An improved sum-product estimate for general finite fields

0.15 MB·English

by  Oliver Roche-Newton

An improved sum-product estimate for general finite fields
Oliver Roche-Newton
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1
0
Abstract
2

n This paper improves on a sum-product estimate obtained by Katz and Shen for subsets of a
a finite field whose order is not prime.
J

7
2
1 Introduction

]
O
C LetA be a subset ofF =Fpn for some prime p andsome n∈N. Consider the sumsetandproductset
of A, defined respectively as
.
h
t
a A+A:={a+b:a,b∈A},
m
A·A:={ab:a,b∈A}.
[

1
An interesting problem is to establish lower bounds on the quantity max{|A+A|,|A·A|}. The
v
existenceofnon-trivialboundswasfirstestablishbyBourgain,KatzandTao[2]. Garaev[3]established
8
4 the first quantitative sum-product estimate for fields of prime order. Garaev’sresult can be stated as
3 follows:
5
. Theorem 1.1. Let A⊂F for some prime p, such that |A|≤p7/13(logp)−4/13. Then
1 p
0
1 |A|15/14
1 max{|A+A|,|A·A|}≫ .
(log|A|)2/7
:
v
i
X This resultwas generalised,for finite fields whose order is not necessarilyprime, by Katz andShen[4]
r in the form of the following theorem.
a
Theorem 1.2. Let F =Fpn be a finite field. Suppose that A is a subset of F so that for any subfield
G⊆F, and any elements c,d∈F,
|A∩(cG+d)|≤max{|G|1/2,|A|18/19}. (1)
Then it must be the case that
|A|20/19
max{|A+A|,|A·A|}≫ ,
(log|A|)α
where 0<α≤1 is some absolute constant.
Since Garaev’s sum-product estimate[3], there have been a number of small improvements made
courtesy of more subtle arguments by Katz-Shen[5], Bourgain-Garaev[1], Shen[9], Li[6] and most
recently Rudnev[7]; so the following result of Rudnev represents the state of the art.
1
Theorem 1.3. Let A⊂F∗ with |A|<p1/2 and p large. Then
p
|A|12/11
max{|A+A|,|A·A|}≫ .
(log|A|)4/11
ThispaperseekstoupdatetheresultofKatzandShenforgeneralfinitefields,byusingtherefinements
that allowed for the improved sum-product estimates in the prime setting in [5], [1], [9] and [7]. As
wellasadoptingthetechniquesfromthemorerecentsum-productestimatesinprimefields,aslightly
different argument is made in what will later be called case 3, which leads to a further improvement
in the final estimate. The outcome is the following result:
Theorem 1.4. Let F =F be a finite field and suppose that A is a subset of F∗ with the following
pn
property. For any subfield G⊂F and any elements c,d∈F,
|A∩(cG+d)|≤max{|G|14/25,|A|6/7}. (2)
Then we get the sum-product inequality
15
|A|14
max{|A+A|,|A·A|}≫ .
log|A|
Note that the final estimate here is aligned with the original quantitative bound Garaev obtained in
Theorem 1.1.
1.1 Some remarks concerning Theorem 1.4
As well as improving the exponent in the final bound, the first term from condition (1) has been
increased in (2), thus relaxing one of the conditions. However, it seems necessary to also tighten one
of the conditions, as the second term from (1) becomes smaller in (2).
The statement of Theorem 1.4 is in fact quite flexible. The tools used do not distinguish between
additionandsubtraction,whichmeansthatthe difference set, A−A, canreplaceA+Ainthe above.
It is also possible to get a sum-ratio estimate, where A·A is replaced by the ratio set A:A, which is
defined as follows:
a
def
A:A = such that a,b∈A,b6=0 .
b
n o
The presentationof a sum-ratio estimate proofis of a very similar nature, but rather more simple, as
Rudnev [7] alluded to. Lemma 3.1 can be replacedby a more straightforwardpigeonholing argument
and there is no need for dyadic pigeonholing, which means less technicalities are required and no
logarithmic factor appears in the final estimate. One more benefit of this simplified argument is that
the conditionfromTheorem1.2whichwasmade stricterforthe mainresultinthis paper (the second
term in (1) and (2)), can be relaxed slightly.
1.2 Notation
Throughoutthispaper,thesymbols≪,≫and≈areusedtosuppressconstants. Forexample,X ≪Y
means that there exists some absolute constant C such that X < CY. X ≈ Y means that X ≪ Y
andY ≪X. Whendescribingsucharoughinequality ingenerallanguage,invertedcommasareused
in an effort to avoid confusion. For example, the statement that “X is ‘at most’ Y” tells us that
X ≪Y. A similar meaning is attached to ‘at least’.
2
1.3 Acknowledgements
The author would like to thank Timothy Jones and Misha Rudnev for some extremely helpful con-
versations and typographical corrections.
2 Preliminary results
Before proving the main theorem, it is necessary to state some previous results. The first is a lemma
from the paper of Katz and Shen[4].
Lemma 2.1. Let A⊂F and let G⊆F be a subfield such that
A−A
⊆G.
A−A
Then there exist c,d∈F such that
A⊆cG+d.
The next two results are well-known within arithmetic combinatorics, and have been crucial to all
known quantitative sum-product estimates over finite fields. The first is due to Plu¨nnecke and Rusza
(see [8]), whilst the second is a generalisationwhich Katz and Shen[5] used to obtain an improve the
original quantitative sum-product estimate of Garaev[3].
Lemma 2.2. Let X,B ,...,B be subsets of F. Then
1 k
|X +B |...|X +B |
1 k
|B +...+B |≤ .
1 k |X|k−1
Lemma 2.3. Let X,B ,...,B besubsets of F. Then for any0<ǫ<1, there exists asubset X′ ⊆X,
1 k
with |X′|≥(1−ǫ)|X|, and some constant C(ǫ), such that
|X +B |...|X +B |
|X′+B +...+B |≤C(ǫ) 1 k .
1 k |X|k−1
Finally, we will need the following covering lemma, which appeared in sum-product estimates for
the first time in [9]. An application of this lemma has been the key to the two of the most recent
improvements to the sum-product estimate over prime fields (see [7], [9],).
Lemma 2.4. Let X and Y be additive sets. Then for any ǫ∈(0,1)there is some constant C(ǫ), such
that at least (1−ǫ)|X| of the elements of X can be covered by C(ǫ)min{|X+Y|,|X−Y|} translates of Y.
|Y|
The rest of this paper is devoted to proving Theorem 1.4.
3 Proof of Theorem 1.4
Let A be a set satisfying the conditions of Theorem 1.4, and suppose that |A+A|,|A·A| ≤ K|A|.
The aim is to show that K ≫ |A|1/14.
log|A|
At the outset, apply Lemma 2.3 to identify some subset A′ ⊂A, with cardinality |A′|≈|A|, so that
3
|A+A|3
|A′+A′+A′+A′|≪ ≪K3|A|.
|A|2
Next apply Lemma 2.3 again to identify a further subset A′′, with cardinality |A′′|≈|A|, such that
|A+A|2
|A′′+A′′+A′′|≪ ≪K2|A|.
|A|
Since manymore refinementsofA areneededthroughoutthe proof,this firstchangeis made without
achangeinnotation. So,throughoutthe restofthe proof,whenthe setAis referredto,we arereally
talking about the large subset A′′, which satisfies each of the above inequalities. In other words, we
assume that
|A+A+A+A|≪K3|A|. (3)
and
|A+A+A|≪K2|A| (4)
hold, for our set A.
Consider the point set A×A ⊂ F ×F. The multiplicative energy, E(A), of A is defined to be the
number of solutions to
a a
1 3
= , (5)
a a
2 4
such that a ,a ,a ,a ∈A. Let L be the set of all lines through the origin. By definition,
1 2 3 4
E(A)= |l∩(A×A)|2.
Xl∈L
Bydyadicpigeonholing,atechniquefamiliarfrommostknownsum-productestimates(see[3],[5],[4],[9],[7]),
wecanidentifyapopulardyadicgroup. Thisisdonebypartitioninglinesthroughtheoriginaccording
to their popularity. Speaking more precisely, let L be the set of all lines l through the origin such
i
that 2i ≤|l∩A×A|<|2i+1|. Therefore the multiplicative energy may be rewritten in the form
log|A|
E(A)= |l∩(A×A)|.
Xj=0 lX∈Lj
Then,byelementarypiegeonholing,onemaychooseaparticulardyadicgroup,whichcontributesmore
than the average for this sum. Therefore, there exists a set of L lines, each supporting ≈ N points
from A×A such that
E(A) |A|4 |A|3
M :=LN2 ≫ ≥ ≥ , (6)
log|A| |A·A|log|A| Klog|A|
where the second inequality comes from the standard Cauchy-Schwarz bound on the multiplicative
energy (see [10]). Refine the point set by now considering only points that lie on these L lines, and
call this set P ⊂A×A. Clearly, |P|≈LN. Denote by Ξ the set of slopes through the origin in this
refined set P. More precisely, define Ξ as follows:
4
b
Ξ:= :(a,b)∈P .
(cid:26)a (cid:27)
The fact that LN ≤|A|2 easily implies that
M
N ≥ . (7)
|A|2
Similarly, since N ≤A, it is clear that
M
L≥ . (8)
|A|2
Let
A ={y :(x,y)∈P}
x
and
B ={x:(x,y)∈P}.
y
So, A is the setofordinatesofP fora fixedabscissax, andB is the setofabscissaeofP forafixed
x y
ordinate y.
A little more notation is required still. For some element ξ ∈Ξ, let P be the projection of points in
ξ
P on the line with equation y =ξx onto the x-axis. So,
P ={x:(x,ξx)∈P},
ξ
and since all lines with slope in Ξ intersect P with cardinality approximately N, it follows that
|P |≈N for all ξ∈Ξ. Note also, since P ⊂A×A, that P is a subset of A.
ξ ξ
The following lemma is taken from Rudnev([7]).
Lemma 3.1. There exists a popular abscissa x and a popular ordinate y (popular here means that
0 0
|A |,|B |≫ LN), as well as a large subset A˜ ⊆A , with
x0 y0 A x0 x0
LM
|A˜ |≫ , (9)
x0 |A|3
such that for every z ∈A˜ ,
x0
LMN
|A˜ :=P ∩B |≫ . (10)
z˜ z/x0 y0 |A|4
Since the sum-product estimate is invariantunder dilation, we may assume without loss of generality
that x =1. This means that elements of A are also the popular slopes described above, i.e.
0 x0
A ⊂Ξ.
x0
5
3.1 Application of the covering lemma
Since Lemma 2.4 is applied in a similar manner severaltimes throughout the remainder of the proof,
it is worthwhile highlighting this in advance, to avoid repetition.
Application 3.2. If A′ ⊂ A and ξ ∈ Ξ, then ±ξA′ can be 90% covered by ‘at most’ K|A| translates
N
of A.
Proof. 90% of ξA′ can be covered by ‘at most’
|ξA′+ξP | |A+A| K|A|
ξ
≪ ≤ ,
|ξP | N N
ξ
translates of ξP , which is a subset of A. Similarly, 90% of −ξA′ can be coveredby ‘at most’
ξ
|−ξA′−ξP | |A+A| K|A|
ξ
≪ ≤ ,
|ξP | N N
ξ
translates of ξP ⊂A.
ξ
3.2 Four Cases
For any set B ⊂F, define R(B) to be the set
b −b
1 2
R(B):= :b ,b ,b ,b ∈B,b 6=b .
1 2 3 4 3 4
(cid:26)b −b (cid:27)
3 4
The remainder of the proof is now divided into four cases corresponding to the nature of the sets
R(A˜ ) and R(B ).
x0 y0
Case 1 R(A˜ ) is a subfield:
x0
Case 1.1: First of all, it is possible that |A˜ |≤ |A|6/7. If so, then combining this bound with (9), it
x0
can be deduced that
|A|27/7 ≫LM.
Using (8), rearranging,and then subsequently applying (6) implies that
|A|3 2
|A|41/7 ≫M2 ≫ ,
(cid:18)Klog|A|(cid:19)
and a simple rearrangementof the above gives
|A|1/14
K ≫ ,
log|A|
as required.
Case 1.2: The other possibility here is that |A˜ | > |A|6/7. In this case, since R(A˜ ) is a subfield,
x0 x0
Lemma 2.1 tells us that A˜ ⊆cR(A˜ )+d for some c,d∈F, and clearly
x0 x0
6
|A∩(cR(A˜ )+d)|≥|A˜ |>|A|6/7.
x0 x0
Then, the hypotheses of Theorem 1.4 imply that
|R(A˜ )|14/25 ≥|A∩(cR(A˜ )+d)|≥|A˜ |,
x0 x0 x0
and therefore
|R(A˜ )|≥|A˜ |25/14.
x0 x0
For some r ∈R(A˜ ), define E (A˜ ) to be the number of solutions to
x0 r x0
a +ra =a +ra ,
1 2 3 4
such that a ,a ,a ,a ∈ A˜ . A solution is considered to be trivial if a = a , and non-trivial
1 2 3 4 x0 2 4
otherwise. Summing over all r ∈R(A˜ ) yields
x0
E (A˜ )=|{all trivial solutions}|+|{all non-trivial solutions}|
r x0
r∈RX(A˜x0)
≤|A˜ |2|R(A˜ )|+|A˜ |4
x0 x0 x0
≤|A˜ |2|R(A˜ )|+|A˜ |31/14|R(A˜ )|.
x0 x0 x0 x0
Since the last term is dominant, it follows that
E (A˜ )≪|A˜ |31/14|R(A˜ )|.
r x0 x0 x0
r∈RX(A˜x0)
Hence, there exists some r = p−q ∈ R(A˜ ), such that E (A˜ ) ≪ |A˜ |31/14. Fix this r and
s−t x0 r x0 x0
corresponding elements p,q,s,t∈A˜ .
x0
Let A˜′ be any subset of A˜ such that |A˜′ | ≈ |A˜ |. This vague subset is introduced at this stage
x0 x0 x0 x0
so that the covering lemma can later be applied effectively. Now, apply Cauchy-Schwarzin the usual
way:
2
|A˜ |4 ≈|A˜′ |4 = ν(x) ≤|A˜′ +rA˜′ ||E (A˜′ )|≪|A˜′ +rA˜′ ||A˜ |31/14,
x0 x0 x0 x0 r x0 x0 x0 x0
x∈A˜′X+rA˜′ 
 x0 x0 
where ν(x) is the number of representations of x as an element of A˜′ +rA˜′ . This implies that
x0 x0
|A˜ |25/14 ≪|A˜′ +rA˜′ |≤|pA˜′ −qA˜′ +sA˜′ −tA˜′ |.
x0 x0 x0 x0 x0 x0 x0
Now,byApplication3.2,eachofpA˜′ ,−qA˜′ ,sA˜′ and−tA˜′ canbe 90%coveredby≪ K|A| trans-
x0 x0 x0 x0 N
lates of A. Therefore, the subset A˜′ is chosen earlier in the proof so that each of pA˜′ ,−qA˜′ ,sA˜′
x0 x0 x0 x0
and −tA˜′ are covered completely by these copies of A. Applying the covering lemma four times
x0
yields
7
K|A| 4 K7|A|5
|A˜ |25/14 ≪|A+A+A+A| ≪ ,
x0 (cid:18) N (cid:19) N4
where the last step follows from (3). Applying bound (9) from Lemma 3.1 and rearranging gives
M50/14N6/14 ≪K7|A|5|A|75/14.
Next, use (7) to obtain
M4 ≪K7|A|5|A|75/14|A|12/14 =K7|A|157/14.
Finally, apply (6) and rearrange to get
|A|1/14
K ≫ ,
(log|A|)4/11
which is slightly better than required.
Case 2 - R(A˜ )6=R(B ):
x0 y0
Case 2.1: There is some element r ∈R(A˜ ) such that r ∈/ R(B ). Fix this r = a1−a2 and elements
x0 y0 a3−a4
a ,a ,a ,a ∈A˜ representing it. Since r ∈/ R(B ), there exist only trivial solutions to
1 2 3 4 x0 y0
b +rb =b +rb , (11)
1 2 3 4
such that b ,b ,b ,b ∈ B . Let B′ be some subset of B , with cardinality |B′ | ≈ |B |. Once
1 2 3 4 y0 y0 y0 y0 y0
again, this subset is required for the benefit of applying the covering lemma, and can be specified
later. The absence of non-trivial solutions to (11) implies that
|B′ |2 ≤|B′ +rB′ |.
y0 y0 y0
So,
NL 2
≤|a B′ −a B′ +a B′ −a B′ |.
(cid:18)|A|(cid:19) 1 y0 2 y0 3 y0 4 y0
By Application 3.2, each of a B ,−a B ,a B and −a B can be 90% coveredby ≪ K|A| trans-
1 y0 2 y0 3 y0 4 y0 N
lates of A. By choosing an appropriate subset B′ , we can ensure that each of the four terms in the
y0
above sumset get fully covered. Therefore, Application 3.2 is used four times in order to deduce that
2 4 4
NL K|A| K|A|
≪ |A+A+A+A|≪ K3|A|.
(cid:18)|A|(cid:19) (cid:18) N (cid:19) (cid:18) N (cid:19)
The above inequality can be rearrangedinto the form
M2N2 ≪K7|A|7.
An application of (7) and then subsequently (6) implies that
8
|A|1/11
K ≫ ,
(log|A|)4/11
which is a better bound than required.
Case 2.2: There is some r ∈ R(B ) such that r ∈/ R(A˜ ). Fix this r = p−q as well as elements
y0 x0 s−t
p,q,s,t∈B representing r. Since (p,y )∈P and all lines through the origin in P support approx-
y0 0
imately N points in P, it can be deduced that the line with gradient y0 supports ≈ N points in P.
p
The same can be said of the lines with gradients y0,y0 and y0 respectively. In other words, each of
q s t
y0,y0,y0 and y0 belong to Ξ.
p q s t
Next, since P is symmetric through the line y =x, it can be observed that p , q , s and t are also
y0 y0 y0 y0
elements of Ξ, the set of slopes supporting ≈N points from P. r can be rewritten as
p − q
r = y0 y0.
s − t
y0 y0
Let A˜′ be a positively proportioned subset of A˜ , to be chosen later in order to apply the covering
x0 x0
lemma. Now, since r ∈/ R(A˜′ ), there exist only trivial solutions to the equation
x0
a +ra =a +ra ,
1 2 3 4
such that a ,a ,a ,a ∈A˜′ . Therefore,
1 2 3 4 x0
|A˜ |2 ≈|A˜′ |2 ≤|A˜′ +rA˜′ |
x0 x0 x0 x0
p q s t
≤ A˜′ − A˜′ + A˜′ − A˜′ .
(cid:12)y x0 y x0 y x0 y x0(cid:12)
(cid:12) 0 0 0 0 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
By Application 3.2, each of p A˜ ,− q A˜ , s A˜ and − t A˜ can be 90% covered by ≪ K|A|
y0 x0 y0 x0 y0 x0 y0 x0 N
translates of A. The subset A˜′ may be chosen earlier so that each of p A˜′ ,− q A˜′ , s A˜′ and
x0 y0 x0 y0 x0 y0 x0
− t A˜′ are covered in their entirety by the translates of A.
y0 x0
Therefore, four applications of the covering lemma, along with (9) and (3) yield
LM 2 K|A| 4 K7|A|5
≪ |A+A+A+A|≪ .
(cid:18)|A|3(cid:19) (cid:18) N (cid:19) N4
This can be rearrangedto give
M4 ≪K7|A|11.
Finally, apply (6) to deduce that
|A|1/11
K ≫ .
(log|A|)4/11
From this point forward, we may assume that R(A˜ )=R(B ).
x0 y0
Case 3(Worst case) - R(A˜ )R(A˜ )*R(A˜ )=R(B ):
x0 x0 x0 y0
9
So, there must exist some a ,a ,b ,b ,c ,c ,d ,d ∈A˜ such that
1 2 1 2 1 2 1 2 x0
a −b a −b
1 1 2 2 ∈/ R(A˜ ).
c −d c −d x0
1 1 2 2
Claim: This implies that there exists some elements a,b,c,d,e,f ∈A˜ ⊆A such that
x0 x0
(a−b)(c−d)
r := ∈/ R(A˜ )=R(B ).
e−f x0 y0
Suppose not. Then certainly
p−q a −b
2 2
=a −b ∈R(A ),
s−t 1 1c −d x0
2 2
for some elements p,q,s,t∈A . Therefore,
x0
1 p−q
∈/ R(A ).
c −d s−t x0
1 1
Note also that R(A ) is closed under reciprocation, and thus
x0
s−t
c −d ∈/ R(A ),
1 1p−q x0
which proves the claim.
Consider subsets A˜ ,A˜ and A˜ of B ; recall that these subsets were defined in the statement of
a˜ c˜ e˜ y0
Lemma 3.1.
Furthermore, let A˜′ ⊂A˜ ,A˜′ ⊆A˜ and A˜′ ⊆A˜ be subsets with cardinality |A˜′|≈|A˜ |,|A˜′|≈|A˜ |,
a˜ a˜ c˜ c˜ e˜ e˜ a˜ a˜ c˜ c˜
and |A˜′|≈|A˜ |. These subsets are to be specified later in order to apply the covering lemma.
e˜ e˜
Since A˜′ and A˜′ are subsets of B , there exist only trivial solutions to the equation
a˜ e˜ y0
a +ra =a +ra ,
1 2 3 4
such that a ,a ∈ A˜′ and a ,a ∈ A˜′, as otherwise r would be an element of R(B ). This implies
1 3 e˜ 2 4 a˜ y0
that
|A˜′||A˜′|≤|A˜′ +rA˜′|. (12)
e˜ a˜ e˜ a˜
On the other hand, Lemma 2.2 can be applied with X = c−dA˜′ to bound the right hand side of the
e−f c˜
above inequality as follows:
|c−dA˜′ +A˜′||A˜′ +(a−b)A˜′|
|A˜′ +rA˜′|≤ e−f c˜ e˜ c˜ a˜
e˜ a˜ |A˜′|
c˜
|cA˜′ −dA˜′ +eA˜′ −fA˜′||aA˜′ −bA˜′ +A˜′|
≤ c˜ c˜ e˜ e˜ a˜ a˜ c˜ .
|A˜′|
c˜
Observe also that crucially, aA˜′ ⊂ aP ⊂ A, and similarly cA˜′ ⊂ A and eA˜′, meaning there is no
a˜ a c˜ e˜
need to apply the covering lemma for these terms of the sum. Also, by definition, A˜′ ⊂A.
c˜
10