General Kloosterman sums over ring of Gaussian integers
The general Kloosterman sum $K(m, n; k; q)$ over $\mathbb{Z}$ was studied by $S$. Kanemitsu, Y. Tanigawa, Yi. Yuan, Zhang Wenpeng in their research of problem of D. H. Lehmer. In this paper, we obtain the similar estimations of $K(\alpha, \beta; k; \gamma)$ over $\mathbb{Z}[i]$. We also consider t...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860509462294429696 |
|---|---|
| author | Varbanets, S. P. Варбанець, С. П. |
| author_facet | Varbanets, S. P. Варбанець, С. П. |
| author_sort | Varbanets, S. P. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-03-18T19:52:51Z |
| description | The general Kloosterman sum $K(m, n; k; q)$ over $\mathbb{Z}$ was studied by $S$. Kanemitsu, Y. Tanigawa, Yi. Yuan, Zhang Wenpeng in their research of problem of D. H. Lehmer.
In this paper, we obtain the similar estimations of $K(\alpha, \beta; k; \gamma)$ over $\mathbb{Z}[i]$.
We also consider the sum $\widetilde{K}(\alpha, \beta; h, q; k)$ which has not an analogue in the ring $\mathbb{Z}$ but it can be used for the inversigation of the second moment of the Hecke zeta-fonction of field $\mathbb{Q}(i)$. |
| first_indexed | 2026-03-24T02:41:29Z |
| format | Article |
| fulltext |
UDC 511.19
S. P. Varbanets (Odessa I. I. Mechnikov Nat. Univ.)
GENERAL KLOOSTERMAN SUMS
OVER RING OF GAUSSIAN INTEGERS
УЗАГАЛЬНЕНI СУМИ КЛОСТЕРМАНА
НАД КIЛЬЦЕМ ЦIЛИХ ГАУССОВИХ ЧИСЕЛ
The general Kloosterman sum K(m, n; k; q) over Z was studied by S. Kanemitsu, Y. Tanigawa, Yi. Yuan,
Zhang Wenpeng in their research of problem of D. H. Lehmer. In this paper, we obtain the similar
estimations of K(α, β; k; γ) over Z[i]. We also consider the sum K̃(α, β; h, q; k) which has not an
analogue in the ring Z but it can be used for the inversigation of the second moment of the Hecke
zeta-function of field Q(i).
Узагальнену суму Клостермана K(m, n; k; q) над Z вивчали S. Kanemitsu, Y. Tanigawa, Yi. Yuan,
Zhang Wenpeng в їх дослiдженнi проблеми D. H. Lehmer. У цiй статтi отримано подiбнi оцiнки
K(α, β; k; γ) над Z[i]. Також розглянуто суму K̃(α, β; h, q; k), що не має аналога в кiльцi Z, але
може бути використана при дослiдженнi другого моменту дзета-функцiї Геке поля Q(i).
1. Introduction. The classic 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:
K(a, b; q) :=
q∑
x=1
(x,q)=1
e2πi
ax+bx′
q , a, b ∈ Z, q > 1 is natural,
here and in seque x′ denote the reciprocal to x modulo q, i.e., xx′ ≡ 1 (mod q). By the
relation for q = q1q2, (q1, q2) = 1,
K(a, b; q) = K(aq′2, bq
′
2; q1)K(aq′1, bq
′
1; q2)
follows that suffices to obtain the estimations K(a, b; q) only for a case q = pn, p be a
prime, n ∈ N.
The greatest difficultly in an estimation of the Kloosterman sums provides the case
q = p. The estimation K(a, b; p) � p
3
4 under a condition (a, b, p) = 1 was obtained in
the named work of Kloosterman, and then Davenport [2] improved on it up to � p
2
3 .
A. Weill [3] proved the Riemann hypothesis for algebraic curves of over finite field and
obtained the best possible estimation � p
1
2 .
Davenport [2] studies the general Kloosterman sums over finite field with the multi-
plicative character ψ of this field
Kψ(a, b; p) =
∑
x∈F∗p
ψ(x)e2πi
ax+bx′
p .
The further generalization of the Kloosterman sums concerned with a substitution
of a prime field Fp on it a finite expansion Fq, q = pn, n ∈ N. The generalization of
the Kloosterman sums concerned with theory of modular forms studies in the works
Kuznetsov [4, 5], Bruggeman [6], Deshoillers, Iwaniec [7], Proskurin [8]. In last years
in connection with the investigation of the D. H. Lemer problem was studied others
generalizations of the Kloosterman sums (see [9, 10]):
K(a, b; q, k, ψ) =
∑
x∈R∗(q)
ψ(x)e2πi
axk+bx′ k
q , xx′ ≡ 1 (mod q),
where ψ is a multiplicative character modulo q.
c© S. P. VARBANETS, 2007
ISSN 1027-3190. Укр. мат. журн., 2007, т. 59, № 9 1179
1180 S. P. VARBANETS
The multiple Kloosterman sums introduced Mordell [11]:
K(a1, . . . , an; q) =
∑
x1,...,xn∈F∗q
x1...xn=1
e2πi
σm(a1x1+...+anxn)
p ,
where a1, . . . , an ∈ F∗q , q = pm, σm(c) is a trace from Fq into Fp.
The multiple Kloosterman sums are a particular case of the trigonometric sums on an
algebraic variate over a finite field. By virtue of the investigations Dwork [12] (which
has proved a rationality of the zeta-function of an algebraic variate over finite field),
Deligne [13] (which has proved the Riemann hypothesis for an algebraic variate over
Fq) and Bombieri [14] (which has estimated in terms of a generative polynomial the
number of characteristic roots of the zeta-function) was obtained the final estimation
(see Deligne [15], Bombieri [14])
K(a1, . . . , an; q) ≤ nq
n−1
2 .
In this paper we obtain the estimations of general Kloosterman sums over the ring
of the Gaussian integers.
Notations. We denote Z[i] the ring of the Gaussian integers
Z[i] :=
{
a+ bi
∣∣a, b ∈ Z, i2 = −1
}
.
For the designation of the Gaussian integers we shall use the Greek letters α, β, γ, ξ, η; a
Gaussian prime number denote through p if p 6∈ Z. For α ∈ Z[i] we put Sp(α) = α+α,
N(α) = αα, where α denotes a complex conjugate with α; Sp(α) and N(α) we name a
trace and a norm (accordingly) of α from Q(i) into Q. Fq denotes a field which contain
just q an element, q = pn, n ∈ N.
For x ∈ Fq we denote through σn(x) a trace x from Fq into Fp, i.e.,
σn(x) := x+ xp + . . .+ xp
n−1
, σ1(x) = σ(x) = x.
The writing a ∈ R(q) (accordingly, a ∈ R(q, i)) denotes that a ∈ Z (accordingly,
a ∈ Z[i]) and a runs a complete residue system modulo q. Analogous, a ∈ R∗(q)
(accordingly, a ∈ R∗(q, i)) denotes a ∈ Z (accordingly a ∈ Z[i]) and runs a reduced
residue system modulo q.
The writing
∑
(U)
denotes that the summation runs over the region U which describe
extra. Moreover, exp (z) = ez, eq(z) = e2πi
z
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, γ we define the Kloosterman sum
K(α, β; γ) =
∑
x∈R∗(γ,i)
exp
(
πiSp
αx+ βx′
γ
)
.
Zanbyrbaeva [16] obtained the estimation
K(α, β; γ) � 2ν(γ)N(γ)
1
2N((α, β, γ))
1
2 ,
where ν(γ) is the number distinct prime divisors of γ; (α, β, γ) denotes the greatest
common divisor of α, β, γ.
We consider two type of general Kloosterman sums over Z[i]
K(α, β; k; γ, ψ) =
∑
x∈R∗(γ,i)
ψ(x) exp
(
πiSp
αxk + βx′
k
γ
)
,
where α, β, γ ∈ Z[i], ψ is multiplicative character modulo γ,
ISSN 1027-3190. Укр. мат. журн., 2007, т. 59, № 9
GENERAL KLOOSTERMAN SUMS OVER RING OF GAUSSIAN INTEGERS 1181
K̃(α, β;h, q; k) =
∑
x,y∈R∗(q,i)
N(xy)≡h (mod q)
eq
(
1
2
Sp(αxk + βyk)
)
,
where α, β ∈ Z[i], h, q ∈ N, (h, q) = 1.
We call K(α, β; k; γ, ψ) the general power Kloosterman sum and K̃(α, β;h, q; k)
call the norm Kloosterman sum.
Our aim is to obtain non trivial estimations for K(α, β; k; γ;ψ) and K̃(α, β;h, q; k).
2. Auxiliary results. For the proofs of our main results some Lemmas are need.
Lemma 2.1. Let p be a Gaussian prime “odd" number,
α1, . . . , αk ∈ Z[i], (α2, p) = . . . = (αk, p) = 1; ν3, ν4, . . . , νk ≥ 2,
are natural numbers.
Then for every natural n ≥ 2 we have∣∣∣∣∣∣
∑
ξ(mod pn)
exp
(
2πiSp
(
α1ξ + α2pξ
2 + α3p
ν3ξ3 + . . .+ αkp
νkξk
pn
))∣∣∣∣∣∣ =
=
{
0 if (α1, p) = 1),
N(p)
n+1
2 if α1 ≡ 0 (mod p).
(2.1)
Lemma 2.2. Let p = 1 + i be a Gaussian “even” number and let αj ∈ Z[i],
j = 1, 2, . . . , k; (α2, p) = . . . = (αk, p) = 1.
Then for any natural numbers νj ≥ 2, j = 2, 3, . . . , k, and any n ≥ 2 the following
estimate:∣∣∣∣∣∣
∑
ξ(mod pn)
exp
(
2πiSp
(
α1ξ + α2pξ
2 + α3p
ν3ξ3 + . . .+ αkp
νkξk
pn
))∣∣∣∣∣∣ ≤ δ · 2n+1,
holds, where
δ =
{
0 if α1 6≡ 0 (mod p2),
2 if α1 ≡ 0 (mod p2).
The assertion of these lemmas are the consequences of the estimates of complete
linear sum and Gauss’sum to which we can reduced the primary sums.
Lemma 2.3. Let p be a prime number, A ∈ Z, (A, p) = 1, f(x) ∈ Z[x],
f(x) = a1x+ a2x
2 + pλ3a3x
3 + . . .+ pλkakx
k,
(ai, p) = 1, i = 2, 3, . . . , k; λj > 0, j = 3, . . . , k.
Then for any n ∈ N the equality
S :=
∑
x(mod pn)
epn(Af(x)) = ε(n)p
n
2 epn
(
AF (a1, . . . , an)
)
holds, where F (x1, . . . , xn) ∈ Z[x1, . . . , xn],
ε(n) =
1 if n is even,(
A
p
)
· (i)(
p−1
2 )2
if n is odd,
(
A
p
)
is a symbol of Legendry.
Proof. We set x = y + pn−1z, y (mod pn−1), z (mod p). Then we have∑
x (mod pn)
epn(Af(x)) =
∑
y (mod pn−1)
∑
z (mod p)
epn
(
A(f(y) + pn−1zf ′(y)
)
.
ISSN 1027-3190. Укр. мат. журн., 2007, т. 59, № 9
1182 S. P. VARBANETS
The sum over z gives zero if f ′(y) 6≡ 0 (mod p).But we have f ′(y) ≡ a1 + 2a2y (mod p).
Let y0 be a root of congruence a1 + 2a2y ≡ 0 (mod p). Then
S = epn(Af(y0))
∑
y (mod pn−1)
epn
(
A(f(y0 + py)− f(y0))
)
=
= epn
(
Af(y0)
) ∑
y (mod pn−1)
epn−2
(
Ag(y)
)
=
= pepn
(
Af(y0)
) ∑
y (mod pn−2)
epn−2
(
Ag(y)
)
,
where
g(y) =
f(y0 + py)− f(y0)
p2
= b1y + b2y
2 + pµ3b3y
3 + . . .+ pµkbky
k,
moreover b1, . . . , bk are linear functions of a1, . . . , ak with the coefficients which
depends on y0, and b2 ≡ a2 (mod p), (bj , p) = 1, µj ≥ 1, j = 3, . . . , k. Thus g(y) is a
polynomial such sort as f(y).
These consideration we continue further. Then for n ≡ 1 (mod 2) we obtain
S = p
n
2 e
2πiA
[
f(y0)
pn +
g(y1)
pn−2 +...
]
,
and for n is even
S = p
n−1
2 e
2πiA
[
f(y0)
pn +
g(y1)
pn−2 +...
] ∑
x (mod p)
e(A(cx+a2x
2))
p =
=
(
A
p
)
i(
p−1
2 )2e
2πiA
[
f(y0)
pn +
g(y1)
pn−2 +...− (2′c)2
p
]
.
The lemma is proved.
Lemma 2.4. Let p be a prime number, p ≡ 3 (mod 4) and let E` be a set of
residue classes mod p` of the ring Z[i], which has norms congruous modulo p` with ±1.
Then E` is a cyclical group of order 2(p+ 1)p`−1.
Proof. From an equality N(αβ) = N(α)N(β) follows that E` is a subgroup of the
group of residue classes modulo p` in Z[i].
At first let ` = 1. Then the residue classes modulo p` organizes a field Fp2 . Let g0
be a generative element of multiplicative group of this field. We denote
gu0 = x(u) + iy(u), (2.2)
where x(u), y(u) ∈ Fp and i is an element of field Fp2 such that i2 = −1. The residue
classes mod p for which norms ≡ ±1 (mod p) be characterized by a condition
x2 + y2 = ±1 (in Fp).
Now from (2.2) we have
gpu0 = x(u)− iy(u).
Hence an element gu0 has a norm ≡ ±1 (mod p) iff g(p+1)u
0 = ±1, i.e., iff
p− 1
2
|u .
Denote u =
p− 1
2
t, t = 0, 1, . . . , 2p + 1, and set g
p−1
2
0 = g. The classes gt,
t = 0, 1, . . . , 2p+ 1, are just those and only those which have a norm ≡ ±1 (mod p).
Let f = g + pλ, λ ∈ Z[i]. Then
fp = gp + pλ1, λ1 ∈ Z[i],
ISSN 1027-3190. Укр. мат. журн., 2007, т. 59, № 9
GENERAL KLOOSTERMAN SUMS OVER RING OF GAUSSIAN INTEGERS 1183
fp+1 ≡ gp+1 + pgpλ1 (mod p2),
f2(p+1) ≡ g2(p+1) + 2pgp+1λ1 (mod p2).
Let g2(p+1) = 1 + pg1. We have
f2(p+1) − 1 ≡ p(g1 + 2gp+1λ1) (mod p2).
Always we can take λ so
f2(p+1) = 1 + ph, h ∈ Z[i], (h, p) = 1.
Thus we can account that a generative element g of the group E` had selected so
g2(p+1) − 1 6≡ 0 (mod p2). Now we easily get
g2(p+1)p`−1
≡ 1 (mod p`), gk 6≡ 1 (mod p`), 0 < k < 2(p+ 1)p`−1,
for every ` = 1, 2, . . . . We must show also that for every ` = 1, 2, . . . there exists g`
such that g` ≡ g (mod p`) and N(g`) ≡ −1 (mod p`).
For ` = 1 we proved already.
Let ` = 2. If g = x+ iy then
x2 + y2 = −1 + λp,
g2(p+1) = 1 + p(h1 + ih2), (h1 + ih2, p) = 1, h1, h2 ∈ Z.
We have for k = k1 + ik2, k1, k2 ∈ Z:
N(x+ iy + p(k1 + ik2)) = −1 + λp+ p(2xk1 + 2yk2) + p2(k2
1 + k2
2) ≡
≡ −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)pλ+ 2xk1 + 2yk2 ≡ 0 (mod p2),
here h(1) and α are the Gaussian integers and co-prome numbers with p.
Next, the congruence −h(1) ≡ α(k1 + ik2) (mod p) holds only for one assembly of
(k0
1, k
0
2) by modulo p. Therefore, if we take k1 6≡ k0
1 (mod p) and define k2 from the
congruence
λ+ 2xk1 + 2yk2 ≡ 0 (mod p2),
then we obtain that f = x+ iy + p(k1 + ik2) has a norm ≡ −1 (mod p2). Moreover, f
belongs to an exponent 2(p+1)p by modulo p2 and f2(p+1)p = 1+Hp2, (H, p) = 1, i.e.,
f ∈ E2 and f belongs to an exponent 2(p+1)p`−1 by modulo p` for every ` = 2, 3, . . . .
Now we note that the g3 = f + p2(m1 + im2) satisfies by the condition g2(p+1)p
3 =
= 1 +H1p
2, (H1, p) = 1, for any m1, m2 ∈ Z. We take m1, m1 ∈ Z, such that
λ2 + 2f1m1 + 2f2m2 ≡ 0 (mod p),
where λ =
N(f)− (−1)
p2
=
N(f) + 1
p2
, f = f1 + if2.
Then g3 = f + p2(m1 + im2) is a generative element of the group E3.
Next, by induction. If we defined already g`−1 then a generative element of E` will
be
ISSN 1027-3190. Укр. мат. журн., 2007, т. 59, № 9
1184 S. P. VARBANETS
g` = g`−1 + p`−1(m1 + im2),
where m1, m2 define from a congruence
λ`−1 + 2g′`−1m1 + 2g′′`−1m2 ≡ 0 (mod p)(
here 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 prime number, p ≡ 3 (mod 4), ` ∈ N. Then every residue
x+ iy a reduced residue system mod p` of the ring of the Gaussian integers has unique
representation in form
x+ iy ≡ gc(u+ iv)d (mod p`),
c = 0, 1, . . . , (p− 1)p`−1 − 1, d = 0, 1, . . . , (p+ 1)p`−1 − 1, (2.3)
where g is a primitive root modulo p` in Z, u+ iv is a generative element of E`.
Proof. Let ϕ(α) denote the Euler function on Z[i]. Then for p ≡ 3 (mod 4) we have
ϕ(p`) = N(p`)
(
1− 1
N(p)
)
= p2(`−1)(p2 − 1).
In the relation (2.3) we have p2(`−1)(p2− 1) the formally distinguishable expressions of
form gc(u+ iv)d. As for any c and d we have (gc(u+ iv)d, p) = 1 then for the proof of
the assertion of lemma sufficiently to show that the expression (2.3) are pairwise disjoint
mod p` for different assemblies of (c, d).
Let us assume
gc1(u+ iv)d1 ≡ gc2(u+ iv)d2 (mod p`), c1 ≥ c2.
Then we have
gc1−c2 ≡ (u+ iv)d2−d1 (mod p`) if d2 ≥ d1
or
gc1−c2(u+ iv)d1−d2 ≡ 1 (mod p`) if d2 < d1.
And now take account that the sets {g`} and
{
(u+ iv)d
}
has only one common element
(it is 1) modulo p` we obtain all once c1 = c2, d1 = d2.
The lemma is proved.
Corollary. All reduced classes x+ iy modulo p`, p ≡ 3 (mod 4) which has equal
norms modulo p` we can write in form
x+ iy ≡ gc(u+ iv)2d,
d = 0, 1, . . . , p`−1(p+ 1)− 1 if N(x+ iy) ≡ g2c (mod p`),
(x+ iy) ≡ gc(u+ iv)2d+1,
d = 0, 1, . . . , p`−1(p+ 1)− 1 if N(x+ iy) ≡ −g2c (mod p`)(
here 0 ≤ c ≤ p− 1
2
p`−1 − 1
)
.
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 the complex-conjugate Gaussian prime numbers (p 6= ±p, ±ip).
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`, p = 1 + i in Z[i].
ISSN 1027-3190. Укр. мат. журн., 2007, т. 59, № 9
GENERAL KLOOSTERMAN SUMS OVER RING OF GAUSSIAN INTEGERS 1185
3. General Kloosterman sum K(α, β; k; γ). We consider the sum K(α, β; k; γ)
defining in Introduction for the trivial character ψ0:
K(α, β; k; γ, ψ0) = K(α, β; k; γ) =
∑
(U)
exp
(
πiSp
αxk + βx′
k
γ
)
, (3.1)
where U =
{
(x, x′) ∈ Z[i]2
∣∣x, x′ (mod γ), xx′ ≡ 1 (mod γ)
}
. Obviously, we have
K(α, β; k; γ) = K(αγ′2, βγ
′
2; k; γ1)K(αγ′1, βγ
′
1; k; γ2) if γ = γ1γ2, (γ1, γ2),
where γ1γ
′
1 ≡ 1 (mod γ2), γ2γ
′
2 ≡ 1 (mod γ1).
Thus we can therefore assume, without loss of generality, that γ = pn, p is a Gaussian
prime number.
In part 1 we had obtained a description of a reduced residue system mod pn, p = p ≡
≡ 3 (mod 4). For p ∈ Z[i], N(p) = p ≡ 1 (mod 4), a reduced residue system mod pn
has a form {
a ∈ Z
∣∣1 ≥ a ≥ pn − 1, (a, p) = 1
}
,
and for 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))
1
2N(p)
1
2 . (3.2)
Proof. If (α, β, p) = p then our assertion is clear. Let (α, β, p) = 1. By a description
of a reduced residue system mod pn we can suppose that α = a, β = b, a, b ∈ Z,
p = c1 + ic2, c1, c2 ∈ Z, (c1, p) = (c2, p) = 1.
Thus we have
K(α, β; k; p) =
∑
u∈R∗(p)
ep
(
1
2
Sp
(
a(c1 − ic2)uk + b(c1 − ic2)u′
k
))
=
=
∑
u∈R∗(p)
ep
(
ac1u
k + bc1u
′k).
The last sum was estimated in [10] but we shall give a calculation in order to make more
pricise an estimation.
We define
Ik(a) := #
{
x ∈ Z
∣∣0 ≥ x ≥ p− 1, xk ≡ a (mod p)
}
.
It is clear that
Ik(a) =
d if d
∣∣ind a
0 otherwise
=
d−1∑
t=0
e2πi
t ind a
d
.
(Here ind a denote an index of integer a, (a, p) = 1, by a radix of some primitive root
modulo p.)
Then we obtain
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1186 S. P. VARBANETS
|K(α, β; k; p)| =
∣∣∣∣∣∣
∑
u∈R∗(p)
Ik(u)ep(au+ bu′)
∣∣∣∣∣∣ ≤
≤
d−1∑
t=0
∣∣∣∣∣∣
∑
u∈R∗(p)
ed(t indu)ep(au+ bu′)
∣∣∣∣∣∣ ≤ 2dp
1
2 . (3.3)
Here we take into account that an inner sum is classical Kloosterman sum weighting by
a character and hence estimates as 2p
1
2 (see Perel’muter [17], Williams [18]).
The theorem is proved.
Theorem 3.2. Let p ≡ 3 (mod 4), k ∈ N, d = (k, p2 − 1). Then∣∣K(α, β; k; p)
∣∣ ≤ 2dN
(
(α, β, p)
) 1
2 N(p)
1
2 . (3.4)
Proof. The residue classes mod p in the ring Z[i] organizes a field Fp2 . Hence,
σ2(x) = x+ xp. But we observed 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) =
∑
x∈R∗(p)
ep
(
1
2
Sp(αxk + βxk)
)
=
∑
x∈F∗
p2
ep
(
σ2
(
2′αxk + 2′βx′k
))
,
where 2 · 2′ ≡ 1 (mod p).
Let g denote a primitive element of the field Fp2 , indg x = indx for x ∈ Fp2 and
let I(u) is a number solutions of equation xk = u in Fp2 . It follows that
|K(α, β; k; p)| =
∣∣∣∣∣∣∣
d−1∑
t=0
∑
x∈F∗
p2
e2πi
t ind x
d ep
(
σ2(2′αx+ 2′βx′)
)∣∣∣∣∣∣∣ ≤
≤
∣∣∣∣∣∣∣
∑
x∈F∗
p2
ep
(
σ2(2′αx+ 2′βx′)
)∣∣∣∣∣∣∣+
∣∣∣∣∣∣∣
d−1∑
t=1
∑
x∈F∗
p2
ψt(x)ep
(
σ2(2′αx+ 2′βx′)
)∣∣∣∣∣∣∣ ,
where ψt(x) = ed(t indx) is a multiplicative character of the field Fp2 .
Again using the estimations of the Kloosterman sums with a character of a finite
field we obtain finally ∣∣K(α, β; k; p)
∣∣ ≤ 2dN
(
(α, β, p)
) 1
2N(p)
1
2 .
The theorem is proved.
For p = 1 + i we have trivially
∣∣K(α, β; k; p)
∣∣ = 1.
Now for γ = pn we make a substitute x (mod pn) = y + pnz, where y (mod pn),
z (mod pn−m), m =
[
n+ 1
2
]
, and then using the standard technique, we easily obtain
the following theorem.
Theorem 3.3. Let γ = (1 + i)n0
∏s
i=1
N(pi)≡1(4)
pni
i
∏t
j=1
pj≡3(4)
p
nj
j . Then
∣∣K(α, β; k; γ)
∣∣ ≤ 2D
√
N((α, β, γ))N(γ)
1
2 , (3.5)
where D =
∏s
i=1
(k, pj − 1)
∏t
j=1
(k, p2
j − 1).
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GENERAL KLOOSTERMAN SUMS OVER RING OF GAUSSIAN INTEGERS 1187
Now we consider a nontrivial multiplicative character ψ of the field Fq, q = pr,
r ∈ N, p be a prime number, α, β ∈ Fq and α 6= 0 or β 6= 0. We define the general
power Kloosterman sum with a character ψ
K(α, β; k; q, ψ) :=
∑
x∈F∗q
ψ(x)ep
(
σ2(αxk + βx′
k)
)
. (3.6)
Let d = (k, q − 1), ψ(x) = e2πi
h ind x
q−1 , where indx take in regard to a some primitive
element for Fq. We have two probable cases: d 6
∣∣h and d
∣∣h.
We shall prove that K(α, β; k; q, ψ) = 0 in first case. We have for β 6= 0:∑
α∈Fq
∣∣K(α, β; k; q, ψ)
∣∣2 =
=
∑
α∈Fq
∑
x,y∈F∗q
xx′=yy′=1
ψ(x)ψ(y′)ep
(
σ2(α(xk − yk) + β(x′k − y′
k))
)
=
=
∑
x∈F∗q
ψ(x)
∑
y∈F∗q
ep
(
σ2(βy′
k(xk − 1))
) ∑
α∈Fq
ep
(
σ2(αyk(xk − 1))
)
=
= q
∑
y∈F∗q
∑
x∈F∗q
xk=1
ψ(x)ep
(
σ2(βy′
k(xk − 1))
)
= q(q − 1)
∑
x∈F∗q
xk=1
ψ(x). (3.7)
In the last sum the summation runs over x ∈ F∗q for which k indx ≡ 0 (mod (q − 1)),
i.e., indx =
q − 1
d
s, s = 0, 1, . . . , d− 1, and thus
∑
α
∣∣K(α, β; k; q, ψ)
∣∣2 = q(q − 1)
d−1∑
s=0
e2πi
hs
d = 0 if h 6≡ 0 (mod d).
If d
∣∣h we have ψ(x) = eq−1(h1d indx) = eq−1(h1 indxd). Hence, setting k1 =
k
d
,
h1 =
h
d
, ψd1 = ψ, we obtain
K(α, β; k; q, ψ) =
∑
x∈F∗q
ψ1(xd)ep
(
σ2(α(xd)k1 + β(x′d)k1)
)
=
=
∑
x∈F∗q
Id(x)ψ1(x)ep
(
σ2(αxk1 + βx′
k1)
)
=
=
d−1∑
s=0
∑
x∈F∗q
ed(s indx)eq−1(h1 indx)ep
(
σ2(αxk1 + βx′
k1)
)
=
=
d−1∑
s=0
∑
x∈F∗q
ψ2(x)ep
(
σ2(αxk1 + βx′
k1)
)
, (3.8)
where ψ2(x) = eq−1(h2 indx), h2 =
s(q − 1) + h
d
.
So then we diminished the exponent k in d-time if d > 1. But if d = 1 then clearly
that
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1188 S. P. VARBANETS
Kk(α, β; q) =
∑
x∈F∗q
eq−1
(
hk′ indxk
)
ep
(
σ2(αxk1 + βx′
k1)
)
=
=
∑
x∈F∗q
eq−1(hk′ indx)ep
(
σ2(αx+ βx′)
)
= K1(α, β; q;ψ3),
where kk′ ≡ 1 (mod (q − 1)), ψ3 = ψk
′
.
The sum K1(α, β; q, ψ3) is the Kloosterman sum over Fq weighting by a multi-
plicative character ψ3 of the field Fq and has a estimation as 2q
1
2 if β 6= 0 (see
Perel’muter [17]). The relation (3.8) show that if (k1, q − 1) = 1 then∣∣Kk(α, β; q, ψ)
∣∣ ≤ 2dq
1
2 .
If (k1, q − 1) = d1 > 1 we again consider two cases
(h1, d1) = d1 or (h2, d1) < d1.
But if (h2, d1) < d1 we have K(α, β; k; q, ψ) = 0.
The case h2
...d1 can execute only for those s, 0 ≤ s ≤ d − 1, for which h2 ≡
≡ 0 (mod d1), i.e., s must satisfy the congruence
s
q − 1
d
+
h
d
≡ 0 (mod d1).
But
(
d1,
q − 1
d
)
= 1 since d1
∣∣k1 and
(
k1,
q − 1
d
)
= 1.
It follows that we have only one value s modulo d1, and hence, at most
[
d
d1
]
+1 the
value of s among 0 ≤ s ≤ d− 1, for which h2
...d1. We apply this reduction and through
ν(k) steps we obtain the estimation∣∣Kk(α, β; q, ψ)
∣∣ ≤ 2ν(k)+1kq
1
2 ,
where ν(k) denote the number a prime divisors of k.
And so we proved the following theorem.
Theorem 3.4. Let α, β ∈ Fq and though one of element α or β is not equal to
zero. Then for any multiplicative character ψ of field Fq the estimation∣∣Kψ(α, β; q; k)
∣∣ ≤ 2kq
1
2
holds.
Corollary. Let p be a Gaussian prime number and let χ is a multiplicative character
of a field of the residue classes mod p. Then∣∣∣∣∣∣
∑
x (mod p)
′χ(x) expπiSp
(
αxk + βx′
k
p
)∣∣∣∣∣∣ ≤ 2ν(k)+1kN(p)
1
2N((α, β, p))
1
2 .
4. General Kloosterman sums over norm. Let α, β ∈ Z[i], h ∈ Z, q ∈ N,
q > 1, (h, q) = 1. We set
K̃(α, β;h, q) :=
∑
x,y (mod q)
N(xy)≡h (mod q)
eq
(
1
2
Sp(αx+ βy)
)
(4.1)
and call the norm Kloosterman sum in Z[i].
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GENERAL KLOOSTERMAN SUMS OVER RING OF GAUSSIAN INTEGERS 1189
For q = q1q2, (q1, q2) = 1 we have
K̃(α, β;h, q) = K̃(α, β;hq′′2, q1)K̃(α, β;hq′′1, q2) =
= K̃(αq2, βq2;h, q1)K̃(αq1, βq1;h, q2).
Thus we shall consider only case q = pn, p is prime rational number, n ∈ N. We denote
mα = maxm≥n {α ≡ 0 (mod pm)}.
Theorem 4.1. Let (h, p) = 1. Then
K̃(α, β;h, pn) � (pmα , pmβ , pn)
1
2 p
3n
2 (4.2)
with an absolute constant in symbol “�”.
Proof. At first let n = 1. The case mα = mβ = 1 is a trivial. Thus we shall
suppose that mα = 0 or mβ = 0. We set α = a1 + ia2, β = b1 + ib2 and, hence,
(a1, a2, b1, b2) = 1.
For p ≡ 1 (mod 4) we have
K̃(α, β;h, p) =
∑
(U)
ep(a1x1 − a2x2 + b1y1 − b2y2), (4.3)
where U =
{
x1, x2, y1, y2 ∈ {0, 1, . . . , p− 1}, (x2
1 + x2
2)(y
2
1 + y2
2) ≡ h (mod p)
}
. Let
ε0 is a solution of congruence x2 ≡ −1 (mod p).
We set
u1 = x1 + ε0x2, u2 = x1 − ε0x2, v1 = y1 + ε0y2, v2 = y1 − ε0y2.
Now by (4.3) we obtain
K̃(α, β;h, p) =
∑
(U)
ep(A1u1 +A2u2 +B1v1 +B2v2),
where U =
{
u1, u2, v1, v2 ∈ {0, 1, . . . , p− 1}, u1u2v1v2 ≡ h (mod p)
}
.
E. Bombieri [14] proved that the last sum can be estimated as� p
3
2 . If p ≡ 3 (mod 4)
then the such estimation holds for the sum (4.3) (the proof is analogous).
The case p = 2 is a trivial.
Now, let n ≥ 2. It is enough to consider only the case (pmα , pmβ , pn) = 1. In this
case though one of number a1, a2, b1, b2 does not divide on p (here α = a1 + ia2,
β = b1 + ib2). We have
K̃(α, β;h, pn) =
=
∑
x,y (mod pn)
1
pn
pn−1∑
k=0
epn
(
k(N(x)N(y)− h) + <(αx) + <(βy)
)
=
=
1
pn
∑
U
epn
(
k((x2
1 + x2
2)(y
2
1 + y2
2)− h) + a1x1 − a2x2 + b1y1 − b2y2
)
, (4.4)
where U :=
{
k (mod pn); x1, x2 (mod pn); y1, y2 (mod pn)
}
.
Though one out of sums over x1, x2, y1, y2 is equal 0 if (k, p) = p (by a rational
analogue of Lemma 2.1).
Thus, supposing (a1, a2, p) = 1, we have
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1190 S. P. VARBANETS
K̃(α, β;h, pn) =
=
1
pn
∑
U
enp (−kh)epn
(
kN(x)(y2
1 + y2
2) + <(αx) + b1y1 − b2y2
)
=
=
1
pn
∑
k (mod pn)
∗ epn(−kh)
∑
x (mod pn)
(N(x),p)=1
+
∑
x (mod pn)
N(x)
...p
=
∑
1
+
∑
2
, (4.5)
say, where U :=
{
k ∈ R∗(pn), x ∈ R(pn, i), y1, y2 ∈ R(pn)
}
. Let N(x)′ and k′ are
the solutions of the congruences
N(x)u ≡ 1 (mod pn), ku ≡ 1 (mod pn),
accordingly. Then∣∣∣∑
1
∣∣∣ =
∣∣∣∣∣∣
∑
k (mod pn)
∗ epn(−kh)
∑
x (mod pn)
epn
(
4′N(x)′k′(b21 + b22) + a1x1 − a2x2
)∣∣∣∣∣∣ .
(4.6)
We set
x1 = x0
1 + pmz1, x2 = x0
2 + pmz2,
0 ≤ x0
1, x
0
2 ≤ pm − 1, 0 ≤ z1, z2 ≤ pn−m − 1, m =
[
n+ 1
2
]
.
It is obvious
N(x)′ =
(
x0
1
2
+ x0
2
2)′(1− 2pm(x0
1
2
+ x0
2
2
)′(x0
1z2 + x0
2z1)
)
and consequently ∣∣∣∑
1
∣∣∣ =
∣∣∣∣∣∣
∑
k (mod pn)
∗ epn(−kh) ×
×
∑
x0
1,x
0
2 (mod pn)
(x0
1
2
+x0
2
2
, p)=1
epn
(
4′k′(x0
1
2
+ x0
2
2
)′(b21 + b22) + a1x
0
1 − a2x
0
2
)
×
×
∑
z1,z2 (mod pn−m)
epn−m
(
(A1 + a1)z1 + (A2 + a2)z2
)∣∣∣∣∣∣ ,
where A1 = 2
(
(x0
1
2 + x0
2
2)′
)2
x0
2, A2 = 2
(
(x0
1
2 + x0
2
2)′
)2
x0
1.
The summation over z1, z2 gives zero if the congruences
A1 + a1 ≡ 0 (mod pn−m), A2 − a2 ≡ 0 (mod pn−m)
or the equivalent congruences
a2x
0
1 + a1x
0
2 ≡ 0 (mod pn−m), 2x0
2 ≡ −a1(x0
1
2
+ x0
2
2
)2 (mod pn−m)
are disturbs.
This system of the congruences has at most three solutions modulo pn−m, and
therefore at most 3pm−(n−m) solutions modulo pm.
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GENERAL KLOOSTERMAN SUMS OVER RING OF GAUSSIAN INTEGERS 1191
Hence,∣∣∣∑
1
∣∣∣ =
∣∣∣∣∣∣p2(n−m)
∑
(U)
epn(a1x
0
1 − a2x
0
1)
∑
k (mod pn)
∗ (kh+ k′B)
∣∣∣∣∣∣ ≤ 8p
3
2n, (4.7)
where
U =
{
x0
1, x
0
2 (mod pm)
∣∣∣a2x
0
1 ≡ −a1x
0
2 (mod pn−m),
2x0
1 ≡ −a1
(
x0
1
2
+ x0
2
2
)2
(mod pn−m)
}
.
At last, if N(x) ≡ 0 (mod p) then
∑
2
= 0 by Lemma 2.1.
The theorem is proved.
For natural k > 1 we set
K̃(α, β;h, q; k) :=
∑
x,y (mod q)
N(xy)≡h (mod q)
eq
(
1
2
Sp(αxk + βyk)
)
. (4.8)
It is obvious that K̃(α, β;h, q; 1) = K̃(α, β;h, q).
The method of investigation of the sum K̃(α, β;h, q; k) towards suffices to consider
the case q = pn, p be a prime. At first we shall account that p ≡ 3 (mod 4).
Theorem 4.2. Let p ≡ 3 (mod 4), h ∈ Z, (h, p) = 1, k ∈ N, d = (k, p− 1). Then
for any Gaussian integers α, β, (α, β, p) = 1 the estimation∣∣∣K̃(α, β;h, p; k)
∣∣∣�
d
2p
3
2 if d− 1 ≤ 4
√
p,
dp2 if d ≥ 4
√
p+ 1
holds.
Proof. Let k = dk1,
(
k1,
p− 1
d
)
= 1. We have
∑
x,y (mod p)
N(xy)≡h (mod p)
ep
(
1
2
Sp(α(xk1)d + β(yk1)d)
)
=
=
∑
x,y (mod p)
N(xk1yk1 )≡hk1 (mod p)
ep
(
1
2
Sp(α(xk1)d + β(yk1)d)
)
=
=
∑
x,y (mod p)
N(xy)≡hk1 (mod p)
ep
(
1
2
Sp(αxd + βyd)
)
= K̃(α, β;hk1 , p; d).
Now, for any multiplicative character χ of field Fp2 we have∑
h∈F∗
p2
χ(h)K̃(α, β;h, p; d) =
=
∑
x,y∈F∗
p2
χ
(
N(x)N(y)
)
ep
(
1
2
Sp(αxd)
)
ep
(
1
2
Sp(βyd)
)
=
=
∑
x∈F∗
p2
χ
(
N(x)ep
(
1
2
Sp(αxd)
))
∑
y∈F∗
p2
χ
(
N(y)
)
ep
(
1
2
Sp(βyd)
), (4.9)
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1192 S. P. VARBANETS
and moreover the sums on the right of (4.9) can be estimate as (d− 1)N(p)
1
2 (see [17]).
Thus we obtain ∣∣∣∣∣∣∣
∑
h∈F∗
p2
χ(h)K̃(α, β;h, p; d)
∣∣∣∣∣∣∣ ≤ (d− 1)2p2.
The application of the Plansherel theorem give∑
h∈F∗
p2
|K(α, β;h, p; d)|2 ≤ (d− 1)4p4.
Now similarly as in the Bombieri work [14] we conclude that the weights of characteristic
roots associating with K̃(α, β;h, p; d) are at most 3 if (d − 1)4 < p. Hence, using the
results of Bombieri [14] and Deligne [13] we infer
K̃(α, β;h, p; d) � (d− 1)2p2 � d2p2 if d− 1 < 4
√
p.
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,∑
x,y (mod p)
N(xy)≡h (mod p)
ep
(
1
2
Sp(αxd + βyd)
)
=
=
∑
x,y (mod p)
(xy)p+1≡h (mod p)
ep
(
1
2
Sp(αxd + βyd)
)
=
=
∑
ε (mod p)
εp+1≡h (mod p)
∑
x (mod p)
ep
(
1
2
Sp(αxd + βyd)
)
. (4.10)
The congruence zp+1 ≡ h (mod p) has exactly p+ 1 solutions mod p. The inner sum in
the right in (4.10) estimates as ≤ 2dp. This completes the proof of the theorem.
Now, let q = pn, p ≡ 3 (mod 4), n ≥ 2. We shall use the description of a reduced
residue system mod pn (see Lemma 2.5).
In farther the following assertion are need.
Lemma 4.1. Let n, k ∈ N, p ≥ 3 be a prime, u ∈ Z, (p, u) = 1. Then for any
natural t we have
(1 + pku)t ≡ 1 + pka1t+ p2ka2t
2 + pλ3a3t
3 + . . .+ pλnant
n (mod pn),
moreover (ai, p) = 1, i = 1, . . . , n, λj > 2k, j = 3, . . . , n.
Proof. By the relation(
t
m
)
=
1
m!
(
tm − m(m− 1)
2
tm−1 + . . .+ (−1)m−1(m− 1)! · t
)
and upper estimation of an exponent with which p enters in m! we obtain
(1 + pku)t ≡ 1 + pka1t+ p2ka2t
2 + pλ3a3t
3 + . . .+ pλnant
n (mod pn),
where (ai, p) = 1, i = 1, . . . , n, λj >
(
k − 1
p− 1
)
j > 2k for j = 3, 4, . . . .
The lemma is proved.
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GENERAL KLOOSTERMAN SUMS OVER RING OF GAUSSIAN INTEGERS 1193
From the proof of Lemma 2.3 it is obvious that a generative element u + iv of the
group E1 can be take that it is a generative 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.
Thus
N(1 + px0 + ipy0) ≡ 1 + 2px0 + p2x2
0 + p2y2
0 ≡ 1 (mod p`).
Hence, 2px0 ≡ 0 (mod p2), x0 = px′0, (y0, p) = 1. So,
(u+ iv)2(p+1) ≡ 1 + p2x0 + ipy0, (x0, p) = (y0, p) = 1.
Now, applying the previous lemma we easy obtain
<((u+ iv)2(p+1)t) ≡ A0 +A1t+A2t
2 + . . .+An−1t
n−1 (mod pn),
=((u+ iv)2(p+1)t) ≡ B0 +B1t+B2t
2 + . . .+Bn−1t
n−1 (mod pn),
(4.11)
where
A0 ≡ 1 (mod p), B0 ≡ 0 (mod p),
A1 ≡ p2x0 + 2′y2
0p
2 (mod p3), i.e., A1 ≡ 0 (mod p3),
A2 ≡ −2′y2
0p
2 (mod p3), i.e., A2 = p2A′2, (A′2, p) = 1,
B1 ≡ py0 (mod p3), i.e., B1 ≡ pB′
1, (B′
1, 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)).
And hence, we have
<
{
(u+ iv)2(p+1)t+z
}
≡ A0(z) +A1(z)t+ . . .+An−1(z)tn−1 (mod pn), (4.12)
where Ai(z) = Aiu(z)−Biv(z).
We clear up for which values z the congruence v(z) ≡ 0 (mod p) holds.
Let v(z) = pv0(z), v0(z) ≡ 0 (mod pk), k ≥ 0. Then
(u+ iv)z = u(z) + ipv0(z),
(u+ iv)z(p−1)pn−k
≡ (u(z))(p−1)pn−k
(mod pn).
The sequences
{
(u+ iv)2β
}
and {gα} can have only two common elements modulo p:
1 or −1. Thus
(u(z))(p−1)pn−k
≡ ±1 (mod pn).
The congruence (u(z))(p−1)pn−k ≡ −1 (mod pn) is impossible, so the other way we
have (−1)p
k−1 ≡ (u(z))(p+1)pn−1 ≡ 1 (mod pn), i.e., −1 ≡ 1 (mod p). Hence
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1194 S. P. VARBANETS
(u(z))(p−1)pn−k
≡ 1 (mod pn),
z(p− 1)pn−k ≡ 0
(
mod2(p+ 1)pn−1
)
.
Since, (p − 1, p + 1) = 2, then z ≡ 0 (mod(p + 1)pk−1). Whence it follows that from
p
∥∥v(z) we have z = p+ 1 and from p2
∣∣v(z) follows z = 0. So we have
p
∥∥A1(z), Ai(z) ≡ 0 (mod p2), i = 2, . . . , n− 1, if z 6= 0, z 6= p+ 1,
A1(0) = A1(p+ 1) ≡ 0 (mod p2), p2
∥∥A2(0), p2
∥∥A2(p+ 1),
Aj(0) ≡ Aj(p+ 1) ≡ 0 (mod p3), j = 3, 4, . . . , n− 1.
We are now in position to prove the following assertion.
Theorem 4.3. Let p be a prime number, p ≡ 3 (mod 4), h ∈ Z, (h, p) = 1, k > 1
is a natural, a, b are the Gaussian integer, (a, p) = (b, p) = 1. Then for n ≥ 2∣∣∣K̃(a, b;h, pn; k)
∣∣∣ ≤ 2p
3
2n+m log pn, (4.13)
where m such that pm
∥∥k.
Proof. Applying Lemma 2.5, we can write a, b in the form
a = gα
′
0(u+ iv)β
′
0 , b = gα
′′
0 (u+ iv)β
′′
0 ,
where g is a primitive root mod pn in Z, u + iv is a generative element of the group
En. Then we obtain
K̃(a, b;h, pn; k) =
=
∑
x,y (mod pn)
N(x)N(y)≡h (mod pn)
epn
(
gα
′
0<((u+ iv)β
′
0xk) + gα
′′
0<((u+ iv)β
′′
0 yk)
)
. (4.14)
Let h ≡ gα (modpn). Then h ≡ ±g2α0 (mod pn), where
2α0 =
α if α is even,
α+
p− 1
2
pn−1 if α is odd.
The sum over x (mod pn) in (4.14) we split into two pairs,
∑
=
∑
1
+
∑
2
.
In sum over
∑
1
we take these x (mod pn) for which N(x) ≡ g2α1 (mod pn), and
in
∑
2
come upon these x (mod pn) for which N(x) ≡ −g2α1 (mod pn). In both cases
α1 runs the values 0, 1, . . . ,
p− 1
2
pn−1 − 1. So
K̃(a, b;h, pn; k) =
∑
1
+
∑
2
. (4.15)
For x from
∑
1
we have
x ≡ gα1(u+ iv)2β1 (mod pn),
α1 = 0, 1, . . . ,
1
2
(p− 1)pn−1 − 1, β1 = 0, 1, . . . , (p+ 1)pn−1 − 1.
Hence,
<
(
(u+ iv)β
′
0xk
)
≡ gkα1<
(
(u+ iv)2kβ1+β
′
0
)
(mod pn).
From the condition N(x)N(y) ≡ h (mod pn) it follows
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GENERAL KLOOSTERMAN SUMS OVER RING OF GAUSSIAN INTEGERS 1195
N(y) ≡ ±g2α2 (mod pn),
where α2 = α0 +
(
(p− 1)pn−1 − 1
)
α1.
And thus we have∑
1
=
∑
(α1)
∑
(β1)
∑
(β2)
epn
(
gα
′
0+α1k<
(
(u+ iv)2kβ1+β
′
0
)
+
+ gα
′′
0 +α2k<
(
(u+ iv)2kβ2+β
′′
0 +δk
))
, (4.16)
here (α1) denotes that α1 runs the value 0, 1, . . . ,
1
2
(p− 1)pn−1− 1; (βi) runs the value
0, 1, . . . , (p+ 1)pn−1 − 1, i = 1, 2; furthermore, δ = 0 if h ≡ g2α0 (mod pn) and δ = 1
if h ≡ −g2α0 (mod pn).
Similarly, ∑
2
=
∑
(α1)
∑
(β1)
∑
(β2)
epn
(
gα
′
0+α1k<
(
(u+ iv)2kβ1+β
′
0+1
)
+
+ gα
′′
0 +α2k<
(
(u+ iv)2kβ2+β
′′
0 +δk
))
. (4.17)
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.
Now by (4.12), (4.13) and Lemma 2.1, it follows that the sums over ti are equal zero if
the congruences
β′0 + 2kz1 ≡ 0 (mod (p+ 1)),
β′′0 + 2kz2 + kδ ≡ 0 (mod (p+ 1)) for a sum
∑
1
,
β′0 + 2kz1 + 1 ≡ 0 (mod (p+ 1)),
β′′0 + 2kz2 + kδ ≡ 0 (mod (p+ 1)) for a sum
∑
2
,
(4.18)
are disturb.
Consequently one from the sums
∑
1
or
∑
2
is equal always zero.
The congruences (4.18) can be true only for (k, p + 1)2 pairs of the value (z1, z2).
Let B be set of these values (z1, z2).
By (4.12) – (4.14) we obtain
K̃(a, b;h, pn; k) =
∑
(α1)
epn(N0g
α1 +M0g
α2)×
×
∑
(z1,z2)∈B
∑
t1,t2 (mod pn−1)
epn−2
(
F1(kt1)gα1 + F2(kt2)gα2
)
,
where Fi(t) = c
(i)
1 t+ c
(i)
2 t2 + pλ3c
(i)
3 t3 + . . .+ pλ`c
(i)
` t`, (c(i)2 , p) = (c(i)3 , p) = . . . = 1,
λj > 0 for j ≥ 3, (N0, p) = (M0, p) = 1.
The sums over t1, t2 calculates equally. Let k = pmk1, (k1, p) = 1. We break
the sum over ti into blocks of the length pn−2−2m (if 2m < n − 2). Then applying
Lemma 2.3, we obtain
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1196 S. P. VARBANETS
K̃(a, b;h, pn; k) = pn+2m
∑
(α1)
epn(N1g
α1 +N2g
α2), (4.19)
where (N1, p) = (N2, p) = 1.
From the definition α2 follows gα2 ≡ gα0(g′)α1 (mod pn).
The sum on the right in (4.19) is the incomplete Kloosterman sum. By a choice of
a primitive root g we have
gp−1 = 1 + pu, (u, p) = 1.
Then g′
p−1
= 1− pu1, (u1, p) = 1, u ≡ u1 (mod p). We set
α1 = (p− 1)t+ z,
t = 0, 1, . . . ,
1
2
(pn−1 − 1), z = 0, 1, . . . , p− 2.
Thus
gα1 = gz(1 + a1pt+ a2p
2t2 + a3p
λ3t3 + . . .) (mod pn),
a1 ≡ −u1, a2 ≡ −2′u2 (mod p), λj ≥ 3.
Similarly
gλ2 ≡ gα0g′
α1 ≡ gα0g′
z(1 + b1pt+ b2p
2t2 + b3p
µ3t3 + . . .) (mod pn),
b1 ≡ −u1, b2 ≡ −2′u2 (mod p), µj ≥ 3.
Hence,
N1g
α1 +N2g
α2 ≡ c0 + c1pt+ c2p
2t2 + c3p
ν3t3 + . . . (mod pn),
where ci = gzaiN1 + gα0g′
z
biN2, i = 1, 2.
Since (N1, p) = (N2, p) = 1 it easy observe that two congruence
c1 ≡ 0 (mod p), c2 ≡ 0 (mod p)
cannot realize simultaneously.
But from c1 ≡ 0 (mod p) follows g2z ≡ gα0N2N
′
1 (mod p). It is possible only one
value z. Let’s designate this value through z0.
Thus from (4.19) we infer
K̃(a, b;h, pn; k) =
= pn+2m
p−2∑
z=0
z 6=z0
1
2 (pn−1−1)∑
t=0
e2πi
c0
pn en−1
p
(
c1t+ c2pt
2 + c3p
ν3−1t3 + . . .
)
+
+
1
2 (pn−1−1)∑
t=0
e2πi
c′0
pn epn−2
(
c′1t+ c′2t
2 + c′3p
ν3−2t3 + . . .
), (4.20)
where (c1, p) = (c′2, p) = 1.
The sums over t are the incomplete rational sums, their estimations we obtain by
way of estimations complete exponent sums.
We have for an arbitrary polynomial Φ(t) ∈ Z[t] :∣∣∣∣∣
T∑
t=0
e2πi
Φ(t)
q − T
q
q−1∑
t=0
e2πi
Φ(t)
q
∣∣∣∣∣ ≤
q∑
r=1
1
min(r, q − r + 1)
∣∣∣∣∣
q−1∑
t=0
e2πi
Φ(t)−t
q
∣∣∣∣∣ . (4.21)
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GENERAL KLOOSTERMAN SUMS OVER RING OF GAUSSIAN INTEGERS 1197
Now, if Φ(t) = c1t+ c2pt
2 + c3p
ν3−1t3 + . . . , (c1, p) = 1, q = pn−1, then the complete
sums in (4.21) are equal to zero for all r except the case r ≡ c1 (mod p). In this special
case we have Ψ(t) = c′1t+ c′2t
2 + c′3p
ν3−1t3 + . . . , (c′2, p) = 1, q = pn−2, and than a
complete sum estimates by a value 2p
n−2
2 .
Hence,
∣∣∣K̃(a, b;h, pn; k)
∣∣∣ ≤ pn+m
p−2∑
z=0
z 6=z0
1
|c1(z)|
+
pn∑
r=1
1
kp
p
n−2
2 + pp
n−2
2
.
At last, we take account that for the distinct values z we have the distinct values
c1(z) (mod p), and thus we obtain∣∣∣K̃(a, b;h, pn; k)
∣∣∣ ≤ p
3
2n+m
(
log p+
log pn
p
)
.
If 2m > n− 2 then the assertion of theorem is trivial.
The theorem is proved.
We go towards a estimation of the norm Kloosterman sum K̃(a, b;h, pn; k) for the
case p ≡ 1 (mod 4), k ≥ 2, (a, p) = (b, p) = 1. For p ≡ 1 (mod 4) we have p = pp,
where p and p are the complex conjugate Gaussian prime number. Then the reduced
residue system mod pn can write as
x = g`1pn + g`2pn, 0 ≤ `1, `2 ≤ (p− 1)pn−1 − 1,
where g is a primitive root mod pn such that
gp−1 = 1 + pH, H ∈ Z, (H, p) = 1.
Thus
N(x) = xx = g2`1pn + g`2pn + g`1+`2p2n + g`1+`2p2n ≡
≡ g`1+`2Sp(p2n) (mod pn). (4.22)
Therefore, if p = c+ id then (c, p) = (d, p) = 1, and by the induction we easily obtain
p2n ≡ cn + idn, n = 1, 2, . . . ,
where
cn ≡
(−1)n−1 · 2m · c · d2(m−1) (mod p),
(−1)m · 2m+2 · d2m,
dn ≡
(−1)2m−1 · 2m · d2m−1 (mod p) if n = 2m− 1,
(−1)m−1 · 2m+2 · c · d2m−1 (mod p) if n = 2m.
Hence, for p ≡ 1 (mod 4) we have
K̃(a, b;h, pn; k) =
∑
(U)
epn(A(g`
′
1k + g`
′
2k) +B(g`
′′
1 k + g`
′′
2 k)) =
=
∑
(U ′)
epn
(
A(xk1 + xk2) +B(yk1 + yk2 )
)
, (4.23)
where
U :=
{
`′1, `
′
2, `
′′
1 , `
′′
2 (mod (p− 1)pn−1)
∣∣g`′1+`′2+`′′1 +`′′2 ≡ H (mod pn)
}
,
ISSN 1027-3190. Укр. мат. журн., 2007, т. 59, № 9
1198 S. P. VARBANETS
U ′ :=
{
x1, x2, y1, y2 (mod pn)
∣∣x1x2y1y2 ≡ H (mod pn)
}
,
A,B,∈ Z, (A, p) = (B, p) = 1.
Theorem 4.4. Let p ≡ 1 (mod 4) is a prime number and let a, b ∈ Z[i], (a, p) =
= (b, p) = 1. Then ∣∣∣K̃(a, b;h, p; k)
∣∣∣� {
d2p
3
2 if (d− 1)4 < p,
d4p2 if (d− 1)4 ≥ p,
where d = (k, p− 1).
Proof. With out loss of generality, we can suppose a, b ∈ Z.
By (4.23) and similarly as in the case p ≡ 3 (mod 4) we obtain
K̃(a, b;h, p; k) =
∑
x2,x2,y1,y2∈F∗p
x1,x2,y1,y2≡Hk
1
ep
(
A(xd1 + xd2) +B(yd1 + yd2)
)
.
Now, for (d− 1)4 < p we obtain by analogy with the case p ≡ 3 (mod 4)
∑
h∈F∗p
χ(h)K̃(α, β;h, p; d) =
∑
x∈F∗p
χ(x)ep(Axd)
2∑
y∈F∗p
χ(y)ep(Byd)
2
.
Hence, ∑∣∣∣k̃(α, β;h, p; d)
∣∣∣2 ≤ (d− 1)4p4 if (d− 1)4 < p.
Then
K̃(a, b;h, p; k) � d2p
3
2 if (d− 1)4 < p.
Let (d − 1)4 ≥ p. Denote through g a primitive element of field Fp and let x = gind x
for x ∈ F∗p.
Let G is a group of multiplicative characters of Fp. For χ ∈ G we have χ(x) =
= ep−1(ν indx) with some ν ∈ Fp. Then using the arguments from Theorem 4.1, we
can obtain on a routine way the following relation:
K̃(a, b;h, p; d) =
=
1
p− 1
∑
χ∈G
χ(H)
d−1∑
s1,...,s4=0
χ(A2B2)ed
(
(s1 + s2) indA+ (s3 + s4) indB
)
×
×
∑
x1,...,x4∈F∗p
ed(s1 indx1 + . . .+ s4 indx4)χ(x1, . . . , x4)ep(x1 + . . .+ x4) =
=
1
p− 1
∑
ν∈Fp
d−1∑
s1,...,s4=0
ep−1(ν indH)ep−1(F1(ν, s))×
×
∑
x1,...,x4∈F∗p
ep−1(F2(ν, s, x))ep(x1 + . . .+ x4),
where
F1(ν, s) :=
(
2ν + (s1 + s2)
p− 1
d
)
indA+
(
2ν + (s3 + s4)
p− 1
d
)
indB,
F2(ν, s, x) :=
(
s1
p− 1
d
+ ν
)
indx1 + . . .+
(
s4
p− 1
d
+ ν
)
indx4.
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GENERAL KLOOSTERMAN SUMS OVER RING OF GAUSSIAN INTEGERS 1199
The last sum over x1, . . . , x4 is the product of the Gauss sums of field Fp. And hence,∣∣∣K̃(a, b;h, p; k)
∣∣∣ ≤ d4p2.
The theorem is proved.
If n ≥ 2 we can use the description of solution of the congruence x1, x2, x3,
x4 ≡ H (mod pn):
xi = yi + pmzi, yi (mod pm), zi (mod pn−m),
(yi, p) = 1, i = 1, 2, 3; m =
[
n+ 1
2
]
,
x4 = Hy′1y
′
2y
′
3
(
1− pmy′1z1 − pmy′2z2 − pmy′3z3
)
, yiy
′
i ≡ 1 (mod pm).
(4.24)
Theorem 4.5. Let p ≡ 1 (mod 4) be a prime number, n ∈ N, n ≥ 2; h ∈ Z,
(h, p) = 1; a, b ∈ Z[i], (a, p) = (b, p) = 1. Then∣∣∣K̃(a, b;h, pn; k)
∣∣∣�
d
4p
3
2n if (d− 1)4 < p,
d4pn+m if (d− 1)4 ≥ p,
where m =
[
n+ 1
2
]
.
Proof. By (4.23), (4.24) we have
K̃(a, b;h, pn; k) =
∑
y1,y2,y3∈R∗(pm)
epn
(
f(y1, y2, y3)
)
×
×
∑
z1,z2,z3 (mod pn−m)
epn−m(F (z1, z2, z3)), (4.25)
where
f(y1, y2, y3) = Ayk1 + ayk2 +Byk3 +BHy′1
k
y′2
k
y′3
k
,
F (z1, z2, z3) = k
[(
Ayk−1
1 −By′1
k+1
y′2
k
y′3
k
)
z1+
+
(
Ayk−1
2 −B(yk+1
1 yk2y
k
3 )′
)
z2 +
(
Ayk−1
3 −B(yk1y
k
2y
k+1
3 )′
)
z3
]
.
Let (k, pn−m) = p`. Then we obtain from (4.25)
K̃(a, b;h, pn; k) = p3(n−m)
∑
(U)
epn(f(y1, y2, y3)),
where U :=
{
y1, y2, y3 (mod pm)
∣∣∣(yi, p) = 1, i = 1, 2, 3; yk1 ≡ y2
k ≡
≡ y3
k (mod pn−m−`), y4k
1 ≡ BA′ (mod pn−m−`)
}
.
Now, for n = 2m we estimate the sum
∑
(U)
by the number triples (y1, y2, y3) ∈ U,
and for n = 2m− 1 we take into account also the Theorem 2.4. Hence, we have finally∣∣∣K̃(a, b;h, pn; k)
∣∣∣�
d
4p
3
2n if (d− 1)4 < p,
d4pn+m if (d− 1)4 ≥ p.
The theorem is proved.
Collection our previous estimations from the Theorems 4.2 – 4.5 we get the following
theorem.
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1200 S. P. VARBANETS
Theorem 4.6. Let α, β ∈ Z[i] and let h, q, k, n ∈ N, k ≥ 2, (k, q) = (h, q) = 1.
Then for (α, q) = (β, q) = 1 we have
K̃(α, β;h, q; k) � D(k, q)q
3
2 ,
where
D(k, q) =
∏
p
∣∣q
p ≡ 1(q)
d6(k, p)
∏
pn
∥∥q
p ≡ 3(q)
d3(k, p) log pn,
d(k, p) = (k, p− 1).
We must note that the norm Kloosterman sum K̃(α, β;h, q; k) has not an analogue
in the ring Z.
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Received 17.02.2006
ISSN 1027-3190. Укр. мат. журн., 2007, т. 59, № 9
|
| id | umjimathkievua-article-3381 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:41:29Z |
| publishDate | 2007 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/e7/793a655632472826705932c8752df8e7.pdf |
| spelling | umjimathkievua-article-33812020-03-18T19:52:51Z General Kloosterman sums over ring of Gaussian integers Узагальнеш суми Клостермана над кільцем цілих гауссових чисел Varbanets, S. P. Варбанець, С. П. The general Kloosterman sum $K(m, n; k; q)$ over $\mathbb{Z}$ was studied by $S$. Kanemitsu, Y. Tanigawa, Yi. Yuan, Zhang Wenpeng in their research of problem of D. H. Lehmer. In this paper, we obtain the similar estimations of $K(\alpha, \beta; k; \gamma)$ over $\mathbb{Z}[i]$. We also consider the sum $\widetilde{K}(\alpha, \beta; h, q; k)$ which has not an analogue in the ring $\mathbb{Z}$ but it can be used for the inversigation of the second moment of the Hecke zeta-fonction of field $\mathbb{Q}(i)$. Узагальнену суму Клостермана $K(m, n; k; q)$ над $\mathbb{Z}$ вивчали $S$. Kanemitsu, Y. Tanigawa, Yi. Yuan, Zhang Wenpeng в їх досліджєнні проблеми D. H. Lehmer. У цій статгі отримано подібні оцінки $K(\alpha, \beta; k; \gamma)$ над $\mathbb{Z}[i]$. Також розглянуто суму $\widetilde{K}(\alpha, \beta; h, q; k)$, що не має аналога в кільці $\mathbb{Z}[i]$, але може бути використана при дослідженні другого моменту дзета-функції Геке поля $\mathbb{Q}(i)$. Institute of Mathematics, NAS of Ukraine 2007-09-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/3381 Ukrains’kyi Matematychnyi Zhurnal; Vol. 59 No. 9 (2007); 1179-1200 Український математичний журнал; Том 59 № 9 (2007); 1179-1200 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/3381/3505 https://umj.imath.kiev.ua/index.php/umj/article/view/3381/3506 Copyright (c) 2007 Varbanets S. P. |
| spellingShingle | Varbanets, S. P. Варбанець, С. П. General Kloosterman sums over ring of Gaussian integers |
| title | General Kloosterman sums over ring of Gaussian integers |
| title_alt | Узагальнеш суми Клостермана над кільцем цілих гауссових чисел |
| title_full | General Kloosterman sums over ring of Gaussian integers |
| title_fullStr | General Kloosterman sums over ring of Gaussian integers |
| title_full_unstemmed | General Kloosterman sums over ring of Gaussian integers |
| title_short | General Kloosterman sums over ring of Gaussian integers |
| title_sort | general kloosterman sums over ring of gaussian integers |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/3381 |
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