Calcified coronary lesions pose significant challenges in percutaneous coronary intervention, which can impede device placement and stent expansion, thus leading to suboptimal clinical outcomes. This paper outlines a comprehensive approach for the safe and effective preparation of calcified lesions. It emphasizes the importance of pre-procedural planning using either coronary artery calcium scoring or computed tomography coronary angiography, as well as advanced intravascular imaging techniques with intravascular ultrasound and optical coherence tomography for accurate lesion assessment. Various lesion modification strategies are reviewed, including balloon angioplasty, rotational atherectomy, orbital atherectomy, laser atherectomy, and lithotripsy. The selection criteria for each technique based on the lesion characteristics, calcium morphology, and operator experience are discussed. Additionally, clinical data is analysed to provide evidence-based recommendations for practice. The paper concludes with a discussion on future directions and innovations alongside a proposed algorithm for the management of calcified coronary lesions, which is aimed at improving patient outcomes through technological advancements and refined procedural techniques.
Citation: Bharat Khialani, Sara Malakouti, Sandeep Basavarajah, Leontin Lazar, Sylwia Iwanczyk, Bernardo Cortese. How to identify and prepare calcified lesions safely and effectively[J]. AIMS Medical Science, 2025, 12(2): 171-192. doi: 10.3934/medsci.2025011
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Calcified coronary lesions pose significant challenges in percutaneous coronary intervention, which can impede device placement and stent expansion, thus leading to suboptimal clinical outcomes. This paper outlines a comprehensive approach for the safe and effective preparation of calcified lesions. It emphasizes the importance of pre-procedural planning using either coronary artery calcium scoring or computed tomography coronary angiography, as well as advanced intravascular imaging techniques with intravascular ultrasound and optical coherence tomography for accurate lesion assessment. Various lesion modification strategies are reviewed, including balloon angioplasty, rotational atherectomy, orbital atherectomy, laser atherectomy, and lithotripsy. The selection criteria for each technique based on the lesion characteristics, calcium morphology, and operator experience are discussed. Additionally, clinical data is analysed to provide evidence-based recommendations for practice. The paper concludes with a discussion on future directions and innovations alongside a proposed algorithm for the management of calcified coronary lesions, which is aimed at improving patient outcomes through technological advancements and refined procedural techniques.
Let Fq be the finite field of q elements, where q=pα with a prime p and a positive integer α. Suppose that
q−1=ef |
with positive integers e and f. Choose a generator γ of the multiplicative cyclic group
F∗q=Fq−{0}. |
For v∈F∗q, let indγ(v) denote the unique non-negative integer m⩽q−2 such that v=γm.
For 0⩽i,j⩽e−1 (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:={v∈Fq−{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,withL≡1(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,witha≡1(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 F∗q by
χe(γm)=ζm(for anym∈Z), |
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 0⩽i,j⩽e−1 (or rather for i,j modulo e), are defined by
J(i,j)=∑v∈Fqχie(v)χje(v+1). |
Jacobi sums and cyclotomic numbers are related by the following finite Fourier series expansions:
J(a,b)=∑0⩽i,j⩽e−1Aijζai+bj,e2Aab=∑0⩽i,j⩽e−1J(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+y3≡1(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,j∈Z/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 v−1=v+1.) Specifically, either e or f is even.
For 0⩽i,j,k⩽e−1 (or rather for i,j,k∈Z/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:={v∈Fq−{0,±1}∣indγ(v−1)≡i(mode), indγv≡j(mode), indγ(v+1)≡k(mode)}. |
This definition relies on the choice of a multiplicative generator γ of F∗q. 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):=∑v∈Fqχie(v−1)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={v∈Fq∣χe(v−1)=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,c∈Z/eZ,
J(a,b,c)=∑i,j,k∈Z/eZAijkζai+bj+ck, | (2.1) |
e3Aabc=∑i,j,k∈Z/eZJ(i,j,k)ζ−(ai+bj+ck). | (2.2) |
Proof. Note that
Fq={0,±1}∪⋃i,j,k∈Z/eZXijk. |
For v∈{0,±1},
χae(v−1)χbe(v)χce(v+1)=0. |
For the Aijk elements v∈Xijk,
χae(v−1)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,k∈Z/eZJ(i,j,k)ζ−(ai+bj+ck)=∑i,j,k∈Z/eZ∑v∈Fqχie(v−1)χje(v)χke(v+1)χe(γ)−(ai+bj+ck)=∑v∈Fq(e−1∑i=0χie(v−1γa))(e−1∑j=0χje(vγb))(e−1∑k=0χke(v+1γc)). |
Note that
e−1∑j=0χje(vγb)={e,if v≠0 and indγ(v)≡b(mode),0,if v=0 or indγ(v)≢b(mode), |
and similar for ∑e−1i=0χie(v−1γa) and ∑e−1j=0χke(v+1γc). So the terms of the above sum are non-zero only when v∈Xabc, and thus
∑i,j,k∈Z/eZJ(i,j,k)ζ−(ai+bj+ck)=∑v∈Xabce3=e3Aabc. |
Following arguments often involve the value of χe(−1). Since
γq−12=−1∈Fq, |
and when e is even,
ζe2=−1∈C, |
we have
χe(−1)−1=χe(−1)=χe(γq−12)=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,k∈Z/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 v∈Fq, let w=−v. Since
χe(−1)=ζef2, |
we have
v∈Xijk⟺(χe(−w−1),χe(−w),χe(−w+1))=(ζi,ζj,ζk)⟺(χe(w−1),χe(w),χe(w+1))=(χe(−1)ζk,χe(−1)ζj,χe(−1)ζi)⟺w∈Xk+ef2,j+ef2,i+ef2. |
So v↦−v induces a bijection from Xijk to Xk+ef2,j+ef2,i+ef2.
Corollary 3. For any a,b,c∈Z/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
ζ−ef2=(−1)f, |
we have
J(a,b,c)=∑i,j,k∈Z/eZAk+ef2,j+ef2,i+ef2ζai+bj+ck=∑l,m,n∈Z/eZAlmnζa(n−ef2)+b(m−ef2)+c(l−ef2)=ζ−ef2(a+b+c)∑l,m,n∈Z/eZAlmnζcl+bm+an=(−1)f(a+b+c)J(c,b,a). |
Classical cyclotomic numbers exhibit another property
Aij=A−i,j−i, |
which implies that
J(a,b)=J(−a−b,b). |
It is similar for ternary cases as follows:
Proposition 4. For any i,j,k∈Z/eZ,
Aijk=Ai−j+ef2,−j,k−j={Ai−j,−j,k−j,if f is even,Ai−j+e2,−j,k−j,if f is odd. |
Proof. For any v∈Fq, let w=v−1. Since
χe(−1)=ζef2, |
we have
v∈Xijk⟺(χe(w−1−1),χe(w−1),χe(w−1+1))=(ζi,ζj,ζk)⟺(χe(w−1)χe(−1)χe(w),χe(w),χe(1+w)χe(w))=(ζi,ζ−j,ζk)⟺(χe(w−1),χe(w),χe(w+1))=(χe(−1)ζi−j,ζ−j,ζk−j)⟺w∈Xi−j+ef2,−j,k−j. |
So v↦v−1 induces a bijection from Xijk to Xi−j+ef2,−j,k−j.
Corollary 5. For any a,b,c∈Z/eZ,
J(a,b,c)=(−1)faJ(a,−a−b−c,c). |
Proof. By Propositions 1 and 4, as
ζ−ef2=(−1)f, |
we have
J(a,b,c)=∑i,j,k∈Z/eZAi−j+ef2,−j,k−jζai+bj+ck=∑l,m,n∈Z/eZAl,m,nζa(l−m−ef2)−bm+c(n−m)=(ζ−ef2)a∑l,m,n∈Z/eZAl,m,nζal+(−a−b−c)m+cn=(−1)faJ(a,−a−b−c,c). |
For classical cyclotomic numbers, we have
∑i∈Z/eZAij={f−1,if j≡0(mode),f,otherwise, and ∑j∈Z/eZAij={f−1,if i≡ef2(mode),f,otherwise. | (2.3) |
We show similar equations for ternary cyclotomic numbers.
Proposition 6. Let g=indγ(2). For any i,j,k∈Z/eZ,
∑t∈Z/eZAtjk={Ajk−1,if j≡0(mode) and k≡g(mode),Ajk,otherwise,∑t∈Z/eZAijt={Aij−1,if i≡g+ef2(mode) and j≡ef2(mode),Aij,otherwise,∑t∈Z/eZAitk={Ai−g,k−g−1,if i ≡ef2(mode) and k≡0(mode),Ai−g,k−g,otherwise. |
Proof. Note that
⋃t∈Z/eZXtjk=Xjk−{1}. |
Also note that 1∈Xjk if and only if
j≡0(mode)andk≡g(mode). |
These Xtjk are pairwise disjoint, which gives the first equation.
For the second equation,
⋃t∈Z/eZXijt={v∈Fq−{0,±1}∣χe(v−1)=ζi, χe(v)=ζj}. |
So
v∈⋃t∈Z/eZXijt |
if and only if
v−1∈Xij−{−2}. |
Also note that −2∈Xij if and only if
i≡g+ef2(mode)andj≡ef2(mode). |
For the third equation,
⋃t∈Z/eZXitk={v∈Fq−{0,±1}∣χe(v−1)=ζi, χe(v+1)=ζk}. |
Then
v∈⋃t∈Z/eZXitk |
if and only if
v−12∈Xi−g,k−g−{−2−1}. |
Here
v↦v−12 |
induces a bijection on Fq. Also note that
−2−1∈Xi−g,k−g |
if and only if
i≡ef2(mode)andk≡0(mode). |
Note that
∑v∈Fqχie(v)=q−2∑m=0χie(γm)=q−2∑m=0(ζi)m={q−1,if i≡0(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)=∑v∈Fqχ0e(v)χie(v+1)=∑v∈Fqχie(v+1)−χie(1)={q−2,if i≡0(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,k∈Z/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 m∈Z, we have
χe(2)=ζgandχe(−1)=(−1)f. |
So
J(0,j,k)=∑v∈Fqχ0e(v−1)χje(v)χke(v+1)=∑v∈Fq−{1}χje(v)χke(v+1)=∑v∈Fqχje(v)χke(v+1)−χje(1)χke(2)=J(j,k)−ζgk,J(i,j,0)=∑v∈Fq−{−1}χie(v−1)χje(v)=∑v∈Fqχie(v−1)χje(v)−χie(−2)χje(−1)=J(i,j)−(−1)f(i+j)ζgi. |
In the following we let
w=v−12. |
Note that
v↦v−12 |
induces a bijection on Fq. Then
J(i,0,k)=∑v∈Fqχie(v−1)χke(v+1)−χie(−1)χke(1)=∑w∈Fqχie(2w)χke(2w+2)−(−1)fi=ζg(i+k)J(i,k)−(−1)fi. |
Corollary 8.
J(0,0,0)=q−3. |
For i≢0(mode), we have
J(i,0,0)=(−1)fi+1(1+ζgi),J(0,i,0)=−1−(−1)fi,J(0,0,i)=−1−ζgi. |
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,c∈Z/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 F∗q. We fix ζ=−1, and let
g=indγ(2). |
By Eq (2.4) and
J2(a,b)=J2(−a−b,b), |
all classical Jacobi sums of order 2 are:
J2(0,0)=q−2,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,j∈Z/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 anym∈Z). |
By definition,
J4(i,j)=∑v∈Fqχi4(v)χj4(v+1). |
A critical insight is that
χ2=χ24 |
on Fq. Also note that
F∗q={γm∣0⩽m⩽2f−1}. |
So
J2(1,1,1)=∑v∈Fqχ2(v−1)χ2(v)χ2(v+1)=∑v∈F∗qχ2(v)χ2(v2−1)=∑v∈F∗qχ4(v2)χ24(v2−1)=2f−1∑m=0χ4(γ2m)χ24(γ2m−1)=2f−1∑m=0χ4(γ2m)χ24(γ2m−1). |
The final expression is a part of the following formula:
J4(2,1)=∑v∈Fqχ24(v)χ4(v+1)=∑w∈Fqχ24(w−1)χ4(w)=∑w∈F∗qχ4(w)χ24(w−1)=2f−1∑n=0χ4(γn)χ24(γn−1)=f−1∑m=0χ4(γ2m)χ24(γ2m−1)+f−1∑m=0χ4(γ2m+1)χ24(γ2m+1−1). |
Since
χ4(γ2m)=(−1)m,χ4(γ2m+1)=i2m+1∈{±i} |
and
χ24(v)∈{0,±1} |
for any v∈Fq,
Re(J4(2,1))=f−1∑m=0χ4(γ2m)χ24(γ2m−1)=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,−m−n)=(−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=pα≡1(mod4) with p prime and α⩾1, and let
s=−Re(J4(1,2)). |
If p≡3(mod4), then α is even,
s=(−p)α/2≡1(mod4) |
and
J4(1,2)=−s=−(−p)α/2. |
If p≡1(mod4), then s is the unique integer coprime to q such that s≡1(mod4) and q−s2 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 p≡3(mod4).
When
f=q−12 |
is even, by Lemma 10,
J2(1,1,1)=2Re(J4(2,1))=2Re((−1)q−14J4(1,2))=(−1)q−14(−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 F∗q are explicitly given as follows, with g=indγ(2) and s defined as in Lemma 10.
J2(0,0,0)=q−3,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 q≡3(mod4),−2s,if q≡1(mod8),2s,if q≡5(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 F∗q, are explicitly evaluated as follows, with g=indγ(2) and s defined as in Lemma 10.
If q≡3(mod4), then
A(2)000=A(2)100=A(2)110=A(2)111=q−5−2(−1)g8,A(2)001=A(2)010=A(2)011=A(2)101=q−1+2(−1)g8. |
If q≡1(mod8), then
A(2)000=q−11−2s−4(−1)g8,A(2)011=A(2)110=q+1−2s8,A(2)101=q−3−2s+4(−1)g8,A(2)001=A(2)100=A(2)010=A(2)111=q−3+2s8. |
If q≡5(mod8), then
A(2)000=A(2)101=q−7+2s8,A(2)011=A(2)110=q+1+2s8,A(2)001=A(2)100=A(2)010=A(2)111=q−3−2s8. |
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=−1−ζ3. |
Also note that
σ2(ζ3)=ζ23=ζ−13=¯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(−a−b,b)=J3(b,−a−b)=J3(−a−b,a)=J3(a,−a−b). |
By Eq (2.4),
J3(0,0)=q−2, |
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.
q=pα=3f+1 |
with p odd prime and α⩾1, we can write
J3(1,1)=L+3M2+3Mζ3, |
for some L,M∈Z with L≡1(mod3).
(1) If p≡2(mod3), then α is even,
M=0,L=−2(−p)α/2,andJ3(1,1)=−(−p)α/2. |
(2) If p≡1(mod3), then (L,M) is the unique solution of the Diophantine system:
{4q=L2+27M2,L≡1(mod3),p∤L,γq−13≡(L+9M)/(L−9M)(modp). | (4.1) |
Proof. For p≡1(mod3), this result is a part of [12, Proposition 1]. For p≡2(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,b∈Z. Write Aij (with 0⩽i,j⩽2) 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(A01−A02)≡0(mod3). |
By Eq (2.3),
f−1=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(A00−A02)≡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 p≡2(mod3) is an Eisenstein prime, which is irreducible in Z[ζ3]. So
a+bζ3=pnu |
for an integer n⩾0 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
b≡0≢±pα/2(mod3) |
requires that u=±1 and thus b=0. Moreover,
J3(1,1)=a=±pα/2≡2≡p(mod 3), |
and hence
J3(1,1)=−(−p)α/2. |
As f is even, by Corollaries 3 and 5, for any a,b,c∈Z/3Z we have
J3(a,b,c)=J3(c,b,a)=J3(a,−a−b−c,c). |
Along with Proposition 7, Corollary 8, and Lemma 9, we obtain:
J3(0,0,0)=q−3,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)=−1−ζg3,J3(2,1,0)=J3(0,1,2)=J3(2,0,0)=J3(0,0,2)=−1−ζ2g3,J3(1,1,0)=J3(0,1,1)=L+3M2+3Mζ3−ζg3,J3(2,2,0)=J3(0,2,2)=L+3M2+3Mζ23−ζ2g3,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
F∗q={γm∣0⩽m⩽3f−1}. |
So
J3(1,2,1)=∑v∈F∗qχ23(v)χ3(v2−1)=3f−1∑m=0χ3(γ2m)χ3(γ2m−1)=232f−1∑m=0χ3(γ2m−1)χ3(γ2m). | (4.2) |
Also note that
J3(1,1)=3f−1∑n=0χ3(γn−1)χ3(γn)=32f−1∑m=0χ3(γ2m−1)χ3(γ2m)+32f−1∑m=0χ3(γ2m+1−1)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=−ζ−26=−ζ−13=−ζ23=1+ζ3. |
Write J6(i,j) (i,j∈Z/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 anym∈Z). |
Note that χ26=χ3 on Fq. By definition,
J6(2,5)=3f−1∑n=0χ26(γn−1)χ56(γn)=3f−1∑n=0χ3(γn−1)χ56(γn)=32f−1∑m=0χ3(γ2m−1)χ56(γ2m)+32f−1∑m=0χ3(γ2m+1−1)χ56(γ2m+1). |
Since
χ56(γ2m)=χ3(γ2m),χ56(γ2m+1)=ζ56χ3(γ2m)=−ζ3χ3(γ2m)=−χ3(γ2m+1), |
we obtain
J6(2,5)=32f−1∑m=0χ3(γ2m−1)χ3(γ2m)−32f−1∑m=0χ3(γ2m+1−1)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 α⩾1. Then
J3(1,2,1)=J3(1,1)+¯J6(1,1)={L,if g≡0(mod3),−L+9M2ζ3,if g≡1(mod3),L+9M2(1+ζ3),if g≡2(mod3). |
Moreover, if p≡2(mod3), then g≡0(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,F∈Z. By finite Fourier series expansions, E and F are indeed Z-linear combinations of classical cyclotomic numbers A(6)ij (with 0⩽i,j⩽5).
For p≡1(mod3), [14, Theorem 2] provides
J6(1,1)=(−E+F)ζ3−(E+F)ζ232=E+F2+Fζ3, |
where
(E,F)={(L,3M),if g≡0(mod3),(−L−9M2,L−3M2),if g≡1(mod3),(−L+9M2,−L−3M2),if g≡2(mod3). | (4.5) |
So we only need to show that Eq (4.5) also holds for p≡2(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 0⩽i,j⩽5). These relations, in fact, are valid for A(6)ij over any finite field Fq with q≡1(mod6). Thus, Dickson's proof of Eq (4.5) is universally applicable to all fields Fq with q≡1(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+E−F2+(3M−F)ζ3={L,if g≡0(mod3),−L+9M2ζ3,if g≡1(mod3),L+9M2(1+ζ3),if g≡2(mod3). |
For the special case of p≡2(mod3), we have
2≡21+2(p−1)(modp) |
by Fermat's little theorem. As
1+2(p−1)≡0(mod3), |
let
1+2(p−1)=3t |
with some t∈Z. Then
g=indγ(2)=indγ(23t)≡3tg(mod(q−1)). |
Since 3∣(q−1), we have g≡0(mod3), and hence
J3(1,2,1)=L=−2(−p)α/2. |
This completes the proof.
Remark 16. For p≡1(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,E≡1(mod3),p∤E,F≡−g(mod3),γq−13≡(−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,b∈Z, whose imaginary part is 0 only if b=0. So J3(1,1,2)=a∈Z.
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 cij∈C (for 0⩽i,j⩽5) such that the equality
J3(1,1,2)=∑0⩽i,j⩽5cijJ6(i,j) |
holds for any finite field Fq with q≡1(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=γn∈F∗q, we note that
2∑i=0χi3(v)=2∑i=0χi3(γn)=2∑i=0(ζn3)i={3,if 3∣n,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)=∑v∈F∗qχ23(v−1)(2∑i=0χi3(v))χ3(v+1)=3f−1∑m=0χ23(γ3m−1)χ3(γ3m+1)=3f−1∑m=1χ3(γ3m+1γ3m−1)=3f−1∑m=1χ3(2γ3m−1+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 F∗q. 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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