A strengthened inequality of Alon-Babai-Suzuki's conjecture on set systems with restricted intersections modulo p

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by  Xin Wang

A strengthened inequality of Alon-Babai-Suzuki’s conjecture
on set systems with restricted intersections modulo p
Wang Xina, Hengjia Weib and Gennian Geb,c,∗
a School ofMathematical Sciences,ZhejiangUniversity,Hangzhou310027, Zhejiang,China
b SchoolofMathematical Sciences, CapitalNormalUniversity,Beijing,100048, China
7 c BeijingCenter forMathematics andInformationInterdisciplinarySciences, Beijing,100048, China
1
0
2

n Abstract
a
J Let K = {k1,k2,...,kr} and L= {l1,l2,...,ls} be disjoint subsets of {0,1,...,p−1},

3 where p is a prime and A = {A1,A2,...,Am} be a family of subsets of [n] such that |Ai|
(mod p)∈K for all Ai ∈A and |Ai∩Aj| (mod p)∈L for i6=j. In 1991, Alon, Babai and
O] Suzukiconjectured that if n≥s+max1≤i≤rki, then |A|≤ ns + s−n1 +···+ s−nr+1 . In
2000, Qian and Ray-Chaudhuri proved the conjecture under the condition n ≥ 2s−r. In
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
C 2015, Hwang and Kim verified theconjecture of Alon, Babai and Suzuki.
h. In this paper, we will provethat if n≥2s−2r+1 or n≥s+max1≤i≤rki, then
t
a n−1 n−1 n−1
m |A|≤ s !+ s−1!+···+ s−2r+1!.
[
This result strengthens theupper bound of Alon, Babai and Suzuki’sconjecture when n≥
1 2s−2.
v
5
8 1 Introduction
5
0
A family A of subsets of [n] is called intersecting if every pair of distinct subsets A ,A ∈ A
0 i j
. haveanonempty intersection. LetLbe a setofs nonnegativeintegers. Afamily Aofsubsets of
1
[n]={1,2,...,n} is L-intersecting if |A ∩A |∈L for every pair of distinct subsets A ,A ∈A.
0 i j i j
7 A family A is k-uniform if it is a collection of k-subsets of [n]. Thus, a k-uniform intersecting
1 family is L-intersecting for L={1,2,...,k−1}.
: The following is an intersection theorem of de Bruijin and Erdo¨s [4].
v
i
X Theorem 1.1 (de Bruijin and Erdo¨s, 1948 [4]). If A is a family of subsets of [n] satisfying
r |Ai∩Ai|=1 for every pair of distinct subsets Ai,Aj ∈A, then |A|≤n.
a
Ayearlater,Bose[2]obtainedthefollowingmoregeneralintersectiontheoremwhichrequires
the intersections to have exactly λ elements.
Theorem 1.2 (Bose, 1949 [2]). If A is a family of subsets of [n] satisfying |A ∩A | = λ for
i i
every pair of distinct subsets A ,A ∈A, then |A|≤n.
i j
∗Corresponding author. Email address: [email protected] (X. Wang), [email protected] (H. Wei),
[email protected](G.Ge).
1
In1961,Erdo¨s,KoandRado[5]provedthefollowingclassicalresultonk-uniformintersecting
families.
Theorem 1.3 (Erd¨os,KoandRado,1961[5]). Letn≥2k and let A beak-uniform intersecting
family of subsets of [n]. Then |A| ≤ n−1 with equality only when A consists of all k-subsets
k−1
containing a common element.
(cid:0) (cid:1)
In 1975, Ray-Chaudhuri and Wilson [11] made a major progress by deriving the following
upper bound for a k-uniform L-intersecting family.
Theorem1.4(Ray-ChaudhuriandWilson,1975[11]). IfAisak-uniformL-intersectingfamily
of subsets of [n], then |A|≤ n .
s
In terms of parameters n(cid:0)a(cid:1)nd s, this inequality is best possible, as shown by the set of all
s-subsets of [n] with L={0,1,...,s−1}.
In 1981, Frankl and Wilson [6] obtained the following celebrated theorem which extends
Theorem 1.4 by allowing different subset sizes.
Theorem 1.5 (Frankl and Wilson, 1981 [6]). If A is an L-intersecting family of subsets of [n],
then A≤ n + n +···+ n .
s s−1 0
The up(cid:0)pe(cid:1)r b(cid:0)ound(cid:1)in Theo(cid:0)rem(cid:1) 1.5 is best possible, as demonstrated by the set of all subsets
of size at most s of [n].
In the same paper, a modular version of Theorem 1.4 was also proved.
Theorem 1.6 (Frankl and Wilson, 1981 [6]). If A is a k-uniform family of subsets of [n] such
that k (mod p)∈/ L and |A ∩A | (mod p)∈L for all i6=j, then |A|≤ n .
i j s
In 1991, Alon, Babai and Suzuki [1] proved the following theorem, w(cid:0)hic(cid:1)h is a generalization
of Theorem1.6 by replacingthe condition ofuniformity with the condition that the members of
A have r different sizes.
Theorem1.7(Alon,BabaiandSuzuki,1991[1]). LetK ={k ,k ,...,k }andL={l ,l ,...,l }
1 2 r 1 2 s
be two disjoint subsets of {0,1,...,p−1}, where p is a prime, and let A be a family of subsets
of [n] such that |A | (mod p) ∈ K for all A ∈ A and |A ∩A | (mod p) ∈ L for i 6= j. If
i i i j
r(s−r+1)≤p−1 and n≥s+max k , then |A|≤ n + n +···+ n .
1≤i≤r i s s−1 s−r+1
In the proof of Theorem 1.7, Alon, Babai and Suzuk(cid:0)i (cid:1)used(cid:0) a v(cid:1)ery elega(cid:0)nt line(cid:1)ar algebra
method together with their Lemma 3.6 which needs the condition r(s −r + 1) ≤ p−1 and
n≥s+max k . Theyconjecturedthattheconditionr(s−r+1)≤p−1inthestatementof
1≤i≤r i
their theorem can be dropped off. However,their approachcannot work for this stronger claim.
In an effort to prove the Alon-Babai-Suzuki’s conjecture, Snevily [12] obtained the following
result.
Theorem 1.8 (Snevily, 1994 [12]). Let p be a prime and K,L be two disjoint subsets of
{0,1,...,p−1}. Let |L|=s and let A be a family of subsets of [n] such that |A | (mod p)∈K
i
for all A ∈A and |A ∩A | (mod p)∈L for i6=j. Then |A|≤ n−1 + n−1 +···+ n−1 .
i i j s s−1 0
Since n−1 + n−1 = n and n > s−2 n−1 when n is(cid:0)suffi(cid:1)cien(cid:0)tly la(cid:1)rge,Theo(cid:0)rem(cid:1)1.8
s s−1 s s−1 i=0 i
not only confirms the conjecture of Alon, Babai and Suzuki in many cases but also strengthens
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) P (cid:0) (cid:1)
the upper bound of their theorem when n is sufficiently large.
In 2000, Qian and Ray-Chaudhuri [10] developed a new linear algebra approach and proved
thenexttheoremwhichshowsthatthesameconclusioninTheorem1.7holdsifthetwoconditions
r(s−r+1) ≤ p−1 and n ≥ s+max k are replaced by a single more relaxed condition
1≤i≤r i
n≥2s−r.
2
Theorem 1.9(QianandRay-Chaudhuri,2000[10]). Letpbeaprimeandlet L={l ,l ,...,l }
1 2 s
and K ={k ,k ,...,k } be two disjoint subsets of {0,1,...,p−1}such that n≥2s−r. Suppose
1 2 r
that A is a family of subsets of [n] such that |A | (mod p) ∈ K for all A ∈ A and |A ∩A |
i i i j
(mod p)∈L for every i6=j. Then |A|≤ n + n +···+ n .
s s−1 s−r+1
Recently, Hwang and Kim [8] verified t(cid:0)he(cid:1)co(cid:0)nject(cid:1)ure of Al(cid:0)on, Ba(cid:1)bai and Suzuki.
Theorem 1.10 (Hwang and Kim, 2015 [8]). Let K = {k ,k ,...,k } and L = {l ,l ,....l }
1 2 r 1 2 s
be two disjoint subsets of {0,1,...,p−1}, where p is a prime, and let A be a family of subsets
of [n] such that |A | (mod p) ∈ K for all A ∈ A and |A ∩A | (mod p) ∈ L for i 6= j. If
i i i j
n≥s+max k , then |A|≤ n + n +···+ n .
1≤i≤r i s s−1 s−r+1
We note here that in some ins(cid:0)ta(cid:1)nces(cid:0)Alo(cid:1)n, Babai(cid:0)and Su(cid:1)zuki’s condition holds but Qian and
Ray-Chaudhuri’sconditiondoes not, while in some otherinstances the later conditionholds but
the former condition does not.
In [3], Chen and Liu strengthened the upper bounds of Theorem 1.8 under the condition
min{k }>max{l }.
i i
Theorem 1.11 (Chen and Liu, 2009 [3]). Let p be a prime and let L = {l ,l ,...,l } and
1 2 s
K = {k ,k ,...,k } be two disjoint subsets of {0,1,...,p−1} such that min{k } > max{l }.
1 2 r i i
Suppose that A is a family of subsets of [n] such that |A | (mod p) ∈ K for all A ∈ A and
i i
|A ∩A | (mod p)∈L for every i6=j. Then |A|≤ n−1 + n−1 +···+ n−1 .
i j s s−1 s−2r+1
In[9], Liu andYang generalizedTheorem1.11un(cid:0)der(cid:1)a re(cid:0)laxed(cid:1) conditio(cid:0)nk >(cid:1)s−r for every
i
i.
Theorem 1.12 (Liu and Yang, 2014 [3]). Let p be a prime and let L = {l ,l ,...,l } and
1 2 s
K = {k ,k ,...,k } be two disjoint subsets of {0,1,...,p−1} such that k > s−r for every
1 2 r i
i. Suppose that A is a family of subsets of [n] such that |A | (mod p) ∈ K for all A ∈ A and
i i
|A ∩A | (mod p)∈L for every i6=j. Then |A|≤ n−1 + n−1 +···+ n−1 .
i j s s−1 s−2r+1
In the same paper, they also obtained the same(cid:0)boun(cid:1)d u(cid:0)nder(cid:1)the cond(cid:0)ition of(cid:1)Theorem 1.7.
Theorem 1.13 (Liu and Yang, 2014 [3]). Let p be a prime and let L = {l ,l ,...,l } and
1 2 s
K ={k ,k ,...,k } be two disjoint subsets of {0,1,...,p−1} such that r(s−r+1)≤p−1 and
1 2 r
n≥s+max k . Suppose that A is a family of subsets of [n] such that |A | (mod p)∈K for
1≤i≤r i i
all A ∈Aand |A ∩A | (mod p)∈L for every i6=j. Then |A|≤ n−1 + n−1 +···+ n−1 .
i i j s s−1 s−2r+1
In this paper, we show that Theorem 1.13 still holds under t(cid:0)he A(cid:1)lon(cid:0), Ba(cid:1)bai and(cid:0)Suzuki(cid:1)’s
condition; that is to say, we can drop the condition r(s−r+1)≤p−1 in Theorem 1.13.
Theorem 1.14. Let p be a prime and let L = {l ,l ,...,l } and K = {k ,k ,...,k } be two
1 2 s 1 2 r
disjoint subsets of {0,1,...,p−1}. Suppose that A is a family of subsets of [n] such that |A |
i
(mod p)∈K for all A ∈A and |A ∩A | (mod p)∈L for every i6=j. If n≥s+max k ,
i i j 1≤i≤r i
then |A|≤ n−1 + n−1 +···+ n−1 .
s s−1 s−2r+1
Notetha(cid:0)t n(cid:1)−1 +(cid:0) n−(cid:1)1 +···+(cid:0) n−1 (cid:1) = n + n +···+ n and n < n for
s s−1 s−2r+1 s s−2 s−2(r−1) s−2i s−i
1≤i≤r−1 when n≥2s−2. Our result strengthens the upper bound of Alon-Babai-Suzuki’s
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
conjecture (Theorems 1.10) when n≥2s−2.
In the proof of Theorem 1.14, we first prove that the bound holds under the condition n ≥
2s−2r+1, which relaxes the condition n≥2s−r in the theorem of Qian and Ray-Chaudhuri.
3
Theorem 1.15. Let p be a prime and let L = {l ,l ,...,l } and K = {k ,k ,...,k } be two
1 2 s 1 2 r
disjoint subsets of {0,1,...,p−1}. Suppose that A is a family of subsets of [n] such that |A |
i
(mod p)∈K for all A ∈A and |A ∩A | (mod p)∈L for every i6=j. If n≥2s−2r+1, then
i i j
|A|≤ n−1 + n−1 +···+ n−1 .
s s−1 s−2r+1
Th(cid:0)eorem(cid:1)s1(cid:0).7,1(cid:1).9,1.12an(cid:0)d1.13h(cid:1)avebeenextendedtok-wiseL-intersectingfamiliesin[7,9].
With a similar idea, our results can also be extended to the k-wise case.
2 Proof of Theorem 1.15
In this section we prove Theorem 1.15, which will be helpful in the proof of Theorem 1.14.
Throughout this section, let X = [n − 1] = {1,2,...,n − 1} be an (n − 1)-element set,
p be a prime, and let L = {l ,l ,...,l } and K = {k ,k ,...,k } be two disjoint subsets of
1 2 s 1 2 r
{0,1,...,p−1}. Suppose that A={A ,A ,...,A } is a family of subsets of [n] such that (1)
1 2 m
|A | (mod p) ∈ K for every 1 ≤ i ≤ m, (2) |A ∩A | (mod p) ∈ L for i 6= j. Without loss of
i i j
generality,assumethatthereexistsapositiveintegertsuchthatn∈/ A for1≤i≤tandn∈A
i i
for i≥t+1. Denote
P (X)={S|S ⊂X and |S|=i}.
i
We associate a variable x for each A ∈A and set x=(x ,x ,...,x ). For each I ⊂X, define
i i 1 2 m
L = x .
I i
i:I⊂XAi∈A
Consider the system of linear equation over the field F :
p
{L =0, where I runs through ∪s P (X)}. (1)
I i=0 i
Proposition 2.1. Assume that L∩K = ∅. If A is a mod p L-intersecting family with |A |
i
(mod p) ∈ K for every i, then the only solution of the above system of linear equations is the
trivial solution.
Proof. Let v = (v ,v ,...,v ) be a solution to the system (1). We will show that v is the zero
1 2 m
solution over the field F . Define
p
s
g(x)= (x−l ),
j
j=1
Y
and
s
h(x)=g(x+1)= (x+1−l ).
j
j=1
Y
Since x , x ,..., x form a basis for the vector space spanned by all the polynomials in F [x]
0 1 s p
of degree at most s, there exist a ,a ,...,a ∈F and b ,b ,...,b ∈F such that
0 1 s p 0 1 s p
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
s
x
g(x)= a ,
i
i
i=0 (cid:18) (cid:19)
X
and
s
x
h(x)= b .
i
i
i=0 (cid:18) (cid:19)
X
4
Let A be an element in A with v 6=0. Next we prove the following identities:
i0 i0
If n∈/ A , then
i0
s
a L = g(|A ∩A |)x ; (2)
i I i i0 i
Xi=0 I∈Pi(XX),I⊂Ai0 AXi∈A
if n∈A , then
i0
s t
b L = h(|A ∩A |)x + h(|A ∩A |−1)x . (3)
i I i i0 i i i0 i
Xi=0 I∈Pi(XX),I⊂Ai0 Xi=1 i≥Xt+1
We provethem by comparingthe coefficients of both sides. For any A ∈A, the coefficient ofx
i i
in the left hand side of (2) is
s s
|A ∩A |
a |{I ∈P (X):I ⊂A ,I ⊂A }|= a i i0 ,
i i i0 i i i
i=0 i=0 (cid:18) (cid:19)
X X
which is equal to g(|A ∩A |) by the definition of a . This proves the identity (2).
i i0 i
For any i≤t, the coefficient of x in the left hand side of (3) is
i
s s
|A ∩A |
b |{I ∈P (X):I ⊂A ,I ⊂A }|= b i i0 ,
i i i0 i i i
i=0 i=0 (cid:18) (cid:19)
X X
for any i≥t+1, the coefficient of x in the left hand side of (3) is
i
s s
|A ∩A |−1
b |{I ∈P (X):I ⊂A ,I ⊂A }|= b i i0 .
i i i0 i i i
i=0 i=0 (cid:18) (cid:19)
X X
This proves the identity (3).
If n6∈A , substituting x with v for all i in the identity (2), we have
i0 i i
s
a L (v)= g(|A ∩A |)v .
i I i i0 i
Xi=0 I∈Pi(XX),I⊂Ai0 AXi∈A
Itisclearthatthelefthandsideis0sincevisasolutionto(1). ForA ∈Awithi6=i ,|A ∩A |
i 0 i i0
(mod p)∈L and so g(|A ∩A |)=0. Thus the right hand side of the above identity is equal to
i i0
g(|A |)v . So g(|A |)v =0. Since L∩K =∅, we have g(|A |)6=0 and so v =0. This is a
i0 i0 i0 i0 i0 i0
contradiction to the definition of v.
If n∈A , substituting x with v for all i in the identity (3), we have
i0 i i
s t
b L (v)= h(|A ∩A |)v + h(|A ∩A |−1)v
i I i i0 i i i0 i
Xi=0 I∈Pi(XX),I⊂Ai0 Xi=1 i≥Xt+1
= h(|A ∩A |−1)v since v =0 for all i≤t.
i i0 i i
i≥t+1
X
Since h(|A ∩A |−1)=g(|A ∩A |), withasimilarargumenttothe abovecase,wecandeduce
i i0 i i0
the same contradiction. Then the proposition follows.
5
As a result of this proposition, we have:
|A|≤dim({L :I ∈∪s P (X)}),
I i=0 i
where dim({L : I ∈ ∪s P (X)}) is defined to be the dimension of the space spanned by
I i=0 i
{L :I ∈∪s P (X)}. In the remaining of this section, we make efforts to give an upper bound
I i=0 i
on this dimension.
Lemma 2.2. For any i∈{0,1,...,s−2r+1} and every I ∈P (X), the linear form
i
L
H
H∈Pi+2Xr(X),I⊂H
is linearly dependent on the set of linear forms {L :i≤|H|≤i+2r−1,H ⊂X} over F .
H p
Proof. Define
r r
f(x)= (x−(k −i)) × (x−(k −1−i)) .
j j
   
j=1 j=1
Y Y
   
We distinguish two cases.
(a) i (mod p) ∈/ K and i+1 (mod p) ∈/ K for all i. In this case ∀ k ∈ K, k −i 6= 0 and
j j
k −i−16=0 in F and so c=(k −i)(k −i)···(k −i)(k −i−1)···(k −i−1)6=0 in
j p 1 2 r 1 r
F . It is clear that there exist a ,a ,...,a ∈F , a =(2r)!∈F −{0} such that
p 1 2 2r−1 p 2r p
x x x
a +a +···+a =f(x)−c,
1 2 2r
1 2 2r
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
since the polynomial in the right hand side has constant term equal to 0.
Next we show that
2r
a L =−cL . (4)
j H I
Xj=1 H∈Pi+Xj(X),I⊂H
In fact both sides are linear forms in x , for A∈A. The coefficient of x in the left hand
A A
side is 2r a |{H|I ⊂H ⊂A,n6∈H,|H|=i+j}|. So it is equal to
j=1 j
P
0, if I 6⊂A;
a |A|−i +a |A|−i +···+a |A|−i , if I ⊂A and n∈/ A;
 1 1 2 2 2r 2r
a1(cid:0)|A|−1i(cid:1)−1 +(cid:0)a2 |A|(cid:1)−2i−1 +···+(cid:0)a2r (cid:1)|A|−2ri−1 , if I ⊂A and n∈A.
By the above(cid:0)polynom(cid:1)ial id(cid:0)entity, (cid:1) (cid:0) (cid:1)
2r
|A|−i
a =f(|A|−i)−c=−c since |A| (mod p)∈K;
j
j
j=1 (cid:18) (cid:19)
X
2r
|A|−i−1
a =f(|A|−i−1)−c=−c since |A| (mod p)∈K.
j
j
j=1 (cid:18) (cid:19)
X
The coefficient of x in the right hand side is obviously the same. This proves (4).
A
6
Writing (4) in a different way, we have
2r−1
1
L =− (cL + a L ).
H I j H
(2r)!
H∈Pi+2Xr(X),I⊂H Xj=1 H∈Pi+Xj(X),I⊂H
This proves the lemma in case (a).
(b) i (mod p) ∈ K or i + 1 (mod p) ∈ K for some i. In this case, the constant term of
(x−(k −i))(x−(k −i))···(x−(k −i))(x−(k −i−1))···(x−(k −i−1)) is 0∈F .
1 2 r 1 r p
So there exists a ,a ,...,a ∈F , a =(2r)!∈F −{0} such that
1 2 2r−1 p 2r p
x x x
a +a +···+a =f(x)
1 2 2r
1 2 2r
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
As a consequence we have
2r
a L =0 ∀I ∈P (X),
j H i
Xj=1 H∈Pi+Xj(X),I⊂H
i.e. we have
2r−1
1
L =− ( a L ).
H j H
(2r)!
H∈Pi+2Xr(X),I⊂H Xj=1 H∈Pi+Xj(X),I⊂H
This finishes the proof of this lemma.
Corollary 2.3. With the same condition as in Lemma 2.2, we have
hL :H ∈∪i+2r−1P (X)i
H j=i j
= L :H ∈∪i+2r−1P (X) + L :I ∈P (X)
H j=i j H i
(cid:10) (cid:11) *H∈Pi+2Xr(X),I⊂H +
Here hL :H ∈∪i+2r−1P (X)i is the vector space spanned by {L :H ∈∪i+2r−1P (X)}.
H j=i j H j=i j
TherestoftheproofissimilartotheproofofTheorem1.9givenbyQianandRay-Chaudhuri
[10]. The next lemma is a restatement of [10, Lemma 2], and is used to prove Lemma 2.5.
Lemma 2.4. For any positive integers u,v with u<v <p and u+v ≤n−1, we have
hL :J ∈P (X)i n−1 n−1
J v
dim ≤ − .
h L :I ∈P (X)i v u
J∈Pv(X),I⊂J J u ! (cid:18) (cid:19) (cid:18) (cid:19)
Here A is the quotientPspace of two vector spaces A and B with B ≤A.
B
Lemma 2.5. For any i∈{0,1,...,s−2r+1},
n−1 n−1 n−1 hLH :H ∈∪sj=iPj(X)i
+ +···+ +dim
(cid:18) i (cid:19) (cid:18)i+1(cid:19) (cid:18)i+2r−1(cid:19) hLH :H ∈∪ij+=2ir−1Pj(X)i!
n−1 n−1 n−1
≤ + +···+ .
s−2r+1 s−2r+2 s
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
7
Proof. We inductons−2r+1−i. Itisclearlytruewhens−2r+1−i=0. Suppose the lemma
holds for s−2r+1−i < l for some positive integer l. Now we want to show that it holds for
s−2r+1−i=l.
We observethati+i+2r ≤(s−2r)+(s−2r)+2r ≤n−1 by the conditionin the theorem.
By Corollary 2.3 and Lemma 2.4, we have
hL :H ∈∪i+2rP (X)i
dim H j=i j
hLH :H ∈∪ij+=2ir−1Pj(X)i!
hL :H ∈∪i+2r−1P (X)i+hL :H ∈P (X)i
H j=i j H i+2r
=dim
hLH :H ∈∪ij+=2ir−1Pj(X)i+h H∈Pi+2r(X),I⊂HLH :I ∈Pi(X)i!
L :H ∈P (X) P
H i+2r
≤dim
L :I ∈P (X)
H∈Pi+2r(X),I⊂H H i !
n−1P n−1
≤ − .
i+2r i
(cid:18) (cid:19) (cid:18) (cid:19)
Now we are ready to prove the lemma.
n−1 + n−1 +···+ n−1 +dim hLH :H ∈∪sj=iPj(X)i
(cid:18) i (cid:19) (cid:18)i+1(cid:19) (cid:18)i+2r−1(cid:19) hLH :H ∈∪ij+=2ir−1Pj(X)i!
n−1 n−1 n−1 hLH :H ∈∪ij+=2irPj(X)i
= + +···+ +dim
(cid:18) i (cid:19) (cid:18)i+1(cid:19) (cid:18)i+2r−1(cid:19) hLH :H ∈∪ij+=2ir−1Pj(X)i!
hL :H ∈∪s P (X)i
H j=i j
+dim
hLH :H ∈∪ij+=2irPj(X)i!
n−1 n−1 n−1 hLH :H ∈∪ij+=2irPj(X)i
= + +···+ +dim
(cid:18) i (cid:19) (cid:18)i+1(cid:19) (cid:18)i+2r−1(cid:19) hLH :H ∈∪ij+=2ir−1Pj(X)i!
hL :H ∈P (X)i+hL :H ∈∪s P (X)i
+dim H i H j=i+1 j
hLH :H ∈Pi(X)i+hLH :H ∈∪ij+=2i+r1Pj(X)i!
n−1 n−1 n−1 hLH :H ∈∪ij+=2irPj(X)i
≤ + +···+ +dim
(cid:18) i (cid:19) (cid:18)i+1(cid:19) (cid:18)i+2r−1(cid:19) hLH :H ∈∪ij+=2ir−1Pj(X)i!
hL :H ∈∪s P (X)i
H j=i+1 j
+dim
hLH :H ∈∪ij+=2i+r1Pj(X)i!
n−1 n−1 n−1 n−1 n−1
≤ + +···+ + −
i i+1 i+2r−1 i+2r i
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
hL :H ∈∪s P (X)i
H j=i+1 j
+dim
hLH :H ∈∪ij+=2i+r1Pj(X)i!
n−1 n hLH :H ∈∪sj=i+1Pj(X)i
= +···+ +dim
(cid:18)i+1(cid:19) (cid:18)i+2r(cid:19) hLH :H ∈∪ij+=2i+r1Pj(X)i!
n−1 n−1
≤ +···+ ,
s−2r+1 s
(cid:18) (cid:19) (cid:18) (cid:19)
where the last step follows from the induction hypothesis since s−2r+1−(i+1)<l.
8
We are now turning to the proof of Theorem 1.15.
Proof.
|A|≤dim(hL :H ∈∪s P (X)i)
H i=0 i
hL :H ∈∪s P (X)i
≤dim(hL :H ∈∪2r−1P (X)i)+dim H i=0 j
H i=0 i hL :H ∈∪2r−1P (X)i
(cid:18) H i=0 j (cid:19)
n−1 n−1 n−1 hL :H ∈∪s P (X)i
≤ + +···+ +dim H i=0 j
0 1 2r−1 hL :H ∈∪2r−1P (X)i
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) H i=0 j (cid:19)
n−1 n−1 n−1
≤ + +···+ by taking i=0 in Lemma 2.5,
s−2r+1 s−2r+2 s
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
which completes the proof of the theorem.
3 Proof of Theorem 1.14
Throughout this section, we let p be a prime and we will use x = (x ,x ,...,x ) to denote a
1 2 n
vectorofnvariableswitheachvariablex takingvalues0or1. Apolynomialf(x)innvariables
i
x , for 1 ≤ i ≤ n, is called multilinear if the power of each variable x in each term is at most
i i
one. Clearly, if each variable x only takes the values 0 or 1, then any polynomial in variable x
i
can be regarded as multilinear. For a subset A of [n], we define the incidence vector v of A to
A
be the vector v =(v ,v ,...,v ) with v =1 if i∈A and v =0 otherwise.
1 2 n i i
Let L = {l ,l ,...,l } and K = {k ,k ,...,k } be two disjoint subsets of {0,1,...,p−1},
1 2 s 1 2 r
where the elements of K are arranged in increasing order. Suppose that A = {A ,...,A } is
1 m
the familyofsubsetsof[n]satisfyingthe conditionsinTheorem1.14. Withoutlossofgenerality,
we may assume that n∈A for j ≥t+1 and n∈/ A for 1≤j ≤t.
j j
For each A ∈A, define
j
s
f (x)= (v x−l ),
Aj Aj i
i=1
Y
where x = (x ,x ,...,x ) is a vector of n variables with each variable x taking values 0 or 1.
1 2 n i
Then each f (x) is a multilinear polynomial of degree at most s.
Aj
Let Q be the family of subsets of [n−1] with sizes at most s−1. Then |Q| = s−1 n−1 .
i=0 i
For each L∈Q, define
P (cid:0) (cid:1)
q (x)=(1−x ) x .
L n i
i∈L
Y
Then each q (x) is a multilinear polynomial of degree at most s.
L
Denote K−1={k −1|k ∈K}. Then |K∪(K−1)|≤2r. Set
i i
n−1
g(x)= x −h .
i
!
h∈KY∪(K−1) Xi=1
Let W be the family of subsets of [n−1] with sizes at most s−2r. Then |W| = s−2r n−1 .
i=0 i
For each I ∈W, define
P (cid:0) (cid:1)
g (x)=g(x) x .
I i
i∈I
Y
Then each g (x) is a multilinear polynomial of degree at most s.
I
9
We want to show that the polynomials in
{f |1≤i≤m}∪{q (x)|L∈Q}∪{g (x)|I ∈W}
Ai(x) L I
are linearly independent over the field F . Suppose that we have a linear combination of these
p
polynomials that equals 0:
m
a f (x)+ b q (x)+ u g (x)=0, (5)
i Ai L L I I
i=1 L∈Q I∈W
X X X
with all coefficients a ,b and u being in F .
i L I p
Claim 1. a =0 for each i with n∈A .
i i
Suppose,to the contrary,that i isa subscriptsuchthat n∈A anda 6=0. Since n∈A ,
0 i0 i0 i0
q (v )=0 for every L∈Q. Recall that f (v )=0 for j 6=i and g(v )=0. By evaluating
L Ai0 Aj i0 0 i0
(5) with x = v , we obtain that a f (v ) = 0 (mod p). Since f (v ) 6= 0, we have
Ai0 i0 Ai0 Ai0 Ai0 Ai0
a =0, a contradiction. Thus, Claim 1 holds.
i0
Claim 2. a =0 for each i with n6∈A . Applying Claim 1, we get
i i
t
a f (x)+ b q (x)+ u g (x)=0. (6)
i Ai L L I I
i=1 L∈Q I∈W
X X X
Suppose,tothecontrary,thati isasubscriptsuchthatn∈/ A anda 6=0. Letv′ =v +
0 i0 i0 i0 i0
(0,0,...,0,1). Then q (v′ )=0 for every L∈Q. Note that f (v′ )=f (v ) for each j with
L i0 Aj i0 Aj i0
n∈/ A andg(v′ )=0. Byevaluating(6)withx=v′ ,weobtaina f (v′ )=a f (v )=0
j i0 i0 i0 Ai0 i0 i0 Ai0 i0
(mod p) which implies a =0, a contradiction. Thus, the claim is verified.
i0
Claim 3. b =0 for each L∈Q.
L
By Claims 1 and 2, we obtain
b q (x)+ u g (x)=0. (7)
L L I I
L∈Q I∈W
X X
Set x =0 in (7), then
n
b x + u g (x)=0.
L i I I
L∈Q i∈L I∈W
X Y X
Subtracting the above equality from (7), we get
b x x =0.
L n i
!
L∈Q i∈L
X Y
Setting x =1, we obtain
n
b x =0.
L i
L∈Q i∈L
X Y
Itisnotdifficulttoseethatthepolynomials x ,L∈Q,arelinearlyindependent. Therefore,
i∈L i
we conclude that b =0 for each L∈Q.
L
Q
By Claims 1-3, we now have
u g (x)=0.
I I
I∈W
X
Thus it is sufficient to prove g ’s are linearly independent.
I
10