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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# Constacyclic codes over 𝔽pm[u1, u2,⋯,uk]/〈 ui2 = ui, uiuj = ujui〉

Xiying Zheng
• Corresponding author
• Institute of Information Engineering, Huanghe Science and Technology College, Zhengzhou 450063, China
• Email
• Other articles by this author:
/ Bo Kong
Published Online: 2018-05-10 | DOI: https://doi.org/10.1515/math-2018-0045

## Abstract

In this paper, we study linear codes over ring Rk = 𝔽pm[u1, u2,⋯,uk]/〈$\begin{array}{}{u}_{i}^{2}\end{array}$ = ui, uiuj = ujui〉 where k ≥ 1 and 1 ≤ i, jk. We define a Gray map from $\begin{array}{}{R}_{k}^{n}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{to}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\mathbb{F}}_{{p}^{m}}^{{2}^{k}n}\end{array}$ and give the generator polynomials of constacyclic codes over Rk. We also study the MacWilliams identities of linear codes over Rk.

MSC 2010: 94B15

## 1 Introduction

Constacyclic codes are an important class of linear codes and have good error-correcting properties as well as have practical applications since they can be encoded with shift registers. have practical applications as they can be encoded with shift registers. Constacyclic codes over finite rings are well-known as they have rich algebraic structures for efficient error detection and correction, which explain their preferred role in engineering. In recent years, due to their rich algebraic structure, constacyclic codes have been studied over finite fields [1, 2, 3, 4]. The class of finite chain rings has been studied, by many authors, [5, 6, 7, 8]. There is a lot of work on constacyclic codes over finite rings of the form 𝔽pm + u 𝔽pm + ⋯ + ue−1𝔽pm by many authors, where ue = 0. For example, Chen et al. in [9] gave the structures of all (a + bu)-constacyclic codes of length 2ps over ring 𝔽pm + u 𝔽pm. Sobhani in [10] completely determined the structure of (δ + α u2)-constacyclic codes of length pk over 𝔽pm + u 𝔽pm + u2𝔽pm. Liu and Xu in [11] gave the structure of cyclic and negacyclic codes of length 2ps over 𝔽pm + u 𝔽pm. Abualrub and Siap in [12] gave the structure of (1 + u)-constacyclic codes of arbitrary length n over 𝔽2 + u𝔽2. Kai et al. in [13] studied the (1 + λu)-constacyclic codes of arbitrary length n over 𝔽p[u]/〈um〉, where (1 + λu) is a unite of 𝔽p[u]/〈um〉. Guenda and Gulliver in [14] gave the structure of repeated root constacyclic codes of length mps over 𝔽pr + u𝔽pr + ⋯ + ue−1𝔽pr.

The class of finite commutative rings of the form R + uR has been studied by many authors, where u2 = 1. For example, in [15] Cengellenmis gave the structure of cyclic codes over 𝔽3 + v𝔽3, where v2 = 1. ¨Qzen et al. in [16] gave the structure of cyclic and some constacyclic codes over the ring ℤ4[u]/〈u2 − 1〉. The class of finite commutative rings of the form 𝔽pm + u 𝔽pm has been studied by many authors, where u2 = u. For example, in [17], Kong and Chang described the structure of cyclic codes and self dual cyclic codes over 𝔽p + u𝔽p, where u2 = u. Cengellenmis et al. in [18] gave the structure of codes over 𝔽2[u1, u2,⋯, uk]/〈$\begin{array}{}{u}_{i}^{2}\end{array}$ = ui, uiuj = ujui〉 with a Gray map. Li et al. in [19] gave the structure of linear codes over ℤ4[u, v]/〈u2 = u, v2 = v, uv = vu〉. In [20], the generators of cyclic codes and (λ1 + λ2u + λ3v + λ4uv)-constacyclic codes over 𝔽p + u𝔽p + v𝔽p + uv𝔽p were given. The purpose of this paper is to continue this line of research. We determine the algebraic structures of all λ-constacyclic codes of 𝔽pm[u1, u2,⋯, uk]/〈$\begin{array}{}{u}_{i}^{2}\end{array}$ = ui, uiuj = ujui〉, where λ is an arbitrary unit of the ring 𝔽pm[u1, u2,⋯, uk]/〈$\begin{array}{}{u}_{i}^{2}\end{array}$ = ui, uiuj = ujui〉.

The remainder of this paper is organized as follows. In section 2, we provide the preliminaries that we need and define a Gray map from $\begin{array}{}{R}_{k}^{n}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{to}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\mathbb{F}}_{{p}^{m}}^{{2}^{k}n}\end{array}$. In section 3, we study the Gray image of linear codes over Rk. In section 4, we give the structure of constacyclic codes of arbitrary length over Rk.

## 2 Preliminaries

An ideal I of a finite commutative ring R is called principal if it is generated by one element. R is a principal ideal ring if its ideals are principal. R is called a local ring if R has a unique maximal ideal. R is called a chain ring if its ideals are linearly ordered by inclusion.

As defined in [18], let

$Rk=Fpm[u1,u2,⋯,uk]/〈ui2=ui,uiuj=ujui〉.$

For any subset A ⊆ {1, 2, ⋯, k}, let

$uA=∏i∈Aui$

with the convention that u = 1. Then any element of Rk can be represented as

$∑A⊆{1,2,⋯,k}cAuA,cA∈Fpm.$

We can easily observe that

$uAuB=uA⋃B.$

Let Pk be the power set of the set {1, 2, ⋯, k}.

It follows that

$(∑A∈PkcAuA)(∑B∈PkcBuB)=∑D∈Pk(∑A∪B=DcAcB)uD.$

By the same method of Theorem 2.3 and Lemma 2.4 in [18] we have the following theorem:

#### Theorem 2.1

The idealw1, w2,⋯, wk〉, where wi ∈ {ui, 1 − ui}, is an ideal of cardinality pm(2k−1) and there are 2k such ideals.

Let ωi = 〈wi1, wi2,⋯, wik〉 be an ideal as described in Theorem 2.1, where wij ∈ {uj, 1 − uj}, 1 ≤ i ≤ 2k. An element e is called an idempotent element if e2 = e. For x, yRk, x, y are called orthogonal if xy = 0. Let ei = wi1wi2wik, where i = 1, 2, ⋯, 2k. We know that ui2 = ui, (1 − ui)2 = 1 − ui, ui(1 − ui) = 0, so e1, e2, ⋯, e2k are pairwise orthogonal non-zero idempotent elements over Rk. By the induction method over Rk, we have 1 = e1 + e2 + ⋯ + e2k. By the Chinese Remainder Theorem, we have that Rk = e1Rk + e2Rk + ⋯ + e2kRk, and for any element rRk, r can be expressed uniquely as r = r1e1 + r2e2 + ⋯ + r2k e2k, where ri ∈ 𝔽pm, i = 1, 2,⋯, 2k.

#### Theorem 2.2

RkRk/ω1 × … × Rk/ω2k.

#### Proof

First, we prove that $\begin{array}{}\bigcap _{i=1}^{{2}^{k}}{\omega }_{i}=\left\{0\right\}.\end{array}$

We use mathematical induction over Rk.

Base case: Setting over R1, we get

$⋂i=12ωi=〈u1〉⋂〈1−u1〉=〈u1−u12〉={0}.$

Induction step: Over Rk−1, suppose that

$⋂i=12k−1ωi′={0},$

where $\begin{array}{}{\omega }_{i}^{{}^{\prime }}\end{array}$ = 〈wi1, wi2, ⋯, wik−1〉, wij ∈ {uj, 1 − uj}, 1 ≤ i ≤ 2k−1, 1 ≤ jk − 1.

Then over Rk

$⋂i=12kωi=⋂i=12k〈wi1,wi2,⋯,wik〉=(⋂i=12k−1〈wi1,wi2,⋯,wik−1,uk〉)⋂(⋂i=12k−1〈wi1,wi2,⋯,wik−1,1−uk〉)=(⋂i=12k−1(ωi′+〈uk〉))⋂(⋂i=12k−1(ωi′+〈1−uk〉))=(⋂i=12k−1ωi′+〈uk〉)⋂(⋂i=12k−1ωi′+〈1−uk〉)=〈uk〉⋂〈1−uk〉=〈uk−uk2〉={0},$

where ωi = 〈wi1, wi2,⋯, wik〉, wij ∈ {uj, 1 − uj}, 1 ≤ i ≤ 2k, 1 ≤ jk.

Secondly, we prove that ω1, ω2, ⋯, ω2k are pairwise coprime. For any two different ideals ωi, ωj, there exist utωi, (1 − ut) ∈ ωj, such that 1 ∈ ωi + ωj, then ωi + ωj = Rk. So ω1, ω2, ⋯, ω2k are pairwise coprime.

By the Chinese Remainder Theorem, we can get that RkRk/ω1 × … × Rk/ω2k. □

#### Theorem 2.3

The ring Rk has cardinality pm2k. The ideal ωi is a maximal ideal of Rk, where i = 1, 2,⋯, 2k. Consequently, Rk$\begin{array}{}{\mathbb{F}}_{{p}^{m}}^{{2}^{k}}\end{array}$.

#### Proof

By Theorem 2.2, we have that $\begin{array}{}|{R}_{k}|=\frac{|{R}_{k}|}{|{\omega }_{1}|}×\cdots ×\frac{|{R}_{k}|}{|{\omega }_{{2}^{k}}|}.\end{array}$ By Theorem 2.1 |ωi| = pm(2k−1), where i = 1, 2,⋯, 2k.

We have that |Rk| = pm2k. Thus $\begin{array}{}\frac{|{R}_{k}|}{|{\omega }_{i}|}\end{array}$ = pm, where i = 1, 2,⋯, 2k. So ωi is a maximal ideal of Rk, we can get that Rk/ωi≅ 𝔽pm, where i = 1, 2,⋯, 2k. So Rk$\begin{array}{}{\mathbb{F}}_{{p}^{m}}^{{2}^{k}}\end{array}$. □

#### Corollary 2.4

There are (pm − 1)2k units in the ring Rk.

#### Proof

There are (pm − 1) units in 𝔽pm. By Theorem 2.3, we know there are (pm − 1)2k units in the ring Rk. □

#### Theorem 2.5

(cf. [21, Theorem 2]). The ring Rk is a principal ideal ring, not a chain ring.

We define the Gray map as follows:

For r = r1e1 + r2e2 + ⋯ + r2ke2kRk, we define ϕ : r ↦ (r1, r2, ⋯, r2k). We expand ϕ as:

$Φ:Rkn→Fpm2kn(c0,c1,⋯,cn−1)↦(r1,0,⋯,r1,n−1,r2,0,⋯,r2,n−1,⋯,r2k,0,⋯,r2k,n−1),$

where ci = r1,ie1 + r2,ie2 + ⋯ + r2k,ie2kRk.

A linear code C of length n over Rk is an Rk-submodule of $\begin{array}{}{R}_{k}^{n}\end{array}$. Every codeword c in such a code C is just an n-tuple of the form c = (c0, c1, ⋯, cn−1) ∈ $\begin{array}{}{R}_{k}^{n}\end{array}$, and can be represented by a polynomial in Rk[x] as follows:

$c=(c0,c1,⋯,cn−1)⟷c(x)=∑i=0n−1cixi∈Rk[x].$

We define a constacyclic shift operator as:

$σλ(c0,c1,⋯,cn−1)=(λcn−1,c0,⋯,cn−2).$

If for any cC, we have σλ(c) ∈ C, then C is called λ-constacyclic code over Rk. Let a = (a0, a1, ⋯, an−1) and b = (b0, b1, ⋯, bn−1) be two elements of $\begin{array}{}{R}_{k}^{n}\end{array}$. Then the usual inner product of a and b is defined as $\begin{array}{}a\cdot b=\sum _{i=0}^{n-1}{a}_{i}{b}_{i}.\end{array}$ If ab = 0, then a and b are said to be orthogonal.

The dual code of C is C = {a| ∀ bC, ab = 0}, which is also a linear code. A code C is self-orthogonal if CC and self dual if C = C.

For all rRk, define the Lee weight of r as follows: wL(r) = wH(ϕ(r)), where let wH(ϕ(r)) denote the Hamming weight of the image of r under ϕ.

For all x = (x1, x2,⋯, xn) ∈ $\begin{array}{}{R}_{k}^{n}\end{array}$, define the Lee weight of x as follows wL(x) = $\begin{array}{}\sum _{i=1}^{n}\end{array}$ wL(xi), the Lee distance of codewords x, y over $\begin{array}{}{R}_{k}^{n}\end{array}$ is defined as dL(x, y) = wL(xy). The Lee distance of C is defined by

$dL(C)=min{dL(x−y),x,y∈C,x≠y}.$

By the definition of the Gray map and the Lee weight of Rk, we can get that Φ is one-to-one and a distance preserving linear map from $\begin{array}{}{R}_{k}^{n}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{to}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\mathbb{F}}_{{p}^{m}}^{{2}^{k}n}.\end{array}$

## 3 Linear codes over Rk

Using the polynomial representation of codewords in $\begin{array}{}{R}_{k}^{n}\end{array}$, we easily have the following.

#### Lemma 3.1

A subset C of $\begin{array}{}{R}_{k}^{n}\end{array}$ is a λ-constacyclic code of length n over Rk if and only if its polynomial representation is an ideal of the ring Rk[x]/〈xn - λ〉.

For any r = (r(0), r(1), ⋯, r(n−1)) ∈ $\begin{array}{}{R}_{k}^{n}\end{array}$, where r(i) = $\begin{array}{}\sum _{j=1}^{{2}^{k}}\end{array}$ rijej, i = 0, 1, ⋯, n − 1. Then r can be uniquely express as r = $\begin{array}{}\sum _{j=1}^{{2}^{k}}\end{array}$ rjej, where rj = (r0j, r1j, ⋯, rn−1,j) ∈ $\begin{array}{}{\mathbb{F}}_{{p}^{m}}^{n},\end{array}$ j = 1, 2,⋯, 2k.

For any r, s$\begin{array}{}{R}_{k}^{n}\end{array}$, where s = $\begin{array}{}\sum _{j=1}^{{2}^{k}}\end{array}$ sjej, sj = (s0j, s1j, ⋯, sn−1,j) ∈ $\begin{array}{}{\mathbb{F}}_{{p}^{m}}^{n},\end{array}$ we can get that

$r⋅s=∑j=12k(rj⋅sj)ej,$

where $\begin{array}{}{r}_{j}\cdot {s}_{j}=\sum _{i=0}^{n-1}\left({r}_{ij}{s}_{ij}\right).\end{array}$

Let C be a linear code over Rk. For j = 1, 2,⋯, 2k, we denote Cj as follows:

$Cj={rj∈Fpmn|∑i=12kriei∈C,ri∈Fpmn,},j=1,2,⋯,2k.$

Clearly, Cj is a linear code of length n over 𝔽pm.

By the definition above we have the following theorems easily.

#### Theorem 3.2

Let C be a linear code over Rk, then $\begin{array}{}C=\sum _{j=1}^{{2}^{k}}{e}_{j}{C}_{j},|C|=\prod _{j=1}^{{2}^{k}}|{C}_{j}|,\end{array}$ where C1, C2, ⋯, C2k are linear codes of length n over 𝔽pm, and the direct sum decomposition is unique.

#### Theorem 3.3

Let C be a linear code over Rk, then $\begin{array}{}{C}^{\mathrm{\perp }}=\sum _{j=1}^{{2}^{k}}{e}_{j}{C}_{j}^{\mathrm{\perp }},\phantom{\rule{thinmathspace}{0ex}}where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{C}_{j}^{\mathrm{\perp }}\end{array}$ is the dual code of Cj, where j = 1, 2,⋯, 2k.

#### Proof

Let $\begin{array}{}\stackrel{~}{C}=\sum _{j=1}^{{2}^{k}}{e}_{j}{C}_{j}^{\mathrm{\perp }}.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}For any}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}c\in C,\stackrel{~}{c}\in \stackrel{~}{C},c\cdot \stackrel{~}{c}=\sum _{j=1}^{{2}^{k}}\left({c}_{j}\stackrel{~}{{c}_{j}}\right){e}_{j},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}where}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}c=\sum _{j=1}^{{2}^{k}}{e}_{j}{c}_{j},\stackrel{~}{c}=\sum _{j=1}^{{2}^{k}}{e}_{j}\stackrel{~}{{c}_{j}},{c}_{j}\in {C}_{j},\stackrel{~}{{c}_{j}}\in {C}_{j}^{\mathrm{\perp }}.\end{array}$ Then c = 0, and thus C. The ring Rk is a principal ideal ring and thus a Frobenius ring, we have |C||C| = |Rk|n. Thus

$|C~|=∏j=12k|Cj⊥|=∏j=12kpn|Cj|=|Rk|n|C|=|C⊥|.$

So C = . □

#### Theorem 3.4

Let C be a linear code over Rk, then C is a self-orthogonal code if and only if Cj is a self-orthogonal over 𝔽pm, where $\begin{array}{}c=\sum _{j=1}^{{2}^{k}}{e}_{j}{c}_{j}.\end{array}$ C is a self-dual code if and only if Cj is a self-dual code over 𝔽pm, where j = 1, 2,⋯, 2k.

#### Proof

By Theorems 3.2 and 3.3, CC if and only if Cj$\begin{array}{}{C}_{j}^{\mathrm{\perp }},\end{array}$ so if C is a self-orthogonal code then Cj is a self-orthogonal code over 𝔽pm, where j = 1, 2,⋯, 2k. Similarly, C is a self-dual code then Cj is a self-dual code over 𝔽pm, where j = 1, 2,⋯, 2k. □

Let C be a linear code of length n over Rk, for any c = c1e1 + c2e2 + ⋯ + c2ke2kC, Φ(c) = (c1, c2, ⋯, c2k) ∈ $\begin{array}{}{\mathbb{F}}_{{p}^{m}}^{{2}^{k}n}\end{array}$. Let C1,C2, ⋯, C2k be linear codes of length n over 𝔽pm, we define

$C1×C2×⋯×C2k={(c1,c2,⋯,c2k),ci∈Ci,i=1,2,⋯,2k}.$

#### Theorem 3.5

Let C = e1C1 + e2C2 + ⋯ + e2kC2k be a linear code of length n over Rk with |C| = pml and the minimum Lee distance dL(C) = d. Then Φ(C) = C1× C2× ⋯ × C2k is a linear code with parameter [2kn, l, d] and Φ(C) = Φ(C). If C is a self-dual code over Rk, then Φ(C) is a self-dual code over 𝔽pm.

#### Proof

By the definition above, we can know that

$C1×C2×⋯×C2k⊆Φ(C)$

and

$|C1×C2×⋯×C2k|=|C1||C2|⋯|C2k|=|C|.$

This gives that

$Φ(C)=C1×C2×⋯×C2k.$

Let $\begin{array}{}c=\sum _{j=1}^{{2}^{k}}{e}_{j}{c}_{j}\in C,d=\sum _{j=1}^{{2}^{k}}{e}_{j}{d}_{j}\in {C}^{\perp },\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{where}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{c}_{j}\in {C}_{j},{d}_{j}\in {C}_{j}^{\perp },\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{then}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}c\cdot d=\sum _{j=1}^{{2}^{k}}{e}_{j}{c}_{j}{d}_{j}=0,\end{array}$ which implies cjdj = 0, so

$Φ(c)⋅Φ(d)=∑j=12kcjdj=0,$

which implies

$Φ(C)⊥⊇Φ(C⊥).$

By Theorem 3.3, we have

$Φ(C⊥)=C1⊥×C2⊥×⋯×C2k⊥.$

Since Φ is one-to-one, we have

$|Φ(C⊥)|=pm2kn|C|=pm2kn|Φ(C)|=|Φ(C)⊥|.$

So

$Φ(C)⊥=Φ(C⊥).$

Let τ be a cyclic shift operator on $\begin{array}{}{\mathbb{F}}_{{p}^{m}}^{n}.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{Let}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}a=\left({a}^{\left(1\right)}|{a}^{\left(2\right)}|\cdots |{a}^{\left({2}^{k}\right)}\right)\in {\mathbb{F}}_{{p}^{m}}^{{2}^{k}n},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}where}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}^{\left(j\right)}\in {\mathbb{F}}_{{p}^{m}}^{n}\end{array}$ for j = 1, 2,⋯, 2k. Let τ2k be the quasi-shift given by

$τ2k(a(1)|a(2)|⋯|a(2k))=(τ(a(1))|τ(a(2))|⋯|τ(a(2k))).$

#### Proposition 3.6

Let σ be a cyclic shift on $\begin{array}{}{R}_{k}^{n}\end{array}$, let Φ be the Gray map from $\begin{array}{}{R}_{k}^{n}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}to\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\mathbb{F}}_{{p}^{m}}^{{2}^{k}n}\end{array}$, and let τ2k be as above. Then Φσ = τ2kΦ.

#### Proof

Let r = (r0, r1, ⋯, rn−1) ∈ $\begin{array}{}{R}_{k}^{n}\end{array}$, where $\begin{array}{}{r}_{i}=\sum _{j=1}^{{2}^{k}}\end{array}$ rijej, i = 0, 1, ⋯, n − 1. We have σ(r) = (rn−1, r0, ⋯, rn−2). If we apply Φ, we have

$Φ(σ(r))=Φ(rn−1,r0,⋯,rn−2)=(r1,n−1,r1,0,⋯,r1,n−2,r2,n−1,r2,0,⋯,r2,n−2,⋯,r2k,n−1,r2k,0,⋯,r2k,n−2).$

On the other hand,

$τ2k(Φ(r))=τ2k(Φ(r0,r1,⋯,rn−1))=τ2k(r1,0,r1,1,⋯,r1,n−1,r2,0,r2,1,⋯,r2,n−1,⋯,r2k,0,r2k,1,⋯,r2k,n−1)=(r1,n−1,r1,0,⋯,r1,n−2,r2,n−1,r2,0,⋯,r2,n−2,⋯,r2k,n−1,r2k,0,⋯,r2k,n−2).$

Therefore, we have

$Φσ=τ2kΦ.$

#### Theorem 3.7

Let C be a cyclic code of length n over Rk. Then Φ(C) is a quasi-cyclic code of index 2k over 𝔽pm with length 2kn.

#### Proof

Since C is a cyclic code, then σ(C) = C. If we apply Φ, we have Φσ(C) = Φ(C). By the Proposition 3.6, Φ(σ(C)) = Φ(C) = τ2k(Φ(C)), so Φ(C) is a quasi-cyclic code of index 2k over 𝔽pm with length 2kn. □

Let C be a linear code of length n over Rk, let A0, A1, ⋯, A2kn denote the number of codewords in C of the Lee weight, and the Lee weight distribution of C is simply the tuple of numbers {A0, A1, ⋯, A2kn}.

Let LeeC(x, y) = $\begin{array}{}\sum _{i=0}^{{2}^{k}n}{A}_{i}{x}^{{2}^{k}n-i}{y}^{i}\end{array}$ denote the Lee weight enumerator of C, we get that

$LeeC(x,y)=∑c∈Cx2kn−wL(c)ywL(c).$

Let WC(x, y) = $\begin{array}{}\sum _{c\in C}{x}^{{2}^{k}n-{w}_{H}\left(c\right)}{y}^{{w}_{H}\left(c\right)}\end{array}$ denote the Hamming weight enumerator of C.

By the results of [22], we have

$WC⊥(x,y)=1|C|WC(x+(|Rk|−1)y,x−y).$

By a proof similar to (cf. [23, Lemma 1]), we obtain the following lemma.

#### Lemma 3.8

Let x and y be two vectors in $\begin{array}{}{R}_{k}^{n}\end{array}$, and let dH(Φ(x), Φ(y)) denote the Hamming distance of Φ(x),Φ(y), where Φ(x), Φ(y) are codewords in $\begin{array}{}{\mathbb{F}}_{{p}^{m}}^{{2}^{k}n}\end{array}$. Let wH(Φ(x)) denote the Hamming weight of Φ, then

1. wL(x) = wH(Φ(x)),

2. dL(x, y) = dH(Φ(x), Φ(y)).

#### Theorem 3.9

Let C be a linear code of length n over Rk, then LeeC(x, y) = $\begin{array}{}\frac{1}{|\mathit{\Phi }\left(C\right)|}\end{array}$ WΦ(C)(x + (pm2k − 1)y, xy).

#### Proof

By Theorem 3.5, we have that

$LeeC⊥(x,y)=WΦ(C⊥)(x,y)=WΦ(C)⊥(x,y).$

So

$LeeC(x,y)=∑c∈Cx2kn−wL(c)ywL(c)=∑Φ(c)⊆Φ(C)x2kn−wH(Φ(c))ywH(Φ(c))=WΦ(C)(x,y).$

As Φ is one-to-one, we have that |Φ(C)| = |C|, hence

$LeeC⊥(x,y)=WΦ(C⊥)(x,y)=1|Φ(C)|WΦ(C)(x+(pm2k−1)y,x−y).$

## 4 λ-Constacyclic codes over Rk

#### Theorem 4.1

Let C = e1C1 + e2C2 + ⋯ + e2kC2k be a linear code over Rk, then C is a1e1 + λ2e2 + ⋯ + λ2ke2k)-constacyclic code over Rk if and only if C1, C2, ⋯, C2k are λi-constacyclic codes over 𝔽pm, where λ1e1 + λ2e2 + ⋯ + λ2ke2k is a unit over Rk.

#### Proof

For any ci = (ci,0, ci,1, ⋯, ci,n−1) ∈ Ci, where i = 1, 2,⋯, 2k. Then

$c=e1c1+e2c2+⋯+e2kc2k=(∑i=12keici,0,∑i=12keici,1,⋯,∑i=12keici,n−1)∈C.$

If λ1e1 + λ2e2 + ⋯ + λ2ke2k is a unit over Rk, it is easy to know that for any element r = r1e1 + r2e2 + ⋯ + r2ke2kRk, r is a unit if and only if ri ≠ 0, where i = 1, 2,⋯, 2k.

For i = 1, 2,⋯, 2k, if Ci is a λi-constacyclic code over 𝔽pm, then

$σλi(ci)=σλi(ci,0,ci,1,⋯,ci,n−1)=(λici,n−1,ci,0,⋯,ci,n−2)∈Ci.$

Then we have

$σλ1e1+λ2e2+⋯+λ2ke2k(c)=((λ1e1+λ2e2+⋯+λ2ke2k)∑i=12keici,n−1,∑i=12keici,0,⋯,∑i=12keici,n−2)=e1σλ1(c1)+e2σλ2(c2)+⋯+e2kσλ2k(c2k)∈C.$

This proves that C is a (λ1e1 + λ2e2 + ⋯ + λ2ke2k)-constacyclic code over Rk.

Conversely, if C is a (λ1e1 + λ2e2 + ⋯ + λ2ke2k)-constacyclic code over Rk, then

$σλ1e1+λ2e2+⋯+λ2ke2k(c)=e1σλ1(c1)+e2σλ2(c2)+⋯+e2kσλ2k(c2k)∈C.$

Thus σλi(ci) ∈ Ci, where i = 1, 2,⋯, 2k.

So Ci is a λi-constacyclic code over 𝔽pm, where i = 1, 2,⋯, 2k. □

#### Theorem 4.2

Let C = e1C1 + e2C2 + ⋯ + e2kC2k be a1e1 + λ2e2 + ⋯ + λ2ke2k)-constacyclic code of length n over Rk, then there exists a polynomial e1g1(x) + e2g2(x) + ⋯ + e2kg2k(x) in Rk[x] that divides xn-(λ1e1 + λ2e2 + ⋯ + λ2ke2k) generates the code, where gi is the generator polynomial of Ci, i = 1, 2,⋯, 2k.

#### Proof

If C = e1C1 + e2C2 + ⋯ + e2kC2k be a (λ1e1 + λ2e2 + ⋯ + λ2ke2k)-constacyclic n over Rk, by Theorem 4.1 we know that Ci is λi-constacyclic code over 𝔽pm, where i = 1, 2,⋯, 2k. Let gi be the generator polynomial of Ci, where i = 1, 2,⋯, 2k. It follows that C has the form

$C=〈e1g1(x),e2g2(x),⋯,e2kg2k(x)〉.$

Let C′ = 〈e1g1(x) + e2g2(x) + ⋯ + e2kg2k(x)〉. We have that C′ ⊆ C.

Note that

$ei[(e1g1(x)+e2g2(x)+⋯+e2kg2k(x)]=eigi(x),$

where i = 1, 2,⋯, 2k.

We get that CC′. So C = C′, and C is generated by a single element g(x) = e1g1(x) + e2g2(x) + ⋯ + e2kg2k(x).

We know that gi divides xni, since gi is the generator polynomial of Ci, where i = 1, 2,⋯, 2k. Let fi(x) be the polynomial such that gi(x)fi(x) = xni, where i = 1, 2,⋯, 2k.

Then we have

$[e1g1(x)+e2g2(x)+⋯+e2kg2k(x)][e1f1(x)+e2f2(x)+⋯+e2kf2k(x)]=λ1e1+λ2e2+⋯+λ2ke2k.$

So we have e1g1(x) + e2g2(x) + ⋯ + e2kg2k(x) in Rk[x] that divides xn-(λ1e1 + λ2e2 + ⋯ + λ2ke2k). □

By Theorem 4.2 we have the following theorem easily:

#### Theorem 4.3

Let C = e1C1 + e2C2 + ⋯ + e2kC2k be a1e1 + λ2e2 + ⋯ + λ2ke2k)-constacyclic code of length n over Rk. Then $\begin{array}{}{C}^{\perp }=〈{e}_{1}{f}_{1}^{\ast }\left(x\right)+{e}_{2}{f}_{2}^{\ast }\left(x\right)+\cdots +{e}_{{2}^{k}}{f}_{{2}^{k}}^{\ast }\left(x\right)〉,|{C}^{\perp }|={p}^{m\left(\sum _{i=1}^{{2}^{k}}\mathrm{d}\mathrm{e}\mathrm{g}\left({g}_{i}\right)\right)},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{f}_{i}^{\ast }\end{array}$ (x) is the reciprocal polynomial of fi(x),i.e., fi(x) = (xni)/gi(x), $\begin{array}{}{f}_{i}^{\ast }\end{array}$ (x) = xdeg(fi) f(x−1), for i = 1, 2,⋯, 2k.

#### Example 4.4

Let n = 10 and R2 = 𝔽3 + u1𝔽3 + u2𝔽3 + u1u2𝔽3, λ = − 1, x10 + 1 = (x2 + 1)(x4 + x3 + 2x + 1)(x4 + 2x3 + x + 1) in 𝔽3(x). Let f1(x) = f2(x) = (x4 + x3 + 2x + 1), f3(x) = f4(x) = (x4 + 2x3 + x + 1), C = 〈(1 + u1 + u2 + u1u2)f1(x),(u1 + u1u2)f2(x),(u2 + u1u2)f3(x),(u1u2)f4(x)〉. C1, C2, C3, C4 are [10, 6,4] linear codes of length 10 with the minimum Lee weight dL = 4. So Φ(C) is a [40, 24,4] linear code.

#### Example 4.5

Let n = 15 and R3 = 𝔽2[u1, u2, u3]/〈$\begin{array}{}{u}_{i}^{2}\end{array}$ = ui, uiuj = ujui〉, x15 − 1 = (x + 1)(x2 + x + 1)(x4 + x + 1)(x4 + x3 + 1)(x4 + x3 + x2 + x + 1) in 𝔽2(x). Let f1(x) = f2(x) = f3(x) = f4(x) = (x4 + x + 1), f5(x) = f6(x) = f7(x) = f8(x) = (x4 + x3 + 1), C = 〈 $\begin{array}{}\prod _{i=1}^{3}\end{array}$ (1 + ui)f1(x), u1(1 + u2)(1 + u3)f2(x), u2(1 + u1)(1 + u3)f3(x), u3(1 + u1)(1 + u2)f4(x), u1u2(1 + u3)f5(x), u1u3(1 + u2)f6(x), u2u3(1 + u1)f7(x), u1u2u3f8(x)〉. Ci is a [15, 11,3] linear code of length 15 with the minimum Lee weight dL = 3, i = 1, 2, ⋯, 8. So Φ(C) is a [120, 88,3] linear code.

## 5 Conclusion

In this paper, we studied the constacyclic codes over Rk = 𝔽pm[u1, u2,⋯, uk]/〈$\begin{array}{}{u}_{i}^{2}\end{array}$ = ui, uiuj = ujui〉. We proved that the (λ1e1 + λ2e2 + ⋯ + λ2ke2k)-constacyclic codes of arbitrary length over Rk can be generated by one polynomial.

## Acknowledgement

This work was supported by the Basic and Advanced Technology Research Project of Henan Province (No.162300410083) and the Science and Technology Developing Project of Henan Province(No.172102210243).

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Accepted: 2018-03-22

Published Online: 2018-05-10

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 490–497, ISSN (Online) 2391-5455,

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