EXISTENCE OF GEOMETRICALLY SIMPLE ORDINARY ABELIAN VARIETIES OVER A FIXED FINITE FIELD

HUI ZHU

Abstract. Given a prime number p and a positive integer d, we prove that there exist geometrically simple d-dimensional ordinary abelian varieties defined over the finite field of p elements. We present an algorithm to determine whether an abelian variety over a finite field k, given with the characteristic polynomial of its Frobenius endomorphism relative to k, is geometrically simple.

1. Introduction

Fix a prime number p. Let F_{p} be the prime field of p elements. For any field k, denote by k^{a} an algebraic closure of k. Let A be an abelian variety over a field k. It is simple if has no proper subabelian varieties over k. It is geometrically simple if it is simple over k^{a}. (It is also called absolutely simple.) In [6], it is shown that every Newton polygon except the supersingular type (i.e., whose Newton polygon is a straight line of slope 1/2) arises as that of a geometrically simple abelian variety over F_{p}^{a}. In this paper we prove the following theorem.

Theorem . Given a prime number p and any positive integer d, there exist geometrically simple d-dimensional ordinary abelian varieties defined over F_{p}.

We would like to remark on some related questions here: (Q1) Given a perfect field k and a positive integer d, does there exist a curve over k of genus d whose Jacobian is geometrically simple? In a recent preprint [1], Chai and Oort gave an affirmative answer when k is an algebraic closure of a finite field. We do not know the general answer in the case where k is a fixed finite field or a number field. In §3 we prove a special case: there exist geometrically simple ordinary Jacobian varieties of dimension 2 over F_{p}. (Q2) Consider the types of Newton polygons for d-dimensional abelian varieties over F_{p}. Our result shows that the ordinary type contains geometrically simple fibres defined over F_{p}. We do not know whether every non-supersingular type contains (geometrically) simple fibres defined over F_{p}.

The organization of the paper is as follows. In §2 we present an algorithm to determine whether an abelian variety over a finite field k is geometrically simple. The proof of the main theorem is reduced to the constructions of certain Weil ppolynomials. Our construction is divided into two parts: §3 deals with the case d = 2 and §4 handles that of d ≥ 3. Of course when d = 1 the problem is trivial.

Here I thank Mike Zieve, who raised the original question which led to the research and results in this paper. I thank Robert Coleman for many conversations and suggestions. Most of all, I am very grateful to Hendrik Lenstra for his contribution of Lemma 1, for fruitful conversations, and some valuable correspondences. Some computations in this paper used the PARI/GP package for computational number theory.

Date: August 9, 1999.

Research supported by the Mathematical Science Research Institute (MSRI) at Berkeley by a postdoctoral fellowship.

2. An algorithm

Let q = p^{c} for some positive integer c. An algebraic integer whose absolute values at all infinite places are \sqrt{q} is called a Weil q-number. Let f = ∑_{i=0}^{n} a_{i}X^{n−i} ∈ Z[X] be a monic polynomial. It is a Weil q-polynomial if its roots are Weil qnumbers and its real roots occur with even multiplicities. Clearly it is of even degree. Consider the following sequence of points in the real coordinate plane: (0, 0), (1, (ord_{p}a_{1})/c), (2, (ord_{p}a_{2})/c), ⋯, (n, (ord_{p}a_{n})/c). The normalized p-adic Newton polygon of f is the “lower convex hull” of the set of these points (see [7, IV, §3]). Let K be the number field generated by a root of f. Each slope s_{j} of the Newton polygon of f corresponds to a prime ideal ℘_{j} in K lying over p. Let e_{f} = 2 if f has real roots, otherwise e_{f} is equal to the least common multiple of the denominators of s_{j}[K_{℘_{j}} : Q_{p}] where s_{j} for all j.

Let k be a finite field of q elements. Let d be a positive integer. There is a 1-1 correspondence between the isogeny classes of simple d-dimensional abelian varieties A over k and the set of Weil q-polynomials f_{A} of degree 2d with f_{A} = g_{A}^{e_{f_{A}}} for an irreducible g_{A} ∈ Z[X]. In this Honda-Tate-correspondence, an abelian variety A over k corresponds to the characteristic polynomial f_{A} of its Frobenius endomorphism relative to k. (See [10, Théorèm 1].) Let k_{n} be a degree n extension of k. Let f_{A/k_{n}} and g_{A/k_{n}} be the characteristic polynomial of the Frobenius of A relative to k_{n}. The roots of f_{A/k_{n}} are equal to π^{n} for some root π of f_{A}. Applying the Honda-Tate theorem to k_{n} for all n ≥ 1, we conclude that A is geometrically simple if and only if f_{A/k_{n}} = g_{A/k_{n}}^{e_{f_{A/kn}}} for all n. But e_{f_{A/kn}} deg(g_{A/k_{n}}) = 2d, where e_{f_{A/kn}} ≤ e_{f_{A/km}} and deg(g_{A/k_{n}}) ≤ deg(g_{A/k_{m}}) for any m | n. Thus A is geometrically simple if and only if f_{A} = g_{A}^{e_{f_{A}}} for some irreducible polynomial g_{A} ∈ Z[X] and Q(π) = Q(π^{n}) for all n > 1.

Algorithm . We have the following algorithm of 5 steps to determine if a given monic polynomial f ∈ Z[X] Honda-Tate-corresponds to a geometrically simple abelian variety over F_{q}. (A1) Check that f is a Weil q-polynomial. (A2) Check that f = g^{e} for some irreducible g ∈ Z[X] and a positive integer e. (A3) Compute e_{f}. Check that e = e_{f}. (A4) Check that g ∉ Z[X^{m}] for all integers m > 1. (A5) Let t be the maximal integer such that ζ_{t} ∈ Q(π), where π is a root of f in

Q^{a}. Check that Q(π) = Q(π^{t}).

Proof. It suffices to check that Steps (A4) and (A5) imply Q(π) = Q(π^{n}) for every integer n > 1. This can be seen from the following lemma, due to Hendrik Lenstra.

Lemma 1. Let π be an algebraic number with irreducible polynomial g ∈ Q[X] such that g ∉ Q[X^{m}] for any integer m > 1. If there exists n > 1 with Q(π) = Q(π^{2}) = ⋯ = Q(π^{n−1}) ≠ Q(π^{n}), then Q(π) = Q(π, ζ_{n}) = Q(π^{n}, ζ_{n}), where ζ_{n} is the primitive n-th root of unity exp(\frac{2π\sqrt{−1}}{n}) in Q^{a}.

Proof. Let σ be a Q(π^{n})-embedding of Q(π) in Q^{a}. Obviously, π^{σ} = πζ_{n}^{l} for some positive integer l ≤ n. Suppose (ζ_{n}^{l})^{s} = 1 for some positive integer s < n. By assumption π ∈ Q(π^{s}), and so π = g_{0}(π^{s}) for some g_{0}(X) ∈ Q[X]. Thus π^{σ} = g_{0}(π^{s})^{σ} = g_{0}((π^{σ})^{s}) and so πζ_{n}^{l} = g_{0}((πζ_{n}^{l})^{s}) = g_{0}(π^{s}) = π, which is absurd. Therefore, ζ_{n}^{l} is primitive. Write K := Q(π) ∩ Q(ζ_{n}). Let g_{1} and g_{1}^{σ} be the minimal polynomials of π and π^{σ} over K and K^{σ}, respectively. But K is Galois over Q, so K^{σ} = K. Write g_{1} = ∑_{i=0}^{e} a_{i}X^{e−i}, then g_{1}^{σ} = ∑_{i=0}^{e} a_{i}(ζ_{n}^{l})^{i}X^{e−i}, where g_{1}, g_{1}^{σ} ∈ K[X]. Let t be the least positive integer such that (ζ_{n}^{l})^{t} ∈ K. For any integer j, we note that a_{j} ≠ 0 implies t | j. Thus g_{1}(X) ∈ Q[X^{t}]. Since Q(ζ_{n})/Q is abelian, K/Q is Galois and so g = ∏_{ρ∈Gal(K/Q)} g_{1}^{ρ}(X) ∈ Q[X^{t}]. Thus t = 1, and hence ζ_{n}^{l} ∈ K and hence Q(π) = Q(π, ζ_{n}). On the other hand, each Q(π^{n})- conjugate π^{σ} of π has its minimal polynomial g_{1}^{σ} ≠ g_{1} over Q(π^{n}, ζ_{n}), which implies that π^{σ} is no longer conjugate to π over Q(π^{n}, ζ_{n}). Thus Q(π, ζ_{n}) = Q(π^{n}, ζ_{n}).

3. Case d = 2

For any monic polynomial H(X) = ∑_{i=0}^{d} c_{i}X^{d−i} ∈ Z[X], define H^{∗}(X) := X^{d}H(X + \frac{p}{X}). We can write

H^{∗}(X) = X^{2d} + c_{1}^{∗}X^{2d−1} + ⋯ + c_{d}^{∗}X^{d} + ⋯ + c_{1}^{∗}p^{d−1}X + p^{d}

for c_{i}^{∗} ∈ Z satisfying the following relation:

(1)  c_{i}^{∗} = c_{i} + p\binom{d−(i−2)}{1}c_{i−2} + p^{2}\binom{d−(i−4)}{2}c_{i−4} + ⋯

+ \{ \begin{matrix}p^{i/2}\binom{d}{i/2}c_{0} & if d is even; \\ p^{(i−1)/2}\binom{d−1}{(i−1)/2}c_{1} & if d is odd. \end{matrix}

Lemma 2. To check if H^{∗} ∈ Z[X] Honda-Tate-corresponds to an ordinary geometrically simple abelian variety over F_{p}, it suffices to apply to H^{∗} the Algorithm where (A1), (A2) and (A3) are replaced by (A1’) Check that H is irreducible, p ∤ c_{d} and H^{∗} is a Weil p-polynomial.

Proof. From (1) we have c_{d}^{∗} ≡ c_{d} mod p. That p ∤ c_{d} implies e_{H^{∗}} = 1. We claim that (A1’) implies H^{∗} is irreducible. Every pair of complex roots π, \overline{π} of H^{∗} satisfy X^{2}−αX +p where α = π+\overline{π} is a root of H. If π ∈ Q(α), i.e., the above quadratic is reducible over Q(α), then α = 2π = ±2\sqrt{p} and so H = X^{2} − 4p. This contradicts our assumption that p ∤ c_{d}. Thus [Q(π) : Q] = 2[Q(α) : Q] = 2d and H^{∗} is irreducible since deg(H^{∗}) = 2d. Then H^{∗} corresponds to some ordinary abelian variety over F_{p} by [2, § 2].

Below we apply Lemma 2 to H^{∗}. Let H = X^{2} + c_{1}X + c_{2} ∈ Z[X] be such that c_{1} ≠ 0 and (A1’) is satisfied. So (A4) is satisfied and Q(π) = Q(π^{2}). Let α denote a root of H in Q^{a} and let π denote a root of H^{∗} in Q^{a} which satisfies X^{2}−αX +p. We claim that N := (c_{2}+4p)^{2}−4pc_{1}^{2} ∉ Q(α)^{2} implies (A5) is satisfied. To show ζ_{2} is the largest root of unity in Q(π), it suffices to show that Q(π) is not Galois since Q(π) is a CM field with totally real subfield Q(α). If Q(π) = Q(α, \sqrt{α^{2} − 4p}) is Galois, then Q(π) = Q(α, \sqrt{−r}) for some r ∈ Q_{>0}. Then N_{Q(α)/Q}(α^{2} −4p) ∈ Q(α)^{2}. This implies N ∈ Q(α)^{2}, which leads to a contradiction. Thus we proved our claim. If additionally Z[α] is the ring of integers in Q(α), then H^{∗} Honda-Tate-corresponds to a geometrically simple ordinary Jacobian variety over F_{p} (see [9, Theorem 1.2]).

Case p = 2, d = 2. Let H = X^{2} − X − 3. Note that N = 17 ∉ Q(α)^{2}, since 17 is a prime number which does not ramify in Q(α) = Q(\sqrt{13}). Case p ≥ 3, p ≢ 1 mod 5, d = 2. Let H = X^{2} + X − 1. Since |α| ≤ 2\sqrt{p}, the roots of X^{2} − αX + p are totally imaginary of absolute value \sqrt{p}. Hence H^{∗} is a Weil p-polynomial. Suppose N = (4p − 1)^{2} − 4p ∈ Q(\sqrt{5})^{2}. Then exist integers x, y such that 16p^{2} − 12p + 1 = (x + \frac{−1+\sqrt{5}}{2}y)^{2}. Comparing both sides, we have 16p^{2} − 12p + 1 = x^{2} or 5x^{2}. In the first possibility, we reduce modulo 3 both sides and get a contradiction. In the second possibility, we reduce modulo 5 both sides then get p ≡ 1 mod 5, which contradicts our assumption. Case p ≥ 3, p ≡ 1 mod 5, d = 2. Let H = X^{2} + 3X + 1. Then Q(α) = Q(\sqrt{5}). It remains to check that N = 16p^{2} − 28p + 1 ∉ Q(\sqrt{5})^{2}. We use an analogous argument as the above case to lead a contradiction.

4. Case d ≥ 3

Denote by Gal(·) the Galois group of a polynomial or a Galois extension. Let f ∈ Q[X] be an irreducible polynomial Let G = Gal(f) operate on the set Σ_{f} of roots of f in Q^{a}. If, for every β ∈ Σ_{f}, the isotropy subgroup G_{β} in G is a maximal subgroup, then G is called a primitive group. (See also [8, Chapter I, Ex. 46]). In such a case, Q(β) is the field fixed by G_{β}, and there is no intermediate subfield in Q(β)/Q.

Lemma 3. To prove the main theorem for d ≥ 3, it suffices to construct a monic polynomial H(X) = ∑_{i=0}^{d} c_{i}X^{d−i} ∈ Z[X] with

(C1) H is irreducible and p ∤ c_{d}; c_{d−1}^{∗} ≠ 0 and Gal(H) contains a (d − 1)-cycle; (C2) H^{∗} is a Weil p-polynomial.

Proof. By Lemma 2 it suffices to show that Q(π) = Q(π^{n}) for all n ≥ 1. Suppose not then there exists an integer n > 1 such that Q(π) = Q(π^{2}) = ⋯ = Q(π^{n−1}) ≠ Q(π^{n}). Suppose α ∈ Σ_{H} is fixed by the given (d − 1)-cycle. The (d − 1)-cycle acts transitively on the complement of α in Σ_{H}, and so Gal(H) is 2-transitive by [8, Chapter I, Ex. 47(a)]. Hence Gal(H) is primitive by [8, Chapter I, Ex. 47(c)]. So there is no intermediate field in the extension Q(π+\overline{π})/Q. If Q(π^{n}+\overline{π}^{n}) = Q(π+\overline{π}) then Q(π) = Q(π^{n}) which contradicts our assumption. So we are forced to have Q(π^{n} + \overline{π}^{n}) = Q. Write L for the Galois closure of Q(π + \overline{π})/Q in Q^{a}. It is totally real since Q(π + \overline{π}) is. Thus Q(π + \overline{π}) = Q(π) ∩ L and Q = Q(π^{n}) ∩ L. The compositum L.Q(π^{n}) is imaginary quadratic over the totally real field L, so it is the Galois closure of Q(π^{n}) in Q^{a} with Gal(L.Q(π^{n})/Q(π^{n})) = Gal(H). On the other hand, (C1) implies that H^{∗}(X) ∉ Z[X^{m}] for any integer m > 1, so Q(π)/Q(π^{n}) is abelian by Lemma 1. Thus Gal(H) is abelian of order d containing a cyclic subgroup of order d − 1, so d ≤ 2, which contradicts our assumption.

Let a = 1 if p ≠ 2, 3 and a = 3/2 if p = 2, 3. Define a function F (X, d, m) := X^{d+1} − X^{d} − 3X^{d−m} − 3(X^{2} − aX − (2 − a)) ∈ Q[X] with domain X ≥ \sqrt{2}, m ≥ 1, and d ≥ 3.

Lemma 4. Let d ≥ 3. If for some non-negative integer m and two distinct primes p_{1}, p_{2} there is a monic H = ∑_{i=0}^{d} c_{i}X^{d−i} ∈ Z[X] such that (C1’) H_{1} := H mod p_{1} ∈ F_{p_{1}}[X] and H_{2} := H mod p_{2} ∈ F_{p_{2}}[X] are polynomials

with one being irreducible and the other a product of a linear polynomial and an irreducible. For all i = 1, ⋯, m we have

c_{i} ≡ −p\binom{d−(i−2)}{1}c_{i−2} − p^{2}\binom{d−(i−4)}{2}c_{i−4} − ⋯

− \{ \begin{matrix}p^{i/2}\binom{d}{i/2}c_{0} mod p_{1}p_{2} & if d is even; \\ p^{(i−1)/2}\binom{d−1}{(i−1)/2}c_{1} mod p_{1}p_{2} & if d is odd. \end{matrix}

(C2’) If p_{1} = 2, p_{2} = 3, we have b := ∑_{i=m+1}^{d−1} \frac{|c_{i}^{∗}|}{\sqrt{p}^{i}} + \frac{|c_{d}^{∗}|}{2\sqrt{p}^{d}} ≤ 1 or F (\sqrt{p}, d, m) ≥ 0.

Then there exists H with (C1) and (C2) satisfied.

Proof. Obviously (C1’) implies (C1). Now suppose p_{1} = 2, p_{2} = 3. While (C1’) determines H mod 6, we may choose H in the following way: By (1) we can construct c_{i}^{∗} from c_{i} mod 6 inductively for i = 1, 2, ⋯, d. It is easy to see that given a c_{i} mod 6 we may choose c_{i} so that |c_{i}^{∗}| ≤ 3 for all 1 ≤ i ≤ d. If c_{d−1}^{∗} = 0 then we replace c_{d−1}^{∗} by 6; If p ≠ 2, 3 and p | c_{d}^{∗} then we replace c_{d}^{∗} by c_{d}^{∗} + 6 or c_{d}^{∗} − 6 so that |c_{d}^{∗}| ≤ 6. If p = 2 (resp., 3) we let the d-th coefficients of H_{1} (resp., H_{2}) be non-zero. Thus p ∤ c_{d}. We get

|c_{i}^{∗}| ≤ 3 for 1 ≤ i ≤ d − 2; c_{d−1}^{∗} ≠ 0; |c_{d−1}^{∗}| ≤ 6; p ∤ c_{d}^{∗};

|c_{d}^{∗}| ≤ \{ \begin{matrix}6 & if p ≠ 2, 3; \\ 3 & otherwise. \end{matrix}

Since (C1’) implies c_{1}^{∗} = c_{2}∗ = ⋯ = c_{m}^{∗} = 0 by (1), we get by the above

b ≤ ∑_{i=m+1}^{d−2} \frac{3}{\sqrt{p}^{i}} + \frac{6}{\sqrt{p}^{d−1}} + \frac{12 − 6a}{2\sqrt{p}^{d}}

= 3\frac{\sqrt{p}^{d−m} + \sqrt{p}^{2} − a\sqrt{p} − (2 − a)}{\sqrt{p}^{d}(\sqrt{p} − 1)},

which is ≤ 1 if F (\sqrt{p}, d, m) ≥ 0. On the other hand, by [3, Lemma 2.5.1], we have b ≤ 1 implies (C2).

To prove the main theorem for d ≥ 3, by Lemmas 3 and 4, it suffices to construct a polynomial H with (C1’) and (C2’), or (C1’) and (C2). The complete proof consists of the following three parts.

4.1. Part I. We shall firstly construct H with (C1’) and (C2’) satisfied for p ≥ 11, d ≥ 3; or p = 7, , d ≥ 4; or p = 5, d ≥ 4; or p = 3, d ≥ 5; or p = 2, d ≥ 10.

Here we check F (\sqrt{p}, d, m) ≥ 0 as in (C2’) by using the following lemma.

Lemma 5. Let F (X, d, m) be defined above Lemma 4. Let G(X, m) := X^{m+1} − X^{m} − 3 where X ≥ \sqrt{2} and m ≥ 1. Suppose for a fixed m we have G(\sqrt{p_{0}}, m) ≥ 0, F (\sqrt{p_{0}}, d_{0}, m) ≥ 0, and \frac{∂F}{∂X} (X, d_{0}, m) ≥ 0 for X ≥ \sqrt{p_{0}}, then F (\sqrt{p}, d, m) ≥ 0 for all p ≥ p_{0} and d ≥ d_{0}.

Proof. This is an exercise in Calculus. Firstly, \frac{∂G}{∂X} = X^{m−1}((m + 1)X − m) > 0 for all X, so G(\sqrt{p_{0}}, m) ≥ 0 implies that G(\sqrt{p}, m) ≥ 0 for all p ≥ p_{0}. But \frac{∂F}{∂d} (X, d, m) = X^{d−m}G(X, m) ln X, so \frac{∂F}{∂d} (\sqrt{p}, d, m) ≥ 0 for p ≥ p_{0}. By our assumptions, we have F (\sqrt{p}, d, m) ≥ F (\sqrt{p_{0}}, d, m) ≥ F (\sqrt{p_{0}}, d_{0}, m) ≥ 0.

Given an arbitrary polynomial f = ∑_{i=0}^{s} a_{i}X^{s−i} over a field, we call a_{i} the i-th coefficient of f. Conditions (C1’) and (C2’) require the prescription of the first m coefficients of H mod 6 (see the following lemma).

Lemma 6. Suppose p_{1} = 2, p_{2} = 3. That m ≤ 6 in Lemma 4 is equivalent to the first m congruences below being satisfied:

c_{1} ≡ 0 mod 6
c_{2} ≡ −pd mod 6
c_{3} ≡ 0 mod 6
c_{4} ≡ −p^{2}(\frac{−d(d−3)}{2}) mod 6
c_{5} ≡ 0 mod 6
c_{6} ≡ −p^{3}(\frac{d(d−4)(d−5)}{6}) mod 6.

By the Chinese remainder theorem, this is equivalent to the prescription of the first m coefficients of H_{1} and H_{2}. Given a prime l and a positive integer n, if l ≠ 2 or n ≠ 2, then there is always an irreducible polynomial over F_{l} of degree n with the first coefficient equal to 0. (This can be seen by examining the size of the kernel of the trace map.) We shall use this for the first case below. Case p ≥ 11, d ≥ 3 or p = 7, d ≥ 4. That m = 1 implies (C2’). We may construct irreducibles H_{1} ∈ F_{2}[X] of degree d ≥ 3 and H_{3} ∈ F_{3}[X] of degree d − 1 ≥ 2 with their first coefficient equal to 0. Let H_{2} = XH_{3}. Thus we get an H with c_{1} ≡ 0 mod 6. So m = 1 by Lemma 6.

Since G(\sqrt{7}, 1) > 0, F (\sqrt{11}, 3, 1) > 0 (and F (\sqrt{7}, 4, 1) > 0), and \frac{∂F}{∂X} (X, 3, 1) > 0 for X ≥ \sqrt{7}, we have (C2’) by Lemma 5.

The following lemma (see [5, Corollary 3.2] and its asymptotic version [4, Theorem 8.1]) implies the existence of irreducible polynomials over a finite field with its first m coefficients prescribed.

Lemma 7. Let l be a prime and m a positive integer. For any n > 2(m+log_{p}(m+ 1)) there exists a monic irreducible polynomial over F_{l} of degree n with the first m coefficients prescribed.

Case p = 5, d ≥ 4. That m = 2 implies (C2’).

(i) d ≥ 8. By Lemma 7, there exist irreducible polynomials H_{1} ∈ F_{2}[X] of degree d ≥ 8 and H_{3} ∈ F_{3}[X] of degree d − 1 ≥ 7 with the first 2 coefficients prescribed. Let H_{2} = XH_{3} we have m = 2. (ii) d = 4, 5, 6, 7. Construct H such that

\begin{matrix}d & H mod 6 \\ 4 & X^{4} − 2X^{2} − X − 1 \\ 5 & X^{5} − X^{3} − 2X + 3 \\ 6 & X^{6} + 2X^{2} − X + 3 \\ 7 & X^{7} + X^{5} + 2X^{3} + X^{2} − X + 3 \end{matrix}

It is easy to verify that c_{1} ≡ 0, c_{2} ≡ d mod 6, hence m = 2 by Lemma 6.

Note that G(\sqrt{5}, 2) > 0, F (\sqrt{5}, 3, 2) > 0 and \frac{∂F}{∂X} (X, 3, 2) > 0 for all X ≥ \sqrt{5}. Thus (C2’) follows by Lemma 5.

Case p = 3, d ≥ 5. That m = 3 implies (C2’).

(i) d ≥ 11. Lemma 7 implies that we may prescribe the first 3 coefficients; (ii) 5 ≤ d ≤ 10. We choose H to be such that

\begin{matrix}d & H mod 6 \\ 5 & X^{5} + 3X^{3} − X + 1 \\ 6 & X^{6} + 3X^{3} + X + 2 \\ 7 & X^{7} + 3X^{5} + 3X^{3} + X^{2} + 2 \\ 8 & X^{8} + X^{2} + 3X + 2 \\ 9 & X^{9} + 3X^{7} + 3X^{5} + 3X^{4} − X^{3} + X^{2} + 3X − 2 \\ 10 & X^{10} − X^{2} + 3X − 2. \end{matrix}

It is easy to verify that c_{1} ≡ 0, c_{2} ≡ 3d, c_{3} ≡ 0 mod 6, hence m = 3 by Lemma 6.

Now we note that G(\sqrt{3}, 3) > 0, F (\sqrt{3}, 5, 3) > 0 and \frac{∂F}{∂X} (X, 5, 3) > 0 for all X ≥ \sqrt{3}. Case p = 2, d ≥ 10. That m = 6 implies (C2’).

(i) d ≥ 18. By Lemma 7, there exist irreducible polynomials H_{1} ∈ F_{2}[X] of degree d ≥ 18 and H_{3} ∈ F_{3}[X] of degree d − 1 ≥ 17 with the first 6 coefficients prescribed. Let H_{2} = (X + 1)H_{3}, we have m = 6. (ii)10 ≤ d ≤ 17. We choose H as the follows

d H mod 6
10 X^{10} − 2X^{8} + 2X^{6} + 2X^{4} − X^{3} − 2X^{2} − 2X + 3
11 X^{11} + 2X^{9} + 2X^{7} − 2X^{5} − 2X^{4} + 3X^{2} + 2X + 1
12 X^{12} − 2X^{6} + X^{3} + 2X + 3
13 X^{13} − 2X^{11} + 2X^{9} + 2X^{7} − 2X^{6} − 2X^{5} + X^{4} + X^{3} − 2X^{2} − X + 3
14 X^{14} + 2X^{12} + 2X^{10} + 2X^{8} + X^{5} − 2X^{4} − 2X^{2} − 2X + 3
15 X^{15} + 2X^{9} + 2X^{8} − 2X^{5} − 2X^{4} − 2X^{3} − 2X^{2} + X + 3
16 X^{16} − 2X^{14} + 2X^{12} + 2X^{9} + 2X^{8} − 2X^{7} − 2X^{6} + 3X^{5} − 2X^{4} + 3X^{3}
−2X^{2} + X + 3
17 X^{17} + 2X^{15} + 2X^{13} + 2X^{10} + 2X^{9} + 2X^{8} − 2X^{7} − 2X^{6} + X^{3} − 2X^{2}
−2X + 3.

It is not hard to check that m = 6 by Lemma 6.

Now we note G(\sqrt{2}, 6) > 0, F (\sqrt{2}, 10, 6) > 0 and \frac{∂F}{∂X} (X, 10, 6) > 0 for all X ≥ \sqrt{2}.

4.2. Part II. We shall secondly construct H with (C1’) and (C2’) satisfied for p = 2, d = 6, 7, 8, 9 and p = 3, d = 4. Here we set m = 0 and check b ≤ 1.

Case p = 2, d = 6, 7, 8, 9. When d = 6, let H^{∗} = X^{12} + 2X^{8} + X^{7} + 3X^{6} + 2X^{5} +

8X^{4}+64, where H ≡ X^{6}+2X^{2}+X +3 mod 6. When d = 7, let H^{∗} = X^{14}+5X^{10}+ X^{7} + 40X^{4} + 128, where H ≡ X^{7} − 2X^{5} + X^{3} − 2X + 1 mod 6. When d = 8, let H^{∗} = X^{16}+3X^{10}+3X^{9}+3X^{8}+6X^{7}+12X^{6}+256, where H ≡ X^{8}+2X^{6}+2X^{4}+X^{2}+ 3X−1 mod 6. When d = 9, let H^{∗} = X^{18}+4X^{12}+5X^{10}+3X^{9}+10X^{8}+32X^{6}+512, where H ≡ X^{9} − 2X^{3} − X + 3 mod 6. Case p = 3, d = 4. Let H^{∗} = X^{8} + X^{6} + X^{5} + X^{4} + 3X^{3} + 9X^{2} + 81, where H ≡ X^{4} + X^{2} + X + 1 mod 6.

4.3. Part III. We shall thirdly construct H with (C1’) and (C2) satisfied for p = 2, d = 3, 4, 5; p = 3, d = 3; p = 5, d = 3; p = 7, d = 3. Here we checked H^{∗} below are all Weil p-polynomials by PARI/GP.

Case p = 2, d = 4, 5.

(i)d = 4. Let H = X^{4} + X^{3} − 7X^{2} − 5X + 5. Note that H mod 2 is irreducible and H mod 3 is a product of a linear factor and an irreducible factor, that is, H ≡ (X + 1)(X^{3} − X − 1) mod 3. Note that H^{∗}(X) = X^{8} + X^{7} + X^{6} + X^{5} + X^{4} + 2X^{3} + 4X^{2} + 8X + 16 is indeed a Weil 2-polynomial. (ii)d = 5. Let H(X) = X^{5} + X^{4} − 9X^{3} − 7X^{2} + 15X + 5. Note that H mod 29 is irreducible, and H mod 23 is a product of a linear factor and an irreducible factor, that is, H(X) ≡ (X + 18)(X^{4} + 6X^{3} + 21X^{2} + 6X + 22) mod 23. Note that H^{∗} = X^{10} + X^{9} + X^{8} + X^{7} + X^{6} + X^{5} + 2X^{4} + 4X^{3} + 8X^{2} + 16X + 32 is indeed a Weil 2-polynomial.

Case p = 2, d = 3. Let H = X^{3} + X^{2} − 5X − 3.

First of all H^{∗} is indeed a Weil 2-polynomial. Note that H mod 5 is irreducible and H ≡ (X + 5)(X^{2} + 9X + 2) mod 13.

Case p = 3, d = 3. Let H = X^{3} + X^{2} − 8X − 5.

First of all H^{∗} is indeed a Weil 3-polynomial. Note that H mod 2 is irreducible and H ≡ (X + 1)(X^{2} + 1) mod 3.

Case p = 5, d = 3. Let H = X^{3} + X^{2} − 14X − 9.

First of all H^{∗} is indeed a Weil 5-polynomial. Note that H mod 2 is irreducible and H(X) ≡ (X + 6)(X^{3} + 2X + 2) mod 7.

Case p = 7, d = 3. Let H = X^{3} + X^{2} − 20X − 13.

First of all H^{∗} is indeed a Weil 7-polynomial. Note that H mod 2 is irreducible and H ≡ (X + 1)(X^{2} + 1) mod 7.

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