VANISHING SUMS OF m TH ROOTS OF UNITY IN FINITE FIELDS

T. Y. LAM AND K. H. LEUNG

Abstract. In an earlier work, the authors have determined all possible weights n for which there exists a vanishing sum ζ_{1} + ⋯ + ζ_{n} = 0 of m th roots of unity ζ_{i} in characteristic 0. In this paper, the same problem is studied in finite fields of characteristic p. For given m and p, results are obtained on integers n_{0} such that all integers n ≥ n_{0} are in the “weight set” W_{p}(m). The main result (1.3) in this paper guarantees, under suitable conditions, the existence of solutions of x_{1}^{d}+⋯+x_{n}^{d} = 0 with all coordinates not equal to zero over a finite field.

1. Introduction

By a vanishing sum of m th roots of unity, we mean an equation α_{1} + ⋯ + α_{n} = 0 where α_{i}^{m} = 1 for each i. The integer n is said to be the weight of this vanishing sum. In [LL], considering m th roots of unity in ℂ, we defined W (m) to be the set of integers n ≥ 0 for which there exists a vanishing sum α_{1} + ⋯ + α_{n} = 0 as above. The principal result in [LL] gives a complete determination of the weight set W (m) (in characteristic 0), as follows.

Theorem 1.1. For any natural number m with prime factorization p_{1}^{a_{1}} ⋯ p_{r}^{a_{r}}, the weight set W (m) is exactly given by ℕ p_{1} + ⋯ + ℕ p_{r}. (Here and in the following, ℕ := {0, 1, 2, ⋯ }).

In this paper, we study vanishing sums of m th roots of unity in characteristic p. In analogy to the characteristic 0 case, we define W_{p}(m) to be the set of weights n for which there exists a vanishing sum α_{1} + ⋯ + α_{n} = 0 where each α_{i} is an m th root of unity in \overline{𝔽}_{p}, the algebraic closure of the prime field 𝔽_{p}. Note that, if m = p^{t}m′ where gcd(p, m′) = 1, we have x^{m} = 1 in \overline{𝔽}_{p} iff x^{m′} = 1; in particular, W_{p}(m) = W_{p}(m′). Therefore, we may assume throughout that gcd(p, m) = 1, i.e. p is not among the prime divisors p_{i} of m. As in the case of charactersitic 0, we have p_{i} ∈ W_{p}(m) for all i. But in characteristic p, we also have p ∈ W_{p}(m) (due to the vanishing sum p · 1 = 0), so now

Lam was supported in part by NSF. Research at MSRI is supported in part by NSF grant DMS-9022140.

Leung’s research was carried out while he was on sabbatical leave at U.C. Berkeley from the National University of Singapore. The hospitality of the former institution is gratefully acknowledged.

(1.2)  W_{p}(m) ⊇ ℕ p + ℕ p_{1} + ⋯ + ℕ p_{r}.

Easy examples (see (2.1)) show that this need not be an equality in general, so we are left with no viable conjecture on the structure of the weight set W_{p}(m) in characteristic p. However, (1.2) does show that, if m > 1, all sufficiently large integers n (in fact all n ≥ (p − 1)(p_{i} − 1)) belong to W_{p}(m). A more tractable problem will then be the determination of more accurate bounds n_{0} such that all integers n ≥ n_{0} belong to W_{p}(m).

In this paper, we will show how such an integer n_{0} can be determined. Our work is divided into three cases, depending on whether gcd(p − 1, m) is 1, 2, or bigger. The estimates on n_{0} differ from case to case, and are given respectively in (5.6), (4.1), and (3.1)-(3.3). Although we have three different estimates on n_{0}, there does exist a (necessarily weaker) uniform estimate for all cases. In the following, we shall try to explain what this uniform estimate is, and why is it a reasonable one.

A guiding principle for our work throughout is the fact that a finite field is a C_{1}-field (see [Gr]). If K = 𝔽_{p^{k}} is a finite field containing all m th roots of unity, then, for d := (p^{k} − 1)/m, the m th roots of unity in \overline{𝔽}_{p} comprise the group K̇^{d}. Therefore, a vanishing sum of m th roots of unity of weight n corresponds precisely to a “good” solution of x_{1}^{d} + ⋯ + x_{n}^{d} = 0 in K, where by a “good” solution we mean one with each x_{i} ≠ 0. If n > d, the fact that K is C_{1} implies that we have a solution (x_{1}, ⋯, x_{n}) ≠ (0, ⋯, 0). It certainly seems tempting to speculate that there exists in fact a “good” solution (in K). If this is indeed the case, then by what we said earlier in this paragraph, any integer n > d will be in the weight set W_{p}(m).

The desired conclusion that, for n > d, x_{1}^{d} + ⋯ + x_{n}^{d} = 0 has a “good” solution in K is, however, not true in general! For instance, if d = p^{k} − 1, then x^{d} = 1 for each x ∈ K̇, so we have a “good” solution for x_{1}^{d} + ⋯ + x_{n}^{d} = 0 in K only when n is a multiple of p. In a similar vein, if k = 1, p is odd, and d = (p − 1)/2, then any nonzero d th power in K is ±1. For any odd integer n ∈ (d, p), the equation x_{1}^{d} +⋯+x_{n}^{d} = 0 again has no “good” solution in 𝔽_{p}. The trouble with these cases is that m ≤ 2, for which we don’t have “enough” m th roots to play with. As it turns out, as soon as we ignore the above cases, we’ll have the following uniform result for getting “good” solutions.

Theorem 1.3. Let K = 𝔽_{p^{k}} and d = (p^{k} −1)/m as above, and assume that m ≠ 1, (m, k) ≠ (2, 1). Then, whenever n > d, the equation x_{1}^{d} + ⋯ + x_{n}^{d} = 0 has a “good” solution in K. In other words, the weight set W_{p}(m) contains all integers ≥ d + 1.

The results in §§3-5 below will cover this theorem in the case m ≥ 3. In the case m = 2, (1.3) is quickly checked as follows. Since we assume in this case that k ≥ 2, we have d ≥ (p^{2} − 1)/2 ≥ p − 1. Given n ≥ d + 1 ≥ p, it is easy to solve the equation α_{1} + ⋯ + α_{n} = 0 with α_{i} = ±1, by considering the parity of n. Having disposed of the trivial cases m = 1, 2, we may assume in §§3-6 of this paper that m ≥ 3.

In the case when p is odd and d = 2, (1.3) says precisely that, for any n > 2, the quadratic form x_{1}^{2} + ⋯ + x_{n}^{2} has a “good” zero over any finite field of more than five elements. This is a special case of a well-known observation of Witt for isotropic diagonal quadratic forms (see [Wi: p.39], or [BS: p.394], [La: p.25, Ex.7]). Thus, (1.3) may be thought of as a generalization of Witt’s result to the higher degree diagonal forms x_{1}^{d} + ⋯ + x_{n}^{d} over finite fields. Note that d | (p^{k} − 1) is not a really essential assumption in (1.3). In dealing with the equation x_{1}^{e} + ⋯ + x_{n}^{e} = 0, we can replace the degree e by d := gcd(p^{k} −1, e), and define m to be (p^{k} −1)/d. Then K̇^{e} = K̇^{d}, so under the assumptions of (1.3), x_{1}^{e} + ⋯ + x_{n}^{e} = 0 will have a “good” solution as long as n > d.

In the literature, there are many results dealing with diagonal equations over finite fields; see, for instance, [LN], [Sch], [Sm], and more recently [QY]. Conventionally, one could apply algebro-geometric methods, or alternatively the method of Gauss and Jacobi sums. As the referee of this paper pointed out, these methods can be utilized to show the existence of “good” solutions to a diagonal equation x_{1}^{d} + ⋯ + x_{n}^{d} = 0 (n > 2) in 𝔽_{q} if q is suitably large compared to d (without the condition n > d). However, these conventional methods do not seem to give enough information if q is “small” in comparison to d. In our setting, working mostly with n > d and taking full advantage of the additive nature of the special equation x_{1}^{d} + ⋯ + x_{n}^{d} = 0, we apply instead the methods of additive number theory. These methods do give fairly precise results, without reference to the size of 𝔽_{q}. In fact, the analysis in §§3-5 will not only prove (1.3), but also show that, in various cases, the equation x_{1}^{d} + ⋯ + x_{n}^{d} = 0 has a “good” solution in K often for much smaller values of n (than n ≥ d + 1). Thus, the more precise results in this paper are to be found in (3.1)-(3.3), (4.1) and (5.6). Theorem 1.3 is only a common denominator of these results giving a convenient and uniform summary of the main work in this paper.

Acknowledgment. We thank the referee of this paper, whose comments enabled us to rewrite more accurately the last paragraph above comparing the use of different methods in treating diagonal equations over finite fields.

2. Some Basic Examples

We shall begin with some examples and computations of the weight sets W_{p}(m). The first couple of examples show that various properties of weight sets in characterictic 0 are no longer valid in characteristic p. For convenience of expressing weight sets, let us use the notation [ n, ∞)_{ℤ} for the set of integers ≥ n.

Example 2.1. Referring to (1.2), the smallest positive element in the set W_{p}(m) may not be min {p, p_{1}, ⋯, p_{r}}. For instance, when p = 11 and m = 5, the 5th roots of unity in \overline{𝔽}_{11} are \{1, 3, 9, 5, 4\}. Observing that 1 + 1 + 9 = 0 in 𝔽_{11}, we see that W_{11}(5) contains 3, which is smaller than 5 and 11. By (2.3) below, we have W_{11}(5) = {0}∪[ 3, ∞)_{ℤ}. Thus, not only (1.2) fails to be an equality, but also W_{11}(5) is not even of the form ∑_{i} ℕ q_{i} for a set of primes q_{i} ’s.

Example 2.2. Contrary to the characteristic 0 case, the set W_{p}(m) may be larger than W_{p}(m_{0}) where m_{0} is the square-free part of m. For instance, let p = 5 and m = 4, so m_{0} = 2. It is easy to see that W_{5}(2) = {0, 2} ∪ [ 4, ∞)_{ℤ}, but W_{5}(4) = {0} ∪ [ 2, ∞)_{ℤ}.

Example 2.3. Let q = p^{a} > 5 where p is an odd prime, and let m = (q − 1)/2. Then d := (q − 1)/m = 2. For any n ≥ 3, the quadratic form X_{1}^{2} + ⋯ + X_{n}^{2} is isotropic over 𝔽_{q}, so by the theorem of Witt referenced before, it has a “good” zero in 𝔽_{q}. Therefore, n ∈ W_{p}(m). It follows that W_{p}(m) = {0}∪[ 2, ∞)_{ℤ} if q ≡ 1 (mod 4), and W_{p}(m) = {0} ∪ [ 3, ∞)_{ℤ} if q ≡ 3 (mod 4).

Example 2.4. (𝔽_{p} contains all m th roots of unity.) Let p = 31, and m = 3. The third roots of unity are {1, 5, 25}, so the equation 25 + 6 · 1 = 0 ∈ 𝔽_{31} shows that 7 ∈ W_{31}(3). A routine computation shows that W_{31}(3) = {0, 3, 6, 7} ∪ [ 9, ∞)_{ℤ}.

Example 2.5. (𝔽_{p} contains no m th roots of unity other than 1.) Let p = 2, and m = 73. We work in K = 𝔽_{2^{9}} which contains all 73 rd roots of unity. By standard tables of irreducible polynomials over finite fields, the trinomial f(X) = X^{9} + X + 1 is irreducible over 𝔽_{2}, so we can take K to be 𝔽_{2}[X]/(f(X)). Let α := \overline{X} ∈ K. We have 0 = (α^{9} + α + 1)^{8} = α^{72} + α^{8} + 1, so α^{73} = α(α^{8} + 1) = α^{9} + α = 1. Thus, the relation α^{9} + α + 1 = 0 shows that 3 ∈ W_{2}(73), and it follows easily that W_{2}(73) = {0} ∪ [ 2, ∞)_{ℤ}.

In the balance of this section, let us consider W_{p}(m) in the case when m is a prime power (not divisible by p). Under a special hypothesis on the cyclotomic polynomial Φ_{m}(X), the weight set W_{p}(m) can be determined explicitly.

Theorem 2.6. Let m = ℓ^{a} where ℓ is a prime different from p, and assume that the cyclotomic polynomial Φ_{m}(X) ∈ ℤ[X] remains irreducible modulo p. Then W_{p}(m) = ℕ p + ℕ ℓ.

Proof. Of course, it suffices to prove the inclusion “⊆”. Let ζ be a primitive m th root of unity in \overline{𝔽}_{p}. Let m′ := m/ℓ, and α := ζ^{m′} (a primitive ℓ th root of unity). Let K = 𝔽_{p}(ζ), and L = 𝔽_{p}(α). Since Φ_{m}(X) is irreducible mod p, [K : 𝔽_{p}] = φ(m) = m′(ℓ − 1). From this, it is easy to see that [K : L] = m′ and [L : 𝔽_{p}] = ℓ − 1.

Any vanishing sum of m th roots of unity can be written in the form ∑_{i=0}^{m′−1} g_{i}ζ^{i} = 0, where each g_{i} is a sum of ℓ th roots of unity. Since the degree of ζ over L is m′, the elements 1, ζ, ⋯, ζ^{m′−1} are linearly independent over L. Therefore, each g_{i} ∈ L is itself a vanishing sum, and it suffices to show that its weight is in ℕ p + ℕ ℓ. Starting over again, we are now down to considering a vanishing sum ∑_{i=0}^{ℓ−1} a_{i}α^{i} = 0, where each a_{i} ∈ ℕ. Let a_{j} be the smallest among the a_{i} ’s. Since the minimal equation of α over 𝔽_{p} is 1 + α + ⋯ + α^{ℓ−1} = 0, it follows easily that a_{0} = a_{1} = ⋯ = a_{ℓ−1} ∈ 𝔽_{p}. The weight of the vanishing sum in question is ∑_{i} a_{i} ≡ ℓ a_{j} (mod p). Since ∑_{i} a_{i} ≥ ℓ a_{j}, it follows that ∑_{i} a_{i} = ℓ a_{j} + bp for some b ∈ ℕ, as desired.

Remark 2.7. A vanishing sum of m th roots of unity is said to be minimal if no proper subsum of it is also vanishing. In general, the problem of determining the minimal vanishing sums is difficult (both in characteristic 0 and in characteristic p). Under the hypothesis of (2.6), however, this problem can be solved. In fact, the argument presented in the proof above can be used to show that, in the setting of (2.6), the minimal vanishing sums of m th roots of unity are, up to multiplication by a power of ζ: (1) p · 1 = 0, and (2) 1 + α + ⋯ + α^{ℓ−1} = 0. (Of course, this implies that W_{p}(m) = ℕ p + ℕ ℓ.) For this conclusion, however, the assumption on the irreducibility of Φ_{m}(X) modulo p is essential, as the examples (2.1), (2.4) and (2.5) show. (In (2.1) and (2.4), Φ_{m}(X) splits completely modulo p, and in (2.5), Φ_{m}(X) splits into the product of eight irreducible factors of degree 9 in 𝔽_{p}[X].)

3. The Case gcd(p − 1, m) ≥ 3

In dealing with W_{p}(m), our main goal is to find good estimates for integers n_{0} such that [n_{0}, ∞)_{ℤ} ⊆ W_{p}(m). We begin our analysis with the case when gcd(p−1, m) ≥ 3. This case turns out to be fairly easy if we use the right tools from additive number theory modulo p. It will be convenient to use the following notations. For a subset A in a field, we shall write |A| for the cardinality of A, and for any integer n ≥ 1, we write n ∗ A for the set A + ⋯ + A with n summands of A.

Theorem 3.1. Assume that m_{0} := gcd(p − 1, m) ≥ 3 and let d_{0} = (p − 1)/m_{0}. Then [ d_{0} + 1, ∞)_{ℤ} ⊆ W_{p}(m_{0}) ⊆ W_{p}(m).

Proof. Since m_{0}|(p − 1), the group H of m_{0} th roots of unity in 𝔽_{p} is exactly 𝔽̇_{p}^{d_{0}} and has exactly m_{0} elements. We claim that |n ∗ H| ≥ nm_{0} for n ≤ d_{0}, and |n ∗ H| = p for n ≥ d_{0} + 1. It suffices to prove this for n = 1, 2, ⋯, d_{0} + 1 (for, once we show that |(d_{0} + 1) ∗ H| = p, then (d_{0} + 1) ∗ H = 𝔽_{p}, and this implies that (d_{0} + i) ∗ H = 𝔽_{p} for any i ≥ 1). We proceed by induction on n, the case n = 1 being clear. Assume that |n ∗ H| ≥ nm_{0} where n < d_{0}. By the Cauchy-Davenport Theorem (see [Ma: Cor.1.2.3]), |(n + 1) ∗ H| is either p (and hence ≥ (n + 1)m_{0}), or else

|(n + 1) ∗ H| ≥ |n ∗ H| + |H| − 1 ≥ (n + 1)m_{0} − 1.

In the latter case, |(n+1)∗H \{0}| ≥ (n+1)m_{0}−2. Since H acts on (n+1)∗H \{0} by multiplication, |(n + 1) ∗ H \ {0}| is a multiple of m_{0}. Since m_{0} ≥ 3, we must therefore have |(n + 1) ∗ H \ {0}| ≥ (n + 1)m_{0}, which gives what we want. This proves our claim for n ≤ d_{0}. In particular, |d_{0} ∗ H| ≥ d_{0}m_{0} = p − 1. By the Cauchy-Davenport Theorem again, (d_{0} + 1) ∗ H must be 𝔽_{p}, for otherwise we would have

|(d_{0} + 1) ∗ H| ≥ |d_{0} ∗ H| + |H| − 1 ≥ d_{0}m_{0} + m_{0} − 1 = p + (m_{0} − 1) > p,

a contradiction. This completes our inductive proof. Thus, for any n ≥ d_{0} + 1, we have 0 ∈ n ∗ H. This means that n ∈ W_{p}(m_{0}), and so [d_{0} + 1, ∞)_{ℤ} ⊆ W_{p}(m_{0}) ⊆ W_{p}(m).

Example 3.2. In many cases Theorem 3.1 gives the best result. For instance, if p ≡ 3 (mod 4) and m = (p − 1)/2 ≥ 3, then d_{0} = 2 and we have d_{0} ∉ W_{p}(m) since m is odd. Even in the case p ≡ 1 (mod 4), the Theorem may still give the best result. For instance, if p = 13 and m = 4, then d_{0} = 3 and G = {±1, ±8}. By a simple calculation, 3 ∗ G = 𝔽̇_{13}, so again d_{0} = 3 ∉ W_{p}(m). On the other hand, if m is divisible by two distinct primes p_{1}, p_{2}, then the fact that p_{1}, p_{2} ∈ W_{p}(m) implies that [n_{0}, ∞)_{ℤ} ⊆ W_{p}(m) for n_{0} := (p_{1} − 1)(p_{2} − 1) (see [LeV: p.22, Ex.4]). In case the number d_{0} in (3.1) is “large”, [n_{0}, ∞)_{ℤ} ⊆ W_{p}(m) will of course give a better result.

We can now derive the first case of Theorem 1.3.

Corollary 3.3. Let K = 𝔽_{p^{k}} ba a finite field containing all m th roots of unity, and let d = (p^{k}−1)/m. If m_{0} := gcd(p−1, m) ≥ 3, then [d+1, ∞)_{ℤ} ⊆ W_{p}(m_{0}) ⊆ W_{p}(m).

Proof. Say p − 1 = m_{0}d_{0} and m = m_{0}m_{1}. Then

d = \frac{(p − 1)(p^{k−1} + ⋯ + p + 1)}{m} = d_{0} · \frac{p^{k−1} + ⋯ + p + 1}{m_{1}}.

Since gcd(d_{0}, m_{1}) = 1, the fraction on the RHS above is an integer. Therefore, we have d_{0} | d, and the desired conclusion follows from Theorem 3.1.

4. The Case gcd(p − 1, m) = 2

We shall assume throughout this section that gcd(p − 1, m) = 2 (and as before m ≥ 3). In particular, p is odd and m is even. In this case, W_{p}(m) contains 2 ℕ and is stable under addition by 2. Thus, once we have an odd integer n ∈ W_{p}(m), we will have automatically [n − 1, ∞)_{ℤ} ⊆ W_{p}(m). This observation will be used without further mention in the following.

Let K = 𝔽_{p^{k}} be any finite field containing the group G of all m th roots of unity. The following result gives a somewhat sharper form of Theorem 1.3 in the case gcd(p − 1, m) = 2 (in that the index [K̇ : G] itself is shown to be a weight, with a minor exception).

Theorem 4.1. Assume that gcd(p − 1, m) = 2, and let d = [K̇ : G] = (p^{k} − 1)/m. Then [ d, ∞)_{ℤ} ⊆ W_{p}(m) unless p = 3 and m = 3^{k} −1 , in which case [ d+1, ∞)_{ℤ} ⊆ W_{3}(m).

Proof. Let us first check the Theorem when m = 4. In this case, the assumption gcd(p − 1, m) = 2 implies that G is not contained in 𝔽_{p}, so k ≥ 2. If p > 3, then d ≥ (p^{2} − 1)/4 ≥ p, and we have [ d, ∞)_{ℤ} ⊆ [ p, ∞)_{ℤ} ⊆ W_{p}(m). If p = 3, then d ≥ (9 − 1)/4 = 2, and we have again [ d, ∞)_{ℤ} ⊆ [ 2, ∞)_{ℤ} ⊆ W_{3}(m) (since W_{3}(m) contains both 2 and 3). In the following, we may therefore assume that m ≥ 6.

Write m = 2m′, so that

(4.2)  d = \frac{(p − 1)(p^{k−1} + ⋯ + p + 1)}{2m′} = \frac{p − 1}{2} · \frac{p^{k−1} + ⋯ + p + 1}{m′}.

Since gcd(m′, (p−1)/2) = 1, we have m′|(p^{k−1}+⋯+p+1). If m′ < p^{k−1}+⋯+p+1, the second factor on the RHS in (4.2) is ≥ 2, so d ≥ p − 1. Since p ∈ W_{p}(m), we have [d, ∞)_{ℤ} ⊆ [p − 1, ∞)_{ℤ} ⊆ W_{p}(m), as desired. Therefore, in the following we may assume that

(4.3)  m′ = p^{k−1} + ⋯ + p + 1, and d = (p − 1)/2.

In this case K̇ = 𝔽̇_{p−1}·G, so any coset of G in K̇ has a “scalar” representative. We fix a generator ζ for the group G, and try to put a lower bound on the cardinality of the set A := 𝔽_{p} ∩ 2 ∗ G.

Recalling that m ≥ 6, write

(ζ − 1)G = a_{1}G , (ζ^{2} − 1)G = a_{2}G , and (ζ^{4} − 1)G = a_{3}G where a_{i} ∈ 𝔽̇_{p}.

Clearly, ±a_{i} ∈ A, since −1 ∈ G. First let us assume that these three G-cosets in K̇ are different. Since a_{i} ≠ −a_{j}, {0, ±a_{i}} are seven different elements of A. (In particular, p ≥ 7 here.) Applying repeatedly the Cauchy-Davenport Theorem in 𝔽_{p}, we see that |n ∗ A| ≥ min{p, 6n + 1}. It follows that

n ≥ (p − 1)/6 ⟹ n ∗ A = 𝔽_{p} ⟹ −1 ∈ n ∗ A ⟹ 2n + 1 ∈ W_{p}(m).

This yields 2⌈\frac{p−1}{6}⌉ + 1 ∈ W_{p}(m) (where ⌈·⌉ denotes the ceiling function). Writing p in the form 6t ± 1, we see easily that 2⌈\frac{p−1}{6}⌉ = ⌈\frac{p−1}{3}⌉. Thus, in this case, we get the stronger conclusion that [ (p − 1)/3, ∞)_{ℤ} ⊆ W_{p}(m).

From now on, we may assume that the three cosets {a_{i}G} above are not all different. If a_{1}G = a_{2}G, then ζ^{2} − 1 = (ζ − 1)ζ^{i} for some i, and so ζ^{i} = ζ + 1. Since m is even, this shows that 3 ∈ W_{p}(m), and so [ 2, ∞)_{ℤ} ⊆ W_{p}(m). We have certainly no problem in this case (except when d = 1, which occurs only when p = 3). If a_{2}G = a_{3}G, we can finish similarly. Now assume a_{1}G = a_{3}G. Here, ζ^{4}−1 = (ζ−1)ζ^{j} for some j, so ζ^{j} = ζ^{3} + ζ^{2} + ζ + 1. As before, this gives 5 ∈ W_{p}(m). If p > 7, then d = (p − 1)/2 ≥ 5, and we have what we want. Thus we are only left with the cases p = 3, 5, 7.

If p = 3, we have d = (p − 1)/2 = 1 (and m = 3^{k} − 1 by (4.3)). In this case the desired conclusion is [ 2, ∞)_{ℤ} ⊆ W_{3}(m), which is true since 2, 3 ∈ W_{3}(m).

If p = 5, then d = 2 ∈ W_{5}(m). In this case we need to show that 3 ∈ W_{5}(m). If a_{1}G = a_{2}G, we are done as before. Otherwise, one of these cosets must be the identity coset G (since [K̇ : G] = d = 2), and this implies again that 3 ∈ W_{5}(m).

Finally, we treat the case p = 7. Here we must show that [K̇ : G] = d = 3 is in the weight set W_{7}(m). We first note that:

(4.4) If (ζ − 1)G = (ζ^{i} − 1)G = (ζ^{i+1} − 1)G for some i ≥ 1, then 3 ∈ W_{7}(m).

Indeed, if we write ζ^{i} − 1 = (ζ − 1)ζ^{r} and ζ^{i+1} − 1 = (ζ − 1)ζ^{s}, then ζ^{r} = ζ^{i−1} + ⋯ + ζ + 1 and ζ^{s} = ζ^{i} + ⋯ + ζ + 1 imply that ζ^{s} = ζ^{i} + ζ^{r}, so 3 ∈ W_{7}(m). Now let C, C′ be the two nonidentity cosets of G in K̇. By reasonings we have used before, we may assume that (ζ^{2} − 1)G = C and (ζ − 1)G = (ζ^{4} − 1)G = C′. Noting that 3 is prime to m (since gcd(p − 1, m) = gcd(6, m) = 2), we may also assume, in view of (4.4), that (ζ^{3} − 1)G = C. Replacing ζ by ζ^{3}, we may further assume that (ζ^{9} − 1)G = C′. Next, note that since m ≥ 6, it cannot divide 10, so ζ^{10} ≠ 1. Thus, in view of (4.4), we may assume that (ζ^{5} − 1)G = C, and hence that (ζ^{10} − 1)G = C′. Now we have C′ = (ζ − 1)G = (ζ^{9} − 1)G = (ζ^{10} − 1)G, so 3 ∈ W_{7}(m) once more by (4.4).

5. The Case gcd(p − 1, m) = 1

Throughout this section, we shall assume that gcd(p − 1, m) = 1 (and as before, m ≥ 3). The analysis of the weight set W_{p}(m) in this case turns out to require the hardest work.

The assumption that gcd(p − 1, m) = 1 means that the only m th root of unity in 𝔽_{p} is 1. Therefore, upon factoring the polynomial X^{m} − 1 modulo p, we have

(5.1)  X^{m} − 1 = (X − 1)g_{1}(X)g_{2}(X) ⋯,

where the g_{i} ’s are irreducible monic polynomials in 𝔽_{p}[X], each of degree ≥ 2. Let ℓ := min{deg(g_{i})}. This integer ℓ will play an important role in finding the estimates on W_{p}(m) in this section, so let us first note a few other characterizations of it.

Recall that the cyclotomic polynomial Φ_{n}(X) ∈ ℤ[X] factors modulo p into a product of irreducible factors each of degree given by the order of the element p in the unit group U(ℤ/n ℤ) (see, e.g. [Gu]). Since X^{m} − 1 = ∏_{n|m} Φ_{n}(X), it follows that ℓ is the minimum of the orders of p in U(ℤ/n ℤ) for n ranging over the divisors of m greater than 1. From this, we see that ℓ is also the minimum of the orders of p in U(ℤ/q ℤ) for q ranging over the prime divisors of m. It is now an easy exercise to check the following:

(5.2)  ℓ = min {e ≥ 1 : gcd(p^{e} − 1, m) > 1}.

This simply means that 𝔽_{p^{ℓ}} is the field with the smallest extension degree over 𝔽_{p} which contains an m th root of unity other than 1. This can also be verified directly from the definition of ℓ.

For the rest of this section, let L := 𝔽_{p^{ℓ}}, m′ := gcd(p^{ℓ} − 1, m), and let H be group of m′ th roots of unity in L. By (5.2), |H| = m′ ≥ 2. It will be important to work with the set T := tr(H) where “tr” denotes the field trace from L to 𝔽_{p}. The next theorem gives a description of W_{p}(m) in terms of ℓ and the cardinality t := |T | (under the standing assumption that gcd(p − 1, m) = 1).

Theorem 5.3. Let ℓ and t be as defined above, and let n := ⌈\frac{p−1}{t−1}⌉. Then [ ℓ n, ∞)_{ℤ} ⊆ W_{p}(m′) ⊆ W_{p}(m).

Proof. Applying the Cauchy-Davenport Theorem to the subset T in 𝔽_{p}, we have |2 ∗ T | ≥ min {p, 2t − 1}, and inductively |i ∗ T | ≥ min {p, it − (i − 1)}. By the definition of n, we have n(t − 1) ≥ p − 1, so nt − (n − 1) ≥ p. Therefore, |n ∗ T | = p. In particular, for every j ≥ 0, there exists an equation t_{1} +⋯+t_{n} = −j ∈ 𝔽_{p}, where all t_{i} ∈ T. Now each t_{i} ∈ tr(H) is a sum of ℓ elements of H, so t_{1} + ⋯ + t_{n} + j · 1 = 0 is a vanishing sum of m′ th roots of unity of weight ℓ n + j. This shows that [ ℓ n, ∞)_{ℤ} ⊆ W_{p}(m′) ⊆ W_{p}(m), as desired.

Note that the above theorem is meaningful only if we know that the trace set T ⊆ 𝔽_{p} has at least two elements. Fortuitously, this is always the case, according to the following result.

Trace Lemma 5.4. In the notations of (5.3), t = |T | ≥ 2.

The proof of this lemma will be postponed to the last section (§6). We shall first assume this lemma and try to get to the main conclusions of this section. Note that the larger the trace set T is, the better bound on W_{p}(m) is given by (5.3). Since t ≥ 2 by (5.4), we have in any case:

Corollary 5.5. [ ℓ (p − 1), ∞)_{ℤ} ⊆ W_{p}(m′) ⊆ W_{p}(m).

Now consider any field K = 𝔽_{p^{k}} containing the group G of all m th roots of unity. Clearly, L ⊆ K, and H = L̇ ∩ G. Let d = (p^{k} − 1)/m and d′ = (p^{ℓ} − 1)/m′. We now proceed to the proof of the following, which is a stronger version of Theorem 1.3 in the case treated in this section.

Theorem 5.6. Assume that gcd(p − 1, m) = 1. Then [ d′, ∞)_{ℤ} ⊆ W_{p}(m′) and [ d, ∞)_{ℤ} ⊆ W_{p}(m) except in the following two special cases: (A) d′ = p − 1; (B) p = 2, d′ = 3, and m′ = 5. In these special cases, we have [ d′ + 1, ∞)_{ℤ} ⊆ W_{p}(m′) and [ d + 1, ∞)_{ℤ} ⊆ W_{p}(m).

Proof. Since d′ = [L̇ : H] = [L̇G : G] divides d = [K̇ : G] and m′ | m, it suffices to prove the theorem for W_{p}(m′). Let

(5.7)  s = (p^{ℓ−1} + ⋯ + p + 1)/m′,

so that d′ = s(p−1). First let us treat the special case (A), where we have s = 1. Here we are supposed to prove that [p, ∞)_{ℤ} ⊆ W_{p}(m′). Since now [L̇ : H] = p − 1 (and H ∩ 𝔽_{p} = {1}), we have L̇ = H · 𝔽̇_{p}. Therefore, fixing a primitive m′ -th root of unity α ∈ H, we have α−1 = b^{−1}α^{i} for some integer i and some b ∈ 𝔽̇_{p}. For convenience, let us think of b as an integer in [1, p − 1]. Multiplying b · 1 + (p − b) · 1 = 0 ∈ 𝔽_{p} by α and using the relation bα = b + α^{i}, we get

0 = bα + (p − b)α = b · 1 + α^{i} + (p − b)α,

which is a vanishing sum of weight p + 1. Multiplying this by α again and repeating the argument, we get vanishing sums (of m′ -th roots of unity) of weight p + i for any i > 0. Coupled with p ∈ W_{p}(m′), this gives [ p, ∞)_{ℤ} ⊆ W_{p}(m′), as desired. For the rest of the proof, we may assume that s > 1. We claim the following:

Lemma 5.8. s > 1 implies that s ≥ ℓ, except perhaps when p = 2 and ℓ = 4, 6, 8, 9.

Thus, leaving aside the four special cases, we have d′ = s(p − 1) ≥ ℓ (p − 1), so the desired conclusion for W_{p}(m′) in (5.6) follows from (5.5). The four special cases will have to be treated later.

Proof of (5.8). We go into the following two cases. Case 1. ℓ is prime. We claim that ℓ ≤ q for any prime q | s (and therefore ℓ ≤ s). In fact, from (5.7), we get p^{ℓ−1} + ⋯ + p + 1 ≡ 0 (mod q), so p^{ℓ} ≡ 1 (mod q). If p ≢ 1 (mod q), then p has order ℓ in U(ℤ/qℤ) (since ℓ is prime), and so ℓ | (q − 1). In this case ℓ ≤ q − 1 < q. If p ≡ 1 (mod q), then from p^{ℓ−1} + ⋯ + p + 1 ≡ ℓ ≡ 0 (mod q), we have in fact ℓ = q. Case 2. ℓ is composite. Let q be the smallest prime divisor of ℓ and write ℓ = qt. Then 1 < t < ℓ and (5.2) implies that gcd(p^{t} − 1, m) = 1. Since (p^{t} − 1) | (p^{ℓ} − 1), we see that p^{t} − 1 divides (p^{ℓ} − 1)/m′ = s(p − 1). Thus, s ≥ (p^{t} − 1)/(p − 1). We shall now exploit the following elementary fact which is easy to prove using calculus:

Lemma 5.9. p^{x} ≥ (p − 1)x^{2} + 1 for every x ∈ [ 2, ∞)_{ℤ} with the exception of p = 2 and x = 2, 3, 4.

Applying this lemma to x = t, we get the desired conclusion

s ≥ \frac{p^{t} − 1}{p − 1} ≥ t^{2} ≥ qt = ℓ,

except when p = 2 and t = 2, 3, 4. If t = 2, we have q = 2 so ℓ = 4. If t = 3, we have q = 2, 3, so ℓ = 6 or 9. Finally, if t = 4, we have q = 2 so ℓ = 8. This proves (5.8), but we still have to complete the proof of (5.6) in the four special cases noted.

In these cases, d′ = (2^{ℓ} − 1)/m′, so both m′, d′ are odd (and > 1). We may assume that d′ < m′. (For, if d′ ≥ m′, we have [ m′, ∞)_{ℤ} ⊆ W_{2}(m′) since m′ is odd, and hence [ d′, ∞)_{ℤ} ⊆ W_{2}(m′).) We simply have to check the four outstanding cases individually.

(1) ℓ = 4. Here 2^{ℓ} − 1 = 15, so d′ = 3, m′ = 5, and we are in the case (B) of (5.6). The desired conclusion in this case is [ 4, ∞)_{ℤ} ⊆ W_{2}(m′), which is true since 5 = m′ ∈ W_{2}(m′). In fact, by (2.6), we have W_{2}(5) = 2 ℕ + 5 ℕ = {0, 2} ∪ [ 4, ∞)_{ℤ}. In particular, 3 ∉ W_{2}(5), so this case is truly exceptional.

(2) ℓ = 6. Here 2^{ℓ} − 1 = 63, so we have either d′ = 7, m′ = 9 or d′ = 3, m′ = 21. In both cases, [ 2, ∞)_{ℤ} ⊆ W_{2}(m′) (since 2, 3 ∈ W_{2}(m′)), so there is no problem. (Actually, in the case d′ = 7, we are in the good case s = d′ ≥ ℓ already.)

(3) ℓ = 8. Here 2^{ℓ} − 1 = 255, so we have either d′ = 5, m′ = 51 or d′ = 15, m′ = 17. The latter case presents no problems, since we are once more in the good case s = d′ ≥ ℓ. In the former case, 3 | m′ implies that W_{2}(m′) = [ 2, ∞)_{ℤ}, so again there is no problem.

(4) ℓ = 9. Here 2^{ℓ} − 1 = 511, so d′ = 7, m′ = 73. We have shown in (2.5) that W_{2}(73) = {0} ∪ [ 2, ∞)_{ℤ}, so there is no problem.

This finally completes the proof of Theorem 5.6.

6. Traces of m th Roots of Unity

In §5, we stated without proof the Trace Lemma 5.4, which was crucial for the proofs of (5.5) and (5.6). In this section, we return to the trace set T = tr(H), and offer a general analysis of T which we believe to be of independent interest. The proof of the Trace Lemma is an easy by-product of this general analysis.

The notations (and hypotheses) introduced at the beginning of §5 will remain in force. In particular, H is the group of m′ th roots of unity in L = 𝔽_{p^{ℓ}}, and “tr” is the field trace from L to 𝔽_{p}. To enumerate the elements in T, let

(6.1)  X^{m′} − 1 = (X − 1)h_{1}(X) ⋯ h_{r}(X)

be the factorization of X^{m′} − 1 into (monic) irreducibles over 𝔽_{p}. Then deg h_{i} ≥ ℓ by the definition of ℓ (and the fact that (X^{m′} − 1) | (X^{m} − 1)). On the other hand, since L contains all m′ th roots of unity, each h_{i}(X) splits completely in L, so deg h_{i} ≤ [L : 𝔽_{p}] = ℓ. Therefore, deg h_{i} = ℓ for all i. Let

(6.2)  h_{i}(X) = X^{ℓ} − a_{i}X^{ℓ−1} + ⋯,

and let {α_{ij}} be all the roots of h_{i}(X) in L. For each h_{i}, we can identify the field 𝔽_{p}[X]/(h_{i}(X)) with L by the correspondence \overline{X} ↔ α_{ij} (for any j). Therefore, tr(α_{ij}) = ∑_{k} α_{ik} = a_{i} for all i, j. We have thus

(6.3)  T = {tr(1), a_{1}, ⋯, a_{r}} = {ℓ, a_{1}, ⋯, a_{r}},

with possible duplications.

It is now easy to prove the Trace Lemma 5.4, which asserted that |T | ≥ 2. Assume, for the moment, that T is a singleton. Then, by (6.3), a_{i} = ℓ for all i. Summing all roots of the polynomial in (6.1) (and recalling that m′ ≥ 2), we get

0 = 1 + a_{1} + ⋯ + a_{r} = 1 + r ℓ = m′ ∈ 𝔽_{p} ,

contradicting the fact that m′ is prime to p.

The equation (6.3) gives an upper bound |T | ≤ 1+r, and this becomes an equality iff the elements listed in (6.3) are distinct. This is the case, for instance, if ℓ = 2. To see this, note that the constant term of each h_{i}(X) in (6.1) is 1, since it is an m th root of unity in 𝔽_{p}, and we are assuming that gcd(p−1, m) = 1. Therefore, if ℓ = 2, we have h_{i}(X) = X^{2} − a_{i}X + 1. Since the h_{i} ’s are distinct, so are the a_{i} ’s, and of course a_{i} ≠ 2 (since otherwise h_{i}(X) = (X − 1)^{2}). Therefore, |T | = 1 + r, and (5.3) gives the pretty good estimate [ 2n, ∞)_{ℤ} ⊆ W_{p}(m′) ⊆ W_{p}(m) with n = ⌈\frac{p−1}{r}⌉. For a simple example of this, let p = 5, and m = 3. Here m′ = 3, ℓ = 2, r = 1, L = 𝔽_{25}, and |T | = 2. By (2.6), W_{5}(3) is the set

3 ℕ + 5 ℕ = {0, 3, 5, 6} ∪ [ 8, ∞)_{ℤ}.

Since 2n = 8, the conclusion in (5.3) is sharp here.

In general, t := |T | may be less than 1+r, since there may be duplications among the elements of T listed in (6.3). For an example where interesting duplications occur, take p = 7 and m = 19. Here m′ = 19 and ℓ = 3, r = 6. Mathematica gives a factorization

X^{19} − 1 = (X − 1)(X^{3} + 2X + 6)(X^{3} + 4X^{2} + X + 6)(X^{3} + 4X^{2} + 4X + 6)

· (X^{3} + 5X^{2} + 6)(X^{3} + 3X^{2} + 3X + 6)(X^{3} + 6X^{2} + 3X + 6) ∈ 𝔽_{7}[X].

Since −4 = 3 = tr(1) in 𝔽_{7}, T has only five (two less than 1 + r = 7) distinct elements \{0, 1, 2, 3, 4\}. In this case, the number n in (5.3) is ⌈\frac{7−1}{5−1}⌉ = 2, and (5.3) shows that [ 6, ∞)_{ℤ} ⊆ W_{7}(19). Note that, in spite of the trace duplications, this is still much sharper than what is given in (5.6).

In general, we cannot hope to improve upon the lower bound |T | ≥ 2. For one thing, 𝔽_{p} may have only two elements to begin with. Also, we may have r = 1, in which case (5.4) and (6.3) show that |T | = 2. Even if p ≥ 3 and r ≥ 2, there are many cases in which T is just a doubleton. Let us illustrate the situation r = 2 by taking m to be an odd prime q (so that m′ = q too), and assuming that p is also odd and has order ℓ = (q − 1)/2 in the group U(ℤ/q ℤ). In this case, r = 2, and (6.1) becomes

(6.4)  X^{q} − 1 = (X − 1)h_{1}(X)h_{2}(X)

where h_{1}, h_{2} are monic irreducible (over 𝔽_{p}) of degree ℓ. Following a standard notation in number theory, let us define q^{∗} to be q if q ≡ 1 (mod 4), and q^{∗} to be −q if q ≡ 3 (mod 4). Then the size of the trace set T = tr(H) is determined as follows.

Proposition 6.5. Under the above assumptions |T | = 2 iff p | (q^{∗}−1) (and |T | = 3 otherwise).

Proof. Let E = ℚ(ζ) where ζ = e^{2πi/q}, and fix a generator t of U(ℤ/q ℤ). Then σ : ζ ↦ ζ^{t} is a generator for Gal(E/ℚ), and σ^{2} : ζ ↦ ζ^{t^{2}} is a generator for Gal(E/F), where F is the fixed field E^{σ^{2}}. Note that X^{q−1} + ⋯ + X + 1 factors into f(X)g(X) over F [X], where

f(X) = X^{ℓ} − a X^{ℓ−1} + ⋯ and g(X) = X^{ℓ} − b X^{ℓ−1} + ⋯

are, respectively, the minimal polynomials of ζ and ζ^{t} over F. We have

a − b = tr_{E/F} (ζ) − tr_{E/F} (ζ^{t}) = ∑_{j=0}^{ℓ−1} ζ^{t^{2j}} − ∑_{j=0}^{ℓ−1} ζ^{t^{2j+1}} ∈ F.

This is precisely the quadratic Gauss sum (with respect to the Legendre character on 𝔽_{q}), so by [IR: (8.2.2)], a − b = \sqrt{q^{∗}}. (Gauss showed that the \sqrt{q^{∗}} here is the one taken in the upper half plane if q ≡ 3 (mod 4), but this will not be needed in the following.) Since we also have a + b = −1, it follows that

a = (\sqrt{q^{∗}} − 1)/2 , b = −(\sqrt{q^{∗}} + 1)/2.

Incidentally, this proves the well-known fact that F = ℚ(\sqrt{q^{∗}}).

Let R be the ring of algebraic integers in F. Since p is unramified in E, it is also unramified in F, so pR = 𝔭 𝔭′, where 𝔭, 𝔭′ are distinct prime ideals of R, both of residue degree 1. Identifying R/𝔭 with 𝔽_{p}, we may take the polynomials h_{1}, h_{2} in (6.4) to be \overline{f} and \overline{g}, where “bar” means reduction modulo 𝔭. In particular, T = \{ℓ, \overline{a}, \overline{b}\} by (6.3). Here the two elements \overline{a}, \overline{b} are always different (for otherwise 𝔭 would contain (a−b)^{2} = q^{∗} as well as p). Therefore, T will have only two elements iff 𝔭 also contains

4(ℓ − a)(ℓ − b) = (q − \sqrt{q^{∗}}) (q + \sqrt{q^{∗}}) = q^{2} − q^{∗} = q^{∗}(q^{∗} − 1).

Since q^{∗} ∉ 𝔭, this happens iff q^{∗} − 1 ∈ 𝔭, that is, iff p | (q^{∗} − 1), as claimed.

Corollary 6.6. Let q = 2ℓ + 1 and p = 2ℓ′ + 1 be distinct primes such that the order of p is ℓ modulo q. If p | (q^{∗} − 1), then [ 2ℓℓ′, ∞)_{ℤ} ⊆ W_{p}(q). Otherwise, [ ℓℓ′, ∞)_{ℤ} ⊆ W_{p}(q).

Proof. This follows from (5.3) and (6.5), since the number n in (5.3) is 2ℓ′ in the first case, and ℓ′ in the second case.

Example 6.7. Let q = 11 (with q^{∗} = −11). Then the primes 3 and 5 both have order ℓ = (q − 1)/2 = 5 modulo q, and according to Mathematica:

X^{11} − 1 = (X − 1)(X^{5} + X^{4} + 2X^{3} + X^{2} + 2)(X^{5} + 2X^{3} + X^{2} + 2X + 2) ∈ 𝔽_{3}[X], X^{11}−1 = (X−1)(X^{5}+2X^{4}+4X^{3}+X^{2}+X+4)(X^{5}+4X^{4}+4X^{3}+X^{2}+3X+4) ∈ 𝔽_{5}[X].

Thus, for p = 3, T = {ℓ, −1, 0} = {2, 0} in 𝔽_{3}. This is consistent with (6.5) since p = 3 divides q^{∗} − 1 = −12. Here, ℓ = 5, ℓ′ = 1, so (6.6) gives [ 10, ∞)_{ℤ} ⊆ W_{3}(11). (In fact, from 0, 2 ∈ T, we see easily that 5, 6 ∈ W_{3}(11), and so 8, 9 ∈ W_{3}(11) also.) On the other hand, if we choose p = 5, then T = {ℓ, −2, −4} = {0, 3, 1} in 𝔽_{5}, consistently with (6.5) since p = 5 does not divide q^{∗} − 1 = −12. Here, ℓ = 5, ℓ′ = 2, so (6.6) gives again [ 10, ∞)_{ℤ} ⊆ W_{5}(11), and 0, 1, 3 ∈ T show further that 5, 7, 9 ∈ W_{5}(11).

The arguments in the proof of (6.5) can be generalized. However, if the order of p modulo q is smaller than (q − 1)/2 (in other words r > 2), the computations of the trace elements in T will involve Gaussian sums with (higher) character values as coefficients. We shall not go into this analysis here. We should point out, however, that if q is fixed, then the prime ideal method (in characteristic 0) used in the proof of (6.5) will suffice to show that the upper bound |T | ≤ 1 + r becomes an equality for sufficiently large p. Therefore, by (5.3), [ ℓ n, ∞)_{ℤ} ⊆ W_{p}(q) with n = ⌈\frac{p−1}{r}⌉, for sufficiently large p.

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VANISHING SUMS OF m TH ROOTS OF UNITY IN FINITE FIELDS 15

T. Y. Lam, Mathematics Department, University of California, Berkeley, CA, 94720

E-mail address: lam@msri.org K. H. Leung, National University of Singapore, Singapore 119260