ON THE p-adic VALUE OF JACOBI SUMS OVER F p

Transcription

1 Kyushu J. Math. 68 (014), 3 38 doi:10.06/kyushujm.68.3 ON THE p-adic VALUE OF JACOBI SUMS OVER F p 3 Takahiro NAKAGAWA (Received 9 November 01 and revised 3 April 014) Abstract. Let p be a prime and q = p s and ζ k a fixed primitive kth root of unity in some extension of Q. Let χ be a multiplicative character over F q of order k and J(χ,χ) the associated Jacobi sum. We give examples of χ which satisfy J(χ,χ) p [s/] Z[ζ k ]. Moreover, for s = 3, we prove that there is only a finite number of k such that J(χ,χ) is an element of pz[ζ k ] except for the case where k is divisible by nine and p 1 ± k/3 (mod k). 1. Introduction Let p be a prime and q = p s and ζ k a fixed primitive kth root of unity in some extension of Q. Let χ be a multiplicative character over F q of order k. We consider the Gauss sum G(χ) and Jacobi sum J(χ,χ). The Gauss sum or Jacobi sum is pure if some non-zero integral power of it is real. It is known that G(χ) or J(χ,χ) is pure if and only if G(χ) or J(χ,χ) is equal to ζ q for some root of unity ζ. The purity of Gauss sum or Jacobi sum has been studied for a long time. Stickelberger showed that if 1 isapowerofp(mod k), then the Gauss sum G(χ) is pure. Moreover, Evans showed that the Jacobi sum J(χ,χ) is also pure (see [6, Theorem 1]) in this case. Aoki proved that there is only a finite number of k such that G(χ) is pure and 1 is not a power of p(mod k) (see [, 3]). Akiyama, Shiratani and Yamada found the condition that J(χ,χ)is pure when s =, and Aoki generalized this result for generalized Jacobi sums [1, 3, 9]. In general, it is known that if J(χ,χ) is pure, then s must be even. In this paper, we consider the condition such that J(χ,χ) is an element of p [s/] Z[ζ k ]. In particular, if s is even and J(χ,χ) has this property, then J(χ,χ)is pure. For s = 3, we prove that there is a finite number of k such that J(χ,χ) is an element of pz[ζ k ] except for the case that k is divisible by nine and p 1 ± k/3 (mod k). This result is similar to the case s =. See Theorems 4.1 and 4... Notation Throughout this paper we use the following notation and conventions. Let p be an odd prime and F q a finite field with q = p s elements. For any positive integer n, ζ n will denote a fixed primitive nth root of unity in some extension of Q. Let ψ : F p Q(ζ p ) be the additive character of F p defined by ψ(t) = ζ t p, and let ψ s : F q Q(ζ p ) denote the additive character of F q defined by ψ s (t) = ψ(tr(t)), where Tr : F q F p is the trace map. For any multiplicative character χ of F q, the Gauss sum G(χ) over F q associated with the characters 010 Mathematics Subject Classification: Primary 11T4; Secondary 11S80. Keywords: Jacobi sums; pure Gauss sums. c 014 Faculty of Mathematics, Kyushu University

2 4 T. Nakagawa ψ s and χ is defined by G(χ) = ψ s (t)χ(t). t F q Let χ,φ be multiplicative characters of F q. The Jacobi sum J (χ, φ) of F q associated with the characters φ and χ is defined by J (χ, φ) = χ(t)φ(1 t). If χ = φ 1, then we have 1=t F q J (χ, φ) = G(χ)G(φ) G(χ, φ). (.1) Let [a, b]={x Z a x b} for integers a, b. For any real number x, [x] denotes the largest integer not greater than x. Let m be a positive integer. For any integers a, b, let [a, b] m ={z [a, b] z is prime to m}. For x Q whose denominator is prime to m, we define R m so that R m (x) [0,m 1] and R m (x) x(mod m), and L m so that L m (x) = [R m (x)/m] {0, 1}. LEMMA.1. Let x, y be real numbers and a, b positive integers which are prime to each other. (1) For any integer n, () For any integer m, we have [x + y] 1 [x]+[y] [x + y], (.) [x + n]=[x]+n. (.3) m Ra (m) R ab (m) = R a (m) + ar b a R a (m) = m a, (.4) m, (.5) a R k (m) = R k (m) kl k (m), (.6) R k (m) = R k (m) + kl k (m). (.7) Moreover R ab (m) < ab/ if and only if R b ((m R a (m))/a) < [(b 1)/] or (R b ((m R a (m))/a) =[(b 1)/] and R a (m) < a/). (3) For all integers m, n, ma + nb is prime to ab if and only if m is prime to b and n is prime to a. Proof. We prove only (.4). Since R ab (m) R a (m) (mod a), we can write R ab (m) R a (m) = na for some integer n. We have directly m Ra (m) n R b (mod b), a 0 n b 1. Therefore, we have n = R b (m R a (m)/a), which implies (.4). Note that for any positive integers a, m, R a (m)/a is the rational number such that 0 R a (m)/a < 1 and m/a R a (m)/a Z.

3 On the p-adic value of Jacobi sums over F p Pure Gauss and Jacobi sums In this section, we review and study the purity of Gauss sums and Jacobi sums. The references for this section are [] and [5]. A Gauss sum or Jacobi sum is called pure if some non-zero, integral power of it is real. Using Stickelberger s theorem [4, Theorem 11..3], we have the following theorem. See also [3] for general Jacobi sums. THEOREM 3.1. Let χ be a multiplicative character over F q of order k =. The Jacobi sum J(χ,χ) p l Z[ζ k ] if and only if s 1 L k (jp i ) l (3.1) for all integer j prime to k. Proof. Note that for all integers j, R k (j/p)p R k (j) k j=0 + R k(j/p )p R k (j/p) k p + + R k(j)p R k (j/p s 1 ) p s 1 k (3.) is the p-adic expansion of R k (j)(p s 1)/k. Indeed R k (j)(p s 1)/k is equal to (3.) and (R k (j/p i+1 )p R k (j/p i ))/k [0, 1,...,p 1] for all integers i. By [4, Theorem 11..3], we have J(χ,χ) p l Z[ζ k ] if and only if, s 1 R k (j/p i s 1 ) R k (j/p i ) kl (for all j [0,k] k ). (3.3) By Lemma.1, the left-hand side of the inequality (3.3) is equal to k s 1 L k(jp i ). Therefore, we have (3.1). Using the result of Stickelberger and Evans, we have the following. THEOREM 3.. (Aoki [3, Proposition.3] and Berndt et al. [4, Theorem 11..3]) Let χ be a multiplicative character over F q of order k. The Gauss sum G(χ) p l Z[ζ p,ζ k ] if and only if s 1 R k (jp i ) lk for all j [0,k] k. (3.4) In particular, the Gauss sum G(χ) is pure if and only if s 1 R k (jp i ) = sk for all j [0,k] k. (3.5) COROLLARY 3.3. Let χ be a multiplicative character over F q of order k =. If the Gauss sum G(χ) is pure, then s 1 L k (jp i )>0 for all j [0,k] k. (3.6) Moreover, if k (mod 4), then J(χ,χ) is pure, that is s 1 L k (jp i ) = s for all j [0,k] k. (3.7) If k (mod 4), then G(χ) is pure if and only if G(χ )/G(χ 4 ) is a root of unity.

4 6 T. Nakagawa Proof. We assume that G(χ) is pure. If i L k(jp i ) = 0 for some j, then s 1 R k(jp i )< sk/. This is a contradiction. Assume k 0 (mod 4). Note that J(χ,χ) p s/ Z[ζ k ] if and only if J(χ,χ) is pure. If j is prime to k, then j + k/ is also prime to k. Therefore, we have s 1 sk = R k ((j + k/)p i s 1 ) = R k (jp i + k/) s 1 = (R k (jp i ) + k/ kl k (jp i )) = sk + sk s 1 kl k (jp i ). Assume that k is odd. If j is prime to k, then j is prime to k. s 1 sk = R k (jp i s 1 ) = (R k (jp i ) kl k (jp i )) = sk s 1 kl k (jp i ). By above equalities, we have (3.7). On the other hand, we assume k (mod 4). If j is prime to k, then j + k/ is prime to k. s 1 G(χ) is pure R k (jp i ) = sk s 1 R k j + k p i = sk for all j [0,k] k for all j [0,k] k s 1 (R k (jp i ) kl k (jp i )) = 0 for all j [0,k] k s 1 (R k (jp i ) R k (4jp i )) = 0 for all j [0,k] k G(χ ) G(χ 4 is a root of unity. ) If k is odd and G(χ) is pure, then the fact that J(χ,χ) is pure is seen easily without the above calculation. LEMMA 3.4. Let χ be a non-trivial multiplicative character over F q of order k =. If k (mod 4), then J(χ,χ) p l Z[ζ k ] if and only if J (χ,χ ) p l Z[ζ k ]. In particular, J(χ,χ) is pure if and only if J (χ,χ ) is pure.

5 On the p-adic value of Jacobi sums over F p 3 7 Proof. Let k = m. First we prove R k (j) < k/ if and only if R m ( j/) < m/. R k (j) < k R k(j) = R m (j) j R m (j) is odd R m ( j) is even R m s 1 < m. Therefore, we see L k (j) = L m ( j/). This implies that L k(jp i ) = s 1 L m ( jp i /) for all j. Since the map j R m ( j/) is an injection from the set [0,k] k to the set [0,m] m, this map is a bijection. We have s 1 min j [0,k] k L k (jp i ) = min j [0,m] m s 1 L m (jp i ). COROLLARY 3.5. Let χ be a non-trivial multiplicative character over F q of order k =. Assume G(χ) is pure. Then: (1) J(χ,χ) pz[ζ k ]; () if k is divisible by four or k is odd, then J(χ,χ) and G(χ ) are pure; (3) if k (mod 4), then J (χ, χ)j (χ,χ ) is pure and G(χ ) pz[ζ k ]. Proof. We use Corollary 3.3. (1) This follows from (3.7). () Since J(χ,χ) = G(χ) /G(χ ) and J(χ,χ) is pure, G(χ ) is also pure. (3) Since J (χ, χ)j (χ,χ ) = G(χ) G(χ )/G(χ 4 ), J (χ, χ)j (χ,χ ) is pure. We see G(χ ) = J (χ,χ )G(χ 4 )/ G(χ ) pz[ζ k,ζ p ] since J (χ,χ ) pz[ζ k ] by Lemma 3.4. By [6, Lemma 6], we have G(χ)/q 1/ Z[ζ k ]. Therefore, G(χ ) = G(χ) /J (χ, χ) Z[ζ k ]. Conversely we have the following corollary. COROLLARY 3.6. If J(χ,χ) p l Z[ζ k ], then G(χ) p l/ Z[ζ k,ζ p ]. Proof. By Lemma.1, we have s 1 R k (jp i s 1 ) = R k/ (jp i ) + k s 1 L k (jp i ) (3.8) for all j [0,k] k. The right-hand side of the above equation is greater than or equal to kl/ for all j [0,k] k by assumption. Therefore, we see G(χ) p l/ Z[ζ k,ζ p ]. We will give an example of k and p such that J(χ,χ) p [s/] Z[ζ k ] and G(χ) is not pure. PROPOSITION 3.7. Let χ be a non-trivial multiplicative character over F q of order k =. Let d be a positive integer such that d s, d k and (k/d) 0 (mod k). If p 1 + mk/d (mod k) for some m [0,d] d, then the Jacobi sum J(χ,χ) is an element of p l Z[ζ k ] with l =[d/](s/d). In particular, if d is even, thenj(χ,χ) is pure. Proof. In the case d = 1, we obtain [d/]=0. Hence, we may assume d>1. Since R k (jp i ) = R k (j + jimk/d) and m is prime to k, we have {R k (jp i ) 0 i s 1}= R k j + jmik 0 i s 1 d ={R k (j + ik/d) 0 i d 1}

6 8 T. Nakagawa for all j [0,k] k. Therefore, we have i [0,s 1] R k (jp i )< k = s d i [0,d 1] R k (j + ik/d) < k. On the other hand, the inequality {i [0,d 1] R k (j + ik/d) < k/} [d/] is seen for all j directly. 4. The Jacobi sum over F p 3 Akiyama, Shiratani and Yamada proved the following. THEOREM 4.1. (Akiyama [1] and Shiratani and Yamada [9]) Let χ be a non-trivial multiplicative character over F p of order k. The Jacobi sum J(χ,χ) is pure if and only if one of the conditions holds: (1) p 1 (mod k); () k is divisible by four and p 1 + k/ (mod k); (3) (k, R k (p)) ={(4, 17), (4, 19), (60, 41), (60, 49)}. In this section, we prove the following. THEOREM 4.. Let χ be a non-trivial multiplicative character over F p 3 of order k. The Jacobi sum J(χ,χ) pz[ζ k ] if and only if one of the conditions holds: (1) k is divisible by 9 and p 1 + k/3 (mod k) or p 1 k/3 (mod k); () (k, R k (p)) ={(7, ), (7, 4), (14, 9), (14, 11), (1, 4), (1, 16), (8, 9), (8, 5), (39, 16), (39, ), (4, 5), (4, 37), (5, 9), (5, 9), (76, 45), (76, 49), (78, 55), (78, 61), (11, 65), (11, 81), (140, 81), (140, 11), (168, 5), (168, 11)}. By Theorem 3.1, the Jacobi sum J(χ,χ) is an element of pz[ζ k ] if and only if L k (jp i ) 1 for all j [0,k] k. This is equivalent to there being an integer j such that R k (jp i ) k/ for any i [0,k] k. Definition 4.3. We say (k, R k (p)) satisfies condition (L) if there exists a j [0,k] k such that R k (jp i )<k/ for all i. We search for (k, R k (p)) which satisfies condition (L). Let a,b,tbe the unique positive integers such that k = 3 t ab, gcd(3, a) = gcd(3, b) = 1 and for any prime factor f of k, f a if and only if p 1 (mod f). We use the following lemma. LEMMA 4.4. We have: (1) p 1 (mod a) and p + p (mod b); () any prime factor of b divides three has a remainder of one; (3) if t 1, then p 1 (mod 3) and if t, then p 1 (mod 3 t 1 ). Proof. (1) We have p (mod b) and p 1 is invertible modulo b. () The assertion is a basic fact of finite field. (3) If t 1, then p p 3 1 (mod 3). If t, then we can write p = 3m + 1 for some m. Since p + p + 1 = 9m + 9m + 3 and p 3 1 (mod 3 t ), p 1 (mod 3 t 1 ).

7 On the p-adic value of Jacobi sums over F p The case b = 1 We assume b = 1. If p 1 (mod k), then R k (p i ) = 1 for all i. Therefore, (k, R k (p)) satisfies condition (L) by Theorem 3.1. If p 1 (mod k), then9 k and there exists a δ Z such that p 1 + ( 1) δ k/3 (mod k) by Lemma 4.4. By Proposition 3.7, (k, R k (p)) does not satisfy condition (L). We may assume b = The case a = 1 and t 1 We assume a = 1 and t 1. We will prove that there exists an integer j such that R k (jp i )<k/ for all i if k is not equal to 7, 1, 39. By Lemma 4.4, we have k is odd and jp + jp + j 0 (mod k) for all integer j. We remark that R k (jp i ) = k or k. In particular R k (p i ) = k. Let k = (k 1)/. For j [1,k ], the set A j is defined by {u [1,k ] u jp i (mod k) for some i}. (4.1) Note that A j ={R k ( j/p), R k ( j/p )} [1,k ]. LEMMA 4.5. (i) For j [0,k ], A j =0 if and only if R k (jp i ) k for all i. In particular, A j =0 for some j [1,k ] k if and only if (k, R k (p)) satisfies condition (L). (ii) If A v ={j, m} for j = m, then A j = A m ={v} and j + m = v. Conversely if A j = {v}, then j<v, A v ={j, v j} and A v j ={v}. Proof. (i) This equivalence is obvious by definition. (ii) We assume A v ={j, m} for j = m. We have j vp l (mod k) and m vp l (mod k) for some l {±1}. Hence, j + m (p l + p l )v v(mod k). This implies j + m = v because j + m, v [0,k 1]. Since v A j j A v,weseev A j and v A m. If v = e A j, then v<v+ e = j = v m<v, which implies A j = A m ={v}. Suppose A j ={v}. By definition, there exists an integer l {±1} such that v jp l (mod k) and R k ( jp l )>k. Since v + R k ( jp l ) j (p l + p l ) j (mod k), we have v + R k ( vp l ) k = j. Hence, j<v. Since we see v j v(1 + p l ) vp l (mod k), A v ={j, v j} (we can see j = v j easily). By the previous argument, we have A v j ={v}. We set x = Min(R k (p), R k (p )) and y = k 1 x. We remark that y = max(r k (p), R k (p )). If R k ((b 1)p i /) = k, then R k ((b 1)p i /) k for all i. Hence, we may assume R k ((b 1)p i /) = k. Using a computer, we have the following fact. LEMMA 4.6. Suppose k < 905. The pair (k, R k (p)) does not satisfy the condition (L). if and only if k = 7, 1, or 39. We may suppose k 905. Then we have the next lemma. LEMMA 4.7. If j [0,...,47], then A k j = 1. Proof. Assume A k j =1. By Lemma 4.5(ii), there exists i [0,j 1] so that k j A k i. Therefore, (k j)/(k i) p l (mod k) for some l {±1}. Since p + p (mod k), we have (j + 1) (j + 1)(i + 1) + (i + 1) 0 (mod k). In particular, k (j + 1) (j + 1)(i + 1) + (i + 1) <(i + 1) 95 = 905. This is a contradiction.

8 30 T. Nakagawa To prove A j =0 for some j [0,k ] k, we suppose A j = for all j [0,k ] k. LEMMA 4.8. Let i be the positive integer such that If i 3, then A k i m =0 for m = 1,...,i. Proof. By (4.), we have inductively A k j = (0 j < i), (4.) j + 1 R k j + 1 R k y A k i <. (4.3) x = j + 1 x k, (4.4) = j + 1 y jk k (4.5) for j = 0,...,i 1. (Indeed we see that if R k (((j 1)/)y) = ((j 1)/)y (j 1)k k, then k < ((j + 1)/)y jk k and if R k (((j 1)/)x) = ((j 1)/)x k, then 0 ((j + 1)/)x k 1). By (4.3), k <R k ((i + 1)x/) or k <R k ((i + 1)y/). To prove k <R k ((i + 1)y/), we suppose R k ((i + 1)y/) k. By this inequality and (4.5), we have R k ((i + 1)y/) = (i + 1)y/ ik 0. Hence, one obtains i + 1 x = i + 1 y + i + 1 (k 1) ik + i + 1 (k 1) k. We see A k i =0. This is a contradiction. Thus, we see k <R k ((i + 1)y/) and R k ((i + 1)y/) = (i + 1)y/ (i 1)k. Finally we shall prove A k i m =0 for m = 1,...,i. For m = 1,...,i, we have (i + m) + 1 (i + m) + 1 y (i + m 1)k = i + 1 y ik + my (m 1)k (i 1)k + k 1 ik + my (m 1)k = m(y k) + k 1 <k 1, y (i + m 1)k = i 1 y (i 1)k + m(y k) + y i 1 y (i 1)k + (i )(y k) + y i 1 = y (i 1)k + k y/ k y/ >k. Hence, k <R k (((i + m) + 1)y/) = ((i + m) + 1)y/ (i + m 1)k. Since x + x (mod k), k 1 x + x. By (4.4), (i 1)x k 1 x + x. Thus, i

9 (x + )/ x. Using this inequality, we have On the p-adic value of Jacobi sums over F p 3 31 (i + m) + 1 x (i 3/)x (i 1)x k 1, (i + m) + 1 x i + 1 x + x = i + 1 (k 1) i + 1 y + x i + 1 (k 1) (i 1)k (k 1) + x = i + x + (k + 1)/ >k, which implies R k (((i + m) + 1)/) = ((i + m) + 1)/ >k for m = 1,...,i. Therefore we see A k i m =0 for m = 1,...,i. LEMMA 4.9. We have A k 1 = and A k 3 =. Proof. By Lemma 4.7, we have A k 1 =0or. We assume A k 1 =0. Note the remainder of any prime factor of b divided by three is one. Hence, we have A k = and A k =. Since A k 1 =0 and A k =, we have 0 5x/ k k, which implies k/5 x. Let i be the positive integer which satisfies A k 3i < and A k 3j = for all 0 j i 1. Inductively we see 0 6j + 5 x (j + 1)k k (4.6) for j = 0,...,i 1. Indeed, assume that the inequality (4.6) is true for j 1. We have 6j + 5 x (j + 1)k = 6j 1 x jk + 3x k k + 3(k 1) k<k 1, (4.7) 6j 1 x jk + 3x k 0 + k > 0. (4.8) 5 Since A k j =, we have (4.6). In particular, we have 6i 1 x ik k, 6k 5 k i (3x k)i k + x < 3k 4. Hence, we have i 3. Since k 3i is prime to k for i = 1,, 3, A k 3i < and + 3i 47, this is a contradiction. Therefore, we have A k 1 =, which means 5x/ k. To prove A k 3 =, we assume A k 3 =0. By Lemma 4.8, A k 4 =0. Since k 5 is prime to k, wesee A k 5 =. Hence, we have 0 11x/ k k. By these inequalities, we have k 11 x k 5. Let i be the positive integer which satisfies A k 5 6i < and A k 5 6j = for all 0 j i 1. Inductively, we see 0 1j + 11 x (j + 1)k k

10 3 T. Nakagawa for all j = 0,...,i 1. In particular, substituting the case j = i 1, we have (6x k)i k + x/. By the inequality ki 1k 11 = 11 k i (6x k)i k + x < 3k 5, we see i 6. Since 5 + 6i 47, A k 5 6i =0 by Lemma 4.7. Therefore, we calculate 1i + 3 x (i + )k = 1i + 3 1i + 11 x (i + 1)k + 6x k>k + 1k 11 k>k, 6(k 1) <k. x (i + )k = 1i 1 x ik + (6x k) k + 5 k Therefore, we see A k 11 6i =0byLemma 4.7. Since for i = 1,...,6, we see k 5 6i or k 11 6i is prime to k, this is a contradiction. We use the following lemma. LEMMA Let i be a positive integer with i 61. Then there exists an integer n such that i + 3 n 4i 3 and n = 5 l 111 l 17 l 33 l 4 for some l 1,l,l 3,l 4 N {0}. Proof. We use induction on i. Since = 5 3, the case i = 61 is true. We assume that the assertion is true for i = k. If k + 3 {5 l 111 l 17 l 33 l 4 l 1,l,l 3,l 4 N {0}}, then there exists an integer n such that k + 5 n 4k 3 and n = 5 l 111 l 17 l 33 l 4 for some l 1,l,l 3,l 4 N {0}. Therefore, the assertion is true for i = k + 1. We suppose that k + 3 = 5 l 111 l 17 l 33 l 4 for some l 1,l,l 3,l 4 N {0}. If l > 0, then we can choose n = 5 l 111 l 1 17 l3+1 3 l 4. If l 3 > 0, then we can choose n = 5 l 111 l 17 l3 1 3 l4+1. If l 4 > 0, then we can choose n = 5 l1+ 11 l 17 l 33 l4 1. If l 1 >, then we can choose n = 5 l l+1 17 l3+1 3 l 4. Let i be the positive integer which satisfies A k i < and A k j = for all 0 j i 1. By Lemma 4.9, i 4. Since k 5 is prime to k, A k 5 = by Lemma 4.7. We assume A k 6 <. By Lemma 4.8, we see A k 8 =0. We have a contradiction since k 8 is prime to k. Therefore, we have i 5. Similarly we see i 61. By Lemma 4.10, there exists a positive integer such that i + 3 n 4i 3 and n = 5 l 111 l 17 l 33 l 4 for some l 1,l,l 3,l 4 N {0}. Then A k (n 1)/ =0 and k (n 1)/ is prime to k. This is a contradiction. Therefore, we see there exists an integer j such that R k (jp i )<k/ for all k > 905 and i The case k (mod 4) To prove this case, we have only to prove the case where k is odd by Lemma 3.4. We may assume k (mod 4). In particular, we assume a The case p 1 (mod 3 t ) and t We assume t and p 1 (mod 3 t ) which means p 1 + ( 1) δ a3 t 1 (mod 3 t a) for some δ {0, 1}. We set x = Min(R b (p), R b (p )), y = b 1 x and a = 3 t a. We remark y = Max(R b (p), R b (p )). Since 1 + p + p 0 (mod b), 1+ R b (p) + R b (p ) = b. By x = y,wehavex < (b 1)/ and y > (b 1)/. We will prove the next proposition.

11 On the p-adic value of Jacobi sums over F p 3 33 PROPOSITION Let n be the integer such that n 1 x [b/6] < n x. For j = ( 1) δ max(0,n) a + (a /3 1)b, R k (jp i )<k/ for any i. Let j = la + (a /3 1)b. We calculate a R a (jp i ) = R a b 3 1 ib( 1) δ a 3 a = b R 3 (i( 1) δ+1 ) a 3 a A such that A Z satisfies 0 (a /3 1)b + R 3 (i( 1) δ+1 )a /3 a A a 1. We easily see (a /3 1)b + R 3 (i( 1) δ+1 )a /3 A =. The equality R k (jp i ) = R a (jp i ) + a R b jp i R a (jp i ) a = R a (jp i ) + a R b lp i + A + R 3 (i( 1) δ+1 ) b 1 3 a 1 = R a (jp i ) + a R b lp i a + R 3(i( 1) δ+1 ) 3 follows from Lemma.1(). To study R k (jp i ), we research 1 L(i) := R b lp i a + R 3(i( 1) δ+1 ) b. 3 By Lemma.1 (), if L(i) < (b 1)/, then we see R k (jp i )<k/. b The case δ = 0 and x>[b/6]. We assume δ = 0 and x>[b/6]. Note [b/6]= (b 1)/6. Since n = 0, we have l = 1. By direct calculation, L(1) = R b p + 1 1a b = R b p + ba, L() = R b p a b. Using the inequality a 9, we see x + ba b 1 1 = b 3, x + ba b > + b By L(1) x + [ b/a ] (mod b) and the above inequalities, we have L(1)<(b 1)/. x b b a b < 1 + = b 1 1, 6 3 x a b 1 b 3 5b b 18 > 1.

12 34 T. Nakagawa Similarly we have L() < (b 1)/. On the other hand, j < (b/3 + 1)a ba / = k/. Therefore, R k (jp i )<k/ for all i The case δ = 0 and x [b/6]. We assume δ = 0 and x [b/6]. Let n be the integer which satisfies n 1 x [b/6] < n x. Choose l = n. Since x, we see l [b/6]: lx + ba b 1 = b < b 3, lx + ba > + b 6 b 1. 9 Since L(1) lx + [ b/a ] (mod b) and the above inequalities, we obtain L(1)< (b 1)/: lx l b b a b < l + = b 1 l, 6 3 lx l b 5b b 5b a b l b 18 > 1. Similarly we see L()<(b 1)/. On the other hand, j<([b/6]+b/3)a a b/ = k/. Therefore, we see R k (jp i )<k/ for all i The case δ = 1 and x>[b/6]. We assume δ = 1 and x>[b/6]. Since n = 0, we have l = 1. By a direct calculation, L(1) = R b p a b, L() = R b p + 1 1a b = R b p + ba. We have x a x a b b < + 6 b b 3 + b 3 5b 9 = b 1, 0. By L(1) x + [ b/a ] (mod b) and the above inequalities, we see L(1)<(b 1)/: 1 + x + ba 1 + b 3 1 = b 3, 1 + x + ba > b 6 b 0. 9 Similarly, we have L()<(b 1)/. Since we can easily see j 0, we have R k (jp i )<k/ for all i.

13 On the p-adic value of Jacobi sums over F p The case δ = 1 and x [b/6]. We assume δ = 1 and x [b/6]. Let n be the integer which satisfies n 1 x [b/6] < n x. We have l = n. Then n x b b a b < + = b 1 6 3, n x b 5b a b Since L(1) n x + n + [ b/a ] (mod b) and the above inequalities, we see L(1)< (b 1)/: n + n x + ba b 3 1 = b 3 6, n + n x + ba b > n + + b Similarly, we have L()<(b 1)/. Since we can easily see j 0, we have R k (jp i )<k/ for all i The case p 1 (mod 3 t ) Suppose p 1 (mod 3 t ). We set x = Min(R b (p), R b (p )), y = b 1 x and a = 3 t a. Let j = αa + βb. We have βb R k (jp i ) = R a (j) + a R b αp i + a. (4.9) We search for integers α and β such that R b (αp i +[βb/a ])<(b 1)/ for all i The case b = 7 and a > 7. We assume b = 7 and a > 7. Let β be an integer such that a b <β a b. (4.10) Choose α = 1 and β = β. By the above inequality, we have 0 bβ + a <a / which implies [bβ/a ]= 1. We see j = a + βb 0 and R k (jp i ) = R a (j) + a R b (p i 1). Since R a (j) = R a (bβ) < a / and R b (p) = or 4, R k (jp i )<k/ for all i by Lemma.1. Therefore, we have only to show that there exists β which is prime to a and satisfies (4.10). If a 350, then we see there exists j [0,k] k such that R k (jp i )<k/ for all i by direct calculation. Therefore, we assume a > 350. Note a (mod 4) by Section 4.3. If a is odd, then we can choose β = n such that a /(b) < n a /b. We assume a 0 (mod 8). Then there exists a positive integer n such that a /(b) < n + a / f a /b where f is a positive integer with f a and f +1 a. Thus, we can choose β = n + a / f. We assume a 4 (mod 8). To prove this case, we have the following lemma. LEMMA 4.1. Let m be a positive integer. If m 350, then there exist non-negative integers i, j such that 3m/56 i 7 j < 5m/56.

14 36 T. Nakagawa Proof. We use inductionon m. The case when m = 350 is true. We assume that the assertion is true for m = l 1. Then there exist i, j such that 3(l 1)/56 i 7 j < 5(l 1)/56. We may assume 3(l 1)/56 i 7 j < 3l/56. If j>0, then i+3 7 j 1 satisfies 3l/56 i+3 7 j 1 < 5l/56. If j = 0, then i 5 follows from 3l/56 > 16. Then we have 3l/56 i 5 7 j+ < 5l/56. By the above lemma, there exist integers γ 1,γ such that 3a 56 γ 1 7 γ < 5a 56. Then β = a /4 γ γ satisfies (4.10) and β is prime to a The case a >b and b = 7. We assume a >b>7. By the case a = 1, there exists an integer α such that 0 <R b (αp i ) (b 1)/ for all i. Choose β = 1. We have R k (jp i ) = R a (j) + a R b (αp i 1)<k/ for all i The case b>a, b 1800 and 4 a 7 We assume b Since b>a, we can calculate directly. As a result, (k, R k (p)) does not satisfy the condition (L) if and only if (k, R k (p)) ={(7, ), (7, 4), (14, 9), (14, 11), (1, 4), (1, 16), (8, 9), (8, 5), (39, 16), (39, ), (4, 37), (4, 5), (5, 9), (5, 9), (76, 45), (76, 49), (78, 55), (78, 61), (11,65), (11, 81), (140, 81), (140, 11), (168, 5), (168, 11) } The case b>a, 1800 <band 4 a 7 We assume b>a and 4 a 7. Let j = αa + βb. We search for α and β which satisfy R k (jp i )<k/ for all i. Set x = Min(R b (p), R b (p )), y = b 1 x. We use the following lemma. LEMMA Let m [0,b] b and ι be integers such that m<r b (mp ι )<b/. If m + R b (mp ι ) [b/a ], then we have R k (jp i )<k/ for j = ma + b and all i. Proof. The equalities b R b (mp ι ) + a b R b (mp ι ) + a b R b (mp ι ) + a b a m< b 1, = b R b (mp ι ) m + <b+ b 1 < 3 b 1 4 b a b, give R k (jp i )<k/ for all i by Lemma.1. By Lemma 4.13, if 1 + x [b/a ], then we have R k (jp i )<k/for j = a + b and all i. Therefore, we may assume 1 + x>[b/a ]. By this inequality, it follows that b x> a 1 100, b a (1 + x) a = + a = a 7. x x x

15 On the p-adic value of Jacobi sums over F p 3 37 Let m 1 be the least positive integer which satisfies m 1 x b x/. We see m 1 = [b/x] or [b/x]+1. In particular, m 1 8. We denote min(r b (m 1 p), R b (m 1 p )) and max(r b (m 1 p), R b (m 1 p )) by x 1 and y 1, respectively. Since m 1 < (b 1)/ and x 1 x/ < (b 1)/4, x 1 + y 1 + m 1 = b. If m 1 + x 1 [b/a ], then R k (jp i )<k/ for j = min(m 1,x 1 )a + b and all i by Lemma Therefore, we may assume m 1 + x 1 > [b/a ]. By this inequality, we see x and [b/x 1 ] [a (m 1 + x 1 )/x 1 ] a 7. This implies [b/x] [b/(x 1 )] [b/x 1 ]/ 7/. In particular, m 1 4. Similarly let m be the least positive integer which satisfies m x 1 b x 1 /. We see m =[b/x 1 ] or [b/x 1 ]+1. We denote min(r b (m x 1 ), R b (m y 1 )) and max(r b (m x 1 ), R b (m y 1 )) by x and y, respectively. We see x x 1 / x/4. Therefore, we have m m 1 + x x b 3 8 b a. By Lemma 4.13, R k (jp i )<k/ for j = min(m 1 m,x )a + b and all i. Note that m 1 m is prime to b The case a <band a 8 We assume b>a and a 8. Choose l {(b 1)/, (b + 1)/} such that b = R b (jp i ) and set σ {±1} such that R b (lp σ )<R b (lp σ ). Note that R k (lp ±δ )<b/. We will prove that there exists a j = αa + βb which satisfies R k (jp i )<k/ for all i. To prove this, we have only to prove R b (αp i +[βb/a ])<(b 1)/ for all i by equation (4.9) The case R b (lp δ )>[b/a ]. For α = l and β = 1, we see R b (αp i +[βb/a ])< (b 1)/ for all i easily The case 1 [b/a ] >R b (lp δ ). For α = l and β = 1, we see R b (αp i +[βb/a ])< (b 1)/ for all i. Then b b 0 R b (lp δ ) + a < a < b 1. Since R b (lp δ ) +[b/a ] R b (αp δ +[βb/a ])(mod b), we see R b (αp δ +[βb/a ])< (b 1)/. One can prove the case i = 0 and δ similarly The case 1 [b/a ] R b (lp δ ) [b/a ]. We suppose α = 3l and β = 1. It is easy to see R b (αp i +[βb/a ])<(b 1)/ except for i = δ, To prove R b (αp δ +[βb/a ])< (b 1)/, we assume a 9 and b 55. Then 3R b (lp δ ) + ba < 3 b 1 + ba < 3 b 1, 3R b (lp δ ) + ba = 3b 3l 3R b (lp δ ) + 3b 3 b b 1 4 b 1 9 ba b 4 a 1 1 = b + b b.

16 38 T. Nakagawa Since 3R b (lp δ ) + [ b/a ] R b (αp δ +[βb/a ])(mod b), R b (αp δ +[βb/a ]) (b 1)/. We also see the case a = 8 similarly. Therefore, we complete the proof of Theorem 4.. REFERENCES [1] S. Akiyama. On the pure Jacobi sums. Acta Arith. 75 (1996), [] N. Aoki. A finiteness theorem on pure Gauss sums. Comment. Math. Univ. St. Pauli 53 (004), [3] N. Aoki. On the purity problem of Gauss sums and Jacobi sums over finite fields. Comm. Math. Univ. St. Pauli 45 (1996), 1 1. [4] B. Berndt, R. Evans and K. Williams. Gauss and Jacobi sums. John Wiley & Sons, New York, [5] R. J. Evans. Generalization of a theorem of Chowla on Gaussian sums. Houston J. Math. 3 (1977), [6] R. Evans. Pure Gauss sums over finite fields. Mathematika 8 (1981), [7] B. Gross and N. Koblitz. Gauss sums and the p-adic G-function. Ann. of Math. () 109 (1979), [8] R. J. Lemke Oliver. Gauss sums over finite fields and roots of unity. Proc. Amer. Math. Soc. 139 (011), [9] K. Shiratani and M. Yamada. On rationarity of Jacobi sums. Col. Math. 73 (1997), [10] P. T. Young. On Jacobi sums, multinomial coefficients, and p-adic hypergeometric functions. J. Number Theory 5 (1995), [11] L. C. Washington. Introduction to Cyclotomic Fields (Graduate Texts in Mathematics, 83). Springer, Berlin, Takahiro Nakagawa Chiba University 1-33 Yayoi-cho, Inage Chiba Japan ( tnakagaw@g.math.s.chiba-u.ac.jp)