# Primitive Idempotents of Irreducible Cyclic Codes of Lengthn

Authors: Yuqian Lin; Qin Yue; Yansheng Wu
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

LetFq be a finite field with q elements and n a positive integer. In this paper, we use matrix method to give all primitive idempotents of irreducible cyclic codes of length n, whose prime divisors divide q-1.


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## Body

## 1. Introduction

LetFq be a finite field with q elements, where q=ps and p is a prime. Let C be a [n,k,d] linear code over Fq, i.e., it is a k-dimensional subspace of Fqn with minimum Hamming distance d. If for each codeword (c0,c1,…,cn-1)∈C, (cn-1,c0,…,cn-2) is also in C, then we call C a cyclic code. In fact, each cyclic code of length n over Fq can be viewed as an ideal in the ring R=Fq[x]/〈xn-1〉 and each irreducible cyclic code of length n over Fq is an ideal of R generated by a primitive idempotent.A lot of papers investigate primitive idempotents ofR. We list some results about the length n.(1)
In [1, 2], n=2,4,lm, and 2lm, where l is an odd prime and p is a primitive root modulo n.(2)
In [3, 4], n=2m,m≥3.(3)
In [5], n=l1ml2, where l1,l2,p are distinct odd primes with gcd(φ(l1m)/2,φ(l2)/2)=1 and p is a common primitive root modulo l1m and l2.(4)
In [6], n=l1m1l2m2, where l1,l2, and p are three distinct odd primes, ordl1m1(p)=φ(l1m1)/2, ordl2m2(p)=φ(l2m2)/2, and gcd(φ(l1m1),φ(l2m2))=2.(5)
In [7, 8], n=tlm,t,m≥1, where l is an odd prime different from the characteristic of Fq, t∣(q-1),gcd(t,l)=1 and ordtlm(q)=φ(lm); n=lm,m≥1, where l is an odd prime and l∣(q-1).(6)
In [9, 10], n=l1m1l2m2, where l1,l2 are two distinct primes with l1l2∣(q-1); n=4lm and 8lm, where l is an odd prime with l∣(q-1).(7)
In [11], n=2ml1m1l2m2, where l1,l2 are two distinct primes with 4l1l2∣(q-1).(8)
In [12], n=l1m1⋯lrmr, where l1,…,lr are distinct odd primes with l1⋯lr∣(q-1).In this paper, suppose thatrad(n)∣(q-1). We shall use matrix method to give all primitive idempotents of the ring R. The rest of paper is organized as follows: in Section 2, we give some basic results, in Section 3, we obtain all primitive idempotents in Fq[x]/〈xn-1〉 under the condition: rad(n)∣(q-1), and in Section 4, we conclude this paper.


## 2. Preliminaries

If a positive integern has a prime factorization, n=p1α1p2α2⋯plαl, where p1,p2,…,pl are distinct primes and positive integers αi≥1 for 1≤i≤l, we denote radn=p1p2⋯pl and vpin=αi, 1≤i≤l, and ord(α) is the order of α∈Fq∗. Through this paper, we always assume that gcd(n,q)=1.Every cyclic code of lengthn over a finite field Fq is identified with exactly one ideal of the quotient algebra Fq[x]/〈xn-1〉. Some explicit factorizations of xn-1 can be found in [7–11, 13–16]. We need the following results about the irreducible factorization of xn-1 over Fq.Lemma 1 ([14, Corollary 1]).
LetFq be a finite field and n a positive integer such that both rad(n)∣(q-1) and either q≢3(mod4) or 8∤n. Let m1=n/gcd⁡n,q-1, l1=(q-1)/gcd⁡n,q-1, and θ be a generator of Fq∗. Then one has the following:
(1) The factorization ofxn-1 into irreducible factors in Fqx is (1)∏tm1∏1≤u≤gcd⁡n,q-1gcd⁡u,t=1xt-θul1.
(2) For eachtm1, the number of irreducible factors of degree t is φt/t·gcd⁡n,q-1, where φ denotes the Euler Totient function, and the number of irreducible factors is (2)N1=gcd⁡n,q-1·∏pm1pprime1+vpm1·p-1p.Lemma 2 ([14, Corollary 2]).
LetFq be a finite field and n a positive integer such that rad(n)∣(q-1), q≡3(mod4), and 8∣n. Let m2=n/gcd⁡n,q2-1, l1=(q-1)/gcd⁡n,q-1, l2=q2-1/gcd⁡n,q2-1, r=min⁡v2n/2,v2q+1, and α be a generator of Fq2∗ satisfying θ=αq+1. Then one has the following:
(1) The factorization ofxn-1 into irreducible factors in Fqx is (3)∏t∣m2todd∏1≤w≤gcd⁡n,q-1gcd⁡w,t=1xt-θwl1·∏t∣m2∏u∈Rtx2t-αul2+αqul2xt+θul2,where Rt is the set (4)u∈N1≤u≤gcdn,q2-1,2r∤u,gcd⁡u,t=1,u<qugcd⁡n,q2-1and ab denotes the remainder of the division of a by b.
(2) For eacht odd with t∣m2, the number of irreducible polynomials of degree t is φt/t·gcd⁡n,q-1, and the number irreducible polynomials of degree 2t is(5)φtt·2r-1·gcd⁡n,q-1iftis  even,φt2t·2r-1·gcdn,q-1iftis  odd.The total number of irreducible factors is (6)N2=gcd⁡n,q-1·12+2r-22+v2m·∏pm2podd  prime1+vpm2·p-1p.Lemma 3 (see [17]).
Letm1,…,mt be positive integers. For a set of integers a1,…,at, the system of congruences y≡ai(modmi),i=1,…,t, has solutions if and only if(7)ai≡ajmodgcd⁡mi,mj,i≠j,1≤i,j≤t.If (7) is satisfied, the solution is unique modulo lcm(m1,…,mt).


## 3. Primitive Idempotents inR

In this section, we shall give all primitive idempotents inR if rad(n)∣(q-1).First, we consider the case q≢3(mod4) or 8∤n.In Lemma1, let t1,…,td be all positive factors of m1=n/gcd⁡n,q-1. For each ti with 1≤i≤d, there are si=φti/ti·gcd⁡n,q-1 positive integers ui1,ui2,…,uisi satisfying 1≤uij≤gcd⁡n,q-1 and gcd⁡uij,ti=1, j=1,…,si. Since l1=(q-1)/gcd⁡n,q-1 and θ=Fq∗, δ=θl1 is of order gcd⁡n,q-1. Then the irreducible factorization of xn-1 over Fq can be rewritten as(8)xn-1=∏1≤i≤d1≤j≤sixti-δuij=∏1≤j≤s1xt1-δu1j⋯∏1≤j≤sdxtd-δudj.Note that the number of primitive idempotents inR coincides with the number of irreducible factors of xn-1 over Fq.Theorem 4.
Letrad(n)∣(q-1) and either q≢3(mod4) or 8∤n. Then there are N1 primitive idempotents in R as follows: (9)θijx=tin∑k=0n/ti-1δ-uijkxkti,corresponding to the irreducible polynomials xti-δuij over Fq, i=1,…,d,j=1,…,si.Proof.
For eachi, 1≤i≤d, let Ri=∏1≤j≤siFq[x]/〈xti-δuij〉 be a ring with si direct summands; for 0≤k≤n-1, k=tiu+v, 0≤u≤n/ti-1, and 0≤v≤ti-1. By (8) and Chinese Remainder Theorem, there is an Fq-algebra isomorphism: (10)ψ=ψ1,ψ2,…,ψd:R⟶R1×R2×⋯×Rd,where each ψi:R→Ri,∑k=0n-1akxk↦Ai,0+Ai,1x+⋯+Ai,ti-1xti-1 is an Fq-algebraic epimorphism and each (11)Ai,v=∑u=0n/ti-1atiu+vδuui1,∑u=0n/ti-1atiu+vδuui2,…,∑u=0n/ti-1atiu+vδuuisi∈Fqsi,0≤v≤ti-1.
Note that∑i=1dsiti=n. Hence there is a Fq-linear space isomorphism: (12)ϕ=ϕ1,ϕ2,…,ϕd:R1×R2×⋯×Rd⟶∏i=1dFqsiti=Fqn,where each ϕi:Ri→Fqsiti,Ai,0+Ai,1x+⋯+Ai,ti-1xti-1↦Ai,0,Ai,1,…,Ai,ti-1 is a Fq-linear space epimorphism. Hence there is a Fq-linear space isomorphism: (13)χ=ϕψ:R⟶Fqn,∑k=0n-1akxk⟼A1,0,…,A1,t1-1,…,Ad,0,…,Ad,td-1,(14)A1,0,…,A1,t1-1,…,Ad,0,…,Ad,td-1=a0,a1,…an-1B,where B is a n×n invertible matrix over Fq. Now we shall determine B and B-1.
In (14), let B:=(B1(δ),…,Bd(δ)) be a n×n matrix, where each Bi(δ)=(Bi(1)(δ),…, Bi(ti)(δ)) is a n×siti matrix and each Bi(v)(δ), 1≤v≤ti, is a n×si matrix: (15)
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We know thatord(δ)=gcd⁡(n,q-1), xti-δuij, 1≤uij≤gcd⁡(n,q-1), and gcd⁡(ti,uij)=1 are an irreducible polynomial of xn-1, so (δuij)n/ti=1. Fix i and ti, 1≤i≤d. If 1≤uij≠uij′≤gcd⁡(n,q-1), gcd⁡(uij,ti)=1, gcd(uij′,ti)=1. Then δuij-uij′≠1 and (δuij-uij′)n/ti=1. Let (16)Bivδ-1T=⋯⋯⋮⋯00⋮0δ-ui10δ-ui20⋮δ-uisi000⋮0⋯⋯⋮⋯δ-ui11δ-ui21⋮δ-uisi100⋮0⋯⋯⋮⋯δ-ui1n/ti-1δ-ui2n/ti-1⋮δ-uisin/ti-100⋮0⋯⋯⋮⋯vti+vnti-1ti+vbe a si×n matrix over Fq. Then, (17)Bivδ-1TBivδ=ntiEsi,Bivδ-1TBiv′δ=0if1≤v≠v′≤ti,i.e., (18)Biδ-1T·Biδ=Bi1δ-1T⋮Bitiδ-1TB11δ,…,Bitiδ=ntiEsiti,where Esi and Esiti are the identity matrices of order si×si and siti×siti, respectively.
Let(19)Biδ-1T=Bi1δ-1T⋮Bitiδ-1Tbe a siti×n matrix. Next, we shall prove that (Bi(δ-1))T·Bi′(δ)=0,1≤i≠i′≤d. In fact, (20)Biδ-1T·Bi′δ=Bi1δ-1T⋮Bitiδ-1TBi′1δ,…,Bi′ti′δ=Bi1δ-1TBi′1δ⋯Bi1δ-1TBi′ti′δ⋮⋮Bitiδ-1TBi′1δ⋯Bitiδ-1TBi′ti′δ.Hence we only need to show that(21)Bivδ-1TBi′v′δ=0,1≤v≤ti,1≤v′≤ti′.We consider the following congruence equations:(22)x≡vmodtix≡v′modti′.
Suppose thatgcd⁡(ti,ti′)∤(v-v′). Then it has no solution in (22) by Lemma 3, so it holds in (21).
Suppose thatgcd(ti,ti′)∣(v-v′). Then this is unique solution x=a0 in (22) with 1≤x≤lcm(ti,ti′). Let lcm(ti,ti′)=c=tiα=ti′β. Then x=a0+cl,l=0,1,…,n/c-1 are all solutions in (22) with 1≤x≤n. Let (Mi(v)(δ-1))TMi′(v′)(δ)=(cjj′) be a si×si′ matrix over Fq. Then for 1≤j≤si,1≤j′≤si′, the (j,j′) entry is (23)cjj′=∑l=0n/c-1δ-uijαlδui′j′βl=∑l=0n/c-1δ-uijα+ui′j′βl,where 1≤uij,ui′j′≤gcd⁡(n,q-1), gcd⁡(uij,ti)=1, and gcd⁡(ui′j′,ti′)=1. Since xti-δuij is an irreducible divisor of xn-1 over Fq, (δuij)n/ti=1; similarly, (δui′j′)n/ti′=1. Hence (24)δ-uijα+ui′j′βn/c=δ-uijn/tiδui′j′n/ti′=1.On the other hand, by ti≠ti′ we assume that there is a prime p such that vp(ti)>vp(ti′). Then p∣β and p∤α by lcm(ti,ti′)=c=tiα=ti′β, so p∤(-uijα+ui′j′β) and p∣gcd(n,q-1). Hence δ-uijα+ui′j′β≠1. Therefore, cjj′=∑l=0n/c-1(δ-uijα+ui′j′β)l=0, and it holds in (21).
In conclusion,(Bi(δ-1))TBi(δ)=(n/ti)Esiti, (Bi(δ-1))TBi′(δ)=0,1≤i≠i′≤d, and (25)B-1=1nt1B1δ-1Tt2B2δ-1T⋮tdBdδ-1T.
In the following, we present all primitive idempotents inR by lifting some primitive idempotents in Fqn through the isomorphism χ.
By Lemma1, the number of irreducible factors of xn-1, which coincides with the number of primitive idempotents in R, is N1. Let {e1,…,en} denote the standard basis of Fqn. Hence the vectors of Fqn, e1,e2,…,es1,et1s1+1,et1s1+2,…,et1s1+s2,…,e∑h=1d-1thsh+1, e∑h=1d-1thsh+2,…,e∑h=1d-1thsh+sd, correspond to all primitive idempotents in R. Hence for i,j, 1≤i≤d,1≤j≤si, let θijx=∑k=0n-1akxk be a primitive idempotent in R, which is corresponding to e∑h=1i-1thsh+j. By (14),(26)χθijx=a0,a1,…,an-1B=0,…,0,1∑h=1i-1thsh+j,0,…,0=e∑h=1i-1thsh+j,and a0,a1,…,an-1=e∑h=1i-1thsh+jB-1. So we have proved the theorem.Remark 5.
In special cases in Theorem4, we can give those results in [8–11].Second, we consider the case q≡3(mod4) and 8∣n.In Lemma2, let t1,t2,…,td be all odd factors of m2=n/gcd⁡n,q2-1 and let td+1,td+2,…,td+d′ be all even factors of m2. For each ti with 1≤i≤d, there are si=φti/ti·gcd⁡n,q-1 positive integers wi1,wi2,…,wisi satisfying 1≤wij≤gcd⁡n,q-1 and gcd⁡wij,ti=1, j=1,2,…,si. For each ti with 1≤i≤d+d′, there are gi positive integers ui1,ui2,…,uigi satisfying 1≤uij≤2rgcd⁡n,q-1,gcd⁡(ti,uij)=1,2r∤uij, j=1,…,gi. In fact, n=∑i=1dsiti+∑i=1d+d′2tigi.Sincel1=(q-1)/gcd⁡n,q-1, l2=(q2-1)/gcd⁡n,q2-1, θ=Fq∗, and α=Fq2∗, there exist δ∈Fq∗ and σ∈Fq2∗ such that θl1=δ and αl2=σ. Then the irreducible factorization of xn-1 over Fq can be rewritten as (27)xn-1=∏1≤i≤d1≤j≤sixti-δwij·∏1≤i≤d+d′1≤j≤gix2ti-σuij+σquijxti+σq+1uij.Theorem 6.
Suppose thatrad(n)∣(q-1), q≡3(mod4), and 8∣n. Then there are N2 primitive idempotents in R as follows:
(1)(28)θij=tin∑k=0n/ti-1δ-wijkxkticorrespond to the irreducible polynomials xti-δwij over Fq,i=1,…,d,j=1,…,si.
(2)(29)ηij=tin∑k=0n/ti-1Trσ-uijkxkticorrespond to the irreducible polynomials x2ti-σuij+σquijxti+σq+1uij over Fq, i=1,…,d+d′,j=1,…,gi, where Tr is the trace map from Fq2 into Fq.Proof.
The factorization ofxn-1 into irreducible factors in Fq2[x] is(30)xn-1=∏1≤i≤d1≤j≤sixti-δwij·∏1≤i≤d+d′1≤j≤gixti-σuijxti-σquij.Similarly to proving Theorem 4, there is a Fq2-linear space isomorphism:(31)χ=χ1,…,χd,λ1,…,λd+d′,λ1q,…,λd+d′q:Fq2xxn-1⟶Fq2n,(32)∑k=0n-1akxk⟼A1,0,…,Ad,td-1,D1,0,…,Dd+d′,td+d′-1,D1,0q,…,Dd+d′,td+d′-1q,where there are Fq2-epimorphisms: for 1≤i≤d,(33)χi:Fq2xxn-1⟶∏1≤j≤siFq2xxti-δwij⟶Fq2siti(34)∑k=0n-1akxk⟼Ai,0,…,Ai,ti-1,for 1≤i≤d+d′,(35)λi:Fq2xxn-1⟶∏1≤j≤giFq2xxti-σuij⟶Fq2giti(36)∑k=0n-1akxk⟼Di,0,…,Di,ti-1,and for 1≤i≤d+d′,(37)λiq:Fq2xxn-1⟶∏1≤j≤giFq2xxti-σquij⟶Fq2giti(38)∑k=0n-1akxk⟼Di,0q,…,Di,ti-1q.Hence there is a n×n invertible matrix B over Fq2 such that(39)a0,a1,…,an-1B=A1,0,…,Ad,td-1,D1,0,…,Dd+d′,td+d′-1,D1,0q,…,Dd+d′,td+d′-1q.Now we shall construct the matrix B. Let (40)B=B1δ,…,Bdδ,B1σ,…,Bd+d′σ,B1σq,…,Bd+d′σq,where Bi(δ) are n×siti matrices over Fq2, 1≤i≤d, and Bi(σ), Bi(σq) are n×giti matrices over Fq2, 1≤i≤d+d′.
(a) For eachi with 1≤i≤d, by (33) we have (41)a0,a1,…,an-1Biδ=Ai,0,Ai,1,…,Ai,ti-1,where Ai,0,Ai,1,…,Ai,ti-1∈Fqsi. Let Biδ=(Bi(1)(δ),…,Bi(ti)(δ)) be a n×siti matrix, and each Bi(v)(δ), 1≤v≤ti be a n×si matrix as shown in Theorem 4.
(b) For eachi with 1≤i≤d+d′, by (35) we have that a0,a1,…,an-1Biσ=Di,0,Di,1,…,Di,ti-1, where Di,0,Di,1,…,Di,ti-1∈Fqgi. Let Biσ=(Bi(1)(σ),…,Bi(ti)(σ)) be a n×giti matrix and each Bi(v)(σ), 1≤v≤ti, a n×gi matrix: (42)
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(c) For eachi with 1≤i≤d+d′, by (37) we have that a0,a1,…,an-1Biσq=Di,0(q),Di,1(q),…,Di,ti-1(q), where Di,0(q),Di,1(q),…,Di,ti-1(q)∈Fqgi. Let Biσq=(Bi(1)(σq),…,Bi(ti)(σq)) be a n×giti matrix, and each Bi(v)(σq), 1≤v≤ti, a n×gi matrix:(43)
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Similarly to proving Theorem4, we obtain that(44)B-1=1nt1B1δ-1T⋮tdBdδ-1Tt1B1σ-1T⋮td+d′Bd+d′σ-1Tt1B1σ-qT⋮td+d′Bd+d′σ-qT,where
(a) for eachi with 1≤i≤d, (Biδ-1)T=(Bi(1)(δ-1))T⋮(Bi(ti)(δ-1))T, and for each v with 1≤v≤ti, (Bi(v)(δ-1))T is a si×n matrix as shown in Theorem 4.
(b) for eachi with 1≤i≤d+d′, (Biσ-1)T=(Bi(1)(σ-1))T⋮(Bi(ti)(σ-1))T, and for each v with 1≤v≤ti, (Bi(v)(σ-1))T is a gi×n matrix: (45)Bivσ-1T=⋯⋯⋮⋯00⋮0σ-ui10σ-ui20⋮σ-uisi000⋮0⋯⋯⋮⋯00⋮0σ-ui11σ-ui21⋮σ-uisi100⋮0⋯⋯⋮⋯σ-ui1n/ti-1σ-ui2n/ti-1⋮σ-uisin/ti-100⋮0⋯⋯⋮⋯vti+vnti-1ti+v;
(c) for eachi with 1≤i≤d+d′, (Biσ-q)T=(Bi(1)(σ-q))T⋮(Bi(ti)(σ-q))T, and for each v with 1≤v≤ti, (Bi(v)(σ-q))T is a gi×n matrix: (46)Bivσ-qT=⋯⋯⋮⋯00⋮0σ-qui10σ-qui20⋮σ-quisi000⋮0⋯⋯⋮⋯00⋮0σ-qui11σ-qui21⋮σ-quisi100⋮0⋯⋯⋮⋯σ-qui1n/ti-1σ-qui2n/ti-1⋮σ-quisin/ti-100⋮0⋯⋯⋮⋯vti+vnti-1ti+v.
In the following, we give all primitive idempotents inFq[x]/〈xn-1〉.
(1) For fixedi and j with 1≤i≤d and 1≤j≤si, δwij∈Fq. Hence the primitive idempotents in Fq2[x]/〈xti-δwij〉 are the same as Fq[x]/〈xti-δwij〉. We have the result.
(2) For fixedi and j with 1≤i≤d+d′ and 1≤j≤gi, the polynomial x2ti-(σuij+σquij)xti+σ(q+1)uij is irreducible over Fq. In fact, the primitive idempotents in Fq2[x]/〈x2ti-(σuij+σquij)xti+σ(q+1)uij〉 are the same as Fq[x]/〈x2ti-(σuij+σquij)xti+σ(q+1)uij〉.
Note that there areFq2-algebra isomorphisms:(47)τi:Fq2xx2ti-σuij+σquijxti+σq+1uij⟶Fq2xxti-σuij×Fq2xxti-σquij,cx=∑k=02ti-1akxk=y+zxti⟼y,zN=y′,z′,where y=∑k=0ti-1aixi,z=∑k=0ti-1ati+kxk∈Fq2[x], and N is a 2×2 matrix over Fq2: (48)N=11σuijσquij.Note that the identity of Fq2[x]/〈x2ti-σuij+σquijxti+σq+1uij〉 is equal to the identity of Fq[x]/〈x2ti-σuij+σquijxti+σq+1uij〉. Now, take cx=1, then (49)τicx=τi1=1,0N=1,1.Let ηijx=∑k=0n-1akxk be a primitive idempotent in R corresponding to the irreducible polynomials x2ti-σuij+σquijxti+σq+1uij. By (31) (50)χηijx=a0,a1,…,an-1B=b0,b1,…,bn-1=A1,0,…,Ad,td-1,D1,0,…,Dd+d′,td+d′-1,D1,0q,…,Dd+d′,td+d′-1q=0,…,0,Di,0,0,…,0,Di,0q,0,…,0,where Di,0=(0,…,0,1j,0,…,0),Di,0q=(0,…,0,1j,0,…,0). Hence (51)a0,a1,…,an-1=b0,b1,…,bn-1B-1=tinTrσ-uij0,0,…,0,Trσ-uij1,0,…,0,Trσ-uijn/ti-1,0,…,0.Hence we complete the proof.


## 4. Concluding Remarks

In this paper, suppose thatrad(n)∣(q-1), we use matrix method to give all primitive idempotents in the ring Fq[x]/〈xn-1〉. Suppose that the order of q modulo rad(n) is w, where w is a positive integer. We can also obtain all primitive idempotents of irreducible cyclic codes by the similar method in Theorems 4 and 6. Hence, all primitive idempotents of simple root irreducible cyclic codes can be presented by the method in Theorems 4 and 6.



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