COORDINATES OF THE j-INVARIANT OF THE CANONICAL LIFTING

LUÍS R. A. FINOTTI

Abstract. Let j_{0} ↦ (j_{0}, J_{1}(j_{0}), J_{2}(j_{0}), …) be the map that takes the j-invariant of an ordinary elliptic curve in characteristic p to the j-invariant of its canonical lifting over the ring of Witt vectors. We have that J_{i} ∈ 𝔽_{p}(X), and in this paper we describe how to derive these rational functions from the modular polynomial in an efficient way and give more precise description of the numerators and denominators of the reduced forms of these functions. In particular, upper bounds are given for the order of their poles.

1. Introduction

Let 𝕜 be a perfect field of characteristic p > 0 and W(𝕜) be the ring of Witt vectors over 𝕜. Then, given an ordinary elliptic curve E/𝕜, there is a unique elliptic curve (up to isomorphism), say E/W(𝕜), which reduces to E modulo p and for which we can lift the Frobenius. E is then called the canonical lifting of E. (See, for instance, [2] or [12].) Hence, given an ordinary j-invariant j_{0} ∈ 𝕜, the canonical lifting gives us a unique j ∈ W(𝕜). Therefore, if 𝕜^{ord} denotes the set of ordinary values of j-invariants in 𝕜, then we have functions J_{i} : 𝕜^{ord} → 𝕜, for i = 1, 2, 3, …, such that the j-invariant of the canonical lifting of an elliptic curve with j-invariant j_{0} ∈ 𝕜^{ord} is (j_{0}, J_{1}(j_{0}), J_{2}(j_{0}), …).

B. Mazur asked about the nature of these functions J_{i}. Partial answers were given in [4], [6], and [7]. Before we can quote the main results of these references, we need a little more notation.

Let

S_{p}(X) =^{def} \frac{ss_{p}(X)}{X^{δ}(X − 1728)^{ϵ}} ,

where

ss_{p}(X) =^{def} ∏_{j supersing.}(X − j)

is the supersingular polynomial (as in, for instance, [3]),

δ =^{def} \begin{cases}0, & if p ≡ 1 (mod 6); \\ 1, & if p ≡ 5 (mod 6); \end{cases}  and  ϵ =^{def} \begin{cases}0, & if p ≡ 1 (mod 4); \\ 1, & if p ≡ 3 (mod 4). \end{cases}

2000 Mathematics Subject Classification. Primary 11G07; Secondary 11Y99.

Key words and phrases. elliptic curves, canonical lifting, j-invariant, modular polynomial.

Hence, S_{p}(X) ∈ 𝔽_{p}[X], and S_{p}(0), S_{p}(1728) ≠ 0. (See, for instance, [3].) Also, let

ι = \begin{cases}1, & if p ≠ 31; \\ 2, & if p = 31. \end{cases}

In [4], [6], and [7], the following results were proven:

Theorem 1.1. We have J_{i}(X) ∈ 𝔽_{p}(X). More precisely, if p ≥ 5, J_{i} = F_{i}/G_{i}, with F_{i}, G_{i} ∈ 𝔽_{p}[X], (F_{i}, G_{i}) = 1, G_{i} monic, and r_{i} = (i−1)p^{i−1}, then, for i ∈ {1, 2, 3}, we have:

(1) deg F_{i} − deg G_{i} = p^{i} − ι;

(2) G_{i} = S_{p}(X)^{ip^{i−1}+(i−1)p^{i−2}} · H_{i}, where H_{1} = 1, H_{2} = (X − 1728)^{ϵr_{2}}, and H_{3} =

X^{δp^{2}} · (X − 1728)^{t} for some t ∈ {0, …, ϵr_{3}}.

Also, in the above references, explicit and “simplified” (in a sense to be made precise later) expressions for J_{i} based on the modular polynomial Φ_{p}(X, Y), for i ∈ {1, 2, 3} are given. (In fact, the above theorem is derived from these expressions.) The main goals here are to extend the theorem above for i > 3 and give a general simplified expression for J_{n}. Although we need some more notation to give this expression, which is done in Section 3, we can state now the generalization of Theorem 1.1:

Theorem 1.2. Let p ≥ 5, J_{i} = F_{i}/G_{i}, with F_{i}, G_{i} ∈ 𝔽_{p}[X], (F_{i}, G_{i}) = 1, and G_{i} monic. Also, let r_{i} = (i − 1)p^{i−1}, s_{i} = ((i − 3)p^{i} + ip^{i−1})/3 and s′_{i} = max{0, s_{i}}. Then, for all i ∈ ℤ_{>0} we have:

(1) deg F_{i} − deg G_{i} = p^{i} − ι;

(2) G_{i} = S_{p}(X)^{ip^{i−1}+(i−1)p^{i−2}} · H_{i}, where H_{i} | X^{δs′_{i}} · (X − 1728)^{ϵr_{i}}.

It should be observed that, unlike for i = 1, 2, we don’t have an exact formula for H_{i}. In fact, it is not hard to see that if δ = 1 and i > 3, then we actually have that the pole of J_{i} at X = 0 of order strictly less than s_{i}. For poles at X = 1728, examples show that the orders of poles seem also to be always strictly less than r_{i} if i > 2 and ϵ = 1. So, there still some room for improvement.

We now give a brief description of the following sections. In Section 2 we review the Greenberg transform and give a formula for it, as derived in [5]. (Quite a bit of notation is introduced in this section.) In Section 3 we apply the formula for the Greenberg transform to obtain the desired formula for J_{n}. In Section 4 we introduce some lemmas that will help with the proof of Theorem 1.2, while in Sections 5 and 6 we prove items 1 and 2, respectively, from the theorem.

COORD. OF THE j-INV. OF THE CAN. LIFT. 3

2. The Greenberg Transform

The Greenberg transform is crucial to the proof of the main results. In this section we briefly review it. (See also [10] and [8].) We will assume throughout that 𝕜 is a perfect field of characteristic p.

Definition 2.1. Let f(x, y) ∈ W(𝕜)[x, y]. If we replace x and y by (x_{0}, x_{1}, …) and (y_{0}, y_{1}, …), seen as Witt vectors of unknowns, and expand the resulting expression using sums and products of Witt vectors, we obtain a Witt vector (f_{0}, f_{1}, …), with f_{i} ∈ 𝕜[x_{0}, …, x_{i}, y_{0}, …, y_{i}]. This resulting vector is called the Greenberg transform of f and will be denoted by 𝒢 (f).

Moreover, if

C/W(𝕜) : f(x, y) = 0,

we define the Greenberg transform 𝒢 (C) of C to be the (infinite dimensional) variety over 𝕜 defined by the zeros of the coordinates of 𝒢 (f).

It is clear from the definition that there is a bijection between C(W(𝕜)) and 𝒢 (C)(𝕜). In [5] a formula for the Greenberg transform is given. Although rather involved, it will

be our main tool. Before we can give this formula, quite a bit of notation is necessary. We start with some notation regarding Witt vectors.

Definition 2.2. (1) We denote by π the reduction modulo p map, i.e., π((a_{0}, a_{1}, …)) =

a_{0}.

(2) Let a ∈ 𝕜. Then, the Teichmüller lift of a is the Witt vector τ(a) =^{def} (a, 0, 0, …). (Hence, τ is a section of π and when restricted to 𝕜^{×} yields a group homomorphism.)

(3) Define τ(𝕜) =^{def} {(a_{0}, 0, 0, …) ∈ W(𝕜) : a_{0} ∈ 𝕜}. (This is a multiplicative set. E.g., if 𝕜 = 𝔽_{q}, than τ(𝕜) is made of all (q − 1)-th roots of unity and zero.)

(4) Let a ∈ W(𝕜). Define ξ_{k}(a), for k ∈ ℤ_{≥0}, as the unique element of τ(𝕜) such that a = ∑_{k=0}^{∞} ξ_{k}(a)p^{k}. (This is well defined since W(𝕜) is a strict p-ring and τ(𝕜) is a complete set of representatives of 𝕜 = W(𝕜)/(p) in W(𝕜).)

With the notation above, we have

a = ∑_{k=0}^{∞} ξ_{k}(a)p^{k} = (π(ξ_{0}(a)), π(ξ_{1}(a))^{p}, π(ξ_{2}(a))^{p^{2}}, …)  (2.1)

and

(a_{0}, a_{1}, …) = ∑_{k=0}^{∞} τ(a_{k})^{1/p^{k}}p^{k}.  (2.2)

(Remember we are assuming that 𝕜 is perfect.)

We shall also need some auxiliary recursively defined functions.

Definition 2.3. Let p be a prime. Define η_{0}(X_{1}, …, X_{r}) =^{def} X_{1} +⋯+X_{r} ∈ ℚ[X_{1}, …, X_{r}], and recursively for k ≥ 1

η_{k}(X_{1}, …, X_{r}) =^{def} \frac{X_{1}^{p^{k}} + ⋯ + X_{r}^{p^{k}}}{p^{k}} − ∑_{i=0}^{k−1} \frac{η_{i}(X_{1}, …, X_{r})^{p^{k−i}}}{p^{k−i}} .  (2.3)

Also, define η_{k}(X_{1}) = 0 for k ≥ 1.

By Corollary 5.7 from [5], we have that η_{i}(X_{1}, …, X_{r}) ∈ ℤ[X_{1}, …, X_{r}] for all i.

Definition 2.4. If g ∈ W(𝕜)[x, y] is given by g = ∑_{i,j} a_{i,j}x^{i}y^{j}, then we write

ξ_{k}(g) =^{def} ∑_{i,j} ξ_{k}(a_{i,j})x^{i}y^{j},

with ξ_{k} as in Definition 2.2. (Hence, g ≡ ∑_{k=0}^{r} ξ_{k}(g)p^{k} (mod p^{r+1}).) Furthermore, define

g^{(i,j)} =^{def} \frac{1}{i!j!} \frac{∂^{i+j}}{∂x^{i}∂y^{j}} g and g_{i,j,k} =^{def} ξ_{k}(g^{(i,j)}).

Definition 2.5. We define D_{k,n}^{i,j} to be the coefficient of t^{k} in

(tx_{1}^{p^{n−1}} + t^{2}x_{2}^{p^{n−2}} + ⋯ + t^{n}x_{n})^{i}(ty_{1}^{p^{n−1}} + t^{2}y_{2}^{p^{n−2}} + ⋯ + t^{n}y_{n})^{j}.

(E.g., if n ≥ 2, then D_{4,n}^{1,2} = 2x_{1}^{p^{n−1}}y_{1}^{p^{n−1}}y_{2}^{p^{n−2}} +x_{2}^{p^{n−2}}y_{1}^{2p^{n−1}}.) Furthermore, we shall denote

D_{k,n,l}^{i,j} =^{def} ξ_{l}(D_{k,n}^{i,j}).

Note that if k < r (and r ≥ i), then D_{k,n}^{i,r−i} = 0 and if k ≠ 0, then D_{k,n}^{0,0} = 0.

Definition 2.6. If v_{1} and v_{2} are finite vectors, we denote by v_{1} ⊙ v_{2} the concatenation of these vectors, i.e.,

(a_{1}, …, a_{m}) ⊙ (b_{1}, …, b_{n}) = (a_{1}, …, a_{m}, b_{1}, …, b_{n}).

Moreover, if f is a polynomial (possibly in many variables), we write vec (f) for the vector that contains the terms of f (after some choice of order for the monomials). It is important to observe that we are assuming that the terms are reduced, i.e., if f = 1 + X + 2X, then vec (f) = (1, 3X), not (1, X, 2X).

Definition 2.7. Let f ∈ W(𝕜)[x_{0}, y_{0}]. Then, for a given n ≥ 0, let

(𝒢_{n,1}, …, 𝒢_{n,N_{n}}) =^{def} ⨀_{r=0}^{n} ⨀_{i=0}^{r} ⨀_{j=r}^{n} ⨀_{k=r}^{j} vec ((f^{σ^{n}})_{i,r−i,n−j}(x_{0}^{p^{n}}, y_{0}^{p^{n}})D_{k,n,j−k}^{i,r−i}) .

(The ordering of the concatenations is not important.) Also, recursively, if n > 1, define

𝒢_{n,N_{n}+i+1} =^{def} η_{n−i}(𝒢_{i,1}, …, 𝒢_{i,N_{i}+i}),

for i ∈ {0, …, (n − 1)}. Then, define,

f_{n} =^{def} ∑_{i=1}^{N_{n}+n} 𝒢_{n,i},

and 𝒢_{n} =^{def} (𝒢_{n,1}, …, 𝒢_{n,N_{n}+n}).

Then, with the notation above, Theorem 6.4 from [5] gives the desire formula for the Greenberg transform:

Theorem 2.8. With the notation above, we have that 𝒢 (f) = (f_{0}, f_{1}, …), where f_{n} is the reduction modulo p of

f_{n} = ∑_{i=1}^{N_{n}+n} 𝒢_{n,i} = ∑_{r=0}^{n} ∑_{i=0}^{r} ∑_{j=r}^{n} ∑_{k=r}^{j}(f^{σ^{n}})_{i,r−i,n−j}(x_{0}^{p^{n}}, y_{0}^{p^{n}})D_{k,n,j−k}^{i,r−i} + ∑_{i=0}^{n−1} η_{n−i}(𝒢_{i}). (2.4)

3. A Simplified Formula for J_{n}

By Theorem 3 of [12], we have that if (j_{0}, J_{1}, J_{2}, …) is the j-invariant of the canonical lifting of the curve with j-invariant j_{0}, then

Φ_{p}((j_{0}, J_{1}, J_{2}, …), (j_{0}^{p}, J_{1}^{p}, J_{2}^{p}, …)) = 0,  (3.1)

where Φ_{p}(X, Y) is the (classical) modular polynomial. Together with the formula for the Greenberg transform (Theorem 2.8 above), one can immediately deduce a formula for J_{n} (in terms of the modular polynomial). On the other hand, we need some improvements in this first immediate formula.

So, from now on we will let f = Φ_{p}. We will use the notation 𝒢_{n,i} and 𝒢_{n} associated with this particular choice of f.

Definition 3.1. For a given n ≥ 0, let ℋ_{n} = (ℋ_{n,1}, …, ℋ_{n,N_{n}}) =^{def} (𝒢_{n,1}, …, 𝒢_{n,N_{n}}) and

h_{n} = ∑_{i=1}^{N_{n}} ℋ_{n,i} = ∑_{r=0}^{n} ∑_{i=0}^{r} ∑_{j=r}^{n} ∑_{k=r}^{j}(f^{σ^{n}})_{i,r−i,n−j}(x_{0}^{p^{n}}, y_{0}^{p^{n}})D_{k,n,j−k}^{i,r−i}.

Also, let h_{n} be the reduction modulo p of h_{n} and we shall use the over bar to denote further reductions modulo p, e.g., ℋ̄_{n} denotes the reduction modulo p of ℋ_{n}.

The following theorem gives the desired simplification.

Theorem 3.2. Let p ≥ 3 and

𝒥̄_{i,1}^{(l)} =^{def} η_{i}(ℋ̄_{l})
𝒥̄_{i,2}^{(l)} =^{def} η_{i}(η_{l−1}(𝒢̄_{1}), …, η_{1}(𝒢̄_{l−1}))
𝒥̄_{i,3}^{(l)} =^{def} η_{i}(h_{l}, η_{l−1}(𝒢̄_{1}) + ⋯ + η_{1}(𝒢̄_{l−1})).

For i > 1, let for j ∈ {1, …, (i − 1)},

𝒥̄_{i,3+j}^{(l)} =^{def} η_{j}(𝒥̄_{i−j,1}^{(l)}, …, 𝒥̄_{i−j,i−j+2}^{(l)}).

Then, η_{i}(𝒢̄_{l}) ≡ ∑_{j=1}^{i+2} 𝒥̄_{i,j}^{(l)} (mod x_{0}^{p} − y_{0}). In particular, η_{i}(𝒢̄_{1}) ≡ η_{i}(ℋ̄_{1}) (mod x_{0}^{p} − y_{0}) for all i ≥ 1.

Note that the congruences above will actually yield equalities when we replace x_{i} = J_{i} and y_{i} = J_{i}^{p}, as necessary when using Eq. (3.1).

This theorem is a consequence of the Proposition 5.4 from [5], which we state below:

Proposition 3.3. Let v = (a_{1}, …, a_{m}) and w = (b_{1}, …, b_{n}). For any i ≥ 1, let

ℳ_{i,1} = η_{i}(v), ℳ_{i,2} = η_{i}(w),
ℳ_{i,3} = η_{i}(a_{1} + ⋯ + a_{m}, b_{1} + ⋯ + b_{n}),

and, recursively for i > 1, define

ℳ_{i,3+j} = η_{j}(ℳ_{i−j,1}, …, ℳ_{i−j,i−j+2})

for j ∈ {1, …, i − 1}. Then,

η_{i}(v ⊙ w) = ∑_{j=1}^{i+2} ℳ_{i,j}.

We can now prove the Theorem:

Proof of Theorem 3.2. Firstly, by Lemma 5.1 from [7], since p ≥ 3 we have that η_{i}(𝒢̄_{0}) ≡ 0 (mod x_{0}^{p} − y_{0}) for all i ≥ 1.

Let v = ℋ̄_{l} and w = (η_{l}(𝒢̄_{0}), …, η_{1}(𝒢̄_{l−1})). Then, we have that 𝒢̄_{l} = v ⊙ w.

By Proposition 3.3, it suffices to prove that 𝒥̄_{j,k}^{(l)} ≡ ℳ_{j,k} (mod x_{0}^{p} − y_{0}) (with ℳ_{j,k} from Proposition 3.3 with our v and w above). We prove this by induction on i.

Since η_{l}(𝒢̄_{0}) ≡ 0 (mod x_{0}^{p} − y_{0}), we have ℳ_{i,j} ≡ 𝒥̄_{i,j}^{(l)} for j ∈ {1, 2, 3} for any i. In particular, the statement immediately follows for i = 1.

Now, if 𝒥̄_{j,k}^{(l)} ≡ ℳ_{j,k} (mod x_{0}^{p} − y_{0}) for j < i, then for j ∈ {1, …, (i − 1)} we have

ℳ_{i,3+j} = η_{j}(ℳ_{i−j,1}, …, ℳ_{i−j,i−j+2}) ≡ η_{j}(𝒥̄_{i−j,1}^{(l)}, …, 𝒥̄_{i−j,i−j+2}^{(l)}) = 𝒥̄_{i,3+j}^{(l)} (mod x_{0}^{p} − y_{0}), which finishes the proof. □

We introduce some more notation.

Definition 3.4. We define:

(1) If g(x_{0}, y_{0}) ∈ W(𝕜)[x_{0}, y_{0}], then we denote by g_{i,j,k} the reduction modulo p of g_{i,j,k} (with the notation of Definition 2.4).

(2) Given a polynomial f ∈ 𝕜[x_{0}, , …, x_{n}, y_{0}, …, y_{n}], we shall write f′ for f(J_{0}, …, J_{n}, J_{0}^{p}, …, J_{n}^{p}). (E.g., 𝒢̄′_{i,j} is the value of 𝒢̄_{i,j} when x_{t} is replaced by J_{t} and y_{t} is replaced by J_{t}^{p} for all t.)

(3) We denote by E_{k,n,l}^{i,j} the evaluation of the reduction modulo p of D_{k,n,l}^{i,j} (as in Definition 2.5) when x_{t} is replaced by J_{t}(X) and y_{t} is replaced by J_{t}(X)^{p}.

(4) Let ℐ̄_{i,j} =^{def} η_{i}(𝒢̄′_{j}).

We can now give the formula for J_{n}.

Theorem 3.5. Let p ≥ 3, f = Φ_{p}, and

I_{n} =^{def} f_{0,0,n}(X^{p^{n}}, X^{p^{n+1}}) + ∑_{j=1}^{n−1} f_{0,1,n−j}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p^{n−j+1}}

∑_{j=1}^{n−1} f_{1,0,n−j}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p^{n−j}} − ∑_{k=2}^{n} E_{k,n,n−k}^{1,1}+

∑_{r=2}^{n−1} ∑_{i=0}^{r} ∑_{j=r}^{n−1} ∑_{k=r}^{j} f_{i,r−i,n−j}(X^{p^{n}}, X^{p^{n+1}})E_{k,n,j−k}^{i,r−i}. (3.2)

Then, we have

J_{n}^{p} = −\frac{1}{(X^{p^{2}} − X)^{p^{n}}} [I_{n} + ∑_{i=0}^{n−1} ℐ̄_{n−i,i}] .  (3.3)

(Note that this formula is simplified if one uses Theorem 3.2 to compute ∑_{i=0}^{n−1} ℐ̄_{n−i,i}.)

Proof. By Eq. (3.1), we just need to apply Theorem 2.8 to Φ_{p}((j_{0}, J_{1}, …), (j_{0}^{p}, J_{1}^{p}, …)).

First, observer that since f has integral coefficients, we have that f^{σ^{n}} = f. Then, evaluating f_{n} (from Theorem 2.8) at x = (X, J_{1}(X), …) and y = (X^{p}, J_{1}(X)^{p}, …) and reducing modulo p, we obtain

∑_{r=0}^{n} ∑_{i=0}^{r} ∑_{j=r}^{n} ∑_{k=r}^{j} f_{i,r−i,n−j}(X^{p^{n}}, X^{p^{n+1}})E_{k,n,j−k}^{i,r−i} + ∑_{i=0}^{n−1} ℐ̄_{n−i,i} = 0.  (3.4)

When r = 0, the first sum of (3.4) gives only the term f_{0,0,n}(X^{p^{n}}, X^{p^{n+1}}), since D_{k,n,j−k}^{0,0} = 0 if k ≠ 0, or if k = 0 and j ≠ k. So, the part of Eq. (3.4) from r = 0 is only f_{0,0,n}(X^{p^{n}}, X^{p^{n+1}}).

We now look at r = 1. We have that D_{k,n}^{0,1} = y_{k}^{p^{n−k}} and D_{k,n}^{1,0} = x_{k}^{p^{n−k}}. So, D_{k,n,0}^{0,1} = y_{k}^{p^{n−k}}, D_{k,n,0}^{1,0} = x_{k}^{p^{n−k}}, and D_{k,n,j−k}^{0,1} = D_{k,n,j−k}^{1,0} = 0 if k < j. So, the terms with r = 1 from the first (nested) sum of Eq. (3.4) are

∑_{j=1}^{n} f_{0,1,n−j}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p^{n−j+1}} + ∑_{j=1}^{n} f_{1,0,n−j}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p^{n−j}}.  (3.5)

For j = n we get the terms involving J_{n}. By Kronecker’s relation (i.e., f(X, Y) = (X^{p} − Y)(X − Y^{p})), we get that f_{0,1,0}(X^{p^{n}}, X^{p^{n+1}}) = (X^{p^{2}} − X)^{p^{n}} and f_{1,0,0}(X^{p^{n}}, X^{p^{n+1}}) = 0. So, Eq. (3.5) is equal to

(X^{p^{2}} − X)^{p^{n}}J_{n}^{p} + ∑_{j=1}^{n−1} f_{0,1,n−j}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p^{n−j+1}} + ∑_{j=1}^{n−1} f_{1,0,n−j}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p^{n−j}}. (3.6)

We now look at r = 2. First, observe that f_{0,2,0} = f_{2,0,0} = 0, and f_{1,1,0} = −1. So, the term with r = 2 in the first sum of Eq. (3.4) is:

∑_{i=0}^{2} ∑_{j=2}^{n−1} ∑_{k=2}^{j} f_{i,2−i,n−j}(X^{p^{n}}, X^{p^{n+1}})E_{k,n,j−k}^{i,2−i} − ∑_{k=2}^{n} E_{k,n,n−k}^{1,1}.  (3.7)

Finally, for r ≥ 3, we have that f_{i,r−i,0} = 0 for all r and i in their given ranges. Thus, the part with r ≥ 3 of Eq. (3.4) is:

∑_{r=3}^{n} ∑_{i=0}^{r} ∑_{j=r}^{n−1} ∑_{k=r}^{j} f_{i,r−i,n−j}(X^{p^{n}}, X^{p^{n+1}})E_{k,n,j−k}^{i,r−i}.  (3.8)

The result now easily follows. □

It should be noted that, with the observation about the simplification in the statement, Theorem 3.5 is the generalization of the formulas for J_{i} for i = 1, 2, 3, given in [4], by Theorem 9.1 from [6], and by Theorem 5.5 from [7] respectively. As observed in this last reference, great gains are indeed obtained in computations when using the simplifications of Theorem 3.2. (On the other hand, note also that, with Theorem 1.2, one can also compute the J_{n}’s using interpolation.)

4. Some Lemmas

In this section we will introduce some lemmas necessary for the proof of Theorem 1.2. The proof of is done by looking at order of zeros and poles of J_{n}. Hence, we need to look

at valuations of the terms that appear in Eq. (3.3). Clearly the most troublesome part is ∑_{i=0}^{n−1} ℐ̄_{n−i,i}. The next two lemmas help us deal with those.

Lemma 4.1. Let v be a valuation on 𝔽_{p}(X) such that v(a) = 0 for all a ∈ 𝔽_{p}^{×}. Then, if v_{0} < 0 and v =^{def} (a_{1}, …, a_{n}) ∈ 𝔽_{p}(X)^{n} with v(a_{i}) ≥ v_{0} for all i, we have that v(η_{j}(v)) > p^{j}v_{0} for j ≥ 1.

Proof. From definition of η_{j}, a simple induction gives that, for j ≥ 1, the polynomial η_{j}(X_{1}, …, X_{r}) has degree on each variable X_{i} strictly less than p^{j}. The lemma then easily follows. □

Lemma 4.2. Let p ≥ 3 and v be a valuation on 𝔽_{p}(X) such that v(a) = 0 for all a ∈ 𝔽_{p}^{×}. Then, if v(ℋ̄′_{l,j}) ≥ p^{l}v_{0}, with v_{0} < 0, for all j, then v(ℐ̄_{i,l}) > p^{l+i}v_{0} for i ≥ 1. (Remember, ℐ̄_{i,l} = η_{i}(𝒢̄′_{l}).)

Proof. By Theorem 3.2, we have that ℐ̄_{i,l} = η_{i}(𝒢̄′_{l}) = ∑_{j=1}^{i+2}(𝒥̄_{i,j}^{(l)})′, it suffices to show that v((𝒥̄_{i,j}^{(l)})′) > p^{l+i}v_{0}.

We first prove that if v((𝒥̄_{t,j}^{(l)})′) > p^{l+t}v_{0} for all t < i (and hence v(ℐ̄_{t,l}) > p^{l+t}v_{0}), we also have, v((𝒥̄_{i,j}^{(l)})′) > p^{l+i}v_{0}.

Indeed, by Lemma 4.1, we have that for j ∈ {1, …, (i − 1)} that

v((𝒥̄_{i,3+j}^{(l)})′) = v(η_{j}((𝒥̄_{i−j,1}^{(l)})′, …, (𝒥̄_{i−j,i−j+2}^{(l)})′)) > p^{l+i}v_{0}.

Moreover, we have v((𝒥̄_{i,1}^{(l)})′) = v(η_{i}(ℋ̄′_{l})) > p^{l+i}v_{0} (by hypothesis),

v((𝒥̄_{i,2}^{(l)})′) = v(η_{i}(η_{l−1}(𝒢̄′_{1}), …, η_{1}(𝒢̄′_{l−1})))) = v(η_{i}(ℐ̄_{l−1,1}, …, ℐ̄_{1,l−1})) > p^{l+i}v_{0},

and since h_{l} = ∑_{j} ℋ̄_{l,i}, we have

v((𝒥̄_{i,3}^{(l)})′) = v(η_{i}(h′_{l}, ℐ̄_{l−1,1} + ⋯ + ℐ̄_{1,l−1})) > p^{l+i}v_{0}.

Thus, by induction, it suffices to prove that v((𝒥̄_{1,j}^{(l)})′) > p^{l+1}v_{0}, i.e., the case i = 1. We prove this by induction on l.

For l = 1, we have v((𝒥̄_{1,1}^{(1)})′) = v(η_{1}(ℋ̄′_{1})) > p^{2}v_{0} by hypothesis and Lemma 4.1. Also, (𝒥̄_{1,2}^{(1)})′ = (𝒥̄_{1,3}^{(1)})′ = 0. Therefore, the statement is true for l = 1.

So, suppose now that v((𝒥̄_{1,j}^{(k)})′) ≥ p^{k+1}v_{0} for all k < l. Then, v((𝒥̄_{1,1}^{(l)})′) = v(η_{l}(ℋ̄′_{1})) > p^{l+1}v_{0} again by hypothesis and Lemma 4.1. Now, since h_{l} = ∑_{j} ℋ̄_{l,i} and η_{i}(𝒢̄′_{k}) = ∑_{j=1}^{i+2}(𝒥̄_{i,j}^{(k)})′, we also have v((𝒥̄_{1,j}^{(l)})′) > p^{l+1}v_{0} for j = 2, 3, finishing the proof. □

Since we also need to look at degrees of the f_{i,j,k}’s (in Eq. (3.2)), we shall need the following lemma.

Lemma 4.3. Let f = Φ_{p}. Then

f(x, y) = x^{p+1} + y^{p+1} − x^{p}y^{p} + ∑_{k=0}^{p−1} g_{k}(x)y^{k},  where deg g_{i} < p,  (4.1)

and

deg f_{i,j}(x^{p^{n}}, x^{p^{n+1}}) ≤ p^{n+2} − (j − 1)p^{n+1} − ip^{n}.

Proof. The first part is well known. For instance, one can use Theorem 5.3 of [11] together with Kronecker’s congruence relation.

The second part follows from the first. We have:

f_{i,j}(x^{p^{n}}, x^{p^{n+1}}) = δ_{j,0}\binom{p + 1}{i}x^{p^{n+1}−(i−1)p^{n}} + δ_{i,0}\binom{p + 1}{j}x^{p^{n+2}−(j−1)p^{n+1}}

− \binom{p}{i}\binom{p}{j}x^{p^{n+2}−(j−1)p^{n+1}−ip^{n}} + ∑_{k=0}^{p−1−j} \binom{k + j}{j}\frac{g_{k+j}^{(i)}(x^{p^{n}})}{i!}x^{kp^{n+1}},

where δ_{i,j} is the Kronecker delta function. The result then immediately follows. □

We also have the following trivial lemma that we need for looking at valuations of terms in E_{k,n,j−k}^{i,r−i}:

Lemma 4.4. We have

D_{k,n}^{i,r−i} = ∑_{k_{1}=r}^{k} (  ∑_{\begin{matrix}i_{1}+i_{2}+···+i_{n}=i \\ i_{1}+2i_{2}+···+ni_{n}=k_{1} \end{matrix}} x_{1}^{p^{n−1}i_{1}}x_{2}^{p^{n−2}i_{2}} ⋯ x_{n}^{i_{n}} ) (  ∑_{\begin{matrix}j_{1}+j_{2}+···+j_{n}=r−i \\ j_{1}+2j_{2}+···+nj_{n}=k−k_{1} \end{matrix}} y_{1}^{p^{n−1}i_{1}}y_{2}^{p^{n−2}i_{2}} ⋯ y_{n}^{i_{n}} ) .

Moreover, if k < n or i, (r − i) ≠ 0, then the terms in x_{n} and y_{n} can be omitted, i.e., we must have i_{n} = j_{n} = 0.

After possible reindexing, notice we may write (with I_{n} as in Eq. (3.2))

I_{n} = ∑_{j=3}^{N_{n}} ℋ̄′_{n,j},

ℋ̄′_{n,1} = −X^{p^{n}}J_{n}^{p}, and ℋ̄′_{n,2} = X^{p^{n+2}}J_{n}^{p}. We shall keep this notation from now on. We can now introduce the main lemma used in the proof of Theorem 1.2.

Lemma 4.5. Let p ≥ 3 and v be a valuation on 𝔽_{p}(X) such that v(a) = 0 for all a ∈ 𝔽_{p}^{×}. Suppose that for given n_{0}, n > n_{0}, and sequence of real numbers {v_{i}}_{i≥n_{0}} the following conditions are satisfied:

(1) If k < n_{0} we have that v(ℋ̄′_{k,i}) ≥ 0 for all i and v(ℋ̄′_{n_{0},i}) ≥ 0 for all i ≥ 3.

(2) v_{n_{0}+1} < 0 and v_{i} < pv_{i−1} for any i > n_{0}.

(3) v(ℋ̄′_{l,t}) ≥ v_{l} for t ∈ {3, …, N_{l}} if n_{0} < l ≤ n.

(4) v(J_{l}) ≥ (v_{l} − p^{l} v(X^{p^{2}} − X))/p if n_{0} ≤ l < n.

(5) p^{n+e} v(X) + p^{n−l}v_{l} − p^{n} v(X^{p^{2}} − X) ≥ v_{n} for e = 0, 2 if n_{0} ≤ l < n.

Then, we have that v(J_{n}) ≥ (v_{n} − p^{n} v(X^{p^{2}} − X))/p.

Moreover, if further there exists a unique t_{0} ∈ {3, …, N_{n}} such that v(ℋ̄′_{n,t_{0}}) = v_{n}, then v(J_{n}) = (v_{n} − p^{n} v(X^{p^{2}} − X))/p.

Proof. By conditions 4 and 5, we have for n_{0} ≤ l < n that

v(X^{p^{l+e}}J_{l}^{p}) ≥ p^{l+e} v(X) + v_{l} − p^{l} v(X^{p^{2}} − X)) ≥ p^{l−n}v_{n}

for e = 0, 2. Since v_{n} < 0, condition 1 together with the previous argument gives us that v(ℋ̄′_{l,t}) ≥ p^{l−n}v_{n} for t = 1, 2 for all l < n. Since by condition 2 we also have v_{l} > p^{l−n}v_{n}, by conditions 1 and 3 and using Lemma 4.2, we have v(ℐ̄_{n−i,i}) > v_{n} for i ∈ {0, …, n − 1}.

Now, Eq. (3.3) can be written as

−(X^{p^{2}} − X)^{p^{n}}J_{n}^{p} = ∑_{t=3}^{N_{n}} ℋ̄′_{n,t} + ∑_{i=0}^{n−1} ℐ̄_{n−i,i}.

By the argument above and condition 3 (with l = n), we have v(J_{n}) ≥ (v_{n} − p^{n} v(X^{p^{2}} − X))/p.

The final observation on the statement also can easily be seen from the analysis above with the due restrictions. □

5. Proof of Item 1 of Theorem 1.2

We are now ready to prove item 1 of Theorem 1.2. To simplify the exposition we introduce one extra simple lemma.

Lemma 5.1. Let 0 ≤ i ≤ r ≤ n and assume that:

j_{1} + ⋯ + j_{n−1} = i  and  j_{1} + ⋯ + j_{n−1} = r − i,

with i_{t}, j_{t} ∈ ℤ_{≥0}. Then, for r ≥ 2 we have that

∑_{t=1}^{n−1} \frac{i_{t}}{p^{t}} + \frac{j_{t}}{p^{t−1}} > \frac{1}{p^{n−1}} .

Proof. We have:

∑_{t=1}^{n−1} \frac{i_{t}}{p^{t}} + \frac{j_{t}}{p^{t−1}} ≥ ∑_{t=1}^{n−1} \frac{i_{t}}{p^{n−1}} + \frac{j_{t}}{p^{n−2}} ≥ ∑_{t=1}^{n−1} \frac{i_{t}}{p^{n−1}} + \frac{j_{t}}{p^{n−1}} = \frac{r}{p^{n−1}} > \frac{1}{p^{n−1}} .

□

Let v_{∞} be the valuation on 𝔽_{p}(X) given by the order of zero at infinity. Then, item 1 of Theorem 1.2, namely deg F_{n} − deg G_{n} = p^{n} − ι, is equivalent to v_{∞}(J_{n}) = −(p^{n} − ι), which is what we shall prove below.

Let v_{n} =^{def} −(p^{n+2}+p^{n+1}−ιp). We will use induction on n to show that v_{∞}(J_{n}) = −(p^{n}−ι) and v(ℋ̄′_{n,t}) ≥ v_{n} for t ∈ {3, …, N_{n}}. For n = 1 the results hold, as observed in [1]. (Note that ℋ̄′_{1} = 𝒢̄′_{1} = vec (f_{0,0,1}(X^{p}, X^{p^{2}})).)

So, suppose that v(ℋ̄′_{l,t}) ≥ v_{l} for t ∈ {3, …, N_{l}} and v_{∞}(J_{l}) = −(p^{l} − ι) if 1 ≤ l < n. With these assumptions and with v_{n} as given, we are in the conditions of Lemma 4.5 with n_{0} = 1.

Thus, by the lemma, we only need to verify the special case of condition 3, i.e., v_{∞}(ℋ̄′_{n,t}) ≥ −(p^{n+2} + p^{n+1} − ιp) for t ∈ {3, …, N_{n}}, and equality occurs for exactly one t.

By Lemmas 4.4 and 5.1 (and using the notation of the former), we have that if r ≥ 2, then the valuation v_{∞} of a non-zero summand of E_{k,n,j−k}^{i,r−i}, for k < n, or for i, (r − i) ≠ 0, is

− ∑_{t=1}^{n−1} i_{t}p^{n−t}(p^{t} − ι) + j_{t}p^{n−t+1}(p^{t} − ι)

= −p^{n} [∑_{t=1}^{n−1} i_{t} + pj_{t}] + ιp^{n} [∑_{t=1}^{n−1} \frac{i_{t}}{p^{t}} + \frac{j_{t}}{p^{t−1}} ]

= −(r − i)p^{n+1} − ip^{n} + ιp^{n} [∑_{t=1}^{n−1} \frac{i_{t}}{p^{t}} + \frac{j_{t}}{p^{t−1}} ]

> −(r − i)p^{n+1} − ip^{n} + ιp.

In particular, if ℋ̄′_{n,t} comes from E_{k,n,n−k}^{1,1} (and hence r = 2, i = (r − i) = 1), we have that

v_{∞}(ℋ̄′_{n,t}) > −p^{n+1} − p^{n} + ιp.

Also, by Lemma 4.3, we have, the valuation of any non-zero term of f_{i,r−i,n−j}(X^{p^{n}}, X^{p^{n+1}}) is greater than or equal to −(p^{n+2} − (r − i − 1)p^{n+1} − ip^{n}). Hence, a term ℋ̄′_{n,t} from f_{i,r−i,n−j}(X^{p^{n}}, X^{p^{n+1}})E_{k,n,j−k}^{i,r−i}, with r ≥ 2 and k < n (as in Eq. (3.2)), is also such that

v_{∞}(ℋ̄′_{n,t}) ≥ −p^{n+1} − p^{n} + ιp.

Finally, we also have:

• v_{∞}(ℋ̄′_{n,t}) ≥ −p^{n+2}, if ℋ̄′_{n,t} comes from f_{0,0,n}(X^{p^{n}}, X^{p^{n+1}}) (by Eq 4.1);

• v_{∞}(ℋ̄′_{n,t}) ≥ −(p^{n+2} + p^{n+1} − ιp^{n−j+1}) ≥ −(p^{n+2} + p^{n+1} − ιp^{2}), if ℋ̄′_{n,t} comes from

(f_{0,1,n−j}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p^{n−j+1}} with j < n;

• v_{∞}(ℋ̄′_{n,t}) ≥ −(p^{n+2} + p^{n+1} − ιp^{n−j}) ≥ −(p^{n+2} + p^{n+1} − ιp), if ℋ̄′_{n,t} comes from

f_{1,0,n−j}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p^{n−j}} with j < n.

Note that the last equality can only occur if j = (n − 1). The corresponding term comes from f_{1,0,1}(X^{p^{n}}, X^{p^{n+1}})J_{n−1}^{p}. But, by Eq. (4.1), we have that deg f_{1,0,1}(X^{p^{n}}, X^{p^{n+1}}) = p^{n+2} + p^{n+1} − p^{n}, and hence v_{∞}(ℋ̄′_{n,t}) = −(p^{n+2} + p^{n+1} − ιp) for exactly one term (coming from f_{1,0,1}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p}), which finishes the proof.

6. Proof of Item 2 of Theorem 1.2

We will now prove the item 2 of Theorem 1.2. First observe that G_{n} cannot have a zero in any ordinary value, as J_{n} must be regular at that value. In particular, the cases when δ or ϵ is zero follows immediately. Thus, from this point on, we will only consider the cases when δ and ϵ are equal to one.

So, let v_{0}, v_{1728}, w be the orders of zero at 0, 1728, and a root of S_{p}(X) respectively. Then, to prove item 2 of Theorem 1.2, we need to prove the following:

• v_{0}(J_{n}) ≥ p^{n−2}(n(−p^{2} − p)/3 + p^{2}) (when δ = 1);

• v_{1728}(J_{n}) ≥ p^{n−2}(−pn + p) (when ϵ = 1);

• w(J_{n}) = p^{n−2}(n(−p − 1) + 1);

To deal with these three cases at once, we let v denote a valuation on 𝔽_{p}(X) such that v(a) = 0 for all a ∈ 𝔽_{p}^{×}, v(X) ≥ 0, v(X^{p^{2}} − X) = 1, and v(J_{k}) ≥ p^{k−2}(αk + β) for some α < 0 and β > 0 independent of k.

We will again rely on Lemma 4.5, with v_{n} =^{def} p^{n−1}(αn+β)+p^{n} in this case. Since α < 0, the condition 2 from the lemma is satisfied, with n_{0} = 3 if v = v_{0}, with n_{0} = 2 if v = v_{1728}, and with n_{0} = 1 if v = w.

With the above choice of n_{0}, Theorem 1.1 shows that condition 1 is satisfied.

Condition 5 is equivalent to p v(X) − α ≥ p, and therefore is satisfied for the three valuations above.

So, let v be one of the valuations above. We then shall prove by induction on n that v(J_{n}) ≥ p^{n−2}(αn + β), with equality in the case of v = w, and that for n > n_{0} we have v(ℋ̄′_{n,t}) ≥ v_{n} for t ∈ {3, …, N_{n}}, with equality for exactly one t in the case v = w.

If n ≤ n_{0} and v = v_{1728} or v = w, then the result holds by Theorem 1.1. If n ≤ n_{0} and v = v_{0}, then the result holds for n = 1 by Theorem 1.2 from [6] (which just restates results from [9]), for n = 2 by Theorem 6.2 from [7], and for n = 3 by Theorem 1.1.

So, let n > n_{0} and suppose that for all l ∈ {1, …, (n−1)} we have v(J_{l}) ≥ −p^{l−2}(αl+β), with equality if v = w, and if n_{0} < l < n we have v(ℋ̄′_{l,t}) ≥ v_{l} for t ∈ {3, …, N_{l}}. Thus, condition 3 for l < n and condition 4 of Lemma 4.5 are satisfied. Therefore, we only need to verify the case l = n of condition 3, with special care when v = w.

By Lemma 4.4 (and using its notation), we have that the valuation v of a non-zero summand of E_{k,n,j−k}^{i,r−i}, with either k < n or i, (r − i) ≠ 0, is

p^{n−2} ∑_{t=1}^{n−1}(αti_{t} + βi_{t}) + p(αtj_{t} + βj_{t})

= p^{n−2}[(αk_{1} + βi) + p(α(k − k_{1}) + β(r − i))]
= p^{n−2}[αpk + βpr − (p − 1)αk_{1} − (p − 1)βi]

Hence, if T is a term from E_{k,n,j−k}^{i,r−i}, then since k_{1} ∈ {i, …, k} and α < 0, we have

v(T) ≥ p^{n−2}[αpk + βpr − (p − 1)(α + β)i].

Since i ∈ {0, …, r}, we have

v(T) ≥ \begin{cases}p^{n−1}[αk + βr], & if α + β ≤ 0; \\ p^{n−2}[αpk + (β − (p − 1)α)r], & if α + β > 0. \end{cases}

If k ∈ {r, …, n − 1}, then since α < 0 we have

v(T) ≥ \begin{cases}p^{n−1}[α(n − 1) + βr], & if α + β ≤ 0; \\ p^{n−2}[αp(n − 1) + (β − (p − 1)α)r], & if α + β > 0. \end{cases}

Finally, if r ∈ {2, …, n − 1}, then, since α < 0 and β > 0, we have

v(T) ≥ \begin{cases}p^{n−1}[α(n − 1) + 2β], & if α + β ≤ 0; \\ p^{n−1}[α(n − 1) + β] + p^{n−2}[2(α + β) − p(2α + β)], & if α + β > 0. \end{cases}

Therefore, since v(X) ≥ 0, for any ℋ̄′_{n,t} coming from either

∑_{r=2}^{n−1} ∑_{i=0}^{r} ∑_{j=r}^{n−1} ∑_{k=r}^{j} f_{i,r−i,n−j}(X^{p^{n}}, X^{p^{n+1}})E_{k,n,j−k}^{i,r−i}

or ∑_{k=2}^{n−1} E_{k,n,n−k}^{1,1}, we have

v(ℋ̄′_{n,t}) ≥ \begin{cases}p^{n−1}[α(n − 1) + 2β], & if α + β ≤ 0; \\ p^{n−1}[α(n − 1) + β] + p^{n−2}[2(α + β) − p(2α + β)], & if α + β > 0. \end{cases}

Moreover, a similar analysis gives that terms coming from E_{n,n,0}^{1,1} have valuation greater than or equal to p^{n−1}[α(n − 1) + β] + p^{n−2}[α + β].

Also, clearly ℋ̄′_{n,t}’s coming from f_{0,0,n}(X^{p^{n}}, X^{p^{n+1}}) have nonnegative valuations. If it comes from f_{0,1,n−j}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p^{n−j+1}}, for j ∈ {1, …, (n − 1)}, then it has valuation greater than or equal to p^{n−1}[αj + β] ≥ p^{n−1}[(n − 1)α + β]. And if it comes from f_{1,0,n−j}(X^{p^{n}}, X^{p^{n+1}})J_{j}^{p^{n−j}}, again for j ∈ {1, …, (n − 1)}, then it has valuation greater than or equal to p^{n−2}[αj + β] ≥ p^{n−2}[(n − 1)α + β].

We now need to deal with the particular valuations. Case 1: Let v = v_{0}, i.e., n_{0} = 3, α = −(p^{2} + p)/3 and β = p^{2}. Then, the analysis above gives for p ≥ 7

v(ℋ̄′_{n,t}) ≥ p^{n−1}[α(n − 1) + β] + p^{n−2}[2(α + β) − p(2α + β)]] ≥ v_{n}

for t ≥ 3. For p = 5, we have

v(ℋ̄′_{n,t}) ≥ p^{n−1}[α(n − 1) + β] ≥ v_{n}

for t ≥ 3. Case 2: Let v = v_{1728}, i.e., n_{0} = 2, α = −p and β = p. Then, the analysis above gives

v(ℋ̄′_{n,t}) ≥ p^{n−1}[α(n − 1) + β] + p^{n−2}[α + β] ≥ v_{n}

for t ≥ 3. Case 3: Let v = w, i.e., n_{0} = 1, α = −(p + 1) and β = 1. Then, the analysis above gives

v(ℋ̄′_{n,t}) ≥ p^{n−1}[α(n − 1) + β] + p^{n−2}[α + β] ≥ v_{n}

for t ≥ 3. Note that in this case, there is only one term of valuation exactly equal to v_{n}, namely, the term J_{1}^{p^{n−1}}J_{n−1}^{p^{2}} from E_{n,n,0}^{1,1}, and by the induction hypothesis (with equality in the valuation of J_{j} in this case) gives that this term has valuation exactly equal to v_{n}.

The three cases above finish the proof.

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http://www.ma.utexas.edu/users/voloch/lst.html.

Department of Mathematics, University of Tennessee, Knoxville, TN 37996

E-mail address: finotti@math.utk.edu
URL: http://www.math.utk.edu/~finotti/