Some specific classes of permutation polynomials over ${\textbf{F}}_{q^3}$

1.   Introduction

Let ${\bf F}_{q}$ be the finite field of characteristic $p$ with $q$ elements ($q = p^{n}, n\in {\textbf{N}}$), and ${\bf F}_{q}^{*}$ be the nonzero elements of ${\bf F}_{q}$. Let ${\bf F}_{q}[x]$ be the ring of polynomials over ${\bf F}_{q}$ in the indeterminate $x$. A polynomial $f(x)\in{\bf F}_{q}[x]$ is called a permutation polynomial if $f$ induces a bijection from ${\bf F}_{q}$ to itself. Recently, several classes of permutation polynomials were studied, which can be referred to ^{[5,8,11,12,20,21]}. More information about properties, constructions and applications of permutation polynomials may be found in the books ^[7,9]. We refer the readers to ^[3,14] for more details of the recent advances.

We found that the permutation polynomials of the form $(x^{q}+x+\delta)^s+x$ were studied wildly. During the research on Kloosterman sums, a class of permutation polynomials found in ^[4] motivated Yuan and Ding to study the permutation polynomials with the form $(x^{2^k}+x+\delta)^s+x$. Tu et al. ^[13] further proposed two classes of permutation polynomials having the form $(x^{2^m}+x+\delta)^s+x$ over ${\bf F}_{2^{2m}}$. Li et al. ^[6] presented a kind of permutation polynomials over ${\bf F}_{q^{2}}$ of the form

Yuan and Zheng ^[19] studied the permutation polynomials over ${\bf F}_{q^{2}}$ having the form

Recently, Zheng et al. ^[22] constructed large classes of permutation polynomials over ${\bf F}_{q^{2}}$, which have a more general form

where $a, b, c, u, v\in {\bf F}_{q^{2}}, r\in {\rm{Z}}^{+}, \phi(x)\in {\bf F}_{q^{2}}[x].$ We notice that these classes of permutation polynomials are in ${\bf F}_{q^{2}}$. Comparing to permutation polynomials over ${\bf F}_{q^{2}}$, there are few classes of permutation polynomials over ${\bf F}_{q^{3}}$ constructed recently. Ding et al. ^[2] presented six classes of permutation polynomials over ${\bf F}_{3^{3m}}$. Wang, Zhang, Bartoli and Wang ^[15] constructed several classes of permutation polynomials and complete permutation polynomials over ${\bf F}_{q^{3}}$. Motivated by recent constructions of permutation polynomials over ${\bf F}_{q^{2}}$ and ${\bf F}_{q^{3}}$, we can use AGW criterion and piecewise method to construct several classes of permutation polynomials over ${\bf F}_{q^{3}}$. In this paper, we will focus on constructing permutation polynomials over ${\bf F}_{q^{3}}$ having the similar forms to

where $L(x)$ is a linearized polynomial over ${\bf F}_{q}$.

The remainder of this paper is organized as follows. In Section 2, we introduce three lemmas which are used in the following sequels. Subsequently, we give several classes of permutation polynomials of the forms similar to $(x^{q^2}+x^q+x+\delta)^{\frac{q^{3}-1}{d}+1}+L(x)$ for $d = 2, 3, 4, 6$, in Section 3–6, respectively.

2.   Preliminary

In this section, we will present some results that will be used in the sequels.

The following lemma was developed by Akbary, Ghioca and Wang [1, Lemma 1.1], which was called AGW criterion in ^[10]. By using AGW criterion, several classes of permutation polynomials were constructed, which could be seen in ^{[16,17,18,19,22]}. The lemma is listed as follows:

Lemma 2.1. ^[1] Let $A, S$ and $\bar{S}$ be finite sets with $\#S = \#\bar{S}$, and let $f:A \rightarrow A, h :S \rightarrow \bar{S}, \varphi:A \rightarrow S$, and $\psi:A \rightarrow \bar{S}$ be maps such that $\psi\circ f = h \circ \varphi$, i.e., the following diagram is commutative:

If both $\psi$ and $\varphi$ are surjective, then the following statements are equivalent:

(i) $f$ is bijective (a permutation of $A$); and

(ii) $h$ is bijective from $S$ to $\bar{S}$ and $f$ is injective on $\varphi^{-1}(s)$ for each $s \in S$.

Lemma 2.2. Let $k, \delta$ be in ${\bf F}_{q}$. Then we have

Proof. Let $\varphi(x) = x^{q^2}+x^q+x+\delta$, then we find that $\varphi(x)^q = \varphi(x)$. Thus for every $\alpha \in {\bf F}_{q^{3}}$, we have $\varphi(\alpha)\in {\bf F}_{q}.$ Namely, $Im(\varphi(x))\subseteq {\bf F}_{q}.$ On the other hand, since $\#(Im\varphi(x))\geq q^3/deg(\varphi(x)) = q^3/q^2 = q$, it follows that $Im(\varphi(x)) = {\bf F}_{q}.$ Similarly, we can easily get that $Im(x^{q^2}+x^q+x+k\delta) = {\bf F}_{q}.$ Therefore, $\#(Im(x^{q^2}+x^q+x+\delta)) = \#(Im(x^{q^2}+x^q+x+k\delta)) = \# {\bf F}_{q}.$

The proof of Lemma 2.2 is completed.

We close this section with the following result that will be used frequently in the sequels.

Lemma 2.3. ^[7] Let $L(x) = \sum_{i = 0}^{m-1}a_{i}x^{q^i}$ be a linearized polynomial of ${\bf F}_{q^{m}}$ for $m\in Z^{+}$. Then $L(x)$ is a permutation polynomial of ${\bf F}_{q^{m}}$ if and only if

3.   Permutation polynomials of the form $a(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{2}+1}-bx^{q^2}-cx^q-dx$

In this section, we suppose that $p\ne 3$ is an odd prime, $\epsilon$ is a primitive element of ${\bf F}_{q^{3}}$. Define $D_0 = < \epsilon^2 >$, which is the multiplicative group generated by $\epsilon^2$, and $D_1 = \epsilon D_0$. Then we get that ${\bf F}_{q^{3}} = \{0\}\cup D_0 \cup D_1.$ Furthermore, if $x\in D_i$, we have

where $i = 0, 1.$

Since $p$ is an odd prime, then $q\equiv 1\pmod 2$. By Lemma 2.2, we know that

In this case, $Im(x^{q^2}+x^q+x+\delta) = Im(x^{q^2}+x^q+x+k\delta) = {\bf F}_{q}$ is the disjoint union of $0$ and $\frac{q-1}{2}$ elements from $D_0$ and $\frac{q-1}{2}$ elements from $D_1$.

By using Lemma 2.3, we give the first main result of this paper.

Theorem 3.1. Let $p\neq3$ be an odd prime and $a, b, c, d, \delta \in {\bf F}_{q}$. If $b, c$ and $d$ satisfy

then

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $\big(3a-(b+c+d), -(3a+b+c+d)\big)$ belongs to one of $\{ D_0\times D_0, D_1\times D_1 \}$.

Proof. Let $f(x) = a(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{2}+1}-bx^{q^2}-cx^q-dx$, $\varphi(x) = x^{q^2}+x^q+x+\delta$, $\psi(x) = x^{q^2}+x^q+x-(b+c+d)\delta$ and $h(x) = x(3ax^{\frac{q^3-1}{2}}-(b+c+d))$. One has

By Lemma 2.3, it follows from

that $bx^{q^2}+cx^q+dx$ is a permutation polynomial of ${\bf F}_{q^{3}}$. Clearly, $as^{\frac{q^3-1}{2}+1}-(bx^{q^2}+cx^q+dx)$ for every $s\in {\bf F}_{q^{3}}$ is also a permutation polynomial of ${\bf F}_{q^{3}}$. For $x\in\varphi^{-1}(s)$, it implies that $f(x) = as^{\frac{q^3-1}{2}+1}-(bx^{q^2}+cx^q+dx)$. Thus for every $s\in Im(\varphi)$, $f(x)$ is injective on $\varphi^{-1}(s)$. By Lemma 2.1, we know that $f(x)$ is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $h(x)$ is a bijection from $Im(\varphi)$ to $Im(\psi)$.

It is easy to check that

Then we can infer that $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$ if and only if $(3a-(b+c+d), -(3a+b+c+d))$ belongs to one of $\{ D_0\times D_0, D_1\times D_1 \}$. Thus

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $\big(3a-(b+c+d), -(3a+b+c+d)\big)$ belongs to one of $\{ D_0\times D_0, D_1\times D_1 \}$.

The proof of Theorem 3.1 is completed.

Corollaries 3.2 and 3.3 are two explicit examples of Theorem 3.1, we omit their proofs, just list them in the following.

Corollary 3.2. Let $p\neq3$ be an odd prime, $\delta \in {\bf F}_{q}$ and let $a, b\in {\bf F}_{q}$ satisfy $3a-b\neq 0$ and $3a+b\neq 0$. Then we get that

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $\big(3a-b, -(3a+b)\big)$ belongs to one of $\{ D_0\times D_0, D_1\times D_1 \}$.

Corollary 3.3. Let $p\ne3, 5$ be an odd prime, $\delta \in {\bf F}_{q}$ and $-5\in D_0$. Then we get that

is a permutation polynomial over ${\bf F}_{q^{3}}$.

Example 3.4. For $q = 19$ and $\delta\in {\bf F}_{19}$, it is easy to check that $(2, -1)\in D_1\times D_1$ by Magma, thus we can imply from Corollary 3.2 that $(x^{q^2}+x^q+x+\delta)^{2287}-x$ is a permutation polynomial over ${\bf F}_{19^{3}}$.

4.   Permutation polynomials of the form $a(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{3}+1}+a(x^{q^2}+x^q+x+\delta)^{\frac{2(q^3-1)}{3}+1}+bx^{q^2}+cx^q+dx$

In this section, we suppose $p\ne 3$ be an odd prime and $q^3\equiv 1\pmod 3$. Let $\epsilon$ be a primitive element of ${\bf F}_{q^{3}}$, we define $D_0 = < \epsilon^3 >$ and $D_i = \epsilon^i D_0$ for $i = 1, 2.$ Then ${\bf F}_{q^{3}} = \{0\}\cup D_0\cup D_1 \cup D_2$. Let $\omega = \epsilon^{\frac{q^3-1}{3}}$, one has $\omega^2+\omega+1 = 0.$

It follows from $q^3\equiv 1\pmod 3$ that $q\equiv 1\pmod3$ and $q^2+q+1\equiv 0 \pmod 3$, so ${\bf F}_{q}\subseteq \{0\}\cup D_0$.

We can get the following result, which is the second main result in this paper.

Theorem 4.1. Let $p\neq 3$ be an odd prime, $q\equiv 1 \pmod 3$ and $a, b, c, d, \delta\in {\bf F}_{q}$. If $6a+b+c+d\ne 0$, $-3a+b+c+d\ne 0$ and

then we have

is a permutation polynomial over ${\bf F}_{q^{3}}$.

Proof. Define

In the following, it is easy to check that

Since

it follows from Lemma 2.3 that $bx^{q^2}+cx^q+dx$ is a permutation polynomial of ${\bf F}_{q^{3}}$. Thus $f(x)$ is injective on $\varphi^{-1}(s)$. By Lemma 2.1, $f(x)$ is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$.

By Lemma 2.2, we get that $Im(\varphi) = Im(\psi) = {\bf F}_{q}$. Hence we need to prove that $h(x)$ is a bijection on ${\bf F}_{q}$. Since $x^{\frac{q^3-1}{3}}+x^{\frac{2(q^3-1)}{3}} = 2$ for $x\in D_0$ and $x^{\frac{q^3-1}{3}}+x^{\frac{2(q^3-1)}{3}} = -1$ for $x\in D_1\cup D_2$, we get that

Since $q\equiv 1 \pmod 3$, then ${\bf F}_{q}\subseteq D_0\cup \{0\}.$ It follows that $6a+b+c+d$ and $-3a+b+c+d$ are in $D_0$. Thus $h(x)$ is a bijection on ${\bf F}_{q}$.

Therefore, we can conclude that $a(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{3}+1}+a(x^{q^2}+x^q+x+\delta)^{\frac{2(q^3-1)}{3}+1}+bx^{q^2}+cx^q+dx$ is a permutation polynomial over ${\bf F}_{q^{3}}$. This completes the proof of Theorem 4.1.

By Theorem 4.1, we can easily give the following results.

Corollary 4.2. Let $p\neq 3$ be an odd prime, $q\equiv 1 \pmod 3$ and $\delta\in {\bf F}_{q}$. Then we get that

is a permutation polynomial over ${\bf F}_{q^{3}}$.

Corollary 4.3. Let $p\neq 3$ be an odd prime, $q\equiv1\pmod 3$ and $\delta\in {\bf F}_{q}$. Then we get that

is a permutation polynomial over ${\bf F}_{q^{3}}$.

5.   Permutation polynomials of the form $a(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{4}+1}+a(x^{q^2}+x^q+x+\delta)^{\frac{3(q^3-1)}{4}+1}+bx^{q^2}+cx^q+dx$

In this section, we suppose that $p\ne 3$ is an odd prime and $q^3\equiv 1\pmod 4$, $\epsilon$ is a primitive element of ${\bf F}_{q^{3}}$. Define $D_0 = < \epsilon^4 >$, which is the multiplicative group generated by $\epsilon^4$, and $D_i = \epsilon^i D_0$ for $i = 1, 2, 3$. Then we get that ${\bf F}_{q^{3}} = \{0\}\cup D_0 \cup D_1 \cup D_2 \cup D_3.$ Note that $x^{\frac{q^3-1}{4}} = \epsilon^{i\frac{q^3-1}{4}},$ for $x\in D_i$, where $i = 1, 2, 3.$ Let $\omega = \epsilon^{\frac{q^3-1}{4}}$, then $\omega^2+1 = 0.$

Since $q^3\equiv 1\pmod4$, we get that $q\equiv 1\pmod 4$, then $q^2+q+1\equiv 3\pmod 4$. By Lemma 2.2, we know that

It follows that ${\bf F}_{q}$ is a disjoint union of $0$ and $\frac{q-1}{4}$ elements from $D_0$ and $\frac{q-1}{4}$ elements from $D_1$ and $\frac{q-1}{4}$ elements from $D_2$ and $\frac{q-1}{4}$ elements from $D_3$.

We present the third main result of this paper in the following.

Theorem 5.1. Let $p\neq 3$ be an odd prime, $q\equiv 1\pmod 4$ and $\delta\in {\bf F}_{q}$. Let $a\in {\bf F}_{q}^{*}, b, c, d\in {\bf F}_{q}$ satisfy $6a+b+c+d\ne 0, b+c+d\ne0$, $-6a+b+c+d\ne 0$ and

Then

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(6a+b+c+d, b+c+d, -6a+b+c+d)$ belongs to one of $\{D_0\times D_0 \times D_0, D_0\times D_2 \times D_0, D_1\times D_1 \times D_1, D_1\times D_3 \times$ $D_1, D_2\times D_2 \times D_2, D_2\times D_0 \times D_2, D_3\times D_3 \times D_3, D_3\times D_1 \times D_3\}.$

Proof. Let

It is easy to check that

Since

it follows from Lemma 2.3 that $bx^{q^2}+cx^q+dx$ is a permutation polynomial over ${\bf F}_{q^{3}}$. Therefore, we can easily get that for every $s\in Im(\varphi)$, $f(x)$ is injective on $\varphi^{-1}(s)$. By Lemma 2.1, $f(x)$ is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$.

Note that

Then $h(x)$ can be rewritten as

In the following, we focus on proving that $h(x)$ is a bijection from $Im(\varphi)$ to $Im(\psi)$ if and only if $(6a+b+c+d, b+c+d, -6a+b+c+d)$ belongs to one of $\{D_0\times D_0 \times D_0, D_0\times D_2 \times D_0, D_1\times D_1 \times D_1, D_1\times D_3 \times D_1, D_2\times D_2$ $\times D_2, D_2\times D_0 \times D_2, D_3\times D_3 \times D_3, D_3\times D_1 \times D_3\}.$

Now we first prove the sufficiency part. Assume that $h(x)$ is a bijection. We consider the following cases:

(1) $6a+b+c+d\in D_0$. Clearly, we have $(6a+b+c+d)x\in D_0$ for $x\in D_0$, then it implies that $(b+c+d)x\not\in D_0$ for $x\in D_1\cup D_3$. It tells us that $b+c+d\not\in D_1\cup D_3$. If $b+c+d\in D_0$, then $(b+c+d)x\in D_1$ and $(b+c+d)x\in D_3$ for $x\in D_1$ and $x\in D_3$, respectively. Since $h(x)$ is a bijection, we can get that $(-6a+b+c+d)x\in D_2$ for $x\in D_2$, thus $-6a+b+c+d\in D_0$. On the other hand, if $b+c+d\in D_2$, then $(b+c+d)x\in D_3$ for $x\in D_1$ and $(b+c+d)x\in D_1$ for $x\in D_3$. It follows from $h(x)$ being a bijection that $(-6a+b+c+d)x\in D_2$ for $x\in D_2$, one has $-6a+b+c+d\in D_0$.

(2) $6a+b+c+d\in D_1$. In this case, we can imply that $b+c+d\not\in D_0\cup D_2$. If $b+c+d\in D_1$, by $h(x)$ being a bijection, we can get that $-6a+b+c+d\in D_1$. Similarly, if $b+c+d\in D_3$, then it follows from $h(x)$ being a bijection that $-6a+b+c+d\in D_1$.

(3) $6a+b+c+d\in D_2$. In this case, we first get that $b+c+d\not\in D_1\cup D_3$. If $b+c+d\in D_2$, since $h(x)$ is a bijection, we can get that $-6a+b+c+d\in D_2$. If $b+c+d\in D_0$, it follows that $-6a+b+c+d\in D_2$.

(4) $6a+b+c+d\in D_3$. In this case, it is easy to imply that $b+c+d\not\in D_0\cup D_2$. If $b+c+d\in D_1$, then we can get that $-6a+b+c+d\in D_3$. If $b+c+d\in D_3$, it follows from $h(x)$ being a bijection that $-6a+b+c+d\in D_3$.

Conversely, we can easily check the necessity part is true.

Thus we can conclude that $a(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{4}+1}+a(x^{q^2}+x^q+x+\delta)^{3\frac{q^3-1}{4}+1}+bx^{q^2}+cx^q+dx$ is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(6a+b+c+d, b+c+d, -6a+b+c+d)$ belongs to one of $\{D_0\times D_0 \times D_0, D_0\times D_2 \times D_0, D_1\times D_1 \times D_1, D_1\times D_3 \times D_1, D_2\times D_2$ $\times D_2, D_2\times D_0 \times D_2, D_3\times D_3 \times D_3, D_3\times D_1 \times D_3\}.$

By Theorem 5.1, we can get the following results.

Corollary 5.2. Let $p\neq 3$ be an odd prime, $q\equiv 1\pmod 4$ and $\delta\in {\bf F}_{q}$. Then

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(-1, 5, 7)$ belongs to one of $\{D_0\times D_0 \times D_0, D_0\times D_2 \times D_2, D_2\times D_0 \times D_2, D_2\times D_2 \times D_0\}.$

Corollary 5.3. Let $p\neq 3$ be an odd prime, $q\equiv 1\pmod 4$ and $\delta\in {\bf F}_{q}$. Then

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(-1, 2)$ belongs to one of $\{ D_0\times D_0, D_0\times D_2, D_2\times D_3\}.$

Example 5.4. For $q = 29$ and $\delta\in {\bf F}_{q}$, we can check that $(-1, 5, 7)\in D_2\times D_2 \times D_0$ by Magma, it follows from Corollary 5.2 that $(x^{q^2}+x^q+x+\delta)^{6098}+(x^{q^2}+x^q+x+\delta)^{18292}-x$ is a permutation polynomial over ${\bf F}_{29^{3}}$.

In what follows, we will give the fourth main result in this paper.

Theorem 5.5. Let $p\neq 3, 5$ be an odd prime, $q\equiv 1\pmod 4$ and $\delta\in {\bf F}_{q}$. Then

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(2, -4, 3\omega-1, -3\omega-1)$ belongs to one of $\{ D_0\times D_0\times D_0\times D_0, D_0\times D_0\times D_2\times D_2, D_1$ $\times D_2\times D_1\times D_0, D_1\times D_2\times D_2\times D_3, D_1\times D_0\times D_3\times D_0, D_1\times D_0\times D_2\times D_1, D_2$ $\times D_2\times D_0\times D_0, D_2\times D_2\times D_2\times D_2, D_3\times D_2\times D_0\times D_3, D_3\times D_2\times D_2\times D_2, D_3\times D_0\times D_3\times D_2, D_3\times D_0\times D_0\times D_1\}.$

Proof. Let $(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{4}+1}-x$, $\varphi(x) = x^{q^2}+x^q+x+\delta$, $\psi(x) = x^{q^2}+x^q+x-\delta$ and $h(x) = x(3x^{\frac{q^3-1}{4}}-1)$. Then we can easily check that

Namely, the diagram is commutative. Furthermore, it is easy to get that for every $s\in Im(\varphi)$, $f(x)$ is injective on $\varphi^{-1}(s)$. Thus it follows from Lemma 2.1 that $f(x)$ is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$.

By Lemma 2.2, we know that $Im(\varphi) = Im(\psi) = {\bf F}_{q}$. For $x\in {\bf F}_{q}$, we can rewrite $h(x)$ as:

Since $p\neq 3, 5$ is an odd prime, we get that $(3\omega-1)(-3\omega-1)\ne 0$. We claim that $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$ if and only if $(2, -4, 3\omega-1, -3\omega-1)$ belongs to one of $\{ D_0\times D_0\times D_0\times D_0, D_0\times D_0\times D_2\times D_2, D_1$ $\times D_2\times D_1\times D_0, D_1\times D_2\times D_2\times D_3, D_1\times D_0\times D_3\times D_0,$ $D_1\times D_0\times D_2\times D_1, D_2\times D_2\times D_0\times D_0, D_2\times D_2$ $\times D_2\times D_2, D_3\times D_2\times D_0\times D_3, D_3\times D_2\times D_2\times D_2, D_3\times D_0\times D_3\times D_2, D_3\times D_0\times D_0\times D_1\}.$

First we assume $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$. Since $q\equiv 1\pmod 4$, it tells us that $-1\in D_0$ or $-1\in D_2$. We consider the following cases:

(1) $2\in D_0$. Then we get that $-4\in D_0$ or $-4\in D_2$. For $x\in D_0$, we have $2x\in D_0$, it follows from $h(x)$ being a bijection that $-4x\not\in D_0$ for $x\in D_2$, hence $-4\in D_0$ and $-4x\in D_2$ for $x\in D_2$. Furthermore, since $2x\in D_0$ for $x\in D_0$ and $-4x\in D_2$ for $x\in D_2$, then $(3\omega-1)x\not \in D_0\cup D_2$ for $x\in D_1$, thus $3\omega-1\in D_0$ or $3\omega-1\in D_2$. If $3\omega-1\in D_0$, by $h(x)$ being a bijection, one has $-3\omega-1\in D_0$. If $3\omega-1\in D_2$, then it implies that $-3\omega-1\in D_2$.

(2) $2\in D_1$. Since $-1\in D_0$ or $-1\in D_2$, thus $-4\in D_2$ or $-4\in D_0$. For the first case, if $-4\in D_2$, then $2x\in D_1$ for $x\in D_0$ and $-4x\in D_0$ for $x\in D_2$. Since $h(x)$ is a bijection, we know that $3\omega-1\in D_1$ or $3\omega-1\in D_2$. If $3\omega-1\in D_1$, it follows that $-3\omega-1\in D_0$. If $3\omega-1\in D_2$, one has that $-3\omega-1\in D_3$. For the second case, if $-4\in D_0$, then $2x\in D_1$ for $x\in D_0$ and $-4x\in D_2$ for $x\in D_2$. Since $h(x)$ is a bijection, we know that $3\omega-1\in D_3$ or $3\omega-1\in D_2$. If $3\omega-1\in D_3$, it follows that $-3\omega-1\in D_0$. If $3\omega-1\in D_2$, we know that $-3\omega-1\in D_1$.

(3) $2\in D_2$. Since $2\in D_2$, then we get that $-4\in D_0$ or $-4\in D_2$. For $x\in D_0$, we have $2x\in D_2$, it follows from $h(x)$ being a bijection that $-4x\not\in D_2$ for $x\in D_2$, hence $-4\in D_2$ and $-4x\in D_0$ for $x\in D_2$. Furthermore, since $2x\in D_2$ for $x\in D_0$ and $-4x\in D_0$ for $x\in D_2$, then $(3\omega-1)x\not \in D_0\cup D_2$ for $x\in D_1$, thus $3\omega-1\in D_0$ or $3\omega-1\in D_2$. If $3\omega-1\in D_0$, it implies that $-3\omega-1\in D_0$. If $3\omega-1\in D_2$, we have $-3\omega-1\in D_2$.

(4) $2\in D_3$. Since $-1\in D_0$ or $-1\in D_2$, thus $-4\in D_2$ or $-4\in D_0$. For the first case, if $-4\in D_2$, then $2x\in D_3$ for $x\in D_0$ and $-4x\in D_0$ for $x\in D_2$. Since $h(x)$ is a bijection, we know that $3\omega-1\in D_0$ or $3\omega-1\in D_1$. If $3\omega-1\in D_0$, it follows that $-3\omega-1\in D_3$. If $3\omega-1\in D_1$, we get that $-3\omega-1\in D_2$. For the second case, if $-4\in D_0$, then $2x\in D_3$ for $x\in D_0$ and $-4x\in D_2$ for $x\in D_2$. Since $h(x)$ is a bijection, one has $3\omega-1\in D_3$ or $3\omega-1\in D_0$. If $3\omega-1\in D_3$, it follows that $-3\omega-1\in D_2$. If $3\omega-1\in D_0$, it implies that $-3\omega-1\in D_1$.

The sufficiency part is proved.

For the necessity part, if $(2, -4, 3\omega-1, -3\omega-1)$ belongs to one of $\{ D_0\times D_0\times D_0\times D_0, D_0\times D_0\times D_2\times D_2, D_1\times D_2\times D_1\times D_0, D_1\times D_2\times D_2$ $\times D_3, D_1\times D_0\times D_3\times D_0, D_1\times D_0\times D_2\times D_1, D_2\times D_2\times D_0\times D_0, D_2\times D_2\times D_2$ $\times D_2, D_3\times D_2\times D_0\times D_3, D_3\times D_2\times D_2\times D_2, D_3\times D_0\times D_3\times D_2, D_3\times D_0\times D_0\times D_1\},$ we easily check that $h(x)$ is a bijection from $Im(\varphi)$ to $Im(\psi)$. Thus the claim is true.

Therefore, we get that

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(2, -4, 3\omega-1, -3\omega-1)$ belongs to one of $\{ D_0\times D_0\times D_0\times D_0, D_0\times D_0\times D_2\times D_2, D_1\times D_2$ $\times D_1\times D_0, D_1\times D_2\times D_2\times D_3, D_1\times D_0\times D_3\times D_0,$ $D_1\times D_0\times D_2\times D_1, D_2\times D_2\times D_0\times D_0, D_2\times D_2\times D_2$ $\times D_2, D_3\times D_2\times D_0\times D_3, D_3\times D_2\times D_2\times D_2, D_3\times D_0\times D_3\times D_2, D_3\times D_0\times D_0\times D_1\}.$

The proof of Theorem 5.5 is completed.

Similarly, we have

Theorem 5.6. Let $p\neq 3, 5$ be an odd prime, $q\equiv 1\pmod 4$ and $\delta\in {\bf F}_{q}$. Then

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(2, -4, -3\omega-1, 3\omega-1)$ belongs to one of $\{ D_0\times D_0\times D_0\times D_0, D_0\times D_0\times D_2\times D_2, D_1\times D_2$ $\times D_1\times D_0, D_1\times D_2\times D_2\times D_3, D_1\times D_0\times D_3\times D_0, D_1\times D_0\times D_2\times D_1, D_2\times D_2$ $\times D_0\times D_0, D_2\times D_2\times D_2\times D_2, D_3\times D_2\times D_0\times D_3, D_3$ $\times D_2\times D_2\times D_2, D_3\times D_0\times D_3\times D_2, D_3\times D_0\times D_0\times D_1\}.$

6.   Permutation polynomials of the form $a(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{6}+1}+bx$

In this section, let $p\ne 3$ be an odd prime and $q^3\equiv 1\pmod 6$. We assume $\epsilon$ is a primitive element of ${\bf F}_{q^{3}}$ and define $D_0 = < \epsilon^6 >$, which is the multiplicative group generated by $\epsilon^6$, and $D_i = \epsilon^i D_0$ for $i = 1, 2, 3, 4, 5$. Then we get that ${\bf F}_{q^{3}} = \{0\}\cup D_0 \cup D_1 \cup D_2 \cup D_3\cup D_4\cup D_5.$ Furthermore, if $x\in D_i$, we notice that $x^{\frac{q^3-1}{6}} = \epsilon^{i\frac{q^3-1}{6}}.$ For simplicity, we define $\omega = \epsilon^{\frac{q^3-1}{6}}$, which satisfies $\omega^2-\omega+1 = 0.$

Since $q^3\equiv 1\pmod6$, we know that $q\equiv 1\pmod 6$, then $q^2+q+1\equiv 3\pmod 6$ and $q\equiv 1\pmod 2$. By Lemma 2.2, we know that

and ${\bf F}_{q}$ is a disjoint union of $0$ and $\frac{q-1}{2}$ elements from $D_0$ and $\frac{q-1}{2}$ elements from $D_3$.

In what follows, we can get the permutation polynomials of the form $a(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{6}+1}+bx$, which is the fifth main result in this paper.

Theorem 6.1. Let $p\neq 3$ be an odd prime, $q\equiv 1\pmod 6$ and $\delta\in {\bf F}_{q}$, and let $a, b\in {\bf F}_{q}$ such that $3a+b\ne0$ and $-3a+b\ne0$. Then

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(3a+b, -3a+b)$ belongs to one of $\{D_0\times D_0, D_3\times D_3\}.$

Proof. Let $f(x) = a(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{6}+1}+bx$, $\varphi(x) = x^{q^2}+x^q+x+\delta$, $\psi(x) = x^{q^2}+x^q+x+b\delta$ and $h(x) = x(3ax^{\frac{q^3-1}{6}}+b)$. It is easy to check that

Furthermore, we can easily check that for every $s\in Im(\varphi)$, $f(x)$ is injective on $\varphi^{-1}(s)$. Then it follows from Lemma 2.1 that $f(x)$ is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$.

Since $Im(\varphi) = Im(\psi) = {\bf F}_{q}$ is a disjoint union of $0$ and $\frac{q-1}{2}$ elements from $D_0$ and $\frac{q-1}{2}$ elements from $D_3$, we get that

Next we prove that $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$ if and only if $(3a+b, -3a+b)$ belongs to one of $\{D_0\times D_0, D_3\times D_3\}$.

Firstly, we give the proof of the sufficiency part. Suppose $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$. If $3a+b\in D_0$, then we have $(3a+b)D_0 = D_0$. Hence $(-3a+b)D_3 = D_3$, it tells us that $-3a+b\in D_0$. On the other hand, if $3a+b\in D_3$, then $(3a+b)D_0 = D_3$ and $(-3a+b)D_3 = D_0$, thus it follows that $-3a+b\in D_3$.

Now we prove the necessity part. If $(3a+b, -3a+b)\in D_0\times D_0$ or $D_3\times D_3$, it is easy to check that $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$. Therefore, $h(x)$ is bijective from $Im(\varphi)$ to $Im(\psi)$ if and only if $(3a+b, -3a+b)$ belongs to one of $\{D_0\times D_0, D_3\times D_3\}$.

We can conclude that

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(3a+b, -3a+b)$ belongs to one of $\{D_0\times D_0, D_3\times D_3\}.$

The proof of Theorem 6.1 is completed.

Corollary 6.2. Let $p\neq 3$ be an odd prime, $q\equiv 1\pmod 6$ and $\delta\in {\bf F}_{q}$. Then

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(2, -4)$ belongs to one of $\{D_0\times D_0, D_3\times D_3\}.$

Proof. Taking $a = 1, b = -1$ in Theorem 6.1, it follows from Theorem 6.1 that $(x^{q^2}+x^q+x+\delta)^{\frac{q^3-1}{6}+1}-x$ is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(2, -4)$ belongs to one of $\{D_0\times D_0, D_3\times D_3\}.$

Example 6.3. For $q = 19$ and $\delta\in {\bf F}_{19}$, we can find that $(2, -4)\in D_3\times D_3$ by Magma, by Corollary 6.2, it follows that $(x^{q^2}+x^q+x+\delta)^{1144}-x$ is a permutation polynomial over ${\bf F}_{19^{3}}$.

Similarly, we have

Theorem 6.4. Let $p\neq 3$ be an odd prime, $q\equiv 1\pmod 6$ and $\delta\in {\bf F}_{q}$, and let $a, b\in {\bf F}_{q}$ such that $3a+b\ne0$ and $-3a+b\ne0$. Then

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(3a+b, -3a+b)$ belongs to one of $\{D_0\times D_0, D_3\times D_3\}.$

Proof. The proof of Theorem 6.4 is similar to that of Theorem 6.1, here we omit it.

Furthermore, we get a permutation polynomial having a more general form.

Theorem 6.5. Let $p\neq 3$ be an odd prime, $q\equiv 1\pmod 6$ and $\delta\in {\bf F}_{q}$, and let $a, b, c\in {\bf F}_{q}$ such that $3a+3b+c\ne0$ and $-3a-3b+c\ne0$. Then

is a permutation polynomial over ${\bf F}_{q^{3}}$ if and only if $(3a+3b+c, -3a-3b+c)$ belongs to one of $\{D_0\times D_0, D_3\times D_3\}.$

Proof. The proof of Theorem 6.5 is similar to that of Theorem 6.1, here we only give that $h(x) = x(3ax^{\frac{q^3-1}{6}}+3bx^{5\frac{q^3-1}{6}}+c)$, and $h(x)$ can be rewritten as

We omit the other details of the proof.

7.   Conclusions

In this paper, motivated by some constructions of permutation polynomials over ${\bf F}_{q^2}$, we used AGW criterion and piecewise method to construct several classes of permutation polynomials over ${\bf F}_{q^3}$ of the forms $(x^{q^2}+x^q+x+\delta)\phi((x^{q^2}+x^q+x+\delta)^{\frac{q^{3}-1}{d}})+L(x)$, for $d = 2, 3, 4, 6,$ where $L(x)$ is a linearized polynomial over ${\bf F}_{q}$, which enrich the permutation polynomials over ${\bf F}_{q^3}$.

Acknowledgments

The Authors express their gratitude to the anonymous reviewers for carefully examining this paper and providing a number of valuable comments and suggestions. This research was supported by the National Science Foundation of China (No. 11926344) and by Science and Technology Research Program of Chongqing Municipal Education Commission (No. KJQN201900506).

Conflict of interest

We declare that we have no conflict of interest.