenius to the affine part of C. Assume that the first part of conjecture 3.14 holds at least for n and p, and that the lift of U to the affine part of C associated to the Frobenius is given by

ν = ((x_{0}, x_{1}, …, x_{n}), (y_{0}, y_{1}, …, y_{n})),

where x_{i}’s are polynomials in x_{0} and with deg x_{i} = d(p^{i} − 1) + 2 and deg y_{i} ≤ [i(d−2)+d]p^{i}−i(d−2)p^{i−1}, for i = 0, …, (n−1). Then, deg x_{n} ≥ d(p^{n}−1)+2.

If we have the equality for deg x_{n}, we must have

\frac{dx_{n}}{dx_{0}} = λf(x_{0})^{(p^{n}−1)/2} − ∑_{i=0}^{n−1} x_{i}^{p^{n−i}−1} \frac{dx_{i}}{dx_{0}} ,

for some λ ∈ k.

Proof. First, we observe that the first part of conjecture 3.14 depends only on the existence of the lift of the Frobenius and on Buium’s polynomials P_{i}, for i ≤ n.

Let ϕ denote the Frobenius map and its lift to C, and let U denote the affine part of C. Then, the conjecture states that the reduction modulo p of ℱ^{n}(dx/y) is given by

ω =^{def} \frac{1}{y_{0}^{p^{n}}} [∑_{i=0}^{n} x_{i}^{(p^{n−i}−1)} \frac{dx_{i}}{dx_{0}} ] dx_{0}.

Let P be a point at infinity of C. We have

ord_{P} (ω) = p^{n} \frac{d}{ϵ} − (\frac{2}{ϵ} + 1) + ord_{P} (∑_{i=0}^{n} x_{i}^{(p^{n−i}−1)} \frac{dx_{i}}{dx_{0}} ) .

To simplify the notation, we will denote the last summand above α. Also, let β denote the number of zeros, counted with multiplicity, of ω in U. Since dx/y is regular on U, ω must be regular on U. Then, by the Riemman-Roch theorem, we have

ϵα + dp^{n} − (2 + ϵ) + β = 2 \frac{d − ϵ}{2} − 2 = d − (ϵ + 2).

Hence,

α ≤ −\frac{d(p^{n} − 1)}{ϵ}.

By hypothesis, we have

\begin{aligned}ord_{P} (x_{i}^{p^{n−i}−1} \frac{dx_{i}}{dx_{0}} ) & = − [(p^{n−i} − 1) \frac{d(p^{i} − 1) + 2}{ϵ} + \frac{d(p^{i} − 1) + 2}{ϵ} − \frac{2}{ϵ}] \\ & = − [\frac{dp^{n} − (d − 2)p^{n−i} + 2}{ϵ}] > −\frac{d(p^{n} − 1)}{ϵ}, \end{aligned}

for i = 1, …, (n − 1). Therefore

ord_{P} (\frac{dx_{n}}{dx_{0}} ) ≤ −\frac{d(p^{n} − 1)}{ϵ},

what implies that deg x_{n} ≥ d(p^{n} − 1) + 2.

We have the equality if, and only if, β = 0. But then, ord_{P} (ω) = d/ϵ − (2/ϵ + 1) = ord_{P} (dx_{0}/y_{0}). So, if we write ω = g dx_{0}/y_{0}, for some g in the function field of C, g has no zeros or poles at infinity. Since ω and dx_{0}/y_{0} have no poles in U, g has no poles at all, and thus λ =^{def} g ∈ k. Thus, we have

ω = \frac{1}{y_{0}^{p^{n}}} [∑_{i=0}^{n} x_{i}^{(p^{n−i}−1)} \frac{dx_{i}}{dx_{0}} ] dx_{0} = λ \frac{dx_{0}}{y_{0}} ,

what implies the formula for the derivative.

The proof of the above theorem, for p = 3, d = 6 and n = 1, that could be easily generalized, was shown to the author by Felipe Voloch.

We observe that curves of genus g > 1 do not have a lift of the Frobenius (see [5]), so the restriction to affine parts is necessary if g > 1. On the other hand, Mochizuki showed in [4] that there is a lifting of the Frobenius in some open subset of an ordinary curve of genus g defined by taking off some (g −1)(p−1) points from the curve. This was what motivated Voloch to prove the above theorem for p = 3, d = 6 and n = 1: in this case, the curve has genus 2 and is necessarily hyperelliptic, and we have two points to be taken off, that maybe can be put at infinity.

We also note that, although the hypothesis of the theorem seem very strong, it is still gives us some nice results. Besides the case of Mochizuki lift just mentioned, we have:

Corollary 4.2. Let ν be a lift from the affine part of C to the affine part of C/W_{2}(k). Then, deg x_{1} ≥ d(p − 1) + 2, and if we have the equality, we must have that the coefficient of x_{0}^{p−1} in f(x_{0})^{(p−1)/2}, say A, is non zero and

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

Proof. Remember that having a lift modulo p^{2} of U gives a lift of the Frobenius modulo p^{2}, and we can assume that x_{1} is a polynomial in x_{0} and y_{1} is y_{0} times a polynomial in x_{0}. Also, the conjecture 3.14 is true for n = 1. Finally deg x_{0} = 2 and deg y_{0} = d. So, we can apply the theorem 4.1 with n = 1. The fact that λ = A^{−1}, comes from the fact that a derivative of a polynomial in characteristic p cannot have a term in x_{0}^{p−1}.

The above corollary gives a better lower bound for the degree of x_{1} than the one stated in [8], theorem IV.2, for the case of hyperelliptic curves, and it is a little more general, since that theorem just applies for d odd. Also, it gives us a necessary condition to achieve the lower bound: in order for A^{−1}f(x_{0})^{(p−1)/2} −x_{0}^{p−1} to be a derivative, f(x_{0})^{(p−1)/2} cannot have a coefficient of x_{0}^{rp−1} different from zero, except for r = 1, in which case such term is necessarily non zero.

The condition also seems to be sufficient: we tried several cases when the condition was satisfied and we were always able to find a lift with the minimal degree.

Also, another interesting corollary of theorem 4.1 is that for elliptic curves, if we proceed finding minimal lifts modulo higher powers of p that give lifts of the Frobenius, then the derivatives of the x̃_{n}, by induction, would have to be the same as the derivative of x_{n}, and thus deg x̃_{n} = 3p^{n} − 1.

4.2 Mochizuki Lifts

We will now analyze in more detail the case of Mochizuki lifts. As mentioned before, for p = 3 and g = 2, we would have to take off two points of the curve to have a lift of the Frobenius. Every curve of genus 2 is hyperelliptic, and if those points to be taken off are invariant by the hyperelliptic involution, we can assume that those points are at infinity. So we will consider the curve

C/k : y_{0}^{2} = f(x_{0}),

where k is a perfect field of characteristic 3 and deg_{x_{0}} f = 6. In order to have the minimal degree, we will assume there exists an A ∈ k^{∗} such that A^{−1}f(x_{0})− x_{0}^{2} is a derivative, i.e., it does not have the terms in x_{0}^{2} and x_{0}^{5}. Therefore, A is the coefficient of x_{0}^{2} in f(x_{0}), and working in some finite extension of k, we may assume A = 1. Also, the coefficient of x_{0}^{5} has to be zero. So, we can assume that

f(x_{0}) = x_{0}^{6} + α_{0}x_{0}^{4} + β_{0}x_{0}^{3} + x_{0}^{2} + γ_{0}x_{0} + δ_{0}.

But, with the linear change of variables

(x_{0}, y_{0}) ↦ (x_{0} + ϵ_{0}, y_{0}),

with ϵ_{0} satisfying 2ϵ_{0}^{3} + α_{0} ϵ_{0} + β_{0} = 0 (again, maybe in some finite extension of k), allows us to consider f given by

f(x_{0}) = x_{0}^{6} + a_{0}x_{0}^{4} + x_{0}^{2} + b_{0}x_{0} + c_{0}.

In this context, such a curve is “ordinary” if a_{0} ≠ 0.

Assuming that the lift will have minimal degrees and using the formula for the derivative of x_{1} from corollary 4.2, we can use the same algorithm described for elliptic curves in section 2.6 to compute the Mochizuki lift. We have a lift of the form

C/W_{2}(k) : y^{2} = x^{6} + ax^{4} + x^{2} + bx + c,

with

x_{1} = x_{0}^{7} + \frac{b_{0}}{a_{0}^{2}} x_{0}^{6} + 2 a_{0} x_{0}^{5} +

\frac{a_{0}^{4} c_{0}^{2} + 2 a_{0}^{4} + 2 a_{0}^{3} b_{0}^{2} + 2 a_{0}^{3} c_{0} + 2 a_{0}^{2} b_{0}^{2} c_{0} + 2 a_{0}^{2} c_{0}^{2} + a_{0}^{2} + b_{0}^{4}}{a_{0}^{2}} x_{0}^{3}

+ 2 b_{0} x_{0}^{2} + c_{0} x_{0} + \frac{a_{0}^{4} b_{0} + 2 a_{0}^{3} b_{0} c_{0} + a_{0}^{2} b_{0} + b_{0}}{a_{0}^{5}}

F_{1} = (2 a_{0}^{3} + a_{0}) x_{0}^{4} + 2 a_{0} b_{0} x_{0}^{3} + (2 a_{0}^{2} + 1) x_{0}^{2} + (2 a_{0}^{2} b_{0} + b_{0}) x_{0} + 2 a_{0}^{2} c_{0} + b_{0}^{2} a_{1} = 2 a_{0}^{5} c_{0}^{2}+2 a_{0}^{5}+a_{0}^{4} b_{0}^{2}+a_{0}^{4} c_{0}+a_{0}^{3} b_{0}^{2} c_{0}+a_{0}^{3} c_{0}^{2}+2 a_{0}^{3}+a_{0}^{2} c_{0}+2 a_{0} b_{0}^{4}+a_{0}+2 b_{0}^{2}+c_{0}

b_{1} = [2 a_{0}^{7} b_{0}^{3} c_{0}^{2} + 2 a_{0}^{7} b_{0}^{3} + a_{0}^{6} b_{0}^{5} + a_{0}^{6} b_{0}^{3} c_{0} + a_{0}^{6} b_{0} c_{0}^{2} + a_{0}^{5} b_{0}^{5} c_{0} + a_{0}^{5} b_{0}^{3} c_{0}^{2} +2 a_{0}^{5} b_{0}^{3} + a_{0}^{5} b_{0} c_{0} + a_{0}^{4} b_{0} + 2 a_{0}^{3} b_{0}^{7} + 2 a_{0}^{3} b_{0} c_{0} + a_{0}^{2} b_{0} + b_{0}] a_{0}^{−5}

c_{1} = \frac{a_{0}^{7} c_{0}^{3} + 2 a_{0}^{5} b_{0}^{2} c_{0}^{2} + 2 a_{0}^{4} b_{0}^{4} + a_{0}^{3} b_{0}^{4} c_{0} + 2 a_{0}^{2} b_{0}^{4} + 2 b_{0}^{4}}{a_{0}^{5}}

(Remember that y_{n} = y_{0} F_{n}, with F_{n} a polynomial in x_{0}.)

I also checked the case a_{0} = 0 (when C is not ordinary), where we have:

x_{1} = x_{0}^{7} + 2 b_{0} x_{0}^{6} + (2 b_{0}^{4} + 2 b_{0}^{2} c_{0} + 2 c_{0}^{2} + 1) x_{0}^{3} + 2 b_{0} x_{0}^{2} + c_{0} x_{0}

F_{1} = x_{0}^{2} + b_{0} x_{0} + b_{0}^{2}
a_{1} = 2 b_{0}^{2} + c_{0}
b_{1} = b_{0}^{7} + b_{0}^{5} c_{0} + b_{0}^{3} c_{0}^{2} + 2 b_{0}^{3} + b_{0} c_{0}
c_{1} = 2 b_{0}^{2} c_{0}^{2}

If the curve with a_{0} ≠ 0 indeed corresponds to the Mochizuki lift, we should be able to also lift the Frobenius modulo 3^{3}. To compute the lift, say

ν = ((x_{0}, x_{1}, x_{2}), (y_{0}, y_{1}, y_{2})),

that will give the lift of the Frobenius, we use proposition 3.9. We then have that

\frac{dx_{2}}{dx_{0}} = y_{0}^{8} − x_{0}^{8} − x_{1}^{2} (y_{0}^{2} − x_{0}^{2}).

Therefore, we have that deg_{x_{0}} x_{2} ≥ 25. We have done the calculations and found that the minimal lift having such derivative has in fact deg_{x_{0}} x_{2} = 25 if a_{0} ≠ 0. Checking the minimal lifts for a_{0} = 0, we found that if b_{0} ≠ 0, then again deg_{x_{0}} x_{2} = 25. If a_{0} = b_{0} = 0, then deg_{x_{0}} x_{2} = 30. But, we still need to guarantee the existence of the lift of the Frobenius. To check that, we construct polynomials P and Q in W_{3}[x] and W_{3}[x, y] respectively, such that

ϕ(x, y) = (x^{p} + px_{1} + p^{2}P , y^{p} + py_{1} + p^{2}Q).

is a lift of the Frobenius. We do that by following the idea behind the proof of proposition 3.8, more precisely formula (3.5) and its analogue for δ^{2}y. We need then, that x_{2} − x_{0}^{p(p−1)}x_{1} minus the reduction modulo p of (x_{1}^{σ}(x^{p}) − x_{1}(x)^{p})/p + (dx_{1}/dx)^{σ}(x^{p}) · x_{1} is a p-power, which we get by our choice of the derivative of x_{2}, and also y_{2} − y_{0}^{p(p−1)}y_{1} minus the reduction modulo p of (y_{1}^{σ}(x^{p}, y^{p}) − x_{1}(x, y)^{p})/p + (∂y_{1}/∂x)^{σ}(x^{p}, y^{p}) · x_{1} + (∂y_{1}/∂y)^{σ}(x^{p}, y^{p}) · y_{1} is a p-power. We actually checked both cases, i.e. a_{0} ≠ 0 and a_{0} = 0, and in both we could find a lift of the Frobenius. The formulas for P and Q when a_{0} ≠ 0 are too long to be put in here, but if a_{0} = 0, we have that P (x)^{3} and Q(x, y)^{3} are lifts of

(b_{0}^{6} + 2 b_{0}^{2} c_{0}^{2} + b_{0}^{8} c_{0}^{3} + 2 b_{0}^{6} c_{0}^{4} + b_{0}^{6} c_{0}^{6} + 2 c_{0}^{9}) x_{0}^{3} + (b_{0}^{3} + 2 b_{0}^{7} + 2 b_{0}^{11} + 2 b_{0} c_{0} + 2 b_{0}^{5} c_{0} + b_{0}^{9} c_{0} + 2 b_{0}^{3} c_{0}^{2} + b_{0}^{5} c_{0}^{3} + 2 b_{0}^{3} c_{0}^{4}) x_{0}^{6} + (1 + 2 b_{0}^{8} + 2 b_{0}^{16} + 2 b_{0}^{24} + b_{0}^{6} c_{0} + 2 b_{0}^{10} c_{0} + 2 b_{0}^{14} c_{0} + b_{0}^{8} c_{0}^{2} + 2 b_{0}^{12} c_{0}^{2} + b_{0}^{2} c_{0}^{3} + b_{0}^{6} c_{0}^{3} + b_{0}^{10} c_{0}^{3} + 2 b_{0}^{14} c_{0}^{3} + 2 c_{0}^{4} + b_{0}^{4} c_{0}^{4} + b_{0}^{8} c_{0}^{4} + b_{0}^{12} c_{0}^{4} + 2 b_{0}^{2} c_{0}^{5} + b_{0}^{6} c_{0}^{5} + c_{0}^{6} + b_{0}^{12} c_{0}^{6} + 2 b_{0}^{6} c_{0}^{9} + c_{0}^{12}) x_{0}^{9} + (2 b_{0}^{5} + 2 b_{0}^{9} + b_{0}^{3} c_{0} + 2 b_{0}^{3} c_{0}^{3}) x_{0}^{12}

+ (b_{0}^{2} + 2 b_{0}^{6} + 2 c_{0}) x_{0}^{15} + (2 b_{0}^{7} + 2 b_{0}^{15} + 2 b_{0} c_{0} + 2 b_{0}^{5} c_{0}
+ 2 b_{0}^{3} c_{0}^{2} + b_{0}^{3} c_{0}^{6}) x_{0}^{18} + (1 + b_{0}^{8} + 2 b_{0}^{6} c_{0} + b_{0}^{6} c_{0}^{3} + 2 c_{0}^{6}) x_{0}^{21}

+ (b_{0}^{5} + b_{0}^{9} + 2 b_{0}^{3} c_{0} + 2 b_{0}^{3} c_{0}^{3}) x_{0}^{24} + b_{0}^{3} x_{0}^{30} and

y_{0}^{3} [2 b_{0}^{10} + b_{0}^{14} + b_{0}^{18} + 2 b_{0}^{4} c_{0} + 2 b_{0}^{8} c_{0} + 2 b_{0}^{12} c_{0} + 2 b_{0}^{6} c_{0}^{2} + c_{0}^{3}

+ b_{0}^{12} c_{0}^{3} + 2 b_{0}^{6} c_{0}^{6} + 2 c_{0}^{9} + (b_{0}^{7} + 2 b_{0}^{15} + b_{0} c_{0} + b_{0}^{5} c_{0} + b_{0}^{3} c_{0}^{2} + 2 b_{0}^{9} c_{0}^{3} + 2 b_{0}^{3} c_{0}^{6}) x_{0}^{3} + (1 + 2 b_{0}^{12} + 2 b_{0}^{6} c_{0}^{3} + 2 c_{0}^{6}) x_{0}^{6} + 2 b_{0}^{9} x_{0}^{9} + (2 b_{0}^{2} + c_{0} + c_{0}^{3}) x_{0}^{12} + 2 b_{0}^{3} x_{0}^{15} + (2 + 2 b_{0}^{12} + 2 b_{0}^{6} c_{0}^{3} + 2 c_{0}^{6}) x_{0}^{18}

+ 2 b_{0}^{3} x_{0}^{27} + x_{0}^{30}]

respectively. The existence of this lift of the Frobenius leads us to believe that indeed what we just obtained is indeed the Mochizuki lift of C.

Bibliography

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[4] S. Mochizuki. A theory of ordinary p-adic curves. Publ. Res. Inst. Math. Sci., 32:957–1152, 1996.

[5] M. Raynaud. Around the Mordell conjecture for function fields and a conjeture of Serge Lang. Lecture Notes in Math., 1016:1–19, 1983.

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Vita

Luís Renato Abib Finotti was born in Uberlândia, Minas Gerais, Brazil, on March 30, 1973, the son of Eliane Espir Abib Finotti and Gilson Finotti. After completing his work at Palmares High School, São Paulo (Brazil), in 1990, he entered University of São Paulo. He received the degree of Bachelor of Science from University of São Paulo in 1994. In 1995, he entered the Masters Program of University of São Paulo, graduating in 1997. In the same year, he entered the Ph.D. Program of the University of Texas at Austin.

Permanent address: 2814 Nueces St. Austin, TX – 78705

This dissertation was typeset with L^{A}TEX^{‡} by the author.

‡L^{A}TEX is a document preparation system developed by Leslie Lamport as a special version of Donald Knuth’s TEX Program.