On the pure Jacobi Sums ^{∗}

Shigeki Akiyama

Let p be an odd prime and F_{q} be the field of q = p^{2} elements. We consider the Jacobi sum over F_{q};

J(χ, ψ) = ∑_{x∈F_{q}} χ(x)ψ(1 − x),

where χ, ψ is a non trivial character of F_{q}^{×}, whose value at 0 is defined to be 0. It is well known that the absolute value of J(χ, ψ) is \sqrt{q} = p, when χψ is not principal. According to [11], [9], call the Jacobi sum J(χ, ψ) pure if J(χ, ψ)/p is a root of unity.

Let ord(χ) be the order of χ in F̂_{q}^{×}. From now on in this paper, we assume that ord(ψ) = 2 and ord(χ) = n ≥ 3. This special type of Jacobi sums play an important role in evaluating the argument of Gauss sum:

G(χ) = ∑_{x∈F_{q}} χ(x)ζ_{p}^{TrF_{q}/F_{p}(x)},

where ζ_{p} is a primitive p-th root of unity (see [3], [4] ). Moreover, recently the rationality of this Jacobi sum is used to characterize the irreducible module of the Terwilliger algebras of cyclotomic association schemes (see [10]).

In this note, we prove

Theorem. J(χ, ψ) is pure if and only if one of the the following four conditions holds:

1. n is a divisor of p + 1

2. n = 2(p − 1)/k with an odd integer k

3. n = 24 and p ≡ 17, 19 (mod 24)

4. n = 60 and p ≡ 41, 49 (mod 60)

∗Only for the private copy. See Acta Arith. Vol.75, no.2 (1996), 97 – 104, for the exact published version.

Further, J(χ, ψ) = ±p in all four cases.

There are a lot of results concerning the determination of Gauss and Jacobi sums. See eg. [3], [4], [8]. A nice historical survey is found in [5]. Same type of ”purity” problem for the case of Gauss sums are treated in [1], [13], [2], [7], [9], [11]. But it seems that this type of concrete result on Jacobi sums is not known. Although the essential tool to prove the Theorem is the theorem of Stickelberger, the argument of ”only if” part of the proof is elementary and rather complicated. The author feels somewhat curious that this simple result is derived by such brute force method.

We first see

Proposition. Let S_{1} = {x ∈ (Z/nZ)^{×} : x ≡ i (mod n) for i ∈ [1, n/2) ∩ Z}, and S_{2} = (Z/nZ)^{×} \ S_{1}. Then J(χ, ψ) is pure if and only if there exist an element x ∈ S_{1} so that xS_{1} = S_{1} and p ≡ −x (mod n) holds.

Proof. By using the Theorem of Hasse-Davenport (see Theorem 5.1 of [12]) and the well known result of the sign determination of Gauss sums of order 2 for the prime field F_{p}, it is shown that G(ψ) = (−1)^{(p+1)/2}p. The theorem of Stickelberger (see Theorem 2.2 of [12]) reads, J(χ, ψ)/p = ±G(χ)/G(χψ) is a unit of the integer ring of Q(ζ_{n}) if and only if

\{\frac{a}{n}\} + \{\frac{pa}{n}\} = \{\frac{a}{n} + \frac{1}{2}\} + \{p(\frac{a}{n} + \frac{1}{2})\} for all a with (a, n) = 1.

Here {x} = x − [x] and [x] is the greatest integer not exceeds x. As p is odd, we have to check the conditions only for a ∈ S_{1}, because the condition is symmetric with respect to a ↔ n − a. Thus we choose a with 1 ≤ a ≤ n/2 and (a, n) = 1. Then we have

\frac{a}{n} + \{\frac{pa}{n}\} = \frac{a}{n} + \frac{1}{2} + \{\frac{pa}{n} + \frac{1}{2}\} for a ∈ S_{1}.  (1)

Note that the condition depends only on p mod n. We see that (1) is equivalent to {pa/n} ∈ [1/2, 1) for a ∈ S_{1}. Put p = ny − x with integers x, y. Then we see x mod n ∈ S_{1} and {xa/n} must lie in the interval [0,1/2). Noting that |J(χ, ψ)| = p and Q(ζ_{n})/Q is an abelian extention, J(χ, ψ)/p is a root of unity under these conditions. This shows the assertion. □

Remark. Consider the Jacobi sum on the general finite field F_{q} with q = p^{f}, for a while. Then, similarly as in the above proof, we can easily show that, if the extention degree f is odd, then there are no χ that J(χ, ψ)/\sqrt{q} is a root of unity. Our Theorem concerned the first non trivial case.

The sufficiency of the conditions of the Theorem follows immediately. Because • 1 × S_{1} = S_{1} ⟷ Condition 1.

• n ≡ 0 (mod 4) and (n/2 − 1)S_{1} = S_{1} ⟷ Condition 2.

• If n = 24 and S_{1} = {˜1, ˜5, ˜7, ˜11}, then ˜5S_{1} = S_{1} and ˜7S_{1} = S_{1} ⟷ Condition 3.

• If n = 60 and S_{1} = {˜1, ˜7, ˜11, ˜13, ˜17, ˜19, ˜23, ˜29}, then ˜11S_{1} = S_{1} and ˜19S_{1} = S_{1} ⟷ Condition 4.

Here we denote by x̃, the coset x (mod n). Our next task is to show that J(χ, ψ) is real in the above four cases. The first two cases are handled easily.

Lemma 1. We have J(χ, ψ) = ±p when n is a divisor of p + 1 or n = 2(p − 1)/k with an odd integer k.

Proof. We see

J(χ, ψ) = ∑_{x∈F_{q}} χ(x^{p})ψ(1 − x^{p}) = ∑_{x∈F_{q}} χ^{p}(x)ψ^{p}(1 − x) = J(χ^{p}, ψ)  (2)

If p ≡ −1 (mod n) then

J(χ^{p}, ψ) = J(\overline{χ}, ψ) = \overline{J(χ, ψ)}

This shows the first case. For the second case, we have p − 1 ≡ nk/2 ≡ n/2 (mod n) and χ^{p−1} = χ^{n/2} = ψ, as n = ord(χ). By using (2) and ψ(−1) = 1, we have

J(χ, ψ) = ∑_{x∈F_{q}} χ^{p}(x)ψ(1 − x) = ∑_{x∈F_{q}} χ(x)ψ(x(1 − x))
= ∑_{x∈F_{q}^{×}} \overline{χ(x)}ψ(x^{−1}(1 − x^{−1})) = ∑_{x∈F_{q}} \overline{χ(x)}ψ(x − 1)
= \overline{J(χ, ψ)}

This shows the assertion. □

The case n = 24 was already proved in [4]. We show this directly for the convenience of the reader.

Lemma 2. We have J(χ, ψ) = ±p when n = 24 and p ≡ 17, 19 (mod 24).

Proof. We have already shown that J(χ, ψ)/p is a root of unity of the field Q(ζ_{24}). Let σ_{k} be the element of Gal(Q(ζ_{24})/Q) with σ_{k}(ζ_{24}) = ζ_{24}^{k}. Then we have

σ_{11}(J(χ, ψ)) = σ_{11}(G(χ)G(ψ)/G(χ^{13})) = G(\overline{χ^{13}})G(ψ)/G(\overline{χ}) = J(χ, ψ).

This shows that J(χ, ψ)/p, the 24-th root of unity, is invariant under σ_{11}. This shows the assertion. □

The first manuscript of this note does not contain the next proof of Lemma 3, which is essentially due to Mieko Yamada. The author tried in vain to do this. She also informed me that more general assertion concerning Lemma 1,2 and 3 is presented in [10] and the detailed version of it will contain its proof.

Lemma 3. We have J(χ, ψ) = ±p when n = 60 and p ≡ 41, 49 (mod 60).

Proof. Similarly as in Lemma 2, we see σ_{29}(J(χ, ψ)) = (J(χ, ψ)). Thus J(χ, ψ) ∈ Q(\sqrt{−1}) and J(χ, ψ)/p is equal to ±1 or ±\sqrt{−1}. Note that

(J(χ, ψ))^{5} ≡ J(χ^{5}, ψ) (mod 5).  (3)

As ord(χ^{5}) = 12, we see that there exist rational integers C, D with p = C^{2} + D^{2} and

J(χ^{5}, ψ) = −(C + D\sqrt{−1})^{2},

by Theorem 4.8 and 4.10 of [4]. We easily see that CD ≡ 0 (mod 5) as p ≡ ±1 (mod 5). This shows the assertion. □.

Remark. There is a remaining problem to determine the sgn of J(χ, ψ) when it is real. See [4], for the first three cases of the Theorem. If n = 60 and p ≡ 41, 49 (mod 60), the congruence relation (3) and Theorem 4.8 and 4.10 of [4] is enough to determine the sgn ambiguity of J(χ, ψ). Summing up, we have

J(χ, ψ) = \{ \begin{matrix}p, & for the case 1 and 3 \\ (−1)^{(p+1)/2}p, & for the case 2 \\ ± (\frac{p}{3}) p, & for the case 4. \end{matrix}

Here ± of the last case is + (resp. −) when A ≡ 0 (mod 5) (resp. ≢), with a positive odd integer A, which is uniquely determined by p = A^{2} + B^{2}.

Now we show the necessity of the conditions of the Theorem. For the convenience, we identify the element of S_{1} (resp. S_{2}) with the integer in [1, n/2) (resp. [n/2, n) ) in the later proof.

Lemma 4. Besides the case 1 of the Theorem, if J(χ, ψ) is pure then n = ord(χ) is divisible by 4.

Proof. Assume n is odd and a ≠ 1 is an integer with aS_{1} = S_{1}. Choose an integer i ≥ 1 such that n/2^{i+1} < a ≤ n/2^{i}. Then a ≠ 1 implies 1 ≤ n/2^{i+1}. Thus 2^{i} ∈ S_{1} and 2^{i}a ∈ S_{2} which contradicts with aS_{1} = S_{1}. Thus by using the proposition, n must be even. Now if n = 2m and m is odd, then similarly we choose a ∈ S_{1} and an integer i. Noting that a is odd, we have

a(\frac{n}{2} − 2^{i}) ≡ \frac{n}{2} − 2^{i}a (mod n).

We see n/2 − 2^{i} ∈ S_{1} and n/2 − 2^{i}a ∈ S_{2}, which contradicts with aS_{1} = S_{1}. □

The next step is the prototype of our following arguments, which seems somewhat curious at first glance.

Lemma 5. Besides the case 1 or 2 of the Theorem, if n is divisible by 8 and n > 14^{2}, then J(χ, ψ) is not pure.

Proof. Let

T (a, b) = {x ∈ (Z/nZ)^{×} : x ≡ i (mod n) for i ∈ [a, b) ∩ Z}

and define

T_{i} = T (\frac{(i − 1)n}{4}, \frac{in}{4}) (i = 1, 2, 3, 4).

We identify the element of T_{i} with the integer in [(i − 1)n/4, in/4). Let n = 2^{e}m (e ≥ 3) and m is odd. Consider a vector;

(A, B, C, D) = (\frac{n}{2^{e}} + 2^{i}, \frac{n}{2^{e}} + 2^{i+1}, \frac{n}{2^{e}} + \frac{n}{4} + 2^{i}, \frac{n}{2^{e}} + \frac{n}{4} + 2^{i+1}) for i ≥ 1.

Assume that 2^{i+4} ≤ n, which implies A, B, C and D ∈ S_{1}. As n/2 is even, the condition aS_{1} = S_{1} is equivalent to (n/2 − a)S_{1} = S_{1} and n/2 − a also lies in S_{1}. Thus if J(χ, ψ) is pure, we may assume a ∈ [1, n/4) and aS_{1} = S_{1}. We first treat the case a ∈ [8, n/4). Let i ≥ 1 be an integer such that a ∈ [n/2^{i+2}, n/2^{i+1}). (We can choose i which satisfies 2^{i+4} ≤ n in this case.) Then we have aA ∈ T_{1} and aB ∈ T_{2}. In fact, B − A = 2^{i} and 2^{i}a (mod n) has an representative in [n/4, n/2) ∩ Z. Thus if aA (∈ S_{1}) lies in T_{2} then aB must be in S_{2}. Noting a is odd, we classify in two cases:

1. If a ≡ 1 (mod 4) then

aD ≡ \frac{n}{4} + aB (mod n)

which implies aD ∈ T_{3}.

2. If a ≡ 3 (mod 4) then

aC ≡ \frac{3n}{4} + aA (mod n)

which implies aC ∈ T_{4}.

Therefore when a ∈ [8, n/4), we have shown that, contradicting with the assumption, that aA, aB, aC and aD can not lies in S_{1} simultaneously.

It remains the case a ∈ [1, 7]. The value 1 corresponds to the case 1 or 2 of the Theorem. (Remember that we used the reflection a ↔ n/2 − a). Thus the only remaining case is a = 3, 5 and 7. But this case is easily handled because if aS_{1} = S_{1} then a^{2}S_{1} = S_{1} and n/4 > 7^{2} implies a^{2} ∈ [8, n/4). □

Lemma 6. Besides the case 1 or 2 of the Theorem, if n > 10^{2} and n = 4m and (m, 3) = 1 then J(χ, ψ) is not pure.

Proof. In this case, consider a vector;

(A, B, C) = (\frac{n}{4} + 2^{i}, \frac{n}{4} + 2^{i+1}, \frac{n}{4} + 3 · 2^{i}) for i ≥ 1.

As A, B and C must be in S_{1}, we assume 3 · 2^{i+2} ≤ n. Using the reflection a ↔ n/2 − a, we may choose a ∈ [1, n/4). First we assume that a ∈ [6, n/4) and aS_{1} = S_{1}. Choose the integer i by a ∈ [n/2^{i+2}, n/2^{i+1}). Then by the same argument as Lemma 5, we have aA ∈ T_{1} and aB ∈ T_{2}. Thus we have aC ∈ S_{2}, which is a contradiction. The case a = 3 and 5 is handled similarly. □

In the following, we proceed similarly. In other words, we first choose four elements in S_{1}. Then we show that the a-multiple of these elements can not lies in S_{1} simultaneously. The later arguments become a little bit complicated, especially in Lemma 8.

Lemma 7. Besides the case 1 or 2 of the Theorem, if n > 46^{2} and n = 4m and m is odd and not square free, then J(χ, ψ) is not pure.

Proof. Let q be an odd prime and m is divisible by q^{2}. In this case, we take a vector;

(A, B, C, D) = (\frac{n}{4q} + 2^{i}, \frac{n}{4q} + 2^{i+1}, \frac{n}{4q} + \frac{kn}{2q} + 2^{i}, \frac{n}{4q} + \frac{kn}{2q} + 2^{i+1}) for i ≥ 1.

Here k is a positive integer smaller than 3q/4, which is taken suitably later. Assume that 3 · 2^{i+4} ≤ n. Then we have

\frac{n}{4q} + \frac{kn}{2q} + 2^{i+1} ≤ \frac{n}{12} + \frac{3n}{8} + 2^{i+1} ≤ \frac{n}{2}.

Thus A, B, C and D are contained in S_{1}. We first prove the case a ∈ [24, n/4). Choose k so that

\frac{1}{4} ≤ \{\frac{ak}{2q} \} ≤ \frac{3}{4}  (4)

holds. This is possible. In fact, let l ∈ [q/2, 3q/2) ∩ Z and solve the congruence for x:

ax ≡ l (mod 2q) and x ∈ [1, 2q] ∩ Z.

Define

k(l) = ( \begin{matrix}x & for x ≤ q \\ 2q − x & for x > q \end{matrix}  (5)

Then (4) is satisfied for k = k(l). It is easily shown that the number of distinct k(l) is (q + 1)/2. So k = k(l) can be taken smaller than, say, 3q/4. Noting 3 · 2^{i+4} ≤ n and a ≥ 24, we can choose the integer i by a ∈ [n/2^{i+2}, n/2^{i+1}). Then, similarly to Lemma 5, we can easily check that aA ∈ T_{1} and aB ∈ T_{2}. By (4), we have akn/2q ∈ T_{2} ∪ T_{3}. We classify in two cases. If akn/2q ∈ T_{2} then aD ∈ S_{2}. And if akn/2q ∈ T_{3} then aC ∈ S_{2}. This completes the proof for a ∈ [24, n/4). Finally we treat the remaining case a ≤ 23. If 23 ≥ a ≥ 5, then a^{2}S_{1} = S_{1} and 24 < a^{2} < n/4, we can prove these cases. If a = 3 then consider 27S_{1} = S_{1}. □

Lemma 8. Besides the case 1 or 2 of the Theorem, if n > 70^{2} and n = 12m and (m, 6) = 1 and m has a prime factor greater than 6, then J(χ, ψ) is not pure.

Proof. Let q be the greatest prime factor of q. We assume (n/q, q) = 1, in light of Lemma 7. In this case, we take a vector;

(A, B, C, D) = (\frac{n}{12} ± 2^{i+1}, \frac{n}{12} ± 3 · 2^{i}, A + \frac{kn}{2q}, B + \frac{kn}{2q}) for i ≥ 1.

The sgn ± of A and B are defined by

\frac{n}{12} ± 2^{i+1} ≢ 0 (mod 3) .

Here k is a positive integer smaller than 3q/4, which is chosen later. Assume that 3^{2}·2^{i+3} ≤ n. Then we see that A, B, C and D are in [1, n/2). Moreover we see A, B ∈ S_{1} and C, D are coprime with 6. But, in this case, C, D may be divisible by q. Let l ∈ [q/2, 3q/2) ∩ Z and k(l) be the an integer defined by (5) in Lemma 7. Then there exist (q + 1)/2 choices of k(l). As q ≥ 7, we can take [q/4] + 3 different k(l) values. Then there exist at least three k = k(l) such that both (4) and k ≤ 3q/4 holds. Thus we can choose k = k(l) so that C, D ∈ S_{1}. ( Here we used the fact (n/q, q) = 1. ) Now consider the case a ∈ [36, n/2), and take an integer i by a ∈ [n/2^{i+2}, n/2^{i+1}). Similarly, as in Lemma 7, we see that aA, aB, aC and aD can not lies in S_{1} simultaneously. Finally we consider the case a ≤ 35. If a ∈ [7, 35] then a^{2} ∈ [36, n/4). If a = 5 then 5^{3}S_{1} = S_{1} and 5^{3} ∈ [36, n/4). This completes the proof. □

Proof of the Theorem. By using the proposition, very suitable for numerical calculations, we can easily check the assertion of the Theorem for n ≤ 70^{2}. Combining Lemma 4,5,6 and 7, if n is counter example of Theorem 1, we see n = 12m and (m, 6) = 1 with a square free integer m. Now in scope of Lemma 8, the greatest prime factor of m is 5, which yields the case 4 of the Theorem. This completes the proof.

Acknowledgments.

I am very grateful to Yoshio Mimura and Akihiro Munemasa, who showed me this problem and the computational phenomena, which is now explained for the time being. Thanks are also due to Mieko Yamada who kindly permitted me to include her proof of Lemma 3 in this article. And I would like to express my best gratitude to my colleague, Teruo Takeuchi, whom I discussed and enjoyed with. The essential part of the proof of Lemma 4 is due to him.

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SHIGEKI AKIYAMA

Faculty of Science Niigata University Niigata, 950-21, Japan email: akiyama@geb.ge.niigata-u.ac.jp