On cyclic codes over the ring $\mathbb Z _p[u]/\langle u^k\rangle$

Abstract

In this paper, we study cyclic codes over the ring $\mathbb Z _p[u]/\langle u^k\rangle .$ We find a set of generators for these codes. We also study the rank and the Hamming distance of these codes.

1 Introduction

Let $R$ be a ring. A linear code of length $n$ over $R$ is a $R$ submodule of $R^{n}.$ A linear code $C$ of length $n$ over $R$ is cyclic if $(c_{n-1},\,c_0,\ldots ,c_{n-2}) \in C$ whenever $(c_0,\,c_1,\ldots ,c_{n-1}) \in C.$ We can consider a cyclic code $C$ of length $n$ over $R$ as an ideal in $R[x]/\langle x^n - 1\rangle$ via the following correspondence

$$\begin{aligned} R^{n} \longrightarrow R[x]/\langle x^n - 1\rangle ,\quad \left( c_0,\,c_1,\ldots , c_{n-1}\right) \mapsto c_0 + c_1x + \cdots + c_{n-1}x^{n-1}. \end{aligned}$$

In recent time, cyclic codes over rings have been studied extensively because of their important role in algebraic coding theory. The structure of cyclic codes of odd length over rings has been discussed in a series of papers [6, 8, 11, 14]. In [7, 9, 13], a complete structure of cyclic codes of odd length over $\mathbb Z _4$ has been presented. In [5], Blackford studied cyclic codes of length $n=2k,$ when $k$ is odd. The cyclic codes of length a power of 2 over $\mathbb Z _4$ are studied in [1, 3]. The structures of cyclic codes of length $n$ over a finite chain ring $R$ has been discussed in [10] when $n$ is not divisible by the characteristic of the residue field $\bar{R}.$ Bonnecaze and Udaya [6] studied cyclic codes of odd length over $R_{2,2}=\mathbb Z _2 + u \mathbb Z _2,\,u^2 = 0.$ In [2], Abualrub and Siap studied cyclic codes of an arbitrary length over $R_{2,2}=\mathbb Z _2+u \mathbb Z _2,\,u^2=0$ and over $R_{3,2}=\mathbb Z _2 + u \mathbb Z _2 + u^2 \mathbb Z _2,\,u^3 =0.$ Al-Ashker and Hamoudeh [4] extended some of the results in [2] to the ring $R_{k,2}=\mathbb Z _2 + u\mathbb Z _2 +\cdots + u^{k-1}\mathbb Z _2,\,u^k=0.$

Let $R_{k,p} = \mathbb Z _p + u\mathbb Z _p + \cdots + u^{k-1}\mathbb Z _p$ where $p$ is a prime number and $u^k = 0.$ Note that the ring $R_{k,p}$ can also be viewed as the quotient ring $\mathbb Z _p[u]/\langle u^k\rangle .$ In this paper, we discuss the structure of cyclic codes of arbitrary length over the ring $R_{k,p}.$ We find a set of generators and a minimal spanning set for these codes. We also discuss about the rank and the Hamming distance of these codes. Recall that the Hamming weight of a codeword $c$ is defined as the number of non-zero entries of $c$ and the Hamming distance of a code $C$ is the smallest possible weight among all its non-zero codewords. The minimum distance of a code is the minimum Hamming distance between two distinct codewords. When the code is linear, the minimum distance of the code is equal to the Hamming distance. The minimum distance determines the maximum number of errors that can be corrected under any decoding algorithm.

Let $C$ be a cyclic code over the ring $R_{k,p} = \mathbb Z _p + u\mathbb Z _p +\cdots + u^{k-1}\mathbb Z _p,\,u^k = 0.$ The line of arguments we have used to find a set of generators and a minimal spanning set of a code $C$ are somewhat similar to those discussed in [2]. The idea to find a set of generators is as follows. We view the cyclic code $C$ as an ideal in $R_{k,p,n}= R_{k,p}/\langle x^n - 1\rangle .$ Then we define the projection map from $R_{i,p,n}\longrightarrow R_{i-1,p,n}$ for each $i >1.$ For $i=2,$ we have the projection map from $R_{2,p,n}\longrightarrow R_{1,p,n}.$ Note that $R_{1,p,n}$ is nothing but $\mathbb Z _p[x]/\langle x^n - 1\rangle$ and an ideal in $R_{1,p,n}$ gives a cyclic code over $\mathbb Z _p.$ The structure of cyclic codes over $\mathbb Z _p$ is well known. By pullback, we find a set of generators for a cyclic code over $R_{2,p}$ and inductively we get a set of generators for a cyclic code over $R_{k,p}$ for all $k.$ Again, the line of arguments we have used to find minimum distance are similar to [2] but slightly different.

This paper is organized as follows. In Sect. 2, we give a set of generators for the cyclic codes $C$ over the ring $R_{k,p} = \mathbb Z _p + u\mathbb Z _p + \cdots + u^{k-1}\mathbb Z _p,\,u^k = 0.$ In Sect. 3, we find a minimal spanning set for these codes and discuss about the rank. In Sect. 4, we find the minimum distance of these codes. In Sect. 5, we discuss some of the examples of these codes.

2 Preliminaries

Let $R$ be a finite commutative ring. We have the following equivalent conditions.

Proposition 2.1

[10, Proposition 2.1] The following conditions are equivalent for a finite commutative ring $R.$

• (1) $R$ is a local ring and the maximal ideal $M$ of $R$ is principal;

• (2) $R$ is a local principal ideal ring;

• (3) $R$ is a chain ring.

Let $R$ be a finite commutative local ring with maximal ideal $M.$ Let $\bar{R} = R/M.$ Let $\mu :\,R[x] \longrightarrow \bar{R}[x]$ denote the natural ring homomorphism that maps $r \mapsto r + M$ and the variable $x$ to $x.$ We define the degree of the polynomial $f(x) \in R[x]$ as the degree of the polynomial $\mu (f(x))$ in $\bar{R}[x],$ i.e., $\text{ deg }(f(x)) =\text{ deg }(\mu (f(x))$ (see, for example, [12]). A polynomial $f(x) \in R[x]$ is called regular if it is not a zero divisor. The following proposition is well known.

Proposition 2.2

(cf. [12, Exercise XIII.2(c)]) Let $R$ be a finite commutative chain ring. Let $f(x) = a_0 + a_1 x + \cdots + a_n x^n$ be in $R[x],$ then the following are equivalent.

• (1) $f(x)$ is regular,

• (2) $\langle a_0,\,a_1,\ldots , a_n\rangle = R,$

• (3) $a_i$ is a unit for some $i,\,0 \le i \le n,$

• (4) $\mu (f(x)) \ne 0.$

The following version of the division algorithm holds true for polynomials over finite commutative local rings.

Proposition 2.3

(cf. [12, Exercise XIII.6]) Let $R$ be a finite commutative local ring. Let $f(x)$ and $g(x)$ be non-zero polynomials in $R[x].$ If $g(x)$ is regular, then there exist polynomials $q(x)$ and $r(x)$ in $R[x]$ such that $f(x) = g(x)q(x) + r(x)$ and $\mathrm{deg} (r(x)) < \mathrm{deg}(g(x)).$

Let $R_{k,p} = \mathbb Z _p + u\mathbb Z _p + \cdots + u^{k-1}\mathbb Z _p,\,u^k = 0.$ It is easy to see that the ring $R_{k,p}$ is a finite chain ring with unique maximal ideal $\langle u\rangle .$ Let $g(x)$ be a non-zero polynomial in $\mathbb Z _p[x].$ By Proposition 2.2, it is also easy to see that the polynomial $g(x) + u p_1(x) + u^2 p_{2}(x) + \cdots + u^{k-1} p_{k-1}(x) \in R_{k,p}[x]$ is regular. Throughout the paper, we repeatedly make use of Proposition 2.3 for the polynomial $g(x) + u p_1(x) + u^2 p_{2}(x) + \cdots + u^{k-1} p_{k-1}(x) \in R_{k,p}[x].$ Note that $\text{ deg }(g(x) + u p_1(x) + u^2 p_{2}(x) + \cdots + u^{k-1} p_{k-1}(x)) = \text{ deg }(g(x)).$

3 A generator for cyclic codes over the ring $R_{k,p}$

Let $p$ be a prime number. Let $R_{k,p} = \mathbb Z _p + u\mathbb Z _p + \cdots + u^{k-1}\mathbb Z _p,\, u^k = 0$ and $R_{k,p,n} = R_{k,p}[x]/\langle x^n - 1\rangle .$ Let $C_k$ be a cyclic code of length $n$ over $R_{k,p}.$ We also consider $C_k$ as an ideal in $R_{k,p,n}.$ We define the map $\psi _{k-1}:\,R_{k,p} \longrightarrow R_{k-1,p}$ by $\psi _{k-1}(b_0+ub_1+ \cdots +u^{k-1}b_{k-1}) = b_0+ub_1+ \cdots +u^{k-2}b_{k-2},$ where $b_i \in \mathbb Z _p.$ The map $\psi _{k-1}$ is a ring homomorphism. We extend it to a homomorphism $\phi _{k-1}:\,C_k \longrightarrow R_{k-1,p,n}$ defined by

$$\begin{aligned} \phi _{k-1}\left( c_0 + c_1x + \cdots + c_{n-1}x^{n-1}\right) = \psi _{k-1}\left( c_0\right) +\psi _{k-1}\left( c_1\right) x + \cdots + \psi _{k-1}\left( c_{n-1}\right) x^{n-1}, \end{aligned}$$

where $c_i \in R_{k,p}.$ Let $J_{k-1}=\{r(x) \in \mathbb Z _p[x]:\,u^{k-1}r(x) \in \text{ ker } \phi _{k-1}\}.$ It is easy to see that $J_{k-1}$ is an ideal in $R_{1,p,n}.$ Since $R_{1,p,n}$ is a principal ideal ring, we have $J_{k-1} = \langle a_{k-1}(x)\rangle$ for some $a_{k-1}(x) \in \mathbb Z _p[x],$ and $\text{ ker }\phi _{k-1} = \langle u^{k-1}a_{k-1}(x)\rangle$ with $a_{k-1}(x)|(x^n-1)$ mod $p.$

Let $C_{k-1}$ be a cyclic code of length $n$ over $R_{k-1,p}.$ We define the map $\psi _{k-2}:\,R_{k-1,p} \longrightarrow R_{k-2,p}$ by $\psi _{k-2}(b_0+ub_1+ \cdots +u^{k-2}b_{k-2})=b_0+ub_1+ \cdots +u^{k-3}b_{k-3},$ where $b_i \in \mathbb Z _p.$ The map $\psi _{k-2}$ is a ring homomorphism. We extend it to a homomorphism $\phi _{k-2}:\, C_{k-1} \longrightarrow R_{k-2,p,n}$ defined by

$$\begin{aligned} \phi _{k-2}\left( c_0 + c_1x + \cdots + c_{n-1}x^{n-1}\right) = \psi _{k-2}\left( c_0\right) +\psi _{k-2}\left( c_1\right) x + \cdots + \psi _{k-2}\left( c_{n-1}\right) x^{n-1}, \end{aligned}$$

where $c_i \in R_{k-1,p}.$ Let $J_{k-2} = \{r(x) \in \mathbb Z _p[x]:\,u^{k-2}r(x) \in \text{ ker } \phi _{k-2}\}.$ We see that $J_{k-2}$ is an ideal in $R_{1,p,n}.$ As above, we have $J_{k-2} = \langle a_{k-2}(x)\rangle$ for some $a_{k-2}(x) \in \mathbb Z _p[x],$ and $\text{ ker }\phi _{k-2} = \langle u^{k-2}a_{k-2}(x)\rangle$ with $a_{k-2}(x)|(x^n-1)$ mod $p.$

We continue in the same way as above and define $\psi _{k-3},\,\psi _{k-4},\ldots ,\psi _2$ and $\phi _{k-3},\, \phi _{k-4},\ldots ,\phi _2.$ We define $\psi _1:\,R_{2,p} \longrightarrow R_{1,p} = \mathbb Z _p$ by $\psi _{1}(b_0 + ub_1) = b_{0},$ where $b_0,\,b_1 \in \mathbb Z _p.$ The map $\psi _1$ is a ring homomorphism. We extend $\psi _1$ to a homomorphism $\phi _1:\, C_2 \longrightarrow R_{1,p,n}$ defined by

$$\begin{aligned} \phi _1\left( c_0 + c_1x + \cdots + c_{n-1}x^{n-1}\right) = \psi _1\left( c_0\right) + \psi _1\left( c_1\right) x + \cdots + \psi _1\left( c_{n-1}\right) x^{n-1}, \end{aligned}$$

where $c_i \in R_{2,p}.$ As above, we have $\text{ ker }\phi _1 = \langle u a_1(x)\rangle$ for some $a_1(x) \in \mathbb Z _p[x],$ with $a_1(x)|(x^n - 1)$ mod $p.$ The image of $\phi _1$ is an ideal in $R_{1,p,n}$ and hence a cyclic code in $\mathbb Z _p.$ Since $R_{1,p,n}$ is a principal ideal ring, the image of $\phi _1$ is generated by some $g(x) \in \mathbb Z _p[x]$ with $g(x)|(x^n - 1).$ Hence, we have $C_2 = \langle g(x) + u p_1(x),\,u a_1(x)\rangle$ for some $p_1(x) \in \mathbb Z _p[x].$ We have

$$\begin{aligned} \phi _1\left( \frac{x^n - 1}{g(x)}\left( g(x) + u p_1(x)\right) \right) = \phi _1\left( u p_1(x) \frac{x^n - 1}{g(x)}\right) = 0. \end{aligned}$$

Therefore, $u p_1(x)\left( \frac{x^n - 1}{g(x)}\right) \in \text{ ker }\phi _1 = \langle u a_1(x)\rangle .$ Hence, $a_1(x)|p_1(x)\left( \frac{x^n -1}{g(x)}\right) .$ Also we have $u g(x) \in \text{ ker }\phi _1.$ This implies that $a_{1}(x)|g(x).$

Lemma 3.1

Let $C_2$ be a cyclic code over $R_{2,p} = \mathbb Z _p + u \mathbb Z _p,\,u^2 = 0.$ If $C_2 = \langle g(x) + u p_1(x),\,u a_1(x)\rangle ,$ and $g(x) = a_1(x)$ with $\mathrm{deg}g(x) = r,$ then

$$\begin{aligned} C_2 = \langle g(x) + u p_1(x)\rangle \quad \text{ and }\; \left( g(x) + u p_1(x)\right) |\left( x^n - 1\right) \;\text{ in }\; R_{2,p}. \end{aligned}$$

Proof

We have $u (g(x) + u p_1(x)) = u g(x)$ and $g(x) = a_1(x).$ It is clear that $C_2 \subset \langle g(x) + u p_1(x)\rangle .$ Hence, $C_2 = \langle g(x) + u p_1(x)\rangle .$ By the division algorithm, we have

$$\begin{aligned} x^n - 1 = \left( g(x) + u p_1(x)\right) q(x) + r(x),\quad \text{ where }\;r(x) = 0\quad \text{ or } \text{ deg }r(x)<r. \end{aligned}$$

This implies that $r(x) = (x^n - 1) - (g(x) + u p_1(x)) q(x).$ This gives, $r(x) \in C_2.$ Thus, we have $r(x) = 0$ and hence $(g(x) + u p_1(x))|(x^n - 1)\;\text{ in }\;R_{2,p}.$ $\square$

Note that the image of $\phi _2$ is an ideal in $R_{2,p,n},$ hence a cyclic code over $R_{2,p}.$ Therefore, we have $\text{ Im }(\phi _2) = \langle g(x) + u p_1(x),\,u a_1(x)\rangle$ with $a_1(x)|g(x)|(x^n - 1)$ and $a_1(x)|p_1(x)\left( \frac{x^n - 1}{g(x)}\right) .$ Also, we have $\text{ ker }\phi _2 = \langle u^2 a_2(x)\rangle$ with $a_2(x)|(x^n - 1)$ mod $p$ and $u^2 a_1(x) \in \text{ ker }\phi _2.$ As above, the cyclic code $C_3$ over $R_{3,p}$ is given by

$$\begin{aligned} C_3 = \Big \langle g + u p_1(x) + u^2 p_{2}(x),\,u a_1(x) + u^2 q_1(x),\,u^2 a_2(x)\Big \rangle , \end{aligned}$$

for some $p_2(x),\,q_1(x) \in \mathbb Z _p[x],$ with $a_2(x)|a_1(x)|g(x)|(x^n - 1),\,a_1(x)|p_1(x)\left( \frac{x^n - 1}{g(x)}\right)$ mod $p,\,a_2(x)|q_1(x) \left( \frac{x^n - 1}{a_1(x)}\right) ,\,a_2(x)|p_1(x) \left( \frac{x^n - 1}{g(x)}\right)$ and $a_2(x)|p_2(x) \left( \frac{x^n - 1}{g(x)}\right) \left( \frac{x^n - 1}{a_1(x)}\right) .$ We may assume that deg$p_2(x) < \text{ deg } a_2(x),\,\text{ deg } q_1(x) < \text{ deg } a_2(x)$ and deg$p_1(x) < \text{ deg } a_1(x)$ because g.c.d. $(a,\,b)=g.c.d.(a,\,b+da)$ for any $d.$ We have the following lemma.

Lemma 3.2

Let $C_3$ be a cyclic code over $R_{3,p} = \mathbb Z _p + u \mathbb Z _p + u^2\mathbb Z _p,\,u^3 = 0.$ If $C_3 = \langle g + u p_1(x) + u^2 p_{2}(x),\,u a_1(x) +u^2 q_1(x),\,u^2 a_2(x)\rangle ,$ and $a_2(x) = g(x),$ then $C_3 = \langle g + up_1(x) + u^2 p_{2}(x)\rangle ,\,(g(x) + u p_1(x))|(x^n - 1)$ in $R_{2,p}$ and $(g + u p_1(x) + u^2 p_{2}(x))|(x^n -1)$ in $R_{3,p}.$

Proof

Since $a_2(x) = g(x),$ we get $a_1(x) = a_2(x) = g(x).$ We have $\phi _2(C_3) = \langle g + u p_1(x),\,u a_1(x)\rangle .$ From Lemma 3.1, we get $(g(x) + u p_1(x))| (x^n - 1)$ in $R_{2,p},$ and $\phi _2(C_3) = \langle g + u p_1(x)\rangle .$ This gives, $C_3 =\langle g + u p_1(x) + u^2 p_{2}(x),\,u^2 a_2(x)\rangle .$ The rest of the proof is similar to Lemma 3.1. $\square$

If we continue in the same way as above, we can see that the image of $\phi _{k-1}$ is an ideal in $R_{k-1,p,n},$ hence a cyclic code over $R_{k-1,p}.$ By induction hypothesis, we can assume that $\text{ Im }(\phi _{k-1}) = \langle g + u p_1(x) + u^2 p_{2}(x) + \cdots + u^{k-2} p_{k-2}(x),\,u a_1(x) + u^2 q_1(x) + \cdots + u^{k-2} q_{k-3}(x),\,u^2 a_2(x) + u^3l_{1}(x) + \cdots + u^{k-2}l_{k-4}(x),\ldots , u^{k-3} a_{k-3}(x)+ u^{k-2} s_1(x),\,u^{k-2} a_{k-2}(x)\rangle$ with $a_{k-2}(x)|\cdots |a_2(x)|a_1(x)|g(x)|(x^n - 1)$ mod $p,\,a_{k-3}(x)|p_1 (x)\left( \frac{x^n - 1}{g(x)}\right) ,\ldots , a_{k-2}|s_1(x)\left( \frac{x^n - 1}{a_{k-3}(x)}\right) ,\ldots ,a_{k-2}|p_{k-2}\left( \frac{x^n - 1}{g(x)}\right) \cdots \left( \frac{x^n - 1}{a_{k-3}(x)}\right) .$ Also, we have $\text{ ker }\phi _{k-1} = \langle u^{k-1} a_{k-1}(x)\rangle$ with $a_{k-1}(x)|(x^n - 1)$ mod $p$ and $u^{k-1} a_{k-2}(x) \in \text{ ker }\phi _{k-1}.$ Following the same process as above and by induction on $k,$ we get the following theorem.

Theorem 3.3

Let $C_k$ be a cyclic code over $R_{k,p} = \mathbb Z _p + u \mathbb Z _p + u^2 \mathbb Z _p+ \cdots + u^{k-1} \mathbb Z _p,\,u^k = 0.$ If $n$ is not relatively prime to $p,$ then

• (1) $C_k = \langle g(x) + u p_1(x) + u^2 p_{2}(x) + \cdots + u^{k-1} p_{k-1}(x)\rangle$ where $g(x)$ and $p_i(x)$ are polynomials in $\mathbb Z _p[x]$ with $g(x)|(x^n - 1)$ mod $p,\,(g(x) + u p_1(x) + u^2 p_{2}(x) + \cdots + u^{k-1} p_{i-1}(x))|(x^n - 1)$ in $R_{i,p}$ and $\mathrm{deg} p_i < \mathrm{deg} p_{i-1}$ for all $1 < i \le k.$ Or

• (2) $C_k = \langle g + u p_1(x) + u^2 p_{2}(x) + \cdots + u^{k-1} p_{k-1}(x),\,u a_1(x) + u^2 q_1(x) + \cdots + u^{k-1} q_{k-2}(x),\,u^2 a_2(x) + u^3 l_{1}(x) + \cdots + u^{k-1}l_{k-3}(x),\ldots , u^{k-2} a_{k-2}(x) + u^{k-1} t_1(x),\,u^{k-1} a_{k-1}(x)\rangle$ with $a_{k-1}(x)|a_{k-2}(x)|\cdots |a_2(x)|a_1(x)|g(x)|(x^n - 1)$ mod $p,\,a_{k-2}(x)|$ $p_1\,(x)\,\left( \frac{x^n - 1}{g(x)}\right) ,\,\,\ldots ,\,\, a_{k-1}|\,t_1\,(x)\left( \frac{x^n - 1}{a_{k-2}(x)}\right) ,\,\,\ldots ,\,\,a_{k-1}|p_{k-1} \left( \frac{x^n - 1}{g(x)}\right) \,\,\cdots \,\,\left( \frac{x^n - 1}{a_{k-2}(x)}\right) .$ Moreover, $\mathrm{deg} p_{k-1}(x) < \mathrm{deg}a_{k-1}(x),\cdots , \mathrm{deg} t_1(x) < a_{k-1}(x),\cdots ,$ and $\mathrm{deg} p_1(x) < \mathrm{deg}a_{k-2}(x).$

Note that if we have $a_{k-1}(x) \!=\! g(x)$ in part 2 of the above theorem then we get part 1. If we have $a_{k-1} \!\ne \! g(x)$ but $a_{k-2} \!=\! g(x)$ then $C_k \!=\! \langle g(x) \!+\! u p_1(x) \!+\! u^2 p_{2}(x) \!+\! \cdots \!+\! u^{k-1} p_{k-1}(x),\,u^{k-1}a_{k-1}(x)\rangle$ where $a_{k-1}(x)|g(x)|(x^n - 1)$ mod $p,\,g(x) \!+\! u p_1(x) \!+\! \cdots \!+\! u^{i-1}p_{i-1}|(x^n - 1)$ in $R_{i,p}$ for $1\!<\! i \!\le \! k-1,\,g(x)|p_1(x)\left( \frac{x^n-1}{g(x)}\right)$ and $a_{k-1}(x)|p_1(x)\left( \frac{x^n-1}{g(x)}\right) ,\,a_{k-1}(x)|p_2(x)\left( \frac{x^n-1}{g(x)}\right) \left( \frac{x^n-1}{g(x)}\right) ,\ldots , a_{k-1}(x)|p_{k-1}(x) \tiny {\underbrace{\left( \frac{x^n-1}{g(x)}\right) \ldots \left( \frac{x^n-1}{g(x)}\right) }_{k-1~\mathrm{times}}}$ and $\mathrm{deg}p_{k-1}(x) < \mathrm{deg}a_{k-1}(x).$ Similarly we get the simpler form for $C_k$ if we have $a_{k-1},\,a_{k-2},\ldots , a_{i} \ne g(x)$ but $a_{i-1} = g(x)$ for $i > 1.$

If $n$ is relatively prime to $p,$ then the following theorem follows from [10, Theorems 3.4–3.6].

Theorem 3.4

Let $C_k$ be a cyclic code over $R_{k,p} = \mathbb Z _p + u \mathbb Z _p + u^2 \mathbb Z _p+ \cdots + u^{k-1} \mathbb Z _p,\,u^k = 0.$ If $n$ is relatively prime to $p,$ then we have $C_k = \langle g(x),\,u a_1(x),\,u^2 a_2(x),\ldots ,u^{k-1} a_{k-1}(x)\rangle$ = $\langle g(x) + u a_1(x) + u^2 a_2(x) + \cdots + u^{k-1} a_{k-1}(x)\rangle$ over $R_{k,p}.$

4 Ranks and minimal spanning sets

Theorem 4.1

Let $n$ is not relatively prime to $p.$ Let $C_2$ be a cyclic code of length $n$ over $R_{2,p} = \mathbb Z _p + u \mathbb Z _p,\,u^2 = 0.$

• (1) If $C_2 = \langle g(x) + u p(x)\rangle$ with deg$g(x)=r$ and $(g(x) + u p(x))|(x^n - 1),$ then $C_2$ is a free module with rank $n-r$ and a basis $B_1 = \{g(x) + u p(x),\,x(g(x) + u p(x)),\ldots ,x^{n-r-1}(g(x) + up(x))\},$ and $|C_2| = p^{2n-2r}.$

• (2) If $C_2 = \langle g(x) + u p(x),\,ua(x)\rangle$ with deg$g(x) = r$ and deg$a(x) =t,$ then $C_2$ has rank $n-t$ and a minimal spanning set $B_2 = \{g(x) + u p(x),\,x(g(x) + u p(x)),\ldots ,x^{n-r-1}(g(x) + up(x)),\,ua(x),\,xua(x),\ldots , x^{r-t-1}ua(x)\},$ and $|C_2| =p^{2n-r-t}.$

Proof

(1) Suppose $x^n - 1= (g(x) + u p(x))(h(x) + u h_1(x))$ over $R_{2,p}.$ Let $c(x) \in C_2 = \langle g(x) + u p(x)\rangle ,$ then $c(x) = (g(x) + up(x))f(x)$ for some polynomial $f(x).$ If deg$f(x) \le n-r-1,$ then $c(x)$ can be written as linear combinations of elements of $B_1.$ Otherwise by the division algorithm there exist polynomials $q(x)$ and $r(x)$ such that

$$\begin{aligned} f(x) = \left( \frac{x^n - 1}{g(x) + u p(x)}\right) q(x) + r(x)\quad \text{ where }\; r(x) = 0\quad ~\text{ or }\;\text{ deg }r(x) \le n-r-1. \end{aligned}$$

This gives,

$$\begin{aligned} (g(x) + u p(x))f(x)&= (g(x) + u p(x))\left( \left( \frac{x^n - 1}{g(x) + u p(x)}\right) q(x) + r(x)\right) \\&= (g(x) + u p(x))r(x). \end{aligned}$$

Since deg$r(x) \le n-r-1,$ this shows that $B_1$ spans $C_2.$ Now we only need to show that $B_1$ is linearly independent. Let $g(x) = g_0 + g_1x + \cdots + g_rx^r$ and $p(x) = p_0 + p_1x + \cdots +p_lx^l,\,g_0 \in \mathbb Z _p^{\times },\,g_i,\,p_{i-1} \in \mathbb Z _p,\,i \ge 1.$ Suppose

$$\begin{aligned} (g(x)+up(x))c_0 + x(g(x)+up(x))c_1 + \cdots + x^{n-r-1}(g(x)+up(x))c_{n-r-1} = 0. \end{aligned}$$

By comparing the coefficients in the above equation, we get

$$\begin{aligned} \left( g_0 + u p_0\right) c_0 = 0~\text{(constant } \text{ coefficient) }. \end{aligned}$$

Since $(g_0 + u p_0)$ is unit, we get $c_0 = 0.$ Thus,

$$\begin{aligned} x(g(x)+up(x))c_1 + \cdots + x^{n-r-1}(g(x)+up(x))c_{n-r-1} = 0. \end{aligned}$$

Again comparing the coefficients, we get

$$\begin{aligned} \left( g_0 + u p_0\right) c_1 = 0~\text{(coefficient } \text{ of }~x). \end{aligned}$$

As above, this gives $c_1 = 0.$ Continuing in this way we get that $c_i = 0$ for all $i = 0,\,1,\ldots , n-r-1.$ Therefore, the set $B_1$ is linearly independent and hence a basis for $C_2.$ (2) If $C_2 = \langle g(x) + u p(x),\,ua(x)\rangle$ with deg$g(x) = r$ and deg$a(x) =t.$ The polynomial $a(x)$ is the lowest degree polynomial such that $ua(x)\!\in \! C_2.$ It is suffices to show that $B_2$ spans $B =\{ g(x) + u p(x),\,x(g(x) + u p(x)),\ldots , x^{n-r-1}(g(x) + up(x)),\,ua(x),\,xua(x),\ldots ,x^{n-t-1}ua(x)\}.$ We first show that $u x^{r-t} a(x) \in \text{ span }(B_2).$ Let the leading coefficients of $x^{r-t} a(x)$ be $a_0$ and of $g(x) + u p(x)$ be $g_0.$ There exists a constant $c_0 \in \mathbb Z _p$ such that $a_0 = c_0 g_0.$ Then we have

$$\begin{aligned} u x^{r-t} a(x) = u c_0(g(x) + u p(x)) + u m(x), \end{aligned}$$

where $m(x)$ is a polynomial of degree less than $r$ such that $um(x)\in C_2.$ Since $C_2 = \langle g(x) + u p(x),\,ua(x)\rangle ,$ any polynomial $p(x)$ such that $up(x)$ is in $C_2$ must have degree greater or equal to deg$a(x) = t.$ Hence, $t \le \text{ deg }m(x) <r$ and

$$\begin{aligned} um(x) = \alpha _0 u a(x) + \alpha _1 x u a(x) + \cdots +\alpha _{r-t-1} x^{r-t-1} u a(x). \end{aligned}$$

Thus, $u x^{r-t} a(x) \in \text{ span }(B_2).$ Inductively, we can show that $u x^{r-t+1} a(x),\ldots , ux^{n-t-1}a(x) \in \text{ span }(B_2).$ Hence, $B_2$ is a generating set. As in (1), by comparing the coefficients we can see that $B_2$ is linearly independent. Therefore, $B_2$ is a minimal spanning set and $|C_2|=p^{2n-r-t}.$ $\square$

Following the same process as in the above theorem, we can find the rank and the minimal spanning set of any cyclic code over the ring $R_{k,p},\,k \ge 1.$

Theorem 4.2

Let $n$ is not relatively prime to $p.$ Let $C_k$ be a cyclic code of length $n$ over $R_{k,p} = \mathbb Z _p + u \mathbb Z _p + \cdots + u^{k-1} \mathbb Z _p,\,u^k = 0.$ We assume the constraints on the generator polynomials of $C_k$ as in Theorem 3.3.

• (1) If $C_k = \langle g(x) + u p_1(x) + u^2 p_2(x) + \cdots + u^{k-1}p_{k-1}(x)\rangle$ with deg$g(x) = r,$ then $C_k$ is a free module with rank $n-r$ and a basis $B_1 = \{ g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x),\,x(g(x) + u p_1(x) + \cdots +u^{k-1}p_{k-1}(x)),\ldots ,x^{n-r-1}(g(x) + u p_1(x) + \cdots +u^{k-1}p_{k-1}(x))\}.$

• (2) If $C_k = < g(x) + u p_1(x) + u^2 p_{2}(x) + \cdots + u^{k-1} p_{k-1}(x), u a_1(x) + u^2 q_1(x) + \cdots + u^{k-1} q_{k-2}(x), u^2 a_2(x) + u^3 l_{1}(x) + \cdots + u^{k-1}l_{k-3}(x), \dots , u^{k-2} a_{k-2}(x) + u^{k-1} t_1(x), u^{k-1} a_{k-1}(x)>$ with deg $g(x) = r_1$, deg $a_1(x) =r_2$, deg $a_2(x) =r_3, \dots ,$ deg $a_{k-1}(x) =r_k$, then $C_k$ has rank $n-r_k$ and a minimal spanning set $B_2 = \{ g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x), x(g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x)), \dots , x^{n-r_1-1}(g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x)), u a_1(x) + u^2 q_1(x) + \cdots + u^{k-1} q_{k-2}(x), x(u a_1(x) + u^2 q_1(x) + \cdots + u^{k-1} q_{k-2}(x)), \dots , x^{r_1-r_2-1}(u a_1(x) + u^2 q_1(x) + \cdots + u^{k-1} q_{k-2}(x)), u^2 a_2(x) + u^3 l_{1}(x) + \cdots + u^{k-1}l_{k-3}(x), x(u^2 a_2(x) + u^3 l_{1}(x) + \cdots + u^{k-1}l_{k-3}(x)), \dots , x^{r_2-r_3-1}(u^2 a_2(x) + u^3 l_{1}(x) + \cdots + u^{k-1}l_{k-3}(x)), \dots , u^{k-1}a_{k-1}(x), xu^{k-1}a_{k-1}(x), \dots , x^{r_{k-1}-r_k-1}u^{k-1}a_{k-1}(x)\}.$

Proof

(1) The proof is same as in Theorem 4.1. Suppose

$$\begin{aligned} x^n-1=\left( g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x)\right) \left( h(x) + u h_1(x) + \cdots + u^{k-1}h_{k-1}(x)\right) , \end{aligned}$$

over $R_{k,p}.$ Suppose $x^n - 1= (g(x) + u p_1(x))(h(x) + u h_1(x))$ over $R_{2,p}.$ Let $c(x) \in C_k = \langle g(x) + u p_1(x) + u^2 p_2(x) + \cdots + u^{k-1}p_{k-1}(x)\rangle ,$ then $c(x) = (g(x) + u p_1(x) + u^2 p_2(x) + \cdots + u^{k-1}p_{k-1}(x))f(x)$ for some polynomial $f(x).$ If deg$f(x) \le n-r-1,$ then $c(x)$ can be written as linear combinations of elements of $B_1.$ Otherwise by the division algorithm there exist polynomials $q(x)$ and $r(x)$ such that

$$\begin{aligned} f(x) = \left( \frac{x^n - 1}{g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x)}\right) q(x) + r(x), \end{aligned}$$

where $r(x) = 0$ or deg$r(x) \le n-r-1.$ This gives,

$$\begin{aligned} \left( g(x) \!+\! u p_1(x) + \cdots + u^{k-1}p_{k-1}(x)\right) f(x) \!=\! \left( g(x) \!+\! u p_1(x) + \cdots + u^{k-1}p_{k-1}(x)\right) r(x). \end{aligned}$$

Since deg$r(x) \le n-r-1,$ this shows that $B_1$ spans $C_k.$ Now we only need to show that $B_1$ is linearly independent. Let $g(x)= g_0 + g_1x + \cdots + g_rx^r$ and $p_1(x) = p_{1,0} + p_{1,1}x + \cdots + p_{1,l_1}x^{l_1},\,p_2(x) = p_{2,0} + p_{2,1}x + \cdots + p_{2,l_2}x^{l_2},\ldots ,p_{k-1}(x) = p_{k-1,0} + p_{k-1,1}x +\cdots + p_{k-1,l_{k-1}}x^{l_{k-1}},\,g_0 \in \mathbb Z _p^{\times },\,g_i,\,p_{j, i-1} \in \mathbb Z _p,\,i,\,j \ge 1.$ Suppose $(g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x))c_0 + x(g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x))c_1 + \cdots +x^{n-r-1}(g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x))c_{n-r-1} =0.$ By comparing the coefficients in the above equation, we get

$$\begin{aligned} \left( g_0 + u p_{1,0} + \cdots + u^{k-1}p_{k-1,0}\right) c_0 = 0~\text{(constant } \text{ coefficient) }. \end{aligned}$$

Since $(g_0 + u p_{1,0} + \cdots + u^{k-1}p_{k-1,0})$ is unit, we get $c_0 = 0.$ Thus, $x(g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x))c_1 + \cdots + x^{n-r-1}(g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x))c_{n-r-1} = 0.$ Again comparing the coefficients, we get

$$\begin{aligned} \left( g_0 + u p_{1,0} + \cdots + u^{k-1}p_{k-1,0}\right) c_1 = 0~\text{(coefficient } \text{ of }~x). \end{aligned}$$

As above, this gives $c_1 = 0.$ Continuing in this way we get that $c_i = 0$ for all $i = 0,\, 1,\ldots , n-r-1.$ Therefore, the set $B_1$ is linearly independent and hence a basis for $C_k.$

(2) If $C_k = \langle g(x) + u p_1(x) + \cdots + u^{k-1} p_{k-1}(x),\,u a_1(x) + u^2 q_1(x) + \cdots + u^{k-1} q_{k-2}(x),\,u^2 a_2(x) + u^3l_{1}(x) + \cdots + u^{k-1}l_{k-3}(x),\ldots , u^{k-2} a_{k-2}(x) +u^{k-1} t_1(x),\,u^{k-1} a_{k-1}(x)\rangle$ with deg$(g(x) + u p_1(x) +\cdots + u^{k-1} p_{k-1}(x)) = r_1,$ deg$(a_1(x)) = r_2,$ deg$(a_2(x)) = r_3,\ldots ,$ and deg$(a_{k-1}(x)) = r_k.$ The polynomial $a(x)$ is the lowest degree polynomial such that $u^{k-1}a_{k-1}{(x)}\in C_k.$ It is suffices to show that $B_2$ spans $B=\{ g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x),\,x(g(x) + u p_1(x)+ \cdots + u^{k-1}p_{k-1}(x)),\ldots ,x^{n-r_1-1}(g(x) + u p_1(x) +\cdots + u^{k-1}p_{k-1}(x)),\,u a_1(x) + u^2 q_1(x) + \cdots +u^{k-1} q_{k-2}(x),\,x(u a_1(x) + u^2 q_1(x) + \cdots + u^{k-1}q_{k-2}(x)),\ldots ,x^{r_1-r_2-1}(u a_1(x) + u^2 q_1(x) + \cdots +u^{k-1} q_{k-2}(x)),\,u^2 a_2(x) + u^3 l_{1}(x) + \cdots +u^{k-1}l_{k-3}(x),\,x(u^2 a_2(x) + u^3 l_{1}(x) + \cdots +u^{k-1}l_{k-3}(x)),\ldots ,x^{r_2-r_3-1}(u^2 a_2(x) + u^3 l_{1}(x)+ \cdots + u^{k-1}l_{k-3}(x)),\ldots ,u^{k-1}a_{k-1}(x),\,xu^{k-1}a_{k-1}(x),\ldots ,x^{n-r_k-1}u^{k-1}a_{k-1}(x)\}.$ As in the proof of part 2 of Theorem 4.1, it is suffices to show that $u^{k-1} x^{r_{k-1}-r_k} a_{k-1}(x) \in \text{ span }(B_2).$ Let the leading coefficients of $x^{r_{k-1}-r_k} a_{k-1}(x)$ be $a_0$ and of $g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x)$ be $g_0.$ There exists a constant $c_0 \in \mathbb Z _p$ such that $a_0 = c_0 g_0.$ Then we have $u^{k-1} x^{r_{k-1}-r_k} a_{k-1}(x) \!=\! u^{k-1} c_0\left( g(x) + u p_1(x) + \cdots + u^{k-1}p_{k-1}(x)\right) + u^{k-1} m(x),$ where $m(x)$ is a polynomial of degree less than $r_{k-1}$ such that $u^{k-1}m(x)\in C_k.$ Any polynomial $p(x)$ such that $u^{k-1} p(x)$ is in $C_k$ must have degree greater or equal to deg$(a_{k-1}(x)) = r_k.$ Hence, $r_k \le \text{ deg }m(x)< r_{k-1}$ and $u^{k-1}m(x) = \alpha _0 u^{k-1}a_{k-1}(x) + \alpha _1 x u^{k-1}a_{k-1}(x) + \cdots + \alpha _{r_{k-1}-r_k-1} x^{r_{k-1}-r_k-1} u^{k-1}a_{k-1}(x).$ Thus, $u^{k-1} x^{r_{k-1}-r_k} a_{k-1}(x) \in \text{ span }(B_2).$ Hence, $B_2$ is a generating set. As in (1), by comparing the coefficients we can see that $B_2$ is linearly independent. Therefore, $B_2$ is a minimal spanning set. $\square$

5 Minimum distance

Let $n$ is not relatively prime to $p.$ Let $C_2 = \langle g(x)+up(x),\,ua(x)\rangle$ be a cyclic code of length $n$ over $R_{2,p} = \mathbb Z _p + u\mathbb Z _p,\,u^2 = 0.$ We define $C_{2,u} = \{k(x) \in R_{2,p,n}:\,u k(x)\in C_2\}.$ It is easy to see that $C_{2,u}$ is a cyclic code over $\mathbb Z _p.$ Let $C_k$ be a cyclic code of length $n$ over $R_{k,p} =\mathbb Z _p + u \mathbb Z _p + \cdots + u^{k-1}\mathbb Z _p,\,u^k = 0.$ We define $C_{k,u^{k-1}} = \{k(x) \in R_{k,p,n}:\,u^{k-1} k(x) \in C_k\}.$ Again it is easy to see that $C_{k, u^{k-1}}$ is a cyclic code over $\mathbb Z _p.$

Theorem 5.1

Let $n$ is not relatively prime to $p.$ If $C_k = \langle g(x) + u p_1(x)+ u^2 p_{2}(x) + \cdots + u^{k-1} p_{k-1}(x),\,u a_1(x) + u^2 q_1(x)+ \cdots + u^{k-1} q_{k-2}(x),\,u^2 a_2(x) + u^3 l_{1}(x) + \cdots +u^{k-1}l_{k-3}(x),\ldots ,u^{k-2} a_{k-2}(x) + u^{k-1} t_1(x),\,u^{k-1} a_{k-1}(x)\rangle$ is a cyclic code of length $n$ over $R_{k,p} = \mathbb Z _p + u \mathbb Z _p + \cdots + u^{k-1}\mathbb Z _p,\,u^k = 0.$ Then $C_{k,u^{k-1}} = \langle a_{k-1}(x)\rangle$ and $w_{H}(C_{k}) = w_{H}(C_{k, u^{k-1}}).$

Proof

We have $u^{k-1} a_{k-1}(x) \in C_k,$ thus $\langle a_{k-1}(x)\rangle \subseteq C_{k, u^{k-1}}.$ If $b(x) \in C_{k,u^{k-1}},$ then $u^{k-1}b(x) \in C_k$ and hence there exist polynomials $b_1(x),\ldots ,b_k(x) \in \mathbb Z _p[X]$ such that $u^{k-1}b(x) = b_1(x)u^{k-1}g(x) +b_2(x)u^{k-1}a_1(x) + b_2(x)u^{k-1}a_2(x)+ \cdots +b_k(x)u^{k-1}a_{k-1}(x).$ Since $a_{k-1}(x)|a_{k-2}(x)|\cdots |a_{2}(x)|a_{1}(x)|g(x),$ we have $u^{k-1}b(x) = m(x)u^{k-1}a_{k-1}(x)$ for some polynomial $m(x) \in \mathbb Z _p[x].$ So, $C_{k, u^{k-1}} \in \langle a_{k-1}(x)\rangle ,$ and hence $C_{k, u^{k-1}} =\langle a_{k-1}(x)\rangle .$ Let $m(x) = m_0(x) + um_1(x) + \cdots + u^{k-1}m_{k-1}(x) \in C_k,$ where $m_0(x),\,m_1(x),\ldots , m_{k-1}(x) \in \mathbb Z _p[x].$ We have $u^{k-1}m(x) = u^{k-1}m_0(x),\,w_{H}(u^{k-1}m(x)) \le w_{H}(m(x))$ and $u^{k-1}C_k$ is subcode of $C_k$ with $w_{H}(u^{k-1}C_k) \le w_{H}(C_k).$ Therefore, it is sufficient to focus on the subcode $u^{k-1}C_k$ in order to prove the theorem. Since $u^{k-1}C_k = \langle u^{k-1}a_{k-1}(x)\rangle ,$ we get $w_{H}(C_{k}) =w_{H}(C_{k, u^{k-1}}).$ $\square$

Definition 5.2

Let $m = b_{l-1}p^{l-1} + b_{l-2}p^{l-2} + \cdots + b_1p + b_0,\,b_i \in \mathbb Z _p,\,0 \le i \le l-1,$ be the $p$-adic expansion of $m.$

• (1) If $b_{l-i} \ne 0$ for all $1 \le i \le q,\,q < l,$ and $b_{l-i} = 0$ for all $i,\,q+1 \le i \le l,$ then $m$ is said to have a $p$-adic length $q$ zero expansion.

• (2) If $b_{l-i} \ne 0$ for all $1 \le i \le q,\,q<l,\,b_{l-q-1} = 0$ and $b_{l-i} \ne 0$ for some $i,\,q+2 \le i\le l,$ then $m$ is said to have $p$-adic length $q$ non-zero expansion.

• (3) If $b_{l-i} \ne 0$ for $1 \le i \le l,$ then $m$ is said to have a $p$-adic length $l$ expansion or $p$-adic full expansion.

Lemma 5.3

Let $C$ be a cyclic code over $R_{k,p}$ of length $p^l$ where $l$ is a positive integer. Let $C = \langle a(x)\rangle$ where $a(x) = (x^{p^{l-1}} -1)^bh(x),\,1 \le b < p.$ If $h(x)$ generates a cyclic code of length $p^{l-1}$ and minimum distance $d$ then the minimum distance $d(C)$ of $C$ is $(b+1)d.$

Proof

For $c \in C,$ we have $c = (x^{p^{l-1}} - 1)^bh(x)m(x)$ for some $m(x) \in \frac{R_{k,p}[x]}{(x^{p^l}-1)}.$ Since $h(x)$ generates a cyclic code of length $p^{l-1},$ we have $w(c) = w((x^{p^{l-1}} - 1)^bh(x)m(x)) = w(x^{p^{l-1}b}h(x)m(x)) + w(^bC_1x^{p^{l-1}(b-1)}h(x)m(x)) + \cdots +w(^bC_{b-1}x^{p^{l-1}}h(x)m(x)) + w(h(x)m(x)).$ Thus, $d(C) = (b+1)d.$ $\square$

Theorem 5.4

Let $C_k$ be a cyclic code over $R_{k,p}$ of length $p^l$ where $l$ is a positive integer. Then, $C_k = \langle g(x) + u p_1(x) + u^2 p_{2}(x) + \cdots + u^{k-1} p_{k-1}(x),\,u a_1(x) + u^2 q_1(x) +\cdots + u^{k-1} q_{k-2}(x),\,u^2 a_2(x) + u^3 l_{1}(x) + \cdots +u^{k-1}l_{k-3}(x),\ldots ,u^{k-2} a_{k-2}(x) + u^{k-1} t_1(x),\,u^{k-1} a_{k-1}(x)\rangle$ where $g(x) = (x-1)^{t_1},\,a_1(x) = (x-1)^{t_2},\cdots ,a_{k-1}(x) = (x-1)^{t_k}.$ for some $t_1 > t_2 > \cdots > t_k > 0.$

• (1) If $t_k \le p^{l-1},$ then $d(C) = 2.$

• (2) If $t_k > p^{l-1},$ let $t_k = b_{l-1}p^{l-1} +b_{l-2}p^{l-2} + \cdots + b_1p + b_0$ be the $p$-adic expansion of $t_k$ and $a_{k-1}(x) = (x-1)^{t_k} = (x^{p^{l-1}} -1)^{b_{l-1}}(x^{p^{l-2}} - 1)^{b_{l-2}}\cdots (x^{p^{1}} -1)^{b_1}(x^{p^0} - 1)^{b_0}.$

  • (a) If $t_k$ has a $p$-adic length $q$ zero expansion or full expansion $(l=q).$ Then, $d(C_k) =(b_{l-1}+1)(b_{l-2}+1)\cdots (b_{l-q}+1).$

  • (b) If $t_k$ has a $p$-adic length $q$ non-zero expansion. Then, $d(C_k) = 2(b_{l-1}+1)(b_{l-2}+1)\cdots (b_{l-q}+1).$

Proof

The first claim easily follows from Theorem 3.3. From Theorem 5.1, we see that $d(C_{k}) = d(u^{k-1}C_{k}) = d((x -1)^{t_k}).$ hence, we only need to determine the minimum weight of $u^{k-1}C_{k} = (x - 1)^{t_k}.$

• (1) If $t_k \le p^{l-1},$ then $(x - 1)^{t_k}(x -1)^{p^{l-1}-t_k} = (x - 1)^{p^{l-1}} = (x^{p^{l-1}} - 1) \in C_k.$ Thus, $d(C_k) = 2.$

• (2) Let $t_k > p^{l-1}.$

  • (a) If $t_k$ has a $p$-adic length $q$ zero expansion, we have $t_k = b_{l-1} p^{l-1} + b_{l-2}p^{l-2} + \cdots +b_{l-q}p^{l-q},$ and $a_{k-1}(x) = (x - 1)^{t_k} = (x^{p^{l-1}}-1)^{b_{l-1}}(x^{p^{l-2}}-1)^{b_{l-2}}\cdots (x^{p^{l-q}}-1)^{b_{l-q}}.$ Let $h(x) = (x^{p^{l-q}}-1)^{b_{l-q}}.$ Then $h(x)$ generates a cyclic code of length $p^{l-q+1}$ and minimum distance $(b_{l-q}+1).$ By Lemma 5.3, the subcode generated by $(x^{p^{l-q+1}}-1)^{b_{l-q+1}}h(x)$ has minimum distance $(b_{l-q+1}+1) (b_{l-q}+1).$ By induction on $q,$ we can see that the code generated by $a_{k-1}(x)$ has minimum distance $(b_{l-1}+1)(b_{l-2}+1)\cdots (b_{l-q}+1).$ Thus, $d(C_k) = (b_{l-1}+1)(b_{l-2}+1)\cdots (b_{l-q}+1).$

  • (b) If $t_k$ has a $p$-adic length $q$ non-zero expansion, we have $t_k = b_{l-1} p^{l-1} + b_{l-2}p^{l-2} + \cdots + b_{1}p +b_0,\,b_{l-q-1}.$ Let $r = b_{l-q-2}p^{l-q-2}+b_{l-q-3}p^{l-q-3}+ \cdots + b_1p + b_0$ and $h(x) = (x-1)^r = (x^{p^{l-q-2}}-1)^{b_{l-q-2}}(x^{p^{l-q-3}}-1)^{b_{l-q-3}}\cdots (x^{p^{1}}-1)^{b_{1}}(x^{p^{0}}-1)^{b_{0}}.$ Since $r < p^{l-q-1},$ we have $p^{l-q-1} = r+j$ for some non-zero $j.$ Thus, $(x-1)^{p^{l-q-1}-j}h(x) = (x^{p^{l-q-1}} - 1) \in C_k.$ Hence, the subcode generated by $h(x)$ has minimum distance 2. By Lemma 5.3, the subcode generated by $(x^{p^{l-q}} -1)^{b_{l-q}}h(x)$ has minimum distance $2(b_{l-q}+1).$ By induction on $q,$ we can see that the code generated by $a_{k-1}(x)$ has minimum distance $2(b_{l-1}+1)(b_{l-2}+1)\cdots (b_{l-q}+1).$ Thus, $d(C_k) = 2(b_{l-1}+1)(b_{l-2}+1)\cdots (b_{l-q}+1).$ $\square$

6 Examples

In this section, we give some examples of cyclic codes of different lengths over the ring $R_{k,p}.$

Example 6.1

Cyclic codes of length 5 over $R_{4,3} = \mathbb Z _3 + u \mathbb Z _3 + u^2 \mathbb Z _3 +u^3 \mathbb Z _3,\,u^4 = 0:$ We have

$$\begin{aligned} x^5-1 = (x-1)\left( x^4 + x^3 + x^2 + x +1\right) ~\text{ over }~ R_{4,3}. \end{aligned}$$

Let $g_1 = x-1$ and $g_2 = x^4 + x^3 + x^2 + x +1.$ The non-zero cyclic codes of length 5 over $R_{4,3}$ with generator polynomial are given in Table 1.

Table 1 Cyclic codes of length 5 over $R_{4,3}=\mathbb Z _3 + u \mathbb Z _{3} + u^{2} \mathbb Z _{3}+ u^{3} \mathbb Z _{3},\,u^4=0$

Example 6.2

Cyclic codes of length 5 over $R_{2,5} = \mathbb Z _5 + u \mathbb Z _5,\,u^2 = 0:$ We have

$$\begin{aligned} x^5-1 = (x-1)^5 = g^5 ~\text{ over }~ R_{2,5}. \end{aligned}$$

The non-zero cyclic codes of length 5 over $R_{2,5}$ with generator polynomial and minimum distance are given in Tables 2 and 3.

Table 2 Non-zero free module cyclic codes of length 5 over $\mathbb Z _5 + u \mathbb Z _5,\,u^2=0$

Table 3 Non-free module cyclic codes of length 5 over $\mathbb Z _5 + u \mathbb Z _5,\,u^2 = 0$

Example 6.3

Cyclic codes of length 9 over $\mathbb Z _3 + u \mathbb Z _3,\,u^2 = 0:$ We have

$$\begin{aligned} x^9-1 = (x-1)^9 = g^9 ~\text{ over }~ R_{2,3}. \end{aligned}$$

The non-zero cyclic codes of length 9 over $R_{2,3}$ with generator polynomial and minimum distance are given in Tables 4 and 5.

Table 4 Non-zero free module cyclic codes of length 9 over $\mathbb Z _3 + u \mathbb Z _3,\,u^2 = 0$

Table 5 Non-free module cyclic codes of length 9 over $\mathbb Z _3 + u \mathbb Z _3,\,u^2 = 0$