Cyclic codes over $\mathbb{Z}_4+u\mathbb{Z}_4$

0.19 MB·English

by  Rama Krishna Bandi

Manuscriptsubmittedto doi:10.3934/xx.xx.xx.xx
AIMS’Journals
VolumeX,Number0X,XX200X pp. X–XX
CYCLIC CODES OVER Z +uZ
4 4
Rama Krishna Bandi
5 DepartmentofMathematics
1 IndianInstituteofTechnologyRoorkee
0 Roorkee-247667, INDIA
2 Maheshanand Bhaintwal

n DepartmentofMathematics
a IndianInstituteofTechnologyRoorkee
J
Roorkee-247667, INDIA

6
(Communicated by the associate editor name)

]
T
I Abstract. In this paper, we have studied cyclic codes over the ring R =
s. Z4+uZ4,u2=0. Wehaveconsideredcycliccodesofoddlengths. Asufficient
c condition for a cyclic code over R to be a Z4-free module is presented. We
[ have provided the general form of the generators of a cyclic code over R and

determinedaformulafortheranksofsuchcodes. Inthispaperwehavemainly
1
focused on principally generated cyclic codes of odd length over R. We have
v
determinedanecessaryconditionandasufficientconditionforcycliccodes of
7
oddlengthsoverRtobeR-free.
2
3
1. Introduction. Cyclic codes are amongst the most studied algebraic codes.
1
0 Their structure is well known over finite fields [7]. Recently codes over rings have
. generatedalotofinterestafterabreakthroughpaperbyHammonsetal. [5]showed
1
0 that some well known binary non-linear codes are actually images of some linear
5 codes over Z4 under the Gray map. Since then, cyclic codes have also been ex-
1 tensively studied over various finite rings. Their structure over finite chain rings is
v: nowwellknown[9]. They havealsobeenstudiedoverother ringssuchasF2+uF2,
i u2 =0,[3]; F +uF +vF +uvF ,u2 =v2 =0,uv =vu,[12];andF +vF ,v2 =v,
X 2 2 2 2 2 2
[13].
r BonnecazeandUdaya[3]havestudiedcycliccodesovertheringF +uF ,u2 =0,
a 2 2
and provided their basic framework. The ring F +uF is useful because it shares
2 2
many properties of Z , and since it has characteristic 2, it also shares properties of
4
the field F . In mostof these studies lengthof the cyclic code is relativelyprime to
4
the characteristic of the ring. A complete structure of cyclic codes over Z of odd
4
length has been given in [10] and [6].
In this paper, we have studied cyclic codes over the ring R=Z +uZ , u2 =0.
4 4
We have considered cyclic codes of odd lengths. Recently, Yildiz and Karadeniz
[11] have studied linear codes over R. A linear code C over R can be expressed
as C = C +uC , where C ,C are linear codes over Z . As usual, a cyclic code
1 2 1 2 4
of length n over R is an ideal of R = R[x] . We have shown that a linear code
n hxn−1i
C =C +uC oflengthnoverRisacycliccodeifandonlyifC ,C arecycliccodes
1 2 1 2
2010 Mathematics Subject Classification. Primary: 94B05,94B15.
Key words and phrases. Cycliccodes,codes overrings,freecodes.
1
2 RAMA KRISHNA BANDI AND MAHESHANAND BHAINTWAL
of length n over Z . We have determined a sufficient condition for a cyclic code of
4
odd length over R to be a Z -free module. We have provided the general form of
4
the generators of a cyclic code over R, from which we have determined a formula
for the ranks of such codes. The ring R is in general not a principal ideal ring,
n
and so a cyclic code over R is in general not principally generated. In this paper
we have mainly focused on cyclic codes of odd length over R which are principally
generated. We havedetermined a necessarycondition anda sufficientconditionfor
principally generated cyclic codes of odd lengths over R to be R-free.
The paper isorganizedasfollows: InSectionII, wepresentthe preliminaries. In
Section III, we have discussed the Galois extensions of R and the ideal structure
of these extensions. In Section IV, we have studied cyclic codes of odd length over
R. The forms of the ranks and minimal spanning sets of these codes are presented.
In Section V, we have mainly focused to principally generated cyclic codes of odd
length over R and determined a necessary condition and a sufficient condition for
cycliccodesoverRtobeR-free. Wehavealsoexpressedprincipallygeneratedcyclic
codes in terms of the nth roots of unity.
2. Preliminaries. Throughout the paper, R denotes the ring Z +uZ = {a+
4 4
ub | a,b∈Z } with u2 =0. R can be viewed as the quotient ring Z [u]/ u2 . The
4 4 (cid:10) (cid:11)
units of R are
1,3,1+u,1+2u,1+3u,3+u,3+2u,3+3u ,
and the non-units are
0,2,u,2u,2+u,2+2u,3u,2+3u .
R has six ideals in all: {0},hui = {0,u,2u,3u},h2i = {0,2,2u,2+ 2u},h2ui =
{0,2u},h2+ui={0,2+u,2u,2+3u}andh2,ui={0,2,2u,3u,2+u,2+2u,2+3u}.
R is a local ring of characteristic 4 with h2,ui as its unique maximal ideal. A
commutativeringRiscalledachainring ifitsidealsformachainundertherelation
of inclusion. From the ideals of R, we can see that they do not form a chain; for
instance, the ideals hui and h2i are not comparable. Therefore, R is a non-chain
extensionof Z . Also R is not a principalideal ring; for example, the ideal h2,ui is
4
not generated by any single element of R.
We denotethe residuefield R ofRbyR. Since {0+h2,ui}∪{1+h2,ui}=R,
h2,ui
therefore R ∼= F . The image of any element a ∈ R under the projection map
2
µ : R → R is denoted by a. The map µ is extended to R[x] → R[x] in the usual
way. The image of an element f(x)∈R[x] in R[x] under this projectionis denoted
by f(x). A polynomial f(x) ∈ R[x] is called basic irreducible (primitive) if f(x) is
an irreducible (primitive) polynomial in R[x]. Basic irreducible polynomials over
finite local rings play approximately the same role as irreducible polynomials play
over finite fields.
Apolynomial f(x)overR is calleda regular polynomial ifit is nota zerodivisor
inR[x],equivalently,f(x)isregulariff(x)6=0. Twopolynomialsf(x),g(x)∈R[x]
are said to be coprime if there exist a(x),b(x)∈R[x] such that
a(x)f(x)+b(x)g(x)=1 .
Now we recall the Hensel’s Lemma and factorization of polynomials in Z [x]. A
4
polynomial f(x) in Z [x] is said to be primary if the principal ideal hfi is primary,
4
i.e., whenever ab∈hfi, then either a∈hfi or bj ∈hfi for positive integer j.
CYCLIC CODES OVER Z4+uZ4 3
Theorem 2.1 (Hensel’s Lemma [10]). Let f be a monic polynomial in Z [x] and
4
assume that f (mod 2) = g g ···g , where g ,g ,...,g are pairwise coprime
1 2 r 1 2 r
monic polynomials over F . Then there exist pairwise coprime monic polynomi-
2
als f ,f ,...,f over Z such that f = f f ···f in Z [x] and f (mod 2) = g ,
1 2 r 4 1 2 r 4 i i
i=1,2,...,r.
A Gray map φ:Rn →Z2n is defined by (see [11])
4
(a+ub)7→(b,a+b) .
The Lee weight is defined on R by
w (a+ub)=w (b,a+b) ,
L L
where w (b,a+b) is the usual Lee weight of (b,a+b) in Z2. This weight is then
L 4
extended componentwise to Rn. The Lee weight of an element x ∈ Rn is the sum
of the Lee weights of the coordinates of x.
Theorem 2.2. [11] The Gray map φ : Rn → Z2n is a distance preserving linear
4
isometry with respect to the Lee weights in Rn and Z2.
4
AlinearcodeC oflengthnoverRisanR-submoduleofRn. C maynotbeanR-
freemodule. WecanexpressRn asRn =Zn+uZn,andsoalinearcodeC oflength
4 4
noverRcanbe expressedasC =C +uC ,whereC ,C arelinearcodesoflength
1 2 1 2
n over Z . The Euclidean inner product of any two elements x = (x ,x ,...,x )
4 1 2 n
and y = (y ,y ,...,y ) of Rn is defined as x · y = x y + x y + ··· + x y ,
1 2 n 1 1 2 2 n n
where the operation is performed in R. The dual of a linear code C is defined as
C⊥ = {y ∈ Rn | x·y = 0 ∀x ∈ C}. It follows immediately that if C = C +uC
1 2
is a linear code over R, then C⊥ = C⊥ +uC⊥. We define the rank of a code C
1 2
as the minimum number of generators for C and the free rank of C is the rank
of C if C is a free module over R. There are two other codes associated with C,
namely Tor(C) and Res(C) and are defined as Tor(C) = {b ∈ Z4[x] : ub ∈ C}
hxn−1i
and Res(C)={a∈ Z4[x] : a+ub∈C for some b∈ Z4[x] }.
hxn−1i hxn−1i
3. Galois extension of R. Let n be an odd integer. We first consider the factor-
ization of xn−1 over R, as it plays a vital role in the study of cyclic codes over R
of length n.
Theorem3.1. Letg(x)∈F [x]beamonicirreducible(primitive)divisorofx2r−1−
2
1. Then there exists a unique monic basic irreducible (primitive) polynomial f(x)
in R[x] such that f(x)=g(x) and f(x) | (x2r−1−1) in R[x].
Proof. Letx2r−1−1=g(x)g′(x)inF [x]. ByHensel’slemma,thereexistf(x),f′(x)∈
2
Z [x]suchthatx2r−1−1=f(x)f′(x)inZ [x]andf(x)(mod2)=g(x),f′(x)(mod2)=
4 4
g′(x). SinceZ isasubringofR,f(x)∈R[x]. Alsof(x)=f(x)(mod h2,ui)=g(x)
4
and f(x) | (x2r−1−1) in R[x].
We call the polynomial f(x) in Theorem (3.1) the Hensel lift of g(x) to R.
Sincenisodd,itfollowsfrom[8,TheoremXIII.11]thatxn−1factorizesuniquely
into pairwise coprime basic irreducible polynomials over R. Let
xn−1=f f ···f
1 2 m
4 RAMA KRISHNA BANDI AND MAHESHANAND BHAINTWAL
be such a factorization of xn −1. Then it follows from the Chinese Remainder
Theorem that
R[x] R[x]
=⊕m .
hxn−1i i=1 hf i
i
Therefore every ideal I of R[x] can be expressed as I = ⊕m I , where I is an
hxn−1i i=1 i i
ideal of the ring R[x]/hf i, i=1,2,...,m.
i
Let us recall the Galois extension of Z . Let h(x) be a monic basic irreducible
4
polynomialofdegreer inZ [x]. ThentheGaloisringGR(4,r)overZ isdefinedas
4 4
the residue class ring Z4[x] . The ring GR(4,r) is a local ring with unique maximl
hh(x)i
ideal h2i and the residue field F .
2r
Let T = {0,1,ξ,ξ2,...,ξ2r−2} be the Teichmu¨ller representatives of GR(4,r),
where ξ is a root of a basic primitive polynomial of degree r in Z [x]. Then each
4
element a of GR(4,r) can be written as a = a +2a , where a ,a ∈ T. This
0 1 0 1
representationis called the 2-adic representationof elements of GR(4,r).
Now we consider the Galois extension of R. Let f(x) be a basic irreducible
polynomial of degree r in R[x]. Then the Galois extension of R is defined as the
quotient ring R[x] and is denoted by GR(R,r). If α is a root of f(x) then the
hf(x)i
elementsofGR(R,r)canuniquelybewrittenasm +m α+m α2+···+m αr−1,
0 1 2 r−1
m ∈ R, i = 0,1,...,r−1, i.e. GR(R,r) is free module of rank r over R with a
i
basis {1,α,α2,...,αr−1} and |GR(R,r)| = 16r. From Theorem (3.5), it follows
thatthe ringGR(R,r) is a localringwith unique maximalideal hh2, ui+hfii and
the residue field F . Furthermore,
2r
GR(4,r)[u]
GR(R,r)≃ ≃GR(4,r)⊕uGR(4,r) ,
hu2i
where GR(4,r) is the Galois ring of degree r over Z and u2 =0.
4
Therefore, an element x of GR(R,r) can be represented as x = a+ub, where
a,b ∈ GR(4,r). Using the 2-adic representation of a = a +2a , b = a +2a ,
0 1 2 3
a ,a ,a ,a ∈ T, the element x ∈ GR(R,r) can further be represented as x =
0 1 2 3
a +2a +ua +2ua .
0 1 2 3
Lemma 3.2. A non-zero element x=a +2a +ua +2ua of GR(R,r) is unit if
0 1 2 3
and only if a is non-zero in T.
0
Proof. Since x4 =a4 for every non-zero x in GR(R,r), the result follows.
0
Thus the group of units of GR(R,r), denoted by GR(R,r)∗, is given by
GR(R,r)∗ ={a +2a +ua +2ua : a ,a ,a ,a ∈T,a 6=0}.
0 1 2 3 0 1 2 3 0
Theorem 3.3. The group of units GR(R,r)∗ is a direct product of two groups G
C
and G , i.e., GR(R,r)∗ = G ×G , where G is a cyclic group of order 2r −1
A C A C
and G is an abelian group of order 8r.
A
Proof. Let ξ be a primitive element of GR(R,r) and G =T∗ ={1,ξ,...,ξ2r−2}.
C
ThenG isamultiplicativecyclicgroupoforder2r. Letx=a +2a +ua +2ua ∈
C 0 1 2 3
GR(R,r)∗. Define a mapping Γ : GR(R,r)∗ −→ G such that Γ(x) = a . It can
C 0
easily be seen that for any α,x,y ∈GR(R,r)∗, Γ(αx+y)=Γ(α)Γ(x)+Γ(y). Γ is
obviously a surjective map. Therefore GR(R,r)∗ ≃ G , where ker Γ = {1+2a +
kerΓ C 1
ua +2ua : a ,a ,a ∈ T}. Denote Ker Γ by G . Then it can easily be seen
2 3 1 2 3 A
that GR(R,r)≃G ×G . Moreover, |GR(R,r)∗|=|G ||G |=8r(2r−1).
C A c A
CYCLIC CODES OVER Z4+uZ4 5
ThesetofallzerodivisorsofGR(R,r)isgivenby{2a +ua +2ua : a ,a ,a ∈
1 2 3 1 2 3
T}, which is maximal ideal generated by h2,ui in GR(R,r).
Now we consider the ideal structure of GR(R,r). We first prove the following
Lemma.
Lemma 3.4. Let f(x),g(x) ∈ R[x]. Then f(x), g(x) are coprime if and only if
their images f(x), g(x) are coprime in R[x].
Proof. If f(x), g(x) are coprime, then it is immediate that f(x) and g(x) are co-
prime. Nowsupposethatf(x), g(x)arecoprime. Thenthereexista(x),b(x)∈R[x]
such that
a(x)f(x)+b(x)g(x)=1 .
Then there exits r(x),s(x) ∈R[x] such that
a(x)f(x)+b(x)g(x)=1+2r(x)+us(x) . (1)
Multiplying (1) by 2r(x) and by us(x), we respectively get equations:
2r(x)a(x)f(x)+2r(x)b(x)g(x)=2r(x)+2ur(x)s(x) . (2)
us(x)a(x)f(x)+us(x)b(x)g(x)=us(x)+2ur(x)s(x) . (3)
On adding (2) and (3), we get
a(x)(2r(x)+us(x))f(x)+b(x)(2r(x)+us(x))g(x)=2r(x)+us(x) . (4)
Putting the value of 2r(x)+us(x) in (1), we get
a(x)(1−2r(x)−us(x))f(x)+b(x)(1−2r(x)−us(x))g(x)=1 .
Therefore f(x) and g(x) are coprime.
NowweconsidertheidealsofR[x]/hfi,wheref isabasicirreduciblepolynomial
over R.
Theorem 3.5. Let f ∈ R[x] be a basic irreducible polynomial. Then the ideals of
R[x]/hfiareprecisely,{0},h1+hfii,h2+hfii,hu+hfii,h2u+hfii,h2+u+hfii
and hh2,ui+hfii.
Proof. Let I be a non-zero ideal of R[x]/hfi. Let h + hfi ∈ R[x]/hfi. Since
f is basic irreducible, f is irreducible in R[x]. Therefore gcd(f,h) = 1 or f. Let
gcd(f,h)=1. Thenf andharecoprimeinR[x],andhencethereexistλ ,λ ∈R[x]
1 2
such that
λ f +λ h=1 .
1 2
Fromthisfollowsthatλ h=1(modf). ThushisaninvertibleelementofR[x]/hfi
2
and so I =h1+hfii=R[x]/hfi.
Now suppose that gcd(f,h) = f. Then there exists polynomials g,f ,f ∈ R[x]
1 2
such that
h=fg+2f +uf ,
1 2
and gcd(f,f ) = 1 or gcd(f,f ) = 1. It follows that h + hfi ∈ hh2,ui+hfii.
1 2
Therefore if I 6= h1+hfii, then I ⊆ hh2,ui+hfii. The non-zero ideals contained
inhh2,ui+hfiiareh2+hfii,hu+hfii,h2u+hfii,h2+u+hfiiandhh2,ui+hfii
itself. The result follows.
The Galois groupGal(GR(R,r)) of Gal(R,r) is a cyclic groupof order (2r−1),
whichisgeneratedbytheFrobeniusautomorphism σonGR(R,r)definedasσ(x)=
a2+2a2+ua2+2ua2, where x=a +2a +ua +2ua ∈R. The automorphism
0 1 2 3 0 1 2 3
σ fixes the ring R.
6 RAMA KRISHNA BANDI AND MAHESHANAND BHAINTWAL
Example 3.6. Considerthebasicirreduciblepolynomialh(x)=x4+3x3+2x2+1,
which is the Hensel lift to R of the polynomial x4+x3+1∈F [x]. Let ξ be a root
2
of h(x). Then
ξ4 =ξ3+2ξ2+3, ξ5 =3ξ3+2ξ2+3ξ+3, ξ6 =ξ3+ξ2+3ξ+1,
ξ7 =2ξ3+ξ2+ξ+3, ξ8 =3ξ3+ξ2+ξ, ξ9 =3ξ2+3,
ξ10 =3ξ3+3ξ, ξ11 =3ξ3+ξ2+1, ξ12 =2ξ2+ξ+1,
ξ13 =2ξ3+ξ2+ξ, ξ14 =3ξ3+ξ2+2ξ, ξ15 =1.
LetT ={0,1,ξ,ξ2,ξ3,ξ3+2ξ2+3,3ξ3+2ξ2+3ξ+3,ξ3+ξ2+3ξ+1,2ξ3+ξ2+ξ+
3,3ξ3+ξ2+ξ,3ξ2+3,3ξ3+3ξ,3ξ3+ξ2+1,2ξ2+ξ+1,2ξ3+ξ2+ξ,3ξ3+ξ2+2ξ}. Then
GR(R,4)={a +2a +ua +2ua : a ∈T,i=0,1,2,3} and |GR(R,4)|=416.
0 1 2 3 i
4. CycliccodesofoddlengthoverZ +uZ . Weassumethatnisoddthrough-
4 4
outthis section. Fora finite chainringR, it is wellknownthat the ring R[x] is a
hxn−1i
principal ideal ring [9]. However, in the present case the ring R is not a chain ring
and the situation is not as straightforward. In fact, the ring R = R[x] is not in
n hxn−1i
generalaprincipalidealring,asthenextresultshows. Theresultisageneralization
of [12, Lemma 2.4].
Theorem 4.1. The ring R = R[x] is not a principal ideal ring.
n hxn−1i
Proof. Consider the augmentation mapping γ : R[x] →R defined by
hxn−1i
γ(a +a x+...+a xn−1)=a + a + ... + a .
0 1 n−1 0 1 n−1
This is a surjective ring homomorphism. Consider now the ideal I = h2,ui of R,
which we know is not a principal ideal. Let J = γ−1(I). It is well known that the
inverse image under a homomorphism of an ideal is an ideal. So J is an ideal of
R[x] . Now if we assume J to be a principal ideal, then its homomorphic image I
hxn−1i
must be principal, a contradiction. Hence J is not a principal ideal and R[x] is
hxn−1i
therefore not a principal ideal ring.
Therefore,acycliccodeoflengthnoverRisingeneralnotprincipallygenerated.
Since n is odd, the ring Z4[x] is a principal ideal ring. Therefore a cyclic code
hxn−1i
oflengthnoverR isofthe formC =C +uC =hg i+uhg i,whereg ,g ∈Z [x]
1 2 1 2 1 2 4
are generator polynomials of the cyclic codes C , C , respectively.
1 2
Let τ be the standard cyclic shift operator on Rn. A linear code C of length n
over R is cyclic if τ(c) ∈ C whenever c ∈ C, i. e., if (c ,c ,...,c ) ∈ C, then
0 1 n−1
(c ,c ,c ,...,c ) ∈ C. As usual, in the polynomial representation, a cyclic
n−1 0 1 n−2
code of length n over R is an ideal of R[x] .
hxn−1i
Theorem 4.2. Let xn −1 = f f ···f , where f , i = 1,2,...,m are basic irre-
1 2 m i
ducible pairwise coprime polynomials in R[x]. Then any ideal in R is the sum of
n
the ideals of R[x]/hf i, i=1,2,...,m.
i
Proof. It follows from the Chinese Remainder Theorem.
Corollary 1. The number cyclic codes over R is 7m.
Proof. Each ideal of R is a direct sum of the ideals of R[x]/hf i, i = 1,2....,m.
n i
From Theorem (3.5) and for each i, R[x]/hf i has 7 ideals. The result follows.
i
CYCLIC CODES OVER Z4+uZ4 7
Theorem 4.3. A linear code C =C +uC of length n over R is cyclic if and only
1 2
if C , C are cyclic codes of length n over Z .
1 2 4
Proof. Let c +uc ∈ C, where c ∈ C and c ∈ C . Then τ(c +uc ) = τ(c )+
1 2 1 1 2 2 1 2 1
uτ(c )∈C, since C is cyclic and τ is a linear map. So, τ(c )∈C and τ(c )∈C .
2 1 1 2 2
Therefore C ,C are cyclic codes. Conversely if C , C are cyclic codes, then for
1 2 1 2
any c +uc ∈C, where c ∈C and c ∈C , we have τ(c )∈C and τ(c )∈C ,
1 2 1 1 2 2 1 1 2 2
and so, τ(c +uc )=τ(c )+uτ(c )∈C. Hence C is cyclic.
1 2 1 2
The following result gives a sufficient condition for a cyclic code C over R to be
a free Z -code.
4
Theorem 4.4. Let C = C +uC be a cyclic code of length n over R. If C , C
1 2 1 2
are free codes over Z , then C is a free Z -module.
4 4
Proof. Suppose that C , C are Z -free codes of ranks k , k , respectively. Let
1 2 4 1 2
{c ,c ,...,c } and {c ,c ,...,c } be Z -bases of C and C , respectively.
11 12 1k1 21 22 2k2 4 1 2
Then the set {c ,c ,...,c ,uc ,uc ,..., uc } spans C, as every element of
11 12 1k1 21 22 2k2
C can be expressed as a linear combination of elements of this set. Now suppose
there exist scalars a ,b ∈Z such that
i j 4
k1 k2
Xaic1i+uXbjc2j =0 .
i=1 j=1
Then k1 a c = 0 and k2 b c = 0. Since the elements c ,c ,...,c are
Pi=1 i 1i Pj=1 j 2j 11 12 1k1
independent and so are the elements c ,c ,...,c , therefore a = 0 and b = 0
21 22 2k2 i j
for all i and j. Hence C is a Z -free module.
4
The converse of the above Theorem is not true in general, i. e., if a cyclic code
C =C +uC is a free Z -module of length n over R, then C or C may not be a
1 2 4 1 2
free code of length n over Z (see example 4.6). However, if C is an R-free module
4
(code) of length n over R then C must be a free code of length n over Z (see
1 4
Theorem 1).
Example 4.5. The polynomial x7−1 factorizes into irreducible polynomials over
F as x7 −1 = (x−1)(x3 +x+1)(x3 +x2 +1). The Hensel lifts of x3 +x+1
2
and x3 +x2 +1 to Z are x3 +2x2 +x−1 and x3 −x2 −2x−1, respectively.
4
Thereforex3+2x2+x−1andx3−x2−2x−1aredivisorsofx7−1overZ . Define
4
C = x3+2x2+x−1 +u x3−x2−2x−1 . Then C is a cyclic code of length
(cid:10) (cid:11) (cid:10) (cid:11)
7 over R, which is also a free Z -module.
4
Example 4.6. Let C = C + uC be a free Z -cyclic code of length 5 over R
1 2 4
generated by g(x) = u+2x+ux2. Then C is a cyclic code of length 5 over Z
1 4
generated by g(x) (mod u)=2x which is not Z - free.
4
Now we consider the general form of the generators of cyclic codes over R.
Define ψ : R → Z such that ψ(a +bu) = a (mod u). It can easily be seen
4
that ψ is a ring homomorphisms with ker ψ = hui = uZ . Extended ψ to the
4
homomorphismφ: R[x] → Z4[x] suchthatφ(a +a x+a x2+...+a xn−1)=
hxn−1i hxn−1i 0 1 2 n−1
ψ(a )+ψ(a )x+ψ(a )x2+...+ψ(a )xn−1. Let C be a cyclic code of length n
0 1 2 n−1
over R. Restrict φ to C and define
Z [x]
4
J ={h(x)∈ : uh(x)∈ker φ} .
hxn−1i
8 RAMA KRISHNA BANDI AND MAHESHANAND BHAINTWAL
Clearly J is an ideal of Z4[x] . So J is a cyclic code over Z and J = ha(x)i for
hxn−1i 4
some a(x) ∈ Z [x]. Therefore ker φ = hua(x)i. Similarly, the image of C under
4
φ is an ideal of Z4[x] and φ(C) = hg(x)i for some g(x) ∈ Z [x]. Hence C =
hxn−1i 4
hg(x)+up(x), ua(x)i for some p(x) ∈ Z [x]. Since ug(x) = u(g(x)+up(x)) ∈ C
4
and φ(ug(x))=0, so a(x)|g(x). Thus a cyclic code C over R has the form
C =hg(x)+up(x),ua(x)i ,
where g(x),p(x),a(x)∈Z [x] and a(x)|g(x). In particularif a(x)=g(x), we have
4
the following result.
Theorem 4.7. Let n be an odd integer and C be a cyclic code of length n over R
such that C =hg(x)+up(x), ug(x)i. Then C =hg(x)+up(x)i.
Proof. Clearlyhg(x)+up(x)i⊆C. Sinceu(g(x)+up(x))=ug(x)andg(x)=a(x),
C ⊆hg(x)+up(x)i. Hence C =hg(x)+up(x)i.
Itmaybenotedherethatunlikeinthecaseoffinitefields,ageneratorpolynomial
ofkerφorφ(C)maynotnecessarilydividexn−1. Theproofofthefollowingresult
is straightforward,as the result is well known for codes over finite fields.
Theorem4.8. LetC beacycliccodeoflengthnoverR. IfC =hg(x)+up(x), ua(x)i
and deg g(x)=k and deg a(x)=k , then C has rank 2n−k −k and a minimal
1 2 1 2
spanningsetA={(g(x)+up(x)),x(g(x)+up(x)),x2(g(x)+up(x)),··· ,xn−k1−1(g(x)+
up(x)), ua(x),xua(x),x2ua(x),··· ,xn−k2−1ua(x)}.
In Theorem (4.8), if we put the restriction on g(x) and a(x) such that they are
regular and monic polynomials, respectively, over Z , then the minimal spanning
4
setofC reducestothatof[1,Theorem3]. Toprovethis,wefirstprovethefollowing
lemma, which appears as an exercise (Exercise XIII.6) in [8, p. 273].
Lemma 4.9. Let f(x) and g(x) be twopolynomials in R[x]. If g(x) is regular, then
there exists polynomials q(x) and r(x) such that f(x)=g(x)q(x)+r(x), deg r(x)<
deg g(x).
Proof. Sinceg(x)isregular,by[8,TheoremXIII.6]thereexistsamonicpolynomial
g∗(x)∈R[x] such that g(x)=v(x)g∗(x), where v(x) is a unit in R[x].
Since g∗(x) is monic, by division algorithm, there exists q′(x) and r(x) in R[x]
such that f(x) = g∗(x)q′(x)+r(x), where deg r(x) < deg g∗(x). On multiplying
bothsidesbyv(x), wegetv(x)f(x)=v(x)g∗(x)q′(x)+v(x)r(x), fromwhichweget
f(x)=g(x)q(x)+r(x), where q(x)=(v(x))−1q′(x).
Since g∗(x) is monic, so deg g(x) ≥ deg g∗(x), as deg g(x) = deg v(x)+ deg
g∗(x). From this follows that deg r(x)< deg g(x).
The following result is a generalization of [1, Theorem 3] in the present setting.
Theorem4.10. LetC =hg(x)+up(x), ua(x)ibeacyclic codeof lengthnover R,
andg(x)isregularanda(x)ismonicinZ [x]withdeg g(x)=k anddeg a(x)=k ,
4 1 2
respectively. Then C has rank n−k and a minimal spanning set B = {(g(x)+
2
up(x)), x(g(x) + up(x)), x2(g(x) + up(x)),··· , xn−k1−1(g(x) + up(x)), ua(x),
xua(x),x2ua(x), ··· ,xk1−k2−1ua(x)}.
Proof. Suppose C = hg(x)+up(x),ua(x)i with deg g(x) = k and deg a(x) = k ,
1 2
whereg(x)isregularanda(x)ismonicinZ [x]. ToproveBistheminimalspanning
4
setof C, it suffices to show that B spans the spanof A={(g(x)+up(x)),x(g(x)+
CYCLIC CODES OVER Z4+uZ4 9
up(x)),x2(g(x)+up(x)), ··· ,xn−k1−1(g(x)+up(x)), ua(x),xua(x),x2ua(x),··· ,
xn−k2−1ua(x)}. For this, we first show that xk1−k2ua(x)∈ Span B.
Since g(x) is regular, then so is (g(x)+up(x)). By Lemma (4), xk1−k2ua(x) =
u(g(x)+up(x))q(x)+ur(x),wherer(x)=0ordegr(x)<k ,andq(x)∈Z [x]. This
1 4
implies that ur(x) ∈ C. Since deg r(x) < deg (g(x)+up(x)), so if r(x) 6= 0, then
it cannot be expressed as a linear combination of (g(x)+up(x)) and its multiples.
Therefore, ur(x)=ua(x)b(x) for some b(x)∈R[x].
Since a(x) is monic,so degur(x)= degua(x)+ degb(x). Fromthis followsthat
degb(x)≤k −k −1. Thus,wegetxk1−k2ua(x)=u(g(x)+up(x))q(x)+ua(x)b(x)
1 2
with deg b(x)≤k −k −1. It follows that xk1−k2ua(x)∈ span B.
1 2
Similarly, we can show that xk1−k2+1ua(x), xk1−k2+2ua(x), ···, xn−k2−1ua(x)
are in span B. Hence B is a minimal generating set of C.
To prove the linear independence of B, assume that s(x)(g(x) + up(x)) = 0
(modxn−1)andut(x)a(x)=0(modxn−1)forsomes(x)=s +s x+s x2+···+
0 1 2
s xn−k1−1 ∈R[x]andt(x)=t +t x+t x2+···+t xk1−k2−1 ∈Z [x].
n−k1−1 0 1 2 k1−k2−1 4
Since g(x)+up(x) is regular, by [8, Theorem XIII.6] there exists a monic poly-
nomial g∗(x) ∈ R[x] such that (g(x)+up(x)) = v(x)g∗(x), where v(x) is a unit
R[x]. Therefore, s(x)v(x)g∗(x) = 0 (mod xn − 1), which implies s(x)g∗(x) =
0 (mod xn−1), as v(x) is a unit in R[x]. Let g∗(x)=g∗+g∗x+g∗x2+···+g∗xt,
0 1 2 t
where t≤n−k −1. Then
1
s +s x+s x2+···+s xn−k1−1 g∗+g∗x+g∗x2+···+g∗xt =0(modxn−1).
(cid:0) 0 1 2 n−k1−1 (cid:1)(cid:0) 0 1 2 t (cid:1)
Bycomparingthecoefficientofhighestpowerofxonbothsides,wegets g∗ =
n−k1−1 t
0, from which follows that s = 0, as g∗ is a unit in R. Again by comparing
n−k1−1 t
the coefficient of next highest power of x, we get s g∗ +s g∗ = 0,
n−k1−1 t−1 n−k1−2 t
which implies that s = 0. On continuing this way, we get s = 0 for i =
n−k1−2 i
0,1,...,s . Similarly we can show that t = 0 for all i = 0,1,...,t .
n−k1−3 i k1−k2−1
Therefore B is linearly independent.
Theorem 4.11. Let C =hg(x)+up(x),ua(x)i be a cyclic code of length n over R.
Then w (C)=w (ker φ), i.e., w (C)=w (hua(x)i).
H H H H
Proof. Let c(x) = c (x) + uc (x) ∈ C. Then uc(x) = uc (x). It is clear that
0 1 0
w (uc(x)) = w (uc (x)) ≤ w (c(x)). So w (uC) ≤ w (C). Also, since uC is a
H H 0 H H H
subcode of C, w (C)≤w (uC). Hence the result.
H H
5. One generator cyclic codes over R. We now consider cyclic codes over R
which are principal ideals in R[x] . For a finite chain ring R, the ring R[x] is a
hxn−1i hxn−1i
principalidealringandtheformofthegeneratorofanidealof R[x] iswellknown
hxn−1i
[9]. Using the form of this generator,a necessary and sufficient condition for cyclic
codes over Z to be free is provided in [2, Proposition 1]. However, in the present
q
case, R is not a chain ring and the form of the generatorof a principally generated
idealof R[x] is not known. Below we generalize[2, Proposition1]for the present
hxn−1i
case and provide a necessary condition (Theorem (5.1)) and a sufficient condition
(Theorem (5.2)) for the cyclic codes over R to be free.
Theorem 5.1. Let C be a principally generated cyclic code of length n over R
generated by g(x)∈R[x]. If g(x)|xn−1, then C is R-free.
Proof. Suppose that g(x) | xn − 1 and xn − 1 = g(x)h(x). Since xn − 1 is a
regularpolynomial,g(x)andh(x)mustalsoberegularpolynomials. By[8,Theorem
10 RAMA KRISHNA BANDI AND MAHESHANAND BHAINTWAL
XIII.6], there exist monic polynomials g′(x),h′(x) such that g(x)=v (x)g′(x) and
1
h(x)=v (x)h′(x)andg(x)=g′(x)andh(x)=h′(x), wherev (x),v (x)∈R[x]are
2 1 2
units. Therefore,xn−1=g(x)h(x)=v (x)v (x)g′(x)h′(x). Sincexn−1,g′(x)and
1 2
h′(x) are all monic, we must have v (x)v (x)=1 and xn−1=g′(x)h′(x). Let deg
1 2
g′(x) = n−k. Then deg h′(x) =k. We have C = hg(x)i = hv (x)g′(x)i = hg′(x)i,
1
as v (x) is a unit. Obviously the set S ={g′(x),xg′(x),...,xk−1g′(x)} spans C.
1
Nowsupposea(x)g′(x)=0(modxn−1)forsomea(x)∈R[x]withdega(x)<k.
Then xn−1 | a(x)g′(x), which implies that xn−1 | a(x), i. e., h′(x) | a(x). Since
g′(x)
h′(x) is monic polynomial of degree k, it cannot divide a non-zero polynomial of
degree less than k. It follows that a(x) = 0. So the set S is linearly independent
and thus forms a basis for C. Hence C is an R-free code.
We have following converse of Theorem (5.1).
Theorem 5.2. Let C be a principally generated cyclic code of length n over R
generated by g(x)∈R[x]. If C is R-free, then there exists a monic generator g′(x)
of C such that g′(x)|xn−1.
Proof. Suppose thatC isanR-freecode. Since g(x) generatesanR-free code, g(x)
must be a regular polynomial. Therefore there exist a monic polynomial g′(x) ∈
R[x] such that g(x) = v(x)g′(x) and g(x) = g′(x), where v(x) is a unit in R[x].
Let the R-rank of C be s and S = {c ,c ,...,c } an R-basis of C. Then the
1 2 s
set {c ,c ,...,c } forms a basis for the cyclic code C over the finite field R. Since
1 2 s
C =hg(x)i,soC =hg(x)i=g′(x). Sinceg′(x)ismonic,thereforeitisthegenerator
polynomialofC. Letdegg′(x)=n−k. Thentheset{g′(x),xg′(x),...,xk−1g′(x)}
forms a basis for C. So we must have s=k.
Now C = hg(x)i = hg′(x)i. Clearly, the elements g′(x),xg′(x),x2g′(x)... span
C. Also, the elements {g′(x),xg′(x),...,xk−1g′(x)} are linearly independent over
R; for if they are not, then they give a dependence relation among the elements
g′(x),xg′(x),...,xk−1g′(x), a contradiction. Now since xkg′(x) is a codeword, we
canwritexkg′(x)asalinearcombinationoftheelementsxig′(x),i=0,1,...,k−1.
Let
k−1
xkg′(x)=Xaixig′(x) ,
i=0
which can be written as k a xig′(x) = 0 with a = 1, or a(x)g′(x) = 0. Then
Pi=0 i k
xn−1 | a(x)g′(x) and since a(x)g′(x) is a monic polynomial of degree n, we must
have xn−1=a(x)g′(x). Therefore, g′(x)|xn−1.
The following result follows from Theorem (5.1) and Theorem (5.2).
Proposition 1. Let C be a principally generated cyclic code of length over R.
Then C is free if and only if there exists a monic generator g(x) in C such that
g(x) | xn −1. Furthermore, C has free rank n−deg g(x) and the elements g(x),
xg(x), ···, xn−deg g(x)−1g(x) forms a basis for C.
Example 5.3. Consider the cyclic code C of length 7 over R generated by the
polynomial g(x)=x3+2x2+x−1. g(x) is the Hensel lift of x3+x+1∈F [x] to
2
R. The cyclic code C = hg(x)i an R-free cyclic code of length 7 and the free rank
4.
Theorem 5.4. If C =C +uC is free cyclic code over R then so is C over Z .
1 2 1 4