2·1. Rational points on absolutely irreducible curves

Let q be a prime power. It is well known that by the Weil bound we have

(2·1) \begin{equation}\# \{(x,y)\in {\mathbb F}_q^2 \;:\; F(x,y) = 0\} = q+ O\left(d^2 q^{1/2}\right)\end{equation}

for any absolutely irreducible polynomial $F(X,Y) \in {\mathbb F}_q[X,Y]$ of degree d (see, for example, [ Reference Lorenzini15 , section X·5, equation (5·2)]). One can see that (2·1) is a genuine asymptotic formula only for $d = O(q^{1/4})$ and is in fact weaker than the trivial bound

\begin{align*} \# \{(x,y)\in {\mathbb F}_q^2 \;:\; F(x,y) = 0\} = O\left(d q\right)\end{align*}

for $d \ge q^{1/2}$ , which is exactly the range of our interest. To obtain nontrivial bounds for such large values of d we recall the following result, which is a combination of [ Reference Voloch27 , theorem (i)] with the Weil bound (2·1) (and the trivial inequality $p + 2d^2p^{1/2} \le 3p$ for $d \le p^{1/4}$ ).

Lemma 2·1. Let p be prime and let $F(X,Y) \in {\mathbb F}_p[X,Y]$ be an absolutely irreducible polynomial of degree $d<p$ . Then

\begin{align*} \# \{(x,y) \in {\mathbb F}_p^2 \;:\; F(x,y) = 0\} \le 4d^{4/3} p^{2/3} + 3p.\end{align*}

2·2. Absolute irreducibility of some polynomials

To apply the bound of Lemma 2·1 we need to establish absolute irreducibility of polynomials relevant to our applications. We present it in a general form for arbitrary finite fields, as it may be useful for other applications. Below we use a natural mapping of integers into elements of a finite field ${\mathbb F}_q$ of q elements of characteristic p via the reduction modulo p.

Lemma 2·2. Given integers $n>m \ge 1$ with $\gcd(m,n)=1$ , there exists a non-zero polynomial $\Delta(U,V)\in {\mathbb Z}[U,V]$ , such that for every prime power q and positive integer s with $\gcd(s,q)=1$ and $A,B \in {\mathbb F}_q$ with $\Delta(A,B) \ne 0$ , the polynomial

\begin{align*} F(X,Y) = (X^{sm} + Y^{sm} -A)^n - (X^{sn} + Y^{sn} -B)^m\in {\mathbb F}_q[X,Y]\end{align*}

is absolutely irreducible.

Proof. Let us begin by considering the case $s=1$ .

We introduce a new variable Z and note that for $s=1$ the curve $F=0$ is isomorphic to

(2·2) \begin{equation} \begin{cases}& X^{m} + Y^{m} -A = Z^m \cr&X^{n} + Y^{n} -B = Z^n. \end{cases}\end{equation}

The isomorphism is the projection to the X, Y-plane, with inverse given by

\begin{align*} (X,Y)\mapsto (X,Y,\left(X^{m} + Y^{m} -A\right)^u\left(X^{n} + Y^{n} -B\right)^v)\end{align*}

with some fixed integers u and v satisfying

\begin{align*} mu+nv=1.\end{align*}

The two equations in (2·2) have gradients

\begin{align*} m(X^{m-1},Y^{m-1},-Z^{m-1}) \qquad\mbox{and}\qquad n (X^{n-1},Y^{n-1},-Z^{n-1}),\end{align*}

respectively. For the curve defined by the system (2·2) to be singular at a point (x, y, z) the corresponding gradients have to be linearly dependent. This condition, when fed back into (2·2) gives a relation between A, B.

More explicitly we proceed as follows.

First, we seek the polynomial $\Delta$ in the form

(2·3) \begin{equation}\Delta = mn \Delta_0\end{equation}

with some $\Delta_0(U,V)\in {\mathbb Z}[U,V]$ , thus $\Delta(A,B) \ne 0$ , guarantees that $mn \ne 0$ in ${\mathbb F}_q$ .

If (x, y, z) is a solution to (2·2) with $x=y=0$ , then $z^m = -A$ , $z^n = -B$ , and so $(\!-\!A)^n = (\!-\!B)^m$ . Thus we also request

(2·4) \begin{equation}\left((\!-\!A)^n - (\!-\!B)^m\right) \mid \Delta(A,B).\end{equation}

The possibilities $x=z=0$ and $y=z=0$ can be similarly treated and lead to the requirement

(2·5) \begin{equation}\left(A^n - B^m\right) \mid \Delta(A,B).\end{equation}

If $x=0$ and $yz \ne 0$ , then the gradient condition gives $y = \zeta z$ with some $\zeta^{n-m}=1$ and (2·2) gives

\begin{align*} \left(\zeta^m-1\right)z^m = A \qquad\mbox{and}\qquad \left(\zeta^n-1\right)z^n = B.\end{align*}

Therefore, $\left(\zeta^n-1\right)^m A^n = \left(\zeta^m-1\right)^n B^m$ , and we also request

(2·6) \begin{equation}\prod_{\zeta^{n-m}=1} \left(\left(\zeta^n-1\right)^m A^n - \left(\zeta^m-1\right)^n B^m\right) \mid \Delta(A,B),\end{equation}

where the product is taken over all roots of unity $\zeta$ with $\zeta^{n-m}=1$ .

If $xyz \ne 0$ , then we see from (2·7) that there has to be a constant $\lambda$ with

(2·7) \begin{equation}x^{n-m} = y^{n-m} = z^{n-m} = \lambda,\end{equation}

so we can write $y = \zeta_1 x,$ and $z = \zeta_2 x$ with $\zeta_1^{n-m}= \zeta_2^{n-m}=1$ , leading to

(2·8) \begin{equation}\prod_{\substack{\zeta_1^{n-m}= 1\cr\zeta_2^{n-m}=1}} \left(\left(1+\zeta_1^n-\zeta_2^n\right)^m A^n - \left(1+\zeta_1^m-\zeta_2^m\right)^n B^m \right) \mid \Delta(A,B),\end{equation}

where the product is taken over all pairs of roots of unity $(\zeta_1, \zeta_2)$ with $\zeta_1^{n-m}= \zeta_2^{n-m}=1$ .

A similar argument works at infinity and shows that, for generic A and B, the curve is smooth.

Given X, Y, there is a unique choice of Z satisfying (2·2), so the projection to the X, Y does not acquire singularities from distinct points in three-space. The only singularities are cusps coming from a vertical tangent line which are unibranched (since the curve in three-space is smooth). However, a reducible plane curve has singular points with more than one branch wherever two components meet. Hence (2·2) is an irreducible curve. (See [ Reference Kunz12 , chapter 16] for a detailed exposition of branches of curve singularities.)

We have shown that, for $s=1$ , the polynomial F is absolutely irreducible. We consider the algebraic curve ${\mathcal C}$ which is a non-singular projective model of $F=0$ (still with $s=1$ ).

Suppose $n>m>1$ . We now use an argument similar to [ Reference Ostafe, Shparlinski and Voloch19 , lemma 4·3].

For a point $P=(0,y_0)$ on the curve $F=0$ we have

\begin{align*} \frac{\partial F}{\partial X} (0,y_0) = 0.\end{align*}

Next we show that the point $P=(0,y_0)$ is a simple point on the curve $F=0$ with

(2·9) \begin{equation}\frac{\partial F}{\partial Y}(0, y_0) \ne 0\end{equation}

(for generic A and B). It now suffices to show that the discriminant D(A, B) of $F(0,Y) \in {\mathbb F}_q[Y]$ (as a polynomial in A, B) is not identically zero, and can be chosen with integer coefficients which depend only on m and n.

Taking $A=1$ and $B = 0$ , the polynomial F specialises to

\begin{align*} (Y^m-1)^n - Y^{mn} = \prod_{\xi^n = 1} (Y^m(1-\xi) -1),\end{align*}

which has a non-zero discriminant as each factor $(Y^m(1-\xi) -1)$ is square-free and these factors are relatively prime. Thus we impose the condition

(2·10) \begin{equation}D(A,B) \mid \Delta(A,B).\end{equation}

It remains to choose $ \Delta(A,B) \in {\mathbb Z}[A,B]$ as an arbitrary fixed polynomial which depends only on m and n and satisfies the divibility conditions (2·3), (2·4), (2·5), (2·6), (2·8) and (2·10).

We now consider the case of arbitrary $s\ge 1$ (and $n>m>1$ ).

So P corresponds to a place of ${\mathcal C}$ . We consider the functions x, y on ${\mathcal C}$ that satisfy the equation $F(x,y)=0$ . The function x has a simple zero at P, hence is not a power of another function on ${\mathcal C}$ . It follows from [ Reference Stichtenoth26 , proposition 3·7·3], that the equation $U^s = x$ is irreducible over the function field of ${\mathcal C}$ and defines a cover ${\mathcal D}$ of ${\mathcal C}$ . Now, consider any point Q on ${\mathcal D}$ above a point $(x_0,0)$ on ${\mathcal C}$ . Since x is not zero at $(x_0,0)$ (for generic A, B), the curve ${\mathcal D}$ is locally isomorphic to ${\mathcal C}$ near Q and we conclude, as above, that the function y on ${\mathcal D}$ has a simple zero at Q and, in particular, is not a power of another function on ${\mathcal D}$ . Again, we conclude that the equation $W^s = y$ is irreducible over the function field of ${\mathcal D}$ and defines a cover ${\mathcal E}$ of ${\mathcal D}$ . In other words, $F(U^s,W^s)=0$ is an absolutely irreducible equation defining the curve ${\mathcal E}$ , which concludes the case $n>m>1$ .

We are now left with $n>m=1$ . It is still true that a point $P=(0,y_0)$ is a simple point on the curve $F=0$ with (2·9) (for generic A, B). Indeed, if

\begin{align*} 0 = \frac{\partial F}{\partial Y} (0,y_0) = n(y_0^{n-1} -(y_0-B)^{n-1}),\end{align*}

then combining this with

\begin{align*} 0=F(0,y_0) = -A+y_0^{n} -(y_0-B)^{n} ,\end{align*}

we derive

\begin{align*} 0= -A+y_0^{n} -(y_0-B)^{n} = -A+y_0^{n} -(y_0-B) y_0^{n-1} = -A+B y_0^{n-1} .\end{align*}

Hence

\begin{align*} 0 = By_0^{n-1} -B(y_0-B)^{n-1} = A -B(y_0-B)^{n-1}.\end{align*}

Considering the resultant of $B(Y-B)^{n-1} -A$ and $BY^{n-1} - A$ (which clear does not vanish for $A=1$ and $B=0$ and thus is a nontrivial polynomial) gives a contradiction for generic A, B. The proof then continues as before in the case $n>m>1$ .

3·1. Exponential sums and the number of solutions to some systems of equations

Here we collect some previous results on exponential sums and also about links between these bounds and the number of solutions to some congruences.

Given an integer vector $\mathbf{n} = \left(n_1, \ldots, n_r\right)\in {\mathbb Z}^r$ with $n_r> \cdots > n_1 \ge 1$ , and a subgroup ${\mathcal G} \subseteq {\mathbb F}_p^*$ , we denote by $Q_{k}(\mathbf{n};\;{\mathcal G})$ the number of solutions to the following system of r equations

(3·1) \begin{equation}\begin{split} g_1^{n_i}+\cdots + g_k^{n_i} & = g_{k+1}^{n_i}+\cdots + g_{2k}^{n_i}, \qquad i =1, \ldots, r, \cr& g_1, \ldots , g_{2k} \in {\mathcal G}.\end{split} \end{equation}

The following link between $S({\mathcal G};\;\; f)$ and $Q_{k}(\mathbf{n};\; G)$ is a slight variation of several previous results of a similar spirit.

Lemma 3·1. Let

\begin{align*} f(X) = a_r X^{n_r} + \cdots + a_1 X^{n_1} \in {\mathbb F}_p[X] \end{align*}

with nonzero coefficients $a_1, \ldots, a_r \in {\mathbb F}_p^*$ and integer exponents $n_r> \cdots > n_1 \ge 1$ . Then, for a subgroup ${\mathcal G} \subseteq {\mathbb F}_p^*$ and for any positive integers k and $\ell$ , we have

\begin{align*} \left|S({\mathcal G};\;\; f)\right|^{2k\ell} \le p^r \tau^{2k\ell - 2k -2 \ell} Q_{k}(\mathbf{n};\; {\mathcal G}) Q_{\ell}(\mathbf{n};\; {\mathcal G}).\end{align*}

Proof. We start by noticing that, for any $h \in {\mathcal G}$ , we have

\begin{align*} S({\mathcal G};\;\; f)= \sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 (hg)^{n_1} + \cdots + a_r (hg)^{n_r}\right).\end{align*}

Hence, for any integer $k \ge 1$ we have

\begin{align*}\tau\left(S({\mathcal G};\;\; f)\right)^k &= \sum_{h \in {\mathcal G}} \left(\sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 (hg)^{n_1} + \cdots + a_r (hg)^{n_r}\right)\right)^k\cr& = \sum_{h \in {\mathcal G}} \sum_{\lambda_1, \ldots, \lambda_r \in {\mathbb F}_p} J_k(\lambda_1, \ldots, \lambda_r) {\mathbf{\,e}}_p\left(\lambda_1 h^{n_1} + \cdots + \lambda_r h^{n_r}\right), \end{align*}

where $J_k(\lambda_1, \ldots, \lambda_r)$ is the number of solutions to the following system of equations:

\begin{align*}a_i\left( g_1^{n_i}+\cdots + g_k^{n_i}\right)& = \lambda_i, \qquad i =1, \ldots, r, \cr g_1&, \ldots , g_{k} \in {\mathcal G}.\end{align*}

Hence, changing the order of summations, we obtain

\begin{align*} \tau\left|S({\mathcal G};\;\; f)\right|^k \le \sum_{\lambda_1, \ldots, \lambda_r \in {\mathbb F}_p} J_k(\lambda_1, \ldots, \lambda_r) \left| \sum_{h \in {\mathcal G}} {\mathbf{\,e}}_p\left(\lambda_1 h^{n_1} + \ldots + \lambda_r h^{n_r}\right) \right|.\end{align*}

Observe that since $a_1, \ldots, a_r \ne 0$ , we have

\begin{align*}& \sum_{\lambda_1, \ldots, \lambda_r \in {\mathbb F}_p} J_k(\lambda_1, \ldots, \lambda_r) = \tau^k, \cr& \sum_{\lambda_1, \ldots, \lambda_r \in {\mathbb F}_p} J_k(\lambda_1, \ldots, \lambda_r)^2 =Q_{k}(\mathbf{n};\; G),\end{align*}

where $\mathbf{n} = \left(n_1, \ldots, n_r\right)$ .

Writing

\begin{align*} J_k(\lambda_1, \ldots, \lambda_r) = J_k(\lambda_1, \ldots, \lambda_r)^{1-1/\ell} \left(J_k(\lambda_1, \ldots, \lambda_r)^2\right)^{1/2\ell} \end{align*}

and applying the Hölder inequality, we derive

\begin{align*}\tau^{2\ell}\left(S({\mathcal G};\;\; f)\right)^{2k\ell} & \le \left( \sum_{\lambda_1, \ldots, \lambda_r \in {\mathbb F}_p} J_k(\lambda_1, \ldots, \lambda_r) \right)^{2\ell-2}\cr& \qquad \quad \times \sum_{\lambda_1, \ldots, \lambda_r \in {\mathbb F}_p} J_k(\lambda_1, \ldots, \lambda_r)^2 \cr & \qquad \qquad \quad \times \sum_{\lambda_1, \ldots, \lambda_r \in {\mathbb F}_p} \left| \sum_{h \in {\mathcal G}} {\mathbf{\,e}}_p\left(\lambda_1 h^{n_1} + \ldots + \lambda_r h^{n_r}\right) \right|^{2\ell}\cr & = p^{r} \tau^{k(2\ell-2)} Q_{k}(\mathbf{n};\; G) Q_{\ell}(\mathbf{n};\;G) \end{align*}

which concludes the proof.

We now establish a link in the opposite direction, that is, from bounds on exponential sums to bounds on $ Q_{k}(\mathbf{n};\; {\mathcal G})$ .

Lemma 3·2. Let $r \ge 2$ and let $\varepsilon\ge 0$ be fixed. Assume that there is some fixed $\eta> 0$ (depending only on r and $\varepsilon$ ) such that for all nonzero vectors $\left(a_1, \ldots, a_r\right)\in {\mathbb F}_p^r$ and for a vector $\mathbf{n} = \left(n_1, \ldots, n_r\right)\in {\mathbb Z}^r$ with $n_r> \ldots > n_1 \ge 1$ , for a subgroup ${\mathcal G} \subseteq {\mathbb F}_p^*$ of order $\tau$ with

(3·2) \begin{equation} p^{3/7+\varepsilon} \le \tau \le p^{3/4} \end{equation}

we have

(3·3) \begin{equation} \sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 g^{n_1} + \cdots + a_r g^{n_r}\right) \ll \tau p^{-\eta}. \end{equation}

Then for any integer $k\ge 3$ we have

\begin{align*} Q_{k}(\mathbf{n};\; {\mathcal G}) \ll \tau^{2k} p^{-\xi}, \end{align*}

where $\xi = \min\{r, \eta(2k-6) +1 +7\varepsilon/3\}$ .

Proof. Using the orthogonality of characters, we write

(3·4) \begin{equation}\begin{split} Q_{k}(\mathbf{n}&, {\mathcal G})\cr & = \frac{1}{p^r} \sum_{a_1, \ldots, a_r \in {\mathbb F}_p} \left|\sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 g^{n_1} + \cdots + a_r g^{n_r}\right) \right|^{2k}\cr & = \frac{\tau^{2k}}{p^r} + \frac{1}{p^r} \sum_{\substack{a_1, \ldots, a_r\in {\mathbb F}_p\cr \left(a_1, \ldots, a_r\right)\ \ne \mathbf{0}}} \left|\sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 g^{n_1} + \cdots + a_r g^{n_r}\right) \right|^{2k}.\end{split} \end{equation}

Now, using our assumption (3·3) we obtain

\begin{align*} \frac{1}{p^r} \sum_{\substack{a_1, \ldots, a_r\in {\mathbb F}_p\cr \left(a_1, \ldots, a_r\right)\ \ne \mathbf{0}}} & \left|\sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 g^{n_1} + \cdots + a_r g^{n_r}\right) \right|^{2k} \cr &\ll \frac{ \left(\tau p^{-\eta} \right)^{2k-6}}{p^r} \sum_{\substack{a_1, \ldots, a_r\in {\mathbb F}_p\cr \left(a_1, \ldots, a_r\right)\ \ne \mathbf{0}}} \left|\sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 g^{n_1} + \cdots + a_r g^{n_r}\right) \right|^{6} . \end{align*}

Dropping the restriction $\left(a_1, \ldots, a_r\right)\ \ne \mathbf{0}$ from the summation, we now obtain

\begin{align*} \frac{1}{p^r} \sum_{\substack{a_1, \ldots, a_r\in {\mathbb F}_p\cr \left(a_1, \ldots, a_r\right)\ \ne \mathbf{0}}} & \left|\sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 g^{n_1} + \cdots + a_r g^{n_r}\right) \right|^{2k} \cr &\ll \frac{ \left(\tau p^{-\eta} \right)^{2k-6}}{p^r} \sum_{a_1, \ldots, a_r\in {\mathbb F}_p} \left|\sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 g^{n_1} + \cdots + a_r g^{n_r}\right) \right|^{6} \cr & = \left(\tau p^{-\eta} \right)^{2k-6} Q_{3}(\mathbf{n};\; {\mathcal G}) . \end{align*}

Since $r \ge 2$ , we obviously have

\begin{align*} Q_{3}(\mathbf{n};\; {\mathcal G}) \le Q_{3}(n_1,n_2;\; {\mathcal G}) . \end{align*}

Thus applying Corollary 3·4 in Section 3·2 below and using that under our assumption (3·2) we have $\tau^{11/3} \ge \tau^5/p$ , we obtain

\begin{align*} \frac{1}{p^r} \sum_{\substack{a_1, \ldots, a_r\in {\mathbb F}_p\cr \left(a_1, \ldots, a_r\right)\ \ne \mathbf{0}}} \left|\sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 g^{n_1} + \cdots + a_r g^{n_r}\right) \right|^{2k} & \ll \left(\tau p^{-\eta} \right)^{2k-6} \tau^{11/3} \cr & = \tau^{2k-7/3} p^{-\eta(2k-6)}. \end{align*}

Recalling (3·2) again we see that

\begin{align*} \frac{1}{p^r} \sum_{\substack{a_1, \ldots, a_r\in {\mathbb F}_p\cr \left(a_1, \ldots, a_r\right)\ \ne \mathbf{0}}} \left|\sum_{g \in {\mathcal G}} {\mathbf{\,e}}_p\left(a_1 g^{n_1} + \cdots + a_r g^{n_r}\right) \right|^{2k} \ll \tau^{2k} p^{-\eta(2k-6) -1 -7\varepsilon/3}, \end{align*}

which together with (3·4) concludes the proof.

3·2. Bounds on the number of solutions to some systems of equations in six variables

We start with an observation that the results of this section are independent of those in Section 3·1 and hence there is no logical problem in our use of them in the proof of Lemma 3·2.

For $r=2$ and $\mathbf{n} = (m,n)$ we write $Q_{k}(m,n;\; {\mathcal G})$ for $Q_{k}(\mathbf{n};\; {\mathcal G})$ .

Here we obtain some bounds on $Q_{3}(m,n;\;{\mathcal G})$ . In fact, it is easier to work with the following system of equations:

(3·5) \begin{equation} \begin{cases}&x_1^{sm}+x_2^{sm}+ x_3^{sm} = x_{4}^{sm}+x_5^{sm} + x_{6}^{sm}\cr& x_1^{sn}+x_2^{sn}+ x_3^{sn} = x_{4}^{sn}+x_5^{sn} + x_{6}^{sn} \end{cases}, \quad x_1, \ldots , x_{6} \in {\mathbb F}_p^*,\end{equation}

instead of the system of the type (3·1) with group elements.

Denoting by $T_{3}(m,n;\;s)$ the number of solutions to (3·5) we see that

(3·6) \begin{equation}Q_{3}(m,n;\;{\mathcal G}) = s^{-6} T_{3}(m,n;\;s),\end{equation}

where

\begin{align*} s = \frac{p-1}{\tau}\end{align*}

and, as before, $\tau = \# {\mathcal G}$ .

Lemma 3·3. For integers $n > m >0$ , we have

\begin{align*} T_{3}(m,n;\;s) \ll s^{7/3} p^{11/3} + sp^4. \end{align*}

Proof. First we note that if $\gcd(m,n)= d$ then

\begin{align*} T_{3}(m,n;\;s) \le e^6 T_{3}(m/d,n/d;\;s),\end{align*}

where $e = \gcd(d,p-1)$ .

Hence we can assume that

(3·7) \begin{equation}\gcd(m,n)=1,\end{equation}

which enables us to apply Lemma 2·2.

We now fix $x_4$ , $x_5$ and $x_6$ and thus we obtain $(p-1)^3$ systems of equations of the form

\begin{align*} \begin{cases}&x_1^{sm}+ x_2^{sm}+ x_3^{sm} = A\cr& x_1^{sn} + x_2^{sn}+ + x_3^{sn} = B \end{cases}, \qquad x_1, x_2, x_3 \in {\mathbb F}_p^*,\end{align*}

where

\begin{align*} A=x_4^{sm}+ x_5^{sm}+ x_6^{sm} \qquad\mbox{and}\qquad B=x_4^{sn}+ x_5^{sn}+ x_6^{sn},\end{align*}

from which we derive

(3·8) \begin{equation}\left(x_1^{sm}+ x_2^{sm} -A\right)^n = \left(x_1^{sn} + x_2^{sn} -B\right)^m.\end{equation}

Under the assumption (3·7), let the polynomials $\Delta(U,V)\in {\mathbb Z}[U,V]$ be as in Lemma 2·2.

Since $\Delta$ depends only on m and n, we see that if p is large enough, $\Delta$ is a non-zero polynomial modulo p.

We assume first that $\Delta(A,B) = 0$ for a pair $(A,B)\in {\mathbb F}_p^2$ as above. Thus

(3·9) \begin{equation} \Delta\left(x_4^{sm}+ x_5^{sm}+ x_6^{sm}, x_4^{sn}+ x_5^{sn}+ x_6^{sn}\right) = 0.\end{equation}

If $\Delta\left(X^{sm}+x_5^{sm}+ x_6^{sm},X^{sn}+ x_5^{sn}+ x_6^{sn}\right)$ , as a polynomial in X, is not identically zero for some $(x_5,x_6)\in{\mathbb F}_p^2$ , then obviously it has O(s) zeros. Thus, in this case, the equation (3·9) has $O(sp^2)$ solutions $(x_4,x_5,x_6)\in{\mathbb F}_p^3$ .

On the other hand, if $\Delta\left(X^{sm}+x_5^{sm}+ x_6^{sm},X^{sn}+ x_5^{sn}+ x_6^{sn}\right)$ , as a polynomial in X, is identically zero, then it also holds for $X=0$ , thus

(3·10) \begin{equation} \Delta\left(x_5^{sm}+ x_6^{sm}, x_5^{sn}+ x_6^{sn}\right)=0. \end{equation}

Now, a similar argument shows that (3·10) holds for O(sp) pairs $(x_5,x_6)$ for which there are at most p values of $x_4$ .

Therefore, the equation (3·9) has $O(sp^2)$ solutions $(x_4,x_5,x_6)\in{\mathbb F}_p^3$ in total.

For each of such $O\left(sp^2\right)$ values of $(x_4,x_5,x_6)$ the corresponding equation (3·8) is nontrivial since it contains a unique term $nx_1^{sm(n-1)} x_2^{sm}$ and hence has O(sp) solutions $(x_1,x_2)$ after which there are O(s) possible values for $x_3$ (we recall that the implied constants may depend on n). Hence, the total contribution from the case $\Delta(A,B) = 0$ is $O\left(s^3 p^3\right)$ .

If $\Delta(A,B) \ne 0$ , then for the corresponding $O(p^3)$ possibilities for $(x_4, x_5, x_6) \in {\mathbb F}_p^3$ , by Lemma 2·2, we can apply Lemma 2·1 to bound the number of solutions to (3·8) (after which we have O(s) possibilities for $x_3$ ). Hence, the total contribution from the case $\Delta(A,B) \ne 0$ is $O\left(s \left(s^{4/3} p^{2/3} + p\right)p^3\right)$ .

Therefore $T_{3}(m,n;\;s) \le s^3 p^3 + s^{7/3} p^{11/3} + sp^4$ . Since $s^3 p^3 \le s^{7/3} p^{11/3}$ for $s \le p$ , the result follows.

Recalling (3·6), we see that Lemma 3·3 implies the following.

Corollary 3·4. For integers $n > m >0$ , we have

\begin{align*} Q_{3}(m,n;\; {\mathcal G}) \ll \tau^{11/3} + \tau^5/p. \end{align*}

3·3. Bounds on monomial and binomial sums

First we recall the following result of Shkredov [ Reference Shkredov21 , theorem 1] (with a slight generalisation and also combined with a direct implication of (1·1)).

Lemma 3·6. Let $f(X) = aX^n \in {\mathbb F}_p[X]$ of degree $n\ge 1$ and with $a \ne 0$ , and let ${\mathcal G}\subseteq {\mathbb F}_p^*$ be a subgroup ${\mathcal G}$ of order $\tau$ . Then

\begin{align*} S({\mathcal G};\;\; f) \ll \min\{p^{1/2}, \tau^{1/2} p^{1/6} (\log p)^{1/6}\}.\end{align*}

Proof. We remark that the result of Shkredov [ Reference Shkredov21 , theorem 1] corresponds to $n=1$ . Otherwise we note that

\begin{align*} S({\mathcal G};\;\; f) = d \sum_{x \in {\mathcal G}^d} {\mathbf{\,e}}_p(ax),\end{align*}

where $d = \gcd(\tau, n)$ and ${\mathcal G}^d= \{g^d \;:\; g \in {\mathcal G}\}$ .

We now derive the following estimate, which improves (1·3) for $\tau \le p^{21/40}$ , remains nontrivial for $\tau \ge p^{3/7+\varepsilon}$ for any fixed $\varepsilon>0$ and which we believe is of independent interest. For this, we apply Lemma 3·1 with $k=\ell=3$ and we use Corollary 3·4.

Lemma 3·7. Let $f(X) = aX^m + bX^n \in {\mathbb F}_p[X]$ with integers $n > m \ge 1$ and $(a,b) \ne (0,0)$ , and let ${\mathcal G}\subseteq {\mathbb F}_p^*$ be a subgroup of order $\tau$ . Then

\begin{align*} S({\mathcal G};\;\; f) \ll \tau^{20/27} p^{1/9}. \end{align*}

Hence we now assume $a,b \in {\mathbb F}_p^*$ . We apply Lemma 3·1 with

\begin{align*} (k, \ell) = (3,3) \end{align*}

and the bound of Corollary 3·4. We also note that for $\tau >p^{21/40}$ we have

\begin{align*} p^{1/2} \le \tau^{20/27} p^{1/9}. \end{align*}

Hence we only need to apply Corollary 3·4 for $\tau \le p^{21/40}$ in which case $ \tau^{11/3} \ge \tau^5/p$ , and thus in this case we simply have $Q_{3}(m,n;\; {\mathcal G}) \ll \tau^{11/3}$ and the result follows.

4·1. Preliminaries and the basis of induction

We prove the result by induction on the number of terms r in the polynomial

\begin{align*} f(X) = a_r X^{n_r} + \cdots + a_1 X^{n_1} \in {\mathbb F}_p[X] \end{align*}

with nonzero coefficients $a_1, \ldots, a_r \in {\mathbb F}_p^*$ and integer exponents $n =n_r> \cdots > n_1 \ge 1$ .

We see from Lemma 3·7 that for $r=1,2$ the condition (3·3) of Lemma 3·2 is satisfied with

\begin{align*} \eta = \eta_1(\varepsilon) \qquad\mbox{and}\qquad \eta = \eta_2(\varepsilon),\end{align*}

respectively, where $\eta_{1}(\varepsilon)$ and $\eta_{2}(\varepsilon)$ are given by (1·5), which form the basis of induction.

4·2. Inductive step

Assume that the result holds for all nontrivial polynomials of degree at most n with at most $r-1$ monomials and we prove it for polynomials with $r \ge 3$ monomials. First we note that we can assume that $\tau \le p^{3/4}$ since otherwise the bound (1·3) is stronger than that of Theorem 1·1.

In particular, the condition (3·2) of Lemma 3·2 is satisfied. We fix some arbitrary positive integers k, $\ell$ , u and v with $u, v \le r$ . Let

\begin{align*} \mathbf{n}_u = (n_1, \ldots, n_u) \qquad\mbox{and}\qquad \mathbf{n}_v= (n_1, \ldots, n_v) . \end{align*}

We now use the trivial bounds

(4·1) \begin{equation} Q_{k}(\mathbf{n};\; {\mathcal G}) \le Q_{k}(\mathbf{n}_u;\; {\mathcal G}) \qquad\mbox{and}\qquad Q_{\ell}(\mathbf{n};\; {\mathcal G})\le Q_{\ell}(\mathbf{n}_v;\; {\mathcal G}).\end{equation}

In fact, we choose $u = r-1$ and $v = 2$ . Furthermore, we recall the definition (1·7) and set

(4·2) \begin{equation} k = \kappa_r(\varepsilon) \qquad\mbox{and}\qquad \ell = 3,\end{equation}

in which case, using the induction assumption, by Lemma 3·2, used with $u = r-1$ instead of r, we have

(4·3) \begin{equation} Q_{k}(\mathbf{n}_{r-1};\; {\mathcal G})\ll \tau^{2k}/p^{r-1},\end{equation}

while by Corollary 3·4, using that $\tau\le p^{3/4}$ , we obtain

(4·4) \begin{equation} Q_{3}(\mathbf{n}_2;\; {\mathcal G}) \ll \tau^{11/3} + \tau^5/p \ll \tau^{11/3}.\end{equation}

Indeed, (4·3) follows from the definition of $\kappa_r(\varepsilon)$ in (1·7), which ensures that $\xi =r-1$ in Lemma 3·2.

Next, substituting the bounds (4·1), (4·3) and (4·4) in the estimate of Lemma 3·1, we obtain

\begin{align*} S({\mathcal G};\;\; f)^{6k} \ll p^r \tau^{4k -6} \frac{\tau^{2k}}{p^{r-1}} \tau^{11/3}= \tau^{6k -7/3} p \le \tau^{6k -7\varepsilon/3} .\end{align*}

Hence

\begin{align*} S({\mathcal G};\;\; f) \ll \tau^{1 -7\varepsilon/18k}.\end{align*}

Recalling the definition (1·6) and the choice of k in (4·2), we conclude the proof.