Zeros of a cyclic code

i∈C_jxⁱ. Furthermore, by examining (4.6), we see thate(x)=e(x)µ2. By Theorem 4.3.13,C=Cµ2. We summarize these results.

Theorem 4.3.16 LetCbe a cyclic code overF4with generating idempotent e(x). Then e(x) has the form given in(4.6). AlsoC=Cµ2is cyclic with generating idempotent e(x)µ2. Exercise 235 Show that ife(x) is the generating idempotent of a cyclic codeCoverF4, thenCis a cyclic code with generating idempotente(x)=

jaj

i∈C_jxⁱ. Show also that

e(x)=e(x)µ2.

Exercise 236 Do the following:

(a) List the 4-cyclotomic cosets modulo 21.

(b) Construct a setS =K ∪L₁∪L₂of distinct 4-cyclotomic coset representatives modulo 21 which can be used to construct idempotents inR21overF4as in the discussion prior to Theorem 4.3.16.

(c) Give the general form of such an idempotent.

(d) How many of these idempotents are there?

(e) Write down four of these idempotents.

Theorem 4.3.13 shows thatµa maps cyclic codes to cyclic codes with the generating idempotent mapped to the generating idempotent; however, the generator polynomial may not be mapped to the generator polynomial of the image code. In fact, the automorphism µqmaps the generator polynomial to itsqth power. See Exercise 232.

A multiplier takes a cyclic code into an equivalent cyclic code. The following theorem, a special case of a theorem of P´alfy (see [150]), implies that, in certain instances, two cyclic codes are permutation equivalent if and only if a multiplier takes one to the other. This is a very powerful result when it applies.

Theorem 4.3.17 Let C1 and C2 be cyclic codes of length n over Fq. Assume that gcd(n, φ(n))=1,whereφis the Eulerφ-function. ThenC1andC2are permutation equiv-alent if and only if there is a multiplier that mapsC1toC2.

Since multipliers send generating idempotents to generating idempotents, we have the following corollary.

Corollary 4.3.18 Let C1 and C2 be cyclic codes of length n over Fq. Assume that gcd(n, φ(n))=1,whereφis the Eulerφ-function. ThenC1andC2are permutation equiva-lent if and only if there is a multiplier that maps the idempotent ofC1to the idempotent ofC2.

4.4 Zeros of a cyclic code

Recall from Section 4.1 and, in particular Theorem 4.1.1, that if t=ordn(q), then Fq^t

is a splitting field of xⁿ−1; so Fq^t contains a primitive nth root of unity α, and xⁿ−1=+n−1

i=0(x−αⁱ) is the factorization of xⁿ−1 into linear factors over Fq^t. Fur-thermore xⁿ−1=+

sM_α^s(x) is the factorization of xⁿ−1 into irreducible factors

over Fq, where s runs through a set of representatives of the q-cyclotomic cosets modulon.

LetCbe a cyclic code inRnwith generator polynomialg(x). By Theorems 4.1.1(i) and 4.2.1(vii), g(x)=+

sM_α^s(x)=+

s

+

i∈C_s(x−αⁱ), wheres runs through some subset of representatives of theq-cyclotomic cosetsCs modulon. LetT =.

sCs be the union of theseq-cyclotomic cosets. The roots of unityZ= {αⁱ |i∈T}are called thezerosof the cyclic codeCand{αⁱ |i ∈T}are thenonzerosofC. The setT is called thedefining setof C. (Note that if you change the primitiventh root of unity, you changeT; soT is computed relative to a fixed primitive root. This will be discussed further in Section 4.5.) It follows thatc(x) belongs toC if and only ifc(αⁱ)=0 for eachi ∈T by Theorem 4.2.1. Notice thatT, and hence either the set of zeros or the set of nonzeros, completely determines the generator polynomialg(x). By Theorem 4.2.1, the dimension ofCisn− |T|as|T|is the degree ofg(x).

Example 4.4.1 In Example 4.3.4 a table giving the dimension, generator polynomialsgi(x), and generating idempotentsei(x) of all the cyclic codesCi of length 7 overF2was given.

We add to that table the defining sets of each code relative to the primitive rootαgiven in Example 3.4.3.

i dim gi(x) ei(x) Defining set

0 0 1+x⁷ 0 {0,1,2,3,4,5,6}

1 1 1+x+x²+ · · · +x⁶ 1+x+x²+ · · · +x⁶ {1,2,3,4,5,6}

2 3 1+x²+x³+x⁴ 1+x³+x⁵+x⁶ {0,1,2,4} 3 3 1+x+x²+x⁴ 1+x+x²+x⁴ {0,3,5,6}

4 4 1+x+x³ x+x²+x⁴ {1,2,4}

5 4 1+x²+x³ x³+x⁵+x⁶ {3,5,6}

6 6 1+x x+x²+ · · · +x⁶ {0}

7 7 1 1 ∅

Exercise 237 What would be the defining sets of each of the codes in Example 4.4.1 if the primitive rootβ=α³were used to determine the defining set rather thanα?

Our next theorem gives basic properties of cyclic codes in terms of their defining sets, summarizing the above discussion.

Theorem 4.4.2 Letαbe a primitive nth root of unity in some extension field ofFq. L etC be a cyclic code of length n overFq with defining set T and generator polynomial g(x).

The following hold.

(i) T is a union of q-cyclotomic cosets modulo n.

(ii) g(x)=+

i∈T(x−αⁱ).

(iii) c(x)∈Rnis inCif and only if c(αⁱ)=0for all i ∈T . (iv) The dimension ofCis n− |T|.

Exercise 238 LetCbe a cyclic code overFqwith defining setT and generator polynomial g(x). LetCebe the subcode of all even-like vectors inC.

(a) Prove thatCeis cyclic and has defining setT ∪ {0}.

(b) Prove thatC=Ceif and only if 0∈T if and only ifg(1)=0.

143 4.4 Zeros of a cyclic code

(c) Prove that ifC=Ce, then the generator polynomial ofCeis (x−1)g(x).

(d) Prove that ifCis binary, thenCcontains the all-one vector if and only if 0∈T. Exercise 239 Let Ci be cyclic codes of length n over Fq with defining sets Ti for i =1, 2.

(a) Prove thatC1∩C2has defining setT1∪T2. (b) Prove thatC1+C2has defining setT₁∩T₂. (c) Prove thatC1⊆C2if and only ifT2⊆T1.

Note: This exercise shows that the lattice of cyclic codes of lengthn overFq, where the join of two codes is the sum of the codes and the meet of two codes is their intersection, is isomorphic to the “upside-down” version of the lattice of subsets ofN = {0,1, . . . ,n−1}

that are unions ofq-cyclotomic cosets modulon, where the join of two such subsets is the set union of the subsets and the meet of two subsets is the set intersection of the

subsets.

The zeros of a cyclic code can be used to obtain a parity check matrix (possibly with dependent rows) as explained in the next theorem. The construction presented in this theorem is analogous to that of the subfield subcode construction in Section 3.8.

Theorem 4.4.3 LetCbe an[n,k]cyclic code overFqwith zerosZin a splitting fieldFq^t

of xⁿ−1overFq. L etα∈Fq^t be a primitive nth root of unity inFq^t,and letZ= {α^j | j ∈Ci₁∪ · · · ∪Ci_w},where Ci₁, . . . ,Ci_w are distinct q-cyclotomic cosets modulo n. Let L be thew×n matrix overFq^t defined by

L =







1 αⁱ¹ α²ⁱ¹ · · · α^(n−1)i1 1 αⁱ² α²ⁱ² · · · α^(n−1)i2

...

1 α^iw α^2iw · · · α^(n−1)iw





.

Thencis inCif and only if Lc^T=0. Choosing a basis ofFq^t overFq,we may represent each element ofFq^t as a t×1column vector overFq. Replacing each entry of Lby its corresponding column vector,we obtain a tw×n matrix H overFqwhich has the property thatc∈Cif and only if Hc^T=0. In particular,k≥n−tw.

Proof: We havec(x)∈C if and only if c(α^j)=0 for all j∈Ci₁∪ · · · ∪Ci_w, which by Theorem 3.7.4 is equivalent toc(α^ij)=0 for 1≤ j ≤w. Clearly, this is equivalent to Lc^T =0, which is a system of homogeneous linear equations with coefficients that are powers ofα. Expanding each of these powers ofαin the chosen basis ofFq^t overFq yields the equivalent systemHc^T=0. As the rows ofHmay be dependent,k≥n−tw. IfCis the code overFq^t with parity check matrixL in this theorem, then the codeCis actually the subfield subcodeC|_Fq.

Exercise 240 Show that the matrixLin Theorem 4.4.3 has rankw. Note: The matrixLis

related to a Vandermonde matrix. See Lemma 4.5.1.

For a cyclic code C inRn, there are in general many polynomials v(x) in Rn such thatC= v(x). However, by Theorem 4.2.1 and its corollary, there is exactly one such

polynomial, namely the monic polynomial inCof minimal degree, which also dividesxⁿ−1 and which we call the generator polynomial ofC. In the next theorem we characterize all polynomialsv(x) which generateC.

Theorem 4.4.4 LetCbe a cyclic code of length n overFqwith generator polynomial g(x).

Letv(x)be a polynomial inRn.

(i) C= v(x)if and only ifgcd(v(x),xⁿ−1)=g(x).

(ii) v(x)generatesCif and only if the nth roots of unity which are zeros ofv(x)are precisely the zeros ofC.

Proof: First assume that gcd(v(x),xⁿ−1)=g(x). Asg(x)|v(x), multiples ofv(x) are multiples of g(x) in Rn and so v(x) ⊆C. By the Euclidean Algorithm there exist polynomials a(x) and b(x) in Fq[x] such that g(x)=a(x)v(x)+b(x)(xⁿ−1). Hence g(x)=a(x)v(x) in Rn and so multiples of g(x) are multiples of v(x) in Rn implying v(x) ⊇C. ThusC= v(x).

For the converse, assume thatC= v(x). Letd(x)=gcd(v(x),xⁿ−1). Asg(x)|v(x) andg(x)|(xⁿ−1) by Theorem 4.2.1,g(x)|d(x) by Exercise 158. Asg(x)∈C= v(x), there exists a polynomiala(x) such thatg(x)=a(x)v(x) inRn. So there exists a polynomial b(x) such thatg(x)=a(x)v(x)+b(x)(xⁿ−1) inFq[x]. Thusd(x)|g(x) by Exercise 158.

Hence as bothd(x) andg(x) are monic and divide each other,d(x)=g(x) and (i) holds.

As the only roots of bothg(x) andxⁿ−1 arenth roots of unity,g(x)=gcd(v(x),xⁿ−1) if and only if thenth roots of unity which are zeros ofv(x) are precisely the zeros ofg(x);

the latter are the zeros ofC.

Corollary 4.4.5 LetC be a cyclic code of length n overFq with zeros{αⁱ |i ∈T}for some primitive nth root of unityαwhere T is the defining set ofC. Let a be an integer such thatgcd(a,n)=1and let a⁻¹be the multiplicative inverse of a in the integers modulo n.

Then{α^a−1i |i ∈T}are the zeros of the cyclic codeCµaand a⁻¹T modn is the defining set forCµa.

Proof: Lete(x) be the generating idempotent of C. By Theorem 4.3.13, the generating idempotent of the cyclic codeCµa ise(x)µa. By Theorem 4.4.4, the zeros ofCandCµa

are thenth roots of unity which are also roots ofe(x) ande(x)=e(x)µa, respectively. As e(x)=e(x)µa ≡e(x^a) (modxⁿ−1),e(x)=e(x^a)+b(x)(xⁿ−1) inFq[x]. The corol-lary now follows from the fact that thenth root of unityα^j is a root ofe(x) if and only if

α^{a j} is a root ofe(x).

Theorem 4.3.13 implies that the image of one special vector, the generating idempotent, of a cyclic code under a multiplier determines the image code. As described in Corollary 4.4.5 a multiplier maps the defining set, and hence the zeros, of a cyclic code to the defining set, and hence the zeros, of the image code. Such an assertion is not true for a general permutation.

Exercise 241 An equivalence classof codes is the set of all codes that are equivalent to one another. Give a defining set for a representative of each equivalence class of the binary cyclic codes of length 15. Example 3.7.8, Theorem 4.3.17, and Corollary 4.4.5 will

be useful.

145 4.4 Zeros of a cyclic code

Exercise 242 Continuing with Exercise 241, do the following:

(a) List the 2-cyclotomic cosets modulo 31.

(b) List the defining sets for all [31,26] binary cyclic codes. Give a defining set for a representative of each equivalence class of the [31,26] binary cyclic codes. Hint: Use Theorem 4.3.17 and Corollary 4.4.5.

(c) Repeat part (b) for [31,5] binary cyclic codes. (Take advantage of your work in part (b).)

(d) List the 15 defining sets for all [31,21] binary cyclic codes, and give a defining set for a representative of each equivalence class of these codes.

(e) List the 20 defining sets for all [31,16] binary cyclic codes, and give a defining set for a representative of each equivalence class of these codes.

IfCis a code of lengthnoverFq, then acomplementofCis a codeCc such thatC+Cc= Fⁿq andC∩Cc= {0}. In general, the complement is not unique. However, Exercise 243 shows that ifCis a cyclic code, there is a unique complement ofCthat is also cyclic. We call this codethe cyclic complementofC. In the following theorem we give the generator polynomial and generating idempotent of the cyclic complement.

Exercise 243 Prove that a cyclic code has a unique complement that is also cyclic.

Theorem 4.4.6 LetCbe a cyclic code of length n overFqwith generator polynomial g(x), generating idempotent e(x),and defining set T . LetCc be the cyclic complement ofC. The following hold.

(i) h(x)=(xⁿ−1)/g(x)is the generator polynomial forCc and1−e(x)is its generating idempotent.

(ii) Cc is the sum of the minimal ideals ofRnnot contained inC. (iii) IfN = {0,1, . . . ,n−1},thenN \T is the defining set ofCc.

Exercise 244 Prove Theorem 4.4.6.

The dual C^⊥ of a cyclic code C is also cyclic as Theorem 4.2.6 shows. The gener-ator polynomial and generating idempotent for C^⊥ can be obtained from the generator polynomial and generating idempotent ofC. To find these, we reintroduce the concept of the reciprocal polynomial encountered in Exercise 192. Let f(x)= f0+ f1x+ · · · + fax^a be a polynomial of degree a in Fq[x]. The reciprocal polynomial of f(x) is the polynomial

f^∗(x)=x^af(x⁻¹)=x^a(f(x)µ₋1)= fa+ fa−1x+ · · · + f0x^a.

So f^∗(x) has coefficients the reverse of those of f(x). Furthermore, f(x) isreversible provided f(x)= f^∗(x).

Exercise 245 Show that a monic irreducible reversible polynomial of degree greater than 1 cannot be a primitive polynomial except for the polynomial 1+x+x²overF2. We have the following basic properties of reciprocal polynomials. Their proofs are left as an exercise.

Lemma 4.4.7 Let f(x)∈Fq[x].

(i) Ifβ1, . . . , βr are the nonzero roots of f in some extension field ofFq,thenβ₁⁻¹, . . . , βr⁻¹are the nonzero roots of f^∗in that extension field.

(ii) If f(x)is irreducible overFq,so is f^∗(x).

(iii) If f(x)is a primitive polynomial,so is f^∗(x).

Exercise 246 Prove Lemma 4.4.7.

Exercise 247 In Example 3.7.8, the factorization ofx¹⁵−1 into irreducible polynomials overF2was found. Find the reciprocal polynomial of each of these irreducible polynomials.

How does this confirm Lemma 4.4.7?

Exercise 248 Prove that if f1(x) and f2(x) are reversible polynomials inFq[x], so is

f₁(x)f₂(x). What about f₁(x)+ f₂(x)?

The connection between dual codes and reciprocal polynomials is clear from the follow-ing lemma.

Lemma 4.4.8 Leta=a₀a₁· · ·an−1andb=b₀b₁· · ·bn−1be vectors inFⁿqwith associated polynomials a(x) and b(x). Then a is orthogonal to b and all its shifts if and only if a(x)b^∗(x)=0inRn.

Proof: Letb⁽ⁱ⁾ =bibi+1· · ·bn+i−1be theith cyclic shift ofb, where the subscripts are read modulon. Then

a·b⁽ⁱ⁾=0 if and only if

n−1

j=0

ajbj+i =0. (4.7)

But a(x)b^∗(x)=0 in Rn if and only if a(x)(x^{n−1−degb(x)})b^∗(x)=0 in Rn. But a(x)(x^{n−1−degb(x)})b^∗(x)=n−1

i=0(n−1

j=0ajbj+ixⁿ⁻¹⁻ⁱ). Thusa(x)b^∗(x)=0 in Rn if and

only if (4.7) holds for 0≤i ≤n−1.

We now give the generator polynomial and generating idempotent of the dual of a cyclic code. The proof is left as an exercise.

Theorem 4.4.9 LetC be an[n,k]cyclic code overFq with generator polynomial g(x), generating idempotent e(x),and defining set T . Let h(x)=(xⁿ−1)/g(x). The following hold.

(i) C^⊥is a cyclic code andC^⊥=Ccµ₋1.

(ii) C^⊥has generating idempotent1−e(x)µ−1and generator polynomial x^k

h(0)h(x⁻¹).

(iii) Ifβ1, . . . , βkare the zeros ofC,thenβ₁⁻¹, . . . , βk⁻¹are the nonzeros ofC^⊥. (iv) IfN = {0,1, . . . ,n−1},thenN \(−1)T modn is the defining set ofC^⊥. (v) Precisely one ofCandC^⊥is odd-like and the other is even-like.

The polynomialh(x)=(xⁿ−1)/g(x) in this theorem is called thecheck polynomialof C. The generator polynomial ofC^⊥in part (ii) of the theorem is the reciprocal polynomial ofh(x) rescaled to be monic.

147 4.4 Zeros of a cyclic code

Exercise 249 Prove Theorem 4.4.9.

Exercise 250 LetCbe a cyclic code with cyclic complementCc. Prove that ifCis MDS

so isCc.

The following corollary determines, from the generator polynomial, when a cyclic code is self-orthogonal.

Corollary 4.4.10 LetCbe a cyclic code overFq of length n with generator polynomial g(x)and check polynomial h(x)=(xⁿ−1)/g(x). ThenCis self-orthogonal if and only if h^∗(x)|g(x).

Exercise 251 Prove Corollary 4.4.10.

Exercise 252 Using Corollary 4.4.10 and Examples 3.7.8, 4.2.4, and 4.3.4 give the gener-ator polynomials of the self-orthogonal binary cyclic codes of lengths 7, 9, and 15.

In the next theorem we show how to decide when a cyclic code is self-orthogonal.

In particular, this characterization shows that all self-orthogonal cyclic codes are even-like. In this theorem we use the observation that ifC is aq-cyclotomic coset modulon, eitherCµ−1 =C orCµ−1=Cfor some differentq-cyclotomic cosetC, in which case Cµ₋1=Casµ²₋₁is the identity.

Theorem 4.4.11 Let C be a self-orthogonal cyclic code overFq of length n with defin-ing set T . Let C1, . . . ,Ck,D1, . . . ,D,E1, . . . ,Ebe all the distinct q-cyclotomic cosets modulo n partitioned so that Ci =Ciµ₋1 for1≤i ≤k and Di =Eiµ₋1for1≤i ≤. The following hold.

(i) Ci ⊆T for1≤i ≤k and at least one of Dior Eiis contained in T for1≤i ≤. (ii) Cis even-like.

(iii) C∩Cµ−1 = {0}.

Conversely,ifCis a cyclic code with defining set T that satisfies (i),thenCis self-orthogonal.

Proof: LetN = {0,1, . . . ,n−1}. LetT^⊥ be the defining set ofC^⊥. By Theorem 4.4.9, T^⊥=N \(−1)T modn. AsC⊆C^⊥,N \(−1)T modn⊆Tby Exercise 239. IfCi ⊆T, thenCi ⊆(−1)T modnbecauseCi =Ciµ−1implying thatCi ⊆N \(−1)T modn⊆T, a contradiction. IfDi ⊆T, then Ei ⊆(−1)T modn because Ei=Diµ−1 implying that Ei ⊆N \(−1)T modn⊆T, proving (i). Part (ii) follows from part (i) and Exercise 238 asCi = {0}for somei. By Corollary 4.4.5,Cµ−1 has defining set (−1)T modn. By (i) T ∪(−1)T modn=N yielding (iii) using Exercise 239.

For the converse, assumeT satisfies (i). We only need to show thatT^⊥⊆T, whereT^⊥= N\(−1)T modnby Exercise 239. AsCi⊆Tfor 1≤i ≤k,Ci⊆(−1)T modnimplying Ci ⊆T^⊥. Hence T^⊥ is a union of some Dis and Eis. If Di ⊆N \(−1)T modn, then Di ⊆(−1)T modnand soEi ⊆T. By (i)Di ⊆T. Similarly ifEi⊆N \(−1)T modn, thenEi⊆T. HenceT^⊥⊆T implyingCis self-orthogonal.

Exercise 253 Continuing with Exercise 242, do the following:

(a) Show that all [31,5] binary cyclic codes are self-orthogonal.

(b) Show that there are two inequivalent [31,15] self-orthogonal binary cyclic codes, and give defining sets for a code in each equivalence class.

Corollary 4.4.12 LetD=C+ 1be a cyclic code of length n overFq,whereC is self-orthogonal. ThenD∩Dµ−1= 1.

Exercise 254 Prove Corollary 4.4.12.

Corollary 4.4.13 Let p1(x), . . . ,pk(x),q1(x), . . . ,q(x),r1(x), . . . ,r(x)be the monic ir-reducible factors of xⁿ−1overFq[x]arranged as follows. For1≤i ≤k,p^∗_i(x)=aipi(x) for some ai∈Fq,and for1≤i≤,r_i^∗(x)=biqi(x)for some bi∈Fq. L etC be a cyclic code of length n over Fq with generator polynomial g(x). Then C is self-orthogonal if and only if g(x) has factors p1(x)· · ·pk(x) and at least one of qi(x) or ri(x) for 1≤i ≤.

Exercise 255 Prove Corollary 4.4.13.

Exercise 256 Using Corollary 4.4.13 and Examples 3.7.8, 4.2.4, and 4.3.4 give the genera-tor polynomials of the self-orthogonal binary cyclic codes of lengths 7, 9, and 15. Compare

your answers to those of Exercise 252.

Exercise 257 Let j(x)=1+x+x²+ · · · +xⁿ⁻¹inRnandj(x)=(1/n)j(x). In Exer-cise 221 we gave properties of j(x) and j(x). LetCbe a cyclic code overFqwith generating idempotenti(x). LetCebe the subcode of all even-like vectors inC. In Exercise 238 we found the generator polynomial ofCe.

(a) Prove that 1− j(x) is the generating idempotent of the [n,n−1] cyclic code overFq

consisting of all even-like vectors inRn.

(b) Prove thati(1)=0 ifC=Ceandi(1)=1 ifC=Ce.

(c) Prove that ifC=Ce, theni(x)−j(x) is the generating idempotent ofCe. We illustrate Theorems 4.3.13, 4.3.17, 4.4.6, and 4.4.9 by returning to Examples 4.3.4 and 4.3.5.

Example 4.4.14 In Examples 4.3.4 and 4.3.5, the following codes are cyclic complemen-tary pairs:C1andC6,C2 andC5, andC3andC4. In both examples, the following are dual pairs:C1 andC6,C2andC4, andC3 andC5. In Example 4.3.4,C2 andC3 are equivalent underµ3, as areC4 andC5. In Example 4.3.5, the same pairs are equivalent underµ2. In both examples, the permutation automorphism group for each ofC1,C6, andC7is the full symmetric group. (In general, the permutation automorphism group of the repetition code of lengthn, and hence its dual, is the symmetric group onnletters.) In Example 4.3.4, the group of order 3 generated byµ2is a subgroup of the automorphism group of the remaining four codes; in Example 4.3.5, the group of order 5 generated byµ3is a subgroup of the

automorphism group of the remaining four codes.

Exercise 258 Verify all the claims in Example 4.4.14.

149 4.4 Zeros of a cyclic code

Exercise 259 LetCbe a cyclic code of lengthnoverFq with generator polynomialg(x).

What conditions ong(x) must be satisfied for the dual ofCto equal the cyclic complement

ofC?

Exercise 260 Identify all binary cyclic codes of lengths 7, 9, and 15 whose duals equal their cyclic complements. (Examples 3.7.8, 4.2.4, and 4.3.4 will be useful.) In Theorem 4.4.9 we found the generating idempotent of the dual of any code. The multiplierµ₋1was key in that theorem. We can also find the generating idempotent of the Hermitian dual of a cyclic code overF4. Hereµ−2will play the role ofµ−1. Recall from Exercise 8 that ifCis a code overF4, thenC^⊥H =C^⊥.

Theorem 4.4.15 LetC be a cyclic code of length n overF4 with generating idempotent e(x)and defining set T . The following hold.

(i) C^⊥H is a cyclic code andC^⊥H =Ccµ₋2,whereCc is the cyclic complement ofC.

(ii) C^⊥H has generating idempotent1−e(x)µ−2.

(iii) IfN = {0,1, . . . ,n−1},thenN \(−2)T modn is the defining set ofC^⊥H. (iv) Precisely one ofCandC^⊥H is odd-like and the other is even-like.

Proof: We leave the fact thatC^⊥H is a cyclic code as an exercise. Exercise 8 shows that C^⊥H =C^⊥. By Theorem 4.4.9 C^⊥=Cc

µ₋1. Theorem 4.3.16 shows thatCc

=Cc=Ccµ2, and (i) follows sinceµ2µ−1=µ−2. By Theorem 4.4.6C^⊥H =Ccµ−2has generating idem-potent (1−e(x))µ₋2=1−e(x)µ₋2, giving (ii).Cµ₋2has defining set (−2)⁻¹T modnby Corollary 4.4.5. However, (−2)⁻¹T =(−2)T modulonbecauseµ²₋₂=µ4andµ4fixes all 4-cyclotomic cosets. By Theorem 4.4.6(iii) and Corollary 4.4.5,C^⊥H =Ccµ₋2has defin-ing set (−2)⁻¹(N \T)=N \(−2)T modn, giving (iii). Part (iv) follows from (iii) and Exercise 238 as precisely one ofN andN \(−2)T modncontains 0.

Exercise 261 Prove that ifCbe a cyclic code overF4, thenC^⊥H is also a cyclic code.

Exercise 262 Using Theorem 4.3.16 find generating idempotents of all the cyclic codesC of length 9 overF4, their ordinary dualsC^⊥, and their Hermitian dualsC^⊥H. We can obtain a result analogous to Theorem 4.4.11 for Hermitian self-orthogonal cyclic codes overF4. Again we simply replaceµ₋1 byµ₋2 and apply Theorem 4.4.15. Notice that ifC is a 4-cyclotomic coset modulo n, then eitherCµ−2 =C orCµ−2=C for a different 4-cyclotomic cosetC, in which caseCµ₋2=C asµ²₋₂=µ4andµ4 fixes all 4-cyclotomic cosets.

Theorem 4.4.16 LetCbe a Hermitian self-orthogonal cyclic code overF4of length n with defining set T . Let C₁, . . . ,Ck,D₁, . . . ,D,E₁, . . . ,E be all the distinct 4-cyclotomic cosets modulo n partitioned so that Ci =Ciµ−2for1≤i ≤k and Di =Eiµ−2for1≤ i ≤. The following hold:

(i) Ci ⊆T for1≤i ≤k and at least one of Dior Eiis contained in T for1≤i ≤. (ii) Cis even-like.

(iii) C∩Cµ−2 = {0}.

Conversely,ifC is a cyclic code with defining set T that satisfies (i),thenCis Hermitian self-orthogonal.

Exercise 263 Prove Theorem 4.4.16.

Corollary 4.4.17 Let D=C+ 1 be a cyclic code of length n overF4 such that C is Hermitian self-orthogonal. ThenD∩Dµ₋2= 1.

Exercise 264 Prove Corollary 4.4.17.

The next theorem shows the rather remarkable fact that a binary self-orthogonal cyclic code must be doubly-even.

Theorem 4.4.18 A self-orthogonal binary cyclic code is doubly-even.

Proof: LetCbe an [n,k] self-orthogonal binary cyclic code with defining setT. By Theo-rem 1.4.5(iv),Chas only even weight codewords and hence 0∈Tby Exercise 238. Suppose thatC is not doubly-even. Then the subcodeC0 ofCconsisting of codewords of weights divisible by 4 has dimensionk−1 by Theorem 1.4.6. Clearly,C0 is cyclic as the cyclic shift of a vector is a vector of the same weight. By Theorem 4.4.2 and Corollary 4.2.5 (or Exercise 239), the defining set ofC0 isT ∪ {a}for somea∈/T. But then{a}must be a 2-cyclotomic coset modulon, which implies that 2a≡a(modn). Hencea=0 asnis odd,

which is impossible as 0∈T.

In Theorem 4.3.8, the minimal cyclic codes inRnare shown to be those with generator polynomialsg(x) where (xⁿ−1)/g(x) is irreducible overFq. So minimal cyclic codes are sometimes calledirreducible cyclic codes. These minimal cyclic codes can be described using the trace function.

Theorem 4.4.19 Let g(x)be an irreducible factor of xⁿ−1 overFq. Suppose g(x)has degree s,and letγ ∈Fq^s be a root of g(x). L etTrs :Fq^s →Fq be the trace map fromFq^s

toFq. Then C_γ =

/_n−1

i=0

Trs(ξγⁱ)xⁱ |ξ ∈Fq^s

0

is the[n,s]irreducible cyclic code with nonzeros{γ^−qi |0≤i<s}.

Proof: By Lemma 3.8.5,C_γ is a nonzero linear code overFq. Ifc_ξ(x)=n−1

i=0 Trs(ξγⁱ)xⁱ, then c_ξγ−1(x)=c_ξ(x)x in Rn implying that Cγ is cyclic. Let g(x)=n−1

i=0gixⁱ. By Lemma 3.8.5, asgi ∈Fq andg(γ)=0,

n−1

i=0

giTrs(ξγⁱ)=Trs

ξ

n−1

i=0

giγⁱ

=Trs(0)=0.

Henceg(x) ⊆C^⊥_γ. By Theorem 4.4.9,C^⊥_γ is a cyclic code not equal toRn asCγ = {0}.

Henceg(x) ⊆C^⊥_γ. By Theorem 4.4.9,C^⊥_γ is a cyclic code not equal toRn asCγ = {0}.

Dans le document Fundamentals of Error-Correcting Codes (Page 160-170)