Let $ \mathbb{N} = \{ 0, 1, 2, \cdots\} $ and write $ \mathbb{N}_+ $ for $ \mathbb{N}\setminus\{0\}. $ For a subset $ A $ of an additive group, we define the difference set $ A-A = \{ a-a':a, a'\in A\}. $ If $ A $ also is finite, we denote by $ |A| $ its cardinality.

In the late 1970s, Furstenberg [1] and S$ \acute{\mathrm{a}} $rk$ \ddot{\mathrm{o}} $zy [2] independently proved the following conclusion. If $ A $ is a subset of positive upper density of $ \mathbb{Z}, $ then there exist two distinct elements of $ A $ whose difference is a perfect square. The latter also provided an explicit estimate, but the former result is not quantitative. S$ \acute{\mathrm{a}} $rk$ \ddot{\mathrm{o}} $zy's theorem was later improved by Pintz, Steiger and Szemer$ \acute{\mathrm{e}} $di in [3], where they obtained the following theorem.

Theorem A There exists a constant $ D>0 $ such that the following holds. Let $ N\in\mathbb{N}_+ $ and $ A\subseteq\mathbb{N}\cap [1, N]. $ If $ (A-A)\cap\{ n^2:n\in\mathbb{N}_+\} = \emptyset, $ then we have

$ |A|\leq DN(\log N)^{-\frac{1}{12}\log\log\log\log N}. $

Remark 1 Balog, Pelik$ \acute{\mathrm{a}} $n, Pintz and Szemer$ \acute{\mathrm{e}} $di [4] showed that one may replace $ \frac{1}{12} $ by $ \frac{1}{4} $ in the above bound. This estimate is the current best known bound.

In 1996, by extending the ideas of Furstenberg, Bergelson and Leibman [5] established a far reaching qualitative result, the so-called Polynomial Szemer$ \acute{\mathrm{e}} $di theorem. It is natural to ask for a quantitative version of the Polynomial Szemer$ \acute{\mathrm{e}} $di theorem. Recently, Lyall and Magyar [6] made some progress towards this problem. They first proved a higher dimensional analogue of S$ \acute{\mathrm{a}} $rk$ \ddot{\mathrm{o}} $zy's theorem.

Theorem B For $ k\in\mathbb{N} $ with $ k\geq 2, $ there exists a constant $ D'>0 $ such that the following holds. Let $ N\in\mathbb{N}_+ $ and $ A\subseteq\mathbb{N}^k\cap [1, N]^k. $ If $ (A-A)\cap\big\{ (n, n^2, \cdots, n^k):n\in\mathbb{Z}\setminus\{0\}\big\} = \emptyset, $ then we have

$ |A|\leq D'N^k\Big(\frac{\log\log N}{\log N}\Big)^{\frac{1}{k-1}}. $

Then by applying Theorem B, they established a quantitative result on the existence of polynomial configurations of the type in the Polynomial Szemer$ \acute{\mathrm{e}} $di theorem in the difference set of sparse subsets of $ \mathbb{Z}. $

Theorem C Let $ l\in\mathbb{N}_+ $ and $ P_1, \cdots, P_l\in\mathbb{Z}[x] $ with $ P_i(0) = 0 $ for $ i = 1, \cdots, l. $ Suppose that $ k = \max\limits_{1\leq i\leq l}\mathrm{deg}P_i\geq 2. $ Then there exists a constant $ D''>0 $ such that the following inequality holds: let $ N\in\mathbb{N}_+ $ and $ A\subseteq\mathbb{N}\cap [1, N]. $ If $ \big\{P_1(n), \cdots, P_l(n)\big\}\nsubseteq A-A $ for any $ n\in\mathbb{Z}\setminus\{0\}, $ then we have

$ |A|\leq D''N\Big(\frac{\log\log N}{\log N}\Big)^{\frac{1}{(k-1)l}}. $

Remark 2 Theorems B and C were quoted from the revised version of [6], where the authors improved the main results in the original edition.

By taking $ l = 1, $ $ P_1 = x^2 $ and $ k = 2, $ S$ \acute{\mathrm{a}} $rk$ \ddot{\mathrm{o}} $zy's theorem follows from Theorem C. Thus, we may consider Theorem C to be S$ \acute{\mathrm{a}} $rk$ \ddot{\mathrm{o}} $zy's theorem for a family of polynomials.

Let $ \mathbb{F}_q $ be the finite field of $ q $ elements. Let $ p $ denote the characteristic of $ \mathbb{F}_q. $ We denote by $ \mathbb{A} = \mathbb{F}_q[t] $ the polynomial ring over $ \mathbb{F}_q $ and write $ \mathbb{A}^\times = \mathbb{F}_q[t]\setminus\{0\}. $ For $ N\in\mathbb{N}, $ let $ \mathbb{G}_N $ be the set of all polynomials in $ \mathbb{A} $ of degree less than $ N. $

By adapting part of the Pintz-Steiger-Szemer$ \acute{\mathrm{e}} $di argument, L$ \hat{\mathrm{e}} $ and Liu [7] obtained an analogue of Theorem A in function fields.

Theorem D If $ p\geq 3, $ then there exists a constant $ D'''>0, $ depending only on $ q, $ such that the following holds: let $ N\in\mathbb{N} $ with $ N\geq 2 $ and $ A\subseteq\mathbb{G}_N. $ If $ (A-A)\cap\{ d^2:d\in\mathbb{A}^\times\} = \emptyset, $ then we have

$ |A|\leq D'''q^N\frac{(\log N)^7}{N}. $

In this paper, for the case $ k = 2, $ we consider the analogues of Theorems B and C in function fields. First, by closely following the approach of Lyall and Magyar, which is explained in detail by Rice [8], we prove a 2-dimensional version of S$ \acute{\mathrm{a}} $rk$ \ddot{\mathrm{o}} $zy's theorem in function fields.

Theorem 1 If $ p\geq 3, $ then there exists a constant $ C>0, $ depending only on $ q, $ such that the following holds: let $ N\in\mathbb{N} $ with $ N\geq 2 $ and $ A\subseteq\mathbb{G}_N^2. $ If $ (A-A)\cap\{ (d, d^2):d\in\mathbb{A}^\times\} = \emptyset, $ then we have

$ |A|\leq Cq^{2N}\frac{\log N}{N}. $

By adapting the lifting argument in [6], we deduce the following analogue of Theorem C from Theorem 1.

Theorem 2 Let $ l\in\mathbb{N}_+ $ and $ P_1, \cdots, P_l\in\mathbb{A}[x] $ with $ P_i(0) = 0 $ for $ i = 1, \cdots, l. $ Suppose that $ \max\limits_{1\leq i\leq l}\mathrm{deg}P_i\leq 2 $ and $ p\geq 3. $ Then there exists a constant $ C'>0, $ depending only on $ q, P_1, \cdots, P_l, $ such that the following inequality holds: let $ N\in\mathbb{N} $ with $ N\geq 2 $ and $ A\subseteq\mathbb{G}_N. $ If $ \big\{P_1(d), \cdots, P_l(d)\big\}\nsubseteq A-A $ for any $ d\in\mathbb{A}^\times, $ then we have $ |A|\leq C'q^N\big(\frac{\log N}{N}\big)^{\frac{1}{l}}. $

In particular, by taking $ l = 1 $ and $ P_1 = x^2 $ in Theorem 2, we obtain a slight improvement of Theorem D.

In the general cases $ k\geq 3, $ it is more difficult to establish a $ k $-dimensional analogue of Theorem B in function fields. The main obstruction is that we are not able to obtain satisfactory exponential sum estimates on the minor arcs (for details of the circle method, see [9]), i.e., suitable generalizations of Proposition 10. We intend to return to this topic in the future.

Let $ \mathbb{K} = \mathbb{F}_q(t) $ be the field of fractions of $ \mathbb{A}. $ For $ a, b\in\mathbb{A} $ with $ b\neq 0, $ we define $ |\frac{a}{b}| = q^{\mathrm{deg}a-\mathrm{deg}b}. $ Then $ |\cdot | $ is a valuation on $ \mathbb{K}. $ The completion of $ \mathbb{K} $ with respect to this valuation is $ \mathbb{K}_\infty = \big\{\sum\limits_{i\leq r}c_it^i:r\in\mathbb{Z}\ \mathrm{and}\ c_i\in\mathbb{F}_q\ (i\leq r)\big\}, $ the field of formal Laurent series in $ \frac{1}{t}. $

For $ \omega = \sum\limits_{i\leq r}c_it^i\in\mathbb{K}_\infty, $ if $ c_r\neq 0, $ we define $ \mathrm{ord}\omega = r. $ Also, we adopt the convention that $ \mathrm{ord} 0 = -\infty. $ Thus, we have $ |\omega | = q^{\mathrm{ord}\omega}. $ We define $ \{ \omega\} = \sum\limits_{i\leq -1}c_it^i $ to be the fractional part of $ \omega $ and we write $ [\omega] $ for $ \sum\limits_{i\geq 0}c_it^i. $ Then it follows that $ \omega = [\omega]+\{ \omega\}. $ We also write $ \mathrm{res}\omega $ for $ c_{-1} $ which is said to be the residue of $ \omega. $

$ \mathbb{K}_\infty $ is a locally compact field and $ \mathbb{T} = \big\{\omega\in\mathbb{K}_\infty :\mathrm{ord}\omega\leq -1\big\} $ is a compact subring of $ \mathbb{K}_\infty. $ Let $ d\omega $ be the Haar measure on $ \mathbb{K}_\infty $ such that $ \int_\mathbb{T}1d\omega = 1. $

Let $ \mathrm{tr}:\mathbb{F}_q\rightarrow\mathbb{F}_p $ be the familiar trace map. For $ c\in\mathbb{F}_q, $ write $ e_q(c) = \exp(\frac{2\pi\sqrt{-1}}{p}\mathrm{tr}(c)). $ The exponential function $ e:\mathbb{K}_\infty\rightarrow\mathbb{C}^\times $ is defined by $ e(\omega ) = e_q(\mathrm{res}\; \omega). $ Using this function, one can establish Fourier analysis in $ \mathbb{A}. $ In particular, $ \mathbb{A}, \mathbb{K}, \mathbb{K}_\infty, \mathbb{T} $ play the roles of $ \mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{R}/\mathbb{Z}, $ respectively.

For $ \omega\in\mathbb{K}_\infty $ and $ \gamma = (\gamma_1, \gamma_2), \gamma' = (\gamma'_1, \gamma'_2) \in\mathbb{K}^2_\infty, $ write $ \omega\gamma = (\omega\gamma_1, \omega\gamma_2) $ and $ \gamma\gamma' = \gamma_1\gamma'_1+\gamma_2\gamma'_2. $

Let $ f, g:\mathbb{A}^2\rightarrow\mathbb{C} $ be functions with finite support sets. The Fourier transform $ \hat{f}:\mathbb{T}^2\rightarrow\mathbb{C} $ of $ f $ is defined by $ \hat{f}(\alpha) = \sum\limits_{m\in\mathbb{A}^2}f(m)e(m\alpha). $ The convolution $ f*g:\mathbb{A}^2\rightarrow\mathbb{C} $ of $ f $ and $ g $ is defined by

$ f*g(n) = \sum\limits_{m\in\mathbb{A}^2}f(m)g(n-m). $

Then it follows that

$ \mathrm{supp}f*g\subseteq\mathrm{supp}f+\mathrm{supp}g\ \mathrm{and}\ \widehat{f*g}(\alpha) = \hat{f}(\alpha)\hat{g}(\alpha). $

Let $ d\alpha $ denote the product of Haar measures. For $ m\in\mathbb{A}^2, $ we have the orthogonal relation

$ \begin{equation} \int_{\mathbb{T}^2}e(\alpha m)d\alpha = \left\{ \begin{array}{ll} 1, & \mathrm{if}\ m = 0, \\ 0, & \mathrm{otherwise.} \end{array} \right. \end{equation} $

(2.1)

Lemma 1 For $ M\in\mathbb{N}_+ $ and $ \omega\in\mathbb{K}_\infty, $ we have

$ \sum\limits_{d\in\mathbb{G}_M}e(\omega d) = \left\{ \begin{array}{ll} q^M, & \mathrm{if}\ \mathrm{ord}\{\omega\}<-M, \\ 0, & \mathrm{otherwise.} \end{array} \right. $

Proof This is [10, Lemma 7].

Let $ a, b\in\mathbb{A} $ with $ b\neq 0 $ and $ \mathrm{gcd}(b, a) = 1. $ For $ m = (m_1, m_2)\in\mathbb{A}^2, $ if $ \mathrm{gcd}(b, m_1, m_2) = 1, $ we define

$ G(\frac{a}{b}, m) = \sum\limits_{d\in\mathbb{G}_{\mathrm{ord}b}}e(\frac{a}{b}m\overrightarrow{d}), $

where $ \overrightarrow{d} = (d, d^2). $

For $ N\in\mathbb{N}_+, $ the exponential sum $ S_N:\mathbb{T}^2\rightarrow\mathbb{C} $ is defined by $ S_N(\alpha) = \sum\limits_{d\in\mathbb{G}_N}e(\alpha\overrightarrow{d}). $

Lemma 2 Let $ N\in\mathbb{N}_+ $ and $ \alpha = (\alpha_1, \alpha_2)\in\mathbb{T}^2. $ Let $ b\in\mathbb{A}^\times $ and $ m = (m_1, m_2)\in\mathbb{A}^2 $ with $ \mathrm{gcd}(b, m_1, m_2) = 1. $ Suppose that $ \mathrm{ord}b\leq N, \big|\alpha_1-\frac{m_1}{b}\big|<|b|^{-1} $ and $ \big|\alpha_2-\frac{m_2}{b}\big|<q^{1-N}|b|^{-1}. $ Then we have

$ S_N(\alpha) = \frac{1}{|b|}G(\frac{1}{b}, m)S_N(\alpha -\frac{1}{b}m). $

Proof Write $ \beta = (\beta_1, \beta_2) = \alpha-\frac{1}{b}m. $ Then

$ S_N(\alpha) = \sum\limits_{t\in\mathbb{G}_{\mathrm{ord}b}}e(\frac{1}{b}m\overrightarrow{t}) \sum\limits_{s\in\mathbb{G}_{N-\mathrm{ord}b}}e(\beta\overrightarrow{sb+t}). $

Let $ s\in\mathbb{G}_{N-\mathrm{ord}b} $ and $ t\in\mathbb{G}_{\mathrm{ord}b}. $ Note that

$ \mathrm{ord}(\beta_1(sb+t)-\beta_1sb) = \mathrm{ord}\beta_1+\mathrm{ord}t\leq (-\mathrm{ord}b-1)+(\mathrm{ord}b-1) = -2, $

we have $ e(\beta_1(sb+t)) = e(\beta_1sb). $ Similarly, since

$ \begin{eqnarray*} \mathrm{ord}(\beta_2(sb+t)^2-\beta_2s^2b^2)&\leq&\mathrm{ord}\beta_2+\mathrm{ord}t+\max\{\mathrm{ord}t, \mathrm{ord}sb\}\\ &\leq&(-N-\mathrm{ord}b)+(\mathrm{ord}b-1)+(N-1)\\ & = &-2, \end{eqnarray*} $

it follows that $ e(\beta_2(sb+t)^2) = e(\beta_2s^2b^2). $ Thus, we obtain

$ \begin{eqnarray*} S_N(\alpha)& = &\sum\limits_{t\in\mathbb{G}_{\mathrm{ord}b}}e(\frac{1}{b}m\overrightarrow{t}) \sum\limits_{s\in\mathbb{G}_{N-\mathrm{ord}b}}e(\beta\overrightarrow{sb})\\ & = &G(\frac{1}{b}, m)\sum\limits_{s\in\mathbb{G}_{N-\mathrm{ord}b}}e(\beta\overrightarrow{sb})\\ & = &\frac{1}{|b|}G(\frac{1}{b}, m) \sum\limits_{t\in\mathbb{G}_{\mathrm{ord}b}}\sum\limits_{s\in\mathbb{G}_{N-\mathrm{ord}b}}e(\beta\overrightarrow{sb+t})\\ & = &\frac{1}{|b|}G(\frac{1}{b}, m)S_N(\beta). \end{eqnarray*} $

This completes the proof of the lemma.

Lemma 3 Let $ r_1, r_2\in\mathbb{N}. $ Then for any $ \alpha = (\alpha_1, \alpha_2)\in\mathbb{T}^2, $ there exists $ (b, m_1, m_2)\in\mathbb{A}^3 $ with the following properties

$ \rm(i) $ $ b $ is monic and $ \mathrm{ord}b\leq r_1+r_2; $

$ \rm(ii) $ $ \mathrm{gcd}(b, m_1, m_2) = 1; $

$ \rm(iii) $ $ \mathrm{ord}m_j<\mathrm{ord}b $ and $ \big|\alpha_j -\frac{m_j}{b}\big|<q^{-r_j}|b|^{-1}\ (1\leq j\leq 2). $

Proof For $ 1\leq j\leq 2, $ let $ \mathbb{T}_j = \big\{\omega\in\mathbb{T}:\mathrm{ord}\omega\leq -r_j-1\big\}. $ Then $ \mathbb{T}_j $ is a subgroup of $ \mathbb{T}. $ Also, $ \big|\mathbb{T}/\mathbb{T}_j\big| = q^{r_j}. $

Note that $ \big|\prod\limits_{j = 1}^2\mathbb{T}/\mathbb{T}_j\big| = q^{r_1+r_2}<|\mathbb{G}_{r_1+r_2+1}|, $ we can find two distinct elements $ d_1, d_2 $ of $ \mathbb{G}_{r_1+r_2+1} $ such that

$ \big(\{d_1\alpha_1\}+\mathbb{T}_1, \{d_1\alpha_2\}+\mathbb{T}_2\big) = \big(\{d_2\alpha_1\}+\mathbb{T}_1, \{d_2\alpha_2\}+\mathbb{T}_2\big). $

Write $ b' = d_2-d_1. $ Then we have $ b'\neq 0 $ and $ \mathrm{ord}b'\leq r_1+r_2. $

Let $ m'_j = [b'\alpha_j]. $ Then $ \mathrm{ord}m'_j\leq\mathrm{ord}(b'\alpha_j) = \mathrm{ord}b'+\mathrm{ord}\alpha_j<\mathrm{ord}b'. $

Since $ \mathrm{ord}(b'\alpha_j-m'_j) = \mathrm{ord}\{ b'\alpha_j\} = \mathrm{ord}(\{ d_2\alpha_j\}-\{ d_1\alpha_j\})\leq -r_j-1, $ we have

$ \big|\alpha_j -\frac{m'_j}{b'}\big|<q^{-r_j}|b'|^{-1}. $

Let $ c $ be the leading coefficient of $ b' $ and let $ a = \mathrm{gcd}(b', m'_1, m'_2). $ By taking $ b = \frac{b'}{ac} $ and $ m_j = \frac{m'_j}{ac}, $ the lemma follows.

In this section, we obtain an estimate for $ G(\frac{a}{b}, m). $ Our arguments run in parallel with the approach of Chen [11].

Lemma 4 Let $ a_1, a_2, b_1, b_2 \in\mathbb{A} $ with $ b_1, b_2\neq 0 $ and $ \mathrm{gcd}(b_1, a_1) = \mathrm{gcd}(b_2, a_2) = 1. $ Let $ m = (m_1, m_2)\in\mathbb{A}^2. $ Suppose that $ \mathrm{gcd}(b_1, m_1, m_2) = \mathrm{gcd}(b_2, m_1, m_2) = 1. $ If $ \mathrm{gcd}(b_1, b_2) = 1, $ then

$ G(\frac{a_1}{b_1}, m)G(\frac{a_2}{b_2}, m) = G(\frac{a_1b_2+a_2b_1}{b_1b_2}, m). $

Proof Since $ \mathrm{gcd}(b_1, b_2) = 1, $ $ b_2+b_1\mathbb{A} $ is invertible in the ring $ \mathbb{H}_1 = \mathbb{A}/b_1\mathbb{A}. $ Thus,

$ G(\frac{a_1}{b_1}, m) = \sum\limits_{d+b_1\mathbb{A}\in\mathbb{H}_1}e(\frac{a_1}{b_1}m\overrightarrow{d}) = \sum\limits_{d+b_1\mathbb{A}\in\mathbb{H}_1}e(\frac{a_1}{b_1}m\overrightarrow{b_2d}) = \sum\limits_{d\in\mathbb{G}_{\mathrm{ord}b_1}}e(\frac{a_1}{b_1}m\overrightarrow{b_2d}). $

Similarly, we have

$ G(\frac{a_2}{b_2}, m) = \sum\limits_{d\in\mathbb{G}_{\mathrm{ord}b_2}}e(\frac{a_2}{b_2}m\overrightarrow{b_1d}). $

Combining the above two equalities, it follows that

$ \begin{eqnarray} G(\frac{a_1}{b_1}, m)G(\frac{a_2}{b_2}, m)& = &\sum\limits_{d_1\in\mathbb{G}_{\mathrm{ord}b_1}, d_2\in\mathbb{G}_{\mathrm{ord}b_2}} e(\frac{a_1}{b_1}m\overrightarrow{b_2d_1})e(\frac{a_2}{b_2}m\overrightarrow{b_1d_2})\\ & = &\sum\limits_{d_1\in\mathbb{G}_{\mathrm{ord}b_1}, d_2\in\mathbb{G}_{\mathrm{ord}b_2}}e(\frac{a_1b_2+a_2b_1}{b_1b_2}m\overrightarrow{b_1d_2+b_2d_1})\\ & = &\sum\limits_{d\in\mathbb{G}_{\mathrm{ord}b_1b_2}}e(\frac{a_1b_2+a_2b_1}{b_1b_2}m\overrightarrow{d}). \end{eqnarray} $

(3.1)

Equality (3.1) follows since $ \mathrm{gcd}(b_1, b_2) = 1. $

Lemma 5 Let $ a, b\in\mathbb{A} $ with $ b\neq 0 $ and $ \mathrm{gcd}(b, a) = 1. $ Let $ m = (m_1, m_2)\in\mathbb{A}^2. $ Suppose that $ \mathrm{gcd}(b, m_1, m_2) = 1. $ If $ p\geq 3 $ and $ b $ is irreducible, then we have

$ \big| G(\frac{a}{b}, m)\big|\leq |b|^{\frac{1}{2}}. $

Proof Since $ b $ is irreducible and $ \mathrm{gcd}(b, a) = 1, $ it follows that $ a\neq 0. $ We divide into two cases.

Case 1 Suppose that $ b\mid m_2. $ Since $ \mathrm{gcd}(b, m_1, m_2) = 1, \ b\nmid m_1. $ By Lemma 1, we have

$ G(\frac{a}{b}, m) = \sum\limits_{d\in\mathbb{G}_{\mathrm{ord}b}}e(\frac{am_1}{b}d) = 0. $

Case 2 Suppose that $ b\nmid m_2. $ Since $ b $ is irreducible, $ \mathbb{H} = \mathbb{A}/b\mathbb{A} $ is a field. Note that $ |\mathbb{H}| = |b|, $ we can find an isomorphism $ T:\mathbb{F}_{|b|}\rightarrow\mathbb{H} $ of fields.

Consider $ \psi:\mathbb{F}_{|b|}\rightarrow\mathbb{C}^\times $ defined by $ \psi(c) = e(\frac{a}{b}T(c)). $ It follows from Lemma 1 that

$ \sum\limits_{c\in\mathbb{F}_{|b|}}\psi(c) = \sum\limits_{d\in\mathbb{G}_{\mathrm{ord}b}}e(\frac{ad}{b}) = 0. $

Thus, $ \psi $ is a non-trivial additive character of $ \mathbb{F}_{|b|}. $ Let $ P(t) = \sum_{j = 1}^2T^{-1}(m_j+b\mathbb{A})t^j. $ Then $ P $ is a polynomial of degree 2 in $ \mathbb{F}_{|b|}[t]. $

Note that

$ G(\frac{a}{b}, m) = \sum\limits_{d\in\mathbb{G}_{\mathrm{ord}b}}\psi(P(T^{-1}(d+b\mathbb{A}))) = \sum\limits_{c\in\mathbb{F}_{|b|}}\psi(P(c)), $

by Weil's theorem in [12], we have $ \big| G(\frac{a}{b}, m)\big|\leq |b|^{\frac{1}{2}}. $

Combining the above two cases, the lemma follows.

Lemma 6 Let $ a, b\in\mathbb{A} $ with $ b\neq 0 $ and $ \mathrm{gcd}(b, a) = 1. $ Let $ m = (m_1, m_2)\in\mathbb{A}^2. $ Suppose that $ \mathrm{gcd}(b, m_1, m_2) = 1. $ If $ p\geq 3 $ and $ b $ is irreducible, then for any $ r\in\mathbb{N}_+, $ we have

$ \big| G(\frac{a}{b^r}, m)\big|\leq |b|^{\frac{r}{2}}. $

Proof We will prove this lemma by induction on $ r. $

For $ r = 1, $ the lemma follows from Lemma 5.

Let $ r\in\mathbb{N} $ with $ r\geq 2. $ Suppose that the lemma holds for all $ r'\in\mathbb{N}_+ $ with $ r'<r. $ We now prove that the statement is true for $ r. $

Note that for $ d\in\mathbb{G}_{\mathrm{ord}b^{r}}, $ there exist $ d_1\in\mathbb{G}_{\mathrm{ord}b^{r-1}} $ and $ d_2\in\mathbb{G}_{\mathrm{ord}b} $ such that $ d = d_2b^{r-1}+d_1. $ This observation allows us to obtain

$ \begin{equation} G(\frac{a}{b^r}, m) = \sum\limits_{d_1\in\mathbb{G}_{\mathrm{ord}b^{r-1}}}e(\frac{a}{b^r}m\overrightarrow{d_1}) \sum\limits_{d_2\in\mathbb{G}_{\mathrm{ord}b}}e\big(\frac{a}{b}(m_1+2m_2d_1)d_2\big). \end{equation} $

(3.2)

There are two cases.

Case 1 Suppose that $ b\mid m_2. $ Since $ b\nmid m_1, $ by Lemma 1, we have

$ \sum\limits_{d_2\in\mathbb{G}_{\mathrm{ord}b}}e\big(\frac{a}{b}(m_1+2m_2d_1)d_2\big) = \sum\limits_{d_2\in\mathbb{G}_{\mathrm{ord}b}}e(\frac{am_1}{b}d_2) = 0. $

By (3.2), we have

$ G(\frac{a}{b^r}, m) = 0. $

Case 2 Suppose that $ b\nmid m_2. $ Then there exists unique $ d_0\in\mathbb{G}_{\mathrm{ord}b} $ such that

$ m_1+2m_2d_0\equiv 0\ (\mathrm{mod}\ b). $

For any $ d_1\in\mathbb{G}_{\mathrm{ord}b^{r-1}}, $ it follows from Lemma 1 that

$ \sum\limits_{d_2\in\mathbb{G}_{\mathrm{ord}b}}e\big(\frac{a}{b}(m_1+2m_2d_1)d_2\big) = \left\{ \begin{array}{ll} |b|, &\mathrm{if}\ d_1\equiv d_0\ (\mathrm{mod}\ b), \\ 0, &\mathrm{otherwise.} \end{array} \right. $

Write

$ \Lambda = \big\{ d\in\mathbb{G}_{\mathrm{ord}b^{r-1}}:d\equiv d_0\ (\mathrm{mod}\ b)\big\}. $

By (3.2), we have

$ G(\frac{a}{b^r}, m) = \sum\limits_{d_1\in\Lambda}|b|e(\frac{a}{b^r}m\overrightarrow{d_1}). $

If $ r = 2, $ then

$ |G(\frac{a}{b^r}, m)| = \big||b|e(\frac{a}{b^2}m\overrightarrow{d_0})\big| = |b|^\frac{r}{2}. $

If $ r\geq 3, $ then

$ \begin{equation} G(\frac{a}{b^r}, m) = \sum\limits_{d\in\mathbb{G}_{\mathrm{ord}b^{r-2}}}|b|e\big(\frac{a}{b^r}m\overrightarrow{db+d_0}\big). \end{equation} $

(3.3)

Let $ m'_1 = \frac{m_1+2m_2d_0}{b}, $ then $ m'_1\in\mathbb{A}. $ Write $ m' = (m'_1, m_2). $ Note that

$ m\overrightarrow{db+d_0}-m\overrightarrow{d_0} = b^2m'\overrightarrow{d}, $

we deduce from (3.3) that

$ G(\frac{a}{b^r}, m) = |b|e\big(\frac{a}{b^r}m\overrightarrow{d_0}\big)G(\frac{a}{b^{r-2}}, m'). $

By the induction hypothesis, it follows that

$ \big| G(\frac{a}{b^r}, m)\big| = |b|\big| G(\frac{a}{b^{r-2}}, m')\big|\leq |b|^{\frac{r}{2}}. $

By combining the above two cases, we complete the proof of the lemma.

Proposition 7 Let $ a, b\in\mathbb{A} $ with $ b\neq 0 $ and $ \mathrm{gcd}(b, a) = 1. $ Let $ m = (m_1, m_2)\in\mathbb{A}^2. $ Suppose that $ \mathrm{gcd}(b, m_1, m_2) = 1. $ If $ p\geq 3, $ then we have

$ \big| G(\frac{a}{b}, m)\big|\leq |b|^{\frac{1}{2}}. $

Proof Without loss of generality, we assume that $ a\neq 0 $ and $ \mathrm{ord}b\geq 1. $ Also, $ b $ is monic. There exist $ \iota, j_1, \cdots, j_\iota\in\mathbb{N}_+ $ and distinct monic irreducible polynomials $ \sigma_1, \cdots, \sigma_\iota $ in $ \mathbb{A} $ such that $ b = \prod\limits_{i = 1}^\iota\sigma_i^{j_i}. $ We prove the lemma by induction on $ \iota. $

For $ \iota = 1, $ the lemma follows from Lemma 6.

Let $ \iota\in\mathbb{N} $ with $ \iota\geq 2. $ Suppose that the lemma is true for $ \iota-1. $ We now prove that the claim holds for $ \iota. $ Since $ \mathrm{gcd}(b, a) = 1, $ we can find $ a_l, a'\in\mathbb{A}^\times $ such that

$ \frac{a}{\prod\limits_{i = 1}^\iota\sigma_i^{j_i}} = \frac{a_l}{\sigma_l^{j_l}}+\frac{a'}{\prod\limits_{i = 1}^{\iota-1}\sigma_i^{j_i}}\ \mathrm{and}\ \mathrm{gcd}(\sigma_l^{j_l}, a_l) = \mathrm{gcd}(\prod\limits_{i = 1}^{\iota-1}\sigma_i^{j_i}, a') = 1. $

By Lemmas 4 and 6, we have

$ \big| G(\frac{a}{\prod\limits_{i = 1}^\iota\sigma_i^{j_i}}, m)\big| = \big| G(\frac{a_l}{\sigma_l^{j_l}}, m)\big|\Big| G(\frac{a'}{\prod\limits_{i = 1}^{\iota-1}\sigma_i^{j_i}}, m)\Big| \leq |\sigma_l|^{\frac{j_l}{2}}\Big| G(\frac{a'}{\prod\limits_{i = 1}^{\iota-1}\sigma_i^{j_i}}, m)\Big|. $

By the induction hypothesis, the proposition follows.

For the present, we fix $ N\in\mathbb{N}_+ $ and $ A\subseteq\mathbb{G}_N\times\mathbb{G}_{2N} $ with $ |A| = \delta q^{3N}. $ Throughout this section, we assume that the following hypothesis holds.

Hypothesis A $ p\geq 3, \ (A-A)\cap\{\overrightarrow{d}:d\in\mathbb{A}^\times\} = \emptyset\ \mathrm{and}\ \delta\geq q^{1-\frac{N}{12}}. $

Take $ \theta\in\mathbb{N}_+ $ with $ q^{-\theta}<\delta\leq q^{1-\theta}. $ Then $ N\geq 12\theta. $ Write $ M = N-6\theta. $

The characteristic function $ 1_A:\mathbb{A}^2\rightarrow\mathbb{R} $ of $ A $ is defined by

$ 1_A(m) = \left\{ \begin{array}{ll} 1, &\mathrm{if}\ m\in A, \\ 0, &\mathrm{otherwise}. \end{array} \right. $

Write $ \Gamma_N = \mathbb{G}_N\times\mathbb{G}_{2N}. $ We define the balanced function $ f_A:\mathbb{A}^2\rightarrow\mathbb{R} $ of $ A $ to be $ f_A = 1_A-\delta 1_{\Gamma_N}. $

Let $ b\in\mathbb{A}^\times $ with $ b $ monic. Write

$ \mathcal{A}_b = \big\{(a_1, a_2)\in\mathbb{A}^2:\mathrm{gcd}(b, a_1, a_2) = 1, \ \mathrm{ord}a_j<\mathrm{ord}b\ (1\leq j\leq 2)\big\}. $

For $ (a_1, a_2)\in\mathcal{A}_b, $ we define the Farey arc $ F(b, a_1, a_2) $ to be

$ F(b, a_1, a_2) = \Big\{(\alpha_1, \alpha_2)\in\mathbb{T}^2:|\alpha_j-\frac{a_j}{b}|<q^{-jM}|b|^{-1}\ (1\leq j\leq 2)\Big\}. $

Also, we define

$ F_b = \bigcup\limits_{(a_1, a_2)\in\mathcal{A}_b}F(b, a_1, a_2). $

We say $ F(b, a_1, a_2) $ is major if $ \mathrm{ord}b\leq 2\theta+3 $ and minor if $ \mathrm{ord}b>2\theta+3. $ Let

$ \mathcal{B} = \{b\in\mathbb{A}^\times:b\ \mathrm{monic}, \ \mathrm{ord}b\leq 2\theta+3\}. $

We define the major arcs $ \mathfrak{M} $ and the minor arcs $ \mathfrak{m} $ as follows:

$ \mathfrak{M} = \bigcup\limits_{b\in\mathcal{B}}F_b\ \mathrm{and}\ \mathfrak{m} = \mathbb{T}^2\setminus\mathfrak{M}. $

Lemma 8 Let $ b, b'\in\mathcal{B}. $ Suppose that $ (a_1, a_2)\in\mathcal{A}_b $ and $ (a'_1, a'_2)\in\mathcal{A}_{b'}. $ If $ (b, a_1, a_2)\neq(b', a'_1, a'_2), $ then we have

$ F(b, a_1, a_2)\cap F(b', a'_1, a'_2) = \emptyset. $

Proof To prove the lemma, we suppose the contrary. Then there exists

$ (\alpha_1, \alpha_2)\in F(b, a_1, a_2)\cap F(b', a'_1, a'_2). $

Let $ 1\leq j\leq 2. $ Since

$ \big|\frac{a_j}{b}-\frac{a'_j}{b'}\big|\leq\max\big\{\big|\alpha_j-\frac{a_j}{b}\big|, \ \big|\alpha_j-\frac{a'_j}{b'}\big|\big\}<q^{-jM}\max\big\{ |b|^{-1}, |b'|^{-1}\big\}, $

it follows that

$ |a_jb'-a'_jb|<q^{-jM}\max\{ |b|, |b'|\}\leq q^{2\theta+3-M}\leq q^{-\theta}<1. $

Thus $ a_jb' = a'_jb. $ Let $ A_j, B_j\in\mathbb{A} $ with $ B_j $ monic such that

$ \mathrm{gcd}(B_j, A_j) = 1\ \mathrm{and}\ \frac{A_j}{B_j} = \frac{a_j}{b} = \frac{a'_j}{b'}. $

It is easy to see that $ b = \mathrm{lcm}(B_1, B_2) = b'. $ It follows that $ a_j = a'_j. $ This leads to a contradiction, and the lemma follows.

Proposition 9 If $ b\in\mathcal{B}, $ then for any $ \alpha\in F_b, $ we have

$ |S_N(\alpha)|\leq q^N|b|^{-1/2}. $

Proof Write $ (\alpha_1, \alpha_2) = \alpha. $ Take $ a = (a_1, a_2)\in\mathcal{A}_b $ such that $ \alpha\in F(b, a_1, a_2). $ Since

$ |\alpha_2-\frac{a_2}{b}|<q^{-2M}|b|^{-1}\leq q^{-N}|b|^{-1}\ \mathrm{and}\ \mathrm{ord}\;b\leq 2\theta+3<N, $

by Lemma 2, we have

$ S_N(\alpha) = \frac{1}{|b|}G(\frac{1}{b}, a)S_N(\alpha -\frac{1}{b}a). $

It follows from Proposition 7 that

$ |S_N(\alpha)|\leq|b|^{-\frac{1}{2}}|S_N(\alpha-\frac{1}{b}a)| \leq|\mathbb{G}_N||b|^{-1/2}. $

Proposition 10 For any $ \alpha\in \mathfrak{m}, $ we have

$ |S_N(\alpha)|\leq\frac{\delta}{4}q^N. $

Proof Write $ \alpha = (\alpha_1, \alpha_2). $ By using Lemma 3 for $ r_1 = 0 $ and $ r_2 = N, $ we can find a monic polynomial $ b $ in $ \mathbb{A}^\times $ and $ a = (a_1, a_2)\in\mathbb{A}^2 $ such that

$ \mathrm{ord}b\leq N, \ \mathrm{gcd}(b, a_1, a_2) = 1, \ \mathrm{ord}a_j<\mathrm{ord}b\ \mathrm{and}\ \big|\alpha_j-\frac{a_j}{b}\big|<q^{-(j-1)N}|b|^{-1} \ (1\leq j\leq 2). $

Write $ \beta = (\beta_1, \beta_2) = \alpha-\frac{1}{b}a. $ If $ \mathrm{ord}b\geq 2\theta+4, $ by Lemma 2 and Proposition 7, we have

$ |S_N(\alpha)|\leq|b|^{-1}|G(\frac{1}{b}, a)||S_N(\beta)|\leq|b|^{-\frac{1}{2}}|S_N(\beta)| \leq q^{-\theta-2}|\mathbb{G}_N|\leq \frac{\delta}{4}q^N. $

In the following, we assume that $ \mathrm{ord}b\leq 2\theta+3. $ Consider the following estimate

$ \begin{eqnarray*} \big|S_N(\beta)\big|^2& = &\sum\limits_{d, d'\in\mathbb{G}_N}e\big(\beta_1(d-d')+\beta_2(d+d')(d-d')\big)\\ & = &\sum\limits_{d, d'\in\mathbb{G}_N}e(\beta_1d+\beta_2dd')\\ &\leq&\sum\limits_{d\in\mathbb{G}_N}\Big|\sum\limits_{d'\in\mathbb{G}_N}e(\beta_2dd')\Big|. \end{eqnarray*} $

For $ d\in\mathbb{G}_N, $ since

$ \mathrm{ord}(\beta_2d) = \mathrm{ord}\beta_2+\mathrm{ord}d\leq(-N-\mathrm{ord}b-1)+(N-1)\leq -2, $

it follows that $ \{\beta_2d\} = \beta_2d. $ By Lemma 1, we have

$ \big|S_N(\beta)\big|^2\leq\sum\limits_{d\in\mathbb{G}_N, \mathrm{ord}(\beta_2d)<-N}q^N\leq|\beta_2|^{-1}. $

Combining Lemma 2 and Proposition 7 with the above inequality, it follows that

$ \begin{equation} |S_N(\alpha)|\leq|b|^{-1}|G(\frac{1}{b}, a)||S_N(\beta)|\leq|b|^{-\frac{1}{2}}|S_N(\beta)|\leq |b|^{-\frac{1}{2}}|\beta_2|^{-\frac{1}{2}}. \end{equation} $

(4.1)

Since $ \alpha\notin \mathfrak{M}, $ there are two cases.

Case 1 Suppose that $ |\beta_2|\geq q^{-2M}|b|^{-1}. $ By (4.1), we have

$ |S_N(\alpha)|\leq q^M = q^{N-6\theta}\leq\frac{\delta}{4}q^N. $

Case 2 Suppose that $ |\beta_1|\geq q^{-M}|b|^{-1} $ and $ |\beta_2|<q^{-2M}|b|^{-1}. $

If $ \mathrm{ord}\beta_2\geq 1-N+\mathrm{ord}\beta_1, $ then by (4.1), we have

$ |S_N(\alpha)|\leq |b|^{-\frac{1}{2}}|\beta_1|^{-\frac{1}{2}}q^{\frac{N-1}{2}}\leq q^{\frac{M+N-1}{2}}\leq q^{N-3\theta}\leq\frac{\delta}{4}q^N. $

Thus, it remains to estimate $ |S_N(\alpha)| $ under the additional assumption $ \mathrm{ord}\beta_2\leq\mathrm{ord}\beta_1-N. $

Write $ L_1 = -\mathrm{ord}\beta_1, $ then $ 1\leq L_1\leq M+\mathrm{ord}b; $ write $ L_2 = -\mathrm{ord}\beta_2, $ then $ L_2\geq 1+2M+\mathrm{ord}b; $ write $ K = \lfloor\frac{L_1+N}{2}\rfloor, $ since $ L_1\leq M+2\theta+3<N, $ we have $ L_1\leq K\leq N-1. $

For $ j\in\mathbb{N}, $ write $ \mathcal{C}_j = \{ d\in\mathbb{A}:\mathrm{ord}d = j\}, $ then

$ S_N(\beta) = \sum\limits_{d\in\mathbb{G}_K}e(\beta\overrightarrow{d})+\sum\limits_{j = K}^{N-1}\sum\limits_{d\in\mathcal{C}_j}e(\beta\overrightarrow{d}). $

Let $ d\in\mathbb{G}_K. $ By the assumption $ \mathrm{ord}\beta_2\leq\mathrm{ord}\beta_1-N, $ we have

$ \mathrm{ord}(\beta_2d^2) = 2\mathrm{ord}d-L_2\leq 2(K-1)+(-N-L_1)\leq-2. $

It follows that $ e(\beta_2d^2) = 1. $ Note that $ \mathrm{ord}\{\beta_1\} = -L_1\geq -K, $ by Lemma 1, we have

$ \sum\limits_{d\in\mathbb{G}_K}e(\beta\overrightarrow{d}) = \sum\limits_{d\in\mathbb{G}_K}e(\beta_1d) = 0. $

Thus

$ \begin{equation} S_N(\beta) = \sum\limits_{I = K}^{N-1}\sum\limits_{d\in\mathcal{C}_I}e(\beta\overrightarrow{d}). \end{equation} $

(4.2)

Take the sequences $ \big\{\mu_i\big\}_{i = -\infty}^{-L_1} $ and $ \big\{\nu_j\big\}_{j = -\infty}^{-L_2} $ in $ \mathbb{F}_q $ such that

$ \beta_1 = \sum\limits_{i\leq -L_1}\mu_it^i\ \mathrm{and}\ \beta_2 = \sum\limits_{j\leq -L_2}\nu_jt^j. $

Let $ K\leq I\leq N-1 $ and $ d\in\mathcal{C}_I. $ Take $ c_0, c_1, \cdots, c_I\in\mathbb{F}_q $ with $ c_I\neq 0 $ such that $ d = \sum\limits_{i = 0}^Ic_it^i. $ Then

$ \mathrm{res}(\beta\overrightarrow{d}) = \sum\limits_{i = L_1-1}^I\mu_{-i-1}c_i+\sum\limits_{l = L_2-1}^{2I}\nu_{-l-1}\sum\limits_{0\leq i, j\leq I, i+j = l}c_ic_j. $

For $ 0\leq i, j\leq I, $ if $ i+j\geq L_2-1, $ by the assumption $ \mathrm{ord}\beta_2\leq\mathrm{ord}\beta_1-N, $ we have

$ \min\{i, j\}\geq L_2-1-I\geq (N+L_1)-1-(N-1) = L_1. $

Thus, there exists the polynomial $ Q_I(t_1, \cdots, t_{I-L_1+1}) $ of $ (I-L_1+1) $ variables over $ \mathbb{F}_q $ such that

$ \mathrm{res}(\beta\overrightarrow{d}) = \mu_{-L_1}c_{L_1-1}+Q_I(c_{L_1}, c_{L_1+1}, \cdots, c_I). $

Substituting this into the definition of the function $ e(\cdot), $ and noting that $ \mu_{-L_1}\neq 0, $ we have

$ \sum\limits_{d\in\mathcal{C}_I}e(\beta\overrightarrow{d}) = \sum\limits_{j\neq L_1-1, 0\leq j\leq I-1}\sum\limits_{c_j\in\mathbb{F}_q}\sum\limits_{c_I\in\mathbb{F}^\times_q} e_q\big(Q_I(c_{L_1}, \cdots, c_I)\big)\sum\limits_{c_{L_1-1}\in\mathbb{F}_q}e_q\big(\mu_{-L_1}c_{L_1-1}\big) = 0. $

It follows from (4.2) that $ S_N(\beta) = 0. $ Finally, by Lemma 2, we have $ S_N(\alpha) = 0. $

Combining the above two cases, we complete the proof of the proposition.

In this section, we continue to fix $ N\in\mathbb{N}_+ $ and $ A\subseteq\Gamma_N $ with $ |A| = \delta q^{3N}. $ Also, we assume that Hypothesis A holds.

Lemma 11

$ \int_{\mathbb{T}^2}|\widehat{f_A}(\alpha)|^2|S_N(\alpha)|d\alpha\geq \frac{1}{2}\delta^2q^{4N}. $

Proof Write $ \mathrm{I} = \sum\limits_{d\in\mathbb{G}_N, m\in\mathbb{A}^2}f_A(m)f_A(m+\overrightarrow{d}). $ By (2.1), we have

$ \begin{equation} \mathrm{I} = \sum\limits_{d\in\mathbb{G}_N, m, n\in\mathbb{A}^2}f_A(m)f_A(n) \int_{\mathbb{T}^2}e(\alpha(m+\overrightarrow{d}-n))d\alpha = \int_{\mathbb{T}^2}|\widehat{f_A}(\alpha)|^2S_N(\alpha)\mathrm{d}\alpha. \end{equation} $

(5.1)

If $ d\in\mathbb{G}_N, $ then $ \overrightarrow{d}\in\Gamma_N. $ Thus $ \Gamma_N+\overrightarrow{d} = \Gamma_N-\overrightarrow{d} = \Gamma_N. $ It follows that $ (A-A)\cap\{\overrightarrow{d}:d\in\mathbb{A}^\times\} = \emptyset $ from Hypothesis A. Thus

$ \begin{eqnarray*} \mathrm{I}& = &\sum\limits_{m\in\mathbb{A}^2}1_A(m)-\delta\sum\limits_{d\in\mathbb{G}_N, m\in\mathbb{A}^2} 1_A(m)\Big(1_{\Gamma_N}(m+\overrightarrow{d})+1_{\Gamma_N}(m-\overrightarrow{d})\Big)\\ &&+\delta^2\sum\limits_{d\in\mathbb{G}_N, m\in\mathbb{A}^2} 1_{\Gamma_N}(m)1_{\Gamma_N}(m+\overrightarrow{d})\\ & = &|A|-\delta\sum\limits_{d\in\mathbb{G}_N}\Big(\big|A\cap(\Gamma_N-\overrightarrow{d})\big| +\big|A\cap(\Gamma_N+\overrightarrow{d})\big|\Big)+\delta^2\sum\limits_{d\in\mathbb{G}_N}\big|\Gamma_N\cap(\Gamma_N-\overrightarrow{d})\big|\\ & = &|A|-2\delta|A|\big|\mathbb{G}_N\big|+\delta^2\big|\mathbb{G}_N\big|\big|\Gamma_N\big|\\ & = &-\delta^2q^{4N}\big(1-\frac{1}{\delta q^N}\big). \end{eqnarray*} $

By Hypothesis A, we have $ \delta q^N\geq q^{1+\frac{11N}{12}}\geq 2. $ It follows that

$ \begin{equation} \mathrm{I}\leq-\frac{1}{2}\delta^2q^{4N}. \end{equation} $

(5.2)

Finally, by (5.1) and (5.2), we obtain

$ \int_{\mathbb{T}^2}|\widehat{f_A}(\alpha)|^2|S_N(\alpha)|d\alpha \geq |\mathrm{I}|\geq\frac{1}{2}\delta^2q^{4N}. $

Lemma 12 There exists a monic polynomial $ b_0 $ in $ \mathbb{G}_{2\theta+4} $ such that

$ \int_{F_{b_0}}|\widehat{f_A}(\alpha)|^2d\alpha\geq c\delta^{3}q^{3N}, $

where $ 0<c<1 $ is a constant depending only on $ q. $

Proof By Proposition 10, we have

$ \begin{eqnarray*} \int_{\mathfrak{m}}|\widehat{f_A}(\alpha)|^2|S_N(\alpha)|d\alpha &\leq&\frac{\delta}{4}q^N\int_{\mathfrak{m}}|\widehat{f_A}(\alpha)|^2d\alpha\\ &\leq&\frac{\delta}{4}q^N\sum\limits_{m\in\mathbb{A}^2}|f_A(m)|^2\\ &\leq&\frac{\delta^2}{4}q^{4N}. \end{eqnarray*} $

Write

$ \mathrm{II} = \int_{\mathfrak{M}}|\widehat{f_A}(\alpha)|^2|S_N(\alpha)|d\alpha. $

Combining the above inequality with Lemma 11, it follows that

$ \begin{equation} \mathrm{II}\geq\int_{\mathbb{T}^2}|\widehat{f_A}(\alpha)|^2|S_N(\alpha)|\mathrm{d}\alpha-\frac{\delta^2}{4}q^{4N} \geq\frac{\delta^2}{4}q^{4N}. \end{equation} $

(5.3)

For $ j\in\mathbb{N}, $ write $ \mathcal{O}_j = \{ b\in\mathbb{A}^\times:b\ \mathrm{monic}, \ \mathrm{ord}b = j\}. $ By Lemma 8 and Proposition 9, we have

$ \mathrm{II} = \sum\limits_{j = 0}^{2\theta+3}\sum\limits_{b\in\mathcal{O}_j} \int_{F_b}|\widehat{f_A}(\alpha)|^2|S_N(\alpha)|d\alpha \leq\sum\limits_{j = 0}^{2\theta+3}q^{N-\frac{j}{2}}\sum\limits_{b\in\mathcal{O}_j} \int_{F_b}|\widehat{f_A}(\alpha)|^2d\alpha. $

Take a monic polynomial $ b_0 $ in $ \mathbb{G}_{2\theta+4} $ such that

$ \ \int_{F_{b_0}}|\widehat{f_A}(\alpha)|^2\mathrm{d}\alpha = \max\limits_{0\leq j\leq 2\theta+3, b\in\mathcal{O}_j}\int_{F_b}|\widehat{f_A}(\alpha)|^2\mathrm{d}\alpha. $

It follows from the above inequality that

$ \mathrm{II}\leq\int_{F_{b_0}}|\widehat{f_A}(\alpha)|^2\mathrm{d}\alpha\sum\limits_{j = 0}^{2\theta+3}\big|\mathcal{O}_j\big|q^{N-\frac{j}{2}} = \int_{F_{b_0}}|\widehat{f_A}(\alpha)|^2\mathrm{d}\alpha\sum\limits_{j = 0}^{2\theta+3}q^{N+\frac{j}{2}}. $

Since $ \delta\leq q^{1-\theta}, $ we can find a constant $ c'>1, $ depending only on $ q, $ such that

$ \mathrm{II}\leq\frac{c'}{\delta}q^N\int_{F_{b_0}}|\widehat{f_A}(\alpha)|^2d\alpha. $

By taking $ c = \frac{1}{4c'}, $ the lemma follows from (5.3).

Lemma 13 There exists $ n_0\in\Gamma_N $ such that

$ |A\cap(n_0+b_0\Gamma_M)|\geq\delta(1+\frac{c}{2}\delta)q^{3M}, $

where $ b_0\Gamma_M = \big\{b_0m:m\in\Gamma_M\big\}. $

Proof Write $ P = b_0\Gamma_M. $ Let $ m = (m_1, m_2)\in\Gamma_M $ and $ 1\leq j\leq 2. $ Since

$ \mathrm{ord}(b_0m_j) = \mathrm{ord}b_0+\mathrm{ord}m_j\leq(2\theta+3)+(jM-1)\leq jN-1, $

we have $ b_0m\in\Gamma_N. $ Thus, $ P\subseteq\Gamma_N. $ Also, we have

$ \mathrm{supp}f_A*1_{-P}\subseteq\mathrm{supp}f_A+\mathrm{supp}1_{-P}\subseteq\Gamma_N+(-P) = \Gamma_N. $

For $ n\in\Gamma_N, $ we have

$ \begin{eqnarray} f_A*1_{-P}(n)& = &\sum\limits_{m\in\mathbb{A}^2}1_A(m)1_P(m-n)-\delta\sum\limits_{m\in\mathbb{A}^2}1_{\Gamma_N}(m)1_P(m-n)\\ & = &|A\cap(n+P)|-\delta\big|\Gamma_N\cap(n+P)\big|\\ & = &|A\cap(n+P)|-\delta|P|. \end{eqnarray} $

(5.4)

If there exists $ n_0\in\Gamma_N $ such that $ f_A*1_{-P}(n_0)\geq\delta|P|, $ then

$ |A\cap(n_0+P)| = f_A*1_{-P}(n_0)+\delta|P|\geq2\delta|P|\geq\delta(1+\frac{c}{2}\delta)q^{3M}. $

Thus, in the following, we assume that $ f_A*1_{-P}(n)\leq\delta|P| $ for all $ n\in\Gamma_N. $ It follows from (5.4) that

$ \begin{equation} \big|f_A*1_{-P}(n)\big|\leq\delta|P|. \end{equation} $

(5.5)

Let $ \alpha = (\alpha_1, \alpha_2)\in F_{b_0}. $ Take $ a = (a_1, a_2)\in\mathcal{A}_{b_0} $ such that $ \alpha\in F(b_0, a_1, a_2). $ Since

$ \begin{eqnarray*} \mathrm{ord}\big(m_j(b_0\alpha_j-a_j)\big)& = &\mathrm{ord}m_j+\mathrm{ord}b_0+\mathrm{ord}(\alpha_j-\frac{a_j}{b_0})\\ &\leq&(jM-1)+\mathrm{ord}b_0+(-jM-\mathrm{ord}b_0-1) = -2, \end{eqnarray*} $

we have $ e(b_0m_j\alpha_j) = e(m_ja_j) = 1. $ Thus, $ \widehat{1_{-P}}(\alpha) = |P|. $ It follows from (5.5) that

$ \begin{eqnarray*} \int_{F_{b_0}}|\widehat{f_A}(\alpha)|^2d\alpha & = &\frac{1}{|P|^2}\int_{F_{b_0}}|\widehat{f_A*1_{-P}}(\alpha)|^2d\alpha\\ &\leq&\frac{1}{|P|^2}\sum\limits_{n\in\mathbb{A}^2}|f_A*1_{-P}(n)|^2\\ &\leq&\frac{\delta}{|P|}\sum\limits_{n\in\mathbb{A}^2}|f_A*1_{-P}(n)|. \end{eqnarray*} $

By Lemma 12, we have

$ \sum\limits_{n\in\mathbb{A}^2}|f_A*1_{-P}(n)|\geq c\delta^2q^{3(M+N)}. $

Note that $ \sum\limits_{n\in\mathbb{A}^2}f_A(n) = 0, $ we have

$ \sum\limits_{n\in\mathbb{A}^2}\big(f_A*1_{-P}\big)_+(n)\geq\frac{c}{2}\delta^2q^{3(M+N)}. $

Take $ n_0\in\Gamma_N $ such that

$ f_A*1_{-P}(n_0) = \max\limits_{n\in\Gamma_N}f_A*1_{-P}(n). $

By (5.4), we have

$ |A\cap(n_0+P)| = \delta|P|+f_A*1_{-P}(n_0)\geq\delta|P|+\frac{1}{|\Gamma_N|}\sum\limits_{n\in\mathbb{A}^2}\big(f_A*1_{-P}\big)_+(n) \geq\delta(1+\frac{c}{2}\delta)q^{3M}. $

Proposition 14 There exist $ N'\in\mathbb{N}_+ $ and $ A'\subseteq\Gamma_{N'} $ with $ |A'| = \delta'q^{3N'} $ such that

$ \rm(i) $ $ (A'-A')\cap\big\{\overrightarrow{d}:d\in\mathbb{A}^\times\big\} = \emptyset; $

$ \rm(ii) $ $ \delta'\geq\delta(1+\frac{c}{2}\delta); $

$ \rm(iii) $ $ N'\geq N-11\log_q\big(\frac{q}{\delta}\big), $ where $ \log_qx = \log x/\log q. $

Proof Write $ L = \mathrm{ord}b_0 $ and $ T = |b_0|. $ Then $ 0\leq L\leq 2\theta+3. $ By taking $ N' = M-L, $ property (iii) follows. Take $ d_1, \cdots, d_T\in\mathbb{G}_M $ and $ d'_1, \cdots, d'_T\in\mathbb{G}_{2M-L} $ such that

$ \begin{equation} \mathbb{G}_M = \bigcup\limits_{i = 1}^T\big(d_i+\mathbb{G}_{N'}\big)\ \mathrm{and}\ \mathbb{G}_{2M-L} = \bigcup\limits_{i = 1}^T\big(d'_i+\mathbb{G}_{2N'}\big). \end{equation} $

(5.6)

For $ d\in\mathbb{G}_L $ and $ 1\leq i, j\leq T, $ write

$ \Upsilon_{d, i, j} = n_0+(0, b_0d)+\overrightarrow{b_0}\odot(d_i, d'_j)+\overrightarrow{b_0}\odot\Gamma_{N'}, $

where

$ \overrightarrow{b_0}\odot(d_i, d'_j) = (b_0d_i, b_0^2d'_j)\ \mathrm{and}\ \overrightarrow{b_0}\odot\Gamma_{N'} = \big\{\overrightarrow{b_0}\odot m:m\in\Gamma_{N'}\big\}. $

Let $ m = (m_1, m_2)\in\Gamma_M. $ Take $ d\in\mathbb{G}_L $ and $ d'\in\mathbb{G}_{2M-L} $ such that $ m_2 = d+b_0d'. $ By (5.6), we can find $ 1\leq i, j\leq T $ such that $ (m_1, d')\in (d_i, d'_j)+\Gamma_{N'}. $ Then we have

$ n_0+b_0m = n_0+(0, b_0d)+\overrightarrow{b_0}\odot (m_1, d')\in\Upsilon_{d, i, j}. $

Thus, we see that

$ n_0+b_0\Gamma_M = \bigcup\limits_{d\in\mathbb{G}_L, 1\leq i, j\leq T}\Upsilon_{d, i, j}. $

Take $ d_0\in\mathbb{G}_L $ and $ 1\leq i_0, j_0\leq T $ such that

$ \big|A\cap\Upsilon_{d_0, i_0, j_0}\big| = \max\limits_{d\in\mathbb{G}_L, 1\leq i, j\leq T}\big|A\cap\Upsilon_{d, i, j}\big|. $

By Lemma 13, we have

$ \big|A\cap\Upsilon_{d_0, i_0, j_0}\big|\geq\frac{1}{T^3}\sum\limits_{d\in\mathbb{G}_L, 1\leq i, j\leq T}\big|A\cap\Upsilon_{d, i, j}\big| = \frac{1}{T^3}|A\cap (n_0+b_0\Gamma_M)|\geq\delta(1+\frac{c}{2}\delta)q^{3N'}. $

Consider the bijection $ f:\Gamma_{N'}\rightarrow \Upsilon_{d_0, i_0, j_0} $ defined by

$ f(t) = n_0+(0, b_0d_0)+\overrightarrow{b_0}\odot(d_{i_0}, d'_{j_0})+\overrightarrow{b_0}\odot t. $

By taking $ A' = f^{-1}(A\cap \Upsilon_{d_0, i_0, j_0}), $ property (ii) follows. To prove property (i), we suppose the contrary. Then there exist $ t_1, t_2\in A' $ and $ d\in\mathbb{A}^\times $ such that $ t_2-t_1 = \overrightarrow{d}. $ It follows that

$ f(t_2)-f(t_1) = \overrightarrow{b_0}\odot\overrightarrow{d} = \overrightarrow{b_0d}\in A-A, $

which contradicts Hypothesis A. This completes the proof of the proposition.

Proposition 15 If $ p\geq 3, $ then there exists a constant $ C_1>0, $ depending only on $ q, $ such that the following inequality holds. Let $ N\in\mathbb{N} $ with $ N\geq 2 $ and $ A\subseteq\mathbb{G}_N\times\mathbb{G}_{2N}. $ If $ (A-A)\bigcap\big\{\overrightarrow{d}:d\in\mathbb{A}^\times\big\} = \emptyset, $ then we have

$ |A|\leq C_1q^{3N}\frac{\log N}{N}. $

Remark 3 Note that $ d\in\mathbb{G}_N\Leftrightarrow d^2\in\mathbb{G}_{2N}, $ the form of Proposition 15 is more natural than of Theorem 1.

Proof Write $ |A| = \delta q^{3N}. $ If $ \delta\leq q^{1-\frac{N}{12}}, $ then by taking

$ C_1 = \sup\limits_{N\geq 2}q^{1-N/12}\frac{N}{\log N}, $

the proposition follows. Thus in the following, we assume that $ \delta\geq q^{1-\frac{N}{12}}. $

Now, we recursively define a sequence of triples $ (N_i, A_i, \delta_i) $ with $ N_i\in\mathbb{N}_+, \ A_i\subseteq\Gamma_{N_i} $ and $ |A_i| = \delta_iq^{3N_i} $ as follows. Take $ (N_0, A_0, \delta_0) = (N, A, \delta). $ Let $ i\in\mathbb{N}. $ Suppose that $ (N_i, A_i, \delta_i) $ is defined. If $ \delta_i<q^{1-\frac{N_i}{12}}, $ we stop the definition. If $ \delta_i\geq q^{1-\frac{N_i}{12}}, $ by Proposition 14, we can find $ N_{i+1}\in\mathbb{N}_+ $ and $ A_{i+1}\subseteq\Gamma_{N_{i+1}} $ with $ |A_{i+1}| = \delta_{i+1}q^{3N_{i+1}} $ such that

(i) $ (A_{i+1}-A_{i+1})\cap\big\{\overrightarrow{d}:d\in\mathbb{A}^\times\big\} = \emptyset; $

(ii) $ \delta_{i+1}\geq\delta_i(1+\frac{c}{2}\delta_i); $

(iii) $ N_{i+1}\geq N_i-11\log_q\big(\frac{q}{\delta_i}\big). $

Write $ c' = \frac{c}{2}. $ It follows from (ii) that $ \delta_{i+1}-\delta_i\geq c'\delta^2. $ Since $ \delta_{i+1}\leq 1, $ this process produces a finite sequence $ \big\{ (N_i, A_i, \delta_i)\big\}_{i = 1}^J. $ Then for any $ 0\leq i\leq J-1, $ the triple $ (N_{i+1}, A_{i+1}, \delta_{i+1}) $ satisfies the above conditions (i)–(iii). Also, we have

$ \begin{equation} \delta_J<q^{1-\frac{N_J}{12}}. \end{equation} $

(6.1)

Claim 1 For $ j\in\mathbb{N}, $ write $ I_j = \lceil\frac{1}{2^jc'\delta}\rceil. $ If $ i\geq\sum\limits_{l = 0}^jI_l, $ then $ \delta_i\geq 2^{j+1}\delta. $

Proof We prove the claim by induction on $ j. $ For $ j = 0, $ we have $ I_j\geq\frac{1}{c'\delta}. $ It follows from (ii) that

$ \delta_i\geq\delta_0+c'i\delta_0^2. $

Thus if $ i\geq I_0, $ then $ \delta_i\geq 2\delta. $

Suppose that the claim holds for $ j. $ We now prove that the statement is true for $ j+1. $

Write $ k = \sum\limits_{l = 0}^jI_l. $ Let $ i>k. $ By (ii), we have $ \delta_i\geq\delta_k+(i-k)c'\delta_k^2. $ Thus, if $ i\geq\sum\limits_{l = 0}^{j+1}I_l, $ it follows from the induction hypothesis that

$ \delta_i\geq 2^{j+1}\delta+c'I_{j+1}(2^{j+1}\delta)^2\geq 2^{j+2}\delta. $

This completes the proof of the claim.

Take $ j_0\in\mathbb{N} $ such that $ 2^{j_0}\delta\leq1<2^{j_0+1}\delta. $ Then we have

$ J<\sum\limits_{0\leq i\leq j_0}I_i\leq\frac{2}{c'\delta}\sum\limits_{i\in\mathbb{N}}2^{-i} = \frac{4}{c'\delta}. $

It follows from (iii) that

$ N_J\geq N-11J\log_q\big(\frac{q}{\delta}\big)\geq N-\frac{44}{c'\delta}\log_q\big(\frac{q}{\delta}\big). $

By (6.1), we have

$ \delta\leq\delta_J\leq q^{1-\frac{N}{12}}\big(\frac{q}{\delta}\big)^{\frac{11}{3c'\delta}}. $

Thus, there exists a constant $ C_1>1, $ depending only on $ q, $ such that

$ 2N\leq\frac{C_1}{\delta}\log\frac{C_1}{\delta}. $

Note that the function $ x\log x $ on $ [1, +\infty) $ is increasing, and the proposition follows since

$ \frac{2N}{\log 2N}\log\big(\frac{2N}{\log 2N}\big)\leq 2N. $

Proof of Theorem 1 Write $ |A| = \delta q^{2N}. $ If $ N\leq 7, $ by taking $ C = \frac{7}{\log 7}, $ the theorem follows. In the following, we assume that $ N\geq 8. $ Write

$ N' = \big\lfloor\frac{N}{4}\big\rfloor, \ S = q^{N-N'}\ \mathrm{and}\ T = q^{N-2N'}. $

For $ 1\leq i\leq S $ and $ 1\leq j\leq T, $ take $ d_i, d'_j\in\mathbb{G}_N $ such that

$ \mathbb{G}_N = \bigcup\limits_{i = 1}^S(d_i+\mathbb{G}_{N'}) = \bigcup\limits_{j = 1}^T(d'_j+\mathbb{G}_{2N'}). $

Then, we have

$ \mathbb{G}^2_N = \bigcup\limits_{1\leq i\leq S, 1\leq j\leq T}(d_i+\mathbb{G}_{N'})\times(d'_j+\mathbb{G}_{2N'}) = \bigcup\limits_{1\leq i\leq S, 1\leq j\leq T}\big((d_i, d'_j)+\Gamma_{N'}\big). $

Write

$ A_{i, j} = A\bigcap\big((d_i, d'_j)+\Gamma_{N'}\big). $

Take $ 1\leq i_0\leq S $ and $ 1\leq j_0\leq T $ such that

$ \big|A_{i_0, j_0}\big| = \max\limits_{1\leq i\leq S, 1\leq j\leq T}\big|A_{i, j}\big|. $

Write $ A' = A_{i_0, j_0}. $ Then we have $ (A'-A')\bigcap\big\{\overrightarrow{d}:d\in\mathbb{A}^\times\big\} = \emptyset $ and

$ |A'|\geq\frac{1}{ST}\sum\limits_{1\leq i\leq S, 1\leq j\leq T}|A_{i, j}|\geq\frac{1}{ST}\Big|\bigcup\limits_{1\leq i\leq S, 1\leq j\leq T}A_{i, j}\Big| = \frac{1}{ST}|A| = \delta q^{3N'}. $

Define $ f:\Gamma_{N'}\rightarrow (d_{i_0}, d'_{j_0})+\Gamma_{N'} $ to be $ f(m) = (d_{i_0}, d'_{j_0})+m. $ Then $ f $ is a bijection. Take $ B = f^{-1}(A'). $ Since $ B-B = A'-A', $ we have $ (B-B)\bigcap\big\{\overrightarrow{d}:d\in\mathbb{A}^\times\big\} = \emptyset. $ It follows from Proposition 15 that

$ |B|\leq C_1q^{3N'}\frac{\log N'}{N'}\leq C_1\frac{N}{N/4-1}q^{3N'}\frac{\log N}{N}. $

Note that $ N\geq 8 $ and $ \delta\leq|B|q^{-3N'}, $ by taking $ C = 8C_1, $ the theorem follows.

For $ 1\leq s\leq l, $ take $ c_{s1}, c_{s2}\in\mathbb{A} $ such that $ P_s(x) = c_{s1}x+c_{s2}x^2. $ Write $ \mathcal{P} = \big(c_{sj}\big)_{1\leq s\leq l, 1\leq j\leq 2}. $ Denote by $ r $ the rank of the matrix $ \mathcal{P}. $ Then $ 1\leq r\leq 2. $ Thus, we divide into two case.

Case 1 Suppose that $ r = 2. $ Without loss of generality, we assume that $ \big(c_{11}, c_{12}\big) $ and $ \big(c_{21}, c_{22}\big) $ are linearly independent. Write $ \mathcal{R} = \big(c_{ij}\big)_{1\leq i, j\leq 2}, \ e_1 = (1, 0) $ and $ e_2 = (0, 1). $ For $ 1\leq i\leq 2, $ take $ \xi'_i\in\mathbb{K}^2 $ such that $ \mathcal{R}\xi'_i = e_i. $ When $ l\geq 3, $ take $ \mathcal{D} = \big(d'_{tj}\big)_{1\leq t\leq l-2, 1\leq j\leq 2} $ such that

$ \big(c_{t'j}\big)_{3\leq t'\leq l, 1\leq j\leq 2} = \mathcal{D}\mathcal{R}. $

Take $ S\in\mathbb{N} $ with $ S\geq 4 $ and $ D\in\mathbb{A}^\times $ such that

$ D, c_{ij}\in\mathbb{G}_S, \ \xi_i = D\xi'_i\in\mathbb{G}^2_S\ \; (1\leq i, j\leq 2). $

If $ l\geq 3, $ we also require

$ d_{tj} = Dd'_{tj}\in\mathbb{G}_S\ (1\leq t\leq l-2, 1\leq j\leq 2). $

If $ N\leq S, $ by taking $ C' = \big(\frac{S}{\log S}\big)^\frac{1}{l}, $ the theorem follows. Thus, we assume that $ N\geq S+1. $

Claim 2 For $ m\in\mathbb{G}_S^2, $ write $ B'_m = \big\{ b\in\mathbb{G}^2_{N+S}:\mathcal{R}b+m\in A^2\big\}. $ Then there exists $ \underline{m}\in\mathbb{G}_S^2 $ such that

$ \big|B'_{\underline{m}}\big|\geq q^{-2S}|A|^2. $

Proof Let $ a = (a_1, a_2)\in A^2. $ For $ 1\leq i\leq 2, $ take $ a'_i\in\mathbb{G}_{N-\mathrm{ord}D} $ and $ a''_i\in\mathbb{G}_{\mathrm{ord}D} $ such that $ a_i = Da'_i+a''_i. $ Write $ b = \sum\limits_{i = 1}^2a'_i\xi_i $ and $ m' = (a''_1, a''_2). $ Then we have

$ b\in\mathbb{G}_{N+S}^2, \ m'\in\mathbb{G}_S^2\ \mathrm{and}\ \mathcal{R}b = a-m'. $

It follows that $ a\in\mathcal{R}\big(\mathbb{G}_{N+S}^2\big)+m'. $ Thus, we see that

$ \begin{equation} A^2\subseteq\bigcup\limits_{m\in\mathbb{G}_S^2}\Big(\mathcal{R}\big(\mathbb{G}_{N+S}^2\big)+m\Big). \end{equation} $

(7.1)

Take $ \underline{m}\in\mathbb{G}_S^2 $ such that $ \big|B'_{\underline{m}}\big| = \max\limits_{m\in\mathbb{G}_S^2}\big|B'_m\big|. $ By (7.1), we have

$ \big|B'_{\underline{m}}\big|\geq\frac{1}{q^{2S}}\sum\limits_{m\in\mathbb{G}_S^2}\big|B'_m\big|\geq\frac{1}{q^{2S}} \Big|\bigcup\limits_{m\in\mathbb{G}_S^2}\Big(\big(\mathcal{R}\big(\mathbb{G}_{N+S}^2\big)+m\big)\cap A^2\Big)\Big| = \frac{1}{q^{2S}}|A^2|. $

This completes the proof of the claim.

Claim 3 Suppose that $ l\geq 3. $ For $ m\in\mathbb{G}_{N+3S}^{l-2}, $ write $ B''_m = \big\{ b\in B'_{\underline{m}}:\mathcal{D}\mathcal{R}b+m\in A^{l-2}\big\}. $ Then there exists $ \underline{m'}\in\mathbb{G}_{N+3S}^{l-2} $ such that

$ \big|B''_{\underline{m'}}\big|\geq q^{-(l-2)N-(3l-4)S}|A|^l. $

Proof Let $ n\in \mathbb{A}^{l-2} $ and $ b\in B'_{\underline{m}}. $ If $ n+\mathcal{D}\mathcal{R}b\in A^{l-2}, $ then $ n\in\mathbb{G}_{N+3S}^{l-2}. $ Thus

$ \begin{equation} \sum\limits_{n\in\mathbb{G}_{N+3S}^{l-2}}\sum\limits_{b\in B'_{\underline{m}}}1_{A^{l-2}}\big(n+\mathcal{D}\mathcal{R}b\big) = \sum\limits_{b\in B'_{\underline{m}}}\sum\limits_{n\in \mathbb{A}^{l-2}}1_{A^{l-2}}\big(n+\mathcal{D}\mathcal{R}b\big) = |B'_{\underline{m}}||A|^{l-2}. \end{equation} $

(7.2)

Take $ \underline{m'}\in\mathbb{G}_{N+3S}^{l-2} $ such that $ \big|B''_{\underline{m'}}\big| = \max\limits_{m\in\mathbb{G}_{N+3S}^{l-2}}\big|B''_m\big|. $ Then we have

$ \big|B''_{\underline{m'}}\big|\geq\frac{1}{q^{(l-2)(N+3S)}}\sum\limits_{m\in\mathbb{G}_{N+3S}^{l-2}}|B''_m| = \frac{1}{q^{(l-2)(N+3S)}}\sum\limits_{m\in\mathbb{G}_{N+3S}^{l-2}}\sum\limits_{b\in B'_{\underline{m}}}1_{A^{l-2}}\big(m+\mathcal{D}\mathcal{R}b\big). $

The claim follows from (7.2) and Claim 2.

Write

$ \overline{m} = \left\{ \begin{array}{ll} \underline{m}, &\mathrm{if}\ l = 2, \\ (\underline{m}, \underline{m'}), &\mathrm{if}\ l\geq 3. \end{array} \right. $

Define $ B = \big\{ b\in\mathbb{G}^2_{N+S}:\mathcal{P}b+\overline{m}\in A^l\big\}. $ Then by Claims 2 and 3, we have

$ \begin{equation} |B|\geq q^{-(l-2)N-(3l-4)S}|A|^l. \end{equation} $

(7.3)

Suppose that there exists $ d\in\mathbb{A} $ suth that $ b'-b = \overrightarrow{d} $ for some $ b, b'\in B. $ Since

$ \mathcal{P}\overrightarrow{d} = \mathcal{P}b'-\mathcal{P}b\in A^l-A^l, $

we have

$ \big\{ P_1(d), \cdots, P_l(d)\big\}\subseteq (A-A), $

from which it follows that $ d = 0. $ Thus, we obtain

$ (B-B)\bigcap\big\{\overrightarrow{d}:d\in\mathbb{A}^\times\big\} = \emptyset. $

By Theorem 1, we have

$ |B|\leq Cq^{2(N+S)}\frac{\log (N+S)}{N+S}\leq Cq^{2(N+S)}\frac{\log N}{N}. $

By taking $ C' = C^{\frac{1}{l}}q^{\frac{(3l-2)S}{l}}, $ the theorem follows from (7.3).

Case 2 Suppose that $ r = 1. $ Without loss of generality, we assume that $ \mathcal{R} = \big(c_{11}, c_{12}\big)\neq 0. $ Take $ \xi'\in\mathbb{K}^2 $ such that $ \mathcal{R}\xi' = 1. $ When $ l\geq 2, $ take $ \mathcal{D} = \big(d'_1, \cdots, d'_{l-1}\big) $ such that $ \big(c_{t'j}\big)_{2\leq t'\leq l, 1\leq j\leq 2} = \mathcal{D}\mathcal{R}. $

Take $ S\in\mathbb{N} $ with $ S\geq 4 $ and $ D\in\mathbb{A}^\times $ such that

$ D, c_{1j}\in\mathbb{G}_S\ (1\leq j\leq 2), \ \xi = D\xi'\in\mathbb{G}^2_S. $

If $ l\geq 2, $ we also require

$ d_t = Dd'_t\in\mathbb{G}_S\ (1\leq t\leq l-1). $

If $ N\leq S, $ by taking $ C' = \big(\frac{S}{\log S}\big)^\frac{1}{l}, $ the theorem follows. Thus we assume that $ N\geq S+1. $

Claim 4 For $ m\in\mathbb{G}_S, $ write $ B'_m = \big\{ b\in\mathbb{G}^2_{N+S}:\mathcal{R}b+m\in A\big\}. $ Then there exists $ \underline{m}\in\mathbb{G}_S $ such that

$ \big|B'_{\underline{m}}\big|\geq q^{N-S}|A|. $

Proof Let $ a\in A. $ Take $ a'\in\mathbb{G}_{N-\mathrm{ord}D} $ and $ a''\in\mathbb{G}_{\mathrm{ord}D} $ such that $ a = Da'+a''. $ Write $ b = a'\xi. $ Then we have

$ b\in\mathbb{G}_{N+S}^2, \ a''\in\mathbb{G}_S\ \mathrm{and}\ \mathcal{R}b = a-a''. $

It follows that $ a\in\mathcal{R}\big(\mathbb{G}_{N+S}^2\big)+a''. $ Thus, we see that

$ \begin{equation} A\subseteq\bigcup\limits_{m\in\mathbb{G}_S}\Big(\mathcal{R}\big(\mathbb{G}_{N+S}^2\big)+m\Big). \end{equation} $

(7.4)

For $ m\in\mathbb{G}_S, $ write $ A_m = A\bigcap\big(\mathcal{R}\big(\mathbb{G}_{N+S}^2\big)+m\big). $ For each $ a\in A_m, $ we fix a $ \hat{a}\in\mathbb{G}_{N+S}^2 $ such that $ \mathcal{R}\hat{a}+m = a. $ Since

$ \big\{\hat{a}+d(-c_{12}, c_{11}):a\in A_m, \ d\in\mathbb{G}_N\big\}\subseteq B'_m, $

it follows that $ |B'_m|\geq q^N|A_m|. $ Take $ \underline{m}\in\mathbb{G}_S $ such that $ \big|B'_{\underline{m}}\big| = \max\limits_{m\in\mathbb{G}_S}\big|B'_m\big|. $ By (7.4), we have

$ \big|B'_{\underline{m}}\big|\geq\frac{1}{q^S}\sum\limits_{m\in\mathbb{G}_S}\big|B'_m\big| \geq q^{N-S}\sum\limits_{m\in\mathbb{G}_S}\big|A_m\big|\geq q^{N-S}\Big|\bigcup\limits_{m\in\mathbb{G}_S}A_m\Big| = q^{N-S}|A|. $

This completes the proof of the claim.

Claim 5 Suppose that $ l\geq 2. $ For $ m\in\mathbb{G}_{N+3S}^{l-1}, $ write $ B''_m = \big\{ b\in B'_{\underline{m}}:\mathcal{D}\mathcal{R}b+m\in A^{l-1}\big\}. $ Then there exists $ \underline{m'}\in\mathbb{G}_{N+3S}^{l-1} $ such that

$ \big|B''_{\underline{m'}}\big|\geq q^{-(l-2)N-(3l-2)S}|A|^l. $

Proof The claim follows from the similar argument as in Claim 3.

Write

$ \overline{m} = \left\{ \begin{array}{ll} \underline{m}, &\mathrm{if}\ l = 1, \\ (\underline{m}, \underline{m'}), &\mathrm{if}\ l\geq 2. \end{array} \right. $

Define $ B = \big\{ b\in\mathbb{G}^2_{N+S}:\mathcal{P}b+\overline{m}\in A^l\big\}. $ Then by Claims 4 and 5, we have

$ \begin{equation} |B|\geq q^{-(l-2)N-(3l-2)S}|A|^l. \end{equation} $

(7.5)

By using similar arguments as in Case 1, we obtain $ |B|\leq Cq^{2(N+S)}\frac{\log N}{N}. $ By taking $ C' = C^{\frac{1}{l}}q^{3S}, $ the theorem follows from (7.5).

Combining the above two cases, the proof of the theorem is completed.