Products of Differences over Arbitrary Finite Fields

There exists an absolute constant $\delta>0$ such that for all $q$ and all subsets $A \subseteq \mathbb{F}_q$ of the finite field with $q$ elements, if $|A|>q^{2/3 - \delta}$, then \[ |(A-A)(A-A)| = |\{ (a -b) (c-d) : a,b,c,d \in A\}|>\frac{q}{2}. \] Any $\delta<1/13,542$ suffices for sufficiently large $q$. This improves the condition $|A|>q^{2/3}$, due to Bennett, Hart, Iosevich, Pakianathan, and Rudnev, that is typical for such questions. Our proof is based on a qualitatively optimal characterisation of sets $A,X \subseteq \mathbb{F}_q$ for which the number of solutions to the equation \[ (a_1-a_2) = x (a_3-a_4) \, , \; a_1,a_2, a_3, a_4 \in A, x \in X \] is nearly maximum. A key ingredient is determining exact algebraic structure of sets $A, X$ for which $|A + XA|$ is nearly minimum, which refines a result of Bourgain and Glibichuk using work of Gill, Helfgott, and Tao. We also prove a stronger statement for \[ (A-B)(C-D) = \{ (a -b) (c-d) : a \in A, b \in B, c \in C, d \in D\} \] when $A,B,C,D$ are sets in a prime field, generalising a result of Roche-Newton, Rudnev, Shkredov, and the authors.

Using standard notation for set addition and multiplication, one may state the last result for (w − x)(y − z) as: If |A| > q 2/3 , then All of these results have straightforward proofs using the completion method, which roughly involves replacing a sum of squares over a large set in a space by the sum of squares over the whole space. This seemingly expensive step allows one to take advantage of properties like character orthogonality and produces non-trivial results for large sets. The completion method has been applied to additive characters, multiplicative characters, and the spectral theory of regular graphs.
A unifying approach for "pinned" versions of the results was presented in [35]. The fact that two points in the two-dimensional vector space F 2 q determine a unique line, combined with a second moment argument, is used as a substitute for the so-called expander mixing lemma [21,1]. The corresponding result for the polynomial (w − x)(y − z) was omitted from [35], but it is straightforward to prove that if |A| > q 2/3 , then there exist a, b ∈ A such that

Main result
Going below the q 2/3 threshold for any of the polynomials listed above has proven to be surprisingly hard, even in prime order fields where there are additional tools. Let us write p for the order of the field when it is a prime. When A ⊆ F p is a multiplicative subgroup, Heath-Brown and Konyagin applied Stepanov's method [25] to show that |A + A| > c min{p, |A| 3/2 }, matching the p 2/3 threshold (note that A + A = A + AA = A(A + A)). Similarly, Pham, Vinh, and de Zeeuw used Rudnev's point-plane incidence bound in F 3 p [46,66] to prove that |f (A, A, A)| > c min{p, |A| 3/2 } for a large class of quadratic polynomials in three variables [42].
The first step in going below the p 2/3 threshold for prime p was taken in [40] for (w − x)(y − z), where it was shown that there exist constants c and C such that if |A| > Cp 5/8 , then |(A − A)(A − A)| > p/2. This argument does not carry through to arbitrary F q , because it relies on non-trivial bounds on the number of ordered solutions to the equation (a 1 −a 2 ) = x(a 3 −a 4 ) with the a i ∈ A and x in an arbitrary set X ⊆ F p . Bourgain [4, Theorem C] proved a non-trivial bound in F p , but such a theorem is false in F q , in general.
Roche-Newton, Rudnev, Shkredov, and the authors [37,Theorem 27] improved this result, showing that |A| > Cp 3/5 implies the "pinned" result |(A − a)(A − b)| > p/2. The proof uses a bound on the number of ordered collinear triples in A×A ⊆ F 2 p that is better than what follows from the completion method. In terms of incidence geometry results, the method of proof combines Vinh's point-line incidence theorem [63] with a point-line incidence theorem of Stevens and de Zeeuw [58]. This method is also specific to prime fields and does not carry over in arbitrary F q either.
Our main theorem breaks the q 2/3 threshold over arbitrary finite fields.
Theorem A. There exist absolute positive constants C, k > 0 such that for all prime powers q, if A ⊆ F q satisfies |A| > Cq The proof of Theorem A uses most of the existing techniques for sum-product questions in arbitrary finite fields at different stages of the proof: the completion method in Section 3, the pivot method in Section 4, and character sum estimates in Section 5.
Before discussing the proof of Theorem A in more detail, we make some remarks.
2. The proof of Theorem A contains a simple example of a set A ⊂ F q of cardinality |A| = q 2/3 with the property |(A−A)(A−A)| = (1+o(1))q/2; see Proposition 11. This shows that (A − A)(A − A) cannot have density greater than 1/2, in general. Hart, Iosevich, and Solymosi [23] conjectured that there exists ε > 0 such that if |A| > Cq 1/2+ε , then (A − A)(A − A) = F q . The example shows that ε must be greater than 1/6. The best known lower bound on |A| for this complementary question is Cq 3/4 and is due to Hart, Iosevich, and Solymosi [23,Theorem 1.4].
3. Vinh has obtained expansion with threshold q 5/8 for the four-variable polynomial wx + (y − z) 2 [64] in finite fields F q with odd characteristic. Vinh's elegant proof is based on ideas of Garaev [14] and Solymosi [57] and strongly relies on having a polynomial where the additive and multiplicative parts are separated.
4. Rudnev, Stevens, and Shkredov have proved the strongest expansion result in F p we are aware of. They worked with the four-variable rational function (wx − y)(w − z) −1 [47] and showed there exist absolute constants c, C, and k such that if A ⊆ F p satisfies |A| > C log(p) k p 25/42 , then |f (A, A, A, A)| ≥ cp. The exponent 25/42 is slightly smaller than the exponent 3/5, which is known for (w −x)(y −z) [37], the three variable rational function (x−y)(x−z) −1 [37], and the four variable rational functions covered by the forthcoming Theorem F.

Proof outline and structure of the paper
Theorem A is proved using a dichotomy argument, carried out in three steps: a completion argument that works for generic sets, an exact structural characterisation of sets for which the completion argument fails, and a character sum argument that deals with the structured sets.
Standard results from arithmetic combinatorics used throughout the paper are given in Section 2. To make the statements of the forthcoming theorems more compact, we make one preliminary definition: for A ⊆ F q and ξ ∈ F * q we use E(A, ξA) to denote the number of ordered solutions to the equation with all variables a 1 , . . . , a 4 in A.

Section 3: The generic sets completion argument
The first step is similar to arguments used to deal with large subsets of finite fields. It mimics the argument in [40] to show that Theorem A holds unless there exists a set X ⊂ F q of cardinality roughly q/|A| such that E(A, ξA) is greater than |A| 3 /K, for all ξ ∈ X and some suitable K. This is is achieved by a careful analysis of a completion argument.
More precisely, we prove the following theorem in Section 3.
Theorem B. Let A be a subset of F q . Suppose that there is a positive constant K such that |A| ≤ q/(4K) and There exists a subset X ⊆ F q such that E(A, ξA) ≥ |A| 3 /K for all ξ ∈ X and It is easy to show, for example using multiplicative characters, that D × (A) = |A| 8 /q + O(q|A| 5 ), yielding the |A| > Cq 2/3 threshold. The |A| 8 /q term, which comes from the principal multiplicative character, is the expected value of D × (A) for large random sets.

Section 4: Inverse sum-product results
The next step in the proof is to show that the conclusion of Theorem B is impossible for K = q δ for some small δ > 0, unless A has a specific structure.
A result of Bourgain implies that the conclusion in Theorem B is never true in F p when K is small power of p and |A||X| is about p. Namely, [4,Theorem C] states that there exists an absolute constant η > 0 such that for sets A, X ⊆ F p satisfying |A||X| < p, we have This suffices to go below p 2/3 in F p [36,40].
However, Bourgain's result is not true over arbitrary finite fields. If X is contained in a proper subfield and A is a vector space over the subfield, then E(A, ξA) = |A| 3 for all ξ ∈ X. Taking q = p 3 , X to be the subfield isomorphic to F p and A to be a two-dimensional vector space over X, produces a pair of sets A and X with |A||X| = q, where the conclusion (and therefore the hypothesis) of Theorem B is false.
Therefore, we must deal with pairs of sets A, X where for a parameter K that will eventually be taken to be a small power of q.
Bourgain and Glibichuk proved in [6, Proposition 2] that if (2) holds, then X must have large intersection with the dilate of a subfield. This characterisation is hard to apply in our context because we are more interested in A than in X. We strengthen this result to show that the example given above is essentially the only way that (2) can hold: a large subset of X must be contained in a dilate of a subfield and A must have a large intersection with a vector space over the same subfield. Though X itself may be small compared to the subfield, A must have comparable size to the vector space. This characterisation is qualitatively best possible.
The next theorem contains a precise statement of this result. For brevity, we write f g if there exist constants C, k such that f ≤ C log k (q) g.
Theorem C. Let A and X be subsets of F * q and let K ≥ 1 be a real number such that There exists an elementā in A, a subfield F ⊆ F q , and an F -vector space V ⊆ F q such that Further, there is an element x 0 in X and subsets A 1 ⊆ A −ā, X ′ ⊆ x −1 0 X such that F is the subfield generated by X ′ , V is the F -vector space generated by A 1 , and Moreover, if E(A, ξA) ≥ |A| 3 /K for all ξ ∈ X we have |X ′ | K −4 |X| and only require K 1504 |X|.
Theorem C constitutes the biggest step towards the proof of Theorem A. Its proof has two main parts. The second part in the proof of Theorem C is to extract exact algebraic structure for A and X when |A + XA| is nearly minimum. Roughly, if A + XA is small, then A has large relative density in the vector space it spans over the field generated by X. In Section 4 we prove the following more precise result.
Theorem D. Let W be a subset of an F q -vector space, let X ⊆ F q be a set of scalars, and let F = X be the subfield generated by X. If In Theorem D we characterise the whole of W , whereas in Theorem C only a large subset of A. This has to do with additive energy being a robust measure of additive structure (perturbing the set a little makes little difference), while cardinality is a delicate measure of additive structure (perturbing the set a little can make a big difference).
In the context of Theorem A, where |A| is q 2/3 (up to small powers of q) and |X| is q 1/3 (up to small powers of q) , we can extract from Theorem C even more information and show that the example mentioned above (q = p 3 , X is a subset of the subfield isomorphic to F p and A is a two-dimensional vector space over F p ), is morally the only way that (2) can occur.
There exists a subfield F of cardinality q 1/3 ,ā in A, and a two-dimensional vector space The proof of Theorem C combines the work of Bourgain and Glibichuk with ideas of Glibichuk and Konyagin [16], Helfgott [27,28,29], Gill and Helfgott [15], and Tao [61]. In the background is the work of Ruzsa [50] and Gowers [17,18].

Section 5: Large subsets of a vector space
By Theorem B and Corollary 1 we are left to deal with the case where q is a cube, F is the subfield of cardinality q 1/3 , and a translate of A has large intersection with a two-dimensional vector space over F . To deal with such sets we combine basic algebra with a character sum argument of Hart, Iosevich, and Solymosi [23].
Theorem E. Let q be a cube. Suppose F ⊂ F q is the subfield of cardinality q 1/3 and One may compare Theorem E, which we prove in Section 5, to the result of Hart, Iosevich and Solymosi [ The former states that if |A| > Cq 3/4 for some absolute constant C, then (A − A)(A − A) = F q . Theorem E is analogous because F q is a one-dimensional vector space over itself and F q F q = F q . The exponent in Theorem E is worse because we work in a two-dimensional vector space.
The latter states that if |A| > Cq 2/3 for some absolute C, then (A − A)(A − A) determines a positive proportion of the elements of F q . Restricting A to subsets of V allows one to get a similar result for smaller sets.
To get the best possible lower bound on |A| in Theorem E we apply Weil's bounds on Kloosterman sums. Kloosterman's bounds, which are easier to derive, could be used at no cost in the required lower bound on |A| in Theorem A.

Structure of the rest of the paper
The derivation of Theorem A from Theorems B, Corollary 1, and Theorem E is conducted in Section 6.
In Section 7 we prove a stronger result over prime order fields.
Bourgain's version of Theorem C [4, Theorem C] over prime finite fields is a powerful result, with many applications in the literature. In Section 8 we discuss what is currently known and highlight the strong bounds that follow from Rudnev's point-plane incidence theorem [46].
Appendix A is dedicated to the proof of a sharper version of a result of Bourgain and Glibichuk from [6]. It roughly asserts that if (2) holds, then there are large subsets In Appendix A, we also prove a version of the Balog-Szemerédi-Gowers theorem [17,18] that shows if E(A, B) ≥ (|A||B|) 3/2 /K, then |A ′ + B ′ | K 3 (|A||B|) 1/2 for subsets A ′ ⊆ A, B ′ ⊆ B of relative density 1/K. This is slightly worse than the bounds achieved by Schoen [53] under the hypothesis E(A) ≥ |A| 3 /K, however it applies to the case of distinct A and B, and improves the bounds of Bourgain and Garaev [5].

Notation
For positive quantities X and Y , we use X ≪ Y to mean that X ≤ CY for some constant C > 0 and we use X Y to mean that X ≪ (log Y ) k Y for some constant k > 0. The letter p always denotes a prime and q a prime power; F q is the finite field with q elements and F d p the d-dimensional vector space over F q . For x, y ∈ F p we sometimes write x y instead of xy −1 and implicitly assume y = 0. We use standard notation on set operations. For example we denote by We use representation functions like r A+B (x), which counts the number of ordered ways x ∈ F q can be expressed as a sum a + b with a ∈ A and b ∈ B. Note that A + B is the

Arithmetic combinatorics
In various stages of the proofs we utilise sumset cardinality inequalities that follow from the work of of Plünnecke and Ruzsa [43,48,49,50]. An excellent reference is [51].
Lemma 2 (Sumset inequalities). Let A, B 1 , . . . , B h be finite non-empty sets in a commutative group. The following inequalities are true.
We also make heavy use of the notion of so-called additive energy.
Definition. Let A, B ⊆ F q and ξ ∈ F * q . The additive energy of A and ξB is defined in the following equivalent ways.
The Cauchy-Schwarz inequality and the identity x r A+ξB (x) = |A||B| (each ordered pair (a, b) ∈ A × B contributes 1 to the sum) imply We will use repeatedly the following simple yet powerful lemma; see [ Lemma 3]. We need a straightforward generalisation of additive energy to F q -vector spaces. Given a set A and a non-zero element ξ in the vector space and a set B ⊆ F * q , we define the additive energy E(A, ξB) like in F q .
Lemma 3. Let V be a vector space over F q . If A and S are finite subsets of V , with 0 ∈ S, and B ⊆ F * q is a collection scalars, then Hence there exists ξ ∈ S such that The lemma applied to V = F q , S = F * q , and A = B implies that E(A, ξA) > |A| 3 /K for at most Kq/|A| many ξ. The proof of this fact is implicit in the proof of Theorem B. The majority of the remainder of the paper is devoted to the task of refining this simple corollary of Lemma 3.

Proof of Theorem B
In this section we prove Theorem B, which we recall here.
Theorem B. Let A be a set in F q and D × (A) be the quantity Suppose that there is a positive constant K such that |A| ≤ q/(4K) and There exists a set X ⊆ F q such that E(A, ξA) ≥ |A| 3 /K for all ξ ∈ X and Proof. Let Then There are at most 4|A| 6 solutions where either side of (4) is zero, thus If we set Now we estimate the second moment of Q ξ . First we replace one power of Q ξ by E ξ and estimate the error using (6).
Hence by (5) Now we estimate the remaining sum over ξ. Let Combining the lower bound on D × (A) in the hypothesis of the theorem, the upper bound |A| ≤ q/(4K) (which implies 4|A| 6 ≤ q|A| 5 /K), and equations (8) and (9), we get By the trivial bound Q ξ E ξ ≤ |A| 6 we derive |X| ≥ q/(|A|K).
The final step is to obtain an upper bound on |X|.

Two auxiliary results
The proof of Theorem C relies on two auxiliary results. The first roughly speaking asserts that if the additive energy E(A, ξA) is large for many ξ in X, then |A + XA| is small.
then there exist elementsā ∈ A and x 0 ∈ X and subsets A 1 ⊆ A −ā and X ′ ⊆ x −1 0 X such that: The second auxiliary result is Theorem D, which we restate here.
Theorem D. Let W be a subset of an F q -vector space, let X ⊆ F q be a set of scalars, and let F = X be the subfield generated by X. If We also record the following specialization of Theorem D, which will be used in the proof of Corollary 1.
Corollary 5. Let q > 2 12 be a prime power, let W, X ⊂ F q be sets and K 1 , K 2 , K 3 be a real numbers. Suppose that q 1/4 < |X| < q 1/2 and Corollary 5 and Theorem D are both proved further down in this section. Let us now deduce Theorem C and Corollary 1, assuming the auxiliary results.

Proof of Theorem C and Corollary 1
Theorem C. Let A and X be subsets of F * q and let K ≥ 1 be a real number such that There exists an elementā in A, a subfield F ⊆ F q , and an F -vector space V ⊆ F q such that Further, there is an element x 0 in X and subsets A 1 ⊆ A −ā, X ′ ⊆ x −1 0 X such that F is the subfield generated by X ′ and V is the F -vector space generated by A 1 , and Proof. We are given By Proposition 4 there exist elementsā ∈ A and x 0 ∈ X and subsets A 1 ⊆ A −ā and Since 1505 = (4 · 226 + 92 + 4 · 126) + 5, the hypothesis K 1505 |X| means and we may apply Theorem D. Letting F be the field generated by X", we get For the second lower bound on The special case where E(A, ξA) ≥ |A| 3 /K for all ξ ∈ X is deduced similarly from the last part of Proposition 4.
There exists a subfield F of cardinality q 1/3 ,ā in A, and a two-dimensional vector space Proof. Proceeding like in the proof of Theorem C, only using the special case of Proposition 4, we get elementsā ∈ A and x 0 ∈ X and subsets A 1 ⊆ A −ā and X ′ ⊆ x −1 0 X such that As observed in the proof of Theorem C, because K 1504 |X|, we may apply Corollary 5 to A ′ and X ′ as long as The condition q > 2 12 is irrelevant in this context. Assuming (11) for now, we conclude that there is a subfield F of cardinality q 1/3 and a two-dimensional vector space V over as desired.

A pivot argument for vector spaces
We now prove Theorem D, which is a specialization of Gill and Helfgott's pivot theorem [15]. Our approach is inspired by Tao's concept of involved elements [61, Section 1.5].
Throughout the remaining of this section F is a finite field and V is a vector space over F.
For a subset X ⊆ F we denote by X the subfield generated by X. We use X + and X × to denote the additive and multiplicative groups generated by X, respectively, so that For a subset W ⊆ V we denote by Span X (W ) the vector space over X generated by the vectors in W . The key definition is the following.
Definition (Pivots and involved elements). Let V be a vector space over a finite field An element is involved with the sets W and X, if it is not a pivot. We denote by I W,X the set of involved elements.
The rest of this section is devoted to showing that the collection of involved elements has an exact algebraic structure (namely, it is a vector space) assuming that W and X have weak algebraic structure (namely, |W + XW | is close to minimum).
The next lemma shows that involved elements have low complexity in terms of W and X.
Lemma 6 (Properties of involved elements). Let V be a vector space over a finite field F, W ⊆ V and X ⊆ F. A vector ξ ∈ V is an involved element for (W, X) if and only if Proof. Both parts follow from the non-injectivity of φ ξ .
For the first part note that there exist distinct (v 1 , For the second part note |W + Xξ| = φ ξ (W, X).
Lemma 6 implies that I W,X ⊆ Span X (W ), since every involved element ξ can be written The following proposition proves the reverse inclusion when |W + XW | and |W + W | are sufficiently small.
Proposition 7. Let V be a vector space over a finite field F, W ⊆ V and X ⊆ F. In the notation used in this section, if then Span X (W ) ⊆ I W,X and therefore Span X (W ) = I W,X .
In the proof of Proposition 7 we will repeatedly use a key observation: if ξ is an involved element, then we may strengthen the bound |W + Xξ| < |X||W | by exploiting the low complexity of ξ.
Lemma 8. Let V be a vector space over a finite field F, W ⊆ V and X ⊆ F. Suppose that |W + XW | ≤ K 1 |W | and |W + xW | ≤ K 3 |W | for all x ∈ X. If ξ is an involved element, then Proof. Since ξ is an involved element, there exist x 1 , x 2 ∈ X, w 1 , w 2 ∈ W such that By the Plünnecke-Ruzsa inequality for different summands in Lemma 2 we have We apply Lemma 8 to prove that the set of involved elements is closed under both addition and scalar multiplication, under suitable conditions on sumset cardinalities.
Lemma 9 (Closure of involved elements under addition). Let V be a vector space over a finite field F, W ⊆ V and X ⊆ F. If |W + XW | ≤ K 1 |W |, |W + W | ≤ K 2 |W |, and |W + xW | ≤ K 3 |W | for all x ∈ X, where K 4 1 K 2 K 4 3 < |X|, then I W,X ± I W,X ⊆ I W,X .
We must also show that ξ ∈ I W,X if and only if −ξ ∈ I W,X . It is enough to prove that if ξ ∈ I W,X , then −ξ ∈ I W,X . This follows from Lemma 2, Lemma 8 and Lemma 6 Lemma 10 (Closure of involved elements under scalar multiplication). Let V be a vector space over a finite field F, W ⊆ V and X ⊆ F. If |W + XW | ≤ K 1 |W |, |W + W | ≤ K 2 |W |, and |W + xW | ≤ K 3 |W | for all x ∈ X, where K 2 1 K 3 3 < |X|, then I W,X is closed under multiplication by X × .
Proof. By Lemma 6 it is enough to show that, if ξ is an involved element and α ∈ X, then Indeed, and, similarly, It follows by induction that if α 1 , . . . , α n ∈ X and ξ ∈ I W,X , then We now prove Proposition 7.
Proof of Proposition 7. We want to show that the set I W,X of involved elements contains Span X (W ), where X is the subfield of F q generated by X.
First, we show that I W,X contains W . If ξ ∈ W , then Hence ξ is an involved element by Lemma 6. Now by Lemma 10, we have X × W ⊆ I W,X , and hence by Lemma 9 we have for sums of any finite length ± X × W ± · · · ± X × W ⊆ I W,X .
This implies that X × + W ⊆ I W,X , hence X W + · · · + X W ⊆ I W,X for sums of any finite length, which implies that Span X (W ) ⊆ I W,X .

Proof of Theorem D and Corollary 5
Proof of Theorem D. Proposition 7 implies that I W,X = Span F (W ). Hence by Lemma 3 applied to S = Span F (W ), there is an involved element ξ 0 such that On the other hand, since ξ 0 is an involved element, Lemma 8 implies that which proves the theorem.
Since W ⊆ Span F (W ), the lower bound on |W | is fairly tight. The inclusion X ⊆ F implies |X| ≤ |F |. The next step is to show that, under suitable cardinality constraints on X and W , W must have additional structure.
Proof of Corollary 5. Let q = p r and F = X , so that |F | = p k for k | r. Also let d = dim(Span F (W )). Our aim is to show that d = 2 and |F | = q 1/3 .
Note next that d > 1, because d = 1 means F = Span F (W ). Therefore The constrains on |W | and |F | immediately prohibit F from being anything but F q . However, F = F q is not possible either because it would imply
We conclude W ⊆ Span F (W ) = F v 1 + F v 2 for linearly independent v 1 , v 2 ∈ W .

Proof of Theorem E
The set up for this section is as follows: q is a cube, F is the subfield of cardinality q 1/3 , V is a two-dimensional vector space over F , and A ⊆ V . Theorem E asserts that is |A| ≫ q 7/12 , then |(A − A)(A − A)| > q/2. It is proved in two steps: first we prove

Basic facts
We begin with some basic facts about F q , V and V , the dual group of V viewed as a commutative group under addition.
Let ξ ∈ V \F . Note that {1, ξ} is a linearly independent set in V and forms a basis of V . Hence V = F + F ξ. Moreover, {1, ξ, ξ 2 } is linearly independent in F q (if ξ 2 were a nontrivial linear combination of {1, ξ}, then V would be closed under multiplication and therefore would be a subfield of cardinality q 2/3 ). This means that F q = F + F ξ + F ξ 2 .
Suppose now that V is a vector space over a field F . Let ψ ∈ F be a non-trivial additive character of F . The dual group V of additive characters of V consists of characters of the form v → ψ(v · m), where m is an element of V . The orthogonality of characters implies that

For a set A ⊆ V we denote by A(v) the indicator function on A and write
for the Fourier coefficients of A.

The product set of the vector space
Proposition 11. Let q be a cube. Suppose F ⊂ F q is the subfield of cardinality q 1/3 and V ⊂ F q is a two-dimensional vector space over F .
Proof. We first note 0 ∈ V V . For non-zero elements a + bξ + cξ 2 ∈ F q we seek u, v ∈ V such that uv = a + bξ + cξ 2 . After scaling we may assume that u = 1 + cz −1 ξ and v = a + zξ for some z ∈ F * . Next we observe that a + bξ + cξ 2 = (1 + cz −1 ξ)(a + zξ) precisely when z 2 − bz + ac = 0. This means that a + bξ + cξ 2 ∈ V V precisely when the quadratic monic polynomial z 2 − bz + ac has a root in F (the case a = b = c = 0 is included here). We treat the cases of even and odd characteristic separately.
Odd characteristic: The quadratic polynomial above has a root precisely when its discriminant b 2 − 4ac is a square. We set S = {x 2 : x ∈ F } to be the set of squares and count the number of ordered triples (a, b, c) ∈ F ×F ×F such that b 2 −4ac ∈ S. To this end, for y ∈ F , we write r(y) = |{(a, c) ∈ F × F : y = 4ac}| for the number of ordered representations of y as a product 4ac with a, c ∈ F . Observe that r(y) = |F | − 1 for non-zero y and r(0) = 2|F | − 1.

The number of ordered triples
Even characteristic: Let |F | = 2 n . Recall that the Galois group of F over F 2 is generated by σ : x → x 2 . The trace function Tr : F → F 2 is defined by Since σ is F 2 -linear, we may view Tr as a F 2 -linear functional on F . Let S = {x ∈ F : Tr(x) = 0} be its kernel. In particular note that |S| = |F |/2 because Tr is non-trivial.
To solve the equation we consider two cases.
Second, suppose that b = 0. Then by replacing z with z/b we may put (13) in the form Note that z 2 + z = z − σ(z). Hilbert's Theorem 90 [34,Theorem 6.3] states that for cyclic extensions, w = z − σ(z) if and only the trace of w is zero. Hence (14) has a solution if and only if ac/b 2 ∈ S.
To summarize: the quadratic polynomial has a root precisely when b = 0 or ac ∈ b 2 S.
Writing r(y) = |{(a, c) ∈ F × F : y = ac}| for the number of ordered representations of y as a product ac with a, c ∈ F and noting that r(y) = |F | − 1 for non-zero y and r(0) = 2|F | − 1 we see that the number of ordered triples (a, b, c

Large subsets of vector spaces
The second step in the proof of Theorem E is to prove that if |A| ≫ V 7/8 = q 7/12 , then We begin with a generalisation of a result of Hart, Iosevich, and Solymosi [23,Theorem 1.4]. Its importance in our setting will become clear in the forthcoming corollary.
Proposition 12. Let V be a vector space over a field F , u, v ∈ V \ {0} be non-zero vectors and A ⊆ V be a set in V . Suppose |A| ≥ √ 2|V |/|F | 1/4 . There exists x ∈ F * such that both xu and x −1 v belong to A − A.
Proof. Writing for the number of ordered ways to express t ∈ V as difference in A − A, we must prove Recall that ψ ∈ F is a non-trivial additive character of F and that, for all m ∈ V , the function v → ψ(m · v) is an additive character of V . By character orthogonality Products of differences in arbitrary fields Therefore, after some rearranging, the left side of (15) equals If m·u = 0 or m ′ ·v = 0, then the bracketed sum is a Kloosterman sum or a Ramanujan sum and its modulus is at most 2|F | 1/2 , see [31,Theorem 11.11] and [11] for even characteristic. Otherwise, when m · u = m ′ · v = 0, the bracketed sum equals |F | − 1. Substituting in (16), we see that the left side of (15) equals The first summand is bounded from below by the contribution coming from m = m ′ = 0 (noting that the remaining terms are all positive), which is (|F | − 1)|A| 4 /|V | 2 . The modulus of the second summand is at most Therefore the left side of (15) is positive if which holds when Corollary 13. Let F q be a finite field, F a subfield of F q , V ⊆ F q be a vector space In particular, if F q is a cubic extension of F and V is a two dimensional vector space over F , then |A| ≥ √ 2|V Proof. 0 belongs to both sets. Given w ∈ V V \ {0}, there exist u, v ∈ V \ {0} such that w = uv. By Proposition 12 there exists x ∈ F * such that both xu, Remark. The proof has the curious characteristic that we use additive characters of V to deduce something about elements of F q that may not belong to V .

Proof of Theorem A
At last we are ready to prove Theorem A.
Theorem A. For all q and all sets A ⊆ F q that satisfy |A| q We consider separately two cases. The first is loosely speaking the case where D × (A) is large and is where Theorem B, Corollary 1, and Theorem E are used. The second case, where D × (A) is small, only requires a simple second moment argument.
A peculiarity of the argument is that the first case leads to a better lower bound for |A|. The explanation for this is that in order to apply Corollary 1 we need D × (A) − |A| 8 /q to be very large. Once we assume this and apply the corollary, we are in the regime of Theorem E and the rest of the argument is fairly efficient. The price for assuming that D × (A) − |A| 8 /q is large is paid when applying the complimentary second moment argument. It is not possible to balance the two parts of the argument because the first part stops working before it can it be balanced with the second.

Case I: Large D × (A)
We assume that To be able easily confirm that the various mild constraints appearing in Theorem B, Corollary 1, and Theorem E are indeed satisfied, we assume that q 3/5 ≤ |A| ≤ q 2/3 (the result of Bennett, Hart, Iosevich, Pakianathan, and Rudnev [3] deals with larger |A|). This gives the generous bound K q 1/3,000 , which is adequate for this purpose.
In particular, we have |A| ≪ q/K. Theorem B gives a set X such that E(A, ξA) ≥ |A| 3 /K for all ξ ∈ X and We now apply Corollary 1. The constraints on |X| and |A| are satisfied because of the constraints on |A| and |X| and K. The, more restrictive, condition K 1504 |X| is satisfied because of the assumption K 1505 q/|A|: Corollary 1 now gives a subfield F of cardinality q 1/3 , a two-dimensional vector space The final step is to apply Theorem E to the set Aā = (A −ā) ∩ V . If |Aā| ≫ |V | 7/8 = q 7/12 , then The condition |Aā| ≫ q 7/12 is satisfied when |A| |A| q 85/1505 q 7/12 ⇐= |A| q 2,311/3,816 .

Proof of Theorem F
We prove a generalisation of [37,Theorem 27].
We use the following generalisation of the quantity D × (A). Given four sets A, B, C, D ⊆ F p we denote by D × (A, B, C, D) the number of ordered solutions to the equation with a 1 , a 2 ∈ A, b 1 , b 2 ∈ B, c 1 , c 2 ∈ C and d 1 , d 2 ∈ D. We continue to write D × (A) in place of D × (A, A, A, A).
The proof is broken down in three steps. We first relate D × (A, B, C, D) to D × (A), D × (B), D × (C), and D × (D); then we use a recent result of Roche-Newton, Rudnev, Shkredov, and the authors on collinear triples to bound D × (A) [37]; the proof is then completed by the standard second moment argument given in Section 6.2. The middle step, where all the non-trivial work lies, depends on Rudnev's points-planes incidence bound [46] and Lemma 3.
Let D × (A) * denote the non-zero solutions to D × (A).

Lemma 14.
Let q be a prime power and A, B, C, D be sets in F q . Suppose |A| ≤ |B|, |C| ≤ |D| and |B| ≤ |D|. We first bound D × (A, B, C, D) 0 : It is at most To bound D × (A, B, C, D) * we use an argument similar to that in the proof of Theorem B. The first step is to show

Products of differences in arbitrary fields
We proceed by showing, say, Indeed, We now state the upper bound on D × (A) given in [37]. This time the bound only holds in F p .
Theorem 15. Let p be a prime and A ⊆ F p be a set.

Theorem C over prime order fields
Over prime finite fields there are quantitatively strong versions of the results of Bourgain [4] and Bourgain-Glibichuk [6]. The first was shown to the authors by Rudnev.
A careful analysis shows that the above implies that if |A| ≥ p 1/2 , then  Lemma 18 (Popularity principle). Fix a number 0 < λ < 1, let S be a finite set, and let f be a function such that f (x) ≥ 0 for all x in S.

Suppose that
Further then there exists a subset P * ⊆ S and a number N such that When w(x) = 1 for all x ∈ S, the second part of this lemma is just a dyadic pigeonholing argument applied after the first part of lemma. The general statement of the second part follows from dyadic pigeonholing applied after a weighted version of the first part.
Lemma 20 (Paths of length three). Let K be a positive number, and let G = G(A, B, E) be a bipartite graph with |B| ≤ |A| and |E| = |A||B|/K. There exist subsets A ′ ⊆ A and B ′ ⊆ B such that and for each a ∈ A ′ , b ∈ B ′ there are |A||B|/(2 12 K 5 ) paths of length 3 with endpoints a and b.
Note that there is a typo in the print version of [62]-the factor K 4 should be K 5 ; the errata can be found on the first author's webpage.
We combine Lemma 20 with Lemma 18 to get a slight improvement of results of Bourgain and Garaev [ Schoen's result relies on the incorrect version of Lemma 20 in [62], so the actual powers of K may be different.
Proof ≤ N ≤ |B| (19) such that the subset G ⊆ A × B defined by satisfies (a,b)∈G Above, we denoted by x ∈ A G + B the statement x = a + b with (a, b) ∈ G. By (20) and (21) we have and Now we apply Lemma 20 with E = G, and K 0 = |A||B|/|G| to find subsets A ′ ⊆ A, B ′ ⊆ B such that and each pair a ∈ A ′ and b ∈ B ′ is connected by at least |A||B|/(2 12 K 5 0 ) paths of length 3.

A.3 Proof of Proposition 4
We now prove Proposition 4, which we restate here. The argument is essentially due to Bourgain [4, Proof of Theorem C]. Quantitative aspects were worked out by Bourgain and Glibichuk [6, Proof of Proposition 2]. We offer further quantitative improvements by a more careful analysis.
If there is a positive real number K > 0 such that then there exist elementsā ∈ A and x 0 ∈ X and subsets A 1 ⊆ A −ā and X ′ ⊆ x −1 0 X such that: Moreover, if E(A, ξA) ≥ |A| 3 /K for all ξ ∈ X, then |X ′ | K −4 |X|.
As a comparison we state what can be read off of the proof of [6, Proposition 2]: The proof of Proposition 4 follows from two lemmata. There exist subsets Moreover, if E(A, bA) ≥ |A| 3 /K for all b ∈ B, then we have the same conclusion with |B ′ | |B|/K 4 . Lemma 23. Let A and B be subsets of F * q and let K, K 0 be a positive constants. Suppose that |A ± bA| ≤ K|A| for all b in B, and that |A ± A| ≤ K 0 |A|. There exists an elementā in A and a subset A 1 ⊆ A −ā such that Proof of Proposition 4. ξ∈X E(A, ξA) ≥ |A| 3 |X| K .
By Lemma 22 there exist subsets A ′ ⊆ A and X ′ ⊆ x −1 0 X such that The second part of (i) has been established.
We now apply Lemma 23 to A = A ′ , B = X ′ , K K 42 and K 0 K 8 . There is an elementā in A ′ and a subset A 1 ⊆ A ′ −ā such that This proves (ii) and the first part of (i).
For (iii) note Similarly, for (iv) note The proof of the special case where E(A, ξA) ≥ |A| 3 /K for all ξ ∈ X is similar, the only difference is in the relative density of X ′ .

A.4 Proof of Lemma 22
Proof of Lemma 22. We are given b∈B E(A, bA) ≥ |A| 3 |B| K .
By the popularity principle ((17) in Lemma 18 with λ = 1/2, S = B, f (b) = E(A, bA), For each b in B 1 , we apply Proposition 21 with A = A, B = bA and K = 2K to find A Now we find a large common subset of the A x We use (24), (25), (29) and (30) to estimate the remaining sumsets: Combined with the previous chain of inequalities, this proves (28).
Setting b 0 = b * and invoking (26) completes the proof of the most general statement.
If E(A, bA) ≥ |A| 3 /K for all b ∈ B, then we may take B 1 = B and so the lower bound on |B 2 | in (26) becomes |B|/K 4 .

A.5 Proof of Lemma 23
Proof of Lemma 23. We begin with a simple observation we use further down: we may without loss of generality assume that |B| ≥ 32K 0 K 3 because otherwise the conclusion follows immediately with A 1 = A: For a number τ ≥ |B|/(2K 2 ) we say that a non-zero ξ = 0 is τ -involved if min B ′ ⊆B,|B ′ |≥τ Y ⊆A,|Y |≥|A|/2 If |Y + B ′ ξ| ≤ 4K 0 K|A| for some B ′ ⊆ B and Y ⊆ A satisfying |B ′ | ≥ τ, |Y | ≥ |A|/2, we say that ξ is witnessed by B ′ .
Let Ω denote the set of τ -involved elements. We obtain an upper bound on Ω. First observe that for any subsets A ′ ⊆ A, B ′ ⊆ B we have We may assume |B| ≥ 32K 0 K 3 , as detailed at the beginning of the proof, and so we have |B ′ | ≥ |B|/(2K 2 ) ≥ 16K 0 K. Thus |B ′ |/(8K 0 K) − 1 ≥ |B ′ |/(16K 0 K) and we get for all involved ξ Hence we get the following upper bound on Ω via Lemma 3 We use this upper bound on |Ω| further down. Our next task is to find elements in Ω. To do so, we must find particular ξ, B ′ , and Y for which (31) holds. Our aim is to identify two ξ for each b ∈ B. We begin with an averaging argument. Since |A + bA| ≤ K|A| for all b in B, we have b∈B E(A, bA) ≥ |A| 3 |B| K .
For each a in A, let B a = {b ∈ B : b(a −ā) ∈ A −ā}.
In this notation we have a∈A |B a | ≥ |A||B| K , and so, by the first part of Lemma 19, (a,a ′ )∈A×A |B a ∩ B a ′ | ≥ |A| 2 |B| K 2 .
We show that both elements ξ ± belong to Ω because they are witnessed by the set B ′ = B a ∩ B a ′ , which, by (33) has cardinality at least τ ≥ |B|/(2K 2 ). To see why ξ ± are witnessed by B ′ , recall that by its definition, B a is the set of b such that b(a −ā) ∈ A −ā. Therefore for any set Y we have |Y + B ′ ξ ± | ≤ |Y + B a ′ (a ′ −ā) ± bB a (a −ā)| ≤ |Y + A ± bA|. This proves that ξ ± ∈ Ω. This is true for all b ∈ B and therefore (a ′ −ā) ± B(a −ā) ⊆ Ω ∀(a, a ′ ) ∈ P.
The final step of the proof is to get an upper bound on |A + A 1 B| for some suitable set A 1 (of relative density 1/K 2 in A) by double-counting the number of representations of elements of A + A 1 B as a sum of elements in A − A 1 and Ω.
Set A 1 = A * −ā and r(x) := r (A−A+ā)+Ω (x). We show that for all x in A + A 1 B r(x) ≥ N ∀x ∈ A + A 1 B.
To see why consider x =ã ± (a −ā)b ∈ A ± A 1 B (so a ∈ A * ). For all a ′ ∈ P (a), inclusion (35) gives Distinct a ′ ∈ P (a) give distinct representations and so we indeed get r(x) ≥ |P (a)| ≥ N.
We are ready to conclude the proof. By (37), |A 1 | = |A * | |A|/K 2 and by the above inequality and (36)