1.4. Equidistribution in finite fields 69
Proposition 1.4.8.
R[D]
is equidistributed in
Σ[D],
thus
|{(x, h1,...,hD) V
d+1
:
R[D](x,
h1,...,hD) = r}|
=
1
|Σ[D]|
+ o(1) |V
|d+1
for all r
Σ[D].
Furthermore, we have the refined bound
|{(h1,...,hD) V
d
:
R[D](x,
h1,...,hD) = r}|
=
pm
|Σ[D]|
+ o(1) |V
|d
for all r
Σ[D]
and all x Sr0 .
Proof. It suffices to prove the second claim. Fix x and r = (rω)ω∈{0,1}D .
From the definition of
Σ[D],
we see that r is uniquely determined by the
component r0 and rd := (rω)ω∈{0,1}D:0 |ω|d. It will thus suffice to show
that
|{(x, h1,...,hD) V
d
:
Rd](x, [D
h1,...,hD) = rd}|
=
pm
|Σ[D]|
+ o(1) |V
|d
for all rd
(Fm){ω∈{0,1}D:0 |ω|d},
where
Rd](x, [D
h1,...,hD)
:= (R(x + ω1h1 + · · · + ωDhD))ω∈{0,1}D:0
|ω|d
.
By Fourier analysis, it suffices to show that
Eh1,...,hD∈V e ξ ·
Rd](x, [D
h1,...,hD) = o(1)
for any non-zero ξ
(Fm){ω∈{0,1}D:0
|ω|d}.
In other words, we need to
show that
(1.39)
Eh1,...,hD∈V e


ω∈{0,1}D:|ω|d
ξω · R(x + ω1h1 + · · · +
ωDhD)⎠

= o(1)
whenever the ξω
Fm
for ω {0,
1}D,
0 |ω| d are not all zero.
Let ω0 be such that ξω0 = 0, and such that |ω| is as large as possible; let
us write d := |ω0|, so that 0 d d. Without loss of generality, we may
take ω0 = (1,..., 1, 0,..., 0). Suppose (1.39) failed, then by the pigeonhole
principle one can find hd +1,...,hD such that
|Eh1,...,hd
∈V
e(
ω∈{0,1}D:|ω|d
ξω · R(x + ω1h1 + · · · + ωDhD))| 1.
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