ON THE SIZE OF KAKEYA SETS IN FINITE FIELDS
MINGFENG ZHAO
March 01, 2010
In this note, we always assume that F is a finite field with |F | = q and n ∈ N.
Definition 1. A subset E ⊆ F n is called a Kakeya set if for any v ∈ F n \{0}, there exists some xv ∈ E such that
{xv + tv : t ∈ F } ⊆ E.
Lemma 1. Let d ∈ N and Vd be the set of all polynomials of degree at most d in F [X1 , · · · , Xn ], then
dimF (Vd ) =
n+d
.
n
Proof. For any multi-index α = (α1 , · · · , αn ) ∈ Nn0 , it’s easy to see that {X α : |α| ≤ d} is an F -basis of Vd . So dimF (Vd )
is just the non-negative integer solutions of α1 + · · · + αn + αn+1 = d. Therefore, we have
divF (Vd ) =
d+n+1−1
n+d
=
.
n+1−1
n
Lemma 2. Let E be any subset of F n , then the set of all functions from E to F is an F -vector space with dimension
|E|.
Proof. Let ΣE be the set of all functions from E to F , it’s easy to see that ΣE is an F -vector space. Let |E| = m and
E = {e1 , · · · , em }, for each i = 1, · · · , m, define vi : E → F as vi (ej ) = δij .
Claim I: {vi : i = 1, · · · , m} is an F -basis of ΣF .
n
X
Let ai ∈ F such that
ai vi = 0, then for each j = 1, · · · , m, we have 0 =
i=1
m
X
a i vi
i=1
ej ) =
m
X
i=1
ai vi (ej ) =
m
X
ai δij =
i=1
aj , which implies that {vi : i = 1, · · · , m} are F -linearly independent. On the other hand, for any function f : E → F ,
m
m
m
X
X
X
for any j = 1, · · · , m, we have f (ej ) =
f (ei )δij =
f (ei )vi (ej ), that is, f =
f (ei )vi . In summary, we know
i=1
i=1
i=1
that {vi : i = 1, · · · , m} is an F -basis of ΣF .
n+d
Lemma 3. Let d ∈ N and E be any subset of F such that |E| <
, then there exists some non-zero polynomial
n
f (X1 , · · · , Xn ) ∈ F [X1 , · · · , Xn ] such that deg f (X) ≤ d and f (P ) = 0 for all P ∈ E.
n
1
2
MINGFENG ZHAO
Proof. Let Vd be the set of all polynomials of degree at most d in F [X1 , · · · , Xn ], and ΣE be the set of all functions
from E to F . Define σ : Vd → ΣE as: for any f (X) ∈ Vd , then σ(f (X1 , · · · , Xn ))(P ) = f (P ) for all P ∈ E. It’s easy to
see that σ is an F -linear map. On the other hand, by Lemma 1 and Lemma 2, then ker(σ) 6= 0, which implies that
there exists some non-zero polynomial f (X1 , · · · , Xn ) ∈ Vd such that σ(f (X1 , · · · , Xn )) = 0, that is, f (P ) = 0 for all
P ∈ E.
Lemma 4. Let d ∈ N and f (X) be a non-zero polynomial in F [X] such that deg f (X) ≤ d, then f (X) at most has d
zeroes in F .
Proof. The statement follows directly from Euclid’s long division.
Lemma 5. There holds that
n+q−1
1
>
· qn
n
n!
Proof. In fact, we have
n+q−1
n
=
(n + q − 1)!
n!(q − 1)!
=
1
· (n + q − 1)(n + q − 2) · · · q
{z
}
n! |
there are n terms
>
1
· qn .
n!
Lemma 6. Let m ∈ N be such that m < q and f (X1 , · · · , Xn ) ∈ F [X1 , · · · , Xn ] be such that f (X1 , · · · , Xn ) at most
has degree m and f (P ) = 0 for all P ∈ F n , then f (X1 , · · · , Xn ) = 0.
Proof. We prove the statement by induction on m + n. For any k ∈ N, let Vk be the set of all polynomials of degree at
most k in F [X1 , · · · , Xn ]. If m = 1 and any n ∈ N, for any f (X1 , · · · , Xn ) ∈ Vm = V1 such that f (P ) = 0 for all P ∈ F n ,
then there are some a0 , a1 , · · · , an ∈ F such that f (X1 , · · · , Xn ) = a0 + a1 X1 + · · · + an Xn . Since f (0, · · · , 0) = 0,
then a0 = 0. For each i = 1, · · · , n, since f (0, · · · , 1, · · · , 0) = 0, then a0 = 0. Hence we have f (X1 , · · · , Xn ) = 0,
that is, the statement is true for m = 1 and all n ∈ N. If n = 1 and any m ∈ N, for any f (X1 ) ∈ Vm such that
f (P ) = 0 for all P ∈ F , since deg f (X1 ) ≤ m < q = |F |, by Lemma 4, then f (X1 ) = 0, that is, the statement is true
for n = 1 and all m ∈ N. Now let’s look at q > m > 2 and n > 2, assume that the statement is true for < m + n.
Now for any f (X1 , · · · , Xn ) ∈ Vm such that f (P ) = 0 for all P ∈ F n , if f (X1 , · · · , Xn ) at most has degree m − 1, by
induction, then f (X1 , · · · , Xn ) = 0. So we can assume that deg f (X1 , · · · , Xn ) = m. Since m > 2, then there are some
ON THE SIZE OF KAKEYA SETS IN FINITE FIELDS
indeterminate, without lose of generality, Xn such that f (X1 , · · · , Xn ) =
t
X
3
gi (X1 , · · · , Xn−1 )X i for some 1 ≤ t ≤ m
i=0
and gi (X1 , · · · , Xn−1 ) ∈ F [X1 , · · · , Xn−1 ] such that deg gi (X1 , · · · , Xn−1 ) ≤ m − i ≤ m. For any v1 , · · · , vn−1 ∈ F ,
t
X
let h(Xn ) = f (v1 , · · · , vn−1 , Xn ) =
gi (v1 , · · · , vn−1 )Xni , then h(v) = f (v1 , · · · , vn−1 , v) = 0 for all v ∈ F . Since
i=0
deg h(Xn ) ≤ t ≤ m < q, by Lemma 4, then h(Xn ) = 0, which implies that gi (v1 , · · · , vn−1 ) = 0 for all i = 0, · · · , t. So
for each i = 0, · · · , t, we have gi (P ) = 0 for all P ∈ F n−1 . By induction hypothesis, then gi (X1 , · · · , Xn−1 ) = 0 for all
i = 0, · · · , t, which implies that f (X1 , · · · , Xn ) = 0, that is, the statement is true for n + m. In summary, the statement
is true for all m, n ∈ N such that m < q.
Theorem 7 (Theorem 1.5 on Page 1094 in Dvir[1]). For any Kakeya set E ⊆ F n , we have
n+q−1
|E| >
.
n
In particular, by Lemma 5, we have
|E| >
1
· qn .
n!
n+q−1
n+d
=
, by Lemma 2, then there
n
n
exists some non-zero polynomial f (X1 , · · · , Xn ) ∈ F [X1 , · · · , Xn ] such that deg f (X1 , · · · , Xn ) ≤ d and f (P ) = 0 for
m
X
all P ∈ E. Let m = deg f (X), then 1 ≤ m ≤ d and f (X) =
fi (X), where fi (X1 , · · · , Xn ) are homogeneous of
Proof. Let d = q − 1, if the statement is not true, that is, |E| <
i=0
degree i, for each i = 0, · · · , d, that is, each fi (X1 , · · · , Xn ) is just the i-th homogeneous component of f (X). Since
deg f (X1 , · · · , Xn ) = m, then fm (X1 , · · · , Xn ) 6= 0.
Claim I: fm (P ) = 0 for all P ∈ F n .
For any v ∈ F n \{0}, since E is a Kakeya set of F n , then there exists some xv ∈ E such that {xv + tv : t ∈ F } ⊆ E.
Since f (P ) = 0 for all P ∈ E, then f (xv + tv) = 0 for all t ∈ F . Let gv (X) = f (xv + Xv), then gv (X) at most has
degree m ≤ d < q. But since gv (t) = f (xv + tv) = 0 for all t ∈ F and |F | = q, by Lemma 4, then gv (X) = 0. Think
v = (v1 , · · · , vn ) as n indeterminates, we know that gv (X) = fm (v)X m + the order of X is less than m (notice that for
any multi-index α ∈ Nn0 , we have (Xv1 )α1 · · · (Xvn )αn = X |α| v1α1 · · · vnαn ). Since gv (X) = 0 in F [X], then fm (v) = 0.
Since fm (X1 , · · · , Xn ) is homogeneous of degree m, then fm (P ) = 0 for all P ∈ F n .
By Claim I and Lemma 6, then fm (X1 , · · · , Xm ) = 0, contradiction. Therefore, we must have |E| >
n+q−1
.
n
References
[1] Zeev Dvir. On the size of kakeya sets in finite fields. Journal of the American Mathematical Society, 2(4):1093–1097, 2009.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009
E-mail address: mingfeng.zhao@uconn.edu