A Survey of Permutation Polynomials and Their Compositional Inverses over Finite Fields

1. Introduction

Let ${\mathbb{F}}_q$ be the finite field with q elements and ${\mathbb{F}}_q^{*}$ denote the multiplicative group with the nonzero element in ${\mathbb{F}}_q$, where q is a prime power. Let ${\mathbb{F}}_q\left[x\right]$ be the ring of polynomials in a single indeterminate x over ${\mathbb{F}}_q$. A polynomial $f\in {\mathbb{F}}_q\left[x\right]$ is called a permutation polynomial (PP) of ${\mathbb{F}}_q$ if its associated polynomial mapping $f:c\mapsto f\left(c\right)$ from ${\mathbb{F}}_q$ to itself is bijective. The unique polynomial denoted by $f^{-1}\left(x\right)$ over ${\mathbb{F}}_q$ such that $f\left(f^{-1}\left(x\right)\right)\equiv f^{-1}\left(f\left(x\right)\right)\equiv x(mod{x}^q-x)$ is called the compositional inverse of $f\left(x\right).$ Furthermore, $f\left(x\right)$ is called an involution when $f^{-1}\left(x\right)=f\left(x\right).$

The study of permutation polynomials and their compositional inverses over finite fields in terms of their coefficients is a classical and difficult subject which attracts people’s interest partially due to their wide applications in coding theory [1,2,3], cryptography [4,5], combinatorial design theory [6], and other areas of mathematics and engineering [7,8].

In general, determining whether a polynomial over a finite field is a permutation polynomial is challenging. Moreover, computing the coefficients of the compositional inverse of a permutation polynomial is even more difficult, except for several well-studied classes such as monomials, linearized polynomials, and Dickson polynomials, which prossess a well-defined structure. Q. Wang [9] has summarized the current methods used in the study of the compositional inverses of permutation polynomials, including the experimental method, the power sum method, the matrix method, the group algebra method, the piecewise method, the decomposition method, the commutative diagram method, and the local method. On the basis of the existing classification system, this paper further supplements and refines the theoretical system of permutation polynomials via targeted in-depth discussions. Specifically, this survey consists of two core parts. It first conducts an in-depth exploration of the AGW criterion and local criterion, and systematically sorts out the algebraic structure-based theoretical methods for permutation polynomial construction. Furthermore, different from the macroscopic classification in previous studies, this paper focuses on the local method, rederiving and reinterpretating all existing formulas for compositional inverses of permutation polynomials. This work standardizes the relevant theoretical derivation logic and complements the existing research system, offering a solid and systematic theoretical basis for subsequent related studies and practical applications.

2. Methods on Permutation Polynomials over Finite Fields

2.1. AGW Criterion, Local Criterion and Local Method

Before discussing the compositional inverse of permutation polynomials, this section first introduces several fundamental theorems for constructing permutation polynomials, laying the groundwork for subsequent analysis.

2.1.1. The AGW Criterion: Theory and Further Understanding

The Akbary–Ghioca–Wang (AGW) criterion [10] [Lemma 1.1] is an important method for constructing PPs. By providing necessary and sufficient conditions for a polynomial to be a permutation via a commutative diagram, this criterion not only unifies many classical constructions of permutation polynomials, but also yields a large number of new ones, thereby significantly advancing the study of permutation polynomials.

Theorem 1.

[10] [AGW criterion] Let $A,S$ and $\overline{S}$ be finite sets with $♯S=♯\overline{S}$, and let $f\left(x\right):A\to A$, $h\left(x\right):S\to \overline{S}$, $\lambda \left(x\right):A\to S,$ and $\overline{\lambda }\left(x\right):A\to \overline{S}$ be maps such that $\overline{\lambda }\left(x\right)\circ f\left(x\right)=h\left(x\right)\circ \lambda \left(x\right).$ If both $\lambda \left(x\right)$ and $\overline{\lambda }\left(x\right)$ are surjective, then the following statements are equivalent:

(i) $f\left(x\right)$ is bijective (a permutation of A); and

(ii) $h\left(x\right)$ is bijective from S to $\overline{S}$ and $f\left(x\right)$ is injective on ${\lambda }^{-1}\left(s\right)$ for each $s\in S.$

Figure 1. The AGW criterion

Figure 1. The AGW criterion

Building on the AGW criterion recalled above, we next present several related characterizations established in [11].

Theorem 2.

[11] [Theorem 2.3] Let A and S be finite sets, $f:A\to A$ a map, and $\phi :A\to S$ be a surjective map. Then f is a bijection if and only if the following conditions hold:

(i) f is injective on each ${\phi }^{-1}\left(s\right)$ for all $s\in S$;

(ii) for each bijection $h:S\to S$, there exists a uniquely determined surjective map $\psi :A\to S$ such that the following diagram commutes:

As a straightforward generalization of Theorem 2, Yuan [11] obtained the following result:

Theorem 3.

[11] [Theorem 2.4] Let A and S be finite sets and let $f\left(x\right):A\to A$ be a map and $\phi \left(x\right):A\to S$ be a surjective map. Then $f\left(x\right)$ is a bijection if and only if the following conditions hold:

(i) $f\left(x\right)$ is injective on each ${\phi }^{-1}\left(s\right)$ for all $s\in S$;

(ii) for each bijection $h\left(x\right):S\to \overline{S}$, where $\overline{S}$ is a set with $♯S=♯\overline{S}$, there exists an uniquely determined surjective map $\psi \left(x\right):A\to \overline{S}$ such that the following diagram commutes

Theorem 3 shows that every permutation polynomial can be characterized by the AGW criterion.

Further equivalent characterizations for bijective self-maps on finite sets are also provided in the same reference, as listed below.

Theorem 4.

[11] [Lemma 2.5] Let $A,S$ be finite sets, $f\left(x\right):A\to A$ a map and $\lambda \left(x\right):A\to S$ a surjective map. Then f is a bijection if and only if

(i) $f\left(x\right)$ is injective on each ${\lambda }^{-1}\left(s\right)$ for all $s\in S$.

(ii) If $\lambda \left(a\right)\ne \lambda \left(b\right)$, then $f\left(a\right)\ne f\left(b\right)$.

Moreover (ii) is equivalent to $f\left({\lambda }^{-1}\left(s_1\right)\right)\cap f\left({\lambda }^{-1}\left(s_2\right)\right)=\varnothing$ for any distinct $s_1,s_2\in S$.

Theorem 5.

[11] [Lemma 2.5] Let $A,S,$ and $\overline{S}$ be finite sets with $♯S=♯\overline{S}$, $f\left(x\right):A\to A$ a map and $\lambda \left(x\right):A\to S$ a surjective map. Then $f\left(x\right):A\to A$ is a bijection if and only if the following two conditions hold:

(i) $f\left(x\right)$ is injective on each ${\lambda }^{-1}\left(s\right)$ for all $s\in S$.

(ii) There exists a pair of maps $(\overline{\lambda },h)$ such that $\overline{\lambda }:A\to \overline{S}$ is a surjective, $h:S\to \overline{S}$ is a bijective and the following diagram commutes.

The following theorem gives a unique surjection characterization together with an enumeration formula for such bijections.

Theorem 6.

[11] [Theorem 2.8] Let A and S be finite sets. Assume that $\phi \left(x\right):A\to S$ and $f\left(x\right):A\to A$ are maps. If $\phi \left(x\right)$ is surjective and $f\left(x\right)$ is injective on ${\phi }^{-1}\left(s\right)$ for each $s\in S,$ then $f\left(x\right)$ is bijective if and only if there exists a unique surjection $\psi \left(x\right):A\to S$ such that $\phi \left(x\right)=\psi \left(x\right)\circ f\left(x\right).$ Moreover, there are ${\prod }_{s\in S}\left(♯{\phi }^{-1}\left(s\right)\right)!$ bijections $f\left(x\right)$ determined by the map $\phi \left(x\right).$

The dual diagram of the AGW criterion was put forward by Yuan in [12]. This dual theoretical framework acts as an important tool to derive compositional inverses of PPs when applying Theorem 14 (See Theorems 27, 33 for more details).

Theorem 7.

[12] [Theorem 2.6] Let the notations be defined as in Lemma 1. If $f\left(x\right):A⟶A$ is a bijection, $f^{-1}\left(x\right)$ and $h^{-1}\left(x\right)$ are the compositional inverses of $f\left(x\right)$ and $h\left(x\right),$ respectively, then we have

$$\lambda \left(x\right)\circ f^{-1}\left(x\right)=h^{-1}\left(x\right)\circ \overline{\lambda }\left(x\right),$$

i.e., the following diagram commutes

Figure 2. The dual diagram of AGW criterion

Figure 2. The dual diagram of AGW criterion

2.1.2. The Group Structure of the Set of Permutation Polynomials: A Perspective Based on the AGW Criterion

In 1991, Wan and Lidl [13] showed that all permutation polynomials of the form $x^{r}h\left(x^{(q-1)/d}\right)$ from a group $G(d,q)$ under composition and this group is isomorphic to a generalized wreath product. Yuan [12,14] obtained a similar result for other permutation polynomials over finite fields and determine their group structure.

Theorem 8.

[12] [Theorem 4.1] Let $A,S$ be finite sets and let $\phi :A\to S$ be a surjective map. Let $G_{\phi }$ be the set of all bijections $f:A\to A$ such that ${\pi }_f\circ \phi =\phi \circ f$, where ${\pi }_f:S\to S$ is a bijection. Then $G_{\phi }$ forms a group under composition.

The following two commutative diagrams hold:

Theorem 9.

[12] [Theorem 4.2] Let the notations be as in Theorem 8, and let $G(\phi ,1)=\{f\in G_{\phi }\mid {\pi }_f=1_S\}$, where $1_S$ denotes the identity map on S. Then $G(\phi ,1)$ is a normal subgroup of $G_{\phi }$.

The following two commutative diagrams holds:

Subsequently, Yuan [14] gave the following conclusion for the case where A is a group and $\phi$ is an epimorphism in Theorem8. We first introduce the relevant notation.

The following definitions can be found in [15] [Chapter 7]. Let K be a subgroup (not necessarily normal) of a group G. Then a subgroup $Q\le G$ is a complement of K in G if $K\cap Q=1$ and $KQ=G$, where $KQ:=\{kq\mid k\in K,q\in Q\}$.

Definition 1.

A group G is a semidirect product of K by Q, which is denoted by $G=K⋊Q$, if $K◃G$ and K has a complement $Q_1\cong Q$. One also says that G splits over K.

Definition 2.

Let D and Q be groups, Ω be a finite Q-set, and $K:={\prod }_{\omega \in \Omega }{D}_{\omega }$, where $D_{\omega }\cong D$ for all $\omega \in \Omega$. Then the wreath product of D by Q, denoted by $D\wr Q$, is the semidirect product of K by Q, where Q acts on K by $q·\left(d_{\omega }\right)=\left(d_{q\omega }\right)$ for $q\in Q$ and $\left(d_{\omega }\right)\in {\prod }_{\omega \in \Omega }{D}_{\omega }$. The normal subgroup K of $D\wr Q$ is called the base of the wreath product.

When the group D is finite, we have $\left|K\right|={\left|D\right|}^{|\Omega |}$; if Q is also finite, then

$$|D\wr Q|=|K⋊Q|=\left|K\right|\left|Q\right|={\left|D\right|}^{|\Omega |}\left|Q\right|.$$

Theorem 10.

[14] [Theorem 3.1] Let A be a finite group of order n. Let $S\subseteq A$ be a subgroup with $\left|S\right|=d$ and $\phi :A\to S$ be an epimorphism. Let $G_{\phi }$ be the set of all bijections $f:A\to A$ such that ${\pi }_f\circ \phi =\phi \circ f$, where ${\pi }_f:S\to S$ is a bijection. Then $G_{\phi }\cong S_{n/d}\wr S_d$ and $|G_{\phi }|={\left(\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d}}\right)\right)}^{d}d!$.

2.1.3. The Local Criterion

Next, we will introduce the Local Criterion. Although the Local Criterion is equivalent to the AGW criterion, it provides a more transparent formulation for our purposes: it reveals the fundamental mechanism of constructing permutation polynomials via map compositions (See Theorem 27, Remark 2, Theorem 33 and Remark 4 for further assistance). We state the criterion as follows.

Theorem 11.

[16] [Lemma 2.1 ](Local Criterion) Let A and S be finite sets, and let $f\left(x\right):A\to A$ be a map. Then $f\left(x\right)$ is a bijection if and only if for any surjection $\psi \left(x\right):A\to S,$ the composition $\phi \left(x\right)=\psi \left(x\right)\circ f\left(x\right)$ is a surjection, and $f\left(x\right)$ is injective on ${\phi }^{-1}\left(s\right)$ for each $s\in S.$

Corollary 1.

Let A and S be finite sets, and let $\phi \left(x\right):A\to S$ be a surjective map. If the map $f\left(x\right):A\to A$ is bijective, then $f\left(x\right)$ is injective on ${\phi }^{-1}\left(s\right)$ for each $s\in S.$

From the local criterion, determining whether $f\left(x\right)$ is a permutation polynomial requires the following two conditions:

(1) $\phi \left(x\right)=\psi \left(x\right)\circ f\left(x\right)$ is a surjection from A to S;

(2) For each $s\in S$, the restriction of $f\left(x\right)$ to ${\phi }^{-1}\left(s\right)$ is injective.

This naturally raises the following question: if only one of the above conditions is satisfied, but we appropriately increase the number of maps satisfying that condition, can we still conclude that $f\left(x\right)$ is a permutation polynomial from A to A? Specifically:

(i) Generalization of the surjectivity condition: Let $S_i\subseteq A$ ($i=1,2,\dots ,t$), and let ${\psi }_i:A\to S_i$ be surjective. Suppose that for each i, the composition ${\phi }_i={\psi }_i\circ f$ is a surjection from A to $S_i$. Does it follow that f is a permutation from A to A?

(ii) Generalization of the injectivity condition: Let $S_i\subseteq A$, and let ${\phi }_i:A\to S_i$ be surjective. Suppose that for each i and each $s\in S_i$, the restriction ${f|}_{{\phi }_i^{-1}\left(s\right)}$ is injective. Does it follow that f is a permutation from A to A?

The above two problems lead respectively to the central concepts in this subsection — surjective systems and injective systems.

1. surjective systems

We introduce surjective systems at first.

Definition 3.

[17] For a prime power $q,$ let $f\left(x\right)$ be a polynomial over ${\mathbb{F}}_q$. Assume that $S_i$ are nonempty finite sets of ${\mathbb{F}}_q$ with $|S_i|\le q/2$ and ${\psi }_i\left(x\right)$ are surjections from ${\mathbb{F}}_q$ to $S_i$$(i=1,2,\dots ,t).$ If $f\left(x\right)$ is a permutation polynomial over ${\mathbb{F}}_q$ if and only if ${\phi }_i\left(x\right)={\psi }_i\left(x\right)\circ f\left(x\right)$$(i=1,2,\dots ,t)$ are surjective from ${\mathbb{F}}_q$ to $S_i$, then the polynomial $f\left(x\right)$ is called the local permutation polynomial over ${\mathbb{F}}_q$ with respect to ${\psi }_1\left(x\right),{\psi }_2\left(x\right),\dots ,{\psi }_t\left(x\right).$

Theorem 12.

[17] Let A and $S_i$$(i=1,2,\dots ,t)$ be finite sets, and let ${\psi }_i\left(x\right):A\to S_i$ be surjective maps and $f\left(x\right):A\to A$ be a map. Suppose that $f\left(x\right)$ is a permutation polynomial if and only if ${\phi }_i\left(x\right)={\psi }_i\left(x\right)\circ f\left(x\right)$$(i=1,2,\dots t)$ are surjective from A to $S_i$. Then for any permutation polynomial $g\left(x\right):A\to A,$$g\left(x\right)\circ f\left(x\right)$ is a permutation polynomial if and only if for $i=1,2,\dots ,t,$${\psi }_i\left(x\right)\circ g^{-1}\left(x\right)$ are surjections from A to $S_i.$

2. injective systems

We recall several definitions and results concerning injective systems for permutation polynomials, which were recently proposed and systematically investigated in [11].

Definition 4.

Let q be a prime power, S a proper non-empty subset of ${\mathbb{F}}_q$, and let $\phi \left(x\right):{\mathbb{F}}_q\to S$ be a surjective map. We say that $f\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ is injective for $\phi \left(x\right)$ if ${f|}_{{\phi }^{-1}\left(s\right)}:{\phi }^{-1}\left(s\right)\to {\mathbb{F}}_q$ is injective for any $s\in \mathrm{Im}\left(\phi \right)$. Let ${\phi }_i\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ for $1\le i\le t$. We say that $f\left(x\right)$ is injective for $\{{\phi }_1\left(x\right),{\phi }_2\left(x\right),\dots ,{\phi }_t\left(x\right)\}$ if $f\left(x\right)$ is injective for each ${\phi }_i\left(x\right),1\le i\le t$.

For further understanding the notation of $f\left(x\right)$ is injective for $\phi \left(x\right)$, the following result was established in [11]. .

Lemma 1.

Let q be a prime power and d be a positive integer with $d\mid (q-1)$. Then $x^r$ is injective for $x^d$ if and only if $gcd(r,d)=1$.

As a direct consequence of Lemma 1, the following corollary holds.

Corollary 2.

Let q be a prime power. Then $x^r$ is injective for $\{x^d:d\mid (q-1)anddisaprime\}$ if and only if $gcd(r,q-1)=1$, i.e., $x^r$ is a PP over ${\mathbb{F}}_q$.

The general equivalence between the permutation property and injectivity with respect to polynomial families was presented in [11] as follows.

Proposition 1.

[11] Let q be a prime power, and let ${\phi }_i\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ for $1\le i\le t$. Suppose that $f\left(x\right)$ is a PP if and only if $f\left(x\right)$ is injective for $\{{\phi }_1\left(x\right),{\phi }_2\left(x\right),\dots ,{\phi }_t\left(x\right)\}$. Then for any two distinct elements $y_1,y_2\in {\mathbb{F}}_q$, there exists some $i(1\le i\le t)$ such that ${\phi }_i\left(y_1\right)={\phi }_i\left(y_2\right)$.

Conversely, if for any two distinct elements $y_1,y_2\in {\mathbb{F}}_q$, there exists some $i(1\le i\le t)$ such that ${\phi }_i\left(y_1\right)={\phi }_i\left(y_2\right)$. Then $f\left(x\right)$ is a PP if and only if $f\left(x\right)$ is injective for $\{{\phi }_1\left(x\right),{\phi }_2\left(x\right),\dots ,{\phi }_t\left(x\right)\}$.

Based on the above equivalent characterization, the formal definition of injective systems for permutation polynomials was introduced in [11].

Definition 5.

[11] Let q be a prime power, and for $1\le i\le t$, let ${\phi }_i\left(x\right)$ be non-constant polynomials in ${\mathbb{F}}_q\left[x\right]$. We say that $\{{\phi }_1\left(x\right),{\phi }_2\left(x\right),\dots ,{\phi }_t\left(x\right)\}$ is an injective system for PP over ${\mathbb{F}}_q$ provided that $f\left(x\right)$ is a PP if and only if $f\left(x\right)$ is injective for $\{{\phi }_1\left(x\right),{\phi }_2\left(x\right),\dots ,{\phi }_t\left(x\right)\}$.

Furthermore, a concrete family of injective systems via trace polynomials was also provided in [11].

Proposition 2.

Let q be a prime power and let $n>1$ be a positive integer. Then $\{\mathrm{Tr}\left(ux\right):u\in {\mathbb{F}}_{q^n}^{*}\}$ is an injective system for PP over ${\mathbb{F}}_{q^n}$.

Motivated by the above results, we propose the following notation.

Definition 6.

[11] Let q be a prime power, S a proper non-empty subset of ${\mathbb{F}}_q$, and let $\phi \left(x\right):{\mathbb{F}}_q\to S$ be a surjective map. We say that $f\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ is injective for $\phi \left(x\right)$ if ${f|}_{{\phi }^{-1}\left(s\right)}:{\phi }^{-1}\left(s\right)\to {\mathbb{F}}_q$ is injective for any $s\in \mathrm{Im}\left(\phi \right)$. Let ${\phi }_i\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ for $1\le i\le t$. We say that $f\left(x\right)$ is injective for $\{{\phi }_1\left(x\right),{\phi }_2\left(x\right),\dots ,{\phi }_t\left(x\right)\}$ if $f\left(x\right)$ is injective for each ${\phi }_i\left(x\right),1\le i\le t$.

First, a lower bound on the number of polynomials contained in any injective system is given as follows.

Proposition 3.

Let $q\ge 7$ be a prime power and ${\phi }_i\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ for $1\le i\le t$ be non-constant maps. If $\{{\phi }_1\left(x\right),{\phi }_2\left(x\right),\dots ,{\phi }_t\left(x\right)\}$ is an injective system for PPs over ${\mathbb{F}}_q$, then $t\ge 3$.

Second, an explicit infinite family of injective systems constructed via trace polynomials over extension fields is characterized below.

Proposition 4.

Let q be a prime power and let $n>1$ be a positive integer. Then

$$\{\mathrm{Tr}\left(ux\right):u\in {\mathbb{F}}_{q^n}^{*}\}$$

is an injective system for PPs over ${\mathbb{F}}_{q^n}$.

Furthermore, Yuan proved that we may drastically shrink the above trace family while still preserving the injective system property, as summarized in the following existence theorem.

Theorem 13.

Let q be a prime power and let $n>1$ be a positive integer. There exists a set $A=\{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_{q+1}\}$ such that $\left\{\mathrm{Tr}\right(ux):u\in A\}$ is an injective system for ${\mathbb{F}}_{q^n}$.

2.1.4. The Local Method

Based on the local criterion, Yuan [16] established a local method that can determine whether a polynomial is a permutation polynomial while simultaneously deriving its compositional inverse.

The local method is a powerful and effective tool for constructing permutation polynomials over finite fields and computing their compositional inverses. Originating from number theory and algebraic theory, this method adopts the local–global principle to characterize polynomials over finite fields. Specifically, it deduces the global permutation property of a polynomial from its local characteristics on subsets of the finite field. The essence of the local method lies in characterizing the inherent functional relationship between the polynomial $f\left(x\right)$ and the variable x, which is achieved by analyzing the composite mappings ${\psi }_i\circ f\left(x\right)$ for a family of properly chosen mappings ${\psi }_i\left(x\right)$.

Theorem 14.

[16] [Theorem 2.2] (Local method) Let q be a prime power and $f\left(x\right)$ be a polynomial over ${\mathbb{F}}_q.$ Then $f\left(x\right)$ is a permutation polynomial over ${\mathbb{F}}_q$ if and only if there exist nonempty finite subsets $S_i$, $i=1,2,\dots ,t$ of ${\mathbb{F}}_q$ and maps ${\psi }_i\left(x\right):{\mathbb{F}}_q\to S_i$, $i=1,2,\dots ,t$ such that ${\psi }_i\left(x\right)\circ f\left(x\right)={\phi }_i\left(x\right),$$i=1,2,\dots ,t$ and $x=F({\phi }_1\left(x\right),{\phi }_2\left(x\right),\dots ,{\phi }_t\left(x\right)),$ where $F(x_1,x_2,\dots ,x_t)\in {\mathbb{F}}_q[x_1,x_2,\dots ,x_t].$ Moreover, the compositional inverse of $f\left(x\right)$ is given by

$$f^{-1}\left(x\right)=F({\psi }_1\left(x\right),{\psi }_2\left(x\right),\dots ,{\psi }_t\left(x\right)).$$

The essence of the local method lies in examining the composition ${\psi }_i\left(x\right)\circ f\left(x\right)$ (for several polynomials ${\psi }_i\left(x\right)$ ) to determine its functional relationship with $x.$

2.2. Algebraic Structure of Permutation Polynomials

It is well known that ${\mathbb{F}}_{q^n}$ and ${\mathbb{F}}_q^n$ are isomorphic as vector spaces over ${\mathbb{F}}_q$. This isomorphism allows us to study permutation polynomials by translating a univariate polynomial $f\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ into a multivariate polynomial map $F=(f_1(x_1,\dots ,x_n),\dots ,f_n(x_1,\dots ,x_n))\in {\mathbb{F}}_q^n[x_1,\dots ,x_n]$ .

Based on the basic properties of the trace, we recall the following criterion for ${\mathbb{F}}_q$-space homomorphisms.

Lemma 2.

[18] [Lemma 4.2] Let $f\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ be a map from ${\mathbb{F}}_{q^n}$ to ${\mathbb{F}}_q$. Then $f\left(x\right)$ is an ${\mathbb{F}}_q$-space homomorphism if and only if $f\left(x\right)=\mathrm{Tr}\left(vx\right)$ for some $v\in {\mathbb{F}}_{q^n}$.

Based on the trace representation above, the full characterization of ${\mathbb{F}}_q$-vector space homomorphisms from ${\mathbb{F}}_{q^n}$ to ${\mathbb{F}}_q^n$ is given in the same reference.

Theorem 15.

[18] Let $\rho :{\mathbb{F}}_{q^n}\to {\mathbb{F}}_q^n$ be an ${\mathbb{F}}_q$-vector space homomorphism, then there are n elements $v_1,\dots ,v_n\in {\mathbb{F}}_{q^n}$ such that

$$\rho \left(x\right)=(\mathrm{Tr}\left(v_{1}x\right),\dots ,\mathrm{Tr}\left(v_{n}x\right)),\forall x\in {\mathbb{F}}_{q^n}.$$

(1)

Moreover, ρ is an ${\mathbb{F}}_q$-vector space isomorphism if and only if $\{v_1,\dots ,v_n\}$ is a basis of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$, and when this occurs, its compositional inverse ${\rho }^{-1}:{\mathbb{F}}_q^n\to {\mathbb{F}}_{q^n}$ is given by:

$${\rho }^{-1}(x_1,\dots ,x_n)=u_1{x}_1+\dots +u_n{x}_n,\forall (x_1,\dots ,x_n)\in {\mathbb{F}}_q^n,$$

(2)

where $\{u_1,\dots ,u_n\}$ is the ordered dual basis of $\{v_1,\dots ,v_n\}$.

Symmetrically, homomorphisms in the reverse direction ${\mathbb{F}}_q^n\to {\mathbb{F}}_{q^n}$ admit a dual description, as recorded below.

Theorem 16.

[18] Let $\eta :{\mathbb{F}}_q^n\to {\mathbb{F}}_{q^n}$ be an ${\mathbb{F}}_q$-vector space homomorphism, then there exist n elements $a_1,\dots ,a_n\in {\mathbb{F}}_{q^n}$ such that

$$\eta (x_1,\dots ,x_n)=a_1{x}_1+\dots +a_n{x}_n,\forall (x_1,\dots ,x_n)\in {\mathbb{F}}_q^n.$$

(3)

Moreover, η is an ${\mathbb{F}}_q$-vector space isomorphism if and only if $\{a_1,\dots ,a_n\}$ is a basis of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$, and when this occurs, its compositional inverse ${\eta }^{-1}:{\mathbb{F}}_{q^n}\to {\mathbb{F}}_q^n$ is given by:

$${\eta }^{-1}\left(x\right)=(\mathrm{Tr}\left(b_{1}x\right),\dots ,\mathrm{Tr}\left(b_{n}x\right)),\forall x\in {\mathbb{F}}_{q^n},$$

(4)

where $\{b_1,\dots ,b_n\}$ is the ordered dual basis of $\{a_1,\dots ,a_n\}$.

Combining the isomorphisms $\rho$ and $\eta$ characterized above with an arbitrary permutation g on ${\mathbb{F}}_q^n$, Yuan derived the following compositional criterion for permutation polynomials over ${\mathbb{F}}_{q^n}$.

Theorem 17.

[18] [Theorem 1.1] Given a PP $f\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$. Let $\rho :{\mathbb{F}}_{q^n}\to {\mathbb{F}}_q^n$, $\eta :{\mathbb{F}}_q^n\to {\mathbb{F}}_{q^n}$ be ${\mathbb{F}}_q$-vector space homomorphisms, and let g be a map from ${\mathbb{F}}_q^n$ to ${\mathbb{F}}_q^n$. Then for any $x\in {\mathbb{F}}_{q^n}$ and $(x_1,\dots ,x_n)\in {\mathbb{F}}_q^n$, we have

$$\rho \left(x\right)=(\mathrm{Tr}\left(v_{1}x\right),\dots ,\mathrm{Tr}\left(v_{n}x\right)),v_i\in {\mathbb{F}}_{q^n},1\le i\le n,$$

$$\eta (x_1,\dots ,x_n)=a_1{x}_1+\dots +a_n{x}_n,a_i\in {\mathbb{F}}_{q^n},1\le i\le n.$$

Moreover, $F\left(x\right)=\eta \circ g\circ \rho \circ f\left(x\right)$ is a PP of ${\mathbb{F}}_{q^n}$ if and only if the following two conditions hold:

(i) Both $\{a_1,\dots ,a_n\}$ and $\{v_1,\dots ,v_n\}$ are bases of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$ ;

(ii) $g\in {\mathbb{F}}_q[x_1,\dots ,x_n]$ permutes ${\mathbb{F}}_q^n$.

Further, we take into account the two fundamental polynomial rings, namely ${\mathbb{F}}_{q^n}\left[x\right]$ and ${\mathbb{F}}_q[x_1,\dots ,x_n]$, defined over the finite field ${\mathbb{F}}_q$, and a profound algebraic property is then exhibited as follows. For simplicity, let $Map(A,B)$ denote the set of all mappings from A to B.

Theorem 18.

[18] Let $\left(Map({\mathbb{F}}_q^n,{\mathbb{F}}_q^n),+,·\right)$ and $\left(Map({\mathbb{F}}_{q^n},{\mathbb{F}}_{q^n}),+,·\right)$ be ${\mathbb{F}}_q$-vector spaces, equipped with addition and scalar multiplication operations. Then ψ given by

$$\begin{array}{cc}\hfill \psi :Map({\mathbb{F}}_q^n,{\mathbb{F}}_q^n)& \to Map({\mathbb{F}}_{q^n},{\mathbb{F}}_{q^n})\hfill \\ \hfill g& \mapsto {\rho }^{-1}g\rho \hfill \end{array}$$

is an ${\mathbb{F}}_q$-vector space isomorphism, where $\rho :{\mathbb{F}}_{q^n}\to {\mathbb{F}}_q^n$ and ${\rho }^{-1}:{\mathbb{F}}_q^n\to {\mathbb{F}}_{q^n}$ are ${\mathbb{F}}_q$-vector isomorphisms defined as (1) and (2), respectively. Additionally, ψ is also a groupoid isomorphism with respect to the composition.

We now shift our attention to equivalence relations for multivariate permutation polynomial systems over ${\mathbb{F}}_q^n$, which were first introduced in [19].

Definition 7.

[19] [Definition 3.1] Let q be a prime power and n be a positive integer. Two polynomial systems $F=(f_1,f_2,\dots ,f_n)$ and $G=(g_1,g_2,\dots ,g_n)$, with $f_i,g_i\in {\mathbb{F}}_q[x_1,\dots ,x_n]$, are called linearly equivalent if there exist nonsingular linear transformations $\rho :{\mathbb{F}}_q^n\to {\mathbb{F}}_q^n$ and $\sigma :{\mathbb{F}}_q^n\to {\mathbb{F}}_q^n$ such that $G=\rho \circ F\circ \sigma$.

From this definition, it immediately follows that F is a permutation of ${\mathbb{F}}_q^n$ if and only if G is a permutation of ${\mathbb{F}}_q^n$. On the other hand, observe that the identity map $I(x_1,\dots ,x_n)=(x_1,\dots ,x_n)$ is trivially a permutation system over ${\mathbb{F}}_q^n$. Moreover, modifying the i component of such polynomial system by adding terms that depend solely on previously determined variables preserves its permutation properties. This motivates the following notion of coordinate shift equivalence:

Definition 8.

[19] [Definition 3.2] Let q be a prime power and n be a positive integer. Given two polynomial systems $F,G\in {\left({\mathbb{F}}_q[x_1,\dots ,x_n]\right)}^n$ defined as:

$$F=(f_1(x_1,\dots ,x_k),f_2(x_1,\dots ,x_k),\dots ,f_k(x_1,\dots ,x_k),x_{k+1},\dots ,x_n),$$

$$G=(f_1(x_1,\dots ,x_k),\dots ,f_k(x_1,\dots ,x_k),x_{k+1}+h_{k+1}(x_1,\dots ,x_k),\dots ,x_n+h_n(x_1,\dots ,x_{n-1})),$$

where $0\le k\le n$ and each $h_i\in {\mathbb{F}}_q[x_1,\dots ,x_{i-1}]$. We say F and G are coordinate shift equivalent (abbreviated as CS equivalent). In the special case when $k=0$, the systems reduce to: $F=(x_1,\dots ,x_n)$ and $G=(x_1,x_2+h_2\left(x_1\right),\dots ,x_n+h_n(x_1,\dots ,x_{n-1}))$.

Their investigation commences with quadratic polynomial systems. Let $F\sim G$ denote the equivalence of two polynomial systems F and G under the combined operations of linear equivalences and CS equivalences. They provide complete characterizations of bivariate quadratic permutations over finite fields.

Theorem 19.

[19] [Theorem 3.1] Let ${\mathbb{F}}_q$ be a finite field of characteristic p. A quadratic polynomial system $F=(f_1,f_2)$ defined by

$$\left\{\begin{array}{c}{f}_1(x,y)=a_1{x}^2+a_{2}xy+a_3{y}^2+a_{4}x+a_{5}y,\hfill \\ f_2(x,y)=b_1{x}^2+b_{2}xy+b_3{y}^2+b_{4}x+b_{5}y,\hfill \end{array}\right.$$

is a permutation over ${\mathbb{F}}_q^2$ if and only if:

(i) for odd p, $F\sim (x,y)$,

(ii) for even p, $F\sim (x,y)$, $(x^2,y)$, $(x,y^2)$, $(x^2,y^2)$ or $(y^2+x,c_1{x}^2+c_2{y}^2+c_{3}x+c_{4}y)$ with $c_1{y}^4+(c_2+c_3)y^2+c_{4}y$ being a linearized permutation over ${\mathbb{F}}_q$.

Use the same method, they provide complete characterizations bivariate 3-homogeneous permutations over finite fields.

Theorem 20.

[19] [Theorem 3.2] Let ${\mathbb{F}}_q$ be a finite field with $q\ne 1(mod3)$. The system of 3-homogeneous polynomials defined by

$$\left\{\begin{array}{c}{f}_1(x,y)=a_1{x}^3+a_2{x}^{2}y+a_{3}x{y}^2=x{Q}_1(x,y),\hfill \\ f_2(x,y)=b_2{x}^{2}y+b_{3}x{y}^2+b_4{y}^3=y{Q}_2(x,y),\hfill \end{array}\right.$$

where $a_1{b}_4\ne 0$, $Q_1=a_1{x}^2+a_{2}xy+a_3{y}^2$ and $Q_2=b_2{x}^2+b_{3}xy+b_4{y}^2$, is a permutation of ${\mathbb{F}}_q^2$ if and only if both $Q_1$ and $Q_2$ are irreducible over ${\mathbb{F}}_q$, and one of the following occurs:

(i) $Q_1(x,y)=k{Q}_2(x,y)$ for some constant $k\in {\mathbb{F}}_q^{*}$.

(ii) $gcd(Q_1(x,y),Q_2(x,y))=1$, and the rational function ${\displaystyle \frac{a_1+a_{2}t+a_3{t}^2}{b_{2}t+b_3{t}^2+b_4{t}^3}}$ is equivalent to $\mu \circ t^3\circ \nu$ (for $q\equiv 2(mod3)$) or $\mu \circ (t^3-\gamma t)\circ \nu$ (for $3\mid q$), where $\mu ,\nu$ are degree-one rational functions in ${\mathbb{F}}_q\left(t\right)$ and $\gamma \in {\mathbb{F}}_q$ is either 0 or a quadratic nonresidue.

They determine all binomial permutations of the form $x^3+a{x}^{2q+1}$ over a finite field ${\mathbb{F}}_q$ with $\mathrm{char}\left({\mathbb{F}}_q\right)\ne 3$, as an application of Theorem 20 and obtain the following result.

Proposition 5.

[19] [Propsotion 3.10] Let q be an odd prime power with characteristic $p\ne 3$, and let $f\left(x\right)=x^3+a{x}^{2q+1}\in {\mathbb{F}}_{q^2}\left[x\right]$ with $a\ne 0$. Then $f\left(x\right)$ permutes ${\mathbb{F}}_{q^2}$ if and only if one of the following holds:

(1) If $a\in {\mathbb{F}}_q$, then

(1.1) $a=3$ and $q\equiv 2(mod3)$;

(1.2) $a=1$ and $q\equiv 11$ or $17(mod24)$;

(1.3) $a=-3$ and $q\equiv -1(mod12)$.

(2) If $a\notin {\mathbb{F}}_q$, let $u\in {\mathbb{F}}_q$ be a quadratic nonresidue, and take ${\mathbb{F}}_{q^2}={\mathbb{F}}_q\left(\alpha \right)$ where ${\alpha }^2=u$. Writing $a=a_1+a_2\alpha$ with $a_1,a_2\in {\mathbb{F}}_q$. Then either:

(2.1) $a_1=-1$, $a_2=2\sigma /r$ (where σ satisfies $z^2-2z-2=0$) and the equation $z^4-4z^2+1=0$ has no roots in ${\mathbb{F}}_q$; or

(2.2) $a_1={\displaystyle \frac{3(r^4+6u{r}^2+u^2)}{{(r^2-u)}^2}}$ and $a_2={\displaystyle \frac{12(r^2+u)r}{{(r^2-u)}^2}}$, for some arbitrary $r\in {\mathbb{F}}_q^{*}$.

3. Methods on the Inverse of Permutation Polynomial over Finite Fields

In this section, permutation polynomials are classified according to their algebraic forms. It should be noted that this classification is not strictly exclusive; for instance, $a{x}^q+x$is both a linearized polynomial and a polynomial of the form $x^{r}h\left(x^s\right)$. The classification thus reflects the different methods historically used to study permutation polynomials, rather than a purely set-theoretic categorisation. For each class, we first present a review of the research history concerning its compositional inverse, and then compute its compositional inverse via the local method.

3.1. Inverses of Monomials

For positive integer $n,$$x^n$ is a PP of ${\mathbb{F}}_q$ if and only if $gcd(n,q-1)=1.$ All invertible monomials form a group of order $\phi (q-1)$ under the operation of composition, where $\phi$ is Euler’s totient function. The inverse of $x^n$ over ${\mathbb{F}}_q$ is $x^m,$ where $mn\equiv 1(modq-1).$ The case of monomials being trivial, we forego the derivation of its compositional inverse using the local method.

3.2. Inverses of Dickson PPs

The Dickson polynomial $D_n(x,a)$ of the first kind of degree n with parameter $a\in {\mathbb{F}}_q$ is given as

$$D_n(x,a)={\displaystyle \sum _{i=0}^{\lfloor n/2\rfloor }}{\displaystyle \frac{n}{n-i}}\left({\displaystyle \genfrac{}{}{0pt}{}{n-i}{i}}\right){(-a)}^i{x}^{n-2i},$$

where $n/2$ denotes the largest integer $\le n/2.$

A fundamental property of $D_n(x,a)$ is that for any $u\ne 0$,

$$\mathrm{if}x=u+{\displaystyle \frac{a}{u}},\mathrm{then}{D}_n(x,a)=u^n+{\displaystyle \frac{a^n}{u^n}}.$$

It is well known that $D_n(x,a)$ is a PP over ${\mathbb{F}}_q$ if and only if $gcd(n,q^2-1)=1$ in [7] [Theorem 7.16] and [20] [Theorem 3.2].

The following result, established by Li et al. [21], provides the compositional inverse of a Dickson polynomial under the permutation condition. For the convenience of the reader, we retain the proof.

Theorem 21.

[21] [Lemma 4.8] Let q be a prime power and $gcd(n,q^2-1)=1$. Let m be a positive integer satisfying $nm\equiv 1(mod(q^2-1))$. Then the compositional inverse of the Dickson polynomial $D_n(x,a)$ over ${\mathbb{F}}_q$ is $D_n^{-1}(x,a)=D_m(x,a^n).$

Proof. 

Let $b=a^n\in {\mathbb{F}}_q$ and $y\in {\mathbb{F}}_{q^2}^{*}$ such that $x=y+{\displaystyle \frac{b}{y}}\in {\mathbb{F}}_q$. Then for any $x\in {\mathbb{F}}_q$,

$$\begin{array}{cc}\hfill D_n(D_m(x,b),a)& =D_n\left(y^m+{\displaystyle \frac{b^m}{y^m}},a\right)\hfill \\ \hfill & =D_n\left(y^m+{\displaystyle \frac{a}{y^m}},a\right)\hfill \\ \hfill & =y^{mn}+{\displaystyle \frac{a^n}{y^{mn}}}\hfill \\ \hfill & =y+{\displaystyle \frac{b}{y}}=x.\hfill \end{array}$$

Therefore, the compositional inverse of the Dickson polynomial $D_n(x,a)$ over ${\mathbb{F}}_q$ is $D_m(x,a^n)$. □

Remark 1.

From the proof of Theorem 21, we may take $\psi \left(x\right)=D_m(x,b)$, which yields $\psi \left(D_n(x,a)\right)=x$. Hence, by Theorem 14, we obtain $D_n^{-1}(x,a)=D_m(x,a^n)$. This is essentially the proof itself, which does not yet reveal the advantage of the local method.

The polynomial $x^5+a{x}^3+5^{-1}{a}^{2}x$ over ${\mathbb{F}}_q$, where $a\in {\mathbb{F}}_q$ and $q\equiv \pm 2(mod5)$, is actually the Dickson PP $D_5(x,-5^{-1}a)$ (see [22]), and its inverse on ${\mathbb{F}}_q$ is $D_m(x,-{\left(5^{-1}a\right)}^5)$ by Theorem 21, where $m=(3q^2-2)/5$.

The polynomial $x^7-7a{x}^5+14a^2{x}^3-7a^{3}x$ over ${\mathbb{F}}_q$ with $a\in {\mathbb{F}}_q^{*}$ and $q\equiv \pm 2,\pm 3(mod7)$ is indeed the Dickson polynomial $D_7(x,a)$ (see [23]), and its inverse is $D_m(x,a^7)$ by Theorem 21.

3.3. Inverses of Linearized PPs

Linearized polynomials are always taken as

$$L\left(x\right)={\displaystyle \sum _{i=0}^{n-1}}{a}_i{x}^{q^i}\in {\mathbb{F}}_{q^n}\left[x\right]/(x^{q^n}-x).$$

(5)

We denote by ${\mathcal{L}}_n\left({\mathbb{F}}_{q^n}\right)$ the set of all linearied polynomials in the form (5). Equipped with the composition of polynomials in ${\mathbb{F}}_{q^n}\left[x\right]/(x^{q^n}-x)$, ${\mathcal{L}}_n\left({\mathbb{F}}_{q^n}\right)$ forms non-commutative ${\mathbb{F}}_q$-algebra. Wu and Liu [24] characterize the algebra ${\mathcal{L}}_n\left({\mathbb{F}}_{q^n}\right)$, that is

$${\mathcal{L}}_n\left({\mathbb{F}}_{q^n}\right)\cong {\mathbb{F}}_{q^n}^{\vee }{\otimes }_{{\mathbb{F}}_q}{\mathbb{F}}_{q^n}\cong {\mathcal{D}}_n\left({\mathbb{F}}_{q^n}\right),$$

(6)

where ${\mathbb{F}}_{q^n}^{\vee }$ is the dual space of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$, ${\mathbb{F}}_{q^n}^{\vee }{\otimes }_{{\mathbb{F}}_q}{\mathbb{F}}_{q^n}$ is the so-called composition algebra on ${\mathbb{F}}_{q^n}$, and ${\mathcal{D}}_n\left({\mathbb{F}}_{q^n}\right)$ is an algebra formed by all $n\times n$ matrices over ${\mathbb{F}}_{q^n}$ of the form

$$\left(\begin{array}{cccc}{a}_0& a_1& \dots & a_{n-1}\\ a_{n-1}^q& a_0^q& \dots & a_{n-2}^q\\ ⋮& ⋮& ⋮& ⋮\\ a_1^{q^{n-1}}& a_2^{q^{n-1}}& \dots & a_0^{q^{n-1}}\end{array}\right),$$

which are called Dickson matrices. It is clear that the set of all non-singular Dickson matrices form a group.

Based on the result from (6), the following identity holds:

$$D_{L^{-1}}=D_L^{-1}={\displaystyle \frac{1}{detL}}{D}_L^{*}.$$

Hence, Wu and Liu [24] were able to derive the compositional inverse of a linearized permutation polynomial over ${\mathbb{F}}_{q^n}.$ Unlike their approach, which relies on various properties of linearized polynomials as theoretical groundwork, we adopt the Local method [17] to directly obtain the compositional inverse without the aforementioned preparation. For the convenience of the reader, a detailed proof using the Local method is provided below.

Theorem 22.

[24] [Theorem 4.5][17] Let $L\left(x\right)={\sum }_{i=0}^{n-1}{a}_i{x}^{q^i}\in {\mathcal{D}}_n\left({\mathbb{F}}_{q^n}\right)$ be a linearized permutation polynomial and $D_L$ be its associated Dickson matrix. Then

$$L^{-1}\left(x\right)={\left(det\left(D_L\right)\right)}^{-1}{\displaystyle \sum _{i=0}^{n-1}}{\overline{a}}_i{x}^{q^i},$$

where ${\overline{a}}_i$ is the $(i,0)-$th cofactor of $D_L$, and the determinant of $D_L$ is $det\left(D_L\right)=a_0{\overline{a}}_0+{\sum }_{i=1}^{n-1}{a}_{n-i}^{q^i}{\overline{a}}_i.$

Proof. 

For $i=0,1,\dots ,n-1$, set ${\psi }_i\left(x\right)=x^{q^i}$ and ${\phi }_i\left(x\right)={\psi }_i\left(L\left(x\right)\right)=L{\left(x\right)}^{q^i}$. Since the associated Dickson matrix $D_L$ is non-singular and

$$\left(\begin{array}{c}{\phi }_0\left(x\right)\\ {\phi }_1\left(x\right)\\ ⋮\\ {\phi }_{n-1}\left(x\right)\end{array}\right)=\left(\begin{array}{c}L\left(x\right)\\ L{\left(x\right)}^q\\ ⋮\\ L{\left(x\right)}^{q^{n-1}}\end{array}\right)=D_L\left(\begin{array}{c}x\\ x^q\\ ⋮\\ x^{q^{n-1}}\end{array}\right),$$

multiplying both sides on the left by $D_L^{-1}$ yields

$$D_L^{-1}\left(\begin{array}{c}{\phi }_0\left(x\right)\\ {\phi }_1\left(x\right)\\ ⋮\\ {\phi }_{n-1}\left(x\right)\end{array}\right)=\left(\begin{array}{c}x\\ x^q\\ ⋮\\ x^{q^{n-1}}\end{array}\right).$$

Recalling that $D_L^{-1}={\displaystyle \frac{1}{detL}}{D}_L^{*}$, where $D_L^{*}$ is the adjugate matrix of $D_L$, we obtain

$${\displaystyle \frac{D_L^{*}}{detL}}\left(\begin{array}{c}{\phi }_0\left(x\right)\\ {\phi }_1\left(x\right)\\ ⋮\\ {\phi }_{n-1}\left(x\right)\end{array}\right)=\left(\begin{array}{c}x\\ x^q\\ ⋮\\ x^{q^{n-1}}\end{array}\right).$$

Taking the first row of this matrix identity gives the scalar equation

$${\displaystyle \frac{1}{detL}}{\displaystyle \sum _{i=0}^{n-1}}{\overline{a}}_i{\phi }_i\left(x\right)=x,$$

where ${\overline{a}}_i$ are the entries from the first row of $D_L^{*}$. Finally, by Theorem 14, the compositional inverse of $L\left(x\right)$ is given by

$$L^{-1}\left(x\right)={\displaystyle \frac{1}{detL}}{\displaystyle \sum _{i=0}^{n-1}}{\overline{a}}_i{x}^{q^i}.\square$$

□

The general theorem above provides a theoretical approach for obtaining the compositional inverse of a linearized permutation polynomial via the determinant and the $(i,0)-$th cofactor of its associated Dickson matrix. Below, we give three examples of linearized polynomials whose compositional inverses are obtained through their Dickson matrices.

Example 3.3.1. Coulter and Henderson characterized the following linearized permutation binomial in [25]:

$$L\left(x\right)=x^{q^r}-ax,$$

and Wu subsequently determined its compositional inverse in [26] via the determinant and the $(i,0)$-th cofactor of its associated Dickson matrix.

Inspired by the idea in Section 9 of Wang [9], which presents a new and simpler proof for the compositional inverse of such a binomial, we reformulate Wang’s proof using the language of the local method.

Theorem 23.

[25] [Theorem 3][26] Let $L_r\left(x\right)=x^{q^r}-ax$, where $a\in {\mathbb{F}}_{q^m}^{*}$, and $1\le r\le m-1$. Then $L_r\left(x\right)$ is a permutation polynomial over ${\mathbb{F}}_{q^m}$ if and only if the norm ${\mathrm{N}}_{q^m/q^d}\left(a\right)\ne 1,$ where $d=gcd(m,r).$ In this case, its inverse on ${\mathbb{F}}_{q^m}$ is

$$L_r^{-1}\left(x\right)={\displaystyle \frac{{\mathrm{N}}_{q^m/q^d}\left(a\right)}{1-{\mathrm{N}}_{q^m/q^d}\left(a\right)}}{\displaystyle \sum _{i=0}^{m/d-1}}{a}^{-{\displaystyle \frac{q^{(i+1)r}-1}{q^r-1}}}{x}^{q^{ir}}.$$

Proof. 

Let $d=gcd(m,r).$ In this case, for $0\le i\le m/d-1$, we choose ${\psi }_i\left(x\right)=a^{-{\sum }_{j=1}^i{q}^{jr}}{x}^{q^{ir}}$ and ${\phi }_i\left(x\right)={\psi }_i\left(f\left(x\right)\right)=a^{-{\sum }_{j=1}^i{q}^{jr}}\left(x^{\left(q^{i+1}\right)r}-a^{q^{ir}}{x}^{q^{ir}}\right).$

Obviously,

$${\displaystyle \sum _{i=1}^{m/d-1}}{\phi }_i\left(x\right)=a^{-{\sum }_{j=1}^{m/d-1}{q}^{jr}}{x}^{q^{mr/d}}-ax=a\left({\mathrm{N}}_{q^m/q^d}{\left(a\right)}^{-1}-1\right)x.$$

Hence, it follows from Theorem 14 that f is a PP if and only if ${\mathrm{N}}_{q^m/q^d}\left(a\right)\ne 1$, and the compositional inverse of f is

$$f^{-1}\left(x\right)={\displaystyle \frac{1}{a\left({\mathrm{N}}_{q^m/q^d}{\left(a\right)}^{-1}-1\right)}}{\displaystyle \sum _{i=1}^{m/d-1}}{\psi }_i\left(x\right)={\displaystyle \frac{{\mathrm{N}}_{q^m/q^d}\left(a\right)}{1-{\mathrm{N}}_{q^m/q^d}\left(a\right)}}{\displaystyle \sum _{i=0}^{m/d-1}}{a}^{-{\displaystyle \frac{q^{(i+1)r}-1}{q^r-1}}}{x}^{q^r}.$$

This completes the proof. □

We summarize in Table the results on permutation polynomials of the form $x^{q^r}-ax$ with degree less than 7, obtained based on Theorem 23.

Table 1. Linearized permutation binomials of the form $x^{q^r}-ax$ and their inverses

Polynomial $\mathit{L}\left(\mathit{x}\right)$	Inverse ${\mathit{L}}^{-1}\left(\mathit{x}\right)$	Conditions
$x^3-ax$	${\displaystyle {\displaystyle \sum _{i=0}^{n-1}}{a}^{-\frac{3^{i+1}-1}{2}}{x}^{3^i}}$	$q=3^n$, a not a square
$x^4+ax$	${\displaystyle a^{\frac{q-1}{3}}{(1+a^{\frac{q-1}{3}})}^{-1}{\displaystyle \sum _{i=0}^{n-1}}{a}^{-\frac{4^{i+1}-1}{3}}{x}^{4^i}}$	$q=2^{2n}$, a not a square
$x^4+3x$	${\displaystyle -(x^4-3x)}$	$q=7$
$x^4-3x$	${\displaystyle -(x^4+3x)}$	$q=7$
$x^5+ax$	${\displaystyle a^{\frac{q-1}{4}}{(1+a^{\frac{q-1}{4}})}^{-1}{\displaystyle \sum _{i=0}^{n-1}}{a}^{-\frac{5^{i+1}-1}{4}}{x}^{5^i}}$	$q=5^n$, $a^2=2$
$x^5-ax$	${\displaystyle a^{\frac{q-1}{4}}{(1+a^{\frac{q-1}{4}})}^{-1}{\displaystyle \sum _{i=0}^{n-1}}{a}^{-\frac{5^{i+1}-1}{4}}{x}^{5^i}}$	$q=5^n$, a not a fourth power
$x^7-ax$	${\displaystyle a^{\frac{q-1}{6}}{(1-a^{\frac{q-1}{6}})}^{-1}{\displaystyle \sum _{i=0}^{n-1}}{a}^{-\frac{7^{i+1}-1}{6}}{x}^{7^i}}$	$q=7^n$, a not a sixth power

Table 1. Linearized permutation binomials of the form $x^{q^r}-ax$ and their inverses

Polynomial $\mathit{L}\left(\mathit{x}\right)$	Inverse ${\mathit{L}}^{-1}\left(\mathit{x}\right)$	Conditions
$x^3-ax$	${\displaystyle {\displaystyle \sum _{i=0}^{n-1}}{a}^{-\frac{3^{i+1}-1}{2}}{x}^{3^i}}$	$q=3^n$, a not a square
$x^4+ax$	${\displaystyle a^{\frac{q-1}{3}}{(1+a^{\frac{q-1}{3}})}^{-1}{\displaystyle \sum _{i=0}^{n-1}}{a}^{-\frac{4^{i+1}-1}{3}}{x}^{4^i}}$	$q=2^{2n}$, a not a square
$x^4+3x$	${\displaystyle -(x^4-3x)}$	$q=7$
$x^4-3x$	${\displaystyle -(x^4+3x)}$	$q=7$
$x^5+ax$	${\displaystyle a^{\frac{q-1}{4}}{(1+a^{\frac{q-1}{4}})}^{-1}{\displaystyle \sum _{i=0}^{n-1}}{a}^{-\frac{5^{i+1}-1}{4}}{x}^{5^i}}$	$q=5^n$, $a^2=2$
$x^5-ax$	${\displaystyle a^{\frac{q-1}{4}}{(1+a^{\frac{q-1}{4}})}^{-1}{\displaystyle \sum _{i=0}^{n-1}}{a}^{-\frac{5^{i+1}-1}{4}}{x}^{5^i}}$	$q=5^n$, a not a fourth power
$x^7-ax$	${\displaystyle a^{\frac{q-1}{6}}{(1-a^{\frac{q-1}{6}})}^{-1}{\displaystyle \sum _{i=0}^{n-1}}{a}^{-\frac{7^{i+1}-1}{6}}{x}^{7^i}}$	$q=7^n$, a not a sixth power

Example 3.3.2. Wu [26] studied the linearized polynomial $L\left(x\right)=x^{q^2}+b{x}^q+ax$ by using recurrences to compute the determinant of its Dickson matrix and the $(i,0)$-cofactors. Based on this, he gave a criterion for $L\left(x\right)$ to be a permutation polynomial over ${\mathbb{F}}_{q^n}$ and an explicit formula for its inverse (see [26] [Theorem 3.2.29]).

Taking $q=2$, the polynomial reduces to $L\left(x\right)=x^4+b{x}^2+ax$ over ${\mathbb{F}}_{2^m}$. To state the result of Zheng et al. [22] for this special case, we need the following recurrence sequence:

$$S_{-1}=0,S_0=1,S_i=b^{2^{i-1}}{S}_{i-1}+a^{2^{i-1}}{S}_{i-2},1\le i\le m,$$

where $a,b\in {\mathbb{F}}_{2^m}^{*}$.

Theorem 24

(Corollary 4, [22]). Let $L\left(x\right)=x^4+b{x}^2+ax$, where $a,b\in {\mathbb{F}}_{2^m}^{*}$ and $m>1$. Then $L\left(x\right)$ is a permutation polynomial over ${\mathbb{F}}_{2^m}$ if and only if $S_m+a{S}_{m-2}^2=1$. Moreover, if $L\left(x\right)$ permutes ${\mathbb{F}}_{2^m}$, its inverse is given by

$$L^{-1}\left(x\right)={\displaystyle \sum _{i=0}^{m-1}}\left(S_{m-2-i}^{2^{i+1}}+a^{1-2^{i+1}}{S}_i\right)x^{2^i}.$$

Example 3.3.3. Using heavy computation of determinants, Wu [27] obtained the following result.

Theorem 25.

[27] [Theorem 2.2] Let n be an odd positive integer, and let $a\in {\mathbb{F}}_{2^n}^{*}$ such that ${\mathrm{Tr}}_{2^n/2}\left(a^{-1}\right)=1.$ Then

$$f\left(x\right)=x^2+x+{\mathrm{Tr}}_{2^n/2}(x/a)$$

is a permutation polynomial over ${\mathbb{F}}_{2^n}$ with compositional inverse

$$f^{-1}\left(x\right)={\mathrm{Tr}}_{2^n/2}\left(x\right)+{\displaystyle \sum _{k=0}^{n-1}}{b}_k{x}^{2^k},$$

where the coefficients $b_k$ for $0\le k\le n-1$ are given by

$${\displaystyle \sum _{\begin{array}{c}j=1\\ j\mathrm{odd}\end{array}}^k}{(1/a)}^{2^j}+{\displaystyle \sum _{\begin{array}{c}l=k+1\\ l\mathrm{even}\end{array}}^{n-1}}{(1/a)}^{2^l}.$$

Tuxanidy and Wang [28] generalized Theorem 25 to the general q-case by decomposition method.

Regarding the compositional inverse of linearized polynomials, there is another result as follows.

Bastos [29] studied the inverses of linearized permutation polynomials over ${\mathbb{F}}_{q^n}$ with coefficients in ${\mathbb{F}}_q$, under the condition that $gcd(q,n)=1$.

The key observation is that, for such a linearized polynomial, its q-associate polynomial is coprime to $x^n-1$. Consequently, the conventional associate polynomial of its compositional inverse can be computed via the extended Euclidean algorithm. Let $R_{q,n}={\mathbb{F}}_q\left[x\right]/\langle x^n-1\rangle$. The primitive idempotents in $R_{q,n}$ are easily described using the ${\mathbb{F}}_q$-algebra isomorphism between $R_{q,n}$ and the group algebra ${\mathbb{F}}_{q}C$ of a cyclic group C of order n. By exploiting the decomposition of $R_{q,n}$ via these idempotents, Bastos derived explicit formulas for the inverses.

Theorem 26.

[29] [Theorem 26] Let $E=\{e_1\left(x\right),\dots ,e_t\left(x\right)\}$ be the set of primitive idempotents of $R_{q,n}.$ Given $F\left(x\right)\in {\mathcal{L}}_n\left({\mathbb{F}}_q\right)$ a linearized permutation over ${\mathbb{F}}_{q^n}$ and $f\left(x\right)$ its conventional q-associate, then $f\left(x\right)$ can be written as

$$f\left(x\right)\equiv {\displaystyle \sum _{i=1}^t}{f}_i\left(x\right)e_i\left(x\right)(mod{x}^n-1),$$

in which $gcd(f_i\left(x\right),x^n-1)=1,$ for $1\le i\le t.$ Furthermore, the conventional $q-$associate of the compositional inverse of $F\left(x\right)$ is given as

$$f^{-1}\left(x\right)\equiv {\displaystyle \sum _{i=1}^t}{f}_i^{-1}\left(x\right)e_i\left(x\right)(mod{x}^n-1).$$

This result theoretically provides a method for constructing the inverse permutation of specific linearized polynomials over ${\mathbb{F}}_q$. However, as the method relies on a special coefficient structure, it lacks general applicability for computing the inverse of arbitrary linearized permutation polynomials. Therefore, it is only briefly mentioned in this paper without an in-depth discussion.

3.4. Inverses of Permutation Polynomials of the Form $h\left(x\right)g\left(\lambda \right(x\left)\right)$

For permutation polynomials of the form $h\left(x\right)g\left(\lambda \right(x\left)\right)$, the inductive construction involves compositional inverses of two related families: those of the form $xg\left(\lambda \right(x\left)\right)$ and those of the form $x^{s}h\left(x^r\right)$. In the Local method (Theorem 14), the associated functions ${\psi }_i$ are chosen differently for these two families. Accordingly, the two cases are treated separately below.

3.4.1. Permutation Polynomials of the Form $xg\left(\lambda \right(x\left)\right)$

A result for polynomials of the form $xg\left(\lambda \right(x\left)\right)$ to be permutations was given by Akbary et al. [10] by AGW criterion.

Lemma 3.

[10] [Theorem 6.3] For a prime power q and a positive integer n, let S be a subset of ${\mathbb{F}}_{q^n}$ containing $0.$ Assume that $g\left(x\right),k\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ be any polynomials such that $g\left(0\right)\ne 0$ and $k\left(0\right)=0.$ Let $\lambda \left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ be a polynomial satisfying

(1) $g\left(\lambda \left({\mathbb{F}}_{q^n}\right)\right)\subseteq S;$ and

(ii) $\lambda \left(a\alpha \right)=k\left(a\right)\lambda \left(\alpha \right)$ for all $a\in S$ and all $\alpha \in {\mathbb{F}}_{q^n}.$

Then the polynomial $f\left(x\right)=xg\left(\lambda \right(x\left)\right)$ is a permutation polynomial over ${\mathbb{F}}_{q^n}$ if and only if $h\left(x\right)=xk\left(g\right(x\left)\right)$ induces a permutation over $\lambda \left({\mathbb{F}}_{q^n}\right).$

The commutative diagram for the above permutation polynomial is as follows.

Using the commutative diagram method (see [9,30] for details), Niu et al. [30] derived an explicit expression for the compositional inverse of the permutation polynomial described in Lemma 3. In what follows, we provide an alternative proof of their result by employing the local method (Theorem 14). This approach not only simplifies the original argument but also yields the following result.

Theorem 27.

[30] [Theorem 22 ] Let the notations be defined as in Lemma 3. Let $f\left(x\right)=xg\left(\lambda \right(x\left)\right)$ be a permutation polynomial over ${\mathbb{F}}_{q^n}$ and $h^{-1}\left(x\right)$ be the compositional inverse of $h\left(x\right)=xk\left(g\right(x\left)\right)$ over $\lambda \left({\mathbb{F}}_{q^n}\right).$ Then the compositional inverse of $f\left(x\right)$ over ${\mathbb{F}}_{q^n}$ is

$$f^{-1}\left(x\right)={\displaystyle \frac{x}{g\left(h^{-1}\left(\lambda \left(x\right)\right)\right)}}.$$

Proof. 

By assumption, we have $\lambda \left(x\right)\circ f\left(x\right)=h\left(x\right)\circ \lambda \left(x\right).$ Then by Lemma 7, we obtain

$$\lambda \left(x\right)=h^{-1}\left(x\right)\circ \lambda \left(x\right)\circ f\left(x\right).$$

Let ${\phi }_1\left(x\right)=\lambda \left(x\right)=h^{-1}\left(x\right)\circ \lambda \left(x\right)\circ f\left(x\right),{\psi }_1\left(x\right)=h^{-1}\left(x\right)\circ \lambda \left(x\right),{\phi }_2\left(x\right)=f\left(x\right),$ and ${\psi }_2\left(x\right)=x.$ Note that

$${\displaystyle \frac{{\phi }_2\left(x\right)}{g\left({\phi }_1\left(x\right)\right)}}=x$$

by $f\left(x\right)=xg\left(\lambda \right(x\left)\right).$ It follows from Lemma 14 that the compositional inverse of $f\left(x\right)$ is

$$f^{-1}\left(x\right)={\displaystyle \frac{x}{g\left(h^{-1}\left(\lambda \left(x\right)\right)\right)}}.$$

We are done. □

Remark 2.

Theorem 27 (namely Theorem 22 in [30]) and Lemma 3 can be revisited from the perspective of the local method. We present two remarks below: one simplifies the expression form of Theorem 27, and the other discusses the conditional constraints of Lemma 3 as well as the intrinsic role of the local criterion.

(1) Simplification of the expression. The compositional inverse given in Theorem 27 is

$$f^{-1}\left(x\right)={\displaystyle \frac{x-\lambda \left(x\right)+k\left(g\left(h^{-1}\left(\lambda \left(x\right)\right)\right)\right)h^{-1}\left(\lambda \left(x\right)\right)}{g\left(h^{-1}\left(\lambda \left(x\right)\right)\right)}}.$$

However, this expression is not in its simplest form. Indeed, since $h\left(x\right)=xk\left(g\right(x\left)\right)$, we have

$$k\left(g\left(h^{-1}\left(\lambda \left(x\right)\right)\right)\right)h^{-1}\left(\lambda \left(x\right)\right)=h\left(h^{-1}\left(\lambda \left(x\right)\right)\right)=\lambda \left(x\right),$$

so the numerator simplifies to x. Consequently,

$$f^{-1}\left(x\right)={\displaystyle \frac{x}{g\left(h^{-1}\left(\lambda \left(x\right)\right)\right)}}.$$

(2) On the conditions of Lemma 3 and the role of the local criterion. The two conditions in Lemma 3 (namely $g\left(\lambda \left({\mathbb{F}}_{q^n}\right)\right)\subseteq S$ and $\lambda \left(a\alpha \right)=k\left(a\right)\lambda \left(\alpha \right)$ for all $a\in S$, $\alpha \in {\mathbb{F}}_{q^n}$) are rather restrictive. Their purpose is to construct a polynomial $h\left(x\right)=xk\left(g\right(x\left)\right)$ such that the commutation relation $h\circ \lambda =\lambda \circ f$ holds.

The local criterion is the essential core for determining permutation polynomials: it reveals how to establish a connection between variables through functional composition, thereby determining whether a given polynomial is a permutation polynomial. From this criterion, the local method (Theorem 7) is derived, which proceeds by finding suitable functions ${\psi }_i$ and ${\phi }_i$ satisfying ${\phi }_i={\psi }_i\circ f$, thereby establishing a direct relationship between ${\phi }_i$ and the variable x, and using this relationship to determine whether f is a permutation polynomial.

Within this framework, the commutation relation $h\circ \lambda =\lambda \circ f$ can be rewritten as $\lambda =h^{-1}\circ \lambda \circ f$, which is precisely a concrete instance of ${\phi }_i={\psi }_i\circ f$. The AGW criterion employed in Lemma 3 serves to provide a specific constructive way of finding such ${\psi }_i$ and ${\phi }_i$ (here, λ and $h^{-1}\circ \lambda$) for the local method. However, the strong additional conditions in that lemma are not required by the local criterion or the local method themselves; they are merely one particular construction that realizes the desired relation. In other words, the local criterion reveals the core mechanism of constructing mappings via functional composition; the local method realizes this mechanism by constructing mappings ${\psi }_i$ and ${\phi }_i$ satisfying ${\phi }_i={\psi }_i\circ f$, while the AGW criterion in Lemma 3 serves as an auxiliary constructive tool.

In light of the above discussion — namely that the local criterion, not the restrictive AGW conditions, is the essential core — we generalize Lemma 3 to a more general form using the local criterion and method. The proof is analogous to that of Theorem 27, and is therefore omitted.

Theorem 28.

For a prime power $q,$ let

$$f\left(x\right)=h\left(x\right)g\left(\lambda \right(x\left)\right),$$

where $h\left(x\right),$$g\left(x\right)$ and $\lambda \left(x\right)$ be polynomials over ${\mathbb{F}}_q$ such that $g\left(\lambda \right(\alpha \left)\right)\ne 0$ for all $\alpha \in {\mathbb{F}}_q.$ If there exists a polynomial $\overline{\lambda }\left(x\right)$ such that $\lambda \left(x\right)=\overline{\lambda }\left(x\right)\circ f\left(x\right),$ then the polynomial $f\left(x\right)$ is a permutation polynomial over ${\mathbb{F}}_q$ if and only if $h\left(x\right)$ is injective on ${\lambda }^{-1}\left(s\right)$ for each $s\in \mathrm{Im}\left(\lambda \right).$

Moreover, if $f\left(x\right)$ and $h\left(x\right)$ permute ${\mathbb{F}}_q$, let $h^{-1}\left(x\right)$ be the compositional inverse of $g\left(x\right)$ over ${\mathbb{F}}_q.$ Then the compositional inverse of $f\left(x\right)$ is given by

$$f^{-1}\left(x\right)=h^{-1}\left({\displaystyle \frac{x}{g\left(\overline{\lambda }\left(x\right)\right)}}\right).$$

Remark 3.

It should be noted that the conclusion of Theorem 28 is derived from the Local Criterion applied to polynomials of the form $h\left(x\right)g\left(\lambda \right(x\left)\right)$. Meanwhile, Theorem 3.3 in [12] obtains exactly the same characterization based on the AGW Criterion.

3.4.2. Permutation Polynomials of the Form $x^{r}h\left(x^s\right)$

The study of permutation polynomials of the form $x^{r}h\left(x^s\right)$ was initiated by Wan et al. [13]. Subsequently, independent results on this family were obtained by Park et al. [31], Wang [32], and Zieve [33]. A unifying result from these works is the following lemma, which provides a useful criterion (or description) for such polynomials to be permutations.

Lemma 4.

[31] [Theorem 2.3][13] [Theorem 1. 2][32] [Theorem 1][33] [Lemma 2.1] Let q be a prime power and $f\left(x\right)=x^{r}h\left(x^s\right)\in {\mathbb{F}}_q\left[x\right]$, where $s=(q-1)/l$ and l is an integer. Then $f\left(x\right)$ permutes ${\mathbb{F}}_q$ if and only if

(i) $gcd(r,s)=1;$ and

(ii) $g\left(x\right)=x^{r}h{\left(x\right)}^s$ permutes ${\mu }_l.$

The commutative diagram for the above permutation polynomial is as follows.

Niu et al. [30] [Theorem11] determined the compositional inverse of the permutation polynomial in Lemma 4 using the commutative diagram method. Yuan [16] revisited this theorem and provided a new proof via the local method. For the convenience of the reader, we present Yuan’s proof in full detail below.

Theorem 29.

[34] [Theorem 11] [16] [Proposition 3.1] With the notation from Lemma 4, let $f\left(x\right)=x^{r}h\left(x^s\right)$ be a permutation polynomial over ${\mathbb{F}}_q$ and $g^{-1}\left(x\right)$ be a compositional inverse of $g\left(x\right)=x^{r}h{\left(x\right)}^s$ over ${\mu }_l.$ Suppose that a and b are two positive integers satisfying $as+br=1.$ Then the compositional inverse of $f\left(x\right)$ in ${\mathbb{F}}_q\left[x\right]$ is given by

$$f^{-1}\left(x\right)=g^{-1}{\left(x^s\right)}^a{x}^{b}h{\left(g^{-1}\left(x^s\right)\right)}^{-b}.$$

Proof. 

By the assumptions, we have the following commutative diagrams

where

$${\psi }_1\left(x\right)=g^{-1}\left(x^s\right),{\phi }_1\left(x\right)=x^s,{\psi }_2\left(x\right)=x,{\phi }_2\left(x\right)=f\left(x\right)=x^{r}h\left(x^s\right).$$

We have

$${\phi }_1{\left(x\right)}^a·{\left({\displaystyle \frac{{\phi }_2\left(x\right)}{h\left({\phi }_1\left(x\right)\right)}}\right)}^b=x.$$

It follows from Theorem 14 that

$$f^{-1}\left(x\right)={\psi }_1{\left(x\right)}^a·{\left({\displaystyle \frac{{\psi }_2\left(x\right)}{h\left({\psi }_1\left(x\right)\right)}}\right)}^b={\left(g^{-1}\left(x^s\right)\right)}^a{x}^b{\left(h\left(g^{-1}\left(x^s\right)\right)\right)}^{-b}.$$

This completes the proof. □

Assume that the notations are as defined in Theorem 29. According to this theorem, the computation of the compositional inverse of a permutation polynomial of the form $x^{r}h\left(x^s\right)$ reduces to determining the inverse map of the polynomial $g\left(x\right)=x^{r}h{\left(x\right)}^s$ on the set ${\mu }_{\ell }$.

As a concrete illustration, consider the family $x{(x^s-a)}^{(q^m-1)/s}$. Zheng et al. [23] explicitly computed $g^{-1}\left(x\right)$ for $g\left(x\right)=x{(x-a)}^{q^m}-1$, and consequently obtained the compositional inverse of $x{(x^s-a)}^{(q^m-1)/s}$.

Theorem 30.

[23] [Theorem 3.4] Let $f\left(x\right)=x{(x^s-a)}^t,$ where $a\in {\mathbb{F}}_{q^n}^{*}$, $st=q^m-1,$ and $s,t,m,n\in \mathbb{N}.$ Let

$$d=gcd(m,n),k={\displaystyle \frac{q^d-1}{gcd(s,q^n-1)}},l={\displaystyle \frac{q^n-1}{gcd(s,q^n-1)}}.$$

If $f\left(x\right)$ is a PP over ${\mathbb{F}}_{q^n},$ then its inverse on ${\mathbb{F}}_{q^n}$ is $f^{-1}\left(x\right)=x{\left(G\left(x\right)H\left(x\right)\right)}^t,$ where

$$\begin{array}{cc}\hfill G\left(x\right)=& {(1-a^l)}^{-1}{\displaystyle \sum _{i=0}^{k-1}}{\left(a^{k-i}{x}^{is}\right)}^{{\displaystyle \frac{q^n-1}{q^d-1}}},\hfill \\ \hfill H\left(x\right)=& {\displaystyle \sum _{j=1}^{n/d}}{a}^{-{\displaystyle \frac{q^{jm}-1}{q^m-1}}}{x}^{{\displaystyle \frac{q^{(j-1)m}-1}{t}}}.\hfill \end{array}$$

When $t=2$, they obtain

Theorem 31.

[23] [Theorem 5.2] Let $f\left(x\right)=x^{q^m}-2a{x}^{{\displaystyle \frac{q^m+1}{2}}}+a^{2}x,$ where $a\in {\mathbb{F}}_{q^n}^{*}$, q is odd and $m,n\in \mathbb{N}.$ Let $d=gcd(m,n),$

$$c=a^{{\displaystyle \frac{q^n-1}{q^d-1}}},H_2\left(x\right)=x{\left({\displaystyle \sum _{i=1}^{n/d}}{a}^{-{\displaystyle \frac{q^{im}-1}{q^m-1}}}{x}^{{\displaystyle \frac{q^{(i-1)m}-1}{2}}}\right)}^2.$$

(i) If $m/d$ is even, then $f\left(x\right)$ is a PP of ${\mathbb{F}}_{q^n}$ if and only if $c\ne 1.$ In this case,

$$f^{-1}\left(x\right)=c^2{(1-c)}^{-2}{H}_2\left(x\right).$$

(ii) If $m/d$ is odd, then $f\left(x\right)$ is a PP of ${\mathbb{F}}_{q^n}$ if and only if $c^2\ne 1.$ In this case,

$$f^{-1}\left(x\right)=c^2{(1-c)}^{-2}\left(2c{x}^{(q^n-1)/2}+c^2+1\right)H_2\left(x\right).$$

Taking $q^m=7$ in Theorem 31 leads to the following corollary.

Corollary 3.

[23] [Corollary 5.3] Let $f\left(x\right)=x^7-2a{x}^4+a^{2}x$, where $a\in {\mathbb{F}}_{7^n}^{*}$ and $n\in \mathbb{N}$. Then f is a PP of ${\mathbb{F}}_{7^n}$ if and only if $a^{(7^n-1)/3}\ne 1$. In this case, the inverse of f on ${\mathbb{F}}_{7^n}$ is

$$f^{-1}\left(x\right)=x\left(2+2a^{{\displaystyle \frac{7^n-1}{3}}}-3a^{{\displaystyle \frac{7^n-1}{6}}}{x}^{{\displaystyle \frac{7^n-1}{2}}}\right){\left({\displaystyle \sum _{i=0}^{n-1}}{a}^{-{\displaystyle \frac{7^{i+1}-1}{6}}}{x}^{{\displaystyle \frac{7^i-1}{2}}}\right)}^2.$$

Applying Theorem 30 to $t=3$, they derive the following result.

Theorem 32.

[23] [Theorem 6.2] Let $f\left(x\right)=x{(x^{{\displaystyle \frac{q^m-1}{3}}}-a)}^3$, where $a\in {\mathbb{F}}_{q^n}^{*}$, $m,n\in \mathbb{N}$ and $q^m\equiv 1(mod3)$. Let $d=(m,n)$,

$$c=a^{{\displaystyle \frac{q^n-1}{q^d-1}}},H_3\left(x\right)=x{\left({\displaystyle \sum _{i=1}^{n/d}}{a}^{-{\displaystyle \frac{q^{im}-1}{q^m-1}}}{x}^{{\displaystyle \frac{q^{(i-1)m}-1}{3}}}\right)}^3,$$

and let $D_0,D_1,D_2$ be defined as follows:

$$\begin{array}{cc}\hfill D_0& =\{3\mid q-1\wedge 3d\mid m,3\mid q+1\wedge 2\mid d\wedge 3d\mid m,3\mid q+1\wedge 2\nmid d\},\hfill \\ \hfill D_1& =\{3\mid q-1\wedge 3d\mid m-d,3\mid q+1\wedge 2\mid d\wedge 3d\mid m-d\},\hfill \\ \hfill D_2& =\{3\mid q-1\wedge 3d\mid m-2d,3\mid q+1\wedge 2\mid d\wedge 3d\mid m-2d\},\hfill \end{array}$$

where $a\mid b$ denotes that a divides b, and $A\wedge B$ denotes that A and B.

(i) If $D_0$ holds, then f is a PP of ${\mathbb{F}}_{q^n}$ if and only if $c\ne 1$. In this case,

$$f^{-1}\left(x\right)=c^3{(1-c)}^{-3}{H}_3\left(x\right).$$

(ii) If $D_j$ holds with $j=1$ or 2, then f is a PP of ${\mathbb{F}}_{q^n}$ if and only if $c^3\ne 1$. In this case,

$$f^{-1}\left(x\right)=c^3{(1-c)}^{-3}{\left(c^2+c{x}^{{\displaystyle \frac{j(q^n-1)}{3}}}+x^{{\displaystyle \frac{2j(q^n-1)}{3}}}\right)}^3{H}_3\left(x\right).$$

Applying Theorem 32 with $q^m=7$ gives the following corollary.

Corollary 4.

[23] [Corollary 6.3] Let $f\left(x\right)=x{(x^2-a)}^3$, where $a\in {\mathbb{F}}_{7^n}^{*}$ and $n\in \mathbb{N}$. Then

$$f\left(x\right)=x^7-3a{x}^5+3a^2{x}^3-a^{3}x,$$

and f is a PP of ${\mathbb{F}}_{7^n}$ if and only if $a^{(7^n-1)/2}=-1$. In this case,

$$f^{-1}\left(x\right)=x\left(3a^{{\displaystyle \frac{7^n-1}{6}}}{x}^{{\displaystyle \frac{2(7^n-1)}{3}}}-3{\left(ax\right)}^{{\displaystyle \frac{7^n-1}{3}}}-2\right){\left({\displaystyle \sum _{i=0}^{n-1}}{a}^{-{\displaystyle \frac{7^{i+1}-1}{6}}}{x}^{{\displaystyle \frac{7^i-1}{3}}}\right)}^3.$$

3.5. Inverses of the Permutation Polynomials of the Form $h\left(x\right)+g\left(\lambda \right(x\left)\right)$

We divide this part into two subsections according to whether $h\left(x\right)$ itself is a permutation polynomial.

3.5.1. Inverses of the Permutation Polynomials of the Form $h\left(x\right)+g\left(\lambda \right(x\left)\right)$ with a permutation polynomial $h\left(x\right)$

This subsection contains three main results. For the first family $h\left(x\right)+g\left(\lambda \right(x\left)\right)$, the permutation criterion established by Yuan and Ding [35] using the AGW criterion is recalled in the following lemma.

Lemma 5.

[35] [Theorem 6.1] For a prime power $q,$ assume that ${\mathbb{F}}_q$ is a finite field and $S,$$\overline{S}$ are subsets of ${\mathbb{F}}_q$ with $♯S=♯\overline{S}$ such that the maps $\lambda :{\mathbb{F}}_q\to S$ and $\overline{\lambda }:{\mathbb{F}}_q\to \overline{S}$ are surjective and $\overline{\lambda }$ is additive,

$$\overline{\lambda }(x+y)=\overline{\lambda }\left(x\right)+\overline{\lambda }\left(y\right),x,y\in {\mathbb{F}}_q.$$

Let $g_0:S\to {\mathbb{F}}_q$ and $g:{\mathbb{F}}_q\to {\mathbb{F}}_q$ be maps such that

$$\overline{\lambda }\circ (g+g_0\circ \lambda )=g\circ \lambda$$

and $\overline{\lambda }\left(g_0\left(\lambda \left(x\right)\right)\right)=0$ for every $x\in {\mathbb{F}}_q.$ Then the map $f\left(x\right)=g\left(x\right)+g_0\left(\lambda \left(x\right)\right)$ permutes ${\mathbb{F}}_q$ if and only if g permutes ${\mathbb{F}}_q.$

The commutative diagram for the above permutation polynomial is as follows.

Niu et al. [30] studied the compositional inverse of the permutation polynomial in Lemma 5 using the commutative diagram method. An alternative proof of their result, based on the local method, is also available and leads to the following conclusion.

Theorem 33.

[30] [Theorem 16] Let the notations be defined as in Lemma 5. Let $f\left(x\right)=g\left(x\right)+g_0\left(\lambda \left(x\right)\right)$ be a permutation over ${\mathbb{F}}_q$ and $g^{-1}\left(x\right)$ be the compositional inverse of $g\left(x\right)$ over ${\mathbb{F}}_q.$ Then the compositional inverse of $f\left(x\right)$ is

$$f^{-1}\left(x\right)=g^{-1}(x-g_0\left(g^{-1}\left(\overline{\lambda }\left(x\right)\right)\right)).$$

Proof. 

Since $\overline{\lambda }\left(x\right)\circ f\left(x\right)=g\left(x\right)\circ \lambda \left(x\right),$ we have

$$\lambda \left(x\right)=g^{-1}\left(x\right)\circ \overline{\lambda }\left(x\right)\circ f\left(x\right)$$

by Theorem 7. Let

$${\phi }_1\left(x\right)=\lambda \left(x\right)=g^{-1}\left(x\right)\circ \overline{\lambda }\left(x\right)\circ f\left(x\right),{\psi }_1\left(x\right)=g^{-1}\left(x\right)\circ \overline{\lambda }\left(x\right),$$

$${\phi }_2\left(x\right)=f\left(x\right),{\psi }_2\left(x\right)=x.$$

Then by $f\left(x\right)=g\left(x\right)+g_0\left(\lambda \left(x\right)\right)$, we have ${\phi }_2\left(x\right)-g_0\left({\phi }_1\left(x\right)\right)=g\left(x\right),$ or

$$g^{-1}\left({\phi }_2\left(x\right)-g_0\left({\phi }_1\left(x\right)\right)\right)=x.$$

It follows from Theorem 14 that the compositional inverse of of $f\left(x\right)$ is

$$f^{-1}\left(x\right)=g^{-1}(x-g_0\left(g^{-1}\left(\overline{\lambda }\left(x\right)\right)\right)).$$

This completes the proof. □

Remark 4.

Theorem 33 shows that the requirements in Lemma 5-namely, that S and $\overline{S}$ are subsets of $\mathbb{F}$ with $♯S=♯\overline{S}$, and that $\overline{\lambda }$ is an additive map satisfying $\overline{\lambda }\left(g\left(\lambda \left(x\right)\right)\right)=0$ for all $x\in \mathbb{F}$—are imposed to construct a mapping $g\left(x\right)$ satisfying $g\circ \lambda =\overline{\lambda }\circ f$. However, these very conditions also restrict the possibility of generalizing permutation properties for such polynomials. Furthermore, the purpose of imposing these two conditions, along with the requirement that $g\left(x\right)$ is a bijection from S to $\overline{S}$, is essentially to show that $\lambda \left(x\right)$ can be represented as a composition of a certain function with $f\left(x\right)$.

Following the first class, the second type of permutation polynomials of the same form and their compositional inverses have also been studied in existing literature. To contextualize the subsequent discussion, the explicit compositional inverses of permutation polynomials involving a b-linear translator are presented. Both examples below are concerned with the notion of a b-linear translator.

For $S\subset {\mathbb{F}}_q$, $\gamma ,b\in {\mathbb{F}}_q$ and a mapping $\lambda :{\mathbb{F}}_q\to {\mathbb{F}}_q$, $\gamma$ is referred to as a b-linear translator [10,36,37] of $\lambda$ with respect to S if $\lambda (x+u\gamma )=\lambda \left(x\right)+ub$ for all $x\in {\mathbb{F}}_q$ and $u\in S$.

The following result characterizes when $f\left(x\right)=x+\gamma G\left(\lambda \right(x\left)\right)$ is a permutation polynomial.

Lemma 6.

[10] [Theorem 6.4] Let $S\subseteq {\mathbb{F}}_q$ and $\lambda :{\mathbb{F}}_q\to S,$ be surjective map. Let $\gamma \in {\mathbb{F}}_q^{*}$ be a $b-$linear translator with respect to S for the map $\lambda .$ Then for any $G\in {\mathbb{F}}_q\left[x\right]$ which maps S into $S,$ we have that $f\left(x\right)=x+\gamma G\left(\lambda \right(x\left)\right)$ is a permutation polynomial of ${\mathbb{F}}_q$ if and only if $g\left(x\right)=x+bG\left(x\right)$ permutes $S.$

It can be illustrated by the following commutative diagram.

Niu et al. [30] derived the compositional inverse of permutation polynomials in Lemma 6 by means of commutative diagrams. In this work, we provide a more concise and intuitive proof via the local method.

Theorem 34.

[30] [Theorem 27] Let the notations be defined as in Lemma 6. Assume that $f\left(x\right)=x+\gamma G\left(\lambda \right(x\left)\right)$ is a permutation polynomial over ${\mathbb{F}}_q$ and $g^{-1}\left(x\right)$ is the compositional inverse of $g\left(x\right)=x+bG\left(x\right).$ Then the compositional inverse of $f\left(x\right)$ is given by

$$f^{-1}\left(x\right)=x-\gamma G\left(g^{-1}\left(\lambda \left(x\right)\right)\right).$$

Proof. 

We have $\lambda \left(x\right)\circ f\left(x\right)=g\left(x\right)\circ \lambda \left(x\right)$ by assumption. Next, applying Theorem 7, we deduce

$$\lambda \left(x\right)=g^{-1}\left(x\right)\circ \lambda \left(x\right)\circ f\left(x\right).$$

Take

$${\phi }_1\left(x\right)=\lambda \left(x\right)=g^{-1}\left(x\right)\circ \lambda \left(x\right)\circ f\left(x\right),{\psi }_1\left(x\right)=g^{-1}\left(x\right)\circ \lambda \left(x\right),$$

$${\phi }_2\left(x\right)=f\left(x\right),{\psi }_2\left(x\right)=x$$

in Theorem 14, then we arrive at

$$x={\phi }_2\left(x\right)-\gamma G\left({\phi }_1\left(x\right)\right).$$

Consequently, the compositional inverse of $f\left(x\right)$ over ${\mathbb{F}}_q$ is expressed as

$$f^{-1}\left(x\right)=x-\gamma G\left(g^{-1}\left(\lambda \left(x\right)\right)\right)$$

in view of Theorem 14. This is the desired result. □

Remark 5.

Theorem 27 in [30] establishes that the compositional inverse of $f\left(x\right)$ is

$$f^{-1}\left(x\right)=(b-\gamma )G\left(g^{-1}\left(\lambda \left(x\right)\right)\right)+g^{-1}\left(\lambda \left(x\right)\right)-\lambda \left(x\right)+x.$$

However, this expression can be simplified. Recalling that $g\left(x\right)=x+bG\left(x\right),$ we find that

$$bG\left(g^{-1}\left(\lambda \left(x\right)\right)\right)+g^{-1}\left(\lambda \left(x\right)\right)=g\left(g^{-1}\left(\lambda \left(x\right)\right)\right)=\lambda \left(x\right),$$

which implies that the terms $bG\left(g^{-1}\left(\lambda \left(x\right)\right)\right)+g^{-1}\left(\lambda \left(x\right)\right)-\lambda \left(x\right)$ vanish identically. Therefore, the inverse simplifies to

$$f^{-1}\left(x\right)=x-\gamma G\left(g^{-1}\left(\lambda \left(x\right)\right)\right).$$

When $b=0$, the compositional inverse of $f\left(x\right)$ is provided in [38] [Proposition 4] as $f^{-1}\left(x\right)=x+(p-1)\gamma h\left(\lambda \left(x\right)\right)$, where p denotes the characteristic of ${\mathbb{F}}_{q^n}$.

Another important class arises when G is taken as the identity, i.e., $f\left(x\right)=x+\gamma g\left(x\right)$. The core framework of [37] for establishing the permutation property and the inverse relies on the notion of linear translators and iterative composition of the permutation f. Specifically, the authors first leveraged the properties of linear translators to construct this class; they then conducted a detailed analysis of the cycle structure of f, and ultimately deduced the explicit expression of the compositional inverse.

Below, we provide a new proof of their result using the local method.

Theorem 35.

[37] [Theorem 3] Let $\gamma \in {\mathbb{F}}_{q^n}$ be a b-linear translator of $g\left(x\right):{\mathbb{F}}_{q^n}\to {\mathbb{F}}_q$ and $b\ne -1$. Then $f\left(x\right)=x+\gamma g\left(x\right)$ is a permutation polynomial of ${\mathbb{F}}_{q^n}$. Moreover, its inverse is given by $f^{-1}\left(x\right)=x-{\displaystyle \frac{\gamma }{b+1}}g\left(x\right)$.

Proof. 

Since $\gamma \in {\mathbb{F}}_{q^n}$ is a b-linear translator of $g\left(x\right)$, we have

$$g\left(x\right)\circ f\left(x\right)=g(x+\gamma g(x\left)\right)=g\left(x\right)+bg\left(x\right)=(b+1)g\left(x\right).$$

If $b=-1$, then $f\left(x\right)$ is not a permutation polynomial over ${\mathbb{F}}_{q^n}.$ Now we assume that $b\ne -1.$ Taking ${\psi }_1\left(x\right)=g\left(x\right),$${\phi }_1\left(x\right)=g\left(x\right)\circ f\left(x\right),$${\psi }_2\left(x\right)=x,$ and ${\phi }_2\left(x\right)=x,$ we have

$${\phi }_2\left(x\right)-{\displaystyle \frac{\gamma }{b+1}}{\phi }_1\left(x\right)=x.$$

It follows from Theorem 14 that the compositional inverse of $f\left(x\right)$ is

$$f^{-1}\left(x\right)=x-{\displaystyle \frac{\gamma }{b+1}}g\left(x\right).$$

This completes the proof □

The research on permutation polynomials of the form $f_3\left(x\right)=x^s+\gamma {\mathrm{Tr}}_{2^n/2}\left(x^t\right)$ over ${\mathbb{F}}_{2^n}$ originated from [36,39], where $1\le s,t\le 2^n-2$, $gcd(s,2^n-1)=1$, and $\gamma \in {\mathbb{F}}_{2^n}^{*}$. A sufficient and necessary condition for $f_3\left(x\right)$ to be a permutation polynomial over ${\mathbb{F}}_{2^n}$ was established in [39,40]. Additionally, if $f_3\left(x\right)$ is a permutation polynomial over ${\mathbb{F}}_{2^n}$ and $t=s(2^i+1)$ for some integer $0\le i\le n-1$ with $i\ne n/2$, its compositional inverse is given in [40] [Theorem 4] as $f_3^{-1}\left(x\right)={(x+\gamma {\mathrm{Tr}}_{2^n/2}\left(x^{2^i+1}\right))}^r$, where r is the multiplicative inverse of s modulo $2^n-1$.

By [40] [Theorem 3], the permutation property of $f_3\left(x\right)$ is equivalent to that of $f_4\left(x\right)=x+\gamma {\mathrm{Tr}}_{2^n/2}\left(x^{s^{-1}t}\right)$, where $\gamma$ is a 0-linear translator (i.e., $b=0$) of ${\mathrm{Tr}}_{2^n/2}\left(x^{s^{-1}t}\right)$. Therefore, this conclusion can be regarded as a special case of Theorem 35.

An application of the local criterion and the local method allows us to present the following generalization of Lemma 5, which also unifies Theorems 33 and 34. We omit the detailed proof as it closely follows the arguments in Theorem 33.

Theorem 36.

Let $g\left(x\right),g_0\left(x\right),\lambda \left(x\right)\in {\mathbb{F}}_q\left[x\right]$ be polynomials. Suppose there exists a polynomial $\overline{\lambda }\left(x\right)$ over ${\mathbb{F}}_q$ such that $\overline{\lambda }\left(x\right)\circ \left(g\left(x\right)+g_0\left(\lambda \left(x\right)\right)\right)=\lambda \left(x\right).$ Then the polynomial

$$f\left(x\right)=g\left(x\right)+g_0\left(\lambda \left(x\right)\right)$$

(7)

permutes ${\mathbb{F}}_q$ if and only if $g\left(x\right)$ is injective on ${\lambda }^{-1}\left(s\right)$ for each $s\in \mathrm{Im}\left(\lambda \right(x\left)\right).$

In particular, if $g\left(x\right)$ and $f\left(x\right)$ are permutation polynomials over ${\mathbb{F}}_q$, let $g^{-1}\left(x\right)$ denote the compositional inverse of $g\left(x\right)$ over ${\mathbb{F}}_q.$ Then the compositional inverse of $f\left(x\right)$ over ${\mathbb{F}}_q$ is given by

$$f^{-1}\left(x\right)=g^{-1}\left(x-g_0\left(\overline{\lambda }\left(x\right)\right)\right).$$

Remark 6.

It should be noted that the conclusion of Theorem 36 is derived from the Local Criterion applied to polynomials of the form $g\left(x\right)+g_0\left(\lambda \left(x\right)\right)$. Meanwhile, Theorem 3.4 in [12] obtains exactly the same characterization based on the AGW Criterion if $g\left(x\right)$ is permutation polynomial over ${\mathbb{F}}_q$.

3.5.2. Inverses of the Permutation Polynomials of the Form $h\left(x\right)+g\left(\lambda \right(x\left)\right)$ where $h\left(x\right)$ is not a permutation polynomial

Different from the foregoing two families of the form $h\left(x\right)+g\left(\lambda \right(x\left)\right)$, where $h\left(x\right)$ itself acts as a permutation polynomial, the third family adopts $h\left(x\right)=x^q-x$. Clearly, $x^q-x$ is not a permutation polynomial over ${\mathbb{F}}_{q^n}$, since it is an ${\mathbb{F}}_q$-linear polynomial possessing a non-trivial kernel. Nevertheless, Yuan [16] conducted systematic research on polynomials of the form

$$f\left(x\right)=x^q-x+g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right),$$

where $g\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ and $n>1$ is an integer. As will be seen in the theorem below, $f\left(x\right)$ is a permutation polynomial over ${\mathbb{F}}_{q^n}$ if and only if $gcd(n,q-1)=1$ and g is invertible over ${\mathbb{F}}_q.$

The main result in [16] was proved using the Local method. Here, we merely present the core function pairs required for the application of Theorem 14. Concretely, one may choose the paired functions

$${\phi }_1\left(x\right)={\mathrm{Tr}}_{q^n/q}\left(x\right),{\psi }_1\left(x\right)=g^{-1}\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right),{\phi }_2\left(x\right)=x,{\psi }_2\left(x\right)=f\left(x\right).$$

Combining with the inherent equalities

$${\phi }_2\left(x\right)-g\left({\phi }_1\left(x\right)\right)=x^q-x,{\displaystyle \sum _{i=1}^n}i(x^q-x)=nx-{\mathrm{Tr}}_{q^n/q}\left(x\right)=nx-{\phi }_1\left(x\right),$$

Yuan derived the desired result.

Theorem 37.

[16] [Theorem 4.2] Let $n>1$ be a positive integer, q a prime power and $g\left(x\right)\in {\mathbb{F}}_q\left[x\right]$. Then the polynomial

$$f\left(x\right)=x^q-x+g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)$$

is a permutation polynomial over ${\mathbb{F}}_{q^n}$ if and only if $gcd(n,q-1)=1$ and $g\left(x\right)$ is a permutation polynomial over ${\mathbb{F}}_q$. Under such circumstances, let $g^{-1}$ stand for the compositional inverse of g; then the compositional inverse of $f\left(x\right)$ is given by

$$f^{-1}\left(x\right)={\displaystyle \frac{1}{n}}\left(g^{-1}\left({\displaystyle \frac{{\mathrm{Tr}}_{q^n/q}\left(x\right)}{n}}\right)-{\displaystyle \frac{n+1}{2}}{\mathrm{Tr}}_{q^n/q}\left(x\right)+{\displaystyle \sum _{i=1}^{n-1}}i{x}^{q^i}\right).$$

The commutative diagram for the above permutation polynomial is as follows.

3.6. Inverses of the Permutation Polynomials of the Form $g\left(\psi \left(x\right)\right)+{\sum }_{i=1}^r({\phi }_i\left(x\right)+{\delta }_i)h_i\left(\psi \left(x\right)\right)$

In this subsection, we summarize the known results on the compositional inverses of permutation polynomials of the form

$$F\left(x\right)=g\left(\psi \left(x\right)\right)+{\displaystyle \sum _{i=1}^r}({\phi }_i\left(x\right)+{\delta }_i)h_i\left(\psi \left(x\right)\right).$$

Depending on whether $r=1$ or $r\ge 2$, the structure of the compositional inverse exhibits distinct features. We therefore discuss these two cases separately.

3.6.1. Inverses of the Permutation Polynomials of the Form $h\left(\psi \right(x\left)\right)\phi \left(x\right)+g\left(\psi \right(x\left)\right)$

We now present known results on the compositional inverses of permutation polynomials of the form

$$f\left(x\right)=h\left(\psi \right(x\left)\right)\phi \left(x\right)+g\left(\psi \right(x\left)\right),$$

(8)

over ${\mathbb{F}}_{q^n}$, where $\psi \left(x\right),\phi \left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ are additive polynomials, $h\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ is arbitrary, and $g\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ satisfies certain conditions.

Two main research streams have emerged for this class of polynomials. The first stream focuses on the case where $\psi \left(x\right)=x^{q^l}+ax$ and $h\left(x\right)=1$, with various choices of g (e.g., $b(x^k+\delta )$, $b{(x+\delta )}^s+x$, or ${\sum }_i{b}_i{(x^{t_i}+{\delta }_i)}^{s_i}$) and additive $\phi \left(x\right)$ [35,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64]. The second stream considers $\psi \left(x\right)$ as the trace function from ${\mathbb{F}}_{q^n}$ to ${\mathbb{F}}_q$ (or the absolute trace when the base field is a prime field), with $h\left(x\right)=1$ and $\phi \left(x\right)=x$ [65,66,67], or more generally, $\phi \left(x\right),h\left(x\right)\in {\mathbb{F}}_p$ where $\phi \left(x\right)\ne x$ is additive [68].

Akbary et al. [10] investigated the permutation properties of polynomials of the form (8) using the AGW criterion and established the following result.

Theorem 38.

[10] [Theorem 5.1] Let $\psi \left(x\right),\phi \left(x\right)\in {\mathbb{F}}_{q^n}$ be additive polynomials and $\overline{\psi }\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ be a q-polynomial satisfying $\phi \left(x\right)\circ \psi \left(x\right)=\overline{\psi }\left(x\right)\circ \phi \left(x\right)$ and $♯\psi \left({\mathbb{F}}_{q^n}\right)=♯\overline{\psi }\left({\mathbb{F}}_{q^n}\right)$. Let $h\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ be any polynomial such that $h\left(\psi \left({\mathbb{F}}_{q^n}\right)\right)\subseteq {\mathbb{F}}_q^{*}$, and let $g\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ be any polynomial. Then

$$f\left(x\right)=h\left(\psi \right(x\left)\right)\phi \left(x\right)+g\left(\psi \right(x\left)\right)$$

permutes ${\mathbb{F}}_{q^n}$ if and only if

(i) $\ker \left(\psi \right(x\left)\right)\cap \ker \left(\phi \right(x\left)\right)=\left\{0\right\};$ and

(ii) $\overline{f}\left(x\right)=h\left(x\right)\phi \left(x\right)+\overline{\psi }\left(g\left(x\right)\right)$ is a bijection from $\psi \left({\mathbb{F}}_{q^n}\right)$ to $\overline{\psi }\left({\mathbb{F}}_{q^n}\right).$

The commutative diagram for the above permutation polynomial is as follows.

In the following content, we summarize the existing results on compositional inverses of permutation polynomials of the form given in Theorem 38. We do not provide new proofs for these known conclusions via the local method, and the main reasons are explained in the final paragraph of this subsection.

We first introduce the earliest relevant research outcome. Tuxanidy and Wang [69] studied the compositional inverses of the permutation polynomials in Theorem 38 by the decomposition method. This method is inspired by Wu and Liu [70]. Wu and Liu considered the compositional inverse of a class of bilinearized PPs over ${\mathbb{F}}_{q^n}$, i.e., $f\left(x\right)=x\left(L\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+a{\mathrm{Tr}}_{q^n/q}\left(x\right)+ax\right)$, where q is even, n is odd, $xL\left(x\right)$ is a bilinear permutation polynomial over ${\mathbb{F}}_q$ for a linearized polynomial $L\left(x\right)\in {\mathbb{F}}_q$ (we also derive the compositional inverses of such polynomials via the local method in this paper, see Theorem 48 for details). The idea of the decomposition method mainly consists of the following two steps. First, the finite field ${\mathbb{F}}_{q^n}$ is decomposed into the direct sum ${\mathbb{F}}_q\oplus ker\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)$ by the map $\varphi :x\to ({\mathrm{Tr}}_{q^n/q}\left(x\right),x-{\mathrm{Tr}}_{q^n/q}\left(x\right)).$ Second, based on this decomposition, the problem of computing the inverse of $f\left(x\right)$ can be converted into the problem of computing the inverse of a bivariate function that permutes ${\mathbb{F}}_q\oplus ker\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right).$ This in turn is equivalent to obtaining the inverses of two permutation polynomials over the subspace ${\mathbb{F}}_q$ and $ker\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)$ respectively, thus simplifying the original problem into two more manageable sub-problems.

The key step in extending this useful idea to computing the compositional inverse of permutation polynomials in Theorem 38 is to extend ${\mathrm{Tr}}_{q^n/q}\left(x\right)$ into arbitrary linearized polynomial $\psi \left(x\right)$. However, even though $\phi \left(x\right)\circ \psi \left(x\right)=\overline{\psi }\left(x\right)\circ \phi \left(x\right)$ and $♯\psi \left({\mathbb{F}}_{q^n}\right)=♯\overline{\psi }\left({\mathbb{F}}_{q^n}\right)$ hold, they still may not have a "nice enough" expression for $\mathrm{Ker}\left(\psi \right).$ Tuxanidy and Wang [71] instead use the map ${\varphi }_{\psi }:{\mathbb{F}}_{q^n}\to \psi \left({\mathbb{F}}_{q^n}\right)\oplus S_{\psi },$ where $S_{\psi }=\{x-\psi \left(x\right)\mid x\in {\mathbb{F}}_{q^n}\}$ is a subspace of ${\mathbb{F}}_{q^n},$ in similarity with $ker\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)$ as above. Thus, to calculate the inverse of $f\left(x\right)$ in Theorem 38, they had to add the conditions $♯S_{\psi }=♯S_{\overline{\psi }}$ and $ker\left(\phi \right)\cap \psi \left(S_{\psi }\right)=\left\{0\right\},$ and got the following result.

Theorem 39.

[71] [Theorem 1.2] Using the same notations and assumptions of Theorem 38, assume that $f\left(x\right)$ is a permutation of ${\mathbb{F}}_{q^n}$, and further assume that $♯S_{\psi }=♯S_{\overline{\psi }}$ and $ker\left(\phi \right)\cap \psi \left(S_{\psi }\right)=\left\{0\right\}.$ Then φ induces a bijective from $S_{\psi }$ to $S_{\overline{\psi }}.$ Let ${\overline{f}}^{-1}$, ${\phi }^{-1}{|}_{S_{\overline{\psi }}}\in {\mathbb{F}}_{q^n}\left[x\right]$ induce the inverses of $\overline{f}{|}_{\psi \left({\mathbb{F}}_{q^n}\right)}$ and ${\phi |}_{S_{\psi }}$, respectively. Then the compositional inverse of $f\left(x\right)$ over ${\mathbb{F}}_{q^n}$ is given by

$$f^{-1}\left(x\right)={\overline{f}}^{-1}\left(\overline{\psi }\left(x\right)\right)+{\phi }^{-1}{|}_{S_{\psi }}\left({\displaystyle \frac{x-\overline{\psi }\left(x\right)-g\left({\overline{f}}^{-1}\left(\overline{\psi }\left(x\right)\right)\right)+\overline{\psi }\left(g\left({\overline{f}}^{-1}\left(\overline{\psi }\left(x\right)\right)\right)\right)}{h\left({\overline{f}}^{-1}\left(\overline{\psi }\left(x\right)\right)\right)}}\right).$$

Furthermore, if φ induces a bijection from $\psi \left({\mathbb{F}}_{q^n}\right)$ to $\overline{\psi }\left({\mathbb{F}}_{q^n}\right),$ then φ permutes ${\mathbb{F}}_{q^n}$ and the compositional inverse of $f\left(x\right)$ over ${\mathbb{F}}_{q^n}$ is given by

$$f^{-1}\left(x\right)={\phi }^{-1}\left({\displaystyle \frac{x-g\left({\overline{f}}^{-1}\left(\overline{\psi }\left(x\right)\right)\right)}{h\left({\overline{f}}^{-1}\left(\overline{\psi }\left(x\right)\right)\right)}}\right).$$

For several explicit families of permutation polynomials deduced from Theorem 39 in [71], we omit the detailed expressions of their compositional inverses and will not enumerate them individually. Instead, we mainly survey relevant results on compositional inverses for permutation polynomials of the form $h\left(\psi \right(x\left)\right)\phi \left(x\right)+g\left(\psi \right(x\left)\right)$ from existing literature.

Under the conditions $\psi \left(x\right)=x^{q^i}-x,$$g\left(x\right)=1$, $\phi \left(x\right)=cx,$ and $h\left(x\right)\in {\mathbb{F}}_{q^m}\left[x\right]$ (where $m,i$ are positive integers such that $1\le i\le m-1$, $l=gcd(i,m),$ and $c\in {\mathbb{F}}_{q^l}^{*}$), Niu et al. [34] applied the commutative diagram method to determine the compositional inverse of the permutation polynomial of the form $f\left(x\right)=g\left(x^{q^i}-x+\delta \right)+cx$ over ${\mathbb{F}}_{q^m}$, obtaining the following result.

Theorem 40.

[34] [Theorem 3.7] Let q be a prime power, $m,i$ be positive integers with $1\le i\le m-1$, $l=gcd(i,m),$$c\in {\mathbb{F}}_{q^l}^{*}$, and $g\left(x\right)\in {\mathbb{F}}_{q^m}\left[x\right]$ such that $h\left(x\right)=g{\left(x\right)}^{q^i}-g\left(x\right)+cx+(1-c)\delta \in {\mathbb{F}}_{q^m}\left[x\right]$ permutes ${\mathbb{F}}_{q^m},$ where $\delta \in {\mathbb{F}}_{q^m}.$ Assume $H\left(x\right)$ is the compositional inverse of $h\left(x\right)$. Then for any $\delta \in {\mathbb{F}}_{q^m}$, $f\left(x\right)=g\left(x^{q^i}-x+\delta \right)+cx\in {\mathbb{F}}_{q^m}\left[x\right]$ is a permutation polynomial over ${\mathbb{F}}_{q^m}$ and the compositional inverse of $f\left(x\right)$ over ${\mathbb{F}}_{q^m}$ is

$$f^{-1}\left(x\right)=c^{-1}{x}^{q^i}+c^{-1}g{\left(H(x^{q^i}-x+\delta )\right)}^{q^i}-H(x^{q^i}-x+\delta )+\delta .$$

The commutative diagram for the above permutation polynomial is as follows.

Remark 7.

By virtue of the relationship between affine q-polynomials and additive polynomials, Theorem 40 is merely a special case of Theorem 39.

Next, we present the second research result in this line of work. In 2021, Reis and Wang [72] investigated the compositional inverses of permutation polynomials of the form (8) under more specific conditions. Specifically, they explored the compositional inverses when $\psi \left(x\right)={\mathrm{Tr}}_{q^n/q}\left(x\right)$ and $\phi \left(x\right)$ is the linearized q-associate of $k\left(x\right)$ with $k\left(x\right)\in {\mathbb{F}}_q\left[x\right]$. Recently, Reis and Wang [73] refined their results by studying $\psi \left(x\right)$ as the q-associate of $g_{t,a}\left(x\right)=(x^n-1)/(x^t-a)$ with $t|n$ and $a\in {\mathbb{F}}_q^{*}$ with $a^{n/t}=1.$ By using the AGW criterion, the compositional inverse of $f\left(x\right)$ was transformed into the compositional inverse of a certain related function over the sub-field, and thus the compositional inverse of $f\left(x\right)$ was constructed. They gave the following result.

Theorem 41.

[73] [Theorem 3.2] Let $n,t,a$ be defined as before and δ be a nonzero root of $x^{q^t}-ax$. Let $U_{t,a}=\delta ·{\mathbb{F}}_{q^t}=\{\delta y\mid y\in {\mathbb{F}}_{q^t}\}.$ The polynomial

$$P\left(x\right)=f\left(L_{g_{t,a}}\left(x\right)\right)+k\left(L_{g_{t,a}}\left(x\right)\right)·L_h\left(x\right)\in {\mathbb{F}}_{q^n}$$

with $h\in {\mathbb{F}}_q\left[x\right]$ and $k\left(U_{t,a}\right)\subseteq {\mathbb{F}}_q^{*}$ is a PP if and only if the following conditions holds:

(1) $gcd(h\left(x\right),(x^n-1)/(x^t-a))=1;$

(2) $Q_{t,a}\left(x\right)=T_{t,a}\left[f\right]\left(x\right)+{\delta }^{-1}k\left(\delta x\right)·L_h\left(\delta x\right)\in {\mathbb{F}}_{q^t}$ is a PP over ${\mathbb{F}}_{q^t},$ where $T_{t,a}\left[f\right]\left(x\right)=a^{-1}{\sum }_{i=0}^d{\mathrm{Tr}}_{q^n/q^t}\left({\delta }^{i-1}{a}_i\right)x^i\in {\mathbb{F}}_{q^t}\left[x\right]$ and $a_i$s are the coefficients of f in $x^i.$

In affirmative case, if $Q_{t,a}^{-1}$ is the inverse of $Q_{t,a}$ over ${\mathbb{F}}_{q^t},$ then the inverse PP of $P\left(x\right)$ over ${\mathbb{F}}_{q^n}$ is given by

$$P^{-1}\left(x\right)=F\left(L_{g_{t,a}}\left(x\right)\right)+k{\left({\delta }^{-1}{Q}_{t,a}^{-1}\left({\delta }^{-1}\left(L_{g_{t,a}}\left(x\right)\right)\right)\right)}^{q-2}·L_H\left(x\right),$$

where $H\in {\mathbb{F}}_q\left[x\right]$ and $F\left(x\right)\in {\mathbb{F}}_{q^n}$ are given as follows:

(i) if $p\mid (n/t),$ then $H\left(x\right)\in {\mathbb{F}}_q$ is the unique polynomial of degree at most $n-1$ such that $h\left(x\right)·H\left(x\right)\equiv 1(mod{x}^n-1)$ and F is any polynomial satisfying

$$F\left(x\right)\equiv -k{\left(\delta Q_{t,a}^{-1}\left({\delta }^{-1}x\right)\right)}^{q-2}·L_H\left(f\left(\delta Q_{t,a}^{-1}\left({\delta }^{-1}x\right)\right)\right)(mod{x}^{q^t}-ax);$$

(ii) if $p\nmid (n/t),$ then $H\left(x\right)\in {\mathbb{F}}_q$ is the unique polynomial of degree at most $n-2$ such that $h\left(x\right)·H\left(x\right)\equiv 1(mod{\displaystyle \frac{x^n-1}{x^t-a}})$ and F is any polynomial satisfying $F\left(x\right)\equiv M\left(Q_{t,a}^{-1}\left({\delta }^{-1}x\right)\right)(mod{x}^{q^t}-ax),$ where

$$M\left(x\right)=-k{\left(\delta x\right)}^{q-2}·L_H\left(f\left(\delta x\right)\right)-(at/n)L_{hH-1}\left(\delta x\right).$$

The commutative diagram for the above permutation polynomial is as follows.

Remark 8.

Let $q=3$, $n=2$, $t=1$, $a=1$, and take $\delta =1$, which is a nonzero root of $x^3-x$. Set $f\left(y\right)=y$, $k\left(y\right)\equiv 1$ and $h\left(x\right)=x-1$ in Theorem 41. Then we have $L_{g_{t,a}}\left(x\right)=x^3+x$ and $L_h\left(x\right)=x^3-x$, so that

$$P\left(x\right)=L_{g_{t,a}}\left(x\right)+L_h\left(x\right)=2x^3=-x^3$$

is a permutation polynomial over ${\mathbb{F}}_9$ according to Theorem 41.

Furthermore, we have

$$\left\{\begin{array}{c}{\mathbb{F}}_3\subseteq Im\left(L_{g_{t,a}}(x-L_{g_{t,a}}\left(x\right))\right),\hfill \\ {\mathbb{F}}_3\subseteq ker{L}_h,\hfill \end{array}\right.$$

which implies that the intersection $Im\left(L_{g_{t,a}}(x-L_{g_{t,a}}\left(x\right))\right)\cap ker{L}_h$ contains nonzero elements.

This example verifies that the condition $ker\left(\phi \right)\cap \psi \left(S_{\psi }\right)=\left\{0\right\}$ in Theorem 39 fails to hold in this case. Consequently, Theorem 41 cannot be regarded as a special case of Theorem 39.

The authors also have established the compositional inverses for several classes of permutation polynomials where, in the setting of Theorem 41, $L_{g_{t,a}}\left(x\right)$ acts as a trace function and $k\left(x\right)=1$. This work represents a shift toward a direct computational paradigm using the local method, in contrast to the approach in Theorems 39, 40, and 41, which typically relies on constructing and verifying a candidate polynomial. Specifically: In [74], the authors explicitly derive the compositional inverses for permutation polynomials of the form ${\sum }_{i=1}^k{b}_i{(x^{p^m}+x+\delta )}^{s_i}-x$ over ${\mathbb{F}}_{p^{2m}}$.

In [75], this constructive approach is extended to the more general family of polynomials over ${\mathbb{F}}_{p^{mn}}$, which is of the form ${\sum }_{i=1}^k{b}_i{\left({\mathrm{Tr}}_{p^{mn}/p^m}{\left(x\right)}^{t_i}+\delta \right)}^{s_i}+f_1\left(x\right)$, where $b_i\in {\mathbb{F}}_{p^m}$, and $f_1\left(x\right)$ satisfies the following two key conditions: (i) ${\mathrm{Tr}}_{p^{mn}/p^m}\left(x\right)\circ f_1\left(x\right)=\phi \left(x\right)\circ {\mathrm{Tr}}_{p^{mn}/p^m}\left(x\right)$ for some polynomial $\phi \left(x\right)$ over ${\mathbb{F}}_{p^m}$; (ii) for any $a\in {\mathbb{F}}_{p^m}$, $f_1\left(x\right)$ is injective on the fiber ${\mathrm{Tr}}_{p^{mn}/p^m}{\left(a\right)}^{-1}$. Rather than relying on verification after construction, these papers shift the focus toward systematic derivation via this local method, offering explicit expressions for the inverses.

An additive analogue of the index of a univariate polynomial over a finite field ${\mathbb{F}}_q$ was introduced by Reis and Wang [76]. We can also write an arbitrary polynomial $f\left(x\right)$ uniquely according to its additive index. Namely, $f\left(x\right)=h\left(L\right(x\left)\right)+M\left(x\right),$ where $L\left(x\right),M\left(x\right)$ are p-linearized polynomials over ${\mathbb{F}}_q,$$deg\left(M\right)<deg\left(L\right),$$L\left(x\right)$ splits completely over ${\mathbb{F}}_q$ and L is of the maximal degree. In this case, the additive index of f is ${\displaystyle \frac{q}{deg\left(L\right)}},$ which is the index of the kernel of L as subgroup of the additive group of ${\mathbb{F}}_q.$

Definition 9.

Let $q=p^m$ and $U_0\subseteq {\mathbb{F}}_q$ be an ${\mathbb{F}}_p$-vector space of dimension $n-k.$ Then ${\mathbb{F}}_q$ can be partitioned into $U_0,\dots ,U_{p^k-1},$ where each $U_i$ is of the form ${\zeta }_i+U_0$ with ${\zeta }_i\in {\mathbb{F}}_q.$ For a p-linearized polynomial $M\in {\mathbb{F}}_q\left[x\right]$ and a sequence ${\left\{a_i\right\}}_{0\le i\le p^k-1}$ in ${\mathbb{F}}_q,$ we can define an M-affine mapping P of index $p^k$ with the subspace $U_0$ by

$$f\left(x\right)=M\left(x\right)+a_i\mathrm{if}x\in U_i,i=0,\dots ,p^k-1.$$

(9)

We observe that each branch function ia sn affine polynomial and thus we can obtain the inverse of $f\left(x\right)$ by putting together the inverses of branch functions. The following result provides an implicit way of obtaining the inverse of a PP based on its additive decomposition as $h\left(L\right(x\left)\right)+M\left(x\right).$

Theorem 42.

[76] [Theorem 5.8] Let L be a monic p-linearized polynomial that divides $x^q-x,$ set $p^{n-k}=deg\left(L\right)$ and $U_L=\{z\in {\overline{\mathbb{F}}}_q\mid L\left(z\right)=0\}.$ Let ${\left\{{\zeta }_i\right\}}_{0\le i\le p^k-1}$ be any complete set of representatives for the quotient ${\mathbb{F}}_q/U_L.$ Suppose that $f\left(x\right)=h\left(L\left(x\right)\right)+M\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ is a PP of ${\mathbb{F}}_q,$ where M is a p-linearized polynomial. Then $f_0\left(x\right)=h_0\left(L_0\left(x\right)\right)+M_0\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ is the inverse PP of $f\left(x\right)$ over ${\mathbb{F}}_q,$ where $h_0\left(x\right),L_0\left(x\right)$ and $M_0\left(x\right)$ are given as follows:

(i) $L_0\left(x\right)={\Pi }_{u\in M\left(U_L\right)}(x-u);$

(ii) $M_0\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ is the unique p-linearized polynomial of degree at most $p^{n-k}-1$ such that $M_0\left(M\left(u\right)\right)=u$ for every $u\in U_L;$

(iii) $h_0\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ is the unique polynomial of degree at most $p^k-1$ such that $h_0\left(L_0\left(f\left({\zeta }_i\right)\right)\right)+M_0\left(f\left({\zeta }_i\right)\right)={\zeta }_i.$

In particular, the additive indices of $f\left(x\right)$ and $f_0\left(x\right)$ coincide.

An application of the local criterion and the local method allows us to present the following generalization of Lemma 5, which also unifies Theorems 33 and 34. We omit the detailed proof as it closely follows the arguments in Theorem 33.

Theorem 43.

Let $g\left(x\right),h\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ be polynomials, and let $\phi \left(x\right),\psi \left(x\right)$ be additive polynomials over ${\mathbb{F}}_{q^n}$ such that $h\left(\psi \left({\mathbb{F}}_{q^n}\right)\right)\subseteq {\mathbb{F}}_q\setminus \left\{0\right\}.$ Assume that

$$f\left(x\right)=h\left(\psi \right(x\left)\right)\phi \left(x\right)+g\left(\psi \right(x\left)\right)$$

is a polynomial over ${\mathbb{F}}_{q^n}.$ If there exists a polynomial $\overline{\psi }\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ such that $\overline{\psi }\left(x\right)\circ f\left(x\right)=\psi \left(x\right),$ then $f\left(x\right)$ permutes ${\mathbb{F}}_{q^n}$ if and only if $ker\left(\psi \right)\cap ker\left(\phi \right)=\left\{0\right\}.$

Remark 9.

Based on Lemma 7, the condition in Theorem 38 that $\overline{f}\left(x\right)=h\left(x\right)\phi \left(x\right)+\overline{\psi }\left(g\left(x\right)\right)$ is a bijection from $\psi \left({\mathbb{F}}_{q^n}\right)$ to $\overline{\psi }\left({\mathbb{F}}_{q^n}\right)$ implies a key relationship among the maps. Specifically, for all $x\in \mathbb{F}{q}^n$, we have the functional identity

$$\psi \left(x\right)={\overline{f}}^{-1}\left(x\right)\circ \overline{\psi }\left(x\right)\circ f\left(x\right),$$

where ${\overline{f}}^{-1}\left(x\right)$ is the inverse of $\overline{f}{\left(x\right)|}_{\psi \left({\mathbb{F}}_{q^n}\right)}$. Furthermore, by Theorem 11, $f\left(x\right)$ is a permutation polynomial over ${\mathbb{F}}_{q^n}$ if and only if for every $s\in \psi \left({\mathbb{F}}_{q^n}\right)$, the function $h\left(\psi \right(x\left)\right)\phi \left(x\right)$ is injective on the preimage set ${\psi }^{-1}\left(s\right)$, and this injectivity condition is equivalent to $ker\left(\psi \right)\cap ker\left(\phi \right)=\left\{0\right\}.$

Employing the local method, the authors have fully solved the construction problem of the compositional inverse of permutation polynomials in Theorem 43. This method avoids utilizing the direct-sum decomposition of finite fields, and the corresponding research findings are in preparation for publication.

3.6.2. Inverses of the Permutation Polynomials $g\left(\psi \left(x\right)\right)+{\sum }_{i=1}^r({\phi }_i\left(x\right)+{\delta }_i)h_i\left(\psi \left(x\right)\right)$ with $r>1$

We consider the following generalized form of $h\left(x\right)+g\left(\lambda \right(x\left)\right),$ which is given by

$$f\left(x\right)=g\left(\psi \left(x\right)\right)+{\displaystyle \sum _{i=1}^r}({\phi }_i\left(x\right)+{\delta }_i)h_i\left(\psi \left(x\right)\right)$$

Yuan [35] generalized Theorem 38 and proved the following theorem, which gives a uniform treatment of some earlier constructions of permutation polynomials and also new permutation polynomials.

Theorem 44.

[35] [Theorem 3.1 ] Let q be a prime power, and let $r\ge 1$ and $n\ge 1$ be positive integers. Let $B\left(x\right),L_1\left(x\right),\dots ,L_r\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ be q-polynomials, $g\left(x\right)\in {\mathbb{F}}_{q^n}$, $h_1\left(x\right),\dots ,h_r\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ and ${\delta }_1,\dots ,{\delta }_r\in {\mathbb{F}}_{q^n}$ such that $B\left({\delta }_i\right)\in {\mathbb{F}}_q$ and $h_i\left(B\left({\mathbb{F}}_{q^n}\right)\right)\subseteq {\mathbb{F}}_q$. Then

$$f\left(x\right)=g\left(B\left(x\right)\right)+{\displaystyle \sum _{i=1}^r}\left(L_i\left(x\right)+{\delta }_i\right)h_i\left(B\left(x\right)\right)$$

is a permutation polynomial of ${\mathbb{F}}_{q^n}$ if and only if

(1) $B\left(g\left(x\right)\right)+{\sum }_{i=1}^r\left(L_i\left(x\right)+B\left({\delta }_i\right)\right)h_i\left(x\right)$ permutes $B\left({\mathbb{F}}_{q^n}\right)$; and

(2) for any $y\in B\left({\mathbb{F}}_{q^n}\right)$, ${\sum }_{i=1}^r{L}_i\left(x\right)h_i\left(y\right)$ permutes $ker\left(B\right)$.

We generalize the result by the local criterion and get the following result.

Theorem 45.

Let q be a prime power and $n,r$ be positive integers. For $1\le i\le r,$ let $g\left(x\right),h_i\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right],$${\phi }_i\left(x\right),\psi \left(x\right)$ be additive polynomials. For ${\delta }_1,\dots ,{\delta }_r\in {\mathbb{F}}_{q^n},$ if there exsits a polynomial $\overline{\psi }\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ such that $\overline{\psi }\left(x\right)\circ f\left(x\right)=\psi \left(x\right)$, then the polynomial

$$f\left(x\right)=g\left(\psi \left(x\right)\right)+{\displaystyle \sum _{i=1}^r}({\phi }_i\left(x\right)+{\delta }_i)h_i\left(\psi \left(x\right)\right)$$

permutes ${\mathbb{F}}_{q^n}$ if and only if for any $y\in \psi \left({\mathbb{F}}_{q^n}\right),$${\phi }_y={\sum }_{i=1}^r{\phi }_i\left(x\right)h_i\left(y\right)$ permutes $ker\left(\psi \right).$

Remark 10.

The compositional inverse computation approach established in Theorem 43 is applicable to the permutation polynomial $f\left(x\right)$ constructed in Theorem 45. This enables us to derive an explicit preimage formula for arbitrary elements under $f\left(x\right)$, which further extends the conclusion of Theorem 3.10 in [71].

For permutation polynomials of this category, we omit separate listings if only preimage representations rather than closed-form compositional inverses are obtainable. Currently, relevant findings on compositional inverses of such permutation polynomials remain limited, and all existing achievements are established via the direct sum decomposition method [9,69,71]. The local method proposed in this subsection can universally verify all reported results in this field.

Accordingly, this subsection does not enumerate all available cases exhaustively. we briefly review the research progress on compositional inverses of this type of permutation polynomials in a specific subclass, and adopt the local method to provide new proofs for the latest published results, while other cases are not discussed separately (they can be proved by similar methods). Based on the above considerations, we focus on a special class concerning compositional inverses and provide new proofs for the most recent results.

We first revisit the explicit inverse formula given by Coulter and Henderson for a family of bilinear permutation polynomials over finite fields of characteristic two (Theorem 46).

Theorem 46.

[77] [Theorem 1] Let $q=2^k$, let n be odd, and let $\alpha \in {\mathbb{F}}_q\setminus \{0,1\}$. Then the inverse of the PP $F_1\left(x\right)=(\alpha +1)x^2+x{Tr}_{q^n/q}\left(x\right)$ over ${\mathbb{F}}_{q^n}$ is as follows.

For $1\le i\le k-1$, let

$$C_i={\displaystyle \frac{{\alpha }^{2^{k-1}+2^{k-1-i}-1}+1}{\alpha +1}}.$$

Set

$$A_{\alpha }\left(x\right)=C_{k-1}\left(x^{2^{nk-1}}+{\alpha }^{2^{k-1}-1}{Tr}_{q^n/q}{\left(x\right)}^{2^{k-1}}\right)$$

and

$$B_{\alpha }\left(x\right)={\displaystyle \sum _{i=1}^{k-1}}{C}_i{Tr}_{q^n/q}{\left(x\right)}^{2^{k-1}-2^{k-1-i}}\left({\displaystyle \sum _{j=1}^{(n-1)/2}}{\left(x{Tr}_{q^n/q}\left(x\right)+x^2\right)}^{2^{jk-2-i}}\right).$$

Then $g\left(x\right)=A_{\alpha }\left(x\right)+B_{\alpha }\left(x\right)$ is the inverse of $F_1\left(x\right)$ over ${\mathbb{F}}_{q^n}$.

Laigle-Chapuy [78] generalized the construction of the permutation in Theorem 46 in a recursive manner to obtain the permutation (47).

Theorem 47.

[78] Let q be a power of 2 and n be odd. Assume $xL\left(x\right)$ is a bilinear permutation polynomial over ${\mathbb{F}}_q$ for a linearized polynomial $L\left(x\right)\in {\mathbb{F}}_q\left[x\right]$. Then the polynomial

$$F_2\left(x\right)=x(L\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+a{\mathrm{Tr}}_{q^n/q}\left(x\right)+ax)$$

As observed by Wu [70], Theorem 47 indeed generalizes Theorem 46: by setting $L\left(x\right)=x$ and $a={\displaystyle \frac{\alpha +1}{\alpha }}$ for some $\alpha \in {\mathbb{F}}_q\setminus {\mathbb{F}}_2$ in $F_2$, one easily verifies that

$$\alpha F_2\left(x\right)=\alpha x\left({\displaystyle \frac{1}{\alpha }}{\mathrm{Tr}}_{q^n/q}\left(x\right)+\left(1+{\displaystyle \frac{1}{\alpha }}\right)x\right)=F_1\left(x\right),$$

where $F_1\left(x\right)$ is the permutation polynomial in Theorem 46. Wu [70] also gave the compositional inverse of $F_2\left(x\right)$ by the decomposition method.

Theorem 48.

[70] Use the same notations as in Theorem 2.2 and let $q=2^m$ for a positive integer m. Assume the compositional inverse of $xL\left(x\right)$ is $g\left(x\right)\in {\mathbb{F}}_q\left[x\right]$. Then

$$\begin{array}{cc}\hfill F_2^{-1}\left(x\right)=& a^{2^{m-1}-1}{x}^{2^{nm-1}}+\left(g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+a^{2^{m-1}-1}{\displaystyle \sum _{k=1}^{{\displaystyle \frac{n-1}{2}}}}{x}^{2^{(2k-1)m-1}}\right)\hfill \\ \hfill & ·{\left({\displaystyle \frac{{\mathrm{Tr}}_{q^n/q}\left(x\right)}{g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}}+ag\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)}^{q-1}\hfill \\ \hfill & +{\displaystyle \sum _{j=0}^{m-2}}{a}^{2^j-1}{\left({\displaystyle \frac{{\mathrm{Tr}}_{q^n/q}\left(x\right)}{g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}}+ag\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)}^{2^m-2^{j+1}}{\left({\displaystyle \sum _{k=0}^{{\displaystyle \frac{n-1}{2}}}}{x}^{q^{2^k}}\right)}^{2^j}.\hfill \end{array}$$

Tuxanidy and Wang, building upon their results in [71](which were also obtained via the direct sum decomposition), further generalized the conclusion of Wu [70].

Theorem 49.

[69] [Theorem 3.3, Corollary 4.3] Let $q=p^m$ be a power of a prime number p, let $G\in {\mathbb{F}}_q\left[x\right]$, and let $n,r$ be positive integers such that $d:=(m,r)=(mn,r),p\nmid n$ and $(n,p^{(m,r)}-1)=1$. Then

$$F_3\left(x\right)=a{x}^{p^r}+x\left(G\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)-a{\mathrm{Tr}}_{q^n/q}{\left(x\right)}^{p^r-1}\right)$$

is a complete permutation polynomial over ${\mathbb{F}}_{q^n}$ for each $a\in {\mathbb{F}}_q^{*}$ if and only if $xG\left(x\right)$ is a complete permutation polynomial over ${\mathbb{F}}_q$. Moreover, If the polynomial f in Theorem 3.3 is a permutation polynomial over ${\mathbb{F}}_{q^n}$, then its compositional inverse over ${\mathbb{F}}_{q^n}$ is given by

$$\begin{array}{cc}\hfill F_3^{-1}\left(x\right)=& \left(1-C{\left(x\right)}^{q-1}\right){\left({\displaystyle \frac{x}{a}}\right)}^{{\displaystyle \frac{q^n}{p^r}}}+C{\left(x\right)}^{q-1}{\left({\mathrm{N}}_{q/p^d}\left(C\left(x\right)\right)-1\right)}^{p^d-1}{\displaystyle \frac{{\mathrm{N}}_{q/p^d}{\left(C\left(x\right)\right)}^n}{1-{\mathrm{N}}_{q/p^d}{\left(C\left(x\right)\right)}^n}}\hfill \\ \hfill & ·{\displaystyle \sum _{i=0}^{{\displaystyle \frac{mn}{d}}-1}}C{\left(x\right)}^{-{\displaystyle \frac{p^{(i+1)r}-1}{p^r-1}}}{\left(a^{-1}x\right)}^{p^{ir}}+\left(1-{\left({\mathrm{N}}_{q/p^d}\left(C\left(x\right)\right)-1\right)}^{p^{d-1}}\right)\hfill \\ \hfill & ·n^{-1}\left(f^{-1}\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+{\displaystyle \sum _{j=0}^{{\displaystyle \frac{m}{d}}-1}}C{\left(x\right)}^{-{\displaystyle \frac{p^{(j+1)r}-1}{p^r-1}}}{\left(a^{-1}{\displaystyle \sum _{k=1}^{n-1}}k{x}^{p^{mk{\displaystyle \frac{r}{d}}}}\right)}^{p^{jr}}\right),\hfill \end{array}$$

where $C\left(x\right):={\overline{f}}^{-1}{\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}^{p^r-1}-a^{-1}G\left({\overline{f}}^{-1}\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)\in {\mathbb{F}}_q\left[x\right],$ and ${\overline{f}}^{-1}$ is the compositional inverse of $xG\left(x\right)$ over ${\mathbb{F}}_q.$

Since the proofs of Theorem 47, Theorem 48, and Theorem 49 follow the same line of reasoning, we only present a detailed proof of Theorem 48 by way of illustration via the Local method. We adopt the following notation . For notational convenience, we denote the compositional inverse of the restricted mapping ${f|}_s$ by ${f|}_s^{-1}$, that is, ${f|}_s^{-1}:={\left(f{|}_s\right)}^{-1}$.

Proof of Theorem 48

Firstly, it is obvious that

$$\begin{array}{cc}\hfill {\mathrm{Tr}}_{q^n/q}\left(F_2\left(x\right)\right)& ={\mathrm{Tr}}_{q^n/q}\left(xL\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+ax{\mathrm{Tr}}_{q^n/q}\left(x\right)+a{x}^2\right)\hfill \\ \hfill & ={\mathrm{Tr}}_{q^n/q}\left(L\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)\hfill \\ \hfill & =xL\left(x\right)\circ {\mathrm{Tr}}_{q^n/q}\left(x\right),\hfill \end{array}$$

which implies that

$$g\left(x\right)\circ {\mathrm{Tr}}_{q^n/q}\left(x\right)\circ F_2\left(x\right)={\mathrm{Tr}}_{q^n/q}\left(x\right),$$

(10)

where $g\left(x\right)$ is the inverse of $xL\left(x\right)$ over ${\mathbb{F}}_{q^n}.$ Moreover, we have

$$\begin{array}{cc}\hfill \left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\circ F_2\left(x\right)& =xL\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+ax{\mathrm{Tr}}_{q^n/q}\left(x\right)+a{x}^2+{\mathrm{Tr}}_{q^n/q}\left(x\right)L\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\hfill \\ \hfill & =a{\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}^2+\left(L\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+a{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right).\hfill \end{array}$$

(11)

Define the set $S=\left\{x|L\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+a{\mathrm{Tr}}_{q^n/q}\left(x\right)\ne 0\right\}$ and $\overline{S}={\mathbb{F}}_{q^n}-S.$

If $L\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+a{\mathrm{Tr}}_{q^n/q}\left(x\right)=0$, then $F_2\left(x\right)=a{x}^2,$ this yields

$$F_2{\mid }_S^{-1}={\left(a^{-1}x\right)}^{q^n/2}.$$

(12)

If $L\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+a{\mathrm{Tr}}_{q^n/q}\left(x\right)\ne 0$, we define $A\left(x\right)=a^{-1}L\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+{\mathrm{Tr}}_{q^n/q}\left(x\right).$ One can verify that $A{\left(x\right)}^q=A\left(x\right)$. Substituting the identity $g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)L\left(g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)={\mathrm{Tr}}_{q^n/q}\left(x\right)$ into the expression of $A\left(x\right)$ , we get

$$\begin{array}{cc}\hfill A\left(x\right)& =a^{-1}L\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+{\mathrm{Tr}}_{q^n/q}\left(x\right)\hfill \\ \hfill & =\left(a^{-1}L\left(g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)+g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)\circ F_2\left(x\right)\hfill \\ \hfill & =\left(a^{-1}{\displaystyle \frac{{\mathrm{Tr}}_{q^n/q}\left(x\right)}{g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}}+g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)\circ F_2\left(x\right).\hfill \end{array}$$

(13)

Then by (11)

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{j=0}^{m-1}}A{\left(x\right)}^{-(2^{j+1}-1)}{\displaystyle \sum _{k=0}^{{\displaystyle \frac{n-1}{2}}}}{\left(\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\circ F_2\left(x\right)\right)}^{2^{2km+j}}\hfill \\ \hfill =& {\displaystyle \sum _{j=0}^{m-1}}A{\left(x\right)}^{-(2^{j+1}-1)}{\displaystyle \sum _{k=0}^{{\displaystyle \frac{n-1}{2}}}}{\left({\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}^2+A\left(x\right)\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)}^{2^{2km+j}}\hfill \\ \hfill =& {\displaystyle \sum _{j=0}^{m-1}}A{\left(x\right)}^{-(2^{j+1}-1)}{\displaystyle \sum _{k=0}^{{\displaystyle \frac{n-1}{2}}}}{\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}^{2^{2km+j+1}}+{\displaystyle \sum _{j=0}^{m-1}}A{\left(x\right)}^{-(2^j-1)}{\displaystyle \sum _{k=0}^{{\displaystyle \frac{n-1}{2}}}}{\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}^{2^{2km+j}}\hfill \\ \hfill =& {\displaystyle \sum _{k=0}^{{\displaystyle \frac{n-1}{2}}}}{\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}^{2^{(2k+1)m}}+{\displaystyle \sum _{k=0}^{{\displaystyle \frac{n-1}{2}}}}{\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}^{2^{2km}}\hfill \\ \hfill =& {\mathrm{Tr}}_{q^n/q}\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)+x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\hfill \\ \hfill =& x+{\mathrm{Tr}}_{q^n/q}\left(x\right).\hfill \end{array}$$

Hence, we have

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{j=0}^{m-1}}{\left(\left(a^{-1}{\displaystyle \frac{{\mathrm{Tr}}_{q^n/q}\left(x\right)}{g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}}+g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)\circ F_2\left(x\right)\right)}^{-(2^{j+1}-1)}{\displaystyle \sum _{k=0}^{{\displaystyle \frac{n-1}{2}}}}{\left(\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\circ F_2\left(x\right)\right)}^{2^{2km+j}}\hfill \\ \hfill & +g\left(x\right)\circ {\mathrm{Tr}}_{q^n/q}\left(x\right)\circ F_2\left(x\right)=x.\hfill \end{array}$$

It follows from Theorem 14 that

$$F_2{\mid }_{\overline{S}}^{-1}={\displaystyle \sum _{j=0}^{m-1}}{\left(a^{-1}{\displaystyle \frac{{\mathrm{Tr}}_{q^n/q}\left(x\right)}{g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}}+g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right)\right)}^{-(2^{j+1}-1)}{\displaystyle \sum _{k=0}^{{\displaystyle \frac{n-1}{2}}}}{\left(x+{\mathrm{Tr}}_{q^n/q}\left(x\right)\right)}^{2^{2km+j}}+g\left({\mathrm{Tr}}_{q^n/q}\left(x\right)\right).$$

(14)

Combining equations (12) and (14), we arrive at the desired conclusion.

In the proof of Theorem 48 constructed using the local method, it becomes evident that we do not need to rely on direct sum decompositions to solve for the compositional inverse of $F_2\left(x\right).$

3.7. Inverses of Permutations of the Form $h\left(x\right)+{\sum }_{i=1}^l{u}_i{\left(L_i\left(x\right)\right)}^{m_i}$, where $L_i$ Are Additive Polynomials

In this subsection we consider the problem of explicitly constructing the compositional inverse of permutation polynomials of the form

$$f\left(x\right)=h\left(x\right)+{\displaystyle \sum _{i=1}^l}{u}_i{\left(L_i\left(x\right)\right)}^{m_i},$$

where each $L_i$ is a linearized polynomial over ${\mathbb{F}}_{q^n}$ whose image is a 1-dimensional ${\mathbb{F}}_q$-vector space.

As a transitional statement, we first assume that the sets $A_i$ (where the index i runs over a suitable collection) are pairwise orthogonal.

Let $d>1$ be a positive integer, q a prime power satisfying $q\equiv 1(modd)$, and $\omega$ a primitive d-th root of unity over ${\mathbb{F}}_q$. We recall the definition of a family of polynomials ${\left\{A_i\left(x\right)\right\}}_{i=0}^{d-1}$ introduced in [16]:

$$A_i\left(x\right)=x^{q^{d-1}}+{\omega }^i{x}^{q^{d-2}}+\dots +{\omega }^{i(d-1)}x,0\le i\le d-1.$$

This family of polynomials enjoys three fundamental properties that are central to the study of permutation polynomials (PPs) over finite fields. These properties are summarized below:

(i) For each $0\le i\le d-1$, $A_i\left(x\right)$ satisfies $A_i^q\left(x\right)={\omega }^i{A}_i\left(x\right)$.

(ii) For any positive integer m and any indices $0\le i,j\le d-1$, the composition of $A_j\left(x\right)$ with $A_i^m\left(x\right)$ is given by:

$$A_j\circ A_i^m\left(x\right)=\left\{\begin{array}{cc}d{\omega }^{-j}{A}_i^m\left(x\right),\hfill & \mathrm{if}j\equiv im(modd),\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(iii) For any integer $1\le j\le d-1$ and any polynomial $g\left(x\right)\in {\mathbb{F}}_q\left[x\right]$, the composition $A_j\circ g\left(A_0\left(x\right)\right)$ vanishes identically over ${\mathbb{F}}_{q^d}$.

Now let $m_1,\dots ,m_{d-1}$ be positive integers, $u_1,\dots ,u_{d-1}\in {\mathbb{F}}_q$, and $g\left(x\right)\in {\mathbb{F}}_q\left[x\right]$. In [16], Yuan considered polynomials of the form:

$$f\left(x\right)=g\left(A_0\left(x\right)\right)+{\displaystyle \sum _{i=1}^{d-1}}{u}_i{A}_i^{m_i}\left(x\right).$$

The permutation behavior of $f\left(x\right)$ can be illustrated by the following commutative diagram:

If $gcd(m_i,q-1)=1$, then the compositional inverse of the monomial $d{u}_i{\omega }^{-j}{x}^{m_i}$ is ${\left(d{u}_i{\omega }^{-j}\right)}^{-r_i}{x}^{r_i}$, where $r_i$ is the positive integer satisfying $m_i{r}_i\equiv 1(modd(q-1))$. This gives rise to the following commutative diagram:

where ${\phi }_i\left(x\right)=A_i\left(x\right)$ for $i=0,1,\dots ,d-1$, ${\psi }_0\left(x\right)=g^{-1}(A_0\left(x\right)/d)$, and ${\psi }_i\left(x\right)={\left(d{u}_i{\omega }^{-j}\right)}^{-r_i}{A}_j^{r_i}\left(x\right)$ for $i=1,\dots ,d-1$.

We also recall the standard decomposition identity:

$$x={\displaystyle \frac{1}{d}}{\displaystyle \sum _{i=0}^{d-1}}{\omega }^i{A}_i\left(x\right)={\displaystyle \frac{1}{d}}{\displaystyle \sum _{i=0}^{d-1}}{\omega }^i{\phi }_i\left(x\right).$$

Combining these commutative diagrams and the decomposition identity, Yuan [16] established the following result.

Theorem 50.

[16] [Theorem 4.1] The polynomial $f\left(x\right)$ is a PP over ${\mathbb{F}}_q$ if and only if the following conditions hold:

(1) $\left\{0\right\}\cup \{i{m}_i,1\le i\le d-1\}$ is a complete residue modulo d;

(2) $u_1,\dots ,u_{d-1}\in {\mathbb{F}}_q^{*}$;

(3) $gcd(m_1\dots m_{d-1},q-1)=1$;

(4) $g\left(x\right)$ is a PP over ${\mathbb{F}}_q$.

Furthermore, if $f\left(x\right)$ is a PP over ${\mathbb{F}}_q,$ let $r_i$ be positive integers with $m_i{r}_i\equiv 1(modq-1)(1\le i\le d-1)$ and let $g^{-1}\left(x\right)$ be the compositional inverse of $g\left(x\right).$ Then the compositional inverse of $f\left(x\right)$ is

$$f^{-1}\left(x\right)={\displaystyle \frac{1}{d}}\left(g^{-1}\left(A_0\left(x\right)/d\right)+{\displaystyle \sum _{\begin{array}{c}i=1\\ j\equiv i{m}_i(modd)\end{array}}^{d-1}}{\omega }^i{\left(d{u}_i{\omega }^{-j}\right)}^{-r_i}{A}_j{\left(x\right)}^{r_i}\right).$$

Remark 11.

Let $A_i\left(x\right)$ be as defined in Theorem 50. In [14] [Theorem 5.3], Yuan et al. gave a full characterization of permutation polynomials over ${\mathbb{F}}_{q^d}$ of the form ${\sum }_{i=0}^{d-1}{u}_i{A}_i^{m_i}\left(x\right)$ with $u_i\in {\mathbb{F}}_q$. This characterization is a special case of Theorem 50.

Next, we turn to cases where the orthogonality condition is not satisfied. The first such example was given by Wu and Yuan [79], who studied a class of PPs of the form $f\left(x\right)={(x^{q^2}+\delta x^q+{\delta }^{1+q}x)}^m+L\left(x\right)$ and their inverses over ${\mathbb{F}}_{q^3}$ using the local method.

Theorem 51.

[79] [Theorem 2] Let q be a prime power and m be a positive integer. Assume that $a,c,\delta \in {\mathbb{F}}_{q^3}$ with ${\delta }^{q^2+q+1}=1$, $a{\delta }^{1+q^2}-c\ne 0$ and ${\delta }^{m{q}^2+mq}+{\delta }^{mq+1}{(a{\delta }^{1+q^2}-c)}^{q^2-q}+{\delta }^{1+q^2}{(a{\delta }^{1+q^2}-c)}^{q^2-1}=0.$ Then the polynomial

$$f\left(x\right)={(x^{q^2}+\delta x^q+{\delta }^{1+q^2}x)}^m+a{x}^{q^2}+a\delta x^q+cx$$

is a permutation polynomial over ${\mathbb{F}}_{q^3}$ if and only if $A\ne 0$ and the compositional inverse of $f\left(x\right)$ is

$$f^{-1}\left(x\right)={(c-a{\delta }^{1+q^2})}^{-1}\left(x-A^{-m}{(x^{q^2}+B{x}^q+Dx)}^m-a{A}^{-1}(x^{q^2}+B{x}^q+Dx)\right),$$

where $B=\delta {(a{\delta }^{1+q^2}-c)}^{q^2-q}$, $D={\delta }^{1+q^2}{(a{\delta }^{1+q^2}-c)}^{q^2-1}$ and $A=c^{q^2}+B{a}^q{\delta }^q+Da.$

In this case, ${\psi }_1\left(x\right)=x^{q^2}+B{x}^q+Dx,$${\phi }_1\left(x\right)=x^{q^2}+\delta x^q+{\delta }^{1+q^2}x,$${\psi }_2\left(x\right)=x,$ and ${\phi }_2\left(x\right)=f\left(x\right)$. Moreover, ${\phi }_2\left(x\right)={\phi }_1^n\left(x\right)+a{\phi }_1\left(x\right)+(c-a{\delta }^{q^2+1})x$.

The second example concerns permutation polynomials over ${\mathbb{F}}_{q^2}$ of the form

$$f\left(x\right)={(x^q+ax)}^{m_1}+u{(x^q+bx)}^{m_2}.$$

By direct computation and under the assumption that $gcd(m_1{m}_2,q-1)=1$ (which will appear as a necessary and sufficient condition in the theorem below), Yuan[14] obtain the following two commutative diagrams:

where

$${\psi }_1\left(x\right)=x^q-a^{-m_1}x,{\psi }_2\left(x\right)=x^q-u^{q-1}{b}^{-m_2}x,$$

$${\phi }_1\left(x\right)=(u^q{b}^{-m_2}-a^{-m_1}u){(x^q+bx)}^{m_2},{\phi }_2\left(x\right)=(a^{-m_1}-b^{-m_2}{u}^{q-1}){(x^q+ax)}^{m_1}.$$

If $gcd(m_1{m}_2,q-1)=1$, then there exist positive integers $r_1,r_2,s_1,s_2$ such that

$$m_1{r}_1\equiv 1+s_1(q-1)(mod{q}^2-1),m_2{r}_2\equiv 1+s_2(q-1)(mod{q}^2-1).$$

It follows that

$${\left({\displaystyle \frac{{\phi }_1\left(x\right)}{(u^q{b}^{-m_2}-a^{-m_1}u)}}\right)}^{r_2}={(x^q+bx)}^{m_2{r}_2}=b^{-s_2}(x^q+bx),$$

$${\left({\displaystyle \frac{{\phi }_2\left(x\right)}{(a^{-m_1}-b^{-m_2}{u}^{q-1})}}\right)}^{r_1}={(x^q+ax)}^{m_1{r}_1}=a^{-s_1}(x^q+ax),$$

and consequently,

$$x={\displaystyle \frac{1}{b-a}}\left(b^{s_2}{\left({\displaystyle \frac{{\phi }_1\left(x\right)}{(u^q{b}^{-m_2}-a^{-m_1}u)}}\right)}^{r_2}-a^{s_1}{\left({\displaystyle \frac{{\phi }_2\left(x\right)}{(a^{-m_1}-b^{-m_2}{u}^{q-1})}}\right)}^{r_1}\right).$$

Yuan[14] got the following characterization.

Theorem 52.

[14] [Theorem 5.5] Let q be a prime power, and let $a\ne b$ be two elements of order $q+1$ in ${\mathbb{F}}_{q^2}$. Then the polynomial $f\left(x\right)={(x^q+ax)}^{m_1}+u{(x^q+bx)}^{m_2}$, where $u\in {\mathbb{F}}_{q^2}^{*}$, is a permutation polynomial of ${\mathbb{F}}_{q^2}$ if and only if $gcd(m_1{m}_2,q-1)=1$ and $b^{m_2}-a^{m_1}{u}^{q-1}\ne 0$.

Building on the idea of Yuan [14], Wu and Yuan [79] generalized Theorem 52. Here we do not present the proof details, but only give the necessary notation and the result.

We give some notations at first. Let q be a prime power and $n>1$ be a positive integer. For a positive integer i with $1\le i\le n$, assume that $a_i\in {\mathbb{F}}_{q^n}$ with $a_i^{q^{n-1}+q^{n-2}+...+q^2+q+1}=1$ and the polynomial

$$\begin{array}{cc}\hfill A_i\left(x\right)=& x^{q^{n-1}}+a_i{x}^{q^{n-2}}+a_i^{1+q^{n-1}}{x}^{q^{n-3}}+...+\hfill \\ \hfill & a_i^{1+q^{n-1}+q^{n-2}+...+q^{n-j}}{x}^{q^{n-j-2}}+...+a_i^{1+q^{n-1}+q^{n-2}+...+q^2}x.\hfill \end{array}$$

Let $m_1,m_2,...,m_n$ be positive integers and $u_2,u_3,...,u_n\in {\mathbb{F}}_{q^n}.$ Put

$$D_1=\left(\begin{array}{cccc}1& u_2& \dots & u_n\\ a_1^{m_{1}q}& u_2^q{a}_2^{m_{2}q}& \dots & u_n^q{a}_n^{m_{n}q}\\ ⋮& ⋮& ⋮& ⋮\\ a_1^{m_1{\sum }_{i=1}^j{q}^i}& u_2^{q^j}{a}_2^{m_2{\sum }_{i=1}^j{q}^i}& \dots & u_n^{q^j}{a}_n^{m_n{\sum }_{i=1}^j{q}^i}\\ ⋮& ⋮& ⋮& ⋮\\ a_1^{m_1{\sum }_{i=1}^{n-1}{q}^i}& u_2^{q^{n-1}}{a}_2^{m_2{\sum }_{i=1}^{n-1}{q}^i}& \dots & u_n^{q^{n-1}}{a}_n^{m_n{\sum }_{i=1}^{n-1}{q}^i}\end{array}\right)$$

$$D_2=\left(\begin{array}{ccccccc}1& a_1& a_1^{1+q^{n-1}}& \dots & a_1^{1+q^{n-1}+q^{n-2}+...+q^{n-j}}& \dots & a_1^{1+q^{n-1}+q^{n-2}+...+q^2}\\ 1& a_2& a_2^{1+q^{n-1}}& \dots & a_2^{1+q^{n-1}+q^{n-2}+...+q^{n-j}}& \dots & a_2^{1+q^{n-1}+q^{n-2}+...+q^2}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 1& a_n& a_n^{1+q^{n-1}}& \dots & a_n^{1+q^{n-1}+q^{n-2}+...+q^{n-j}}& \dots & a_n^{1+q^{n-1}+q^{n-2}+...+q^2}\end{array}\right)$$

Let ${\psi }_1\left(x\right)=x,{\psi }_2\left(x\right)=x^q,{\psi }_3\left(x\right)=x^{q^2},...,{\psi }_n\left(x\right)=x^{q^{n-1}},$ and

$$\left\{\begin{array}{cc}{\phi }_1\left(x\right)\hfill & ={\displaystyle \sum _{i=1}^n}{u}_i{A}_i^{m_i}\left(x\right)=f\left(x\right)\hfill \\ {\phi }_2\left(x\right)\hfill & ={\displaystyle \sum _{i=1}^n}{u}_i^q{a}_i^{m_{i}q}{A}_i^{m_i}\left(x\right)=f{\left(x\right)}^q\hfill \\ \hfill & ⋮\hfill \\ {\phi }_j\left(x\right)\hfill & ={\displaystyle \sum _{i=1}^n}{u}_i^{q^{j-1}}{a}_i^{m_i{q}^{j-1}+m_i{q}^{j-2}+\dots +m_{i}q}{A}_i^{m_i}\left(x\right)=f{\left(x\right)}^{q^{j-1}}\hfill \\ \hfill & ⋮\hfill \\ {\phi }_n\left(x\right)\hfill & ={\displaystyle \sum _{i=1}^n}{u}_i^{q^{n-1}}{a}_i^{m_i{q}^{n-1}+m_i{q}^{n-2}+\dots +m_{i}q}{A}_i^{m_i}\left(x\right)=f{\left(x\right)}^{q^{n-1}}.\hfill \end{array}\right.$$

Then the authors have the following results by applying the local method.

Theorem 53.

[79] [Theorem 3] Let q be a prime power and $n>1$ be a positive integer. And let the notations $D_1$, $D_2$, and $A_i\left(x\right)$ be as above. Assume that $m_1,m_2,...,m_n$ are positive integers and $u_2,u_3,...,u_n\in {\mathbb{F}}_{q^n}.$ Then the polynomial

$$f\left(x\right)=A_1^{m_1}\left(x\right)+{\displaystyle \sum _{i=2}^n}{u}_i{A}_i^{m_i}\left(x\right)$$

is a permutation polynomial over ${\mathbb{F}}_{q^n}$ if and only if $gcd(m_1{m}_2...m_n,q-1)=1$ and the determinant of $D_1{D}_2$ is not $0.$ Moreover, if $f\left(x\right)$ is a PP over ${\mathbb{F}}_{q^n},$ then

$$f^{-1}\left(x\right)={\theta }_n{\left({\lambda }_1\left(x\right),{\lambda }_2\left(x\right),\dots ,{\lambda }_n\left(x\right)\right)}^T,$$

where ${\lambda }_i\left(x\right)=a_i^{-q{s}_i}{\left({\eta }_i{\left(x,x^q,\dots ,x^{q^{n-1}}\right)}^T\right)}^{r_i},$${\eta }_i$ is the $i-$th row of $D_1^{-1},$ and ${\theta }_n$ is the last row vector of $D_2^{-1}.$

Based on Theorem 53, six new classes of permutation polynomials of the form $L\left(x\right)+{\mathrm{Tr}}_{p^{3m}/p^m}{\left(x\right)}^s$ over ${\mathbb{F}}_{q^3}$ are constructed in [80]; we do not list them in detail here.

As we well know, the vector space ${\mathbb{F}}_q^n$ and the finite field ${\mathbb{F}}_{q^n}$ are isomorphic as ${\mathbb{F}}_q$-vector spaces, and therefore every permutation of ${\mathbb{F}}_q^n$ can be equivalently represented as a PP over ${\mathbb{F}}_{q^n}$.

Let $\{u_1,\dots ,u_n\}$ and $\{v_1,\dots ,v_n\}$ be a pair of dual bases of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$, and let $f\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ be a polynomial. Then for any $c\in {\mathbb{F}}_{q^n}$, $f\left(c\right)=u_1{a}_1+u_2{a}_2+\dots +u_n{a}_n$ with $a_i\in {\mathbb{F}}_q,$$1\le i\le n,$ we defined

$$f_i:{\mathbb{F}}_{q^n}\to {\mathbb{F}}_q\mathrm{by}{f}_i\left(c\right)=a_i,1\le i\le n.$$

It is easy to see that $f_i\left(x\right)={\mathrm{Tr}}_{q^n/q}\left(v_{i}f\left(x\right)\right)\in {\mathbb{F}}_{q^n},1\le i\le n$ are well-defined and are uniquely determined by $f\left(x\right)$ and the base $\{v_1,\dots ,v_n\}.$ Moreover, we have

$$f\left(x\right)=u_1{f}_1\left(x\right)+u_2{f}_2\left(x\right)+\dots +u_n{f}_n\left(x\right)=u_1{\mathrm{Tr}}_{q^n/q}\left(v_{1}f\left(x\right)\right)+\dots +{\mathrm{Tr}}_{q^n/q}\left(v_{n}f\left(x\right)\right).$$

Theorem 54.

[81] [Theorem 1.2] If $f\left(x\right)\in {\mathbb{F}}_{q^n}$ is a PP over ${\mathbb{F}}_{q^n}$ and $h_i\left(x\right)\in {\mathbb{F}}_q\left[x\right],1\le i\le n$ are maps from ${\mathbb{F}}_q$ to ${\mathbb{F}}_q,$ then the polynomial

$$F\left(x\right)=a_1{h}_1\left(f_1\left(x\right)\right)+\dots a_n{h}_n\left(f_n\left(x\right)\right),\mathrm{with}{a}_i\in {\mathbb{F}}_{q^n},1\le i\le n,$$

where $f_i\left(x\right)={\mathrm{Tr}}_{q^n/q}\left(v_{i}f\left(x\right)\right)\in {\mathbb{F}}_{q^n},1\le i\le n$ are defined above, is a PP over ${\mathbb{F}}_{q^n}$ if and only if the following two conditions hold:

(i) $\{a_1,\dots ,a_n\}$ is a base of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q;$

(ii) $h_i\in {\mathbb{F}}_q\left[x\right]$ are PPs over ${\mathbb{F}}_q.$

Moreover, if $f^{-1}\left(x\right)$ and $h_i^{-1}\left(x\right)$ are the compositional inverses of $f\left(x\right)$ and $h_i\left(x\right),1\le i\le n,$ respectively, and $\{b_1,\dots ,b_n\}$ is the dual base of $\{a_1,\dots ,a_n\}$, then we have

$$F^{-1}\left(x\right)=f^{-1}\left({\displaystyle \sum _{i=1}^n}{u}_i{h}_i^{-1}\left({\mathrm{Tr}}_{q^n/q}\left(b_{i}x\right)\right)\right).$$

As a consequence of Theorem 54, Yuan [81] gave the following results.

Corollary 5.

[81] [Corollary 1.1 ]Let ${\theta }_1,{\theta }_2,\dots ,{\theta }_n$ be n elements of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$, and let

$$F\left(x\right)=a_1{\mathrm{Tr}}_{q^n/q}{\left({\theta }_{1}x\right)}^{m_1}+\dots +a_n{\mathrm{Tr}}_{q^n/q}{\left({\theta }_{n}x\right)}^{m_n},a_i\in {\mathbb{F}}_{q^n},m_i\in \mathbb{N}.$$

Then $F\left(x\right)$ is a $PP$ over ${\mathbb{F}}_{q^n}$ if and only if the following three conditions hold

(i) $gcd(m_1\dots m_n,q-1)=1$,

(ii) $a_1,a_2,\dots ,a_n$ is a basis of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$.

(iii) ${\theta }_1,{\theta }_2,\dots ,{\theta }_n$ is a basis of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$.

Remark 12.

The following characterization of linear permutation polynomials over finite fields is established in [82].

(1) Let $\{{\omega }_1,{\omega }_2,\dots ,{\omega }_n\}$ be any given basis of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$, and let $L\left(x\right)={\sum }_{i=0}^{n-1}{a}_i{x}^{q^i}$ be a linear polynomial over ${\mathbb{F}}_{q^n}$. Then there exist n elements ${\theta }_1,{\theta }_2,\dots ,{\theta }_n\in {\mathbb{F}}_{q^n}$ such that

$$L\left(x\right)=\mathrm{Tr}\left({\theta }_{1}x\right){\omega }_1+\dots +\mathrm{Tr}\left({\theta }_{n}x\right){\omega }_n.$$

Moreover, $L\left(x\right)$ is a permutation polynomial if and only if $\{{\theta }_1,{\theta }_2,\dots ,{\theta }_n\}$ forms a basis of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$. There are exactly $(q^n-1)(q^n-q)\dots (q^n-q^{n-1})$ distinct linear permutation polynomials, all of which admit the above representation with a basis $\{{\theta }_1,{\theta }_2,\dots ,{\theta }_n\}$.

(2) Let $\{{\theta }_1,{\theta }_2,\dots ,{\theta }_n\}$ be any given basis of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$, and let

$$L\left(x\right)=\mathrm{Tr}\left({\theta }_{1}x\right){\omega }_1+\dots +\mathrm{Tr}\left({\theta }_{n}x\right){\omega }_n,{\omega }_i\in {\mathbb{F}}_{q^n}.$$

Then $L\left(x\right)$ is a permutation polynomial if and only if $\{{\omega }_1,{\omega }_2,\dots ,{\omega }_n\}$ is a basis of ${\mathbb{F}}_{q^n}$ over ${\mathbb{F}}_q$.

With the aid of this result, one can observe that Theorem 54 actually generalizes the results given in Theorem 50, 51, Theorem 52 and Theorem 53.

In [36], Charpin and Kyureghyan characterized and constructed permutation polynomials of the form $F\left(x\right)=G\left(x\right)+\gamma \mathrm{Tr}\left(H\right(x\left)\right),$ where $G\left(x\right),H\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$. Moreover, they proposed the following problem.

Problem 1.

[36] [Open Problem 1] Characterize a class of permutation polynomials of type $G\left(x\right)+\gamma \mathrm{Tr}\left(H\right(x\left)\right)$, where $G\left(x\right)$ is neither a permutation nor a linearized polynomial.

Theorem 55.

[81] [Theorem 4.1] If $H\left(x\right)\in {\mathbb{F}}_{q^n}\left[x\right]$ is a polynomial such that $\mathrm{Tr}\left(H\right(x\left)\right)$ is not a constant map, $\gamma \in {\mathbb{F}}_{q^n}$, then there are at least $(q^{n-1}!)q-(q^n-q)\dots (q^n-q^{n-1})$ permutation polynomials of the form

$$u_1{f}_1\left(x\right)+\dots +u_{n-1}{f}_{n-1}\left(x\right)+\gamma \mathrm{Tr}\left(H\left(x\right)\right)$$

such that $G\left(x\right)=u_1{f}_1\left(x\right)+\dots +u_{n-1}{f}_{n-1}\left(x\right)+\gamma \mathrm{Tr}\left(H\left(x\right)\right)$ is neither a permutation nor a linearized polynomial.

In 2008, Charpin and Kyureghyan [83] studied permutation polynomials of the form $G\left(x\right)+\gamma \mathrm{Tr}\left(H\right(x\left)\right)$ over finite fields with even characteristic using linear structures, and obtained six classes of such permutations. They [36] subsequently generalized this work to finite fields with odd characteristic. Meanwhile, in [37], Kyureghyan also constructed large families of permutation polynomials of the form $L\left(x\right)+L\left(\gamma \right)h\left(f\right(x\left)\right)$, where $L\left(x\right)={\sum }_{i=0}^{n-1}{a}_i{x}^{q^i}$ is an ${\mathbb{F}}_q$-linear permutation over ${\mathbb{F}}_{q^n}$.

Theorem 56.

[37] [Theorem 1] Let $f:{\mathbb{F}}_{q^n}\to {\mathbb{F}}_q$, $h:{\mathbb{F}}_q\to {\mathbb{F}}_q$ be arbitrary mappings. Let $L:{\mathbb{F}}_{q^n}\to {\mathbb{F}}_{q^n}$ be an ${\mathbb{F}}_q$-linear permutation of ${\mathbb{F}}_{q^n}$. If $b\in {\mathbb{F}}_q$, and $\gamma \in {\mathbb{F}}_{q^n}^{*}$ is a b-linear translator of f, then

$$G\left(x\right)=L\left(x\right)+L\left(\gamma \right)h\left(f\right(x\left)\right)$$

permutes ${\mathbb{F}}_{q^n}$ if and only if $g\left(x\right)=x+bh\left(x\right)$ permutes ${\mathbb{F}}_q$.

Pang et al. [84] provide the following definition of linear translators, which is somewhat different from [10,36,37].

Definition 10.

[84] [Theorem 3.4] Let $f:{\mathbb{F}}_{q^n}\to {\mathbb{F}}_q$, and $A\left(x\right)$ be an additive permutation polynomial over ${\mathbb{F}}_q$. A non-zero element $\gamma \in {\mathbb{F}}_{q^n}$ is called a $(b,A)$-linear translator of the function f if

$$f(x+u\gamma )-f\left(x\right)=bA\left(u\right)$$

holds for any $x\in {\mathbb{F}}_{q^n},u\in {\mathbb{F}}_q$ and a fixed $b\in {\mathbb{F}}_q$.

Let $A\left(x\right)$ or $A_i\left(x\right)$ be an additive permutation polynomial over ${\mathbb{F}}_q$ in the following two results.

Pang et al. [84] give a generalized result for permutations of ${\mathbb{F}}_{2^n}$ in [85] and [86] based on Definition 10 and give their compositional inverses.

Theorem 57.

[84] [Theorem 3.1 ] Let $f:{\mathbb{F}}_{2^n}\to {\mathbb{F}}_{2^m}$ be an arbitrary mapping, and $L:{\mathbb{F}}_{2^n}\to {\mathbb{F}}_{2^n}$ an ${\mathbb{F}}_{2^m}$-linear permutation of ${\mathbb{F}}_{2^n}$. Let $g:{\mathbb{F}}_{2^m}\to {\mathbb{F}}_{2^m}$ be a permutation of ${\mathbb{F}}_{2^m}$, and $A^{-1}\left(x\right)$ the compositional inverse of $A\left(x\right)$. If $\gamma \in {\mathbb{F}}_{2^n}^{*}$ is a $(b,A)$-linear translator of f, then

$$\varphi \left(x\right)=L\left(x\right)+L\left(\gamma \right)\left(g\left(f\left(x\right)\right)+A^{-1}\left({\displaystyle \frac{f\left(x\right)}{b}}\right)\right)$$

is a permutation polynomial over ${\mathbb{F}}_{2^n}$, and its compositional inverse is

$${\varphi }^{-1}\left(x\right)=L^{-1}\left(x\right)+\gamma A^{-1}\left({\displaystyle \frac{f\left(L^{-1}\left(x\right)\right)}{b}}+{\displaystyle \frac{g^{-1}\left(A^{-1}\left(\frac{f\left(L^{-1}\left(x\right)\right)}{b}\right)\right)}{b}}\right).$$

The commutative diagram for the permutation polynomial $\varphi \left(x\right)$ is as follows:

Using the classic definition of linear translators, Qin and Yan [87] [Theorem 2.1] constructed a class of permutation polynomials of the shape

$$F\left(x\right)=x+{\displaystyle \sum _{i=1}^m}{\gamma }_i{h}_i\left(f_i\left(x\right)\right).$$

Pang et al. [84] show an analogous result with minor modifications and give its compositional inverse.

Theorem 58.

[84] [Theorem 3.3] Let $f_1,f_2,\dots ,f_m:{\mathbb{F}}_{q^n}\to {\mathbb{F}}_q$, $h_1,h_2,\dots ,h_m:{\mathbb{F}}_q\to {\mathbb{F}}_q$. Let $b_1,b_2,\dots ,b_m\in {\mathbb{F}}_q$, and ${\gamma }_1,{\gamma }_2,\dots ,{\gamma }_m\in {\mathbb{F}}_q^{*}$ be linearly independent over ${\mathbb{F}}_q$. If ${\gamma }_i$ is a $(b_i,A_i)$-linear translator of $f_i$ for $i\in \{1,2,\dots ,m\}$, and a $(0,A_j)$-linear translator of $f_j$ when $j\ne i$, then

$$F\left(x\right)=x+{\displaystyle \sum _{i=1}^m}{\gamma }_i{h}_i\left(f_i\left(x\right)\right)$$

is a permutation polynomial over ${\mathbb{F}}_{q^n}$ if and only if $g_i\left(x\right)=x+b_i{A}_i\left(h_i\left(x\right)\right)$ is a permutation polynomial over ${\mathbb{F}}_q$, where $i\in \{1,2,\dots ,m\}$. Moreover, the compositional inverse of $F\left(x\right)$ is given by

$$F^{-1}\left(x\right)=x-{\displaystyle \sum _{i=1}^m}{\gamma }_i{h}_i\left(g_i^{-1}\left(f_i\left(x\right)\right)\right).$$

The commutative diagram for the permutation polynomial $F\left(x\right)$ is as follows:

3.8. Inverses of Piecewise-Constructed Polynomials

The Lagrange interpolation formula is well known for calculating polynomial expressions in a point-to-point manner. For any $f\left(x\right)\in {\mathbb{F}}_q\left[x\right]$, applying the one-variable Lagrange interpolation formula, we can write the polynomial $f\left(x\right)$ as the linear combination

$$f\left(x\right)={\displaystyle \sum _{i=0}^{q-1}}{\alpha }_i\left(1-{(x-a_i)}^{q-1}\right),$$

where $\{a_0,a_1,\dots ,a_{q-1}\}={\mathbb{F}}_q$ and $f\left(a_i\right)={\alpha }_i$. According to the Lagrange interpolation formula, its inverse can be expressed as

$$f^{-1}\left(x\right)=\sum _{c\in {\mathbb{F}}_q}c\left(1-{(x-f\left(c\right))}^{q-1}\right).$$

The basic idea of piecewise constructions of PPs is to partition a finite field into subsets and study the permutation property via the functions’ behavior on these subsets. This idea, summarized in [88] by Cao, Hu and Zha, also applies to finite rings.

The following lemma provides a more general piecewise description of polynomials over finite fields.

Lemma 7.

[88] [Proposition 3] Let $D_1,\dots ,D_m$ be a partition of ${\mathbb{F}}_q$, and let $f_1\left(x\right),\dots ,f_m\left(x\right)\in {\mathbb{F}}_q\left[x\right]$. Define

$$f\left(x\right)={\displaystyle \sum _{i=1}^m}{f}_i\left(x\right)I_{D_i}\left(x\right),$$

(15)

where $I_{D_i}\left(x\right)$ is the characteristic function of $D_i$, i.e., $I_{D_i}\left(x\right)=1$ if $x\in D_i$ and $I_{D_i}\left(x\right)=0$ otherwise. Then $f\left(x\right)$ is a permutation polynomial (PP) of ${\mathbb{F}}_q$ if and only if

(i) $f_i$ is injective on $D_i$ for each $1\le i\le m$; and

(ii) $f_i\left(D_i\right)\cap f_j\left(D_j\right)=\varnothing$ for all $1\le i\ne j\le m$.

From the perspective of construction principles and application purposes, the piecewise method shares similarities with the Lagrange interpolation formula. In this sense, the piecewise method can be viewed as a generalized form of the Lagrange interpolation formula. Specifically, when each $D_i$ is a singleton set, the piecewise expression reduces exactly to the standard Lagrange interpolation formula. Inspired by the lemma above, Zheng et al. [89] present the following piecewise interpolation method for constructing inverses of all PPs of finite fields.

Theorem 59.

[89] [Lemma 2.2] If $f\left(x\right)$ in (15) is a PP of ${\mathbb{F}}_q$, then the compositional inverse of $f\left(x\right)$ over ${\mathbb{F}}_q$ is given by

$$f^{-1}\left(x\right)={\displaystyle \sum _{i=1}^m}{\overline{f}}_i\left(x\right)I_{f_i\left(D_i\right)}\left(x\right),$$

where ${\overline{f}}_i\left(f_i\left(c\right)\right)=c$ for any $c\in D_i,$ and $I_{f_i\left(D_i\right)}\left(x\right)$ is the characteristic function of $f_i\left(D_i\right).$

According to the above lemma, to obtain the compositional inverse of $f\left(x\right)$, we need to accomplish the following two steps: (1) for each i, compute the local inverse map ${\overline{f}}_i$ of $f_i\left(x\right)$ restricted to $D_i$, satisfying ${\overline{f}}_i\left(f_i\left(c\right)\right)=c$ for any $c\in D_i$; and (2) derive an explicit expression for the characteristic function $I_{f_i\left(D_i\right)}\left(x\right)$. We only introduce one typical example of piecewise permutation polynomial along with its compositional inverse based on Theorem 59, which is presented as follows.

Let $\gamma$ be a fixed primitive element of ${\mathbb{F}}_q$ and $\omega ={\gamma }^s$ a fixed primitive l-th root of unity. Let $l\mid (q-1)$ and the set of all nonzero l-th be $C_0.$ Then $C_0$ is a subgroup of ${\mathbb{F}}_q^{*}$ of index $l.$ The elements of the factor group ${\mathbb{F}}_q^{*}\setminus C_0$ are the cyclotomic cosets

$$C_i={\gamma }^i{C}_0,i=0,1,\dots ,l-1.$$

For any $a_0,a_1,\dots ,a_{l-1}\in {\mathbb{F}}_q$ and a positive integer r, the r-th order cyclotomic mapping $f_{a_0,a_1,\dots ,a_{l-1}}^r$ of index l from ${\mathbb{F}}_q$ to itself (see Niederreiter and Winterhof [90] for $r=1$ or Wang [32] for general r ) is defined by

$$f_{a_0,a_1,\dots ,a_{l-1}}^r\left(x\right)=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}x=0;\hfill \\ a_i{x}^r,\hfill & \mathrm{if}x\in C_i,0\le i\le l-1.\hfill \end{array}\right.$$

(16)

Clearly, an r-th order cyclotomic mapping of index l produces a polynomial of the form $x^{r}h\left(x^s\right)$ where $s=(q-1)/l.$ Indeed, the polynomial representation of (16) is given by

$$g\left(x\right)={\displaystyle \frac{1}{l}}{\displaystyle \sum _{j=0}^{l-1}}\left({\displaystyle \sum _{i=0}^{l-1}}{a}_i{\omega }^{-ji}\right)x^{js+r}.$$

(17)

More generally, a simple class of generalized cyclotomic mapping PPs of ${\mathbb{F}}_q$ was defined in [91] as

$$f\left(x\right)={\displaystyle \frac{1}{l}}{\displaystyle \sum _{j=0}^{l-1}}{\displaystyle \sum _{i=0}^{l-1}}{a}_i{\omega }^{-ji}{x}^{js+r_i}.$$

(18)

Several equivalent criteria for f permuting ${\mathbb{F}}_q$ were given in [91], and one is that $f\left(x\right)$ is a PP of ${\mathbb{F}}_q$ if and only if $gcd({\prod }_{i=0}^{l-1}{r}_i,s)=1$ and $\{a_i^s{\omega }^{i{r}_i}:i=0,1,\dots ,l-1\}={\mu }_l.$ In fact, the polynomial defined in (18) can be rewritten as

$$f\left(x\right)=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}x=0;\hfill \\ a_i{x}^{r_i},\hfill & \mathrm{if}x\in C_i,0\le i\le l-1.\hfill \end{array}\right.$$

(19)

Because each branch function is a monomial $a_i{x}^{r_i}$ that maps one coset to another, it is straightforward to find its inverse, which is also a monomial. Putting them together, we obtain the compositional inverse of $f\left(x\right)$.

Theorem 60.

[91,92] The inverse of a generalized cyclotomic mapping PP $f\left(x\right)$ on ${\mathbb{F}}_q$ defined by

$$f\left(x\right)={\displaystyle \frac{1}{l}}{\displaystyle \sum _{j=0}^{l-1}}{\displaystyle \sum _{i=0}^{l-1}}{a}_i{\omega }^{-ji}{x}^{js+r_i}.$$

is given by

$$f^{-1}\left(x\right)={\displaystyle \frac{1}{l}}{\displaystyle \sum _{j=0}^{l-1}}{\displaystyle \sum _{i=0}^{l-1}}{\omega }^{i(t_i-j{r}_i)}{(x/a_i)}^{js+\widetilde{r_i}},$$

where $1\le \widetilde{r_i}<s$ and $r_i\widetilde{r_i}+s{t}_i=1.$

A more generalized notation of coset-wise affine permutation functions and their cycle types were studied in [93]. The most general piecewise construction of permutations using characteristic functions of any partition of a finite field was summarized in [88,94]. Piecewise constructions of inverses of these piecewise PPs were also studied in [89,92,95].

3.9. Coefficients of Inverses of General Permutation Polynomials

In the following, we present two different methods for studying inverse permutation polynomials. The first approach, based on matrix representation, reduces polynomial composition and inversion to matrix multiplication and matrix inversion; the second approach, by contrast, employs the orthogonality of the discrete Fourier transform to express the coefficients of the inverse polynomial as weighted sums of the values of the original polynomial.

3.9.1. Method 1: Matrix Representation

Let ${\mathbb{F}}_q$ be the finite field of order $q=p^n$ where p is a prime number and n is a positive integer. Let $f\left(x\right)={\sum }_{i=0}^{q-1}{a}_i{x}^i$ be a polynomial over ${\mathbb{F}}_q$ with degree at most $q-1$. To compute its composition with another polynomial $g\left(x\right)={\sum }_{i=0}^{q-1}{d}_i{x}^i$, we can either use interpolation to obtain its expression directly, or calculate all the powers $f{\left(x\right)}^i(mod{x}^q-x)$ in the expression $(g\circ f)\left(x\right)={\sum }_{i=0}^{q-1}{d}_i{\left(f\left(x\right)\right)}^i$.

Denote by

$${\left(f\left(x\right)\right)}^k={\displaystyle \sum _{i=0}^{q-1}}{a}_{ik}{x}^i(mod(x^q-x))$$

the k-th power of the polynomial $f\left(x\right)$ for $k=1,2,\dots ,q-1$. Denote by $f_0$ the zero polynomial in ${\mathbb{F}}_q\left[x\right]$. If $f\ne f_0$ we will define ${\left(f\left(x\right)\right)}^0=1$ and $f_0{\left(x\right)}^0=0$.

For any polynomial $f\left(x\right)={\sum }_{i=0}^{q-1}{a}_i{x}^i$ we associate a coefficient vector $v_f$ with it, namely,

$$v_f={(a_0,a_1,\dots ,a_{q-1})}^T.$$

A $q\times q$ matrix associated with $f\left(x\right)\ne f_0\left(x\right)$ is defined by

$$A\left(f\right)=\left[\begin{array}{cccccc}1& a_{01}& a_{02}& \dots & a_{0,q-2}& a_{0,q-1}\\ 0& a_{11}& a_{12}& \dots & a_{1,q-2}& a_{1,q-1}\\ ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& a_{q-2,1}& a_{q-2,2}& \dots & a_{q-2,q-2}& a_{q-2,q-1}\\ 0& a_{q-1,1}& a_{q-1,2}& \dots & a_{q-1,q-2}& a_{q-1,q-1}\end{array}\right],$$

where the k-th column consists of the coefficients of the $(k-1)$-th power of $f\left(x\right)$. With this matrix notation, the composition of polynomials admits a remarkably simple algebraic description.

Theorem 61.

[96] Let $g\left(x\right)={\sum }_{i=0}^{q-1}{d}_i{x}^i\in {\mathbb{F}}_q\left[x\right]$ and $f\left(x\right)={\sum }_{j=0}^{q-1}{a}_j{x}^j\in {\mathbb{F}}_q\left[x\right]$. Let $(g\circ f)\left(x\right)=g\left(f\right(x\left)\right)$ be the composition of g with f. Then

$$A(g\circ f)=A\left(f\right)A\left(g\right).$$

In the special case where f is a permutation polynomial, its compositional inverse also admits a simple expression in terms of this matrix.

Theorem 62.

[96] Let f be a permutation polynomial of ${\mathbb{F}}_q$. Let P be the antidiagonal permutation matrix, i.e. P is defined by $P_{i,(q-i)}=1$ for $i=1,2,\dots ,q$ and zero otherwise. Then

$$A\left(f^{(-1)}\right)={\left(A\left(f\right)\right)}^{-1}=PA\left(f\right)P.$$

3.9.2. Method 2: Discrete Fourier Transform Method

The orthogonality of the discrete Fourier transform over finite fields provides a weighted-sum representation for the coefficients of the inverse polynomial.

Let $f\left(x\right)$ be a permutation polynomial (PP) of ${\mathbb{F}}_q$ and let $Q\left(x\right)={\sum }_{i=0}^{q-2}{b}_i{x}^i$ be its inverse modulo $x^q-x$. For each $0\le n\le q-2$, it is well known that

$$\sum _{s\in {\mathbb{F}}_q}{s}^{q-1-n}Q\left(s\right)=\sum _{s\in {\mathbb{F}}_q}{s}^{q-1-n}{\displaystyle \sum _{i=0}^{q-2}}{b}_i{s}^i={\displaystyle \sum _{i=0}^{q-2}}{b}_i\sum _{s\in {\mathbb{F}}_q}{s}^{q-1+i-n}=-b_n.$$

Since $f\left(x\right)$ is a permutation polynomial of ${\mathbb{F}}_q$, we have

$$b_n=-\sum _{f\left(s\right)\in {\mathbb{F}}_q}{\left(f\left(s\right)\right)}^{q-1-n}Q\left(f\left(s\right)\right)=-\sum _{s\in {\mathbb{F}}_q}sf{\left(s\right)}^{q-1-n}.$$

Set

$$f{\left(x\right)}^{q-1-n}(mod{x}^q-x)=c_0+c_{1}x+\dots +c_{q-1}{x}^{q-1}.$$

Then

$$b_n=-\sum _{s\in {\mathbb{F}}_q}sf{\left(s\right)}^{q-1-n}=-\sum _{s\in {\mathbb{F}}_q}s{\displaystyle \sum _{i=0}^{q-1}}{c}_i{s}^i=c_{q-2}.$$

Therefore, finding any coefficient $b_n$ of the inverse $Q\left(x\right)$ is equivalent to finding the coefficient of $x^{q-2}$ in the expansion of the power $f{\left(x\right)}^{q-1-n}$. The following theorem, is a direct consequence of this observation and provides an explicit formula for the coefficients of the inverse in a special but important class of permutation polynomials.

Theorem 63.

[97] Let $f\left(x\right)=x^{r}h\left(x^s\right)$ be a permutation polynomial of ${\mathbb{F}}_q$, where $r\ge 1$, $s={\displaystyle \frac{q-1}{\ell }}$, $\ell \ge 2$, and $\ell \mid q-1$. Denote by

$$Q\left(x\right)=b_0+b_{1}x+\dots +b_{q-2}{x}^{q-2}$$

the inverse polynomial of $f\left(x\right)$ modulo $x^q-x$. Then the following hold.

(i) There are at most ℓ nonzero $b_n$’s such that $n=is+\overline{r}$, where $i=0,\dots ,\ell -1$ and $\overline{r}\equiv r^{-1}(mods)$.

(ii) Let $\overline{a}\equiv {\displaystyle \frac{r\overline{r}-1}{s}}(mod\ell )$. Then

$$b_{is+\overline{r}}={\displaystyle \frac{1}{\ell }}{\displaystyle \sum _{t=0}^{\ell -1}}{\omega }^{-(ir+\overline{a})t}h{\left({\omega }^t\right)}^{q-1-\overline{r}-is},i=0,\dots ,\ell -1,$$

where ω is a primitive ℓ-th root of unity.

(iii) Let

$$h{\left(x^s\right)}^{q-1-\overline{r}-is}\equiv {\displaystyle \sum _{j=0}^{\ell }}{d}_{i,j}{x}^{js}(mod{x}^q-x)$$

and let $m_i=ir+\overline{a}(mod\ell )$. Then $b_{is+\overline{r}}=d_{i,m_i}$ if $m_i\ne 0,$ and $b_{is+\overline{r}}=d_{i,0}+d_{i,\ell }$ if $m_i=0.$

Although we present explicit coefficient formulas for compositional inverses of permutation polynomials within this section, these identities are of limited utility in practice. For arbitrary q, one cannot extract concrete closed-form expressions for individual coefficients directly from the given formulas.

3.10. Compositional Inverses of Known Permutation Polynomials Not Obtained from the AGW Criterion

The local method is a valuable tool for constructing permutation polynomials over finite fields and for calculating their composition inverses. In the past, we often demonstrated that a polynomial is a permutation over ${\mathbb{F}}_q$ by using the equation $f\left(x\right)=a$ (any $a\in {\mathbb{F}}_q$), which has at most one solution in ${\mathbb{F}}_q$. If we can identify the unique solution of the equation $f\left(x\right)=a$ as $x=g\left(a\right),$ we can then utilize local methods to derive the composition inverse of $f\left(x\right)$ as $g\left(x\right)$. We present the composition inverses for trinomials and quartics at first.

We introduce the resultant of two polynomials.

Definition 11.

Let $f\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+\dots +a_n\in {\mathbb{F}}_q\left[x\right]$ and $g\left(x\right)=b_0{x}^m+b_1{x}^{m-1}+\dots +b_m\in {\mathbb{F}}_q\left[x\right]$ be two polynomials of degree n and m, respectively, with $n,m\in \mathbb{N}.$ Then the resultant $R(f,g,x)$ of the two polynomials with respect to x is defined by the determinant

$$R(f,g,x)=\left|\begin{array}{cccccccc}{a}_0& a_1& \dots & a_n& 0& 0& \dots & 0\\ 0& a_0& a_1& \dots & a_n& 0& \dots & 0\\ ⋮& & & & & & & ⋮\\ 0& \dots & \dots & \dots & a_0& a_1& \dots & a_n\\ b_0& b_1& \dots & b_m& 0& 0& \dots & 0\\ 0& b_0& b_1& \dots & b_m& 0& \dots & 0\\ ⋮& & & & & & & ⋮\\ 0& \dots & \dots & \dots & b_0& b_1& \dots & b_m\end{array}\right|$$

of order $m+n.$

If $f\left(x\right)=a_0(x-{\alpha }_1)(x-{\alpha }_2)\dots (x-{\alpha }_n),$ where $a_0\ne 0,$ in the splitting field of f over ${\mathbb{F}}_q,$ then $R(f,g,x)$ is also given by the formula

$$R(f,g,x)=a_0^m{\displaystyle \prod _{i=1}^n}g\left({\alpha }_i\right).$$

Then $R(f,g,x)=0$ if and only if f and g have a common divisor in ${\mathbb{F}}_q$ of positive degree.

Definition 12.

[98] [Definition 2.2] Two permutation polynomials $f\left(x\right)$ and $g\left(x\right)$ in ${\mathbb{F}}_q\left[x\right]$ are called quasi-multiplicative (QM, for short) equivalence if there exists an integer $1<d<q-1$ with $gcd(d,q-1)=1$ and $f\left(x\right)=ag\left(c{x}^d\right)$, where $a,c\in {\mathbb{F}}_q^{*}$.

The following three results were initially established by Gupta et al. [99] through the multivariate method and polynomial resultants. While Wu et al. [100] confirmed these results using the local method, their work also provided the explicit compositional inverses of the permutation polynomials.

Theorem 64.

[100] [Theorem 3.1] For a positive integer m, let $q=2^m$ and $A\in {\mathbb{F}}_q^{*}$ with $A^3=1.$ Then the polynomial $f_1\left(x\right)=x+A{x}^{q^2-q+1}+x^{q^2+q-1}$ is a permutation polynomial over ${\mathbb{F}}_{q^3}$ if and only if $m\neg \equiv 2(mod3).$ Moreover, if $f_1\left(x\right)$ is a permutation polynomial over ${\mathbb{F}}_{q^3}$, then the compositional inverse of $f_1\left(x\right)$ is

$$f_1^{-1}\left(x\right)=\left\{\begin{array}{cc}{\displaystyle \frac{{(A{x}^{q^2+1}+x^{2q^2}+A{x}^{2q})}^{q+1}(x+A{x}^q+A^2{x}^{q^2})}{(x+A{x}^q+A^2{x}^{q^2})\circ {(A{x}^{q^2+1}+x^{2q^2}+A{x}^{2q})}^{q+1}}},\hfill & \mathrm{if}x+A{x}^q+A^2{x}^{q^2}\ne 0;\hfill \\ {\displaystyle \frac{x^{q^2+2q+1}}{x^{q^2+2q}+A{x}^{2q+1}+x^{2q^2+1}}},\hfill & \mathrm{if}x+A{x}^q+A^2{x}^{q^2}=0\mathrm{and}x\ne 0;\hfill \\ 0,\hfill & \mathrm{if}x=0.\hfill \end{array}\right.$$

Remark 13.

Based on the character function of finite field ${\mathbb{F}}_{q^3}$, the inverse function $f^{-1}\left(x\right)$ originally defined in piecewise form can be converted into an integrated algebraic expression without segmented judgment, whose specific formula is given by

$$\begin{array}{cc}\hfill f^{-1}\left(x\right)=& {(x+A{x}^q+A^2{x}^{q^2})}^{q^3-1}{(A{x}^{q^2+1}+x^{2q^2}+A{x}^{2q})}^{q+1}(x+A{x}^q+A^2{x}^{q^2})\hfill \\ \hfill & ·{\left((x+A{x}^q+A^2{x}^{q^2})\circ {(A{x}^{q^2+1}+x^{2q^2}+A{x}^{2q})}^{q+1}\right)}^{q^3-2}\hfill \\ \hfill & +\left(1-{(x+A{x}^q+A^2{x}^{q^2})}^{q^3-1}\right)x^{q^2+2q+1}{\left(x^{q^2+2q}+A{x}^{2q+1}+x^{2q^2+1}\right)}^{q^3-2}.\hfill \end{array}$$

On this basis, setting $A=1$ in Theorem 64 yields the permutation polynomial $f_1\left(x\right)=x+x^{q^2-q+1}+x^{q^2+q-1}$. This polynomial is consistent with the research object described in Theorem 2.3 of [101]. Furthermore, Gupta et al. pointed out in Corollary 3.3 of [99] that $f_1\left(x\right)$ possesses QM equivalence with permutation polynomial $h_1\left(x\right)=x^{q+1}+{(x+x^q)}^{2q^2}$. The equivalence holds under the condition $gcd\left(q^3-1,{\displaystyle \frac{q^3+q^2-q}{2}}\right)=1$, satisfying the relation $h_1\left(x^{{\displaystyle \frac{q^3+q^2-q}{2}}}\right)=f_1\left(x\right)$. It is also found that the research findings of $h_1\left(x\right)$ can generalize the theoretical results proposed in Theorem 3.1 of [102]. For the above reasons, we will not repeat the derivation of compositional inverses for polynomials involved in the two aforementioned classic theorems.

Theorem 65.

[100] [Theorem 3.2] For a positive integer m, let $q=2^m$ and $A\in {\mathbb{F}}_q^{*}$ with $A^3=1.$ Then the polynomial $f_2\left(x\right)=x+A{x}^{q^3-q^2+q}+x^{q^2+q-1}$ is a permutation polynomial over ${\mathbb{F}}_{q^3}$ if and only if $m\neg \equiv 1(mod3).$ Moreover, if $f_2\left(x\right)$ permutes ${\mathbb{F}}_{q^3}$, then the compositional inverse of $f_2\left(x\right)$ is

$$f_2^{-1}\left(x\right)=\left\{\begin{array}{cc}{\displaystyle \frac{{(x^2+A{x}^{q^2+1}+A{x}^{2q})}^{q+1}(x+A^2{x}^q+A{x}^{q^2})}{(x+A^2{x}^q+A{x}^{q^2})\circ {(x^2+A{x}^{q^2+1}+A{x}^{2q})}^{q+1}}},\hfill & \mathrm{if}x+A^2{x}^q+A{x}^{q^2}\ne 0;\hfill \\ {\displaystyle \frac{x^{q^2+2q+1}}{x^{q^2+2q}+A{x}^{2q^2+1}+x^{2q+1}}},\hfill & \mathrm{if}x+A^2{x}^q+A{x}^{q^2}=0\mathrm{and}x\ne 0;\hfill \\ 0,\hfill & \mathrm{if}x=0.\hfill \end{array}\right.$$

Remark 14.

Setting $A=1$ in Theorem 65 yields the permutation polynomial $f_2\left(x\right)=x+x^{q^3-q^2+q}+x^{q^2+q-1},$ which corresponds to Theorem 4.3 in [103]. This polynomial is QM-equivalent to another permutation polynomial, $h_2\left(x\right)=x^{q+1}+{(x+x^q)}^{2q}$, as established in [99] [Corollary 3.5] via the relation $h_2\left(x^{(q^3+q^2-q)/2}\right)=f_2\left(x\right)$ (given $gcd(q^3-1,(q^3+q^2-q)/2)=1$). The polynomial $h_2\left(x\right)$ itself generalizes the one presented in Theorem 3.2 of the earlier work [102]. Therefore, the compositional inverses for the polynomials in Theorem 4.3 of [103] and Theorem 3.2 of [102] are not reiterated here.

Theorem 66.

[100] [Theorem 3.3] Let q be a prime power and $A\in {\mathbb{F}}_q*$. Then the polynomial

$$f_3\left(x\right)=x+A{x}^{q^2-q+1}+A^2{x}^{q^2}$$

is a permutation polynomial over ${\mathbb{F}}_{q^3}$ if and only if $A^3\ne 1.$ Moreover, if $f_3\left(x\right)$ permutes ${\mathbb{F}}_{q^3}$, the compositional inverse of $f_3\left(x\right)$ is

$$f_3^{-1}\left(x\right)={(A^{2}x+x^q+A{x}^{q^2})}^{q^3-2}{x}^{q+1}.$$

Employing the same multivariate and resultant techniques, Wang et al. [104] established the following six related results.

Theorem 67.

[104] [Theorem 3.1] For a positive integer k with $k\neg \equiv 1(mod3).$ Let $q=3^k.$ Then

$$f\left(x\right)=x^{q^2+q-1}-x^{q^2-q+1}+x$$

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

Wang et al. [104] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=a$ is $x=0,$ and if $a\ne 0,$ then the unique solution of $f\left(x\right)=a$ is

$$x=-{\displaystyle \frac{a^{3}b-a^{3}c-a^{2}bc-a{b}^3+a{b}^{2}c+b^{3}c+b^2{c}^2-b{c}^3-c^4}{a^3-a^{2}c-a{b}^2+abc+b^3-b{c}^2+c^3}},$$

where $b=a^q=f^q\left(x\right),$$c=b^q=f^{q^2}\left(x\right).$ Applying the local method, we have the following result.

Theorem 68.

Let the notations be defined as Theorem 67. Then the compositional inverse of $f\left(x\right)=x^{q^2+q-1}-x^{q^2-q+1}+x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{cc}\hfill f^{-1}\left(x\right)=& -\left(x^{q+3}-x^{q^2+3}-x^{q^2+q+2}-x^{3q+1}+x^{q^2+2q+1}+x^{q^2+3q}+x^{2q^2+2q}-x^{3q^2+q}-x^{4q^2}\right)·\hfill \\ & {\left(x^3-x^{q^2+2}-x^{q+2}+x^{q^2+q+1}+x^{3q}-x^{2q^2+q}+x^{3q^2}\right)}^{q^3-2}.\hfill \end{array}$$

Theorem 69.

[104] [Theorem 3.2] For a positive integer k with $k\neg \equiv 1(mod3).$ Let $q=3^k.$ Then

$$f\left(x\right)=-x^{q^2+q-1}+x^{q^2}+x$$

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

Wang et al. [104] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=a$ is $x=0,$ and if $a\ne 0,$ then the unique solution of $f\left(x\right)=a$ is

$$x={\displaystyle \frac{a^2{b}^2-a^{2}bc+a{b}^3-a{b}^{2}c+a^{3}c-b^{3}c-b^2{c}^2}{a^3-a{b}^2-abc-a{c}^2+b^3-b^{2}c+c^3}},$$

where $b=a^q=f^q\left(x\right),$$c=b^q=f^{q^2}\left(x\right).$ Using the local method, we have the following result.

Theorem 70.

Let the notations be defined as in Theorem 69. Then the compositional inverse of $f\left(x\right)=-x^{q^2+q-1}+x^{q^2}+x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{cc}\hfill f^{-1}\left(x\right)=& \left(x^{2q+2}-x^{q^2+q+2}+x^{3q+1}-x^{q^2+2q+1}+x^{q^2+3}-x^{q^2+3q}-x^{2q^2+2q}\right)·\hfill \\ & {\left(x^3-x^{2q+1}-x^{q^2+q+1}-x^{2q^2+1}+x^{3q}-x^{q^2+2q}+x^{3q^2}\right)}^{q^3-2}.\hfill \end{array}$$

Theorem 71.

[104] [Theorem 3.3] For a positive integer k with $k\neg \equiv 1(mod3).$ Let $q=3^k.$ Then

$$f\left(x\right)=x^{q^2+q-1}+x^q-x$$

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

Wang et al. [104] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=a$ is $x=0,$ and if $a\ne 0,$ then the unique solution of $f\left(x\right)=a$ is

$$x={\displaystyle \frac{-a^3+a^{2}c+a{b}^2-abc-b^3+b{c}^2-c^3}{-a^{2}bc-a^2{c}^2+a{b}^3-ab{c}^2+a{c}^3-b^2{c}^2+b{c}^3}},$$

where $b=a^q=f^q\left(x\right),$$c=b^q=f^{q^2}\left(x\right).$ According to the local method, we have the following result.

Theorem 72.

Let the notations be defined as in Theorem 71. Then the compositional inverse of $f\left(x\right)=x^{q^2+q-1}+x^q-x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{cc}\hfill f^{-1}\left(x\right)=& \left(-x^3+x^{q^2+2}+x^{2q+1}-x^{q^2+q+1}-x^{3q}+x^{2q^2+q}-x^{3q^2}\right)·\hfill \\ & {\left(-x^{q^2+q+2}-x^{2q^2+2}+x^{3q+1}-x^{2q^2+q+1}+x^{3q^2+1}-x^{2q^2+2q}+x^{3q^2+q}\right)}^{q^3-2}.\hfill \end{array}$$

Theorem 73.

[104] [Theorem 3.4] For a positive integer k with $k\neg \equiv 2(mod3).$ Let $q=3^k.$ Then

$$f\left(x\right)=x^{q^2+q-1}-x^{q^3-q^2+q}+x$$

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

Wang et al. [104] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=a$ is $x=0,$ and if $a\ne 0,$ then the unique solution of $f\left(x\right)=a$ satisfies

$$x^q={\displaystyle \frac{a^{3}b-a^{3}c-a^{2}bc-a^2{c}^2-a{b}^3+a{b}^{2}c+a{c}^3+b^{3}c+c^4}{a^3-a^{2}b+abc-a{c}^2+b^3-b^{2}c+c^3}},$$

or

$$x={\displaystyle \frac{c^{3}a-c^{3}b-c^{2}ab-c^2{b}^2-c{a}^3+c{a}^{2}b+c{b}^3+a^{3}b+b^4}{c^3-c^{2}a+cab-c{b}^2+a^3-a^{2}b+b^3}},$$

where $b=a^q=f^q\left(x\right),$$c=b^q=f^{q^2}\left(x\right).$ According to the local method, we have the following result.

Theorem 74.

Let the notations be defined as in Theorem 73. Then the compositional inverse of $f\left(x\right)=x^{q^2+q-1}-x^{q^3-q^2+q}+x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{cc}\hfill f^{-1}\left(x\right)=& \left(x^{3q^2+1}-x^{3q^2+q}-x^{2q^2+q+1}-x^{2q^2+2q}-x^{q^2+3}+x^{q^2+q+2}+x^{q^2+3q}+x^{q+3}+x^{4q}\right)·\hfill \\ & {\left(x^{3q^2}-x^{2q^2+1}+x^{q^2+q+1}-x^{q^2+2q}+x^3-x^{q+2}+x^{3q}\right)}^{q^3-2}.\hfill \end{array}$$

Theorem 75.

[104] [Theorem 3.5] For a positive integer k with $k\neg \equiv 2(mod3).$ Let $q=3^k.$ Then

$$f\left(x\right)=-x^{q^2+q-1}+x^q+x$$

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

Wang et al. [104] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=a$ is $x=0,$ and if $a\ne 0,$ then the unique solution of $f\left(x\right)=a$ is

$$x={\displaystyle \frac{-a^{2}bc+a^2{c}^2+a{b}^3-ab{c}^2+a{c}^3-b^2{c}^2-b{c}^3}{a^3-a^{2}c-a{b}^2-abc+b^3-b{c}^2+c^3}},$$

where $b=a^q=f^q\left(x\right),$$c=b^q=f^{q^2}\left(x\right).$ It follows from the local method that we have the following result.

Theorem 76.

Let the notations be defined as in Theorem 75. Then the compositional inverse of $f\left(x\right)=-x^{q^2+q-1}+x^q+x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{cc}\hfill f^{-1}\left(x\right)=& \left(-x^{q^2+q+2}+x^{2q^2+2}+x^{3q+1}-x^{2q^2+q+1}+x^{3q^2+1}-x^{2q^2+2q}-x^{3q^2+q}\right)·\hfill \\ & {\left(x^3-x^{q^2+2}-x^{2q+1}-x^{q^2+q+1}+x^{3q}-x^{2q^2+q}+x^{3q^2}\right)}^{q^3-2}.\hfill \end{array}$$

Theorem 77.

[104] [Theorem 3.6] For a positive integer k with $k\neg \equiv 2(mod3).$ Let $q=3^k.$ Then

$$f\left(x\right)=x^{q^2+q-1}+x^{q^2}-x$$

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

Wang et al. [104] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=a$ is $x=0,$ and if $a\ne 0,$ then the unique solution of $f\left(x\right)=a$ is

$$x={\displaystyle \frac{a^2{b}^2+a^{2}bc-a{b}^3+a{b}^{2}c-a{c}^3-b^{3}c+b^2{c}^2}{a^3-a^{2}b+abc-a{c}^2+b^3-b^{2}c+c^3}},$$

where $b=a^q=f^q\left(x\right),$$c=b^q=f^{q^2}\left(x\right).$ It follows from the local method that we have the following result.

Theorem 78.

Let the notations be defined as in Theorem 77. Then the compositional inverse of $f\left(x\right)=x^{q^2+q-1}+x^{q^2}-x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{cc}\hfill f^{-1}\left(x\right)=& \left(x^{2q+2}+x^{q^2+q+2}-x^{3q+1}+x^{q^2+2q+1}-x^{3q^2+1}-x^{q^2+3q}+x^{2q^2+2q}\right)·\hfill \\ & {\left(x^3-x^{q+2}+x^{q^2+q+1}-x^{2q^2+1}+x^{3q}-x^{q^2+2q}+x^{3q^2}\right)}^{q^3-2}.\hfill \end{array}$$

Theorem 79.

[105] [Theorem 2] For a positive integer m, let $q=2^m$ and $a\in {\mathbb{F}}_{q^3}.$ If m is even with $m\neg \equiv 2(mod3)$ and $a^{2(q^2+q+1)}+a^{q^2+q+1}=1,$ then

$$f\left(x\right)=x^{q^2-q+1}+a^{q+2}{x}^q+ax$$

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

Pang et al. [105] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=A$ is $x=0,$ and if $A\ne 0,$ then the unique solution of $f\left(x\right)=A$ is

$$x={({\theta }_2/{\theta }_1)}^{1/2},$$

where ${\theta }_1=A^3+A^{2}B{a}^{q^2+q+3}+A^{2}C{a}^{q^2+3q+3}+A{B}^2{a}^{q^2+q+5}+ABC{a}^{2q^2+4q+6}+A{C}^2{a}^{2q^2+6q+6}+B^3{a}^{q^2+q+7}+B^{2}C{a}^{2q^2+4q+8}+B{C}^2{a}^{2q^2+6q+8}+C^3{a}^{2q^2+8q+8},$${\theta }_2=A^4+A^2{C}^2{a}^{4q+4}+B^2{C}^2{a}^{4q+8},$$B=A^q=f^q\left(x\right),$$C=B^q=f^{q^2}\left(x\right).$ It follows from the local method that we have the following result.

Theorem 80.

Let the notations be defined as in Theorem 79. Then the compositional inverse of $f\left(x\right)=x^{q^2-q+1}+a^{q+2}{x}^q+ax$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{c}{f}^{-1}\left(x\right)=(x^{{\displaystyle \frac{3}{2}}}+a^{{\displaystyle \frac{q^2+q+3}{2}}}{x}^{{\displaystyle \frac{q}{2}}+1}+a^{{\displaystyle \frac{q^2+3q+3}{2}}}{x}^{{\displaystyle \frac{q^2}{2}}+1}+a^{{\displaystyle \frac{q^2+q+5}{2}}}{x}^{q+{\displaystyle \frac{1}{2}}}+a^{q^2+2q+3}{x}^{{\displaystyle \frac{q^2+q+1}{2}}}+a^{q^2+3q+3}{x}^{q^2+{\displaystyle \frac{1}{2}}}\hfill \\ \hfill +a^{{\displaystyle \frac{q^2+q+7}{2}}}{x}^{{\displaystyle \frac{3q}{2}}}+a^{q^2+2q+4}{x}^{{\displaystyle \frac{q^2}{2}}+q}+a^{q^2+3q+4}{x}^{q^2+{\displaystyle \frac{q}{2}}}+a^{q^2+4q+4}{x}^{{\displaystyle \frac{3q^2}{2}}}{)}^{q^3-2}·\left(x^2+a^{2q+2}{x}^{q^2+1}+a^{2q+4}{x}^{q^2+q}\right)\end{array}$$

Theorem 81.

[105] [Theorem 3] For a positive integer m, let $q=2^m$ and $a\in {\mathbb{F}}_{q^3}.$ If m is even with $m\neg \equiv 1(mod3)$ and $a^{2(q^2+q+1)}+a^{q^2+q+1}=1,$ then

$$f\left(x\right)=x^{q^2-q+1}+a^{q^2+2}{x}^q+a{x}^{q^2}$$

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

Pang et al. [105] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=A$ is $x=0,$ and if $A\ne 0,$ then the unique solution of $f\left(x\right)=A$ is

$$x={({\theta }_2/{\theta }_1)}^{1/2},$$

where ${\theta }_1=A^3+A^{2}B{a}^{3q^2+q+3}+A^{2}C{a}^{q^2+q+3}+A{B}^2{a}^{6q^2+2q+6}+ABC{a}^{4q^2+2q+6}+A{C}^2{a}^{q^2+q+5}+B^3{a}^{8q^2+2q+8}+B^{2}C{a}^{6q^2+2q+8}+B{C}^2{a}^{4q^2+2q+8}+C^3{a}^{q^2+q+7},$${\theta }_2=A^4+A^2{B}^2{a}^{4q^2+4}+B^2{C}^2{a}^{4q^2+8},$$B=A^q=f^q\left(x\right),$$C=B^q=f^{q^2}\left(x\right).$ It follows from the local method that we have the following result.

Theorem 82.

Let the notations be defined as in Theorem 81. Then the compositional inverse of $f\left(x\right)=f\left(x\right)=x^{q^2-q+1}+a^{q^2+2}{x}^q+a{x}^{q^2}$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{c}{f}^{-1}\left(x\right)=(x^{{\displaystyle \frac{3}{2}}}+a^{{\displaystyle \frac{3q^2+q+1}{2}}}{x}^{{\displaystyle \frac{q}{2}}+1}+a^{{\displaystyle \frac{q^2+q+3}{2}}}{x}^{{\displaystyle \frac{q^2}{2}}+1}+a^{3q^2+q+3}{x}^{q+{\displaystyle \frac{1}{2}}}+a^{q^2+2q+3}{x}^{{\displaystyle \frac{q^2+q+1}{2}}}+a^{{\displaystyle \frac{q^2+q+5}{2}}}{x}^{q^2+{\displaystyle \frac{1}{2}}}\hfill \\ \hfill +a^{4q^2+q+4}{x}^{{\displaystyle \frac{3q}{2}}}+a^{3q^2+q+4}{x}^{{\displaystyle \frac{q^2}{2}}+q}+a^{2q^2+q+4}{x}^{q^2+{\displaystyle \frac{q}{2}}}+a^{{\displaystyle \frac{q^2+q+7}{2}}}{x}^{{\displaystyle \frac{3q^2}{2}}}{)}^{q^3-2}·\left(x^2+a^{2q^2+2}{x}^{q+1}+a^{2q^2+4}{x}^{q^2+q}\right).\end{array}$$

Theorem 83.

[105] [Theorem 4] For a positive integer m, let $q=2^m$ and $a,b\in {\mathbb{F}}_{q^3}.$ Then

$$f\left(x\right)=b{x}^{q(q^2-q+1)}+a{x}^{q^2-q+1}+x$$

permutes ${\mathbb{F}}_{q^3}$ if one of the following conditions holds:

(1) $m\equiv 0(mod3),$$a^{q^2+q+1}=1,$$ab\in {\mathbb{F}}_8^{*}$, and ${\mathrm{Tr}}_1^3\left(ab\right)=0;$

(2) $m\equiv 0(mod3),$$a=\beta c,$$b=c^{q^2+q}$ with $c^{q^2+q+1}=1,$$\beta \in {\mathbb{F}}_8^{*}$, and ${\mathrm{Tr}}_1^3\left(\beta \right)=0;$

(3) m is even, $a^{2(q^2+q+1)}+a^{q^2+q+1}=1,$ and $ab=1.$

Pang et al. [105] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=A$ is $x=0,$ if $A\ne 0,$ then the unique solution of $f\left(x\right)=A$ with condition (1) is

$$\begin{array}{c}x=(b^{2q^2+2q}{A}^2+a{b}^{q}AB+a^{2q^2+2}{B}^2+a^2{b}^{2q}{C}^2)(a^{2q^2+2q}{A}^2+a^{q^2}bAC+a^{2q^2}{b}^2{B}^2+b^{2q+2}{C}^2)·\hfill \\ \hfill {\left(ab\right)}^{-3}{(A+a^{q^2+1}bB+a^{q^2+3}{b}^{2}C)}^{-1}{(A+a^{q^2+1}bB+a^{q^2+7}{b}^{6}C)}^{-1}{(A+a^{q^2+5}{b}^{5}B+a^{q^2+7}{b}^{6}C)}^{-1},\end{array}$$

the unique solution of $f\left(x\right)=A$ with condition (2) is

$$\begin{array}{c}x=(A^2+\beta c^{q^2}AB+{\beta }^4{c}^{2q^2}{B}^2+{\beta }^2{c}^{2q^2+2}{C}^2)({\beta }^4{A}^2+\beta c^{q^2+1}AC+{\beta }^2{c}^{2q^2}{B}^2+c^{2q^2+2}{C}^2)·\hfill \\ \hfill {\beta }^{-3}{(A+{\beta }^2{c}^{q^2}B+\beta c^{q^2+1}C)}^{-1}{(A+{\beta }^6{c}^{q^2}B+\beta c^{q^2+1}C)}^{-1}{(A+{\beta }^6{c}^{q^2}B+{\beta }^5{c}^{q^2+1}C)}^{-1},\end{array}$$

and the unique solution of $f\left(x\right)=A$ with condition (3) is

$$\begin{array}{c}x=(A^2+a^{2q^2+q+2}AC+a^{3q^2+q+1}{B}^2+a^{4q^2+2q+4}{C}^2)(A^2+a^{2q^2+q+1}AB+a^{4q^2+2q+2}{B}^2+a^{2q^2+2}{C}^2)·\hfill \\ \hfill {(A+a^{2q^2+q+1}B+a^{q^2+1}C)}^{-1}{(A+a^{3q^2+2q+2}B+a^{q^2+1}C)}^{-1}{(A+a^{3q^2+2q+2}B+a^{2q^2+q+2}C)}^{-1},\end{array}$$

where $A=f\left(x\right),B=f{\left(x\right)}^q,C=f{\left(x\right)}^{q^2}.$ It follows from the local method that we have the following result.

Theorem 84.

Let the notations be defined as in Theorem 83.

(1) If $m\equiv 0(mod3),$$a^{q^2+q+1}=1,$$ab\in {\mathbb{F}}_8^{*}$, and ${\mathrm{Tr}}_1^3\left(ab\right)=0,$ then the compositional inverse of $f\left(x\right)=b{x}^{q(q^2-q+1)}+a{x}^{q^2-q+1}+x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{c}{f}^{-1}\left(x\right)=(b^{2q^2+2q}{x}^2+a{b}^q{x}^{q+1}+a^{2q^2+2}{x}^{2q}+a^2{b}^{2q}{x}^{2q^2})(a^{2q^2+2q}{x}^2+a^{q^2}b{x}^{q^2+1}+a^{2q^2}{b}^2{B}^2+b^{2q+2}{x}^{q^2})·\hfill \\ \hfill {\left(ab\right)}^{-3}{\left((x+a^{q^2+1}b{x}^q+a^{q^2+3}{b}^2{x}^{q^2})(x+a^{q^2+1}b{x}^q+a^{q^2+7}{b}^6{x}^{q^2})(x+a^{q^2+5}{b}^5{x}^q+a^{q^2+7}{b}^6{x}^{q^2})\right)}^{q^3-2}.\end{array}$$

(2) If $m\equiv 0(mod3),$$a=\beta c,$$b=c^{q^2+q}$ with $c^{q^2+q+1}=1,$$\beta \in {\mathbb{F}}_8^{*}$, and ${\mathrm{Tr}}_1^3\left(\beta \right)=0,$ then the compositional inverse of $f\left(x\right)=b{x}^{q(q^2-q+1)}+a{x}^{q^2-q+1}+x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{c}{f}^{-1}\left(x\right)=(x^2+\beta c^{q^2}{x}^{q+1}+{\beta }^4{c}^{2q^2}{x}^{2q}+{\beta }^2{c}^{2q^2+2}{x}^{2q^2})({\beta }^4{x}^2+\beta c^{q^2+1}{x}^{q^2+1}+{\beta }^2{c}^{2q^2}{x}^{2q}+c^{2q^2+2}{x}^{2q^2})·\hfill \\ \hfill {\beta }^{-3}{\left((x+{\beta }^2{c}^{q^2}{x}^q+\beta c^{q^2+1}{x}^{q^2})(x+{\beta }^6{c}^{q^2}{x}^q+\beta c^{q^2+1}{x}^{q^2})(x+{\beta }^6{c}^{q^2}{x}^q+{\beta }^5{c}^{q^2+1}{x}^{q^2})\right)}^{q^3-2}.\end{array}$$

(3) If m is even, $a^{2(q^2+q+1)}+a^{q^2+q+1}=1,$ and $ab=1,$ then the compositional inverse of $f\left(x\right)=b{x}^{q(q^2-q+1)}+a{x}^{q^2-q+1}+x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{c}{f}^{-1}\left(x\right)={\left((x+a^{2q^2+q+1}{x}^q+a^{q^2+1}{x}^{q^2})(x+a^{3q^2+2q+2}{x}^q+a^{q^2+1}{x}^{q^2})(x+a^{3q^2+2q+2}{x}^q+a^{2q^2+q+2}{x}^{q^2})\right)}^{q^3-2}·\hfill \\ \hfill (x^2+a^{2q^2+q+2}{x}^{q^2+1}+a^{3q^2+q+1}{x}^{2q}+a^{4q^2+2q+4}{x}^{2q^2})(x^2+a^{2q^2+q+1}{x}^{q+1}+a^{4q^2+2q+2}{x}^{2q}+a^{2q^2+2}{x}^{2q^2}).\end{array}$$

Theorem 85.

[105] [Theorem 5] For a positive integer m with $m\equiv 3(mod9)$, let $q=2^m,$$a,b,c\in {\mathbb{F}}_{q^3}$ and $\beta \in {\mathbb{F}}_8^{*}$ with $a=\beta c,$$b={\beta }^2{c}^{2q+1},c^{q^2+q+1}=1,$ and ${\beta }^3+{\beta }^2+1=0.$ Then

$$f\left(x\right)=b{x}^{q^2(q^2-q+1)}+a{x}^{q(q^2-q+1)}+x$$

permutes ${\mathbb{F}}_{q^3}.$

Pang et al. [105] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=A$ is $x=0,$ if $A\ne 0,$ then the unique solution of $f\left(x\right)=A$ is

$$x={({\theta }_2/{\theta }_1)}^{1/2},$$

where ${\theta }_1=c^{2q^2+2}{A}^3+c^{q^2+2}{\beta }^5{A}^{2}B+c^{q^2+1}{A}^{2}C+c^{2}A{B}^2+c{\beta }^{3}ABC+{\beta }^{5}A{C}^2+c^{q+3}{B}^3+c^{q+2}{\beta }^5{B}^{2}C+c^{q+1}B{C}^2+c^q{C}^3,$${\theta }_2=c^{2q^2+3}{A}^4+c{\beta }^3{A}^2{C}^2+c^{2q+5}{\beta }^4{B}^4+c^{2q+1}{\beta }^4{C}^4,$ $A=f\left(x\right),B=f{\left(x\right)}^q,C=f{\left(x\right)}^{q^2}.$ It follows from the local method that we have the following result.

Theorem 86.

Let the notations be defined as in Theorem 85. Then the compositional inverse of $f\left(x\right)=b{x}^{q^2(q^2-q+1)}+a{x}^{q(q^2-q+1)}+x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{c}{f}^{-1}\left(x\right)={\left(c^{2q^2+3}{x}^4+c{\beta }^3{x}^{2q^2+2}+c^{2q+5}{\beta }^4{x}^{4q}+c^{2q+1}{\beta }^4{x}^{4q^2}\right)}^{1/2}·\hfill \\ (c^{2q^2+2}{x}^3+c^{q^2+2}{\beta }^5{x}^{q+2}+c^{q^2+1}{x}^{q^2+2}+c^2{x}^{2q+1}+c{\beta }^3{x}^{q^2+q+1}+\\ \hfill {\beta }^5{x}^{2q^2+1}+c^{q+3}{x}^{3q}+c^{q+2}{\beta }^5{x}^{q^2+2q}+c^{q+1}{x}^{2q^2+q}+c^q{x}^{3q^2}{)}^{{\displaystyle \frac{q^3}{2}}-1}.\end{array}$$

Theorem 87.

[105] [Theorem 6] For a positive integer m with $m\equiv 3(mod9)$, let $q=2^m,$$a,b,c\in {\mathbb{F}}_{q^3}$ and $\beta \in {\mathbb{F}}_8^{*}$ with $a={\beta }^4{c}^{2q^2+1},$$b=\beta c,c^{q^2+q+1}=1,$ and ${\beta }^3+\beta +1=0.$ Then

$$f\left(x\right)=a{x}^{q^2(q^2-q+1)}+b{x}^{q^2-q+1)}+x$$

permutes ${\mathbb{F}}_{q^3}.$

Pang et al. [105] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=A$ is $x=0,$ if $A\ne 0,$ then the unique solution of $f\left(x\right)=A$ is

$$x={({\theta }_2/{\theta }_1)}^{1/2},$$

where ${\theta }_1=c^q{\beta }^5{A}^3+c^{q^2+q}{\beta }^3{A}^{2}B+{\beta }^6{A}^{2}C+c^{2q^2+q}{\beta }^{6}A{B}^2+c^{q^2}{\beta }^{4}ABC+c^{q^2+1}{\beta }^{3}A{C}^2+c^{3q^2+q}{\beta }^5{B}^3+c^{2q^2}{\beta }^3{B}^{2}C+c^{2q^2+1}{\beta }^{6}B{C}^2+c^{2q^2+2}{\beta }^5{C}^3,$${\theta }_2=c^{q^2+2q}{\beta }^2{A}^4+c^{q^2}{\beta }^4{A}^2{C}^2+c^{5q^2+2q}{\beta }^6{B}^4+c^{3q^2+2}\beta C^4,$ $A=f\left(x\right),B=f{\left(x\right)}^q,C=f{\left(x\right)}^{q^2}.$ It follows from the local method that we have the following result.

Theorem 88.

Let the notations be defined as in Theorem 87. Then the compositional inverse of $f\left(x\right)=b{x}^{q^2(q^2-q+1)}+a{x}^{q(q^2-q+1)}+x$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{c}{f}^{-1}\left(x\right)={\left(c^{q^2+2q}{\beta }^2{x}^4+c^{q^2}{\beta }^4{x}^{2q^2+2}+c^{5q^2+2q}{\beta }^6{x}^{4q}+c^{3q^2+2}\beta x^{4q^2}\right)}^{1/2}·\hfill \\ (c^q{\beta }^5{x}^3+c^{q^2+q}{\beta }^3{x}^{q+2}+{\beta }^6{x}^{q^2+2}+c^{2q^2+q}{\beta }^6{x}^{2q+1}+c^{q^2}{\beta }^4{x}^{q^2+q+1}+\\ \hfill c^{q^2+1}{\beta }^3{x}^{2q^2+1}+c^{3q^2+q}{\beta }^5{x}^{3q}+c^{2q^2}{\beta }^3{x}^{q^2+2q}+c^{2q^2+1}{\beta }^6{x}^{2q^2+q}+c^{2q^2+2}{\beta }^5{x}^{3q^2}{)}^{{\displaystyle \frac{q^3}{2}}-1}.\end{array}$$

Bartoli [106] considered four classes of permutation trinomials over ${\mathbb{F}}_{q^3}.$

Let

$$\begin{array}{cc}\hfill h\left(x\right)=& x(x^{22}+9x^{19}+18x^{18}+x^{17}+63x^{16}+78x^{15}+72x^{14}+165x^{13}-215x^{12}\hfill \\ \hfill & -64x^{11}+300x^{10}-108x^9+15x^8+45x^7+7x^6+36x^5+6x^4+3x^2+1)·\hfill \\ \hfill & (x^{44}-x^{42}+3x^{41}-18x^{40}-7x^{39}+22x^{38}-50x^{37}+118x^{36}+145x^{35}\hfill \\ \hfill & -254x^{34}+218x^{33}-112x^{32}-726x^{31}+627x^{30}-217x^{29}-16x^{28}+258x^{27}\hfill \\ \hfill & +996x^{26}-611x^{25}-161x^{24}-691x^{23}-392x^{22}+1189x^{21}+252x^{20}-645x^{19}\hfill \\ \hfill & +458x^{18}-475x^{17}-141x^{16}+237x^{15}+72x^{14}+298x^{13}-327x^{12}-121x^{11}\hfill \\ \hfill & +140x^{10}+47x^9+27x^8-59x^7-10x^6+22x^5+3x^4-3x^2+1).\hfill \end{array}$$

Theorem 89.

[106] [Theorem 3.4] Let $q\equiv 1(mod3).$ If $A,B\in {\mathbb{F}}_q,$$B^2+B+1=0,$$h\left(A\right)\ne 0,$$A^3\ne -1,$ then the polynomial

$$f\left(x\right)=x^{q^2+q-1}+A{x}^{q^2}-Bx$$

is a permutation polynomial of ${\mathbb{F}}_{q^3}.$

Bartoli [106] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=A$ is $x=0,$ and if $A\ne 0,$ then the unique solution of $f\left(x\right)=A$ is

$$x={\displaystyle \frac{(A^{2}B+A^2)a^{q^2+2q+1}+(AB-A)a^{2q^2+q+1}+AB{a}^{q^2+3q}-A{a}^{2q+2}-B{a}^{q^2+q+2}-(B+1)(a^{3q+1}+a^{3q^2+1})-a^{2q^2+2q}}{(A^{3}B+A^3-3B-3)a^{q^2+q+1}+A^{2}B(a^{q+2}+a^{2q^2+1}+a^{q^2+2q})+(AB-A)(a^{q^2+2}+a^{2q+1}+a^{2q^2+q})-a^3-a^{3q}-a^{3q^2}}}.$$

Theorem 90.

Let the notations be defined as in Theorem 89. Then the compositional inverse of $f\left(x\right)=x^{q^2+q-1}+A{x}^{q^2}-Bx$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{c}{f}^{-1}\left(x\right)=\left((A^{2}B+A^2)x^{q^2+2q+1}+(AB-A)x^{2q^2+q+1}+AB{x}^{q^2+3q}-A{x}^{2q+2}-B{x}^{q^2+q+2}-(B+1)(x^{3q+1}+x^{3q^2+1})-x^{2q^2+2q}\right)·\hfill \\ \hfill {\left((A^{3}B+A^3-3B-3)x^{q^2+q+1}+A^{2}B(x^{q+2}+x^{2q^2+1}+x^{q^2+2q})+(AB-A)(x^{q^2+2}+x^{2q+1}+x^{2q^2+q})-x^3-x^{3q}-x^{3q^2}\right)}^{q^3-2}.\end{array}$$

Theorem 91.

[106] [Theorem 3.5] If $A,B\in {\mathbb{F}}_q$ with $B^2+B+1=0$ and $A^3\ne -1,$ are such that $B{x}^3+A^2{x}^2+(A-AB)x-B$ has no roots in ${\mu }_{q^2+q+1}=\{x\mid x^{q^2+q+1}=1\},$ then the polynomial

$$f\left(x\right)=x^{q^2+q-1}+A{x}^q-Bx$$

is a permutation polynomial of ${\mathbb{F}}_{q^3}.$

Bartoli [106] showed that for any $a\in {\mathbb{F}}_{q^3}$, if $a=0,$ then the unique solution of $f\left(x\right)=A$ is $x=0,$ and if $A\ne 0,$ then the unique solution of $f\left(x\right)=A$ is

$$x={\displaystyle \frac{(A^{2}B+A^2)a^{2q^2+q+1}+(AB-A)a^{q^2+2q+1}+AB{a}^{3q^2+q}-A{a}^{2q^2+2}-(B+1)(a^{3q+1}+a^{3q^2+1})-a^{2q^2+2q}-B{a}^{q^2+q+2}}{(A^{3}B+A^3-3B-3)a^{q^2+q+1}+A^{2}B(a^{q^2+2}+a^{2q+1}+a^{2q^2+q})+(AB-A)(a^{q^2+2q}+a^{q+2}+a^{2q^2+1})-a^3-a^{3q}-a^{3q^2}}}.$$

Theorem 92.

Let the notations be defined as in Theorem 91. Then the compositional inverse of $f\left(x\right)=x^{q^2+q-1}+A{x}^q-Bx$ over ${\mathbb{F}}_{q^3}$ is

$$\begin{array}{c}{f}^{-1}\left(x\right)=\left((A^{2}B+A^2)x^{2q^2+q+1}+(AB-A)x^{q^2+2q+1}+AB{x}^{3q^2+q}-A{x}^{2q^2+2}-(B+1)(x^{3q+1}+x^{3q^2+1})-x^{2q^2+2q}-B{x}^{q^2+q+2}\right)·\hfill \\ \hfill {\left((A^{3}B+A^3-3B-3)x^{q^2+q+1}+A^{2}B(x^{q^2+2}+x^{2q+1}+x^{2q^2+q})+(AB-A)(x^{q^2+2q}+x^{q+2}+x^{2q^2+1})-x^3-x^{3q}-x^{3q^2}\right)}^{q^3-2}.\end{array}$$

Theorem 93.

[101] [Theorem 2.4] Let $k,n$ be positive integers with $k\neg \equiv 2(mod3)$ and $n=3k+1.$ Then

$$f\left(x\right)=x^{2^{2k+1}+2^{k+1}+1}+x^{2^{k+1}+1}+x$$

is a permutation polynomial over ${\mathbb{F}}_{2^n}.$

Wang et al. [101] showed that for any $a\in {\mathbb{F}}_{2^n}$, if $a=0,$ then the unique solution of $f\left(x\right)=a$ is $x=0,$ and if $a\ne 0,$ then the unique solution of $f\left(x\right)=a$ is

$$x=a+{\displaystyle \frac{a{b}^2(a^2+b^2+c^2+1)}{a^2+a^2{b}^2+b^4+c^4+1}},$$

where $b=a^{2^k}=f{\left(x\right)}^{2^k},$$c=b^{2^k}=f{\left(x\right)}^{2^{2k}}.$ It follows from the local method that we have the following result.

Theorem 94.

Let the notations be defined as in Theorem 93. Then the compositional inverse of $f\left(x\right)=f\left(x\right)=x^{2^{2k+1}+2^{k+1}+1}+x^{2^{k+1}+1}+x$ is

$$f^{-1}\left(x\right)=x+{\displaystyle \frac{x^{2^{k+1}+1}(x^2+x^{2^{k+1}}+x^{2^{2k+1}}+1)}{x^2+x^{2^{k+1}+2}+x^{2^{k+2}}+x^{2^{2k+2}}+1}}.$$

Theorem 95.

[101] [Theorem 2.5] Let $k,n$ be positive integers $n=3k-1.$ Then

$$f\left(x\right)=x^{2^{3k-1}-2^{2k}+2^k}+x^{2^k-1}+x$$

is a permutation polynomial over ${\mathbb{F}}_{2^n}.$

Wang et al. [101] showed that for any $a\in {\mathbb{F}}_{2^n}$, if $a=0,$ then the unique solution of $f\left(x\right)=a$ is $x=0,$ and if $a\ne 0,$ then the unique solution of $f\left(x\right)=a$ is

$$x=\left\{\begin{array}{cc}a+1,\hfill & \mathrm{if}{a}^2+b^2+bc+c+1=0\hfill \\ \left(abc\right)/(a^2+b^2+bc+c+1),\hfill & \mathrm{if}{a}^2+b^2+bc+c+1\ne 0.\hfill \end{array}\right.$$

where $b=a^{2^k}=f{\left(x\right)}^{2^k},$$c=b^{2^k}=f{\left(x\right)}^{2^{2k}}.$ It follows from the local method that we have the following result.

Theorem 96.

Let the notations be defined as in Theorem 95. Then the compositional inverse of $f\left(x\right)=x^{2^{3k-1}-2^{2k}+2^k}+x^{2^k-1}+x$ over ${\mathbb{F}}_{2^n}$ is

$$f^{-1}\left(x\right)=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}x=0,\hfill \\ x+1,\hfill & \mathrm{if}{x}^2+x^{2^{k+1}}+x^{2^{2k}+2^k}+x^{2^{2k}}+1=0,\hfill \\ {\displaystyle \frac{x^{2^{2k}+2^k+1}}{x^2+x^{2^{k+1}}+x^{2^{2k}+2^k}+x^{2^{2k}}+1}},\hfill & \mathrm{if}{x}^2+x^{2^{k+1}}+x^{2^{2k}+2^k}+x^{2^{2k}}+1\ne 0.\hfill \end{array}\right.$$

Theorem 97.

[98] [Proposition 2.6] Let $m>1$ be an odd integer and $k=(m+1)/2.$ Then the polynomial

$$f\left(x\right)=x+x^{2^k-1}+x^{2^k+1}$$

is a permutation polynomial over ${\mathbb{F}}_{2^m}^{*}.$

As noted in [98], the permutation polynomial $f\left(x\right)=x+x^{2^k-1}+x^{2^k+1}$ is **quasi-multiplicatively (QM) equivalent to the permutation polynomials in Theorems 5.1 and 5.2 of [107], namely:

$$f_1\left(x\right)=x+u^{2^{k-1}-1}{x}^{2^k-1}+u^{2^{k-1}}{x}^{2^k+1}$$

and

$$f_2\left(x\right)=x+u{x}^{2^k-1}+u^{2^k}{x}^{2^m-2^{k+1}+2}.$$

Therefore, these results will not be restated here.

For any $c\in {\mathbb{F}}_{2^m}^{*}$, Wu et al. [98] showed that the unique solution of the equation $f\left(x\right)=x+x^{2^k-1}+x^{2^k+1}=c$ is

$$x={\displaystyle \frac{cd}{1+c^2+d}},$$

where $d=c^{2^k}.$ Then we have the following result.

Theorem 98.

Let the notations be defined as in Theorem 97. Then the compositional inverse of $f\left(x\right)=x+x^{2^k-1}+x^{2^k+1}$ over ${\mathbb{F}}_{2^m}$ is

$$f^{-1}={\displaystyle \frac{x^{2^k+1}}{1+x+x^{2^k}}}.$$

Lemma 8.

[108] [Theorem 3.4] Let k be a positive integer and q be a prime power with $q\neg \equiv 0(mod3),$ and let m be an even positive integer. Then

$$f\left(x\right)=x+x^{k{q}^{m/2}-(k-1)}+x^{(k+1)-k{q}^{m/2}}$$

is a permutation polynomial of ${\mathbb{F}}_{q^m}$ if and only if one of the following three conditions holds:

(1) $m\equiv 0(mod4);$

(2) $q\equiv 1(mod3);$

(3) $m\equiv 2(mod4),$$q\equiv 1(mod3),$ and ${\exp }_3\left(k\right)\ge {\exp }_3(q^{m/2}+1),$ where ${\exp }_3\left(i\right)$ denotes the exponent of 3 in cannonical factorization of $i.$

Ding et al. [108] have show that for any $a\ne 0,$ the equation $f\left(x\right)=a$ has a unique root

$$x={\displaystyle \frac{a^{k{q}^{m/2}+k+1}}{a^{2k}+a^{k+k{q}^{m/2}}+a^{2q^{m/2}}}}(\mathrm{see}\left(3.22\right)\mathrm{in}\mathrm{bad}\mathrm{hbox}$$

and for $a=0,$ the equation $f\left(x\right)=0$ has the only solution $x=0.$ Then we have the following result by Lemma 14.

Theorem 99.

Let the notations be defined in Lemma 8, and

$$f\left(x\right)=x+x^{k{q}^{m/2}-(k-1)}+x^{(k+1)-k{q}^{m/2}}$$

is a permutation polynomial over ${\mathbb{F}}_{q^m}.$ Then the compositional inverse of $f\left(x\right)$ is

$$f^{-1}\left(x\right)=x^{k{q}^{m/2}+k+1}{\left(x^{2k}+x^{k+k{q}^{m/2}}+x^{2q^{m/2}}\right)}^{q^3-2}.$$

For $i=1,2,3,$ let ${\psi }_i\left(x\right)=x^{q^{i-1}},$${\phi }_i\left(x\right)={\psi }_i\left(f\left(x\right)\right)$. The following result is derived from the local method.

Table 2. Known classes of non-linearized permutation trinomial over ${\mathbb{F}}_{p^{3m}}$ for odd characteristic and its inverse

No.	$\mathit{f}\left(\mathit{x}\right)\in {\mathbb{F}}_{{\mathit{q}}^3}(\mathit{q}={\mathit{p}}^{\mathit{m}})$	conditions on m and p	Refs.	its inverse
1	$x^{q^2+q-1}-x^{q^2-q+1}+x$	$p=3$, $m\not\equiv 1(mod3)$	[104] [Theorem 3.1]	Theorem 68
2	$-x^{q^2+q-1}+x^{q^2}+x$	$p=3$, $m\not\equiv 1(mod3)$	[104] [Theorem 3.2]	Theorem 70
3	$x^{q^2+q-1}+x^q-x$	$p=3$, $m\not\equiv 1(mod3)$	[104] [Theorem 3.3]	Theorem 72
4	$x^{q^2+q-1}-x^{q^3-q^2+q}+x$	$p=3$, $m\not\equiv 2(mod3)$	[104] [Theorem 3.4]	Theorem 74
5	$-x^{q^2+q-1}+x^q+x$	$p=3$, $m\not\equiv 2(mod3)$	[104] [Theorem 3.5]	Theorem 76
6	$x^{q^2+q-1}+x^{q^2}-x$	$p=3$, $m\not\equiv 2(mod3)$	[104] [Theorem 3.6]	Theorem 78

Table 2. Known classes of non-linearized permutation trinomial over ${\mathbb{F}}_{p^{3m}}$ for odd characteristic and its inverse

No.	$\mathit{f}\left(\mathit{x}\right)\in {\mathbb{F}}_{{\mathit{q}}^3}(\mathit{q}={\mathit{p}}^{\mathit{m}})$	conditions on m and p	Refs.	its inverse
1	$x^{q^2+q-1}-x^{q^2-q+1}+x$	$p=3$, $m\not\equiv 1(mod3)$	[104] [Theorem 3.1]	Theorem 68
2	$-x^{q^2+q-1}+x^{q^2}+x$	$p=3$, $m\not\equiv 1(mod3)$	[104] [Theorem 3.2]	Theorem 70
3	$x^{q^2+q-1}+x^q-x$	$p=3$, $m\not\equiv 1(mod3)$	[104] [Theorem 3.3]	Theorem 72
4	$x^{q^2+q-1}-x^{q^3-q^2+q}+x$	$p=3$, $m\not\equiv 2(mod3)$	[104] [Theorem 3.4]	Theorem 74
5	$-x^{q^2+q-1}+x^q+x$	$p=3$, $m\not\equiv 2(mod3)$	[104] [Theorem 3.5]	Theorem 76
6	$x^{q^2+q-1}+x^{q^2}-x$	$p=3$, $m\not\equiv 2(mod3)$	[104] [Theorem 3.6]	Theorem 78

Table 3. Known classes of non-linearized permutation trinomial over ${\mathbb{F}}_{2^{3m}}$ and its inverse

No.	$\mathit{f}\left(\mathit{x}\right)\in {\mathbb{F}}_{{\mathit{q}}^3}\mathbf{where}\mathit{q}=2^{\mathit{m}}$	Conditions on m	Refs.	Its inverse
1	$\begin{array}{cc}& ax+a^{q+2}{x}^q+x^{q^2-q+1};\hfill \\ & a^{2(q^2+q+1)}+a^{q^2+q+1}=1\hfill \end{array}$	$m\not\equiv 2(mod3)$   m is even	[105] [Theorem 2]	Theorem 80
2	$\begin{array}{cc}& a^{q^2+2}{x}^q+a{x}^{q^2}+x^{q^2-q+1};\hfill \\ & a^{2(q^2+q+1)}+a^{q^2+q+1}=1\hfill \end{array}$	$m\not\equiv 1(mod3)$   m is even	[105] [Theorem 3]	Theorem 82
3	$\begin{array}{cc}& x+a{x}^{q^2+q-1}+b{x}^{q(q^2-q+1)};\hfill \\ & a^{q^2+q+1}=1,ab\in {\mathbb{F}}_{2^3}^{*}\hfill \\ & \mathrm{and}{\mathrm{Tr}}_1^3\left(ab\right)=0\hfill \end{array}$	$m\equiv 0\left(\mathrm{mod}3\right)$	[105] [Theorem 4]	Theorem 84
	$\begin{array}{cc}& x+a{x}^{q^2+q-1}+b{x}^{q(q^2-q+1)};\hfill \\ & a=\beta c,b=c^{q^2+q}\hfill \\ & \mathrm{with}{c}^{q^2+q+1}=1,\hfill \\ & \beta \in {\mathbb{F}}_{2^3}^{*},{\mathrm{Tr}}_1^3\left(\beta \right)=0\hfill \end{array}$
	$\begin{array}{cc}& x+a{x}^{q^2+q-1}+b{x}^{q(q^2-q+1)};\hfill \\ & a^{q^2+q+1}+a^{q^2+q+1}=1\hfill \\ & \mathrm{and}ab=1\hfill \end{array}$	m is even
4	$\begin{array}{cc}& x+a{x}^{q(q^2-q+1)}+b{x}^{q^2(q^2-q+1)},\hfill \\ & b={\beta }^2{c}^{2q+1},a=\beta c\hfill \\ & \mathrm{with}\beta \in {\mathbb{F}}_{2^3}^{*}\setminus \left\{1\right\},\hfill \\ & {\beta }^3+{\beta }^2+1=0\mathrm{and}{c}^{q^2+q+1}=1\hfill \end{array}$	$m\equiv 3(mod9)$	[105] [Theorem 5]	Theorem 86
5	$\begin{array}{cc}& x+a{x}^{q^2(q^2-q+1)}+b{x}^{q^2-q+1},\hfill \\ & a={\beta }^4{c}^{2q^2+1},b=\beta c\hfill \\ & \mathrm{with}\beta \in {\mathbb{F}}_{2^3}^{*},\hfill \\ & {\beta }^3+\beta +1=0\mathrm{and}{c}^{q^2+q+1}=1\hfill \end{array}$	$m\equiv 3(mod9)$	[105] [Theorem 5]	Theorem 88
6	$x+A{x}^{q^2-q+1}+x^{q^2+q-1},$$A\in {\mathbb{F}}_q$ with $A^3=1$	$m\not\equiv 2(mod3)$	[99] [Theorem 3.2]	Theorem 64
7	$x+A{x}^{q^3-q^2+q}+x^{q^2+q-1},$$A\in {\mathbb{F}}_q$ with $A^3=1$	$m\not\equiv 1(mod3)$	[99] [Theorem 3.4]	Theorem 65

Table 3. Known classes of non-linearized permutation trinomial over ${\mathbb{F}}_{2^{3m}}$ and its inverse

No.	$\mathit{f}\left(\mathit{x}\right)\in {\mathbb{F}}_{{\mathit{q}}^3}\mathbf{where}\mathit{q}=2^{\mathit{m}}$	Conditions on m	Refs.	Its inverse
1	$\begin{array}{cc}& ax+a^{q+2}{x}^q+x^{q^2-q+1};\hfill \\ & a^{2(q^2+q+1)}+a^{q^2+q+1}=1\hfill \end{array}$	$m\not\equiv 2(mod3)$   m is even	[105] [Theorem 2]	Theorem 80
2	$\begin{array}{cc}& a^{q^2+2}{x}^q+a{x}^{q^2}+x^{q^2-q+1};\hfill \\ & a^{2(q^2+q+1)}+a^{q^2+q+1}=1\hfill \end{array}$	$m\not\equiv 1(mod3)$   m is even	[105] [Theorem 3]	Theorem 82
3	$\begin{array}{cc}& x+a{x}^{q^2+q-1}+b{x}^{q(q^2-q+1)};\hfill \\ & a^{q^2+q+1}=1,ab\in {\mathbb{F}}_{2^3}^{*}\hfill \\ & \mathrm{and}{\mathrm{Tr}}_1^3\left(ab\right)=0\hfill \end{array}$	$m\equiv 0\left(\mathrm{mod}3\right)$	[105] [Theorem 4]	Theorem 84
	$\begin{array}{cc}& x+a{x}^{q^2+q-1}+b{x}^{q(q^2-q+1)};\hfill \\ & a=\beta c,b=c^{q^2+q}\hfill \\ & \mathrm{with}{c}^{q^2+q+1}=1,\hfill \\ & \beta \in {\mathbb{F}}_{2^3}^{*},{\mathrm{Tr}}_1^3\left(\beta \right)=0\hfill \end{array}$
	$\begin{array}{cc}& x+a{x}^{q^2+q-1}+b{x}^{q(q^2-q+1)};\hfill \\ & a^{q^2+q+1}+a^{q^2+q+1}=1\hfill \\ & \mathrm{and}ab=1\hfill \end{array}$	m is even
4	$\begin{array}{cc}& x+a{x}^{q(q^2-q+1)}+b{x}^{q^2(q^2-q+1)},\hfill \\ & b={\beta }^2{c}^{2q+1},a=\beta c\hfill \\ & \mathrm{with}\beta \in {\mathbb{F}}_{2^3}^{*}\setminus \left\{1\right\},\hfill \\ & {\beta }^3+{\beta }^2+1=0\mathrm{and}{c}^{q^2+q+1}=1\hfill \end{array}$	$m\equiv 3(mod9)$	[105] [Theorem 5]	Theorem 86
5	$\begin{array}{cc}& x+a{x}^{q^2(q^2-q+1)}+b{x}^{q^2-q+1},\hfill \\ & a={\beta }^4{c}^{2q^2+1},b=\beta c\hfill \\ & \mathrm{with}\beta \in {\mathbb{F}}_{2^3}^{*},\hfill \\ & {\beta }^3+\beta +1=0\mathrm{and}{c}^{q^2+q+1}=1\hfill \end{array}$	$m\equiv 3(mod9)$	[105] [Theorem 5]	Theorem 88
6	$x+A{x}^{q^2-q+1}+x^{q^2+q-1},$$A\in {\mathbb{F}}_q$ with $A^3=1$	$m\not\equiv 2(mod3)$	[99] [Theorem 3.2]	Theorem 64
7	$x+A{x}^{q^3-q^2+q}+x^{q^2+q-1},$$A\in {\mathbb{F}}_q$ with $A^3=1$	$m\not\equiv 1(mod3)$	[99] [Theorem 3.4]	Theorem 65