| dc.description.abstract | Abstract
Let us fix a prime p and a homogeneous system of m linear equations
$$a_{j,1}x_1+\dots +a_{j,k}x_k=0$$
a
j
,
1
x
1
+
⋯
+
a
j
,
k
x
k
=
0
for
$$j=1,\dots ,m$$
j
=
1
,
⋯
,
m
with coefficients
$$a_{j,i}\in \mathbb {F}_p$$
a
j
,
i
∈
F
p
. Suppose that
$$k\ge 3m$$
k
≥
3
m
, that
$$a_{j,1}+\dots +a_{j,k}=0$$
a
j
,
1
+
⋯
+
a
j
,
k
=
0
for
$$j=1,\dots ,m$$
j
=
1
,
⋯
,
m
and that every
$$m\times m$$
m
×
m
minor of the
$$m\times k$$
m
×
k
matrix
$$(a_{j,i})_{j,i}$$
(
a
j
,
i
)
j
,
i
is non-singular. Then we prove that for any (large) n, any subset
$$A\subseteq \mathbb {F}_p^n$$
A
⊆
F
p
n
of size
$$|A|> C\cdot \Gamma ^n$$
|
A
|
>
C
·
Γ
n
contains a solution
$$(x_1,\dots ,x_k)\in A^k$$
(
x
1
,
⋯
,
x
k
)
∈
A
k
to the given system of equations such that the vectors
$$x_1,\dots ,x_k\in A$$
x
1
,
⋯
,
x
k
∈
A
are all distinct. Here, C and
$$\Gamma $$
Γ
are constants only depending on p, m and k such that
$$\Gamma <p$$
Γ
<
p
. The crucial point here is the condition for the vectors
$$x_1,\dots ,x_k$$
x
1
,
⋯
,
x
k
in the solution
$$(x_1,\dots ,x_k)\in A^k$$
(
x
1
,
⋯
,
x
k
)
∈
A
k
to be distinct. If we relax this condition and only demand that
$$x_1,\dots ,x_k$$
x
1
,
⋯
,
x
k
are not all equal, then the statement would follow easily from Tao’s slice rank polynomial method. However, handling the distinctness condition is much harder, and requires a new approach. While all previous combinatorial applications of the slice rank polynomial method have relied on the slice rank of diagonal tensors, we use a slice rank argument for a non-diagonal tensor in combination with combinatorial and probabilistic arguments. | en_US |
