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Review Special Issues

One world, no longer: the past, the present, and the future of global value chains

  • Received: 05 October 2020 Accepted: 06 January 2021 Published: 11 January 2021
  • JEL Codes: B27, D24, F15, F23, F54, F63

  • Global value chains (GVCs) are both a product and a facilitator of the model of globalization that dominated for almost two decades following Soviet collapse in 1991. The North Atlantic Financial Crisis of 2007 onwards undermined that dominance, as did the subsequent economic stagnation and associated rising political and social discord. The reversion to more nationalist modes of discourse and policy marks the return of a more visibly geopolitical dimension to the global political economy. The Covid-19 pandemic has accelerated and accentuated these trends. This paper charts the emergence and consolidation of the era of "one world, ready or not", and employs the work of various critical authors, most prominently William Greider. Greider's extensive critique of US-led globalization, offshoring, and what has since become known as "supply chain capitalism" not only appears prescient by comparison with the work of contemporaneous, high profile representatives of the economics discipline who were its champions, but helps us to locate the sources of its unravelling. The implications of this for GVCs are outlined in the final section, which foresees a fragmentation of the world into spheres of influence dominated by regional powers, each of varying strength and cohesion. This will most likely result in the reconfiguration of many GVCs along more regional lines, as the dictates of efficiency clash with the requirements of supply chain resilience and the associated prerogatives of national security, as defined by those states at the centre of the new regional power blocs. Common to all phases of development discussed in this paper is the subordination of the peoples of the Global South, as the mechanisms of imperialism are adjusted and adapted to the changing conditions arising from the irreconcilable contradictions of global capitalism.

    Citation: Michael Keaney. One world, no longer: the past, the present, and the future of global value chains[J]. National Accounting Review, 2021, 3(1): 1-49. doi: 10.3934/NAR.2021001

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  • Global value chains (GVCs) are both a product and a facilitator of the model of globalization that dominated for almost two decades following Soviet collapse in 1991. The North Atlantic Financial Crisis of 2007 onwards undermined that dominance, as did the subsequent economic stagnation and associated rising political and social discord. The reversion to more nationalist modes of discourse and policy marks the return of a more visibly geopolitical dimension to the global political economy. The Covid-19 pandemic has accelerated and accentuated these trends. This paper charts the emergence and consolidation of the era of "one world, ready or not", and employs the work of various critical authors, most prominently William Greider. Greider's extensive critique of US-led globalization, offshoring, and what has since become known as "supply chain capitalism" not only appears prescient by comparison with the work of contemporaneous, high profile representatives of the economics discipline who were its champions, but helps us to locate the sources of its unravelling. The implications of this for GVCs are outlined in the final section, which foresees a fragmentation of the world into spheres of influence dominated by regional powers, each of varying strength and cohesion. This will most likely result in the reconfiguration of many GVCs along more regional lines, as the dictates of efficiency clash with the requirements of supply chain resilience and the associated prerogatives of national security, as defined by those states at the centre of the new regional power blocs. Common to all phases of development discussed in this paper is the subordination of the peoples of the Global South, as the mechanisms of imperialism are adjusted and adapted to the changing conditions arising from the irreconcilable contradictions of global capitalism.


    Let Fq be the finite field of q elements, where q=pα with a prime p and a positive integer α. Suppose that

    q1=ef

    with positive integers e and f. Choose a generator γ of the multiplicative cyclic group

    Fq=Fq{0}.

    For vFq, let indγ(v) denote the unique non-negative integer mq2 such that v=γm.

    For 0i,je1 (or rather for i,j modulo e), the e2 cyclotomic numbers of order e, denoted by Aij (or A(e)ij to indicate the order e), are defined as the cardinality of the set Xij, where

    Xij:={vFq{0,1}indγ(v)i(mode), indγ(v+1)j(mode)}.

    Introduced by Gauss [1,2] over two hundred years ago, cyclotomic numbers are a significant concept in number theory with deep connections to various mathematical areas. They have been extensively applied in combinatorial designs, coding theory, cryptography, and information theory (see [3,4,5,6]). For both theoretical and practical purposes, it is intriguing to determine all cyclotomic numbers of a given order e for all finite fields, a task usually called the cyclotomic problem.

    In the case q=p, Gauss [1, §358] evaluated A(3)ij in terms of (L,M) satisfying the Diophantine system

    4p=L2+27M2,withL1(mod3).

    This system determines L uniquely and M up to signs. Similarly, Gauss [2] evaluated A(4)ij in terms of (a,b) satisfying

    p=a2+b2,witha1(mod4),

    which fixes a uniquely and b up to signs. His results indicated that solving the cyclotomic problem over Fq generally requires more than just the value of q and order e; it also needs a quadratic partition of q.

    Therefore, classical solutions to the cyclotomic problem are typically expressed using an appropriately chosen solution of a relevant Diophantine system (consisting of equations and congruences), often with the sign ambiguity. In this sense, many mathematicians have investigated the cyclotomic problem for various small orders 22 (see Dickson's early foundational work [7,8,9] and a good recent survey [10]), as well as for special orders such as [11,12,13] for l, [13,14] for 2l, [15] for l2, and [16,17] for 2l2 (with an odd prime l).

    While Gauss initially approached cyclotomy via Gauss sums, Dickson's use of Jacobi sums [7] laid the groundwork for modern cyclotomy. The cyclotomic problem is, in fact, equivalent to the explicit evaluation of Jacobi sums of the same order. Let us recall the definiton of Jacobi sums. Let ζ be a primitive complex e-th root of unity fixed once for all. We define a multiplicative character χe of order e on Fq by

    χe(γm)=ζm(for anymZ),

    and extend χe to a character on Fq by taking χe(0)=0. For convenience, we assume

    χme(0)=0

    for any integer m. The Jacobi sums J(i,j) (or Je(i,j) to indicate the order e) of order e, for 0i,je1 (or rather for i,j modulo e), are defined by

    J(i,j)=vFqχie(v)χje(v+1).

    Jacobi sums and cyclotomic numbers are related by the following finite Fourier series expansions:

    J(a,b)=0i,je1Aijζai+bj,e2Aab=0i,je1J(i,j)ζ(ai+bj).

    To calculate all cyclotomic numbers of order e, it suffices to calculate all Jacobi sums of order e, and vice versa.

    Jacobi made significant contributions to mathematics, including the Jacobi symbol, the Jacobi triple product, the Jacobi elliptic functions, and the Jacobian in variable transformations. Among his notable discoveries are Jacobi sums, which he proposed in 1827 in a letter to Gauss and published ten years later. These sums were later extended by Cauchy, Gauss, and Eisenstein. While Gauss sums are pivotal in proving quadratic reciprocity, Jacobi sums are essential for proving cubic reciprocity and were generalized by Eisenstein for biquadratic reciprocity. Jacobi sums are also used to estimate the number of integer solutions to congruences like

    x3+y31(modp),

    which are crucial for developing the Weil conjectures [18]. In modern mathematics, Jacobi sums have found applications in primality testing [19].

    Wamelen [20] has provided an inductive arithmetic approach to characterize all Jacobi sums of any order, thereby solving the cyclotomic problem in theory. However, the Diophantine system he employed is notably large and intricate in general. In recent years, there has been growing interest in efficiently computing Jacobi sums [21], driven by their importance in applications such as primality testing, cryptosystems, combinatorial designs, and advanced number theory problems [19,22].

    For given Fq, γ, and ζ, classical cyclotomic numbers and Jacobi sums are binary functions depending on two variables i,jZ/eZ. The purpose of this paper is to investigate their trivariate analogs, so-called "ternary cyclotomic numbers" and "ternary Jacobi sums", defined in Section 2. As classical cyclotomic numbers and Jacobi sums are important both theoretically and practically, we want to study their ternary counterparts in order to explore theoretically interesting problems or potential applications. In Section 2, we obtain some ternary properties, which are analogous to those of classical cyclotomic numbers and Jacobi sums, with proofs similar to those of classical ones. Then we use these properties to solve the cyclotomic problem for ternary cyclotomic numbers. Section 3 provides explicit evaluations of all ternary cyclotomic numbers and ternary Jacobi sums for order e=2, and Section 4 nearly completes the evaluation for order e=3, except for an integer J3(1,1,2), which remains unknown. Our calculations show that ternary Jacobi sums, which are some kind of character sums, cannot generally be transformed into classical Jacobi sums. So the cyclotomic problem for ternary cyclotomic numbers has its own interests in theory.

    Let Fq, γ, ζ, χe and

    q=pα=ef+1

    be as described in Section 1. We further assume

    q=ef+1

    is odd for convenience. (The upcoming definitions will be meaningless for F2α, where v1=v+1.) Specifically, either e or f is even.

    For 0i,j,ke1 (or rather for i,j,kZ/eZ), we define the ternary cyclotomic numbers of order e, denoted by Aijk (or A(e)ijk to indicate the order e), as the cardinality of the set

    Xijk:={vFq{0,±1}indγ(v1)i(mode), indγvj(mode), indγ(v+1)k(mode)}.

    This definition relies on the choice of a multiplicative generator γ of Fq. We also define the ternary Jacobi sums of order e, denoted by J(i,j,k) (or Je(i,j,k) to indicate the order e), as

    J(i,j,k):=vFqχie(v1)χje(v)χke(v+1).

    This definition depends on the character χe on Fq, which is determined by the selections of γ and ζ (as χe(γm)=ζm and χe(0)=0). Clearly, we have

    Xijk={vFqχe(v1)=ζi, χe(v)=ζj, χe(v+1)=ζk}.

    Our definition of the ternary Jacobi sum can be viewed as a special case of the character sums studied in [23].

    To calculate all ternary cyclotomic numbers of order e, it suffices to calculate all ternary Jacobi sums of order e, and vice versa, by the following finite Fourier series expansions.

    Proposition 1. (Finite Fourier series expansions) The ternary cyclotomic numbers and ternary Jacobi sums of the same order e are related by: for any a,b,cZ/eZ,

    J(a,b,c)=i,j,kZ/eZAijkζai+bj+ck, (2.1)
    e3Aabc=i,j,kZ/eZJ(i,j,k)ζ(ai+bj+ck). (2.2)

    Proof. Note that

    Fq={0,±1}i,j,kZ/eZXijk.

    For v{0,±1},

    χae(v1)χbe(v)χce(v+1)=0.

    For the Aijk elements vXijk,

    χae(v1)χbe(v)χce(v+1)=ζai+bj+ck.

    Summing them all together, we obtain Eq (2.1). For the Eq (2.2), noting that ζ=χe(γ), we have

    i,j,kZ/eZJ(i,j,k)ζ(ai+bj+ck)=i,j,kZ/eZvFqχie(v1)χje(v)χke(v+1)χe(γ)(ai+bj+ck)=vFq(e1i=0χie(v1γa))(e1j=0χje(vγb))(e1k=0χke(v+1γc)).

    Note that

    e1j=0χje(vγb)={e,if v0 and indγ(v)b(mode),0,if v=0 or indγ(v)b(mode),

    and similar for e1i=0χie(v1γa) and e1j=0χke(v+1γc). So the terms of the above sum are non-zero only when vXabc, and thus

    i,j,kZ/eZJ(i,j,k)ζ(ai+bj+ck)=vXabce3=e3Aabc.

    Following arguments often involve the value of χe(1). Since

    γq12=1Fq,

    and when e is even,

    ζe2=1C,

    we have

    χe(1)1=χe(1)=χe(γq12)=ζef2={1,if f is even,ζe2=1,if f is odd,=(1)f.

    Classical cyclotomic numbers exhibit a symmetry property in their two variables:

    Aij=Aj+ef2,i+ef2={Aji,if f is even,Aj+e2,i+e2,if f is odd,

    which implies that

    J(a,b)=(1)f(a+b)J(b,a).

    It is similar for ternary cyclotomic numbers and ternary Jacobi sums.

    Proposition 2. For any i,j,kZ/eZ,

    Aijk=Ak+ef2,j+ef2,i+ef2={Akji,if f is even,Ak+e2,j+e2,i+e2,if f is odd.

    Proof. For any vFq, let w=v. Since

    χe(1)=ζef2,

    we have

    vXijk(χe(w1),χe(w),χe(w+1))=(ζi,ζj,ζk)(χe(w1),χe(w),χe(w+1))=(χe(1)ζk,χe(1)ζj,χe(1)ζi)wXk+ef2,j+ef2,i+ef2.

    So vv induces a bijection from Xijk to Xk+ef2,j+ef2,i+ef2.

    Corollary 3. For any a,b,cZ/eZ,

    J(a,b,c)=(1)f(a+b+c)J(c,b,a).

    As a consequence, when both f and b are odd, we have J(a,b,a)=0.

    Proof. By Propositions 1 and 2, as

    ζef2=(1)f,

    we have

    J(a,b,c)=i,j,kZ/eZAk+ef2,j+ef2,i+ef2ζai+bj+ck=l,m,nZ/eZAlmnζa(nef2)+b(mef2)+c(lef2)=ζef2(a+b+c)l,m,nZ/eZAlmnζcl+bm+an=(1)f(a+b+c)J(c,b,a).

    Classical cyclotomic numbers exhibit another property

    Aij=Ai,ji,

    which implies that

    J(a,b)=J(ab,b).

    It is similar for ternary cases as follows:

    Proposition 4. For any i,j,kZ/eZ,

    Aijk=Aij+ef2,j,kj={Aij,j,kj,if f is even,Aij+e2,j,kj,if f is odd.

    Proof. For any vFq, let w=v1. Since

    χe(1)=ζef2,

    we have

    vXijk(χe(w11),χe(w1),χe(w1+1))=(ζi,ζj,ζk)(χe(w1)χe(1)χe(w),χe(w),χe(1+w)χe(w))=(ζi,ζj,ζk)(χe(w1),χe(w),χe(w+1))=(χe(1)ζij,ζj,ζkj)wXij+ef2,j,kj.

    So vv1 induces a bijection from Xijk to Xij+ef2,j,kj.

    Corollary 5. For any a,b,cZ/eZ,

    J(a,b,c)=(1)faJ(a,abc,c).

    Proof. By Propositions 1 and 4, as

    ζef2=(1)f,

    we have

    J(a,b,c)=i,j,kZ/eZAij+ef2,j,kjζai+bj+ck=l,m,nZ/eZAl,m,nζa(lmef2)bm+c(nm)=(ζef2)al,m,nZ/eZAl,m,nζal+(abc)m+cn=(1)faJ(a,abc,c).

    For classical cyclotomic numbers, we have

    iZ/eZAij={f1,if j0(mode),f,otherwise, and jZ/eZAij={f1,if ief2(mode),f,otherwise. (2.3)

    We show similar equations for ternary cyclotomic numbers.

    Proposition 6. Let g=indγ(2). For any i,j,kZ/eZ,

    tZ/eZAtjk={Ajk1,if j0(mode) and kg(mode),Ajk,otherwise,tZ/eZAijt={Aij1,if ig+ef2(mode) and jef2(mode),Aij,otherwise,tZ/eZAitk={Aig,kg1,if i ef2(mode) and k0(mode),Aig,kg,otherwise.

    Proof. Note that

    tZ/eZXtjk=Xjk{1}.

    Also note that 1Xjk if and only if

    j0(mode)andkg(mode).

    These Xtjk are pairwise disjoint, which gives the first equation.

    For the second equation,

    tZ/eZXijt={vFq{0,±1}χe(v1)=ζi, χe(v)=ζj}.

    So

    vtZ/eZXijt

    if and only if

    v1Xij{2}.

    Also note that 2Xij if and only if

    ig+ef2(mode)andjef2(mode).

    For the third equation,

    tZ/eZXitk={vFq{0,±1}χe(v1)=ζi, χe(v+1)=ζk}.

    Then

    vtZ/eZXitk

    if and only if

    v12Xig,kg{21}.

    Here

    vv12

    induces a bijection on Fq. Also note that

    21Xig,kg

    if and only if

    ief2(mode)andk0(mode).

    Note that

    vFqχie(v)=q2m=0χie(γm)=q2m=0(ζi)m={q1,if i0(mode),0,otherwise.

    So classical Jacobi sums J(i,0) and J(0,i) can be easily evaluated by

    (1)fiJ(i,0)=J(0,i)=vFqχ0e(v)χie(v+1)=vFqχie(v+1)χie(1)={q2,if i0(mode),1,otherwise. (2.4)

    Similarly, ternary Jacobi sums J(i,j,k), with either i, j or k divided by e, can be evaluated in terms of classical Jacobi sums of the same order e.

    Proposition 7. Let g=indγ(2). For i,j,kZ/eZ, we have

    J(0,j,k)=J(j,k)ζgk,J(i,j,0)=J(i,j)(1)f(i+j)ζgi,J(i,0,k)=ζg(i+k)J(i,k)(1)fi.

    Proof. Recall that

    χme(0)=0

    for any mZ, we have

    χe(2)=ζgandχe(1)=(1)f.

    So

    J(0,j,k)=vFqχ0e(v1)χje(v)χke(v+1)=vFq{1}χje(v)χke(v+1)=vFqχje(v)χke(v+1)χje(1)χke(2)=J(j,k)ζgk,J(i,j,0)=vFq{1}χie(v1)χje(v)=vFqχie(v1)χje(v)χie(2)χje(1)=J(i,j)(1)f(i+j)ζgi.

    In the following we let

    w=v12.

    Note that

    vv12

    induces a bijection on Fq. Then

    J(i,0,k)=vFqχie(v1)χke(v+1)χie(1)χke(1)=wFqχie(2w)χke(2w+2)(1)fi=ζg(i+k)J(i,k)(1)fi.

    Corollary 8.

    J(0,0,0)=q3.

    For i0(mode), we have

    J(i,0,0)=(1)fi+1(1+ζgi),J(0,i,0)=1(1)fi,J(0,0,i)=1ζgi.

    To conclude this section, it is important to highlight a key result that will facilitate our subsequent calculations, as it is self-evident from the definition.

    Lemma 9. Let k be an integer coprime to e, and σk denote the Q-automorphism of the field Q(ζ) with

    σk(ζ)=ζk.

    Then for any a,b,cZ/eZ, we have

    J(ka,kb)=σk(J(a,b)),J(ka,kb,kc)=σk(J(a,b,c)).

    In this section, assuming

    e=2andq=pα=2f+1,

    we calculate all ternary cyclotomic numbers and ternary Jacobi sums of order 2 for a generator γ of Fq. We fix ζ=1, and let

    g=indγ(2).

    By Eq (2.4) and

    J2(a,b)=J2(ab,b),

    all classical Jacobi sums of order 2 are:

    J2(0,0)=q2,J2(1,0)=(1)f+1,J2(1,1)=J2(0,1)=1. (3.1)

    First, let us calculate J2(1,1,1) when f is even (which is this section's most challenging part). We will see that it is related to classical Jacobi sums of order 4. Let us take the imaginary unit

    i=1

    as the primitive 4-th root ζ4 of unity. We write J4(i,j) (with i,jZ/4Z) for the {classical} Jacobi sums of order 4 with respect to γ and ζ4. Let χ4 be the character of Fq defined by

    χ4(0)=0,χ4(γm)=ζm4=im(for anymZ).

    By definition,

    J4(i,j)=vFqχi4(v)χj4(v+1).

    A critical insight is that

    χ2=χ24

    on Fq. Also note that

    Fq={γm0m2f1}.

    So

    J2(1,1,1)=vFqχ2(v1)χ2(v)χ2(v+1)=vFqχ2(v)χ2(v21)=vFqχ4(v2)χ24(v21)=2f1m=0χ4(γ2m)χ24(γ2m1)=2f1m=0χ4(γ2m)χ24(γ2m1).

    The final expression is a part of the following formula:

    J4(2,1)=vFqχ24(v)χ4(v+1)=wFqχ24(w1)χ4(w)=wFqχ4(w)χ24(w1)=2f1n=0χ4(γn)χ24(γn1)=f1m=0χ4(γ2m)χ24(γ2m1)+f1m=0χ4(γ2m+1)χ24(γ2m+11).

    Since

    χ4(γ2m)=(1)m,χ4(γ2m+1)=i2m+1{±i}

    and

    χ24(v){0,±1}

    for any vFq,

    Re(J4(2,1))=f1m=0χ4(γ2m)χ24(γ2m1)=J2(1,1,1)2,

    where Re(z) represents the real component of a complex number z.

    From [24], we extract the evaluation required. Take special note that the Jacobi sums as defined in [24, §2] are distinct from our own definitions. In fact, their Jacobi sums R(m,n) equal our

    J(n,mn)=(1)fmJ(m,n).

    So their finding for R(1,1) [24, Propositions 1, 2] can be reinterpreted as a result for our J4(1,2) as follows.

    Lemma 10. [24, Propositions 1, 2] Let q=pα1(mod4) with p prime and α1, and let

    s=Re(J4(1,2)).

    If p3(mod4), then α is even,

    s=(p)α/21(mod4)

    and

    J4(1,2)=s=(p)α/2.

    If p1(mod4), then s is the unique integer coprime to q such that s1(mod4) and qs2 is a perfect square.

    Remark 11. The earlier literature [25, p.298] also presented the results of Lemma 10, but erred in the sign when p3(mod4).

    When

    f=q12

    is even, by Lemma 10,

    J2(1,1,1)=2Re(J4(2,1))=2Re((1)q14J4(1,2))=(1)q14(2s).

    When f is odd, by Corollary 3, J2(1,1,1)=0.

    Evaluating the ternary Jacobi sums of order 2, excluding J2(1,1,1), is now straightforward by Proposition 7, Corollary 8, and Eq (3.1). Combining all these, we formulate the results into the following theorem.

    Theorem 12. All ternary Jacobi sums of order 2 over Fq with respect to a generator γ of Fq are explicitly given as follows, with g=indγ(2) and s defined as in Lemma 10.

    J2(0,0,0)=q3,J2(1,0,0)=(1)f+1(1+(1)g),J2(0,0,1)=J2(0,1,1)=J2(1,1,0)=1(1)g,J2(0,1,0)=J2(1,0,1)=1(1)f,J2(1,1,1)={0,if q3(mod4),2s,if q1(mod8),2s,if q5(mod8).

    Using finite Fourier series expansions from Proposition 1, along with Theorem 12, we derive a complete and explicit evaluation of all ternary cyclotomic numbers of order 2 as follows.

    Theorem 13. All ternary cyclotomic numbers of order 2 over Fq, corresponding to a generator γ of Fq, are explicitly evaluated as follows, with g=indγ(2) and s defined as in Lemma 10.

    If q3(mod4), then

    A(2)000=A(2)100=A(2)110=A(2)111=q52(1)g8,A(2)001=A(2)010=A(2)011=A(2)101=q1+2(1)g8.

    If q1(mod8), then

    A(2)000=q112s4(1)g8,A(2)011=A(2)110=q+12s8,A(2)101=q32s+4(1)g8,A(2)001=A(2)100=A(2)010=A(2)111=q3+2s8.

    If q5(mod8), then

    A(2)000=A(2)101=q7+2s8,A(2)011=A(2)110=q+1+2s8,A(2)001=A(2)100=A(2)010=A(2)111=q32s8.

    This section aims to compute as many ternary Jacobi sums of order 3 as feasible. Let

    e=3andq=pα=3f+1

    be an odd prime power. Clearly, f is even. Let

    g=indγ(2).

    We choose a cube root of unity ζ3{e±2π3i}. Then

    ζ23=1ζ3.

    Also note that

    σ2(ζ3)=ζ23=ζ13=¯ζ3,

    for the Q-automorphism σ2: Q(ζ)Q(ζ). So σ2 is {just} the complex conjugation.

    First, recall the evaluation of classical Jacobi sums of order 3. Since f is even,

    J3(a,b)=J3(b,a)=J3(ab,b)=J3(b,ab)=J3(ab,a)=J3(a,ab).

    By Eq (2.4),

    J3(0,0)=q2,

    and

    J3(0,1)=J3(1,0)=J3(2,1)=J3(1,2)=J3(2,0)=J3(0,2)=1.

    The value of

    J3(1,1)=¯J3(2,2)

    is also known as the following lemma.

    Lemma 14. [12,25] For

    q=pα=3f+1

    with p odd prime and α1, we can write

    J3(1,1)=L+3M2+3Mζ3,

    for some L,MZ with L1(mod3).

    (1) If p2(mod3), then α is even,

    M=0,L=2(p)α/2,andJ3(1,1)=(p)α/2.

    (2) If p1(mod3), then (L,M) is the unique solution of the Diophantine system:

    {4q=L2+27M2,L1(mod3),pL,γq13(L+9M)/(L9M)(modp). (4.1)

    Proof. For p1(mod3), this result is a part of [12, Proposition 1]. For p2(mod3), [25, p.297] provided the value of J3(1,1) but erred in the sign. To rectify this mistake, we present an elementary proof below.

    By definition,

    J3(1,1)=a+bζ3

    with some a,bZ. Write Aij (with 0i,j2) for the classical cyclotomic numbers of order 3. Note that f is even. So

    A01=A10=A22andA02=A20=A11.

    By definition,

    b=A01+A10+A22(A02+A11+A20)=3(A01A02)0(mod3).

    By Eq (2.3),

    f1=A00+A01+A02,f=A22+A02+A12=A01+A02+A12.

    So

    A12=A00+1.

    By definition,

    a=A00+A12+A21(A02+A11+A20)=2+3(A00A02)2(mod 3).

    By [26, Theorem 2.1.3], if none of i, j, and i+j are multiples of e, then

    |Je(i,j)|=q.

    Therefore,

    pα=q=(a+bζ3)(¯a+bζ3)=(a+bζ3)(a+bζ23).

    The Eisenstein integer ring Z[ζ3] is a unique factorization domain (UFD), where the rational prime p2(mod3) is an Eisenstein prime, which is irreducible in Z[ζ3]. So

    a+bζ3=pnu

    for an integer n0 and an Eisenstein unit

    u{±1,±ζ3,±(1+ζ3)}.

    Since

    pn=|a+bζ3|=pα/2,

    we have

    a+bζ3=pα/2u{±pα/2,±pα/2ζ3,±pα/2(1+ζ3)}.

    The condition

    b0±pα/2(mod3)

    requires that u=±1 and thus b=0. Moreover,

    J3(1,1)=a=±pα/22p(mod 3),

    and hence

    J3(1,1)=(p)α/2.

    As f is even, by Corollaries 3 and 5, for any a,b,cZ/3Z we have

    J3(a,b,c)=J3(c,b,a)=J3(a,abc,c).

    Along with Proposition 7, Corollary 8, and Lemma 9, we obtain:

    J3(0,0,0)=q3,J3(2,0,1)=J3(1,0,2)=J3(0,2,0)=J3(0,1,0)=2,J3(1,2,0)=J3(0,2,1)=J3(1,0,0)=J3(0,0,1)=1ζg3,J3(2,1,0)=J3(0,1,2)=J3(2,0,0)=J3(0,0,2)=1ζ2g3,J3(1,1,0)=J3(0,1,1)=L+3M2+3Mζ3ζg3,J3(2,2,0)=J3(0,2,2)=L+3M2+3Mζ23ζ2g3,J3(1,1,1)=J3(1,0,1)=ζ2g3(L+3M2+3Mζ3)1,J3(2,2,2)=J3(2,0,2)=ζg3(L+3M2+3Mζ23)1.

    Next, we calculate

    J3(1,2,1)=¯J3(2,1,2),

    following a similar approach to that for J2(1,1,1) in Section 3. Note that f is even and

    Fq={γm0m3f1}.

    So

    J3(1,2,1)=vFqχ23(v)χ3(v21)=3f1m=0χ3(γ2m)χ3(γ2m1)=232f1m=0χ3(γ2m1)χ3(γ2m). (4.2)

    Also note that

    J3(1,1)=3f1n=0χ3(γn1)χ3(γn)=32f1m=0χ3(γ2m1)χ3(γ2m)+32f1m=0χ3(γ2m+11)χ3(γ2m+1). (4.3)

    Let us choose the primitive complex 6-th root of unity ζ6 such that ζ26=ζ3. As ζ36=1,

    ζ6=ζ26=ζ13=ζ23=1+ζ3.

    Write J6(i,j) (i,jZ/6Z) for the {classical} Jacobi sums of order 6 with respect to γ and ζ6. Let χ6 be the character of Fq defined by χ6(0)=0 and

    χ6(γm)=ζm6(for anymZ).

    Note that χ26=χ3 on Fq. By definition,

    J6(2,5)=3f1n=0χ26(γn1)χ56(γn)=3f1n=0χ3(γn1)χ56(γn)=32f1m=0χ3(γ2m1)χ56(γ2m)+32f1m=0χ3(γ2m+11)χ56(γ2m+1).

    Since

    χ56(γ2m)=χ3(γ2m),χ56(γ2m+1)=ζ56χ3(γ2m)=ζ3χ3(γ2m)=χ3(γ2m+1),

    we obtain

    J6(2,5)=32f1m=0χ3(γ2m1)χ3(γ2m)32f1m=0χ3(γ2m+11)χ3(γ2m+1). (4.4)

    Theorem 15. Let g=indγ(2), and (L,M) be defined as in Lemma 14 for

    q=pα=3f+1

    with p odd prime and α1. Then

    J3(1,2,1)=J3(1,1)+¯J6(1,1)={L,if g0(mod3),L+9M2ζ3,if g1(mod3),L+9M2(1+ζ3),if g2(mod3).

    Moreover, if p2(mod3), then g0(mod3) and

    J3(1,2,1)=L=2(p)α/2.

    Proof. By Eqs (4.2)–(4.4), we have

    J3(1,2,1)=J3(1,1)+J6(2,5).

    We know

    J3(1,1)=L+3M2+3Mζ3.

    Also note that

    J6(2,5)=J6(5,5)=σ5(J6(1,1))=¯J6(1,1).

    To determine J6(1,1), we note that

    J6(1,1)Z[ζ6]=Z[ζ3]

    by definition. Let

    J6(1,1)=E+F2+Fζ3

    for some E,FZ. By finite Fourier series expansions, E and F are indeed Z-linear combinations of classical cyclotomic numbers A(6)ij (with 0i,j5).

    For p1(mod3), [14, Theorem 2] provides

    J6(1,1)=(E+F)ζ3(E+F)ζ232=E+F2+Fζ3,

    where

    (E,F)={(L,3M),if g0(mod3),(L9M2,L3M2),if g1(mod3),(L+9M2,L3M2),if g2(mod3). (4.5)

    So we only need to show that Eq (4.5) also holds for p2(mod3).

    In an earlier work, Dickson [7, §17–19] established Eq (4.5) in the setting of q=p, utilizing certain linear relations among cyclotomic numbers A(6)ij (with 0i,j5). These relations, in fact, are valid for A(6)ij over any finite field Fq with q1(mod6). Thus, Dickson's proof of Eq (4.5) is universally applicable to all fields Fq with q1(mod6).

    Note that the Jacobi sum R(11) in [7, §17–19] corresponds to our (1)fJ6(1,1). Consequently, the pair (E,F) in [7, §18] (with f even) matches our pair, while the pair in §19 (with f odd) is our (E,F).

    Using Eq (4.5), we obtain

    J3(1,2,1)=J3(1,1)+J6(2,5)=J3(1,1)+¯J6(1,1)=L+3M2+3Mζ3+E+F2+Fζ23=L+3M+EF2+(3MF)ζ3={L,if g0(mod3),L+9M2ζ3,if g1(mod3),L+9M2(1+ζ3),if g2(mod3).

    For the special case of p2(mod3), we have

    221+2(p1)(modp)

    by Fermat's little theorem. As

    1+2(p1)0(mod3),

    let

    1+2(p1)=3t

    with some tZ. Then

    g=indγ(2)=indγ(23t)3tg(mod(q1)).

    Since 3(q1), we have g0(mod3), and hence

    J3(1,2,1)=L=2(p)α/2.

    This completes the proof.

    Remark 16. For p1(mod3) and q=pα, Acharya and Katre [14, Theorem 2] proved that (E,F) is the unique solution of the Diophantine system

    {4q=E2+3F2,E1(mod3),pE,Fg(mod3),γq13(E+F)/(E+F)(modp).

    The remaining issue now is to evaluate

    J3(1,1,2)=J3(2,1,1)=J3(2,2,1)=J3(1,2,2).

    Here, the second equality stems from Corollary 5, while the other two follow from Corollary 3. First, proving it to be an integer is straightforward.

    Proposition 17. J3(1,1,2) is an integer.

    Proof. By Corollaries 3, 5 and Lemma 9,

    J3(1,1,2)=J3(2,1,1)=J3(2,2,1)=σ2(J3(1,1,2))=¯J3(1,1,2).

    So J3(1,1,2)R. By definition,

    J6(1,1)Z[ζ6]=Z[ζ3].

    Let

    J3(1,1,2)=a+bζ3

    for some a,bZ, whose imaginary part is 0 only if b=0. So J3(1,1,2)=aZ.

    We have explored various approaches, but the exact value of J3(1,1,2) still eludes us. Finally, let us elaborate on two unsuccessful ideas regarding its calculation:

    (1) Drawing from the prior computation of J3(1,2,1), we might guess: Can J3(1,1,2) be expressed as a linear combination of J6(i,j), with coefficients independent of Fq? More precisely, let us consider 36 absolute constants cijC (for 0i,j5) such that the equality

    J3(1,1,2)=0i,j5cijJ6(i,j)

    holds for any finite field Fq with q1(mod6). For each q, this equality yields a linear relation among the coefficients cij. Unfortunately, computational solutions (by a computer program) to these linear equations (for a sufficient number of q) reveal that such constants cij do not exist.

    (2) For v=γnFq, we note that

    2i=0χi3(v)=2i=0χi3(γn)=2i=0(ζn3)i={3,if 3n,0,otherwise.

    Also note that

    J(1,1,2)=J(2,1,1)=J(2,2,1),J(2,0,1)=2.

    So

    2J(1,1,2)2=J(2,1,1)+J(2,2,1)+J(2,0,1)=vFqχ23(v1)(2i=0χi3(v))χ3(v+1)=3f1m=0χ23(γ3m1)χ3(γ3m+1)=3f1m=1χ3(γ3m+1γ3m1)=3f1m=1χ3(2γ3m1+1).

    Similar computations for J(1,1,2)+J(0,1,2)+J(2,1,2) and J(1,1,2)+J(1,1,0)+J(1,1,1) yields similar character sums involving cubic elements of Fq. To evaluate J3(1,1,2), it suffices to evaluate any one of them. Currently, we have not found a good way to compute them in general.

    In this paper, we introduce the trivariate counterparts of classical cyclotomic numbers and Jacobi sums, named "ternary cyclotomic numbers" and "ternary Jacobi sums". We present their basic properties that mirror those of the classical cyclotomic numbers and Jacobi sums. In Section 3 we provide explicit evaluations for all ternary Jacobi sums (Theorem 12) and ternary cyclotomic numbers (Theorem 13) of order e=2. Section 4 delivers near-complete results for order e=3, with the exception of the elusive integer J3(1,1,2) for us. To solve the cyclotomic problem for ternary cyclotomic numbers of order 3, one only needs to calculate J3(1,1,2). {Determining the precise value of the integer J3(1,1,2) stands as our initial objective for upcoming endeavors. In the future, we will investigate more general methods for the ternary cyclotomic problem, as well as its potential applications in other fields.}

    Zhichao Tang: writing–original draft, conceptualization, investigation, software, validation; Xiang Fan: writing–review & editing, methodology, supervision, funding acquisition. All authors have read and agreed to the published version of the manuscript.

    The authors gratefully acknowledge the editor and the anonymous reviewers for their insightful feedback, greatly enhancing the manuscript. This work is funded by Guangzhou Science and Technology Programme (No. 202102021218). The second author was also sponsored by the National Natural Science Foundation of China (No. 11801579) and Guangdong Basic and Applied Basic Research Foundation (No. 2020B1515310017).

    All authors declare no conflicts of interest in this paper.



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