Constructing permutation polynomials is a hot topic in finite fields. Recently, huge kinds of permutation polynomials over Fq2 have been studied. In this paper, by using AGW criterion and piecewise method, we construct several classes of permutation polynomials over Fq3 of the forms similar to (xq2+xq+x+δ)q3−1d+1+L(x), for d=2,3,4,6, where L(x) is a linearized polynomial over Fq.
Citation: Xiaoer Qin, Li Yan. Some specific classes of permutation polynomials over Fq3[J]. AIMS Mathematics, 2022, 7(10): 17815-17828. doi: 10.3934/math.2022981
[1] | Qian Liu, Jianrui Xie, Ximeng Liu, Jian Zou . Further results on permutation polynomials and complete permutation polynomials over finite fields. AIMS Mathematics, 2021, 6(12): 13503-13514. doi: 10.3934/math.2021783 |
[2] | Varsha Jarali, Prasanna Poojary, G. R. Vadiraja Bhatta . A recent survey of permutation trinomials over finite fields. AIMS Mathematics, 2023, 8(12): 29182-29220. doi: 10.3934/math.20231495 |
[3] | Kaimin Cheng . Permutational behavior of reversed Dickson polynomials over finite fields. AIMS Mathematics, 2017, 2(2): 244-259. doi: 10.3934/Math.2017.2.244 |
[4] | Kaimin Cheng . Permutational behavior of reversed Dickson polynomials over finite fields II. AIMS Mathematics, 2017, 2(4): 586-609. doi: 10.3934/Math.2017.4.586 |
[5] | Guanghui Zhang, Shuhua Liang . On the construction of constacyclically permutable codes from constacyclic codes. AIMS Mathematics, 2024, 9(5): 12852-12869. doi: 10.3934/math.2024628 |
[6] | Weihua Li, Chengcheng Fang, Wei Cao . On the number of irreducible polynomials of special kinds in finite fields. AIMS Mathematics, 2020, 5(4): 2877-2887. doi: 10.3934/math.2020185 |
[7] | Xiaofan Xu, Yongchao Xu, Shaofang Hong . Some results on ordinary words of standard Reed-Solomon codes. AIMS Mathematics, 2019, 4(5): 1336-1347. doi: 10.3934/math.2019.5.1336 |
[8] | Raziuddin Siddiqui . Geometry of configurations in tangent groups. AIMS Mathematics, 2020, 5(1): 522-545. doi: 10.3934/math.2020035 |
[9] | Weitao Xie, Jiayu Zhang, Wei Cao . On the number of the irreducible factors of $ x^{n}-1 $ over finite fields. AIMS Mathematics, 2024, 9(9): 23468-23488. doi: 10.3934/math.20241141 |
[10] | Waseem Ahmad Khan, Kottakkaran Sooppy Nisar, Dumitru Baleanu . A note on (p, q)-analogue type of Fubini numbers and polynomials. AIMS Mathematics, 2020, 5(3): 2743-2757. doi: 10.3934/math.2020177 |
Constructing permutation polynomials is a hot topic in finite fields. Recently, huge kinds of permutation polynomials over Fq2 have been studied. In this paper, by using AGW criterion and piecewise method, we construct several classes of permutation polynomials over Fq3 of the forms similar to (xq2+xq+x+δ)q3−1d+1+L(x), for d=2,3,4,6, where L(x) is a linearized polynomial over Fq.
Let Fq be the finite field of characteristic p with q elements (q=pn,n∈N), and F∗q be the nonzero elements of Fq. Let Fq[x] be the ring of polynomials over Fq in the indeterminate x. A polynomial f(x)∈Fq[x] is called a permutation polynomial if f induces a bijection from Fq to itself. Recently, several classes of permutation polynomials were studied, which can be referred to [5,8,11,12,20,21]. More information about properties, constructions and applications of permutation polynomials may be found in the books [7,9]. We refer the readers to [3,14] for more details of the recent advances.
We found that the permutation polynomials of the form (xq+x+δ)s+x were studied wildly. During the research on Kloosterman sums, a class of permutation polynomials found in [4] motivated Yuan and Ding to study the permutation polynomials with the form (x2k+x+δ)s+x. Tu et al. [13] further proposed two classes of permutation polynomials having the form (x2m+x+δ)s+x over F22m. Li et al. [6] presented a kind of permutation polynomials over Fq2 of the form
(xq−x+δ)q2−13+1+x. |
Yuan and Zheng [19] studied the permutation polynomials over Fq2 having the form
(xpk+ax+δ)q2−1d+1−ax. |
Recently, Zheng et al. [22] constructed large classes of permutation polynomials over Fq2, which have a more general form
(axq+bx+c)rϕ((axq+bx+c)q2−1d)+uxq+vx, |
where a,b,c,u,v∈Fq2,r∈Z+,ϕ(x)∈Fq2[x]. We notice that these classes of permutation polynomials are in Fq2. Comparing to permutation polynomials over Fq2, there are few classes of permutation polynomials over Fq3 constructed recently. Ding et al. [2] presented six classes of permutation polynomials over F33m. Wang, Zhang, Bartoli and Wang [15] constructed several classes of permutation polynomials and complete permutation polynomials over Fq3. Motivated by recent constructions of permutation polynomials over Fq2 and Fq3, we can use AGW criterion and piecewise method to construct several classes of permutation polynomials over Fq3. In this paper, we will focus on constructing permutation polynomials over Fq3 having the similar forms to
(xq2+xq+x+δ)q3−1d+1+L(x),d=2,3,4,6, |
where L(x) is a linearized polynomial over Fq.
The remainder of this paper is organized as follows. In Section 2, we introduce three lemmas which are used in the following sequels. Subsequently, we give several classes of permutation polynomials of the forms similar to (xq2+xq+x+δ)q3−1d+1+L(x) for d=2,3,4,6, in Section 3–6, respectively.
In this section, we will present some results that will be used in the sequels.
The following lemma was developed by Akbary, Ghioca and Wang [1, Lemma 1.1], which was called AGW criterion in [10]. By using AGW criterion, several classes of permutation polynomials were constructed, which could be seen in [16,17,18,19,22]. The lemma is listed as follows:
Lemma 2.1. [1] Let A,S and ˉS be finite sets with #S=#ˉS, and let f:A→A,h:S→ˉS,φ:A→S, and ψ:A→ˉS be maps such that ψ∘f=h∘φ, i.e., the following diagram is commutative:
Af⟶A↓φ↓ψSh⟶ˉS |
If both ψ and φ are surjective, then the following statements are equivalent:
(i) f is bijective (a permutation of A); and
(ii) h is bijective from S to ˉS and f is injective on φ−1(s) for each s∈S.
Lemma 2.2. Let k,δ be in Fq. Then we have
#Im(xq2+xq+x+δ)=#Im(xq2+xq+x+kδ)=#Fq. |
Proof. Let φ(x)=xq2+xq+x+δ, then we find that φ(x)q=φ(x). Thus for every α∈Fq3, we have φ(α)∈Fq. Namely, Im(φ(x))⊆Fq. On the other hand, since #(Imφ(x))≥q3/deg(φ(x))=q3/q2=q, it follows that Im(φ(x))=Fq. Similarly, we can easily get that Im(xq2+xq+x+kδ)=Fq. Therefore, #(Im(xq2+xq+x+δ))=#(Im(xq2+xq+x+kδ))=#Fq.
The proof of Lemma 2.2 is completed.
We close this section with the following result that will be used frequently in the sequels.
Lemma 2.3. [7] Let L(x)=∑m−1i=0aixqi be a linearized polynomial of Fqm for m∈Z+. Then L(x) is a permutation polynomial of Fqm if and only if
det(a0a1a2⋯am−1aqm−1aq0aq1⋯aqm−2aq2m−2aq2m−1aq20⋯aq2m−3⋮⋮⋮⋯⋮aqm−11aqm−12aqm−13⋯aqm−10)≠0. |
In this section, we suppose that p≠3 is an odd prime, ϵ is a primitive element of Fq3. Define D0=<ϵ2>, which is the multiplicative group generated by ϵ2, and D1=ϵD0. Then we get that Fq3={0}∪D0∪D1. Furthermore, if x∈Di, we have
xq3−12=ϵiq3−12=(−1)i, |
where i=0,1.
Since p is an odd prime, then q≡1(mod2). By Lemma 2.2, we know that
#Im(xq2+xq+x+δ)=#Im(xq2+xq+x+kδ)=#Fq. |
In this case, Im(xq2+xq+x+δ)=Im(xq2+xq+x+kδ)=Fq is the disjoint union of 0 and q−12 elements from D0 and q−12 elements from D1.
By using Lemma 2.3, we give the first main result of this paper.
Theorem 3.1. Let p≠3 be an odd prime and a,b,c,d,δ∈Fq. If b,c and d satisfy
det(bcddbccdb)≠0, |
then
a(xq2+xq+x+δ)q3−12+1−bxq2−cxq−dx |
is a permutation polynomial over Fq3 if and only if (3a−(b+c+d),−(3a+b+c+d)) belongs to one of {D0×D0,D1×D1}.
Proof. Let f(x)=a(xq2+xq+x+δ)q3−12+1−bxq2−cxq−dx, φ(x)=xq2+xq+x+δ, ψ(x)=xq2+xq+x−(b+c+d)δ and h(x)=x(3axq3−12−(b+c+d)). One has
ψ∘f=ψ(f(x))=f(x)q2+f(x)q+f(x)−(b+c+d)δ=a(xq2+xq+x+δ)q3−12+1−bxq−cx−dxq2+a(xq2+xq+x+δ)q3−12+1−bx−cxq2−dxq+a(xq2+xq+x+δ)q3−12+1−bxq2−cxq−dx−(b+c+d)δ=3aφ(x)q3−12+1−(b+c+d)φ(x)=h(φ(x))=h∘φ. |
By Lemma 2.3, it follows from
det(bcddbccdb)≠0, |
that bxq2+cxq+dx is a permutation polynomial of Fq3. Clearly, asq3−12+1−(bxq2+cxq+dx) for every s∈Fq3 is also a permutation polynomial of Fq3. For x∈φ−1(s), it implies that f(x)=asq3−12+1−(bxq2+cxq+dx). Thus for every s∈Im(φ), f(x) is injective on φ−1(s). By Lemma 2.1, we know that f(x) is a permutation polynomial over Fq3 if and only if h(x) is a bijection from Im(φ) to Im(ψ).
It is easy to check that
h(x)={(3a−(b+c+d))x,x∈D0;−(3a+(b+c+d))x,x∈D1. |
Then we can infer that h(x) is bijective from Im(φ) to Im(ψ) if and only if (3a−(b+c+d),−(3a+b+c+d)) belongs to one of {D0×D0,D1×D1}. Thus
a(xq2+xq+x+δ)q3−12+1−bxq2−cxq−dx |
is a permutation polynomial over Fq3 if and only if (3a−(b+c+d),−(3a+b+c+d)) belongs to one of {D0×D0,D1×D1}.
The proof of Theorem 3.1 is completed.
Corollaries 3.2 and 3.3 are two explicit examples of Theorem 3.1, we omit their proofs, just list them in the following.
Corollary 3.2. Let p≠3 be an odd prime, δ∈Fq and let a,b∈Fq satisfy 3a−b≠0 and 3a+b≠0. Then we get that
a(xq2+xq+x+δ)q3−12+1−bx |
is a permutation polynomial over Fq3 if and only if (3a−b,−(3a+b)) belongs to one of {D0×D0,D1×D1}.
Corollary 3.3. Let p≠3,5 be an odd prime, δ∈Fq and −5∈D0. Then we get that
(xq2+xq+x+δ)q3−12+1−x−xq |
is a permutation polynomial over Fq3.
Example 3.4. For q=19 and δ∈F19, it is easy to check that (2,−1)∈D1×D1 by Magma, thus we can imply from Corollary 3.2 that (xq2+xq+x+δ)2287−x is a permutation polynomial over F193.
In this section, we suppose p≠3 be an odd prime and q3≡1(mod3). Let ϵ be a primitive element of Fq3, we define D0=<ϵ3> and Di=ϵiD0 for i=1,2. Then Fq3={0}∪D0∪D1∪D2. Let ω=ϵq3−13, one has ω2+ω+1=0.
It follows from q3≡1(mod3) that q≡1(mod3) and q2+q+1≡0(mod3), so Fq⊆{0}∪D0.
We can get the following result, which is the second main result in this paper.
Theorem 4.1. Let p≠3 be an odd prime, q≡1(mod3) and a,b,c,d,δ∈Fq. If 6a+b+c+d≠0, −3a+b+c+d≠0 and
det(bcddbccdb)≠0, |
then we have
a(xq2+xq+x+δ)q3−13+1+a(xq2+xq+x+δ)2(q3−1)3+1+bxq2+cxq+dx |
is a permutation polynomial over Fq3.
Proof. Define
f(x)=a(xq2+xq+x+δ)q3−13+1+a(xq2+xq+x+δ)2(q3−1)3+1+bxq2+cxq+dx,φ(x)=xq2+xq+x+δ,ψ(x)=xq2+xq+x+(b+c+d)δ,h(x)=x(3axq3−13+3ax2(q3−1)3+b+c+d). |
In the following, it is easy to check that
ψ∘f=ψ(f(x))=f(x)q2+f(x)q+f(x)+(b+c+d)δ=a(xq2+xq+x+δ)q3−13+1+a(xq2+xq+x+δ)2(q3−1)3+1+bxq+cx+dxq2+a(xq2+xq+x+δ)q3−13+1+a(xq2+xq+x+δ)2(q3−1)3+1+bx+cxq2+dxq+a(xq2+xq+x+δ)q3−13+1+a(xq2+xq+x+δ)2(q3−1)3+1+bxq2+cxq+dx+(b+c+d)δ=3aφ(x)q3−13+1+3aφ(x)2(q3−1)3+1+(b+c+d)φ(x)=h(φ(x))=h∘φ. |
Since
det(bcddbccdb)≠0, |
it follows from Lemma 2.3 that bxq2+cxq+dx is a permutation polynomial of Fq3. Thus f(x) is injective on φ−1(s). By Lemma 2.1, f(x) is a permutation polynomial over Fq3 if and only if h(x) is bijective from Im(φ) to Im(ψ).
By Lemma 2.2, we get that Im(φ)=Im(ψ)=Fq. Hence we need to prove that h(x) is a bijection on Fq. Since xq3−13+x2(q3−1)3=2 for x∈D0 and xq3−13+x2(q3−1)3=−1 for x∈D1∪D2, we get that
h(x)={(6a+b+c+d)x,x∈D0;(−3a+b+c+d)x,x∈D1;(−3a+b+c+d)x,x∈D2. |
Since q≡1(mod3), then Fq⊆D0∪{0}. It follows that 6a+b+c+d and −3a+b+c+d are in D0. Thus h(x) is a bijection on Fq.
Therefore, we can conclude that a(xq2+xq+x+δ)q3−13+1+a(xq2+xq+x+δ)2(q3−1)3+1+bxq2+cxq+dx is a permutation polynomial over Fq3. This completes the proof of Theorem 4.1.
By Theorem 4.1, we can easily give the following results.
Corollary 4.2. Let p≠3 be an odd prime, q≡1(mod3) and δ∈Fq. Then we get that
(xq2+xq+x+δ)q3−13+1+(xq2+xq+x+δ)2(q3−1)3+1−3x |
is a permutation polynomial over Fq3.
Corollary 4.3. Let p≠3 be an odd prime, q≡1(mod3) and δ∈Fq. Then we get that
(xq2+xq+x+δ)q3−13+1+(xq2+xq+x+δ)2(q3−1)33+1+xq+x |
is a permutation polynomial over Fq3.
In this section, we suppose that p≠3 is an odd prime and q3≡1(mod4), ϵ is a primitive element of Fq3. Define D0=<ϵ4>, which is the multiplicative group generated by ϵ4, and Di=ϵiD0 for i=1,2,3. Then we get that Fq3={0}∪D0∪D1∪D2∪D3. Note that xq3−14=ϵiq3−14, for x∈Di, where i=1,2,3. Let ω=ϵq3−14, then ω2+1=0.
Since q3≡1(mod4), we get that q≡1(mod4), then q2+q+1≡3(mod4). By Lemma 2.2, we know that
#Im(xq2+xq+x+δ)=#Im(xq2+xq+x+kδ)=#Fq. |
It follows that Fq is a disjoint union of 0 and q−14 elements from D0 and q−14 elements from D1 and q−14 elements from D2 and q−14 elements from D3.
We present the third main result of this paper in the following.
Theorem 5.1. Let p≠3 be an odd prime, q≡1(mod4) and δ∈Fq. Let a∈F∗q,b,c,d∈Fq satisfy 6a+b+c+d≠0,b+c+d≠0, −6a+b+c+d≠0 and
det(bcddbccdb)≠0. |
Then
a(xq2+xq+x+δ)q3−14+1+a(xq2+xq+x+δ)3(q3−1)4+1+bxq2+cxq+dx |
is a permutation polynomial over Fq3 if and only if (6a+b+c+d,b+c+d,−6a+b+c+d) belongs to one of {D0×D0×D0,D0×D2×D0,D1×D1×D1,D1×D3× D1,D2×D2×D2,D2×D0×D2,D3×D3×D3,D3×D1×D3}.
Proof. Let
f(x)=a(xq2+xq+x+δ)q3−14+1+a(xq2+xq+x+δ)3(q3−1)4+1+bxq2+cxq+dx,φ(x)=xq2+xq+x+δ,ψ(x)=xq2+xq+x+(b+c+d)δ,h(x)=x(3axq3−14+3ax3(q3−1)4+b+c+d). |
It is easy to check that
ψ∘f=ψ(f(x))=f(x)q2+f(x)q+f(x)+(b+c+d)δ=a(xq2+xq+x+δ)q3−14+1+a(xq2+xq+x+δ)3(q3−1)4+1+bxq+cx+dxq2+a(xq2+xq+x+δ)q3−14+1+a(xq2+xq+x+δ)3(q3−1)4+1+bx+cxq2+dxq+a(xq2+xq+x+δ)q3−14+1+a(xq2+xq+x+δ)3(q3−1)4+1+bxq2+cxq+dx+(b+c+d)δ=3aφ(x)q3−14+1+3aφ(x)3(q3−1)4+1+(b+c+d)φ(x)=h(φ(x))=h∘φ. |
Since
det(bcddbccdb)≠0, |
it follows from Lemma 2.3 that bxq2+cxq+dx is a permutation polynomial over Fq3. Therefore, we can easily get that for every s∈Im(φ), f(x) is injective on φ−1(s). By Lemma 2.1, f(x) is a permutation polynomial over Fq3 if and only if h(x) is bijective from Im(φ) to Im(ψ).
Note that
xq3−14+x3(q3−1)4={2,x∈D0;0,x∈D1;−2,x∈D2;0,x∈D3. |
Then h(x) can be rewritten as
h(x)={(6a+b+c+d)x,x∈D0;(b+c+d)x,x∈D1;(−6a+b+c+d)x,x∈D2;(b+c+d)x,x∈D3. |
In the following, we focus on proving that h(x) is a bijection from Im(φ) to Im(ψ) if and only if (6a+b+c+d,b+c+d,−6a+b+c+d) belongs to one of {D0×D0×D0,D0×D2×D0,D1×D1×D1,D1×D3×D1,D2×D2 ×D2,D2×D0×D2,D3×D3×D3,D3×D1×D3}.
Now we first prove the sufficiency part. Assume that h(x) is a bijection. We consider the following cases:
(1) 6a+b+c+d∈D0. Clearly, we have (6a+b+c+d)x∈D0 for x∈D0, then it implies that (b+c+d)x∉D0 for x∈D1∪D3. It tells us that b+c+d∉D1∪D3. If b+c+d∈D0, then (b+c+d)x∈D1 and (b+c+d)x∈D3 for x∈D1 and x∈D3, respectively. Since h(x) is a bijection, we can get that (−6a+b+c+d)x∈D2 for x∈D2, thus −6a+b+c+d∈D0. On the other hand, if b+c+d∈D2, then (b+c+d)x∈D3 for x∈D1 and (b+c+d)x∈D1 for x∈D3. It follows from h(x) being a bijection that (−6a+b+c+d)x∈D2 for x∈D2, one has −6a+b+c+d∈D0.
(2) 6a+b+c+d∈D1. In this case, we can imply that b+c+d∉D0∪D2. If b+c+d∈D1, by h(x) being a bijection, we can get that −6a+b+c+d∈D1. Similarly, if b+c+d∈D3, then it follows from h(x) being a bijection that −6a+b+c+d∈D1.
(3) 6a+b+c+d∈D2. In this case, we first get that b+c+d∉D1∪D3. If b+c+d∈D2, since h(x) is a bijection, we can get that −6a+b+c+d∈D2. If b+c+d∈D0, it follows that −6a+b+c+d∈D2.
(4) 6a+b+c+d∈D3. In this case, it is easy to imply that b+c+d∉D0∪D2. If b+c+d∈D1, then we can get that −6a+b+c+d∈D3. If b+c+d∈D3, it follows from h(x) being a bijection that −6a+b+c+d∈D3.
Conversely, we can easily check the necessity part is true.
Thus we can conclude that a(xq2+xq+x+δ)q3−14+1+a(xq2+xq+x+δ)3q3−14+1+bxq2+cxq+dx is a permutation polynomial over Fq3 if and only if (6a+b+c+d,b+c+d,−6a+b+c+d) belongs to one of {D0×D0×D0,D0×D2×D0,D1×D1×D1,D1×D3×D1,D2×D2 ×D2,D2×D0×D2,D3×D3×D3,D3×D1×D3}.
By Theorem 5.1, we can get the following results.
Corollary 5.2. Let p≠3 be an odd prime, q≡1(mod4) and δ∈Fq. Then
(xq2+xq+x+δ)q3−14+1+(xq2+xq+x+δ)3(q3−1)4+1−x |
is a permutation polynomial over Fq3 if and only if (−1,5,7) belongs to one of {D0×D0×D0,D0×D2×D2,D2×D0×D2,D2×D2×D0}.
Corollary 5.3. Let p≠3 be an odd prime, q≡1(mod4) and δ∈Fq. Then
(xq2+xq+x+δ)q3−14+1+(xq2+xq+x+δ)3(q3−1)4+1+xq+x |
is a permutation polynomial over Fq3 if and only if (−1,2) belongs to one of {D0×D0,D0×D2,D2×D3}.
Example 5.4. For q=29 and δ∈Fq, we can check that (−1,5,7)∈D2×D2×D0 by Magma, it follows from Corollary 5.2 that (xq2+xq+x+δ)6098+(xq2+xq+x+δ)18292−x is a permutation polynomial over F293.
In what follows, we will give the fourth main result in this paper.
Theorem 5.5. Let p≠3,5 be an odd prime, q≡1(mod4) and δ∈Fq. Then
(xq2+xq+x+δ)q3−14+1−x |
is a permutation polynomial over Fq3 if and only if (2,−4,3ω−1,−3ω−1) belongs to one of {D0×D0×D0×D0,D0×D0×D2×D2,D1 ×D2×D1×D0,D1×D2×D2×D3,D1×D0×D3×D0,D1×D0×D2×D1,D2 ×D2×D0×D0,D2×D2×D2×D2,D3×D2×D0×D3,D3×D2×D2×D2,D3×D0×D3×D2,D3×D0×D0×D1}.
Proof. Let (xq2+xq+x+δ)q3−14+1−x, φ(x)=xq2+xq+x+δ, ψ(x)=xq2+xq+x−δ and h(x)=x(3xq3−14−1). Then we can easily check that
ψ∘f=h∘φ. |
Namely, the diagram is commutative. Furthermore, it is easy to get that for every s∈Im(φ), f(x) is injective on φ−1(s). Thus it follows from Lemma 2.1 that f(x) is a permutation polynomial over Fq3 if and only if h(x) is bijective from Im(φ) to Im(ψ).
By Lemma 2.2, we know that Im(φ)=Im(ψ)=Fq. For x∈Fq, we can rewrite h(x) as:
h(x)={2x,x∈D0;(3ω−1)x,x∈D1;−4x,x∈D2;(−3ω−1)x,x∈D3. |
Since p≠3,5 is an odd prime, we get that (3ω−1)(−3ω−1)≠0. We claim that h(x) is bijective from Im(φ) to Im(ψ) if and only if (2,−4,3ω−1,−3ω−1) belongs to one of {D0×D0×D0×D0,D0×D0×D2×D2,D1 ×D2×D1×D0,D1×D2×D2×D3,D1×D0×D3×D0, D1×D0×D2×D1,D2×D2×D0×D0,D2×D2 ×D2×D2,D3×D2×D0×D3,D3×D2×D2×D2,D3×D0×D3×D2,D3×D0×D0×D1}.
First we assume h(x) is bijective from Im(φ) to Im(ψ). Since q≡1(mod4), it tells us that −1∈D0 or −1∈D2. We consider the following cases:
(1) 2∈D0. Then we get that −4∈D0 or −4∈D2. For x∈D0, we have 2x∈D0, it follows from h(x) being a bijection that −4x∉D0 for x∈D2, hence −4∈D0 and −4x∈D2 for x∈D2. Furthermore, since 2x∈D0 for x∈D0 and −4x∈D2 for x∈D2, then (3ω−1)x∉D0∪D2 for x∈D1, thus 3ω−1∈D0 or 3ω−1∈D2. If 3ω−1∈D0, by h(x) being a bijection, one has −3ω−1∈D0. If 3ω−1∈D2, then it implies that −3ω−1∈D2.
(2) 2∈D1. Since −1∈D0 or −1∈D2, thus −4∈D2 or −4∈D0. For the first case, if −4∈D2, then 2x∈D1 for x∈D0 and −4x∈D0 for x∈D2. Since h(x) is a bijection, we know that 3ω−1∈D1 or 3ω−1∈D2. If 3ω−1∈D1, it follows that −3ω−1∈D0. If 3ω−1∈D2, one has that −3ω−1∈D3. For the second case, if −4∈D0, then 2x∈D1 for x∈D0 and −4x∈D2 for x∈D2. Since h(x) is a bijection, we know that 3ω−1∈D3 or 3ω−1∈D2. If 3ω−1∈D3, it follows that −3ω−1∈D0. If 3ω−1∈D2, we know that −3ω−1∈D1.
(3) 2∈D2. Since 2∈D2, then we get that −4∈D0 or −4∈D2. For x∈D0, we have 2x∈D2, it follows from h(x) being a bijection that −4x∉D2 for x∈D2, hence −4∈D2 and −4x∈D0 for x∈D2. Furthermore, since 2x∈D2 for x∈D0 and −4x∈D0 for x∈D2, then (3ω−1)x∉D0∪D2 for x∈D1, thus 3ω−1∈D0 or 3ω−1∈D2. If 3ω−1∈D0, it implies that −3ω−1∈D0. If 3ω−1∈D2, we have −3ω−1∈D2.
(4) 2∈D3. Since −1∈D0 or −1∈D2, thus −4∈D2 or −4∈D0. For the first case, if −4∈D2, then 2x∈D3 for x∈D0 and −4x∈D0 for x∈D2. Since h(x) is a bijection, we know that 3ω−1∈D0 or 3ω−1∈D1. If 3ω−1∈D0, it follows that −3ω−1∈D3. If 3ω−1∈D1, we get that −3ω−1∈D2. For the second case, if −4∈D0, then 2x∈D3 for x∈D0 and −4x∈D2 for x∈D2. Since h(x) is a bijection, one has 3ω−1∈D3 or 3ω−1∈D0. If 3ω−1∈D3, it follows that −3ω−1∈D2. If 3ω−1∈D0, it implies that −3ω−1∈D1.
The sufficiency part is proved.
For the necessity part, if (2,−4,3ω−1,−3ω−1) belongs to one of {D0×D0×D0×D0,D0×D0×D2×D2,D1×D2×D1×D0,D1×D2×D2 ×D3,D1×D0×D3×D0,D1×D0×D2×D1,D2×D2×D0×D0,D2×D2×D2 ×D2,D3×D2×D0×D3,D3×D2×D2×D2,D3×D0×D3×D2,D3×D0×D0×D1}, we easily check that h(x) is a bijection from Im(φ) to Im(ψ). Thus the claim is true.
Therefore, we get that
(xq2+xq+x+δ)q3−14+1−x |
is a permutation polynomial over Fq3 if and only if (2,−4,3ω−1,−3ω−1) belongs to one of {D0×D0×D0×D0,D0×D0×D2×D2,D1×D2 ×D1×D0,D1×D2×D2×D3,D1×D0×D3×D0, D1×D0×D2×D1,D2×D2×D0×D0,D2×D2×D2 ×D2,D3×D2×D0×D3,D3×D2×D2×D2,D3×D0×D3×D2,D3×D0×D0×D1}.
The proof of Theorem 5.5 is completed.
Similarly, we have
Theorem 5.6. Let p≠3,5 be an odd prime, q≡1(mod4) and δ∈Fq. Then
(xq2+xq+x+δ)3(q3−1)4+1−x |
is a permutation polynomial over Fq3 if and only if (2,−4,−3ω−1,3ω−1) belongs to one of {D0×D0×D0×D0,D0×D0×D2×D2,D1×D2 ×D1×D0,D1×D2×D2×D3,D1×D0×D3×D0,D1×D0×D2×D1,D2×D2 ×D0×D0,D2×D2×D2×D2,D3×D2×D0×D3,D3 ×D2×D2×D2,D3×D0×D3×D2,D3×D0×D0×D1}.
In this section, let p≠3 be an odd prime and q3≡1(mod6). We assume ϵ is a primitive element of Fq3 and define D0=<ϵ6>, which is the multiplicative group generated by ϵ6, and Di=ϵiD0 for i=1,2,3,4,5. Then we get that Fq3={0}∪D0∪D1∪D2∪D3∪D4∪D5. Furthermore, if x∈Di, we notice that xq3−16=ϵiq3−16. For simplicity, we define ω=ϵq3−16, which satisfies ω2−ω+1=0.
Since q3≡1(mod6), we know that q≡1(mod6), then q2+q+1≡3(mod6) and q≡1(mod2). By Lemma 2.2, we know that
#Im(xq2+xq+x+δ)=#Im(xq2+xq+x+kδ)=#Fq, |
and Fq is a disjoint union of 0 and q−12 elements from D0 and q−12 elements from D3.
In what follows, we can get the permutation polynomials of the form a(xq2+xq+x+δ)q3−16+1+bx, which is the fifth main result in this paper.
Theorem 6.1. Let p≠3 be an odd prime, q≡1(mod6) and δ∈Fq, and let a,b∈Fq such that 3a+b≠0 and −3a+b≠0. Then
a(xq2+xq+x+δ)q3−16+1+bx |
is a permutation polynomial over Fq3 if and only if (3a+b,−3a+b) belongs to one of {D0×D0,D3×D3}.
Proof. Let f(x)=a(xq2+xq+x+δ)q3−16+1+bx, φ(x)=xq2+xq+x+δ, ψ(x)=xq2+xq+x+bδ and h(x)=x(3axq3−16+b). It is easy to check that
ψ∘f=ψ(f(x))=f(x)q2+f(x)q+f(x)+bδ=a(xq2+xq+x+δ)q3−16+1+bxq2+a(xq2+xq+x+δ)q3−16+1+bxq+a(xq2+xq+x+δ)q3−16+1+bx+bδ=3aφ(x)q3−16+1+bφ(x)=h(φ(x))=h∘φ. |
Furthermore, we can easily check that for every s∈Im(φ), f(x) is injective on φ−1(s). Then it follows from Lemma 2.1 that f(x) is a permutation polynomial over Fq3 if and only if h(x) is bijective from Im(φ) to Im(ψ).
Since Im(φ)=Im(ψ)=Fq is a disjoint union of 0 and q−12 elements from D0 and q−12 elements from D3, we get that
h(x)={(3a+b)x,x∈D0;(−3a+b)x,x∈D3. |
Next we prove that h(x) is bijective from Im(φ) to Im(ψ) if and only if (3a+b,−3a+b) belongs to one of {D0×D0,D3×D3}.
Firstly, we give the proof of the sufficiency part. Suppose h(x) is bijective from Im(φ) to Im(ψ). If 3a+b∈D0, then we have (3a+b)D0=D0. Hence (−3a+b)D3=D3, it tells us that −3a+b∈D0. On the other hand, if 3a+b∈D3, then (3a+b)D0=D3 and (−3a+b)D3=D0, thus it follows that −3a+b∈D3.
Now we prove the necessity part. If (3a+b,−3a+b)∈D0×D0 or D3×D3, it is easy to check that h(x) is bijective from Im(φ) to Im(ψ). Therefore, h(x) is bijective from Im(φ) to Im(ψ) if and only if (3a+b,−3a+b) belongs to one of {D0×D0,D3×D3}.
We can conclude that
a(xq2+xq+x+δ)q3−16+1+bx |
is a permutation polynomial over Fq3 if and only if (3a+b,−3a+b) belongs to one of {D0×D0,D3×D3}.
The proof of Theorem 6.1 is completed.
Corollary 6.2. Let p≠3 be an odd prime, q≡1(mod6) and δ∈Fq. Then
(xq2+xq+x+δ)q3−16+1−x |
is a permutation polynomial over Fq3 if and only if (2,−4) belongs to one of {D0×D0,D3×D3}.
Proof. Taking a=1,b=−1 in Theorem 6.1, it follows from Theorem 6.1 that (xq2+xq+x+δ)q3−16+1−x is a permutation polynomial over Fq3 if and only if (2,−4) belongs to one of {D0×D0,D3×D3}.
Example 6.3. For q=19 and δ∈F19, we can find that (2,−4)∈D3×D3 by Magma, by Corollary 6.2, it follows that (xq2+xq+x+δ)1144−x is a permutation polynomial over F193.
Similarly, we have
Theorem 6.4. Let p≠3 be an odd prime, q≡1(mod6) and δ∈Fq, and let a,b∈Fq such that 3a+b≠0 and −3a+b≠0. Then
a(xq2+xq+x+δ)5(q3−1)6+1+bx |
is a permutation polynomial over Fq3 if and only if (3a+b,−3a+b) belongs to one of {D0×D0,D3×D3}.
Proof. The proof of Theorem 6.4 is similar to that of Theorem 6.1, here we omit it.
Furthermore, we get a permutation polynomial having a more general form.
Theorem 6.5. Let p≠3 be an odd prime, q≡1(mod6) and δ∈Fq, and let a,b,c∈Fq such that 3a+3b+c≠0 and −3a−3b+c≠0. Then
a(xq2+xq+x+δ)q3−16+1+b(xq2+xq+x+δ)5(q3−1)6+1+cx |
is a permutation polynomial over Fq3 if and only if (3a+3b+c,−3a−3b+c) belongs to one of {D0×D0,D3×D3}.
Proof. The proof of Theorem 6.5 is similar to that of Theorem 6.1, here we only give that h(x)=x(3axq3−16+3bx5q3−16+c), and h(x) can be rewritten as
h(x)={(3a+3b+c)x,x∈D0;(−3a−3b+c)x,x∈D3. |
We omit the other details of the proof.
In this paper, motivated by some constructions of permutation polynomials over Fq2, we used AGW criterion and piecewise method to construct several classes of permutation polynomials over Fq3 of the forms (xq2+xq+x+δ)ϕ((xq2+xq+x+δ)q3−1d)+L(x), for d=2,3,4,6, where L(x) is a linearized polynomial over Fq, which enrich the permutation polynomials over Fq3.
The Authors express their gratitude to the anonymous reviewers for carefully examining this paper and providing a number of valuable comments and suggestions. This research was supported by the National Science Foundation of China (No. 11926344) and by Science and Technology Research Program of Chongqing Municipal Education Commission (No. KJQN201900506).
We declare that we have no conflict of interest.
[1] |
A. Akbary, D. Ghioca, Q. Wang, On constructing permutations of finite fields, Finite Fields Appl., 17 (2011), 51–67. https://doi.org/10.1016/j.ffa.2010.10.002 doi: 10.1016/j.ffa.2010.10.002
![]() |
[2] |
C. S. Ding, Q. Xiang, J. Yuan, P. Z. Yuan, Explicit classes of permutation polynomials of F33m, Sci. China Ser. A: Math., 52 (2009), 639–647. https://doi.org/10.1007/s11425-008-0142-8 doi: 10.1007/s11425-008-0142-8
![]() |
[3] |
X. D. Hou, Permutation polynomials over finite fields–A survey of recent advances, Finite Fields Appl., 32 (2015), 82–119. https://doi.org/10.1016/j.ffa.2014.10.001 doi: 10.1016/j.ffa.2014.10.001
![]() |
[4] |
T. Helleseth, V. Zinoviev, New Kloosterman sums identities over F2m for all m, Finite Fields Appl., 9 (2003), 187–193. https://doi.org/10.1016/S1071-5797(02)00028-X doi: 10.1016/S1071-5797(02)00028-X
![]() |
[5] |
K. Q. Li, L. J. Qu, Q. Wang, Compositional inverses of permutation polynomials of the form xrh(xs) over finite fields, Cryptogr. Commun., 11 (2019), 279–298. https://doi.org/10.1007/s12095-018-0292-7 doi: 10.1007/s12095-018-0292-7
![]() |
[6] |
N. Li, T. Helleseth, X. H. Tang, Further results on a class of permutation polynomials over finite fields, Finite Fields Appl., 22 (2013), 16–23. https://doi.org/10.1016/j.ffa.2013.02.004 doi: 10.1016/j.ffa.2013.02.004
![]() |
[7] | R. Lidl, H. Niederreiter, Finite fields, 2 Eds, Cambridge: Cambridge University Press, 1997. |
[8] |
Q. Liu, Y. J. Sun, W. G. Zhang, Some classes of permutation polynomials over finite fields with odd characteristic, AAECC, 29 (2018), 409–431. https://doi.org/10.1007/s00200-018-0350-6 doi: 10.1007/s00200-018-0350-6
![]() |
[9] | G. L. Mullen, D. Panario, Handbook of finite fields, Chapman and Hall/CRC, 2013. https://doi.org/10.1201/b15006 |
[10] | G. L. Mullen, Q. Wang, Permutation polynomials in one variable, In: Handbook of finite fields, Chapman and Hall/CRC, 2013,215–229. |
[11] |
X. E. Qin, S. F. Hong, Constructing permutation polynomials over finite fields, Bull. Aust. Math. Soc., 89 (2014), 420–430. https://doi.org/10.1017/S0004972713000646 doi: 10.1017/S0004972713000646
![]() |
[12] |
X. E. Qin, G. Y. Qian, S. F. Hong, New results on permutation polynomials over finite fields, Int. J. Number Theory, 11 (2015), 437–449. https://doi.org/10.1142/S1793042115500220 doi: 10.1142/S1793042115500220
![]() |
[13] |
Z. R. Tu, X. Y. Zeng, Y. P. Jiang, Two classes of permutation polynomials having the form (x2m+x+δ)s+x, Finite Fields Appl., 31 (2015), 12–24. https://doi.org/10.1016/j.ffa.2014.09.005 doi: 10.1016/j.ffa.2014.09.005
![]() |
[14] | Q. Wang, Polynomials over finite fields: An index approach, In: Combinatorics and finite fields: Difference sets, polynomials, pseudorandomness and applications, 2019,319–348. https://doi.org/10.1515/9783110642094-015 |
[15] | Y. P. Wang, W. G. Zhang, D. Bartoli, Q. Wang, Permutation polynomials and complete permutation polynomials over Fq3, 2018, arXiv: 1806.05712v1. |
[16] |
P. Z. Yuan, C. S. Ding, Permutation polynomials over finite fields from a powerful lemma, Finite Fields Appl., 17 (2011), 560–574. https://doi.org/10.1016/j.ffa.2011.04.001 doi: 10.1016/j.ffa.2011.04.001
![]() |
[17] |
P. Z. Yuan, C. S. Ding, Further results on permutation polynomials over finite fields, Finite Fields Appl., 27 (2014), 88–103. https://doi.org/10.1016/j.ffa.2014.01.006 doi: 10.1016/j.ffa.2014.01.006
![]() |
[18] |
P. Z. Yuan, C. S. Ding, Permutation polynomials of the form L(x)+S2ak+S2bk over Fq3k, Finite Fields Appl., 29 (2014), 106–117. https://doi.org/10.1016/j.ffa.2014.04.004 doi: 10.1016/j.ffa.2014.04.004
![]() |
[19] |
P. Z. Yuan, Y. B. Zheng, Permutation polynomials from piecewise functions, Finite Fields Appl., 35 (2015), 215–230. https://doi.org/10.1016/j.ffa.2015.05.001 doi: 10.1016/j.ffa.2015.05.001
![]() |
[20] |
Z. B. Zha, L. Hu, Z. Z. Zhang, New results on permutation polynomials of the form (xpm−x+δ)s+xpm+x over Fp2m, Cryptogr. Commun., 10 (2018), 567–578. https://doi.org/10.1007/s12095-017-0234-9 doi: 10.1007/s12095-017-0234-9
![]() |
[21] |
D. B. Zheng, Z. Chen, More classes of permutation polynomial of the form (xpm−x+δ)s+L(x), AAECC, 28 (2017), 215–223. https://doi.org/10.1007/s00200-016-0305-8 doi: 10.1007/s00200-016-0305-8
![]() |
[22] |
Y. B. Zheng, P. Z. Yuan, D. Y. Pei, Large classes of permutation polynomials over Fq2, Des. Codes Cryptogr., 81 (2016), 505–521. https://doi.org/10.1007/s10623-015-0172-5 doi: 10.1007/s10623-015-0172-5
![]() |