Deep reinforcement learning with emergent communication for coalitional negotiation games

1.   Introduction

Let $\mathbb{F}_{q}$ be the finite field of $q$ elements with characteristic $p$, where $q = p^r$, $p$ is a prime number. Let $\mathbb{F}_{q}^* = \mathbb{F}_{q}\backslash \{0\}$ and $\mathbb{Z}^+$ denote the set of positive integers. Let $s\in \mathbb{Z}^+$ and $b\in \mathbb{F}_{q}$. Let $f(x_1, \ldots, x_s)$ be a diagonal polynomial over $\mathbb{F}_{q}$ of the following form

where $a_{i}\in \mathbb{F}_{q}^*,$ $m_{i}\in \mathbb{Z}^{+},$ $i = 1, \ldots, s.$ Denote by $N_{q}(f = b)$ the number of $\mathbb{F}_{q}$-rational points on the affine hypersurface $f = b$, namely,

In 1949, Hua and Vandiver ^[1] and Weil ^[2] independently obtained the formula of $N_{q}(f = b)$ in terms of character sum as follows

where the sum is taken over all $s$ multiplicative characters of $\mathbb{F}_{q}$ that satisfy $\psi_{i}^{m_{i}} = \varepsilon,$ $\psi_{i}\ne \varepsilon,$ $i = 1, \dots, s$ and $\psi_{1}\cdots \psi_{s} = \varepsilon$. Here $\varepsilon$ is the trivial multiplicative character of $\mathbb{F}_{q}$, and $J_q^{0}(\psi _1, \dots, \psi _s)$ is the Jacobi sum over $\mathbb{F}_{q}$ defined by

Though the explicit formula for $N_q(f = b)$ are difficult to obtain in general, it has been studied extensively because of their theoretical importance as well as their applications in cryptology and coding theory; see^{[3,4,5,6,7,8,9]}. In this paper, we use the Jacobi sums, Gauss sums and the results of quadratic form to deduce the formula of the number of $\mathbb{F}_{q^2}$-rational points on a class of hypersurfaces over $\mathbb{F}_{q^2}$ under certain conditions. The main result of this paper can be stated as

Theorem 1.1. Let $q = 2^r$ with $r\in \mathbb{Z}^+$ and $\mathbb{F}_{q^2}$ be the finite field of $q^2$ elements. Let $f(X) = a_{1}x_{1}^{m_{1}}+a_{2}x_{2}^{m_{2}}+\dots+a_{s}x_{s}^{m_{s}},$ $g(Y) = y_{1}y_{2}+y_{3}y_{4}+\dots+y_{n-1}y_{n}+y_{n-2t-1}^{2}+\dots$ $+y_{n-3}^2+y_{n-1}^{2}+b_{t}y_{n-2t}^{2}+\dots +b_{1}y_{n-2}^{2}+b_{0}y_{n}^{2}$, and $l(X, Y) = f(X)+g(Y)$, where $a_{i}, b_j\in \mathbb{F}_{q^{2}}^{*},$ $m_i\ne 1,$ $(m_{i}, m_{k}) = 1,$ $i\ne k,$ $m_{i}|(q+1),$ $m_{i}\in \mathbb{Z}^{+},$ $2|n$, $n > 2$, $0\leq t\leq \frac{n}{2}-2$, ${\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(b_{j}) = 1$ for $i, k = 1, \dots, s$ and $j = 0, 1, \dots, t$. For $h\in \mathbb{F}_{q^2}$, we have

(1) If $h = 0$, then

(2) If $h\in \mathbb{F}_{q^2}^{*}$, then

Here,

2.   Prerequisites

To prove Theorem 1.1, we need the lemmas and theorems below which are related to the Jacobi sums and Gauss sums.

Definition 2.1. Let $\chi$ be an additive character and $\psi$ a multiplicative character of $\mathbb{F}_{q}$. The Gauss sum $G_q(\psi, \chi)$ in $\mathbb{F}_{q}$ is defined by

In particular, if $\chi$ is the canonical additive character, i.e., $\chi(x) = e^{2 \pi i {\mathrm{Tr}}_{\mathbb{F}_{q}/ \mathbb{F}_{p}}(x)/p}$ where ${\mathrm{Tr}}_{\mathbb{F}_{q}/ \mathbb{F}_{p}}(y) = y + y^{p} + \dots + y^{p^{r - 1}}$ is the absolute trace of $y$ from $\mathbb{F}_{q}$ to $\mathbb{F}_{p}$, we simply write $G_q(\psi): = G_q(\psi, \chi)$.

Let $\psi$ be a multiplicative character of $\mathbb{F}_{q}$ which is defined for all nonzero elements of $\mathbb{F}_{q}$. We extend the definition of $\psi$ by setting $\psi(0) = 0$ if $\psi\neq \varepsilon$ and $\varepsilon(0) = 1$.

Definition 2.2. Let $\psi_1, \ldots, \psi_s$ be $s$ multiplicative characters of $\mathbb{F}_{q}$. Then, $J_q(\psi _1, \dots, \psi _s)$ is the Jacobi sum over $\mathbb{F}_{q}$ defined by

The Jacobi sums $J_q(\psi_1, \ldots, \psi_s)$ as well as the sums $J_q^0(\psi_1, \ldots, \psi_s)$ can be evaluated easily in case some of the multiplicative characters $\psi_i$ are trivial.

Lemma 2.3. ([10,Theorem $5.19$,p. 206]) If the multiplicative characters $\psi_1, \ldots, \psi_s$ of $\mathbb{F}_{q}$ are trivial, then

If some, but not all, of the $\psi_i$ are trivial, then

Lemma 2.4. ([10,Theorem $5.20$,p. 206]) If $\psi_1, \ldots, \psi_s$ are multiplicative characters of $\mathbb{F}_{q}$ with $\psi_s$ nontrivial, then

if $\psi_1\cdots\psi_s$ is nontrivial and

if $\psi_1\cdots\psi_s$ is trivial.

If all $\psi_i$ are nontrivial, there exists an important connection between Jacobi sums and Gauss sums.

Lemma 2.5. ([10,Theorem $5.21$,p. 207]) If $\psi_1, \ldots, \psi_s$ are nontrivial multiplicative characters of $\mathbb{F}_{q}$ and $\chi$ is a nontrivial additive character of $\mathbb{F}_{q}$, then

if $\psi_1\cdots \psi_s$ is nontrivial and

if $\psi_1\cdots \psi_s$ is trivial.

We turn to another special formula for Gauss sums which applies to a wider range of multiplicative characters but needs a restriction on the underlying field.

Lemma 2.6. ([10,Theorem $5.16$,p. 202]) Let q be a prime power, let $\psi$ be a nontrivial multiplicative character of $\mathbb{F}_{q^2}$ of order m dividing $q+1$. Then

For $h\in \mathbb{F}_{q^2}$, define $v(h) = -1$ if $h\in \mathbb{F}_{q^2}^{*}$ and $v(0) = q^2-1$. The property of the function $v(h)$ will be used in the later proofs.

Lemma 2.7. ([10,Lemma $6.23$,p. 281]) For any finite field $\mathbb{F}_{q}$, we have

for any $b\in \mathbb{F}_{q}$,

where the sum is over all $c_1, \ldots, c_m\in \mathbb{F}_{q}$ with $c_1+\dots+c_m = b.$

The quadratic forms have been studied intensively. A quadratic form $f$ in $n$ indeterminates is called nondegenerate if $f$ is not equivalent to a quadratic form in fewer than $n$ indeterminates. For any finite field $\mathbb{F}_q$, two quadratic forms $f$ and $g$ over $\mathbb{F}_q$ are called equivalent if $f$ can be transformed into $g$ by means of a nonsingular linear substitution of indeterminates.

Lemma 2.8. ([10,Theorem $6.30$,p. 287]) Let $f\in \mathbb{F}_{q}[x_1, \ldots, x_n]$, $q$ even, be a nondegenerate quadratic form. If $n$ is even, then $f$ is either equivalent to

or to a quadratic form of the type

where $a\in \mathbb{F}_{q}$ satisfies ${\mathrm{Tr}}_{\mathbb{F}_{q}/\mathbb{F}_{p}}(a) = 1.$

Lemma 2.9. ([10,Corollary $3.79$,p. 127]) Let $a\in \mathbb{F}_{q}$ and let $p$ be the characteristic of $\mathbb{F}_{q}$, the trinomial $x^p-x-a$ is irreducible in $\mathbb{F}_{q}$ if and only if ${\mathrm{Tr}}_{\mathbb{F}_{q}/\mathbb{F}_{p}}(a)\ne 0.$

Lemma 2.10. ([10,Lemma $6.31$,p. 288]) For even $q$, let $a\in \mathbb{F}_{q}$ with ${\mathrm{Tr}}_{\mathbb{F}_{q}/\mathbb{F}_{p}}(a) = 1$ and $b\in \mathbb{F}_{q}.$ Then

Lemma 2.11. ([10,Theorem $6.32$,p. 288]) Let $\mathbb{F}_{q}$ be a finite field with $q$ even and let $b\in \mathbb{F}_{q}$. Then for even $n$, the number of solutions of the equation

in $\mathbb{F}_{q}^n$ is $q^{n-1}+v(b)q^{(n-2)/2}$. For even $n$ and $a\in \mathbb{F}_{q}$ with ${\mathrm{Tr}}_{\mathbb{F}_{q}/\mathbb{F}_{p}}(a) = 1$, the number of solutions of the equation

in $\mathbb{F}_{q}^n$ is $q^{n-1}-v(b)q^{(n-2)/2}$.

Lemma 2.12. Let $q = 2^r$ and $h\in \mathbb{F}_{q^2}$. Let $g(Y)\in \mathbb{F}_{q^2}[y_{1}, y_2, \dots, y_n]$ be a polynomial of the form

where $b_j\in \mathbb{F}_{q^{2}}^{*},$ $2|n$, $n > 2$, $0\leq t\leq \frac{n}{2}-2$, ${\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(b_{j}) = 1,$ $j = 0, 1, \dots, t.$ Then

Proof. We provide two proofs here. The first proof is as follows. Let $q_{1} = q^2$. Then by Lemmas 2.7 and 2.10, the number of solutions of $g(Y) = h$ in $\mathbb{F}_{q^2}$ can be deduced as

By Lamma 2.7 and (2.2), we have

Next we give the second proof. Note that if $f$ and $g$ are equivalent, then for any $b\in \mathbb{F}_{q^2}$ the equation $f(x_1, \ldots, x_n) = b$ and $g(x_1, \ldots, x_n) = b$ have the same number of solutions in $\mathbb{F}_{q^2}.$ So we can get the number of solutions of $g(Y) = h$ for $h\in \mathbb{F}_{q^2}$ by means of a nonsingular linear substitution of indeterminates.

Let $k(X)\in \mathbb{F}_{q^2}[x_1, x_2, x_3, x_4]$ and $k(X) = x_1x_2+x_1^2+Ax_2^2+x_3x_4+x_3^2+Bx_4^2$, where ${\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(A) = {\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(B) = 1.$ We first show that $k(x)$ is equivalent to $x_1x_2 + x_3x_4$.

Let $x_3 = y_1+y_3$ and $x_i = y_i$ for $i\neq 3$, then $k(X)$ is equivalent to $y_1y_2+y_1y_4+y_3y_4+Ay_2^2+y_3^2+By_4^2$.

Let $y_2 = z_2+z_4$ and $y_i = z_i$ for $i\ne 2$, then $k(X)$ is equivalent to $z_1z_2+z_3z_4+Az_2^2+z_3^2+Az_4^2+Bz_4^2.$

Let $z_1 = \alpha_1+A\alpha_2$ and $z_i = \alpha_i$ for $i\ne 1$, then $k(X)$ is equivalent to $\alpha_1\alpha_2+\alpha_3^2+\alpha_3\alpha_4+(A+B)\alpha_4^2.$

Since ${\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(A+B) = 0$, we have $\alpha_3^2+\alpha_3\alpha_4+(A+B)\alpha_4^2$ is reducible by Lemma 2.9. Then $k(X)$ is equivalent to $x_1x_2+x_3x_4$. It follows that if $t$ is odd, then $g(Y)$ is equivalent to $x_{1}x_{2}+x_{3}x_{4}+\cdots+x_{n-1}x_{n}$, and if $t$ is even, then $g(Y)$ is equivalent to $x_{1}x_{2}+x_{3}x_{4}+\cdots+x_{n-1}x_{n}+x_{n-1}^{2}+ax_{n}^{2}$ with ${\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(a) = 1.$ By Lemma 2.11, we get the desired result. □

3.   Proof of Theorem 1.1

From (1.1), we know that the formula for the number of solutions of $f(X) = 0$ over $\mathbb{F}_{q^2}$ is

where $d_{i} = (m_{i}, q^{2}-1)$ and $\psi_i$ is a multiplicative character of $\mathbb{F}_{q^2}$ of order $d_i$. Since $m_{i}|q+1,$ we have $d_i = m_i.$ Let $H = \{(j_{1}, \dots, j_{s}) \mid 1 \leq j_{i} < m_{i},$ $1 \leq i \leq s\}$. It follows that $\psi_{1}^{j_{1}} \cdots \psi_{s}^{j_{s}}$ is nontrivial for any $\left(j_{1}, \dots, j_{s}\right) \in H$ as $(m_{i}, m_{j}) = 1$. By Lemma 2, we have $J_{q^2}^0\left(\psi _{1}^{j_1}, \dots, \psi _{s}^{j_s} \right) = 0$ and hence $N_{q^2}(f(X) = 0) = q^{2(s-1)}.$

Let $N_{q^2}(f(X) = c)$ denote the number of solutions of the equation $f(X) = c$ over $\mathbb{F}_{q^2}$ with $c\in\mathbb{F}_{q^2}^*$. Let $V = \{(j_1, \ldots, j_s)|0\leq j_i < m_i, \; 1\leq i\leq s\}$. Then

Since $\psi_i$ is a multiplicative character of $\mathbb{F}_{q^2}$ of order $m_i$, we have

By Lemma 2.3,

By Lemma 2.5,

Since $m_i|q+1$ and $2\nmid m_i$, by Lemma 2.6, we have

Then

It follows that

where

For a given $h\in \mathbb{F}_{q^2}$. We discuss the two cases according to whether $h$ is zero or not.

Case 1: $h = 0$. If $f(X) = 0$, then $g(Y) = 0$; if $f(X)\ne 0$, then $g(Y)\ne 0$. Then

By Lemma 2.12, (3.1) and (3.2), we have

Case 2: $h\in \mathbb{F}_{q^2}^{*}$. If $f(X) = h$, then $g(Y) = 0$; if $f(X) = 0$, then $g(Y) = h$; if $f(X)\notin \{0, h\}$, then $g(Y)\notin \{0, h\}$. So we have

By Lemma 2.12, (3.1) and (3.4),

It follows that

By (3.3) and (3.5), we get the desired result. The proof of Theorem 1.1 is complete.□

4.   Corollary and examples

There is a direct corollary of Theorem 1.1 and we omit its proof.

Corollary 4.1. Under the conditions of Theorem 1.1, if $a_1 = \cdots = a_s = h\in \mathbb{F}_{q^2}^*$, then we have

where

Finally, we give two examples to conclude the paper.

Example 4.2. Let $\mathbb{F}_{2^{10}} = \left < \alpha \right > = \mathbb{F}_2[x]/{(x^{10}+x^3+1)}$ where $\alpha$ is a root of $x^{10}+x^3+1.$ Suppose $l(X, Y) = \alpha^{33}x_{1}^{3}+x_{2}^{11}+y_{3}^{2}+\alpha^{10} y_{4}^{2}+y_{1}y_{2}+y_{3}y_{4}$. Clearly, ${\mathrm{Tr}}_{\mathbb{F}_{2^{10}}/\mathbb{F}_{2}}(\alpha^{10}) = 1$, $m_{1} = 3,$ $m_2 = 11$, $s = 2,$ $n = 4$, $t = 0,$ $a_{2} = 1$. By Theorem 1.1, we have

Example 4.3. Let $\mathbb{F}_{2^{12}} = \left < \beta \right > = \mathbb{F}_2[x]/{(x^{12}+x^6+x^4+x+1)}$ where $\beta$ is a root of $x^{12}+x^6+x^4+x+1.$ Suppose $l(X, Y) = x_{1}^{5}+x_{2}^{13}+y_{3}^{2}+\beta^{10} y_{4}^{2}+y_{1}y_{2}+y_{3}y_{4}$. Clearly, ${\mathrm{Tr}}_{\mathbb{F}_{2^{12}}/\mathbb{F}_{2}}(\beta^{10}) = 1$, $m_{1} = 5,$ $m_2 = 13$, $s = 2,$ $n = 4$, $t = 0,$ $a_{1} = a_{2} = 1$. By Corollary 1, we have

Acknowledgments

This work was jointly supported by the Natural Science Foundation of Fujian Province, China under Grant No. 2022J02046, Fujian Key Laboratory of Granular Computing and Applications (Minnan Normal University), Institute of Meteorological Big Data-Digital Fujian and Fujian Key Laboratory of Data Science and Statistics.

Conflict of interest

The authors declare there is no conflicts of interest.