On growth of the set $A(A+1)$ in arbitrary finite fields

Let $\mathbb{F}_q$ be a finite field of order $q$, where $q$ is a power of a prime. For a set $A \subset \mathbb{F}_q$, under certain structural restrictions, we prove a new explicit lower bound on the size of the product set $A(A + 1)$. Our result improves on the previous best known bound due to Zhelezov and holds under more relaxed restrictions.


Introduction
Let p denote a prime, F q the finite field consisting of q = p m elements and F * q = F q \{0}. For sets A, B ⊂ F q , we define the sum set A + B = {a + b : a ∈ A, b ∈ B} and the product set AB = {ab : a ∈ A, b ∈ B}. Similarly, we define the difference set A − B and the ratio set A/B.
The sum-product phenomenon in finite fields is the assertion that for A ⊂ F q , the sets A + A and AA cannot both simultaneously be small unless A closely correlates with a coset of a subfield. A result in this direction is due to Li and Roche-Newton [6], who showed that if |A ∩ cG| |G| 1/2 for all subfields G and elements c in F q , then max{|A + A|, |AA|} (log |A|) −5/11 |A| 1+1/11 .
In the realm of small sets A ⊂ F q , with |A| p 5/8 , Stevens and de Zeeuw [9] obtained Warren [11], further improved this bound to (log |A|) −7/6 |A| 1+2/9 under the constraint |A| p 1/4 . Both of these results are based on a bound on incidences between lines and Cartesian products, proved in [9], which in turn relies on a bound on incidences between points and planes due to Rudnev [8]. We point out that the main result of [8] has led to many quantitatively strong sum-product type estimates, however these estimates are restricted to sets which are bounded in size in terms of the characteristic p.
Our main result, stated below, relies on a somewhat more primitive approach towards the sum-product problem in finite fields, often referred to as the additive pivot technique. Specifically, we adopt our main tools and ideas from [4] and [6].
for all proper subfields G of F q and elements c ∈ F q . Then for all α ∈ F * q , we have Theorem 1 provides a quantitative improvement over the relevant estimates implied by (1) and holds under a more relaxed condition than those given by (2). It also improves on (5) in the range q 1/2 |A| q 1/2+1/102 . Given a set A ⊂ F q , we define the additive energy of A as the quantity E + (A) = |{(a 1 , a 2 , a 3 , a 4 ) ∈ A 4 : a 1 + a 2 = a 3 + a 4 }|.
As an application of Theorem 1, we give a bound on the additive energy of subsets of F q .
for all proper subfields G of F q and elements c ∈ F q . Then for any α ∈ F * q , we have Consequently, under restriction (7), we have

Asymptotic notation
We use standard asymptotic notation. In particular, for positive real numbers X and Y , we use X = O(Y ) or X Y to denote the existence of an absolute constant c > 0 such that X cY . If X Y and Y X, we write X = Θ(Y ) or X ≈ Y . We also use X Y to denote the existence of an absolute constant c > 0, such that X (log Y ) c Y .

Preparations
For X ⊂ F q , let R(X) denote the quotient set of X, defined by We present a basic extension of [10, Lemma 2.50].
Lemma 3. Let X ⊂ F q and r ∈ F * q . If r ∈ R(X), for any nonempty subsets X 1 , X 2 ⊆ X, we have Proof. Consider the mapping φ : Then we get which contradicts the assumption that r ∈ R(X). We deduce that φ is injective, which in turn implies the required result.
The next lemma, which appeared in [10, Corollary 2.51], is a simple corollary of Lemma 3.
We have extracted Lemma 5, stated below, from the proof of the main result in [6].
Then R(X) is the subfield of F q generated by X.
The next result has been stated and proved in the proof of [7, Theorem 1].
Then there exists r ∈ R(X) such that for any subset X ⊂ X with |X | ≈ |X|, we have The following lemma enables us to extend our main result to sets which are larger than q 1/2 . See [1, Lemma 3] for a proof. .
Lemma 8. Let X, B 1 , B 2 be nonempty subsets of an abelian group. We have In particular, for A ⊂ F * q , by a multiplicative application of Lemma 8, we have the useful inequality In the next two lemmas we state variants of the Plünnecke-Ruzsa inequality, which can also be found in [5].
Lemma 9. Let X, B 1 , . . . , B k be nonempty subsets of an abelian group. Then Lemma 10. Let X, B 1 , . . . , B k be nonempty subsets of an abelian group. For any 0 < < 1, there exists a subset X ⊆ X, with |X | (1 − )|X| such that The following two lemmas are due to Jones and Roche-Newton [4].
translates of Y . Similarly, (1− )|X| elements of X can be covered by this many translates of −Y .
Next, we record a popularity pigeonholing argument. A proof is provided in [3, Lemma 13. Let X be a finite set and let f be a function such that f (x) > 0 for all Additionally, if f (x) M for all x ∈ X, then |Y | K/(2M ).
For sets X, Y ⊆ F q , we define the multiplicative energy between X and Y as the quantity Then, we have the identities the electronic journal of combinatorics 27(4) (2020), #P4.3 By a simple application of the Cauchy-Schwarz inequality we have The remaining two lemmas together form the basis for the proof of Theorem 1. Lemma 15 is a slight generalisation of [7, Lemma 3].
Lemma 14. Let X, Y ⊂ F q , with |Y | |X|. There exists a set D ⊆ Y /X and an integer N |Y | such that E × (X, Y ) (log |X|)|D|N 2 and |D|N < |X||Y |. Also, for ξ ∈ D we have r Y :X (ξ) ≈ N . Namely, the set of points is supported on |D| lines through the origin, with each line containing Θ(N ) points of P.
Proof. For j 0, let L j = {ξ ∈ Y /X : 2 j r Y :X (ξ) < 2 j+1 }. Then, by (11), we have By the pigeonhole principle there exists some N 1 such that, letting Furthermore, by (10), we have Lemma 15. Let X, Y ⊂ F q . Suppose P ⊂ X × Y is a set of points supported on L lines through the origin, with each line containing Θ(N ) points of P , so that |P | ≈ LN . For x * ∈ X and y * ∈ Y , we write Y x * = {y ∈ Y : (x * , y) ∈ P } and X y * = {x ∈ X : (x, y * ) ∈ P }. There exists a popular abscissa x 0 and a popular ordinate y 0 , so that such that for every z ∈ Y x 0 , we have Proof. Observing that y∈Y |X y | = |P | ≈ LN, by Lemma 13, there exists a subset Y ⊆ Y such that, for all y ∈ Y , we have |X y | LN/|Y |. Let P = {(x, y) ∈ P : y ∈ Y } so that |P | LN . Then By Lemma 13, there exists a subset X ⊆ X such that for all x ∈ X we have Letting P = {(x, y) ∈ P : x ∈ X }, we have |P | LN. Let D = {y/x : (x, y) ∈ P } and let D ⊆ D denote the set of elements ξ such that the lines l ξ , determined by ξ, each contain Ω(N ) points of P . It follows by Lemma 13 that |D | L. Now, we proceed to establish a lower bound on the sum We For a fixed ξ ∈ D , the inner sum may be bounded by the observation that Recall that |D | L and that for ξ ∈ D , we have |P ξ | N . Then, by (15), we have By the pigeonhole principle, applied to (16), there exist (x 0 , y 0 ) ∈ X × Y such that By our assumption, that every line through the origin contains O(N ) points of P , it follows that for all z ∈ Y , we have |P z/x 0 | N . Then, letting Y x 0 ⊆ Y x 0 to denote the set of z ∈ Y x 0 with the property that by Lemma 13, we have

Proof of Theorem 1
It suffices to prove the required result for α = 1. Then the general statement immediately follows since under condition (6) the set A can be replaced by any of its dilates cA, for c ∈ F * q . Without loss of generality assume 0 ∈ A. By Lemma 12, combined with (9), there exists a subset A ⊆ A, with |A | ≈ |A|, such that By Lemma 10 there exists a further subset A ⊆ A , with |A | ≈ |A |, such that Since |A | ≈ |A|, we reset the notation A back to A and henceforth assume the inequalities We apply Lemma 14 to identify a set D ⊆ A/(A + 1) and an integer N 1 such that for ξ ∈ D we have r A:(A+1) (ξ) ≈ N . Additionally, letting L = |D|, in view of (12), we have We define P ⊆ (A + 1) × A by Then |P | ≈ LN . Now, since LN < |A| 2 and N < |A|, we get the electronic journal of combinatorics 27(4) (2020), #P4.3 For ξ ∈ D, we define the projection onto the x-axis of the line with slope ξ as Similarly for λ ∈ D −1 let Q λ = {y : (λy, y) ∈ P } ⊂ A.
By Lemma 15, with X = A + 1 and Y = A, there exists a pair of elements (x 0 , y 0 ) ∈ (A + 1) × A such that the sets A x 0 ⊆ A and B y 0 ⊆ A + 1 satisfy Moreover, there exists a further subsetÃ such that for all z ∈Ã x 0 , letting We require the following corollary of Lemma 11 throughout the remainder of the proof.
Claim 16. For n 4 let a 1 , . . . , a n denote arbitrary elements ofÃ x 0 . Given any set C ⊂ A + 1, there exists a subset C ⊂ C, with |C | ≈ |C|, such that the sets a i C can each be covered by There exists a subset A ⊆Ã x 0 , with |A | ≈ |Ã x 0 |, such that for 1 i 4 the sets b i A can each be covered by O(Γ) translates of ±y 0 A.
Proof. We apply Lemma 11, with X = a i C, Y = a i P a i /x 0 , Z = A, x = a i , y = a i and 0 < < 1/16. Then there exist sets C a i ⊆ C with |C a i | (1 − )|C| such that each of a i C a i can be covered by translates of a i P a i /x 0 ⊂ x 0 A and by at most as many translates of −x 0 A. Let C = C a 1 ∩ · · · ∩ C an , so that |C | (1 − n )|C| (3/4)|C|. Then, by (9) and (21), it follows that (25) denotes the number of translates of ±x 0 A required to cover the sets a i C for 1 i n. Next, we apply Lemma 11, with X = b iÃx 0 , Y = b i Q b i /y 0 , Z = A, x = b i and y = 0. Recalling (21), (22), (23) and proceeding similarly as above, we can identify a subset A ⊆Ã x 0 , with |A | ≈ |Ã x 0 |, such that the sets b i A are each fully contained in O(Γ) translates of ±y 0 A.
We split the proof into four cases based on the nature of the quotient set R(Ã x 0 ).
By Lemma 3, for any subset Y ⊆ B y 0 with |Y | ≈ |B y 0 |, we have Then for any subset Y ⊆Ã x 0 with |Y | ≈ |Ã x 0 |, by Lemma 3, we have By the second part of Claim 16, there exists a subset A ⊂Ã x 0 , with |A | ≈ |Ã x 0 |, such that the sets aA , bA and cA are each fully contained in O(Γ) translates of y 0 A and dA can be covered by O(Γ) translates of −y 0 A. Thus, setting Y = A , by (28), we have Applying (18) yields M 10 L 2 |A| 9 |A(A + 1)| 40 .
. There exist elements a, b, c, d ∈Ã x 0 such that Let Y 1 ⊆ B y 0 and Y 2 ⊆ S a be any sets with |Y 1 | ≈ |B y 0 | and |Y 2 | ≈ |S a |. By Lemma 10, Recall that Y 1 ⊆ B y 0 and Y 2 ⊆ S a ⊆ B y 0 . Then Lemma 3 gives Thus, by (29) we have Since Y 1 , Y 2 ⊆ B y 0 ⊆ A + 1, we have Recall that |Y 1 | ≈ |Y 1 | ≈ |B y 0 | and |Y 2 | ≈ |S a |. Then by (22), (24) and noting that Now, by Claim 16, there exist positively proportioned subsets B y 0 ⊆ B y 0 and S a ⊆ S a such that cB y 0 and dB y 0 can be covered by Using (17) and (18) . There exist elements a, b, c, d, e ∈Ã x 0 such that Given any set Y 1 ⊆ B y 0 , recalling that S a ⊆ B y 0 , it follows from Lemma 3 that For an arbitrary set Y 2 , we apply Lemma 9, with X = b−c d−e Y 2 , to get By Claim 16, we can identify sets C 1 ⊆ S d and C 2 ⊆ P c/x 0 with |C 1 | ≈ |S d | and |C 2 | ≈ |P c/x 0 | ≈ N , such that eC 1 is covered by translates of x 0 A and bC 2 is covered by translates of −x 0 A. We set Y 1 = C 1 and Y 2 = C 2 . Then, by (21), (24) and particularly noting that dY 1 , cY 2 ⊂ x 0 A and Y 2 ⊂ A + 1, we have