Ukrainian Mathematical Journal, Vol. 59, No. 9, 2007
GENERAL KLOOSTERMAN SUMS OVER THE RING OF GAUSSIAN INTEGERS S. P. Varbanets
UDC 511.19
The general Kloosterman sum K(m, n; k; q) over Z was studied by S. Kanemitsu, Y. Tanigawa, Yuan Yi, and Wenpeng Zhang in their research of the problem of D. H. Lehmer. In the present paper, we obtain β; h, q; k), which does similar estimates for K(α, β; k; γ) over Z [i]. We also consider the sum K(α, not have an analog in the ring Z but can be used for the investigation of the second moment of the Hecke zeta function of the field Q (i).
1. Introduction The classical Kloosterman sums appeared first in the work of Kloosterman [1] in connection with the representation of natural numbers by binary quadratic forms. The Kloosterman sum is an exponential sum over a reduced residue system modulo q : q
K(a, b; q) :=
2πi ax+bx q
e
,
a, b ∈ Z,
q > 1 is natural;
x=1 (x,q)=1
here and in what follows, x denotes the reciprocal to x modulo q, i.e., xx ≡ 1 (mod q). It follows from the relation K(a, b; q) = K(aq2 , bq2 ; q1 )K(aq1 , bq1 ; q2 ),
q = q1 q2 ,
(q1 , q2 ) = 1,
that it suffices to estimate K(a, b; q) only in the case q = pn , where p is a prime and n ∈ N. The greatest difficulty in the estimation of the Kloosterman sums is encountered in the case q = p. The 3 estimate K(a, b; p) p 4 under the condition (a, b, p) = 1 was obtained in the cited work of Kloosterman; later, 2 Davenport [2] improved it up to p 3 . A. Weil [3] proved the Riemann hypothesis for algebraic curves over a 1 finite field and obtained the best possible estimate p 2 . Davenport [2] studied the general Kloosterman sums over a finite field with multiplicative character ψ of this field, namely, Kψ (a, b; p) =
2πi ax+bx p
ψ(x)e
.
∗ x∈Fp
The further generalization of Kloosterman sums was associated with the replacement of a prime field Fp in it by a finite expansion Fq , q = pn , n ∈ N. The generalization of Kloosterman sums associated with the theory of modular forms was studied by Kuznetsov [4, 5], Bruggeman [6], Deshouillers and Iwaniec [7], and Proskurin [8]. Mechnikov Odessa National University, Odessa. Published in Ukrains’kyi Matematychnyi Zhurnal, Vol. 59, No. 9, pp. 1179–1200, September, 2007. Original article submitted February 17, 2006. 0041–5995/07/5909–1313
c 2007
Springer Science+Business Media, Inc.
1313
1314
S. P. VARBANETS
In the last years, in connection with the investigation of the Lehmer problem, other generalizations of Kloosterman sums have been studied (see [9, 10]), namely, K(a, b; q, k, ψ) =
k +bx k
2πi ax
ψ(x)e
q
xx ≡ 1 (mod q),
,
x∈R∗ (q)
where ψ is a multiplicative character modulo q. Multiple Kloosterman sums were introduced by Mordell [11]:
K(a1 , . . . , an ; q) =
2πi
e
σm (a1 x1 +...+an xn ) p
,
∗ x1 ,...,xn ∈Fq x1...xn =1
where a1 , . . . , an ∈ F∗q , q = pm , and σm (c) is the trace from Fq into Fp . Multiple Kloosterman sums are a particular case of trigonometric sums on an algebraic variety over a finite field. Due to the investigations of Dwork [12] (who proved the rationality of the zeta function of an algebraic variety over a finite field), Deligne [13] (who proved the Riemann hypothesis for an algebraic variety over Fq ), and Bombieri [14] (who estimated the number of characteristic roots of the zeta function in terms of a generating polynomial), the following final estimate was obtained (see Deligne [15] and Bombieri [14]): K(a1 , . . . , an ; q) ≤ nq
n−1 2
.
In this paper, we obtain estimates for general Kloosterman sums over the ring of Gaussian integers. Notation. Denote the ring of Gaussian integers by Z[i], i.e., Z[i] := a + bi | a, b ∈ Z, i2 = −1 . We denote Gaussian integers by the Greek letters α, β, γ, ξ, and η; a Gaussian prime number is denoted by p if p ∈ Z. For α ∈ Z[i], we set Sp(α) = α + α and N (α) = αα, where α denotes the complex conjugate of α; Sp(α) and N (α) stand for the trace and the norm (respectively) of α from Q(i) into Q. Let Fq denote a field that contain just q as an element, q = pn , n ∈ N. For x ∈ Fq , we denote by σn (x) the trace of x from Fq into Fp , i.e., n−1
σn (x) := x + xp + . . . + xp
,
σ1 (x) = σ(x) = x.
The notation a ∈ R(q) respectively, a ∈ R(q, i) means that a ∈ Z respectively, a ∈ Z[i] and a runs through a complete residue system modulo q. Analogously, a ∈ R∗ (q) respectively, a ∈ R∗ (q, i) means that a ∈ Z respectively, a ∈ Z[i] and a runs through a reduced residue system modulo q. The notation means that the summation is carried out over the region U. Moreover, exp (z) = ez and 2πi z
(U )
eq (z) = e q for q ∈ N; the Vinogradov symbol as in f (x) g(x) means that f (x) = O(g(x)). For Gaussian integers α, β, and γ, we define the Kloosterman sum K(α, β; γ) =
x∈R∗ (γ,i)
αx + βx . exp πi Sp γ
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1315
Zanbyrbaeva [16] obtained the estimate 1
1
K(α, β; γ) 2ν(γ) N (γ) 2 N ((α, β, γ)) 2 , where ν(γ) is the number of different prime divisors of γ, and (α, β, γ) denotes the greatest common divisor of α, β, and γ. We consider the following two types of general Kloosterman sums over Z[i]:
K(α, β; k; γ, ψ) =
x∈R∗ (γ,i)
αxk + βx k ψ(x) exp πi Sp , γ
where α, β, γ ∈ Z[i] and ψ is a multiplicative character modulo γ, and
K(α, β; h, q; k) =
x,y∈R∗ (q,i) N (xy)≡h (mod q)
eq
1 k k Sp(αx + βy ) , 2
where α, β ∈ Z[i], h, q ∈ N, and (h, q) = 1. We call K(α, β; k; γ, ψ) the general power Kloosterman sum and K(α, β; h, q; k) the norm Kloosterman sum. Our aim is to obtain nontrivial estimates for K(α, β; k; γ; ψ) and K(α, β; h, q; k). 2. Auxiliary Results To prove our main results, we need several lemmas. Lemma 2.1. Suppose that p is a Gaussian prime “odd” number, α1 , . . . , αk ∈ Z[i],
(α2 , p) = . . . = (αk , p) = 1,
and
ν3 , ν4 , . . . , νk ≥ 2
are natural numbers. Then, for every natural n ≥ 2, we have 2 ν 3 ν k 3 k α1 ξ + α2 pξ + α3 p ξ + . . . + αk p ξ exp 2πi Sp n p ξ(mod pn ) 0 =
N (p)
if (α1 , p) = 1), n+1 2
if α1 ≡ 0 (mod p).
(2.1)
Lemma 2.2. Let p = 1 + i be a Gaussian “even” number, let αj ∈ Z[i], j = 1, 2, . . . , k, and let (α2 , p) = . . . = (αk , p) = 1.
1316
S. P. VARBANETS
Then, for any natural numbers νj ≥ 2, j = 2, 3, . . . , k, and any n ≥ 2, the following estimate is true: 2 ν 3 ν k 3 k α1 ξ + α2 pξ + α3 p ξ + . . . + αk p ξ ≤ δ · 2n+1 , exp 2πi Sp n p ξ(mod pn ) where
0 δ=
2
if α1 ≡ 0 (mod p2 ), if α1 ≡ 0 (mod p2 ).
The assertions of these lemmas are consequences of estimates for a complete linear sum and a Gauss sum to which the primary sums can be reduced. Lemma 2.3. Suppose that p is a prime number, A ∈ Z, (A, p) = 1, f (x) ∈ Z[x], f (x) = a1 x + a2 x2 + pλ3 a3 x3 + . . . + pλk ak xk , (ai , p) = 1, i = 2, 3, . . . , k, and λj > 0, j = 3, . . . , k. Then, for any n ∈ N, the following equality is true: S :=
n epn (Af (x)) = ε(n)p 2 epn AF (a1 , . . . , an ) ,
x(mod pn )
where F (x1 , . . . , xn ) ∈ Z[x1 , . . . , xn ], 1 ε(n) = A p−1 2 · (i)( 2 ) p
if n
is even,
if n
is odd,
A is the Legendre symbol. and p Proof. We set x = y + pn−1 z, y (mod pn−1 ), z (mod p). Then
epn (Af (x)) =
x (mod pn )
epn A(f (y) + pn−1 zf (y) .
y (mod pn−1 ) z (mod p)
The sum over z gives zero if f (y) ≡ 0 (mod p). However, we have f (y) ≡ a1 + 2a2 y (mod p). Let y0 be a root of the congruence a1 + 2a2 y ≡ 0 (mod p). Then S = epn (Af (y0 ))
epn A(f (y0 + py) − f (y0 ))
y (mod pn−1 )
= epn Af (y0 )
y (mod pn−1 )
epn−2 Ag(y) = pepn Af (y0 )
y (mod pn−2 )
epn−2 Ag(y) ,
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1317
where g(y) =
f (y0 + py) − f (y0 ) = b 1 y + b 2 y 2 + pµ3 b 3 y 3 + . . . + p µk b k y k , p2
b1 , . . . , bk are linear functions of a1 , . . . , ak with coefficients dependent on y0 , b2 ≡ a2 (mod p), (bj , p) = 1, and µj ≥ 1, j = 3, . . . , k. Thus, g(y) is a polynomial of the same sort as f (y). We continue these considerations further. Then, for n ≡ 1 (mod 2), we obtain n 2
2πiA
S=p e
f (y0 ) g(y1 ) + n−2 +... pn p
and, for even n,
S=p
n−1 2
2πiA
e
f (y0 ) g(y1 ) + n−2 +... pn p
2 e(A(cx+a p
x2 ))
x (mod p)
f (y ) g(y1 ) (2 c)2 +...− p A ( p−1 )2 2πiA pn0 + pn−2 i 2 e = . p
The lemma is proved. Lemma 2.4. Let p be a prime number, p ≡ 3 (mod 4), and let E be the set of residue classes mod p of the ring Z[i] that has norms congruous modulo p with ±1. Then E is a cyclic group of order 2(p + 1)p−1 . Proof. It follows from the equality N (αβ) = N (α)N (β) that E is a subgroup of the group of residue classes modulo p in Z[i]. First, let * = 1. Then the residue classes modulo p form a field Fp2 . Let g0 be the generating element of the multiplicative group of this field. We denote g0u = x(u) + iy(u),
(2.2)
where x(u), y(u) ∈ Fp and i is an element of the field Fp2 such that i2 = −1. The residue classes mod p with norms ≡ ±1 (mod p) can be characterized by the condition x2 + y 2 = ±1
(in Fp ).
It now follows from (2.2) that g0pu = x(u) − iy(u). Hence, the element
g0u
has the norm ≡ ±1 (mod p) if and only if
(p+1)u g0
p − 1 = ±1, i.e., if and only if . 2 u
p−1 p−1 t, t = 0, 1, . . . , 2p + 1, and set g0 2 = g. The classes g t , t = 0, 1, . . . , 2p + 1, are 2 just those and only those that have the norm ≡ ±1 (mod p). Let f = g + pλ, λ ∈ Z[i]. Then
We denote u =
f p = g p + pλ1 ,
λ1 ∈ Z[i],
f p+1 ≡ g p+1 + pg p λ1 (mod p2 ),
1318
S. P. VARBANETS
f 2(p+1) ≡ g 2(p+1) + 2pg p+1 λ1 (mod p2 ). Let g 2(p+1) = 1 + pg1 . We have f 2(p+1) − 1 ≡ p(g1 + 2g p+1 λ1 ) (mod p2 ). We can always take λ such that f 2(p+1) = 1 + ph,
h ∈ Z[i],
(h, p) = 1.
Thus, we can assume that the generating element g of the group E has been selected so that g 2(p+1) − 1 ≡ 0 (mod p2 ). We now easily get −1
g 2(p+1)p
≡ 1 (mod p ),
g k ≡ 1 (mod p ),
0 < k < 2(p + 1)p−1 ,
for every * = 1, 2, . . . . We must also show that, for every * = 1, 2, . . . , there exists g such that g ≡ g (mod p ) and N (g ) ≡ −1 (mod p ). For * = 1, this has already been proved. Let * = 2. If g = x + iy, then x2 + y 2 = −1 + λp, g 2(p+1) = 1 + p(h1 + ih2 ),
(h1 + ih2 , p) = 1,
h1 , h2 ∈ Z.
For k = k1 + ik2 , k1 , k2 ∈ Z, we have N (x + iy + p(k1 + ik2 )) = −1 + λp + p(2xk1 + 2yk2 ) + p2 (k12 + k22 ) ≡ −1 + p(λ + 2xk1 + 2yk2 ) (mod p2 ), (x + iy + p(k1 + k2 ))2(p+1)p ≡ (x + iy)2(p+1)p + 2p2 (x + iy)2(p+1)p−1 (k1 + ik2 ) (mod p3 ). Hence, ((x + iy) + p(k1 + ik2 ))2(p+1)p λ + 2xk1 + 2yk2 ≡ 0 (mod p2 ), where h(1) and α are the Gaussian integers and co-prime numbers with p. Next, the congruence −h(1) ≡ α(k1 + ik2 ) (mod p) holds only for one collection of (k10 , k20 ) modulo p. Therefore, if we take k1 ≡ k10 (mod p) and define k2 from the congruence λ + 2xk1 + 2yk2 ≡ 0 (mod p2 ), then we establish that f = x + iy + p(k1 + ik2 ) has the norm ≡ −1 (mod p2 ). Moreover, f belongs to the exponent 2(p + 1)p modulo p2 and f 2(p+1)p = 1 + Hp2 , (H, p) = 1, i.e., f ∈ E2 and f belongs to the exponent 2(p + 1)p−1 modulo p for every * = 2, 3, . . . .
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1319 2(p+1)p
Now note that, by condition, g3 = f +p2 (m1 +im2 ) satisfies the relation g3 for any m1 , m2 ∈ Z. We take m1 , m1 ∈ Z, such that
= 1+H1 p2 , (H1 , p) = 1,
λ2 + 2f1 m1 + 2f2 m2 ≡ 0 (mod p), where λ=
N (f ) + 1 N (f ) − (−1) = , 2 p p2
f = f1 + if2 .
Then g3 = f + p2 (m1 + im2 ) is the generating element of the group E3 . We proceed by induction. If we have already determined g−1 , then the generating element of E is g = g−1 + p−1 (m1 + im2 ), where m1 and m2 are determined from the congruence m1 + 2g−1 m2 ≡ 0 (mod p), λ−1 + 2g−1
where λ−1 =
N (g−1 ) + 1 , p−1
g−1 = g−1 + ig−1 ,
g−1 , g−1 ∈ Z.
The lemma is proved. Lemma 2.5. Let p be a prime number, p ≡ 3 (mod 4), and let * ∈ N. Then every residue x + iy of a reduced residue system mod p of the ring of Gaussian integers admits a unique representation of the form x + iy ≡ g c (u + iv)d (mod p ), c = 0, 1, . . . , (p − 1)p−1 − 1,
(2.3)
d = 0, 1, . . . , (p + 1)p−1 − 1,
where g is a primitive root modulo p in Z and u + iv is the generating element of E . Proof. Let ϕ(α) denote the Euler function on Z[i]. Then, for p ≡ 3 (mod 4), we have
1 ϕ(p ) = N (p ) 1 − N (p)
= p2(−1) (p2 − 1).
In relation (2.3), we have p2(−1) (p2 − 1) formally distinguishable expressions of the form g c (u + iv)d . Since, for any c and d, we have (g c (u + iv)d , p) = 1, for the proof of the assertion of the lemma it suffices to show that expressions (2.3) are pairwise disjoint mod p for different collections of (c, d). Assume that g c1 (u + iv)d1 ≡ g c2 (u + iv)d2 (mod p ),
c1 ≥ c2 .
1320
S. P. VARBANETS
Then g c1 −c2 ≡ (u + iv)d2 −d1 (mod p ) if d2 ≥ d1 or g c1 −c2 (u + iv)d1 −d2 ≡ 1 (mod p ) if d2 < d1 . Taking into account that the sets {g } and (u + iv)d have only one common element (namely 1) modulo p , we get c1 = c2 and d1 = d2 . The lemma is proved. Corollary 2.1. All reduced classes x + iy modulo p , p ≡ 3 (mod 4), that have equal norms modulo p can be written in the form x + iy ≡ g c (u + iv)2d , d = 0, 1, . . . , p−1 (p + 1) − 1
if N (x + iy) ≡ g 2c (mod p ),
(x + iy) ≡ g c (u + iv)2d+1 , d = 0, 1, . . . , p−1 (p + 1) − 1
if N (x + iy) ≡ −g 2c (mod p ),
where 0≤c≤
p − 1 −1 − 1. p 2
Let p be a prime number, p ≡ 1 (mod 4). Then, in the ring Z[i], we have p = p · p, where p and p are complex-conjugate Gaussian prime numbers (p = ±p, ±ip). It is well known that a+bi | a, b =0, 1, . . . , p −1 is a complete residue system mod p . Similarly, for p = 2, we have 2 = −i(1 + i)2 , and a + bi | a, b = 0, 1, . . . , 2 − 1 is a complete system mod p , where p = 1 + i in Z[i]. 3. General Kloosterman Sum K(α, β; k; γ) Consider the sum K(α, β; k; γ) defined in Introduction for the trivial character ψ0 :
K(α, β; k; γ, ψ0 ) = K(α, β; k; γ) =
(U )
αxk + βx k , exp πi Sp γ
where U = (x, x ) ∈ Z[i]2 | x, x (mod γ), xx ≡ 1 (mod γ) . Obviously, we have K(α, β; k; γ) = K(αγ2 , βγ2 ; k; γ1 )K(αγ1 , βγ1 ; k; γ2 ) if γ = γ1 γ2 , (γ1 , γ2 ), where γ1 γ1 ≡ 1 (mod γ2 ) and γ2 γ2 ≡ 1 (mod γ1 ).
(3.1)
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1321
Thus, we can assume, without loss of generality, that γ = pn , where p is a Gaussian prime number. In Sec. 1, we have obtained a description of the reduced residue system mod pn , p = p ≡ 3 (mod 4). For p ∈ Z[i], N (p) = p ≡ 1 (mod 4), the reduced residue system mod pn has the form {a ∈ Z | 1 ≥ a ≥ pn − 1, (a, p) = 1} , and, for a Gaussian prime “even” number p = 1 + i, a + bi | a, b ∈ {0, 1, . . . , 2n − 1} , a ≡ 1 (mod 2), b ≡ 0 (mod 2) . Theorem 3.1. Let p be Gaussian prime number, N (p) = p ≡ 1 (mod 4), and let d = (k, p − 1). Then K(α, β; k; p) ≤ 2dN ((α, β, p)) 12 N (p) 12 .
(3.2)
Proof. If (α, β, p) = p, then our assertion is obvious. Let (α, β, p) = 1. According to the description of a reduced residue system mod pn , we can assume that α = a, β = b, a, b ∈ Z, p = c1 + ic2 , c1 , c2 ∈ Z, and (c1 , p) = (c2 , p) = 1. Thus, K(α, β; k; p) =
u∈R∗ (p)
ep
1 k k = Sp a(c1 − ic2 )u + b(c1 − ic2 )u 2
k ep ac1 uk + bc1 u .
u∈R∗ (p)
The last sum was estimated in [10], but we give a calculation in order to make an estimate more precise. We define Ik (a) := # x ∈ Z | 0 ≥ x ≥ p − 1, xk ≡ a (mod p) . It is clear that
d Ik (a) =
0
if d ind a = otherwise
d−1 t=0
e2πi
t ind a , d
where ind a denotes the index of an integer a, (a, p) = 1, by the radix of some primitive root modulo p. Then d−1 1 K(α, β; k; p) = Ik (u)ep (au + bu ) ≤ ed (t ind u)ep (au + bu ) ≤ 2dp 2 . t=0 u∈R∗ (p) u∈R∗ (p)
(3.3)
Here, we have taken into account that the inner sum is the classical Kloosterman sum weighted by a character, and, 1 hence, it is estimated as 2p 2 (see Perel’muter [17] and Williams [18]). The theorem is proved. Theorem 3.2. Let p ≡ 3 (mod 4), k ∈ N, and d = (k, p2 − 1). Then 1 K(α, β; k; p) ≤ 2d N (α, β, p) 2 N (p) 12 .
(3.4)
1322
S. P. VARBANETS
Proof. The residue classes mod p in the ring Z[i] form a field Fp2 . Hence, σ2 (x) = x + xp . However, we have noted that x ≡ xp (mod p) for x ∈ Z[i]. Thus, Sp(x) = x + x ≡ x + xp ≡ σ2 (x) (mod p). Hence, K(α, β; k; p) =
ep
x∈R∗ (p)
1 k k k ep σ2 2 αxk + 2 βx Sp(αx + βx ) = , 2 ∗ x∈F
p2
where 2 · 2 ≡ 1 (mod p). Let g denote a primitive element of the field Fp2 , indg x = ind x for x ∈ Fp2 , and let I(u) be the number of solutions of the equation xk = u in Fp2 . Then d−1 t ind x K(α, β; k; p) = e2πi d ep σ2 (2 αx + 2 βx ) t=0 x∈F∗p2 d−1 ep σ2 (2 αx + 2 βx ) + ψt (x)ep σ2 (2 αx + 2 βx ) , ≤ ∗ x∈Fp2 t=1 x∈F∗p2 where ψt (x) = ed (t ind x) is a multiplicative character of the field Fp2 . Using again the estimates for the Kloosterman sums with a character of a finite field, we finally obtain 1 K(α, β; k; p) ≤ 2dN (α, β, p) 2 N (p) 12 . The theorem is proved. For p = 1 + i, we trivially get K(α, β; k; p) = 1. n n n n n−m ), and For γ =p , we perform the substitution x (mod p ) = y + p z, where y (mod p ), z (mod p n+1 . Then, using the standard technique, we easily obtain the following theorem: m= 2 Theorem 3.3. Let n0
γ = (1 + i)
s i=1 N (pi )≡1(4)
pni i
t
n
pj j .
j=1 pj ≡3(4)
Then K(α, β; k; γ) ≤ 2D N ((α, β, γ))N (γ) 12 ,
(3.5)
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1323
where D=
s t (k, pj − 1) (k, p2j − 1). i=1
j=1
Now consider a nontrivial multiplicative character ψ of the field Fq , where q = pr , r ∈ N, p is a prime number, α, β ∈ Fq , and either α = 0 or β = 0. We define a general power Kloosterman sum with character ψ as follows:
K(α, β; k; q, ψ) :=
k ψ(x)ep σ2 (αxk + βx ) .
(3.6)
∗ x∈Fq
2πi h ind x
q−1 , where ind x is taken with respect to some primitive element for F . Let d = (k, q − 1) and ψ(x) = e q We have two probable cases: d h and d h. We prove that K(α, β; k; q, ψ) = 0 in the first case. For β = 0, we have
α∈Fq
α∈Fq
=
∗
x,y∈Fq xx =yy =1
k ep σ2 (βy (xk − 1)) ep σ2 (αy k (xk − 1))
y∈Fq∗
α∈Fq
ψ(x)
x∈Fq∗
=q
k k ψ(x)ψ(y )ep σ2 (α(xk − y k ) + β(x − y ))
K(α, β; k; q, ψ) 2 =
y∈Fq∗
k ψ(x)ep σ2 (βy (xk − 1)) = q(q − 1) ψ(x).
∗ x∈Fq k x =1
(3.7)
∗ x∈Fq k x =1
In the last sum, the summation is carried out over x ∈ F∗q for which k ind x ≡ 0 (mod (q − 1)), i.e., ind x = q−1 s, s = 0, 1, . . . , d − 1, and, thus, d d−1 hs K(α, β; k; q, ψ) 2 = q(q − 1) e2πi d = 0 if α
h ≡ 0 (mod d).
s=0
If d h, then ψ(x) = eq−1 (h1 d ind x) = eq−1 (h1 ind xd ). Hence, setting k1 = we obtain
k , d
h1 =
h , d
and
ψ1d = ψ,
1324
S. P. VARBANETS
K(α, β; k; q, ψ) =
d ψ1 (xd )ep σ2 (α(xd )k1 + β(x )k1 )
x∈Fq∗
=
k Id (x)ψ1 (x)ep σ2 (αxk1 + βx 1 )
x∈Fq∗
=
d−1
k ed (s ind x)eq−1 (h1 ind x)ep σ2 (αxk1 + βx 1 )
s=0 x∈Fq∗
=
d−1 s=0
k ψ2 (x)ep σ2 (αxk1 + βx 1 ) ,
(3.8)
x∈Fq∗
where ψ2 (x) = eq−1 (h2 ind x) and
h2 =
s(q − 1) + h . d
Thus, we have diminished the exponent k in d-time if d > 1. However, if d = 1, then it is clear that Kk (α, β; q) =
k eq−1 hk ind xk ep σ2 (αxk1 + βx 1 )
x∈F∗q
=
eq−1 (hk ind x)ep σ2 (αx + βx ) = K1 (α, β; q; ψ3 ),
x∈F∗q
where kk ≡ 1 (mod (q − 1)) and ψ3 = ψ k . The sum K1 (α, β; q, ψ3 ) is the Kloosterman sum over Fq weighted by a multiplicative character ψ3 of the 1 field Fq and is estimated as 2q 2 if β = 0 (see Perel’muter [17]). Relation (3.8) shows that if (k1 , q − 1) = 1, then Kk (α, β; q, ψ) ≤ 2dq 12 . If (k1 , q − 1) = d1 > 1, then we again consider two cases (h1 , d1 ) = d1
or
(h2 , d1 ) < d1 .
However, if (h2 , d1 ) < d1 , then K(α, β; k; q, ψ) = 0. . The case h2 ..d1 can be realized only for those s, 0 ≤ s ≤ d − 1, for which h2 ≡ 0 (mod d1 ), i.e., s must satisfy the congruence q−1 h + ≡ 0 (mod d1 ). d d q−1 q−1 = 1 because d1 k1 and k1 , = 1. However, d1 , d d s
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1325
It follows that we have only one value of s modulo d1 and, hence, at most
d + 1 values of s among d1
. 0 ≤ s ≤ d − 1 for which h2 ..d1 . We use this reduction and, in ν(k) steps, obtain the estimate Kk (α, β; q, ψ) ≤ 2ν(k)+1 kq 12 , where ν(k) denotes the number of prime divisors of k. Thus, we have proved the following theorem:
Theorem 3.4. Suppose that α, β ∈ Fq and at least one of the elements α and β is not equal to zero. Then, for any multiplicative character ψ of the field Fq , the following estimate is true: Kψ (α, β; q; k) ≤ 2kq 12 . Corollary 3.1. Let p be a Gaussian prime number and let χ be a multiplicative character of the field of residue classes mod p. Then
k + βx k αx ≤ 2ν(k)+1 kN (p) 12 N ((α, β, p)) 12 . χ(x) exp πi Sp p x (mod p) 4. General Kloosterman Sums over Norm Let α, β ∈ Z[i], h ∈ Z, q ∈ N, q > 1, and (h, q) = 1. We set
K(α, β; h, q) :=
x,y (mod q) N (xy)≡h (mod q)
eq
1 Sp(αx + βy) 2
(4.1)
and call it the norm Kloosterman sum in Z[i]. For q = q1 q2 , (q1 , q2 ) = 1, we have β; hq 1 , q2 ) = K(αq K(α, β; h, q) = K(α, β; hq 2 , q1 )K(α, 2 , βq2 ; h, q1 )K(αq1 , βq1 ; h, q2 ). Therefore, we consider only the case q = pn , where p is a prime rational number and n ∈ N. Denote mα = max α ≡ 0 (mod pm ) . m≥n
Theorem 4.1. Let (h, p) = 1. Then 1 3n K(α, β; h, pn ) (pmα , pmβ , pn ) 2 p 2
with an absolute constant in the symbol “.”
(4.2)
1326
S. P. VARBANETS
Proof. First, let n = 1. The case mα = mβ = 1 is trivial. Therefore, we assume that mα = 0 or mβ = 0. We set α = a1 + ia2 and β = b1 + ib2 ; hence, (a1 , a2 , b1 , b2 ) = 1. For p ≡ 1 (mod 4), we have K(α, β; h, p) =
ep (a1 x1 − a2 x2 + b1 y1 − b2 y2 ),
(4.3)
(U )
where U = x1 , x2 , y1 , y2 ∈ {0, 1, . . . , p − 1}, (x21 + x22 )(y12 + y22 ) ≡ h (mod p) . Let ε0 be a solution of the congruence x2 ≡ −1 (mod p). We set u2 = x1 − ε0 x2 ,
u1 = x1 + ε0 x2 ,
v1 = y1 + ε0 y2 ,
v2 = y1 − ε0 y2 .
By using (4.3), we obtain K(α, β; h, p) =
ep (A1 u1 + A2 u2 + B1 v1 + B2 v2 ),
(U )
where U = u1 , u2 , v1 , v2 ∈ {0, 1, . . . , p − 1}, u1 u2 v1 v2 ≡ h (mod p) . 3
Bombieri [14] proved that the last sum can be estimated as p 2 . If p ≡ 3 (mod 4), then this estimate holds for sum (4.3) (the proof is analogous). The case p = 2 is trivial. Now let n ≥ 2. It suffices to consider the case (pmα , pmβ , pn ) = 1. In this case, at least one of the numbers a1 , a2 , b1 , and b2 is not divided by p (here α = a1 + ia2 and β = b1 + ib2 ). We have K(α, β; h, pn ) =
x,y (mod pn )
=
pn −1 1 n k(N (x)N (y) − h) + (αx) + (βy) e p pn k=0
1 2 2 2 2 n k((x + x )(y + y ) − h) + a1 x1 − a2 x2 + b1 y1 − b2 y2 , e p 1 2 1 2 pn
(4.4)
U
where U := k (mod pn ); x1 , x2 (mod pn ); y1 , y2 (mod pn ) . At least one of the sums over x1 , x2 , y1 , and y2 is equal to 0 if (k, p) = p (by virtue of a rational analog of Lemma 2.1). Thus, assuming that (a1 , a2 , p) = 1, we get
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1327
1 n ep (−kh)epn kN (x)(y12 + y22 ) + (αx) + b1 y1 − b2 y2 K(α, β; h, pn ) = n p U
=
1 pn
∗
k (mod pn )
epn (−kh)
+
x (mod pn ) (N (x),p)=1
x (mod..pn ) N (x).p
= + , 1
(4.5)
2
where, say, U := k ∈ R∗ (pn ), x ∈ R(pn , i), y1 , y2 ∈ R(pn ) . Let N (x) and k be solutions of the congruences N (x)u ≡ 1 (mod pn ),
ku ≡ 1 (mod pn ),
respectively. Then ∗ epn (−kh) epn 4 N (x) k (b21 + b22 ) + a1 x1 − a2 x2 . = 1 k (mod pn ) x (mod pn ) We set x1 = x01 + pm z1 ,
x2 = x02 + pm z2 , n+1 . m= 2
0≤ It is obvious that
x01 , x02
≤ p − 1, m
0 ≤ z1 , z2 ≤ p
n−m
− 1,
2 2 2 2 N (x) = x01 + x02 1 − 2pm (x01 + x02 ) (x01 z2 + x02 z1 )
and, consequently, ∗ 02 02 2 2 0 0 n n 4 = e (−kh) e k (x + x ) (b + b ) + a x − a x p p 1 1 2 2 1 2 1 2 1 k (mod pn ) 0 ,x0 (mod pn ) x 1 2 2 2 (x01 +x02 , p)=1 × epn−m (A1 + a1 )z1 + (A2 + a2 )z2 , z1 ,z2 (mod pn−m )
(4.6)
1328
S. P. VARBANETS
where 2 2 2 0 A1 = 2 x01 + x02 x2 ,
2 2 2 0 A2 = 2 x01 + x02 x1 .
The summation over z1 and z2 gives zero if the congruences A1 + a1 ≡ 0 (mod pn−m ),
A2 − a2 ≡ 0 (mod pn−m ),
or the equivalent congruences a2 x01 + a1 x02 ≡ 0 (mod pn−m ),
2 2 2 2x02 ≡ −a1 x01 + x02 (mod pn−m ),
are violated. This system of congruences has at most three solutions modulo pn−m and, therefore, at most 3pm−(n−m) solutions modulo pm . Hence, 2(n−m) ∗ 0 0 ≤ 8p 32 n , n = p e (a x − a x ) (kh + k B) p 1 1 2 1 1 (U ) k (mod pn )
(4.7)
where 2 2 2 U = x01 , x02 (mod pm ) a2 x01 ≡ −a1 x02 (mod pn−m ), 2x01 ≡ −a1 x01 + x02 (mod pn−m ) . Finally, if N (x) ≡ 0 (mod p), then The theorem is proved.
2
= 0 according to Lemma 2.1.
For natural k > 1, we set K(α, β; h, q; k) :=
x,y (mod q) N (xy)≡h (mod q)
eq
1 k k Sp(αx + βy ) . 2
(4.8)
It is obvious that K(α, β; h, q; 1) = K(α, β; h, q). The method for the investigation of the sum K(α, β; h, q; k) reduces to considering the case q = pn , where p is a prime. First, we assume that p ≡ 3 (mod 4). Theorem 4.2. Let p ≡ 3 (mod 4), h ∈ Z, (h, p) = 1, k ∈ N, and d = (k, p − 1). Then, for any Gaussian integers α and β, (α, β, p) = 1, the following estimate is true: 3 d2 p 2 K(α, β; h, p; k) 2 dp
if d − 1 ≤ if d ≥
√ 4
√ 4
p,
p + 1.
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
Proof. Let k = dk1 and
p−1 k1 , d
ep
x,y (mod p) N (xy)≡h (mod p)
= 1. We have
1 Sp(α(xk1 )d + β(y k1 )d ) 2
=
ep
x,y (mod p) N (xk1 y k1 )≡hk1 (mod p)
=
1329
ep
x,y (mod p) N (xy)≡hk1 (mod p)
1 k1 d k1 d Sp(α(x ) + β(y ) ) 2
1 Sp(αxd + βy d ) 2
= K(α, β; hk1 , p; d).
For any multiplicative character χ of the field Fp2 , we have h∈F∗2 p
χ(h)K(α, β; h, p; d) =
χ N (x)N (y) ep
x,y∈F∗2
1 1 d d Sp(αx ) ep Sp(βy ) 2 2
p
χ N (x)ep = x∈F∗2
1 1 χ N (y) ep Sp(αxd ) Sp(βy d ) , (4.9) 2 2 ∗ y∈F
p2
p
1
and, moreover, the sums on the right-hand side of (4.9) can be estimated as (d − 1)N (p) 2 (see [17]). Thus, χ(h)K(α, β; h, p; d) ≤ (d − 1)2 p2 . ∗ h∈Fp2 The application of the Plancherel theorem gives
|K(α, β; h, p; d)|2 ≤ (d − 1)4 p4 .
h∈F∗2 p
By analogy with [14], we can now conclude that the weights of the characteristic roots associated with K(α, β; h, p; d) do not exceed 3 if (d − 1)4 < p. Hence, using the results of Bombieri [14] and Deligne [13], we get K(α, β; h, p; d) (d − 1)2 p2 d2 p2
if d − 1 <
√ 4
p.
1330
S. P. VARBANETS
Further, for x = x1 +ix2 , x1 , x2 ∈ Z, we have x1 −ix2 ≡ (x1 +ix2 )p (mod p) and, thus, N (x) ≡ xp+1 (mod p). Hence,
ep
x,y (mod p) N (xy)≡h (mod p)
1 d d Sp(αx + βy ) = 2
=
ep
x,y (mod p) (xy)p+1 ≡h (mod p)
1 d d Sp(αx + βy ) 2
ε (mod p) x (mod p) εp+1 ≡h (mod p)
ep
1 d d Sp(αx + βy ) . 2
(4.10)
The congruence z p+1 ≡ h (mod p) has exactly p + 1 solutions mod p. The inner sum on the right-hand side of (4.10) is estimated as ≤ 2dp. This completes the proof of the theorem. Now let q = pn , p ≡ 3 (mod 4), n ≥ 2. We use the description of the reduced residue system mod pn (see Lemma 2.5). We need the following assertion: Lemma 4.1. Suppose that n, k ∈ N, p ≥ 3 is a prime, u ∈ Z, and (p, u) = 1. Then, for any natural t, we have (1 + pk u)t ≡ 1 + pk a1 t + p2k a2 t2 + pλ3 a3 t3 + . . . + pλn an tn (mod pn ); moreover, (ai , p) = 1, i = 1, . . . , n, and λj > 2k, j = 3, . . . , n. Proof. Using the relation
t m
1 = m!
m(m − 1) m−1 t − + . . . + (−1)m−1 (m − 1)! · t t 2
m
and an upper bound for an exponent with which p enters m!, we obtain (1 + pk u)t ≡ 1 + pk a1 t + p2k a2 t2 + pλ3 a3 t3 + . . . + pλn an tn (mod pn ), where (ai , p) = 1, i = 1, . . . , n, and λj >
1 k− p−1
j > 2k
for j = 3, 4, . . . .
The lemma is proved. It is obvious from the proof of Lemma 2.3 that the generating element u + iv of the group E1 can be taken so that it is the generating element of the group E for any fixed *, * = 2, 3, . . . . Let * = max(5, n). We have N ((u + iv))2 ≡ 1 (mod p ), (u + iv)2(p+1) = 1 + p(x0 + iy0 ),
(x0 + iy0 , p) = 1.
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1331
Thus, N (1 + px0 + ipy0 ) ≡ 1 + 2px0 + p2 x20 + p2 y02 ≡ 1 (mod p ). Hence, 2px0 ≡ 0 (mod p2 ), x0 = px0 , (y0 , p) = 1. Therefore, (u + iv)2(p+1) ≡ 1 + p2 x0 + ipy0 ,
(x0 , p) = (y0 , p) = 1.
Using the previous lemma, we easily obtain
((u + iv)2(p+1)t ) ≡ A0 + A1 t + A2 t2 + . . . + An−1 tn−1 (mod pn ), (4.11) ((u + iv)2(p+1)t ) ≡ B0 + B1 t + B2 t2 + . . . + Bn−1 tn−1 (mod pn ), where A0 ≡ 1 (mod p), A1 ≡ p2 x0 + 2 y02 p2 (mod p3 ), A2 ≡ −2 y02 p2 (mod p3 ), B1 ≡ py0 (mod p3 ),
i.e.,
B0 ≡ 0 (mod p), i.e., A1 ≡ 0 (mod p3 ), A2 = p2 A2 ,
i.e., B1 ≡ pB1 ,
(A2 , p) = 1,
(B1 , p) = 1,
B2 ≡ A3 ≡ B3 ≡ . . . ≡ An−1 ≡ Bn−1 ≡ 0 (mod p3 ). We set β = 2(p + 1)t + z,
0 ≤ t ≤ pn−1 − 1,
0 ≤ z ≤ 2p + 1,
and denote (u + iv)z = u(z) + iv(z),
z = 0, 1, . . . , 2p + 1.
Then (u + iv)β = (u + iv)2(p+1)t (u(z) + iv(z)). Hence,
(u + iv)2(p+1)t+z ≡ A0 (z) + A1 (z)t + . . . + An−1 (z)tn−1 (mod pn ), where Ai (z) = Ai u(z) − Bi v(z).
(4.12)
1332
S. P. VARBANETS
We clarify for what values of z the congruence v(z) ≡ 0 (mod p) is true. Let v(z) = pv0 (z), v0 (z) ≡ 0 (mod pk ), k ≥ 0. Then (u + iv)z = u(z) + ipv0 (z), n−k
(u + iv)z(p−1)p
n−k
≡ (u(z))(p−1)p
(mod pn ).
The sequences (u + iv)2β and {g α } can have only two common elements modulo p, namely, 1 or −1. Thus, n−k
(u(z))(p−1)p
≡ ±1 (mod pn ).
n−k
k−1
≡ −1 (mod pn ) is impossible, because otherwise we have (−1)p The congruence (u(z))(p−1)p n−1 (p+1)p n (u(z)) ≡ 1 (mod p ), i.e., −1 ≡ 1 (mod p). Hence, n−k
(u(z))(p−1)p
≡
≡ 1 (mod pn ),
z(p − 1)pn−k ≡ 0 mod 2(p + 1)pn−1 . ! k−1 ). Hence, the relation p!v(z) yields z = p + 1, and Since (p − 1, p + 1) = 2, we have z ≡ 0 (mod(p + 1)p p2 v(z) yields z = 0. Therefore, ! p!A1 (z),
Ai (z) ≡ 0 (mod p2 ),
i = 2, . . . , n − 1,
A1 (0) = A1 (p + 1) ≡ 0 (mod p2 ),
! p2 !A2 (0),
Aj (0) ≡ Aj (p + 1) ≡ 0 (mod p3 ),
if z = 0, z = p + 1, ! p2 !A2 (p + 1),
j = 3, 4, . . . , n − 1.
We are now in a position to prove the following assertion: Theorem 4.3. Suppose that p is a prime number, p ≡ 3 (mod 4), h ∈ Z, (h, p) = 1, k > 1 is a natural number, and a and b are Gaussian integers, (a, p) = (b, p) = 1. Then, for n ≥ 2, one has 3 n K(a, b; h, p ; k) ≤ 2p 2 n+m log pn ,
(4.13)
! where m is such that pm !k. Proof. Using Lemma 2.5, we can represent a and b in the form
a = g α0 (u + iv)β0 ,
b = g α0 (u + iv)β0 ,
where g is a primitive root mod pn in Z and u + iv is the generating element of the group En . Then b; h, pn ; k) = K(a,
x,y (mod pn ) N (x)N (y)≡h (mod pn )
epn g α0 ((u + iv)β0 xk ) + g α0 ((u + iv)β0 y k ) .
(4.14)
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1333
Let h ≡ g α (modpn ). Then h ≡ ±g 2α0 (mod pn ), where
2α0 =
α
if
α
is even,
α + p − 1 pn−1 2
if
α
is odd.
= + . We split the sum over x (mod pn ) in (4.14) into two parts: 1 2 In , we take x (mod pn ) for which N (x) ≡ g 2α1 (mod pn ). In , we take x (mod pn ) for which 1 2 p − 1 n−1 N (x) ≡ −g 2α1 (mod pn ). In both cases, α1 runs through the values 0, 1, . . . , − 1. Thus, p 2 b; h, pn ; k) = K(a,
For x from
1
1
+
2
.
(4.15)
, we have x ≡ g α1 (u + iv)2β1 (mod pn ), 1 α1 = 0, 1, . . . , (p − 1)pn−1 − 1, 2
β1 = 0, 1, . . . , (p + 1)pn−1 − 1.
Hence,
(u + iv)β0 xk ≡ g kα1 (u + iv)2kβ1 +β0 (mod pn ). It follows from the condition N (x)N (y) ≡ h (mod pn ) that N (y) ≡ ±g 2α2 (mod pn ), where α2 = α0 + (p − 1)pn−1 − 1 α1 . Thus, 1
=
epn g α0 +α1 k (u + iv)2kβ1 +β0 + g α0 +α2 k (u + iv)2kβ2 +β0 +δk ,
(4.16)
(α1 ) (β1 ) (β2 )
1 where (α1 ) means that α1 runs through the values 0, 1, . . . , (p − 1)pn−1 − 1, and (βi ) means running through 2 the values 0, 1, . . . , (p + 1)pn−1 − 1, i = 1, 2; furthermore, δ = 0 if h ≡ g 2α0 (mod pn ) and δ = 1 if h ≡ −g 2α0 (mod pn ). Similarly, 2
=
(α1 ) (β1 ) (β2 )
epn g α0 +α1 k (u + iv)2kβ1 +β0 +1 + g α0 +α2 k (u + iv)2kβ2 +β0 +δk .
(4.17)
1334
S. P. VARBANETS
Again we have βi = (p + 1)ti + zi ,
ti (mod pn−1 ),
zi = 0, 1, . . . , p,
i = 1, 2.
Then kβi = 2(p + 1)kti + kzi ,
i = 1, 2.
It now follows from relations (4.12) and (4.13) and Lemma 2.1 that the sums over ti are equal to zero if the following congruences are violated: β0 + 2kz1 ≡ 0 (mod (p + 1)), β0 + 2kz2 + kδ ≡ 0 (mod (p + 1)) for a sum
1
, (4.18)
β0
+ 2kz1 + 1 ≡ 0 (mod (p + 1)),
β0 + 2kz2 + kδ ≡ 0 (mod (p + 1)) for a sum Consequently, one of the sums
1
and
2
2
.
is always equal to zero.
Congruences (4.18) can only be true for (k, p + 1)2 pairs (z1 , z2 ). Let B be the set of these (z1 , z2 ). Using (4.12)–(4.14). we obtain b; h, pn ; k) = K(a,
epn (N0 g α1 + M0 g α2 )
(i)
epn−2 F1 (kt1 )g α1 + F2 (kt2 )g α2 ,
(z1 ,z2 )∈B t1 ,t2 (mod pn−1 )
(α1 )
(i)
(i)
(i)
(i)
(i)
where Fi (t) = c1 t + c2 t2 + pλ3 c3 t3 + . . . + pλ c t , (c2 , p) = (c3 , p) = . . . = 1, λj > 0 for j ≥ 3, and (N0 , p) = (M0 , p) = 1. The sums over t1 and t2 are calculated in the same way. Let k = pm k1 , (k1 , p) = 1. We split the sum over ti into blocks of length pn−2−2m (if 2m < n − 2). Then, using Lemma 2.3, we obtain b; h, pn ; k) = pn+2m K(a,
epn (N1 g α1 + N2 g α2 ),
(4.19)
(α1 )
where (N1 , p) = (N2 , p) = 1. It follows from the definition of α2 that g α2 ≡ g α0 (g )α1 (mod pn ). The sum on the right-hand side of (4.19) is an incomplete Kloosterman sum. By a choice of a primitive root g, we get g p−1 = 1 + pu,
(u, p) = 1.
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
Then g
p−1
1335
= 1 − pu1 , (u1 , p) = 1, u ≡ u1 (mod p). We set α1 = (p − 1)t + z, 1 t = 0, 1, . . . , (pn−1 − 1), 2
z = 0, 1, . . . , p − 2.
Thus, g α1 = g z (1 + a1 pt + a2 p2 t2 + a3 pλ3 t3 + . . .) (mod pn ), a2 ≡ −2 u2 (mod p),
a1 ≡ −u1 ,
λj ≥ 3.
Similarly, g λ2 ≡ g α0 g
α1
≡ g α0 g (1 + b1 pt + b2 p2 t2 + b3 pµ3 t3 + . . .) (mod pn ), z
b2 ≡ −2 u2 (mod p),
b1 ≡ −u1 ,
µj ≥ 3.
Hence, N1 g α1 + N2 g α2 ≡ c0 + c1 pt + c2 p2 t2 + c3 pν3 t3 + . . .
(mod pn ),
where ci = g z ai N1 + g α0 g z bi N2 , i = 1, 2. Since (N1 , p) = (N2 , p) = 1, it is easy to see that the two congruences c1 ≡ 0 (mod p),
c2 ≡ 0 (mod p)
cannot be realized simultaneously. On the other hand, the relation c1 ≡ 0 (mod p) yields g 2z ≡ g α0 N2 N1 (mod p). This is possible for only one value of z. Denote this value by z0 . Thus, using (4.19), we get b; h, p ; k) = p K(a, n
p−2
1 n−1 (p −1) 2
z=0 z=z0
t=0
n+2m
0 2πi pn c1 t en−1 p c
e
1 n−1 (p −1) 2
+
t=0
where (c1 , p) = (c2 , p) = 1.
+ c2 pt2 + c3 pν3 −1 t3 + . . .
c0 2πi pn
e
epn−2 c1 t + c2 t2 + c3 pν3 −2 t3 + . . . , (4.20)
1336
S. P. VARBANETS
The sums over t are incomplete rational sums. We obtain their estimates by estimating complete exponent sums. For an arbitrary polynomial Φ(t) ∈ Z[t], we have q−1 q−1 q T Φ(t) Φ(t) Φ(t)−t 1 T 2πi q 2πi q 2πi q e − e e ≤ . q min(r, q − r + 1) t=0
t=0
r=1
(4.21)
t=0
If Φ(t) = c1 t + c2 pt2 + c3 pν3 −1 t3 + . . . , (c1 , p) = 1, q = pn−1 , then the complete sums in (4.21) are equal to zero for all r except r ≡ c1 (mod p). In this special case, we have Ψ(t) = c1 t + c2 t2 + c3 pν3 −1 t3 + . . . , n−2 (c2 , p) = 1, and q = pn−2 , and then the complete sum is estimated by the value 2p 2 . Hence, p−2 n n+m ; k) ≤ p K(a, b; h, p z=0 z=z0
1 + |c1 (z)|
pn r=1
n−2 1 n−2 p 2 + pp 2 . kp
Finally, taking into account that, for different values of z, we have different values of c1 (z) (mod p), we obtain 3 log pn n n+m 2 . log p + K(a, b; h, p ; k) ≤ p p If 2m > n − 2, then the assertion of the theorem is trivial. The theorem is proved. b; h, pn ; k) in the case where p ≡ We now pass to the estimation of the norm Kloosterman sum K(a, 1 (mod 4), k ≥ 2, and (a, p) = (b, p) = 1. For p ≡ 1 (mod 4), we have p = pp, where p and p are complex-conjugate Gaussian prime numbers. Then the reduced residue system mod pn can be written as follows: x = g 1 pn + g 2 pn ,
0 ≤ *1 ,
*2 ≤ (p − 1)pn−1 − 1,
where g is a primitive root mod pn such that g p−1 = 1 + pH,
H ∈ Z,
(H, p) = 1.
Thus, N (x) = xx = g 21 pn + g 2 pn + g 1 +2 p2n + g 1 +2 p2n ≡ g 1 +2 Sp(p2n ) (mod pn ). Therefore, if p = c + id, then (c, p) = (d, p) = 1, and, by induction, we easily obtain p2n ≡ cn + idn ,
n = 1, 2, . . . ,
(4.22)
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1337
where
cn ≡
dn ≡
n−1 · 2m · c · d2(m−1) (mod p), (−1) (−1)m · 2m+2 · d2m ,
2m−1 · 2m · d2m−1 (mod p) if n = 2m − 1, (−1) (−1)m−1 · 2m+2 · c · d2m−1 (mod p) if n = 2m.
Hence, for p ≡ 1 (mod 4), we have b; h, pn ; k) = K(a,
epn (A(g 1 k + g 2 k ) + B(g 1 k + g 2 k ))
(U )
=
epn A(xk1 + xk2 ) + B(y1k + y2k ) ,
(4.23)
(U )
where U := *1 , *2 , *1 , *2 (mod (p − 1)pn−1 ) g 1 +2 +1 +2 ≡ H (mod pn ) , U := x1 , x2 , y1 , y2 (mod pn ) x1 x2 y1 y2 ≡ H (mod pn ) , A, B, ∈ Z,
(A, p) = (B, p) = 1.
Theorem 4.4. Suppose that p ≡ 1 (mod 4) is a prime number, a, b ∈ Z[i], and (a, p) = (b, p) = 1. Then 3 d2 p 2 K(a, b; h, p; k) d4 p2
if (d − 1)4 < p, if (d − 1)4 ≥ p,
where d = (k, p − 1). Proof. Without loss of generality, we can assume that a, b ∈ Z. By analogy with the case p ≡ 3 (mod 4), using (4.23), we obtain b; h, p; k) = K(a,
ep A(xd1 + xd2 ) + B(y1d + y2d ) .
∗
x2 ,x2 ,y1 ,y2 ∈Fp x1 ,x2 ,y1 ,y2 ≡H1k
1338
S. P. VARBANETS
For (d − 1)4 < p, by analogy with the case p ≡ 3 (mod 4), we obtain
χ(h)K(α, β; h, p; d) =
h∈Fp∗
2 χ(x)ep (Axd )
x∈Fp∗
2 χ(y)ep (By d ) .
y∈Fp∗
Hence, 2 k(α, β; h, p; d) ≤ (d − 1)4 p4
if (d − 1)4 < p.
Then b; h, p; k) d2 p 32 K(a,
if (d − 1)4 < p.
Assume that (d − 1)4 ≥ p. Let g denote a primitive element of the field Fp and let x = g ind x for x ∈ F∗p . Let G be the group of multiplicative characters of Fp . For χ ∈ G, we have χ(x) = ep−1 (ν ind x) with some ν ∈ Fp . Then, using the arguments of Theorem 4.1, we can routinely obtain the following relation: b; h, p; d) = K(a,
1 χ(H) p−1 χ∈G
d−1 s1 ,...,s4 =0
×
χ(A2 B 2 )ed (s1 + s2 ) ind A + (s3 + s4 ) ind B
ed (s1 ind x1 + . . . + s4 ind x4 )χ(x1 , . . . , x4 )ep (x1 + . . . + x4 )
∗ x1 ,...,x4 ∈Fp
=
1 p−1
d−1
ep−1 (ν ind H)ep−1 (F1 (ν, s))
ν∈Fp s1 ,...,s4 =0
×
ep−1 (F2 (ν, s, x))ep (x1 + . . . + x4 ), ∗
x1 ,...,x4 ∈Fp
where F1 (ν, s) :=
p−1 2ν + (s1 + s2 ) d
F2 (ν, s, x) :=
p−1 ind A + 2ν + (s3 + s4 ) d
ind B,
p−1 p−1 s1 + ν ind x1 + . . . + s4 + ν ind x4 . d d
The last sum over x1 , . . . , x4 is the product of the Gauss sums of the field Fp . Hence, K(a, b; h, p; k) ≤ d4 p2 . The theorem is proved.
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
1339
For n ≥ 2, we can use the description of a solution of the congruence x1 , x2 , x3 , x4 ≡ H (mod pn ): xi = yi + pm zi ,
yi (mod pm ),
zi (mod pn−m ),
(yi , p) = 1,
i = 1, 2, 3,
m=
n+1 , 2
x4 = Hy1 y2 y3 1 − pm y1 z1 − pm y2 z2 − pm y3 z3 ,
(4.24)
yi yi ≡ 1 (mod pm ).
Theorem 4.5. Suppose that p ≡ 1 (mod 4) is a prime number, n ∈ N, n ≥ 2, h ∈ Z, (h, p) = 1, a, b ∈ Z[i], and (a, p) = (b, p) = 1. Then 3 d4 p 2 n K(a, b; h, pn ; k) 4 n+m d p
if (d − 1)4 < p, if (d − 1)4 ≥ p,
where n+1 . m= 2
Proof. By virtue of (4.23) and (4.24), we have
b; h, pn ; k) = K(a,
epn f (y1 , y2 , y3 )
y1 ,y2 ,y3 ∈R∗ (pm )
epn−m (F (z1 , z2 , z3 )),
z1 ,z2 ,z3 (mod pn−m )
where f (y1 , y2 , y3 ) = Ay1k + ay2k + By3k + BHy1 y2 y3 , k
F (z1 , z2 , z3 ) = k
Ay1k−1 − By1 +
k+1 k k y2 y 3
Ay2k−1
−
k
k
z1
B(y1k+1 y2k y3k )
z2 +
Ay3k−1
−
Let (k, pn−m ) = p . Then relation (4.25) yields b; h, pn ; k) = p3(n−m) K(a,
(U )
where
epn (f (y1 , y2 , y3 )),
B(y1k y2k y3k+1 )
z3 .
(4.25)
1340
S. P. VARBANETS
U := y1 , y2 , y3 (mod pm ) (yi , p) = 1, i = 1, 2, 3; y1k ≡ y2 k ≡ y3 k (mod pn−m− ), y14k ≡ BA (mod pn−m− ) . We now estimate the sum
(U )
by the number triples (y1 , y2 , y3 ) ∈ U for n = 2m and take into account
Theorem 2.4 for n = 2m − 1. Finally, we get 4 3n d p 2 K(a, b; h, pn ; k) 4 n+m d p
if
(d − 1)4 < p,
if
(d − 1)4 ≥ p.
The theorem is proved. Combining the estimates established in Theorems 4.2–4.5, we get the following theorem: Theorem 4.6. Suppose that α, β ∈ Z[i], h, q, k, n ∈ N, k ≥ 2, and (k, q) = (h, q) = 1. Then, for (α, q) = (β, q) = 1, we have 3 K(α, β; h, q; k) D(k, q)q 2 ,
where D(k, q) =
p|q p≡1(q)
d6 (k, p)
d3 (k, p) log pn ,
pn q p≡3(q)
d(k, p) = (k, p − 1). We must note that the norm Kloosterman sum K(α, β; h, q; k) does not have analogs in the ring Z. REFERENCES 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
H. D. Kloosterman, “On the representation of numbers in the form ax2 + by 2 + cz 2 + dt2 ,” Acta Math., 49, 407–464 (1926). H. Davenport, “On certain exponential sums,” I. Reine Angew. Math., 169, 158–176 (1933). A. Weil, “On some exponential sums,” Proc. Nat. Acad. Sci. USA, 34, 204–207 (1948). N. V. Kuznetsov, Petersson’s Conjecture for Cusp Forms of Weight Zero and Linnik’s Conjecture [in Russian], Preprint 2, Khabarovsk (1977). N. V. Kuznetsov, “Petersson’s conjecture for cusp forms of weight zero and Linnik’s conjecture. Sums of Kloosterman sums,” Mat. Sb., 3, 334–383 (1980). R. W. Bruggeman, “Fourier coefficients of cusp forms,” Invent. Math., 45, 1–18 (1978). J.-M. Deshouillers and H. Iwaniec, “Kloosterman sums and Fourier coefficients of cusp forms,” Invent. Math., 70, 219–288 (1982). N. V. Proskurin, On General Kloosterman Sums [in Russian], Preprint, Leningrad (1980). Yuan Yi and Wenpeng Zhang, “On the generalization of a problem of D. H. Lehmer,” Kyushu J. Math., 56, 235–241 (2002). S. Kanemitsu, Y. Tanigawa, Yuan Yi, and Wenpeng Zhang, “On general Kloosterman sums,” Ann. Univ. Sci. Budapest. Sec. Comp., 22, 151–160 (2003). L. Y. Mordell, “On a special polynomial congruence and exponential sums,” Calcutta Math. Soc. Golden Iub., Pt. 1, 29–32 (1963).
G ENERAL K LOOSTERMAN S UMS OVER THE R ING OF G AUSSIAN I NTEGERS
12. 13. 14. 15.
1341
B. Dwork, “On the zeta function of a hypersurface. III,” Ann. Math., 83, 457–579 (1966). P. Deligne, “La conjecture de Weil. I, II,” Publ. Math. IHES, 43, 273–307 (1974); 52, 137–252 (1980). E. Bombieri, “On exponential sums in finite fields. II,” Invent. Math., 47, Fasc. 1, 29–39 (1978). ´ P. Deligne, “Applications de la formula des traces aux sommes trigonom´etriques,” in Cohomologie Etale: Lect. Notes Math., Springer, Berlin (1977), pp. 168–232. 16. U. B. Zanbyrbaeva, Asymptotic Problems of Number Theory in Sectorial Regions [in Russian], Thesis, Odessa (1993). 17. G. I. Perel’muter, “On some sums with characters,” Usp. Mat. Nauk, 18, No. 2, 143–149 (1963). 18. K. S. Williams, “A class of character sums,” J. London Math. Soc., 3, No. 2, 61–72 (1971).
