/compile/output.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 9, No. 1, 2016, 39-47

ISSN 1307-5543 – www.ejpam.com

(1− 2u2)-Constacyclic Codes over Fp + uFp + u2
Fp

Hojjat Mostafanasab∗and Negin Karimi

Department of Mathematics and Applications, University of Mohaghegh Ardabili, P.O. Box 179,

Ardabil, Iran

Abstract. Let Fp be a finite field, where p is an odd prime, and let u be an indeterminate. This article

studies (1 − 2u2)-constacyclic codes over the ring Fp + uFp + u2
Fp, where u3 = u. We describe gen-

erator polynomials of this kind of codes and investigate the structural properties of these codes by a

decomposition theorem.

2010 Mathematics Subject Classifications: 94B05, 94B15, 11T71, 13M99

Key Words and Phrases: Finite fields, Cyclic codes, Constacyclic codes

1. Introduction

Error-Correcting codes play important roles in applications ranging from data networking

to satellite communication to compact disks. Most coding theory concerns on linear codes

since they have clear structure that makes them simpler to discover, to understand and to

encode and decode. Codes over finite rings have been studied since the early 1970s. Recently

codes over rings have generated a lot of interest after a breakthrough paper by Hammons et al.

[9] showed that some well known binary non-linear codes are actually images of some linear

codes over Z4 under the Gray map. Cyclic codes are amongst the most studied algebraic

codes. Their structure is well known over finite fields [13]. Constacyclic codes over finite

fields form a remarkable class of linear codes, as they include the important family of cyclic

codes. Constacyclic codes also have practical applications as they can be efficiently encoded

using simple shift registers. They have rich algebraic structures for efficient error detection

and correction, which explains their preferred role in engineering. In general, due to their

rich algebraic structure, constacyclic codes have been studied over various finite chain rings

(see [1, 3–7, 14, 15]). In [15], Zhu and Wang investigated (1− 2u)-constacyclic codes over

Fp + vFp, where v2 = v. In [8, 11, 12], some kind of codes over Fp + uFp + u2
Fp, where

u3 = u, have been studied. The present paper is devoted to a class of constacyclic codes over

Fp + uFp + u2
Fp, i.e., (1− 2u2)-constacyclic codes over Fp + uFp + u2

Fp.

∗Corresponding author.

Email addresses: h.mostafanasab@gmail.com (H. Mostafanasab), neginkarimi8834@gmail.com (N.

Karimi)

http://www.ejpam.com 39 c© 2016 EJPAM All rights reserved.



H. Mostafanasab, N. Karimi / Eur. J. Pure Appl. Math, 9 (2016), 39-47 40

Let σ, γ and ̺ be maps from Rn to Rn given by

σ(r0, r1, . . . , rn−1) =(rn−1, r0, r1, . . . , rn−2),

γ(r0, r1, . . . , rn−1) =(−rn−1, r0, r1, . . . , rn−2), and

̺(r0, r1, . . . , rn−1) =((1− 2u2)rn−1, r0, r1, . . . , rn−2),

respectively. LetC be a linear code of lenght n overR . ThenC is said to be cyclic ifσ(C ) = C ,

negacyclic if γ(C ) = C and (1− 2u2)-constacyclic if ̺(C ) = C .

Let C be a code of length n over R , and P(C ) be its polynomial representation, i.e.,

P(C ) =
¦

n−1
∑

i=0

ri x
i |(r0, . . . , rn−1) ∈ C

©

.

It is easy to see that:

Theorem 1. A code C of length n overR is (1−2u2)-constacyclic if and only if P(C ) is an ideal

of R[x]/〈xn − (1− 2u2)〉.

Let x = (x0, x1, . . . , xn−1) and y = (y0, y1, . . . , yn−1) be two elements ofRn. The Euclidean

inner product of x and y in Rn is defined as x · y = x0 y0 + x1 y1 + . . .+ xn−1 yn−1, where the

operation is performed in R . The dual code of C is defined as

C⊥ = {x ∈ Rn|x · y = 0 for every y ∈ C}.

We define the Gray map Φ :R → F2
p by a+ bu+ cu2 7→ (−c, 2a+ c). This map can be extended

to Rn in a natural way:

Φ :Rn→ F2n
p

(r0, r1, . . . , rn−1) 7→(−c0,−c1, . . . ,−cn−1, 2a0 + c0, 2a1 + c1, . . . , 2an−1 + cn−1)

where ri = ai + biu+ ciu
2, 0≤ i ≤ n− 1.

We denote by η1, η2, η3 respectively the following elements of R:

η1 = 1− u2, η2 = 2−1(u+ u2), η3 = 2−1(−u+ u2).

Note that η1, η2 and η3 are mutually orthogonal idempotents over R and η1 + η2 + η3 = 1.

Let C be a linear code of length n over R . Define

C1 ={x ∈ F
n
p | ∃y, z ∈ Fn

p,η1 x +η2 y +η3z ∈ C},

C2 ={y ∈ F
n
p | ∃x , z ∈ Fn

p,η1 x +η2 y +η3z ∈ C},

C3 ={z ∈ F
n
p | ∃x , y ∈ Fn

p,η1 x +η2 y +η3z ∈ C}.

Then C1,C2 and C3 are all linear codes of length n over Fp. Moreover, the code C of length

n over R can be uniquely expressed as C = η1C1 ⊕η2C2 ⊕η3C3.



H. Mostafanasab, N. Karimi / Eur. J. Pure Appl. Math, 9 (2016), 39-47 41

2. Main Results

Theorem 2. Let ̺ denote the (1− 2u2)-constacyclic shift of Rn and σ the cyclic shift of F2n
p . If

Φ is the Gray map of Rn into F2n
p , then Φ̺ = σΦ.

Proof. Let r̄ = (r0, r1, . . . , rn−1) ∈ R
n where ri = ai + biu + ciu

2 with ai , bi , ci ∈ Fp for

0≤ i ≤ n− 1. Taking (1− 2u2)-constacyclic shift on r̄, we have

̺(r̄) =
�

(1− 2u2)rn−1, r0, r1, . . . , rn−2

�

=
�

an−1 − bn−1u+ (−2an−1 − cn−1)u
2, a0 + b0u+ c0u2

, a1 + b1u+ c1u2, . . . , an−2 + bn−2u+ cn−2u2
�

.

Now, using the definition of Gray map Φ, we can deduce that

Φ(̺(r̄)) =
�

2an−1 + cn−1,−c0,−c1, . . . ,−cn−2, 2an−1 + (−2an−1 − cn−1)

, 2a0 + c0, 2a1 + c1, . . . , 2an−2 + cn−2

�

.

On the other hand,

σ(Φ(r̄)) =σ(−c0,−c1, . . . ,−cn−1, 2a0 + c0, 2a1 + c1, . . . , 2an−1 + cn−1)

=
�

2an−1 + cn−1,−c0,−c1, . . . ,−cn−1, 2a0 + c0, 2a1 + c1, . . . , 2an−2 + cn−2

�

.

Therefore,

Φ̺ = σΦ.

Theorem 3. The Gray image of a (1− 2u2)-constacyclic code over R of lenght n is a cyclic code

over Fp of lenght 2n.

Proof. Let C be a (1 − 2u2)-constacyclic code over R . Then ̺(C ) = C , and therefore,

(Φ̺)(C ) = Φ(C ). It follows from Theorem 2 that σ(Φ(C )) = Φ(C ), which means that Φ(C )
is a cyclic code.

Notice that (1− 2u2)n = 1− 2u2 if n is odd and (1− 2u2)n = 1 if n is even.

Proposition 1. Let C be a code of lenght n over R . Then C is a (1− 2u2)-constacyclic code if

and only if C ⊥ is a (1− 2u2)-constacyclic code.

Proof. The “only if” part follows from Proposition 2.4 of [6]. For the converse, note the

fact that (C ⊥)⊥ = C .

Recall that a code C is said to be self-orthogonal provided C ⊆ C ⊥.

Proposition 2. Let C be a code of length n over R such that C ⊂
�

Fp + u2
Fp

�n
. If C is self-

orthogonal, then so is Φ(C ).



H. Mostafanasab, N. Karimi / Eur. J. Pure Appl. Math, 9 (2016), 39-47 42

Proof. Assume that C is self-orthogonal. Let r1 = a1 + c1u2, r2 = a2 + c2u2 ∈ C , where

ai , ci ∈ F
n
p for i = 1,2. Now by Euclidean inner product of r1 and r2, we have

r1 · r2 =(a1 + c1u2) · (a2 + c2u2)

=a1a2 + (a1c2 + c1a2 + c1c2)u
2.

If r1 · r2 = 0, then a1a2 = a1c2 + c1a2 + c1c2 = 0. Therefore

Φ(r1) ·Φ(r2) =(−c1, 2a1 + c1) · (−c2, 2a2 + c2)

=4a1a2 + 2(c1c2 + a1c2 + c1a2) = 0.

Hence Φ(C ⊥) ⊆ Φ(C )⊥. Consequently Φ(C ) ⊆ Φ(C )⊥.

Theorem 4. Let C = η1C1⊕η2C2⊕η3C3 be a code of length n overR . Then C is a (1−2u2)-

constacyclic code of length n over R if and only if C1 is cyclic and C2, C3 are negacyclic codes of

length n over Fp.

Proof. First of all notice that (1−2u2)η1 = η1, (1−2u2)η2 = −η2 and (1−2u2)η3 = −η3.

Let r̄ = (r0, r1, . . . , rn−1) ∈ C . Then ri = η1ai+η2 bi+η3ci , where ai , bi , ci ∈ Fp, 0≤ i ≤ n−1.

Let a = (a0, a1, . . . , an−1), b = (b0, b1, . . . , bn−1) and c = (c0, c1, . . . , cn−1). Then a ∈ C1, b ∈ C2

and c ∈ C3. Assume that C1 is cyclic and C2, C3 are negacyclic codes. Therefore σ(a) ∈ C1,

γ(b) ∈ C2 and γ(c) ∈ C3. Thus ̺(r̄) = η1σ(a) + η2γ(b) + η3γ(c) ∈ C . Consequently C
is a (1 − 2u2)-constacyclic codes over R . For the converse, let a = (a0, a1, . . . , an−1) ∈ C1,

b = (b0, b1, . . . , bn−1) ∈ C2 and c = (c0, c1, . . . , cn−1) ∈ C3. Set ri = η1ai +η2 bi +η3ci , where

0 ≤ i ≤ n− 1. Hence r̄ = (r0, r1, . . . , rn−1) ∈ C . Therefore ̺(r̄) = η1σ(a) +η2γ(b) +η3γ(c),

̺(r̄) ∈ C which shows that σ(a) ∈ C1, γ(b) ∈ C2 and γ(c) ∈ C3. So C1 is cyclic and C2, C3

are negacyclic codes.

Theorem 5. Let C = η1C1 ⊕η2C2 ⊕η3C3 be a (1− 2u2)-constacyclic code of length n over R
such that g1(x), g2(x), g3(x) are the monic generator polynomials of C1, C2, C2, respectively.

Then C = 〈η1 g1(x),η2 g2(x),η3 g3(x)〉 and |C |= p3n−
∑3

i=1 deg(gi).

Proof. By Theorem 4, C1 = 〈g1(x)〉 ⊆ Fp[x]/〈x
n − 1〉,C2 = 〈g2(x)〉 ⊆ Fp[x]/〈x

n + 1〉 and

C3 = 〈g3(x)〉 ⊆ Fp[x]/〈x
n + 1〉. Since C = η1C1 ⊕η2C2 ⊕η3C3, then

C = {c(x)|c(x) = η1 f1(x) +η2 f2(x) +η3 f3(x), f1(x) ∈ C1, f2(x) ∈ C2 and f3(x) ∈ C3}.

Hence

C ⊆ 〈η1 g1(x),η2 g2(x),η3 g3(x)〉 ⊆ Rn =R[x]/〈x
n − (1− 2u2)〉.

Suppose that η1 g1(x)h1(x) + η2 g2(x)h2(x) + η3 g3(x)h3(x) ∈ 〈η1 g1(x),η2 g2(x),η3 g3(x)〉,
where h1(x),h2(x),h3(x) ∈ Rn. There exist q1(x) ∈ Fp[x]/〈x

n − 1〉,q2(x) ∈ Fp[x]/〈x
n + 1〉

and q3(x) ∈ Fp[x]/〈x
n + 1〉 such that η1h1(x) = η1q1(x), η2h2(x) = η2q2(x) and

η3h3(x) = η3q3(x). Therefore 〈η1 g1(x),η2 g2(x),η3 g3(x)〉 ⊆ C . Consequently

C = 〈η1 g1(x),η2 g2(x),η3 g3(x)〉.

On the other hand |C |= |C1| · |C2| · |C3|= p3n−
∑3

i=1 deg(gi).



H. Mostafanasab, N. Karimi / Eur. J. Pure Appl. Math, 9 (2016), 39-47 43

Theorem 6. LetC be a (1−2u2)-constacyclic code of length n overR . Then there exists a unique

polynomial g(x) such that C = 〈g(x)〉 where g(x) = η1 g1(x) +η2 g2(x) +η3 g3(x).

Proof. Suppose that g1(x), g2(x), and g3(x) are the monic generator polynomials of C1,

C2, and C3, respectively. By Theorem 5, we have C = 〈η1 g1(x),η2 g2(x),η3 g3(x)〉. Let

g(x) = η1 g1(x) + η2 g2(x) + η3 g3(x). Clearly, 〈g(x)〉 ⊆ C . However, η1 g1(x) = η1 g(x),

η2 g2(x) = η2 g(x) and η3 g3(x) = η3 g(x), whence C ⊆ 〈g(x)〉. Thus C = 〈g(x)〉. The

uniqueness of g(x) is followed by that of g1(x), g2(x), and g3(x).

Lemma 1. Let xn − (1− 2u2) = g(x)h(x) in R[x] and let C be the (1− 2u2)-constacyclic code

generated by g(x). If f (x) is relatively prime with h(x) then C = 〈g(x) f (x)〉.

Proof. The proof is similar to that of [2, Lemma 2].

Theorem 7. Let C = η1C1 ⊕η2C2 ⊕η3C3 be a (1− 2u2)-constacyclic code of length n over R
such that g1(x), g2(x), g3(x) are the monic generator polynomials of C1, C2, C2, respectively.

Suppose that g1(x)h1(x) = xn − 1 and g2(x)h2(x) = g3(x)h3(x) = xn + 1 and set

g(x) = η1 g1(x) +η2 g2(x) +η3 g3(x), h(x) = η1h1(x) +η2h2(x) +η3h3(x). Then

(i) g(x)h(x) = xn − (1− 2u2).

(ii) If GC D( fi(x),hi(x)) = 1 for 1 ≤ i ≤ 3, then GC D( f (x),h(x)) = 1 and g(x) = g(x) f (x)

where f (x) = η1 f1(x) +η2 f2(x) +η3 f3(x).

Proof. (i) By assumptions we have that

g(x)h(x) =g(x)
�

η1h1(x) +η2h2(x) +η3h3(x)
�

=η1 g1(x)h1(x) +η2 g2(x)h2(x) +η3 g3(x)h3(x)

=η1(x
n − 1) +η2(x

n + 1) +η3(x
n + 1)

=(η1 +η2 +η3)x
n − (η1 −η2 −η3)

=xn − (1− 2u2).

Hence, g(x)h(x) = xn − (1− 2u2).

(ii) Suppose that GC D( fi(x),hi(x)) = 1 for 1≤ i ≤ 3 and let

f (x) = η1 f1(x)+η2 f2(x)+η3 f3(x). Then for every 1≤ i ≤ 3 there exist ai(x), bi(x) ∈ R[x]
such that ai(x) fi(x) + bi(x)hi(x) = 1. Set a(x) := η1a1(x) +η2a2(x) +η3a3(x) and

b(x) := η1 b1(x) +η2 b2(x) +η3 b3(x). Notice that η1 +η2 +η3 = 1, η2
i
= 1 and ηiη j = 0 for

every 1≤ i 6= j ≤ 3. Thus

a(x) f (x) + b(x)h(x) =η1[a1(x) f1(x) + b1(x)h1(x)] +η2[a2(x) f2(x) + b2(x)h2(x)]

+η3[a3(x) f3(x) + b3(x)h3(x)] = η1 +η2 +η3 = 1.

It follows that GC D( f (x),h(x)) = 1. Now, by part (i) and Lemma 1, C = 〈g(x) f (x)〉. So, the

uniqueness of g(x) implies that g(x) = g(x) f (x).

Similar to [8, Theorem 3], we have the following theorem.



H. Mostafanasab, N. Karimi / Eur. J. Pure Appl. Math, 9 (2016), 39-47 44

Theorem 8. Let C be a (1− 2u2)-constacyclic code of length n over R . Then

C ⊥ = η1C
⊥
1 ⊕η2C

⊥
2 ⊕η3C

⊥
3 .

As a consequence of the previous theorems and [10, Theorem 3.3]we have the next result.

Corollary 1. Let C = 〈η1 g1(x),η2 g2(x),η3 g3(x)〉 be a (1− 2u2)-constacyclic code of length n

over R and g1(x), g2(x), g3(x) be the monic generator polynomials of C1, C2, C3, respectively.

Suppose that g1(x)h1(x) = xn − 1 and g2(x)h2(x) = g3(x)h3(x) = xn + 1 and let

h(x) = η1h1(x) +η2h2(x) +η3h3(x). The following conditions hold:

(i) C ⊥ = 〈η1h⊥1 (x),η2h⊥2 (x),η3h⊥3 (x)〉 and |C ⊥|= p
∑3

i=1 deg(gi).

(ii) C ⊥ = 〈h⊥(x)〉, h⊥(x) = η1h⊥1 (x) +η2h⊥2 (x) +η3h⊥3 (x),

where for 1 ≤ i ≤ 3, h⊥
i
(x) is the reciprocal polynomial of hi(x), and h⊥(x) is the reciprocal

polynomial of h(x).

Theorem 9. Let µ :R[x]/〈xn − 1〉 → R[x]/〈xn − (1− 2u2)〉 be defined as

µ
�

c(x)
�

= c
�

(1− 2u2)x
�

.

If n is odd, then µ is a ring isomorphism.

Proof. Suppose that a(x) ≡ b(x) (mod xn − 1). Then there exists h(x) ∈ R[x] such that

a(x)− b(x) = (xn − 1)h(x). Therefore

a
�

(1− 2u2)x
�

− b
�

(1− 2u2)x
�

=
�

(1− 2u2)n xn − 1
�

h
�

(1− 2u2)x
�

=
�

(1− 2u2)xn − (1− 2u2)2
�

h
�

(1− 2u2)x
�

=(1− 2u2)
�

xn − (1− 2u2)
�

h
�

(1− 2u2)x
�

,

which means if a(x)≡ b(x) (mod xn − 1), then

a
�

(1− 2u2)x
�

≡ b
�

(1− 2u2)x
� �

mod xn − (1− 2u2)
�

.

Now, assume that a
�

(1 − 2u2)x
�

≡ b
�

(1 − 2u2)x
� �

mod xn − (1 − 2u2)
�

. Then there exists

q(x) ∈ R[x] such that

a
�

(1− 2u2)x
�

− b
�

(1− 2u2)x
�

=
�

xn − (1− 2u2)
�

q(x).

Hence

a(x)− b(x) =a
�

(1− 2u2)2 x
�

− b
�

(1− 2u2)2 x
�

=
�

(1− 2u2)n xn − (1− 2u2)
�

q
�

(1− 2u2)x
�

=
�

(1− 2u2)xn − (1− 2u2)
�

q
�

(1− 2u2)x
�

=(1− 2u2)(xn − 1)q
�

(1− 2u2)x
�

,



H. Mostafanasab, N. Karimi / Eur. J. Pure Appl. Math, 9 (2016), 39-47 45

which means if a
�

(1− 2u2)x
�

≡ b
�

(1− 2u2)x
� �

mod xn − (1− 2u2)
�

,

then a(x)≡ b(x) (mod xn − 1). Consequently

a(x)≡ b(x) (mod xn − 1)⇔ a
�

(1− 2u2)x
�

≡ b
�

(1− 2u2)x
� �

mod xn − (1− 2u2)
�

.

Note that one side of the implication tells us that µ is well defined and the other side tells us

that it is injective, but since the rings are finite this proves that µ is an isomorphism.

Corollary 2. Let n be an odd natural number. Then I is an ideal of R[x]/〈xn−1〉 if and only if

µ(I) is an ideal of R[x]/〈xn − (1− 2u2)〉.

Corollary 3. Let µ be the permutation of Rn with n odd such that

µ̄(c0, c1, . . . , cn−1) = (c0, (1− 2u2)c1, (1− 2u2)2c2, . . . , (1− 2u2)ici , . . . , (1− 2u2)n−1cn−1),

and D be a subset of Rn. Then D is a cyclic code if and only if µ̄(D) is a (1− 2u2)-constacyclic

code.

Definition 1. Let τ be the following permutation of {0,1, . . . , 2n− 1} with n odd:

τ= (1, n+ 1)(3, n+ 3) · · · (2i + 1, n+ 2i + 1) · · · (n− 2,2n− 2).

The Nechaev permutation is the permutation π of F2n
p defined by

π(c0, c1, . . . , c2n−1) = (cτ(0), cτ(1), . . . , cτ(2n−1)).

Proposition 3. Let µ be defined as above. If π is the Nechaev permutation and n is odd, then

Φµ̄ = πΦ.

Proof. Let r̄ = (r0, r1, . . . , ri , . . . , rn−1) ∈ R
n where ri = ai + biu+ ciu

2, 0≤ i ≤ n−1. From

µ̄(r̄) = (r0, (1− 2u2)r1, . . . , (1− 2u2)i ri , . . . , (1− 2u2)n−1rn−1)

it follows that

(Φµ̄)(r̄) =(−c0, 2a1 + c1,−c2, 2a3 + c3, . . . , 2an−2 + cn−2,−cn−1

, 2a0 + c0,−c1, 2a2 + c2,−c3, . . . ,−cn−2, 2an−1 + cn−1),

is equal to (πΦ)(r̄).

Corollary 4. Let π be the Nechaev permutation and n be odd. If Γ is the Gray image of a cyclic

code over R , then π(Γ) is a cyclic code.

Proof. Let Γ be such that Γ = Φ(D) where D is a cyclic code over R . From Proposition 3,

(Φµ̄)(D) = (πΦ)(D) = π(Γ). We know from Corollary 3 that µ̄(D) is a (1− 2u2)-constacyclic

code. Thus (Φµ̄)(D) = π(Γ) is a cyclic code, by Theorem 3.

Recall that two codes C1 and C2 of length n overR are said to be equivalent if there exists

a permutation w of {0,1, . . . , n− 1} such that C2 = w̄(C1) where w̄ is the permutation of Rn

such that w̄(c0, c1, . . . , ci , . . . , cn−1) = (cw(0), cw(1), . . . , cw(i), . . . , cw(n−1)).



REFERENCES 46

Corollary 5. The Gray image of a cyclic code over R of odd length is equivalent to a cyclic code.

Example 1. For n = 7, x7 − 1 = (x − 1)(x3 + x + 1)(x3 + x2 + 1) in R[x]. Applying the ring

isomorphism µ, we have

x7 − (1− 2u2) = (x − (1− 2u2))(x3 + x + (1− 2u2))(x3 + (1− 2u2)x2 + (1− 2u2)).

Let f1 = x − (1− 2u2) and f2 = x3 + x + (1− 2u2). If C = ( f1 f2), then by Theorem 3, we know

that the Gray image of the (1− 2u2)-constacyclic code C is a cyclic code.

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