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Research article

Some specific classes of permutation polynomials over Fq3

  • Received: 22 June 2022 Revised: 24 July 2022 Accepted: 28 July 2022 Published: 03 August 2022
  • MSC : 11T06, 12E20

  • 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)q31d+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

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  • 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)q31d+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,nN), and Fq 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

    (xqx+δ)q213+1+x.

    Yuan and Zheng [19] studied the permutation polynomials over Fq2 having the form

    (xpk+ax+δ)q21d+1ax.

    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)q21d)+uxq+vx,

    where a,b,c,u,vFq2,rZ+,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)q31d+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)q31d+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:AA,h:SˉS,φ:AS, and ψ:AˉS be maps such that ψf=hφ, i.e., the following diagram is commutative:

    AfAφψShˉS

    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 φ1(s) for each sS.

    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 αFq3, 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)=m1i=0aixqi be a linearized polynomial of Fqm for mZ+. Then L(x) is a permutation polynomial of Fqm if and only if

    det(a0a1a2am1aqm1aq0aq1aqm2aq2m2aq2m1aq20aq2m3aqm11aqm12aqm13aqm10)0.

    In this section, we suppose that p3 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}D0D1. Furthermore, if xDi, we have

    xq312=ϵiq312=(1)i,

    where i=0,1.

    Since p is an odd prime, then q1(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 q12 elements from D0 and q12 elements from D1.

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

    Theorem 3.1. Let p3 be an odd prime and a,b,c,d,δFq. If b,c and d satisfy

    det(bcddbccdb)0,

    then

    a(xq2+xq+x+δ)q312+1bxq2cxqdx

    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)q312+1bxq2cxqdx, φ(x)=xq2+xq+x+δ, ψ(x)=xq2+xq+x(b+c+d)δ and h(x)=x(3axq312(b+c+d)). One has

    ψf=ψ(f(x))=f(x)q2+f(x)q+f(x)(b+c+d)δ=a(xq2+xq+x+δ)q312+1bxqcxdxq2+a(xq2+xq+x+δ)q312+1bxcxq2dxq+a(xq2+xq+x+δ)q312+1bxq2cxqdx(b+c+d)δ=3aφ(x)q312+1(b+c+d)φ(x)=h(φ(x))=hφ.

    By Lemma 2.3, it follows from

    det(bcddbccdb)0,

    that bxq2+cxq+dx is a permutation polynomial of Fq3. Clearly, asq312+1(bxq2+cxq+dx) for every sFq3 is also a permutation polynomial of Fq3. For xφ1(s), it implies that f(x)=asq312+1(bxq2+cxq+dx). Thus for every sIm(φ), f(x) is injective on φ1(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,xD0;(3a+(b+c+d))x,xD1.

    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)q312+1bxq2cxqdx

    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 p3 be an odd prime, δFq and let a,bFq satisfy 3ab0 and 3a+b0. Then we get that

    a(xq2+xq+x+δ)q312+1bx

    is a permutation polynomial over Fq3 if and only if (3ab,(3a+b)) belongs to one of {D0×D0,D1×D1}.

    Corollary 3.3. Let p3,5 be an odd prime, δFq and 5D0. Then we get that

    (xq2+xq+x+δ)q312+1xxq

    is a permutation polynomial over Fq3.

    Example 3.4. For q=19 and δF19, 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)2287x is a permutation polynomial over F193.

    In this section, we suppose p3 be an odd prime and q31(mod3). Let ϵ be a primitive element of Fq3, we define D0=<ϵ3> and Di=ϵiD0 for i=1,2. Then Fq3={0}D0D1D2. Let ω=ϵq313, one has ω2+ω+1=0.

    It follows from q31(mod3) that q1(mod3) and q2+q+10(mod3), so Fq{0}D0.

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

    Theorem 4.1. Let p3 be an odd prime, q1(mod3) and a,b,c,d,δFq. If 6a+b+c+d0, 3a+b+c+d0 and

    det(bcddbccdb)0,

    then we have

    a(xq2+xq+x+δ)q313+1+a(xq2+xq+x+δ)2(q31)3+1+bxq2+cxq+dx

    is a permutation polynomial over Fq3.

    Proof. Define

    f(x)=a(xq2+xq+x+δ)q313+1+a(xq2+xq+x+δ)2(q31)3+1+bxq2+cxq+dx,φ(x)=xq2+xq+x+δ,ψ(x)=xq2+xq+x+(b+c+d)δ,h(x)=x(3axq313+3ax2(q31)3+b+c+d).

    In the following, it is easy to check that

    ψf=ψ(f(x))=f(x)q2+f(x)q+f(x)+(b+c+d)δ=a(xq2+xq+x+δ)q313+1+a(xq2+xq+x+δ)2(q31)3+1+bxq+cx+dxq2+a(xq2+xq+x+δ)q313+1+a(xq2+xq+x+δ)2(q31)3+1+bx+cxq2+dxq+a(xq2+xq+x+δ)q313+1+a(xq2+xq+x+δ)2(q31)3+1+bxq2+cxq+dx+(b+c+d)δ=3aφ(x)q313+1+3aφ(x)2(q31)3+1+(b+c+d)φ(x)=h(φ(x))=hφ.

    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 φ1(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 xq313+x2(q31)3=2 for xD0 and xq313+x2(q31)3=1 for xD1D2, we get that

    h(x)={(6a+b+c+d)x,xD0;(3a+b+c+d)x,xD1;(3a+b+c+d)x,xD2.

    Since q1(mod3), then FqD0{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)q313+1+a(xq2+xq+x+δ)2(q31)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 p3 be an odd prime, q1(mod3) and δFq. Then we get that

    (xq2+xq+x+δ)q313+1+(xq2+xq+x+δ)2(q31)3+13x

    is a permutation polynomial over Fq3.

    Corollary 4.3. Let p3 be an odd prime, q1(mod3) and δFq. Then we get that

    (xq2+xq+x+δ)q313+1+(xq2+xq+x+δ)2(q31)33+1+xq+x

    is a permutation polynomial over Fq3.

    In this section, we suppose that p3 is an odd prime and q31(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}D0D1D2D3. Note that xq314=ϵiq314, for xDi, where i=1,2,3. Let ω=ϵq314, then ω2+1=0.

    Since q31(mod4), we get that q1(mod4), then q2+q+13(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 q14 elements from D0 and q14 elements from D1 and q14 elements from D2 and q14 elements from D3.

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

    Theorem 5.1. Let p3 be an odd prime, q1(mod4) and δFq. Let aFq,b,c,dFq satisfy 6a+b+c+d0,b+c+d0, 6a+b+c+d0 and

    det(bcddbccdb)0.

    Then

    a(xq2+xq+x+δ)q314+1+a(xq2+xq+x+δ)3(q31)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)q314+1+a(xq2+xq+x+δ)3(q31)4+1+bxq2+cxq+dx,φ(x)=xq2+xq+x+δ,ψ(x)=xq2+xq+x+(b+c+d)δ,h(x)=x(3axq314+3ax3(q31)4+b+c+d).

    It is easy to check that

    ψf=ψ(f(x))=f(x)q2+f(x)q+f(x)+(b+c+d)δ=a(xq2+xq+x+δ)q314+1+a(xq2+xq+x+δ)3(q31)4+1+bxq+cx+dxq2+a(xq2+xq+x+δ)q314+1+a(xq2+xq+x+δ)3(q31)4+1+bx+cxq2+dxq+a(xq2+xq+x+δ)q314+1+a(xq2+xq+x+δ)3(q31)4+1+bxq2+cxq+dx+(b+c+d)δ=3aφ(x)q314+1+3aφ(x)3(q31)4+1+(b+c+d)φ(x)=h(φ(x))=hφ.

    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 sIm(φ), f(x) is injective on φ1(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

    xq314+x3(q31)4={2,xD0;0,xD1;2,xD2;0,xD3.

    Then h(x) can be rewritten as

    h(x)={(6a+b+c+d)x,xD0;(b+c+d)x,xD1;(6a+b+c+d)x,xD2;(b+c+d)x,xD3.

    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+dD0. Clearly, we have (6a+b+c+d)xD0 for xD0, then it implies that (b+c+d)xD0 for xD1D3. It tells us that b+c+dD1D3. If b+c+dD0, then (b+c+d)xD1 and (b+c+d)xD3 for xD1 and xD3, respectively. Since h(x) is a bijection, we can get that (6a+b+c+d)xD2 for xD2, thus 6a+b+c+dD0. On the other hand, if b+c+dD2, then (b+c+d)xD3 for xD1 and (b+c+d)xD1 for xD3. It follows from h(x) being a bijection that (6a+b+c+d)xD2 for xD2, one has 6a+b+c+dD0.

    (2) 6a+b+c+dD1. In this case, we can imply that b+c+dD0D2. If b+c+dD1, by h(x) being a bijection, we can get that 6a+b+c+dD1. Similarly, if b+c+dD3, then it follows from h(x) being a bijection that 6a+b+c+dD1.

    (3) 6a+b+c+dD2. In this case, we first get that b+c+dD1D3. If b+c+dD2, since h(x) is a bijection, we can get that 6a+b+c+dD2. If b+c+dD0, it follows that 6a+b+c+dD2.

    (4) 6a+b+c+dD3. In this case, it is easy to imply that b+c+dD0D2. If b+c+dD1, then we can get that 6a+b+c+dD3. If b+c+dD3, it follows from h(x) being a bijection that 6a+b+c+dD3.

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

    Thus we can conclude that a(xq2+xq+x+δ)q314+1+a(xq2+xq+x+δ)3q314+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 p3 be an odd prime, q1(mod4) and δFq. Then

    (xq2+xq+x+δ)q314+1+(xq2+xq+x+δ)3(q31)4+1x

    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 p3 be an odd prime, q1(mod4) and δFq. Then

    (xq2+xq+x+δ)q314+1+(xq2+xq+x+δ)3(q31)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 δFq, 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)18292x is a permutation polynomial over F293.

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

    Theorem 5.5. Let p3,5 be an odd prime, q1(mod4) and δFq. Then

    (xq2+xq+x+δ)q314+1x

    is a permutation polynomial over Fq3 if and only if (2,4,3ω1,3ω1) 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)q314+1x, φ(x)=xq2+xq+x+δ, ψ(x)=xq2+xq+xδ and h(x)=x(3xq3141). Then we can easily check that

    ψf=hφ.

    Namely, the diagram is commutative. Furthermore, it is easy to get that for every sIm(φ), f(x) is injective on φ1(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 xFq, we can rewrite h(x) as:

    h(x)={2x,xD0;(3ω1)x,xD1;4x,xD2;(3ω1)x,xD3.

    Since p3,5 is an odd prime, we get that (3ω1)(3ω1)0. We claim that h(x) is bijective from Im(φ) to Im(ψ) if and only if (2,4,3ω1,3ω1) 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 q1(mod4), it tells us that 1D0 or 1D2. We consider the following cases:

    (1) 2D0. Then we get that 4D0 or 4D2. For xD0, we have 2xD0, it follows from h(x) being a bijection that 4xD0 for xD2, hence 4D0 and 4xD2 for xD2. Furthermore, since 2xD0 for xD0 and 4xD2 for xD2, then (3ω1)xD0D2 for xD1, thus 3ω1D0 or 3ω1D2. If 3ω1D0, by h(x) being a bijection, one has 3ω1D0. If 3ω1D2, then it implies that 3ω1D2.

    (2) 2D1. Since 1D0 or 1D2, thus 4D2 or 4D0. For the first case, if 4D2, then 2xD1 for xD0 and 4xD0 for xD2. Since h(x) is a bijection, we know that 3ω1D1 or 3ω1D2. If 3ω1D1, it follows that 3ω1D0. If 3ω1D2, one has that 3ω1D3. For the second case, if 4D0, then 2xD1 for xD0 and 4xD2 for xD2. Since h(x) is a bijection, we know that 3ω1D3 or 3ω1D2. If 3ω1D3, it follows that 3ω1D0. If 3ω1D2, we know that 3ω1D1.

    (3) 2D2. Since 2D2, then we get that 4D0 or 4D2. For xD0, we have 2xD2, it follows from h(x) being a bijection that 4xD2 for xD2, hence 4D2 and 4xD0 for xD2. Furthermore, since 2xD2 for xD0 and 4xD0 for xD2, then (3ω1)xD0D2 for xD1, thus 3ω1D0 or 3ω1D2. If 3ω1D0, it implies that 3ω1D0. If 3ω1D2, we have 3ω1D2.

    (4) 2D3. Since 1D0 or 1D2, thus 4D2 or 4D0. For the first case, if 4D2, then 2xD3 for xD0 and 4xD0 for xD2. Since h(x) is a bijection, we know that 3ω1D0 or 3ω1D1. If 3ω1D0, it follows that 3ω1D3. If 3ω1D1, we get that 3ω1D2. For the second case, if 4D0, then 2xD3 for xD0 and 4xD2 for xD2. Since h(x) is a bijection, one has 3ω1D3 or 3ω1D0. If 3ω1D3, it follows that 3ω1D2. If 3ω1D0, it implies that 3ω1D1.

    The sufficiency part is proved.

    For the necessity part, if (2,4,3ω1,3ω1) 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)q314+1x

    is a permutation polynomial over Fq3 if and only if (2,4,3ω1,3ω1) 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 p3,5 be an odd prime, q1(mod4) and δFq. Then

    (xq2+xq+x+δ)3(q31)4+1x

    is a permutation polynomial over Fq3 if and only if (2,4,3ω1,3ω1) 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 p3 be an odd prime and q31(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}D0D1D2D3D4D5. Furthermore, if xDi, we notice that xq316=ϵiq316. For simplicity, we define ω=ϵq316, which satisfies ω2ω+1=0.

    Since q31(mod6), we know that q1(mod6), then q2+q+13(mod6) and q1(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 q12 elements from D0 and q12 elements from D3.

    In what follows, we can get the permutation polynomials of the form a(xq2+xq+x+δ)q316+1+bx, which is the fifth main result in this paper.

    Theorem 6.1. Let p3 be an odd prime, q1(mod6) and δFq, and let a,bFq such that 3a+b0 and 3a+b0. Then

    a(xq2+xq+x+δ)q316+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)q316+1+bx, φ(x)=xq2+xq+x+δ, ψ(x)=xq2+xq+x+bδ and h(x)=x(3axq316+b). It is easy to check that

    ψf=ψ(f(x))=f(x)q2+f(x)q+f(x)+bδ=a(xq2+xq+x+δ)q316+1+bxq2+a(xq2+xq+x+δ)q316+1+bxq+a(xq2+xq+x+δ)q316+1+bx+bδ=3aφ(x)q316+1+bφ(x)=h(φ(x))=hφ.

    Furthermore, we can easily check that for every sIm(φ), f(x) is injective on φ1(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 q12 elements from D0 and q12 elements from D3, we get that

    h(x)={(3a+b)x,xD0;(3a+b)x,xD3.

    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+bD0, then we have (3a+b)D0=D0. Hence (3a+b)D3=D3, it tells us that 3a+bD0. On the other hand, if 3a+bD3, then (3a+b)D0=D3 and (3a+b)D3=D0, thus it follows that 3a+bD3.

    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)q316+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 p3 be an odd prime, q1(mod6) and δFq. Then

    (xq2+xq+x+δ)q316+1x

    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)q316+1x 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 δF19, we can find that (2,4)D3×D3 by Magma, by Corollary 6.2, it follows that (xq2+xq+x+δ)1144x is a permutation polynomial over F193.

    Similarly, we have

    Theorem 6.4. Let p3 be an odd prime, q1(mod6) and δFq, and let a,bFq such that 3a+b0 and 3a+b0. Then

    a(xq2+xq+x+δ)5(q31)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 p3 be an odd prime, q1(mod6) and δFq, and let a,b,cFq such that 3a+3b+c0 and 3a3b+c0. Then

    a(xq2+xq+x+δ)q316+1+b(xq2+xq+x+δ)5(q31)6+1+cx

    is a permutation polynomial over Fq3 if and only if (3a+3b+c,3a3b+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(3axq316+3bx5q316+c), and h(x) can be rewritten as

    h(x)={(3a+3b+c)x,xD0;(3a3b+c)x,xD3.

    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)q31d)+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.



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