On the pure Jacobi sums

Transcription

1 ACTA ARITHMETICA LXXV.2 (1996) O the pure Jacobi sums by Shigeki Akiyama (Niigata) Let p be a odd prime ad F q be the field of q = p 2 elemets. We cosider the Jacobi sum over F q : J(χ, ψ) = x F q χ(x)ψ(1 x), where χ ad ψ are o-trivial characters of F q, whose value at 0 is defied to be 0. It is well kow that the absolute value of J(χ, ψ) is q = p wheever χψ is ot pricipal. Followig [9], [11], call the Jacobi sum J(χ, ψ) pure if J(χ, ψ)/p is a root of uity. Let ord(χ) be the order of χ i F q. We assume that ord(ψ) = 2 ad ord(χ) = 3. This special type of Jacobi sums plays a importat role i evaluatig the argumet of the Gauss sum G(χ) = x F q χ(x)ζ Tr Fq /Fp (x) p, where ζ p is a primitive pth root of uity (see [2], [3]). Moreover, recetly the ratioality of this Jacobi sum is used to characterize the irreducible module of the Terwilliger algebras of cyclotomic associatio schemes (see [10]). I this ote, we prove Theorem. J(χ, ψ) is pure if ad oly if oe of the followig four coditios holds: 1. is a divisor of p + 1, 2. = 2(p 1)/k with a odd iteger k, 3. = 24 ad p 17, 19 (mod 24), 4. = 60 ad p 41, 49 (mod 60). Further, J(χ, ψ) = ±p i all four cases. There are umerous results cocerig the determiatio of Gauss ad Jacobi sums. See e.g. [2], [3], [8]. A ice historical survey is foud i [4]. The same type purity problems for the case of Gauss sums are treated i [6], [97]

2 98 S. Akiyama [13], [1], [7], [9], [11]. But it seems that o such cocrete result o Jacobi sums is kow. Although our essetial tool is the theorem of Stickelberger, the argumet for the oly if part is elemetary ad rather complicated. The author feels somewhat curious that this simple result is derived by such a brute force method. We first see Propositio. Let S 1 = {x (Z/Z) : x i (mod ) for i [1, /2) Z}, ad S 2 = (Z/Z) \ S 1. The J(χ, ψ) is pure if ad oly if there exists x S 1 such that xs 1 = S 1 ad p x (mod ). P r o o f. By the Theorem of Hasse Daveport (see Theorem 5.1 of [12]) ad the well kow result o the sig determiatio of Gauss sums of order 2 for the prime field F p, we get G(ψ) = ( 1) (p+1)/2 p. The theorem of Stickelberger (see Theorem 2.2 of [12]) states that J(χ, ψ)/p = ±G(χ)/G(χψ) is a uit of the iteger rig of Q(ζ ) if ad oly if { a } + { pa } = { a } + { p ( a + 1 )} 2 for all a with (a, ) = 1. Here {x} = x [x] ad [x] is the greatest iteger ot exceedig x. As p is odd, we have to check the coditios for a S 1 oly, because the coditio is symmetric with respect to a a. Thus we choose a with 1 a /2 ad (a, ) = 1. The (1) a + { pa } = a + 1 { pa } 2 for a S 1. Note that the coditio depeds oly o p (mod ). We see that (1) is equivalet to {pa/} [1/2, 1) for a S 1. Put p = y x with itegers x, y. The x (mod ) S 1 ad {xa/} must lie i the iterval [0,1/2). Notig that J(χ, ψ) = p ad Q(ζ )/Q is a abelia extetio, J(χ, ψ)/p is a root of uity uder these coditios. This shows the assertio. R e m a r k. Cosider the Jacobi sum o the geeral fiite field F q with q = p f, for a while. The, similarly to the above proof, we ca easily show that, if the extesio degree f is odd, the there are o χ for which J(χ, ψ)/ q is a root of uity. Our Theorem cocers the first o-trivial case. The sufficiecy of the coditios of the Theorem follows immediately sice: 1 S 1 = S 1 Coditio 1. 0 (mod 4) ad (/2 1)S 1 = S 1 Coditio 2. If = 24 ad S 1 = { 1, 5, 7, 11}, the 5S 1 = S 1 ad 7S 1 = S 1 Coditio 3.

3 Pure Jacobi sums 99 If = 60 ad S 1 = { 1, 7, 11, 13, 17, 19, 23, 29}, the 11S 1 = S 1 ad 19S 1 = S 1 Coditio 4. Here we deote by x the coset x (mod ). Our ext task is to show that J(χ, ψ) is real i the above four cases. The first two cases are hadled easily. Lemma 1. We have J(χ, ψ) = ±p wheever is a divisor of p + 1 or = 2(p 1)/k with a odd iteger k. P r o o f. We have (2) 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). If p 1 (mod ) the J(χ p, ψ) = J(χ, ψ) = J(χ, ψ). This shows the first case. For the secod case, we have p 1 k/2 /2 (mod ) ad χ p 1 = χ /2 = ψ, as = ord(χ). By usig (2) ad 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)) = χ(x)ψ(x 1 (1 x 1 )) = x F q χ(x)ψ(x 1) x F q = J(χ, ψ). This shows the assertio. The case = 24 was already proved i [3]. We show this directly for the coveiece of the reader. Lemma 2. We have J(χ, ψ) = ±p wheever = 24 ad p 17, 19 (mod 24). P r o o f. We have already show that J(χ, ψ)/p is a root of uity of the field Q(ζ 24 ). Let σ k be the elemet of Gal(Q(ζ 24 )/Q) with σ k (ζ 24 ) = ζ k 24. The σ 11 (J(χ, ψ)) = σ 11 (G(χ)G(ψ)/G(χ 13 )) = G(χ 13 )G(ψ)/G(χ) = J(χ, ψ). This shows that J(χ, ψ)/p, the 24-th root of uity, is ivariat uder σ 11, provig the assertio. The first mauscript of this ote did ot cotai the followig proof of Lemma 3, which is essetially due to Mieko Yamada. The author tried i vai to do this. She also iformed me that a more geeral assertio cocerig Lemmas 1 3 is preseted i [10] ad the detailed versio of it will cotai its proof.

4 100 S. Akiyama Lemma 3. We have J(χ, ψ) = ±p wheever = 60 ad p 41, 49 (mod 60). P r o o f. As i Lemma 2, we see σ 29 (J(χ, ψ)) = (J(χ, ψ)). Thus J(χ, ψ) Q( 1) ad J(χ, ψ)/p is equal to ±1 or ± 1. Note that (3) (J(χ, ψ)) 5 J(χ 5, ψ) (mod 5). As ord(χ 5 ) = 12, we see that there exist ratioal itegers C, D with p = C 2 + D 2 ad J(χ 5, ψ) = (C + D 1) 2, by Theorems 4.8 ad 4.10 of [3]. We easily see that CD 0 (mod 5) as p ±1 (mod 5). This shows the assertio. R e m a r k. There is a remaiig problem to determie the sig of J(χ, ψ) whe it is real. See [3] for the first three cases of the Theorem. If = 60 ad p 41, 49 (mod 60), the cogruece relatio (3) ad Theorems 4.8 ad 4.10 of [3] are eough to determie the sig ambiguity of J(χ, ψ). Summig up, we have p, for cases 1 ad 3, J(χ, ψ) = ( 1) (p+1)/2 p, for case 2, ± ( ) p 3 p, for case 4. Here ± of the last case is + (resp. ) whe A 0 (mod 5) (resp. ), with a positive odd iteger A, which is uiquely determied by p = A 2 + B 2. Now we show the ecessity of the coditios of the Theorem. For coveiece, we idetify the elemet of S 1 (resp. S 2 ) with the iteger i [1, /2) (resp. [/2, )) i the later proof. Lemma 4. Besides case 1 of the Theorem, if J(χ, ψ) is pure the = ord(χ) is divisible by 4. P r o o f. Assume is odd ad a 1 is a iteger with as 1 = S 1. Choose a iteger i 1 such that /2 i+1 < a /2 i. The a 1 implies 1 /2 i+1. Thus 2 i S 1 ad 2 i a S 2, which cotradicts as 1 = S 1. Thus by the Propositio, must be eve. Now if = 2m ad m is odd, the similarly we choose a S 1 ad a iteger i. Notig that a is odd, we have ( ) a 2 2i 2 2i a (mod ). We get /2 2 i S 1 ad /2 2 i a S 2, which cotradicts as 1 = S 1. The ext step is a prototype of our followig argumets, which seems somewhat curious at first glace. Lemma 5. Besides cases 1 or 2 of the Theorem, if is divisible by 8 ad > 14 2, the J(χ, ψ) is ot pure.

5 P r o o f. Let Pure Jacobi sums 101 T (a, b) = {x (Z/Z) : x i (mod ) for i [a, b) Z} ad defie ( (i 1) T i = T, i ) (i = 1, 2, 3, 4). 4 4 We idetify each elemet of T i with a iteger i [(i 1)/4, i/4). Let = 2 e m (e 3) with m odd. Cosider a vector ( (A, B, C, D) = 2 e + 2i, 2 e + 2i+1, 2 e i, 2 e + ) 4 + 2i+1 for i 1. Assume that 2 i+4, which implies A, B, C, D S 1. As /2 is eve, the coditio as 1 = S 1 is equivalet to (/2 a)s 1 = S 1 ad /2 a also lies i S 1. Thus if J(χ, ψ) is pure, we may assume a [1, /4) ad as 1 = S 1. We first treat the case a [8, /4). Let i 1 be a iteger such that a [/2 i+2, /2 i+1 ). (We ca choose i which satisfies 2 i+4 i this case.) The we have aa T 1 ad ab T 2. I fact, B A = 2 i ad 2 i a (mod ) has a represetative i [/4, /2) Z. Thus if aa ( S 1 ) lies i T 2 the ab must be i S 2. Notig a is odd, we cosider two cases: 1. If a 1 (mod 4) the which implies ad T If a 3 (mod 4) the which implies ac T 4. ad /4 + ab (mod ), ac 3/4 + aa (mod ), Therefore whe a [8, /4), we have show that, cotrary to the assumptio, aa, ab, ac ad ad caot lie i S 1 simultaeously. There remais the case a [1, 7]. The value 1 correspods to case 1 or 2 of the Theorem. (Remember that we used the reflectio a /2 a.) Thus the oly remaiig case is a = 3, 5 ad 7. But this case is easily hadled because if as 1 = S 1 the a 2 S 1 = S 1 ad /4 > 7 2 implies a 2 [8, /4). Lemma 6. Besides cases 1 or 2 of the Theorem, if > 10 2 ad = 4m ad (m, 3) = 1 the J(χ, ψ) is ot pure. P r o o f. I this case, cosider a vector ( (A, B, C) = 4 + 2i, 4 + 2i+1, ) i for i 1. As A, B ad C must be i S 1, we assume 3 2 i+2. Usig the reflectio a /2 a, we may choose a [1, /4). First we assume that

6 102 S. Akiyama a [6, /4) ad as 1 = S 1. Choose the iteger i with a [/2 i+2, /2 i+1 ). The by the same argumet as i Lemma 5, we have aa T 1 ad ab T 2. Thus ac S 2, which is a cotradictio. The case a = 3 ad 5 is hadled similarly. I the followig, we proceed similarly. I other words, we first choose four elemets i S 1. The we show that the a-multiple of these elemets caot lie i S 1 simultaeously. The later argumets become a little bit complicated, especially i Lemma 8. Lemma 7. Besides cases 1 or 2 of the Theorem, if > 46 2 ad = 4m ad m is odd ad ot square free, the J(χ, ψ) is ot pure. P r o o f. Let q be a odd prime ad m is divisible by q 2. I this case, we take a vector ( ) (A, B, C, D) = 4q +2i, 4q +2i+1, 4q +k 2q +2i, 4q +k 2q +2i+1 for i 1. Here k is a positive iteger smaller tha 3q/4, which is take suitably later. Assume that 3 2 i+4. The 4q + k 2q + 2i i+1 2. Thus A, B, C ad D are cotaied i S 1. We first prove the case a [24, /4). Choose k so that { } 1 ak (4) 4 3 2q 4. This is possible. I fact, let l [q/2, 3q/2) Z ad solve the cogruece for x: Defie ax l (mod 2q) ad x [1, 2q] Z. k(l) = { x for x q, 2q x for x > q. The (4) is satisfied for k = k(l). It is easily show that the umber of distict k(l) is (q + 1)/2. So k = k(l) ca be take smaller tha, say, 3q/4. Notig 3 2 i+4 ad a 24, we ca choose a iteger i satisfyig a [/2 i+2, /2 i+1 ). The, similarly to Lemma 5, we ca easily check that aa T 1 ad ab T 2. By (4), we have ak/(2q) T 2 T 3. We cosider two cases. If ak/(2q) T 2 the ad S 2. Ad if ak/(2q) T 3 the ac S 2. This completes the proof for a [24, /4). Fially, we treat the remaiig case a 23. If 23 a 5, the a 2 S 1 = S 1 ad 24 < a 2 < /4; these cases ca be proved similarly. If a = 3 the cosider 27S 1 = S 1.

7 Pure Jacobi sums 103 Lemma 8. Besides cases 1 or 2 of the Theorem, if > 70 2 ad = 12m ad (m, 6) = 1 ad m has a prime factor greater tha 6, the J(χ, ψ) is ot pure. P r o o f. Let q be the greatest prime factor of q. We assume (/q, q) = 1, i light of Lemma 7. I this case, we take a vector ( (A, B, C, D) = 12 ± 2i+1, 12 ± 3 2i, A + k 2q, B + k ) for i 1. 2q The sigs i A ad B are defied by /12 ± 2 i+1 0 (mod 3). Here k is a positive iteger smaller tha 3q/4, which is chose later. Assume that i+3. The we see that A, B, C ad D are i [1, /2). Moreover, A, B S 1 ad C, D are coprime with 6. But, i this case, C, D may be divisible by q. Let l [q/2, 3q/2) Z ad k(l) be the iteger defied by (5) i Lemma 7. The there exist (q + 1)/2 choices of k(l). As q 7, we ca take [q/4] + 3 differet k(l) values. The there exist at least three k = k(l) such that both (4) ad k 3q/4 hold. Thus we ca choose k = k(l) so that C, D S 1. (Here we used the fact (/q, q) = 1.) Now cosider the case a [36, /4), ad take the iteger i with a [/2 i+2, /2 i+1 ). Similarly to Lemma 7, we see that aa, ab, ac ad ad caot lie i S 1 simultaeously. Fially, we cosider the case a 35. If a [7, 35] the a 2 [36, /4). If a = 5 the 5 3 S 1 = S 1 ad 5 3 [36, /4). This completes the proof. P r o o f o f t h e T h e o r e m. By usig the Propositio, very suitable for umerical calculatios, we ca easily check the assertio of the Theorem for Combiig Lemmas 4 7, if is a couterexample, we see that = 12m ad (m, 6) = 1 with a square free iteger m. Now i view of Lemma 8, the greatest prime factor of m is 5, which yields case 4 of the Theorem. This completes the proof. Ackowledgmets. I am very grateful to Yoshio Mimura ad Akihiro Muemasa who showed me this problem ad the relevat computatioal pheomea. Thaks are also due to Mieko Yamada who kidly permitted me to iclude her proof of Lemma 3 i this article. I would also like to express my best gratitude to my colleague, Teruo Takeuchi, with whom I discussed the subject. The essetial part of the proof of Lemma 4 is due to him. Refereces [1] L. D. Baumert, W. H. Mills ad R. L. Ward, Uiform cyclotomy, J. Number Theory 14 (1982), [2] B. C. B e r d t ad R. J. E v a s, Sums of Gauss, Jacobi, Jacobsthal, ibid. 11 (1979),

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