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In-Depth Information
We put
d
=
n
−
k
and get
N−
1
χ
(
ψ
d
(
ϑ
)
ϑ
q−
2
)
S
χ,a,b
(
N
)
≤
N
(
q
−
1) + 2
(
N
−
d
)
.
(6)
d
=1
ϑ∈
F
q
ε
d
)
q−
2
X
q−
2
.Thenwehaveby(1)
Put
F
d
(
X
)=((
b
−
ε
d
)
X
+
a
)(
X
−
χ
(
ψ
d
(
ϑ
)
ϑ
q−
2
)
−
χ
(
F
d
(
ϑ
))
≤
2(
d
−
1)
ϑ∈
F
q
ϑ∈
F
q
since for
ϑ
=
ε
1
= 0 the two sums agree.
Now, we need to verify that the polynomial
F
d
(
X
) is not, up to a multiplicative
constant, an
s
th power where
s
1 is the order of the multiplicative character
χ
.Inthecase
b
=
ε
d
, the polynomial
F
d
(
X
)equals
a
(
X
|
q
−
b
)
q−
2
X
q−
2
. It can only
be an
s
th power, up to the constant
a
,if
b
=
ε
d
= 0 but this implies
T
a,b
=2.
In the case
b
−
ε
d
)
X
)
q−
2
.Itis
easily seen that this polynomial can not be an
s
th power when we consider the
cases
ε
d
=0and
ε
d
a
=
ε
d
,
F
d
(
X
)isequalto(
b
−
ε
d
)(
X
+
b−ε
d
)((
X
−
= 0 separately. Therefore our sum satisfies all the necessary
conditions for the Weil bound, see [6, Theorem 5.41], and we get
2
q
1
/
2
.
χ
(
F
d
(
ϑ
))
≤
ϑ∈
F
q
Hence
χ
(
ψ
d
(
ϑ
)
ϑ
q−
2
)
2
q
1
/
2
+2
d
≤
−
2
ϑ∈
F
q
and with Equation (6) we have
N−
1
d
)(2
q
1
/
2
+2
d
S
χ,a,b
(
N
)
≤
N
(
q
−
1) + 2
(
N
−
−
2)
d
=1
=
O
(
Nq
+
N
2
q
1
/
2
+
N
3
)
.
The term
N
2
q
1
/
2
never dominates so we have
S
χ,a,b
(
N
)=
O
(
Nq
+
N
3
)
.
