Self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n

Abstract

For any odd positive integer n, we express cyclic codes over ${\mathbb {Z}}_4$ of length 4n in a new way. Based on the expression of each cyclic code ${\mathcal {C}}$, we provide an efficient encoder and determine the type of ${\mathcal {C}}$. In particular, we give an explicit representation and enumeration for all distinct self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n and correct a mistake in the paper “Concatenated structure of cyclic codes over ${\mathbb {Z}}_4$ of length 4n” (Cao et al. in Appl Algebra Eng Commun Comput 10:279–302, 2016). In addition, we obtain 50 new self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 28.

1 Introduction

In [13], it was shown that many interesting binary linear and nonlinear codes are in fact the images under a Gray map of special linear codes over the ring ${\mathbb {Z}}_4$. This important discovery caused an enormous amount of activity led to the study of codes in this ambient space and linear codes over ${\mathbb {Z}}_4$ has become one of the most widely studied areas of algebraic coding theory.

The class of self-dual codes is an interesting topic in coding theory due to their connections with other fields of mathematics such as lattices, cryptography, invariant theory, block designs, etc. In particular, self-dual codes over ${\mathbb {Z}}_4$ relate to combinatorial designs and unimodular lattices (cf. [4, 12, 14,15,16,17]). A common theme for the construction of self-dual codes is the use of computational tools and computer search. To make this search feasible, adding an algebraic structure to the codes considered is an effective way.

We begin with the necessary definitions for codes over rings. Let A be a commutative finite ring with identity $1\ne 0$, and $A^{\times }$ be the multiplicative group of invertible elements of A. For any $f,g\in A$, the ideal of A generated by f and g is denoted by $\langle f,g\rangle$, i.e., $\langle f,g\rangle =Af+Ag=\{af+bg\mid a,b\in A\}$.

A code over A of length N is a nonempty subset ${{{\mathcal {C}}}}$ of $A^N$. The code ${{{\mathcal {C}}}}$ is said to be linear if ${{{\mathcal {C}}}}$ is an A-submodule of $A^N$. Especially, ${{{\mathcal {C}}}}$ is called a ${\mathbb {Z}}_4$-linear code when $A={\mathbb {Z}}_4$. All codes in this paper are assumed to be linear. The ambient space $A^N$ is equipped with the usual Euclidean inner product, i.e. $[a,b]=\sum _{j=0}^{N-1}a_jb_j$, where $a=(a_0,a_1,\ldots ,a_{N-1}), b=(b_0,b_1,\ldots ,b_{N-1})\in A^N$, and the (Euclidean) dual code is defined by ${{{\mathcal {C}}}}^{\bot }=\{a\in A^N\mid [a,b]=0, \forall b\in {{{\mathcal {C}}}}\}$. If ${{{\mathcal {C}}}}^{\bot }={{{\mathcal {C}}}}$, ${{{\mathcal {C}}}}$ is called a (Euclidean) self-dual code over A.

The linear code ${{{\mathcal {C}}}}$ is said to be cyclic if $(c_{N-1}, c_0,c_1,\ldots , c_{N-2})\in {{{\mathcal {C}}}}$ for all $(c_0,c_1,\ldots , c_{N-1})\in {{{\mathcal {C}}}}$. We use the natural connection of cyclic codes to polynomial rings, where $(c_0,c_1,\ldots , c_{N-1})$ is viewed as $c(x)=\sum _{j=0}^{N-1}c_jx^j$ and the cyclic code ${{{\mathcal {C}}}}$ is an ideal in the polynomial residue ring $A[x]/\langle x^N-1\rangle$.

Let ${\mathcal {C}}$ be a nonzero ${\mathbb {Z}}_4$-linear code of length N. Then ${\mathcal {C}}$ has a generator matrix of the form: $G_{{\mathcal {C}}}=\left( \begin{array}{ccc}I_{k_0} &{} A &{} B\\ 0 &{} 2I_{k_1} &{} 2C\end{array}\right) U,$ where U is a suitable $N\times N$ permutation matrix, $I_{k_0}$ and $I_{k_1}$ denotes the $k_0\times k_0$ and $k_1\times k_1$ identity matrices, respectively, A and C are ${\mathbb {Z}}_2$-matrices, and B is a ${\mathbb {Z}}_4$-matrix. Then ${\mathcal {C}}$ is an abelian group of type $4^{k_0}2^{k_1}$ and contains $2^{2k_0+k_2}$ codewords (cf. Wan [23, Proposition 1.1]).

Cyclic codes over ${\mathbb {Z}}_4$ of odd length n followed from results in [3] and also appeared in more detail in [20] and [21]. Abualrub and Oehmk in [1] determined the generators for cyclic codes over ${\mathbb {Z}}_4$ for lengths of the form $2^k$, and Blackford in [2] presented the generators for cyclic codes over ${\mathbb {Z}}_4$ of length 2n.

Let k and n be any integers such that $k\ge 1$ and n is odd. In 2006, Dougherty and Ling [11] gave a representation for cyclic codes over ${\mathbb {Z}}_4$ of length $2^kn$, described the number and dual codes of all these cyclic codes and obtained a rough description of cyclic codes that are self-dual. In 2012, Kiah et al. [18] determined the number of Euclidean self-dual codes over the Galois ring $\mathrm{GR}(4,s)$ of length $2^k$, for any positive integers s and k. In 2016, Jitman et al. [19] pointed out that the determination of the Euclidean dual of a cyclic code in [2, Lemma 9] and [11, Proposition 5.8] is not correct. Further, Jitman et al. found an incorrect statement about the number of Euclidean self-dual cyclic codes ([11, Corollaries 5.7, 5.9 and Proposition 5.8]), and gave the correct statement by [19, Corollary 4.8].

In 2016, Cao et al. [5] gave a concatenated structure for every cyclic code over ${\mathbb {Z}}_4$ of length 4n from a way different that was used in [2, 11, 18, 19]. However, there was a mistake in [5, Corollary 4.6]. In fact, it gives only a part of self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n, not all these self-dual codes.

There are two problems that need to be improved or addressed in [11, 19]:

• 1. For arbitrary positive integer k, by [11, Theorem 5.3], Dougherty and Ling divided 29 cases to express cyclic codes over $\mathrm{GR}(4,s)$ of length $2^k$ and their annihilators. This would lead to a lengthy description for the results of self-dual codes with length $2^kn$, if use [19, Proposition 4.5].

• 2. It is not convenient to express the result for (self-dual) cyclic codes over ${\mathbb {Z}}_4$ of length $2^kn$, if use Discrete Fourier Transform. In fact, it is not easy to apply [19, Proposition 4.5] for constructing and designing self-dual cyclic codes over ${\mathbb {Z}}_4$ of length $2^kn$, for any given concrete integers k and n.

There are similar problems in [5], it is some inconvenient to construct cyclic codes over ${\mathbb {Z}}_4$ of length 4n as well. First, all these cyclic codes were divided into 20 cases to express by [5, Theorem 4.5], which is still a bit lengthy. Second, the expressions for these cyclic codes need to be computed to get them (see [5, Example 3.4 and Theorem 4.5]).

Therefore, it is necessary and meaningful to find a more direct and simple method to express cyclic codes over ${\mathbb {Z}}_4$ of length $2^kn$ and to determine the self-dual cyclic codes accurately.

Recently, Cao et al. [6] gave an explicit representation for cyclic codes over ${\mathbb {Z}}_4$ of length 2n from a new way different from that was used in [2, 5, 11, 18, 19]. Using this representation, we provided an efficient encoder and the type for every code, and determined the (Euclidean) self-dual codes in this class of cyclic codes precisely. However, the methods used in [6] can not be directly used to the case of length 4n, there are many key computational problems that require to develop new methods.

The present paper is organized as follows. In Sect. 2, we introduce necessary notations and provide necessary conclusions. In Sect. 3, we give the main results of this paper by four theorems: give an explicit representation for all distinct cyclic codes over ${\mathbb {Z}}_4$ of length 4n (Theorem 3.1); provide an efficient encoder for each of these cyclic code (Theorem 3.2); determine the dual code for every cyclic code (Theorem 3.3); give an explicit representation and enumeration for all distinct self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n (Theorem 3.4). In addition, we correct a mistake in [5, Corollary 4.6]. In Sect. 4, we describe how to construct self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n by two examples: when $n=3$ and when $n=7$. In particular, we obtain 50 new good self-dual cyclic ${\mathbb {Z}}_4$-codes ${\mathcal {C}}$ with basic parameters $(28,|{\mathcal {C}}|=2^{28},d_H=4,d_L=8,d_E=8)$, where $d_H,d_L$ and $d_E$ are the minimum Hamming distance, Lee distance and Euclidean distance of the codes respectively. In Sect. 5, we give detail proofs for Theorems 3.1–3.4 in Sect. 3. Section 6 concludes the paper.

2 Preliminaries

In this section, we introduce necessary notations and provide necessary conclusions for the following sections.

Let ${\mathbb {F}}_2=\{0,1\}$ in which the arithmetic is done modulo 2, and let ${\mathbb {Z}}_4=\{0,1,2,3\}$ in which the arithmetic is done modulo 4. Denote ${\mathbb {Z}}_4^\times =\{1,3\}=\{1,-1\}$. Let x, y be indeterminate over ${\mathbb {Z}}_4$ and ${\mathbb {F}}_2$. In this paper, we regard ${\mathbb {F}}_2$ as a subset of the ring ${\mathbb {Z}}_4$, though ${\mathbb {F}}_2$ is not a subfield of ${\mathbb {Z}}_4$. By this view, every element $a\in {\mathbb {Z}}_4$ has a unique 2-adic expansion: $a=a_0+2a_1, \ a_0,a_1\in {\mathbb {F}}_2.$ Denote ${\overline{a}}=a_0=a$ (mod 2). Then $^{-}: a\mapsto {\overline{a}}$ ($\forall a\in {\mathbb {Z}}_4$) is a surjective ring homomorphism from ${\mathbb {Z}}_4$ onto ${\mathbb {F}}_2$, and $^{-}$ can be extended to a surjective ring homomorphism from ${\mathbb {Z}}_4[y]$ onto ${\mathbb {F}}_2[y]$ by: ${\overline{f}}(y)=\overline{f(y)}=\sum _{i=0}^d{\overline{b}}_iy^i$, for any $f(y)=\sum _{i=0}^db_iy^i\in {\mathbb {Z}}_4[y]$.

Let $f(y)=\sum _{j=0}^dc_jy^j\in {\mathbb {Z}}_4[y]$ of degree $d\ge 1$. Then f(y) is said to be a monic basic irreducible polynomial if ${\overline{f}}(y)$ is an irreducible polynomial in ${\mathbb {F}}_2[y]$ and $c_d\in {\mathbb {Z}}_4^\times$ (cf. [24, Sect. 13.4]). The reciprocal polynomial of f(y) is defined as ${\widetilde{f}}(y)=\widetilde{f(y)}=y^df(\frac{1}{y})=\sum _{j=0}^dc_jy^{d-j}$. Then f(y) is said to be self-reciprocal if ${\widetilde{f}}(y)=\delta f(y)$ for some $\delta \in {\mathbb {Z}}_4^{\times }$. It is known that $\widetilde{{\widetilde{f}}(y)}=f(y)$ if $f(0)\ne 0$, and $\widetilde{f(y)g(y)}={\widetilde{f}}(y){\widetilde{g}}(y)$ for any monic polynomials $f(y), g(y)\in {\mathbb {Z}}_4[y]$ with positive degrees satisfying $f(0),g(0)\in {\mathbb {Z}}_4^{\times }$.

Throughout this paper, we assume the following factorization of $y^n-1$:

$$\begin{aligned} y^n-1=f_1(y)f_2(y)\ldots f_r(y), \end{aligned}$$

(1)

where $f_1(y)=y-1, f_2(y),\ldots , f_r(y)$ are pairwise coprime monic basic irreducible polynomials in ${\mathbb {Z}}_4[y]$ with degree $\mathrm{deg}(f_i(y))=m_i$ for all $i=1,\ldots ,r$. Then we have $y^n\equiv 1$ (mod $f_i(y)$).

This implies that $x^{-1}\equiv x^{n-1}$ (mod ${\overline{f}}_i(x))$ and $x^{-1}\equiv x^{ln-1} \ (\mathrm{mod} \ ({\overline{f}}_i(x))^l) \ \text{ in } \ {\mathbb {F}}_2[x]$ for any integer $l\ge 2$.

Additionally, we will adopt the following notation.

1.:

The ring ${\mathcal {B}}=\frac{{\mathbb {Z}}_4[x]}{\langle x^{4n}-1\rangle }=\{\sum _{j=0}^{4n-1}b_jx^j\mid b_j\in {\mathbb {Z}}_4, \ j=0,1,\ldots , 4n-1\}$ where the arithmetic is done modulo $x^{4n}-1$.

Then cyclic codes over ${\mathbb {Z}}_4$ of length 4n are viewed as ideals of the ring ${\mathcal {B}}$.

2.:

The ring ${\mathcal {R}}_i=\frac{{\mathbb {Z}}_4[x]}{\langle f_i(x^4)\rangle }=\{\sum _{j=0}^{4m_i-1}b_jx^j\mid b_j\in {\mathbb {Z}}_4, \ j=0,1,\ldots , 4m_i-1\}$ where the arithmetic is done modulo $f_i(x^4)$.

We regard ${\mathcal {R}}_i$ as a subset of the ring ${\mathcal {B}}$ in this paper.

3.:

The set ${\mathcal {T}}_i=\{\sum _{j=0}^{m_i-1}t_jx^j\mid t_j\in {\mathbb {F}}_2=\{0,1\}, \ j=0, \ldots , m_i-1\}$. Then $|{\mathcal {T}}_i|=2^{m_i}$. We regard ${\mathcal {T}}_i$ as a subset of the ring ${\mathcal {R}}_i$.

Hereafter, the set ${\mathcal {T}}_i$ will appear frequently in the succeeding contents.

4.:

Denote by $F_i(y)=\frac{y^n-1}{f_i(y)}\in {\mathbb {Z}}_4[y]$. Then there are polynomials $u_i(y), v_i(y)\in {\mathbb {Z}}_4[y]$ such that (cf. [6, Sect. 2])

$$\begin{aligned} u_i(y)F_i(y)+v_i(y)f_i(y)=1. \end{aligned}$$

(2)

Then we define $\varepsilon _i(x)\in {\mathcal {B}}$ by the following equation:

$$\begin{aligned} \varepsilon _i(x)\equiv u_i(x^4)F_i(x^4)=1-v_i(x^4)f_i(x^4) \ (\mathrm{mod} \ x^{4n}-1). \end{aligned}$$

(3)

5.:

Let $w_{i}(x)\in {\mathcal {T}}_i$ satisfying $f_i(x)^4\equiv 2{\overline{f}}_i(x)^2w_{i}(x)^2$ (mod $f_i(x^4)$) in ${\mathbb {Z}}_4[x]$. Further, let $(w_{i,0}(x),w_{i,1}(x))$ be the unique ordered pair of elements in ${\mathcal {T}}_i$ satisfying

$$\begin{aligned} w_i(x)^2=w_{i,0}(x)+w_{i,1}(x){\overline{f}}_i(x) \ (\text{ mod } \ ({\overline{f}}_i(x))^2) \ \text{ and } \ w_{i,0}(x)\ne 0 \ \text{ in } \ {\mathbb {F}}_2[x]. \end{aligned}$$

$\checkmark$:

The elements $w_{i}(x), w_{i,0}(x),w_{i,1}(x)$ of ${\mathcal {T}}_i$ play key roles in this paper, and they will be determined later by Theorem 2.5.

6.:

After a rearrangement of $f_2(y),\ldots ,f_r(y)$, there are integers $\lambda ,\epsilon$ such that

$\diamond$:

$\lambda \ge 1$, $\epsilon \ge 0$ and $\lambda +2\epsilon =r$.

$\diamond$:

$f_i(y)$ is self-reciprocal, $i=1,\ldots ,\lambda$.

Then $m_i=\mathrm{deg}(f_i(y))$ is even, when $2\le i\le \lambda$ (cf. [8, Lemma 3.2]).

$\diamond$:

${\widetilde{f}}_{\lambda +j}(y)=c_{\lambda +j}f_{\lambda +\epsilon +j}(y)$ for some $c_{\lambda +j}\in {\mathbb {Z}}_4^\times$, $j=1,\ldots ,\epsilon$.

7.:

For any integer i, $2\le i\le \lambda$, we define the following subsets of ${\mathcal {T}}_i$:

• ${\mathcal {V}}_i=\left\{ h(x)\in {\mathcal {T}}_i\mid h(x)+x^{3m_i}h(x^{-1})\equiv 0 \ ({\mathrm{mod}} \ {\overline{f}}_i(x)) \ \text{ in } \ {\mathbb {F}}_2[x]\right\}$.

• ${\mathcal {W}}_i^{(0)}=\{h_0(x)\in {\mathcal {T}}_i\mid h_0(x)+x^{2m_i}(w_i(x^{-1})^2+h_0(x^{-1}))\equiv 0 \ ({\mathrm{mod}} \ {\overline{f}}_i(x)) \text{ in } \ {\mathbb {F}}_2[x]\}$.

• For any $h_0(x)\in {\mathcal {W}}_i^{(0)}$, define

  $$\begin{aligned} {\mathcal {W}}_{i,h_0(x)}^{(1)}= & {} \{h_1(x)\in {\mathcal {T}}_i\mid h_0(x)+h_1(x){\overline{f}}_i(x)\equiv {\widehat{\delta }}_{i,0}(x)+{\widehat{\delta }}_{i,1}(x){\overline{f}}_i(x) \\&\ (\text{ mod } \ ({\overline{f}}_i(x))^2) \ \text{ in } \ {\mathbb {F}}_2[x] \}, \end{aligned}$$

  where ${\widehat{\delta }}_{i,0}(x)=x^{2m_i}(w_i(x^{-1})^2+h_0(x^{-1}))$ and ${\widehat{\delta }}_{i,1}(x)=x^{m_i}h_1(x^{-1})$.

$\checkmark$:

The subsets ${\mathcal {V}}_i, {\mathcal {W}}_i^{(0)}, {\mathcal {W}}_{i,h_0(x)}^{(1)}$ of ${\mathcal {T}}_i$ play key roles in this paper, and they will be determined later by Theorem 2.7.

Then we provide necessary conclusions for the following sections.

First, by substituting $x^4$ for y in Eqs. (1) and (2), we obtain

$$\begin{aligned} x^{4n}-1=f_1(x^4)f_2(x^4)\ldots f_r(x^4) \quad \mathrm{and} \quad u_i(x^4)F_i(x^4)+v_i(x^4)f_i(x^4)=1 \end{aligned}$$

in ${\mathbb {Z}}_4[x]$ respectively, where $F_i(x^4)=\frac{x^{4n}-1}{f_i(x^4)}\in {\mathbb {Z}}_4[x]$. From this, by Eq. (3) and Chinese remainder theorem for commutative rings with identity, one can easily verify the following conclusions. Here we omit the proofs.

Lemma 2.1

• (i) $\varepsilon _1(x)+\cdots +\varepsilon _r(x)=1$, $\varepsilon _i(x)^2=\varepsilon _i(x)$ and $\varepsilon _i(x)\varepsilon _j(x)=0$ for all $1\le i\ne j\le r$ in the ring ${{{\mathcal {B}}}}$.

• (ii) ${{{\mathcal {B}}}}={{{\mathcal {B}}}}_1\oplus \ldots \oplus {{{\mathcal {B}}}}_{r}$, where ${{{\mathcal {B}}}}_i=\langle \varepsilon _i(x)\rangle ={{{\mathcal {B}}}}\varepsilon _i(x)$ is the ideal of ${{{\mathcal {B}}}}$ generated by $\varepsilon _i(x)$, and ${{{\mathcal {B}}}}_i$ is a commutative ring with $\varepsilon _i(x)$ as its multiplicative identity for all $i=1,\ldots , r$. Moreover, ${{{\mathcal {B}}}}$ is a direct sum of rings ${{{\mathcal {B}}}}_1, \ldots , {{{\mathcal {B}}}}_{r}$ in that ${{{\mathcal {B}}}}_i{{{\mathcal {B}}}}_j=\{0\}$ for all $i\ne j$.

• (iii) Define the map $\psi _i: a(x)\mapsto \varepsilon _i(x)a(x) \ (\mathrm{mod} \ x^{4n}-1)$ $(\forall a(x)\in {{{\mathcal {R}}}}_i)$. Then $\psi _i$ is a ring isomorphism from ${{{\mathcal {R}}}}_i$ onto ${{{\mathcal {B}}}}_i$, for $i=1,\ldots ,r$.

• (iv) Define a map $\psi$ by: for any $a_i(x)\in {{{\mathcal {R}}}}_i$, $1\le i\le r$, let

  $$\begin{aligned} \psi (a_1(x),\ldots ,a_r(x))=\sum _{i=1}^r\psi _i(a_i(x))=\sum _{i=1}^r\varepsilon _i(x)a_i(x) \ (\mathrm{mod} \ x^{4n}-1). \end{aligned}$$

  Then $\psi$ is a ring isomorphism from ${{{\mathcal {R}}}}_1\times \cdots \times {{{\mathcal {R}}}}_r$ onto ${{{\mathcal {B}}}}$.

To determine all cyclic codes over ${\mathbb {Z}}_4$ of length 4n (as ideals of the ring ${{{\mathcal {B}}}}$), by Lemma 2.1, we need to give all ideals of the ring ${\mathcal {R}}_i$ for all i.

Now, let i be an integer, $1\le i\le r$. Since $f_i(x)$ is a minic basic irreducible polynomial in ${\mathbb {Z}}_4[x]$ of degree $m_i$, ${\overline{f}}_i(x)$ is an irreducible polynomial in ${\mathbb {F}}_2[x]$ of degree $m_i$. By $\overline{{\mathbb {Z}}}_4={\mathbb {F}}_2$, we have $\overline{f_i(x^4)}={\overline{f}}_i(x)^4$ as polynomials in ${\mathbb {F}}_2[x]$. Then, we investigate the structure and properties of the ring ${\mathcal {R}}_i$. To do this, we introduce the following notation:

$\diamond$:

$\overline{{\mathcal {R}}}_i=\frac{{\mathbb {F}}_2[x]}{\langle {\overline{f}}_i(x)^4\rangle } ={\mathbb {F}}_2[x]/\langle {\overline{f}}_i(x)^4\rangle =\{\sum _{j=0}^{4m_i-1}a_jx^j\mid a_j\in {\mathbb {F}}_2, \ j=0,1,\ldots$, $4m_i-1\}$ in which the arithmetic is done modulo ${\overline{f}}_i(x)^4$ in ${\mathbb {F}}_2[x]$.

$\diamond$:

${\mathcal {F}}_i=\frac{{\mathbb {F}}_{2}[x]}{\langle {\overline{f}}_i(x)\rangle }={\mathbb {F}}_2[x]/\langle {\overline{f}}_i(x)\rangle =\{\sum _{j=0}^{m_i-1}a_jx^j\mid a_j\in {\mathbb {F}}_2, \ j=0, \ldots , m_i-1\}$ in which the arithmetic is done modulo ${\overline{f}}_i(x)$ in ${\mathbb {F}}_2[x]$. Then ${\mathcal {F}}_i$ is a finite field of $2^{m_i}$ elements. As a set, we see ${\mathcal {T}}_i$ and ${\mathcal {F}}_i$ as the same in this paper.

Lemma 2.2

(cf. [7, Lemma 3.7])

• (i) $\overline{{\mathcal {R}}}_i$ is a finite chain ring with the maximal ideal $\langle {\overline{f}}_i(x)\rangle ={\overline{f}}_i(x)\overline{{\mathcal {R}}}_i$, the nilpotency index of ${\overline{f}}_i(x)$ in $\overline{{\mathcal {R}}}_i$ is equal to 4 and that $\overline{{\mathcal {R}}}_i/\langle {\overline{f}}_i(x)\rangle \cong {\mathcal {F}}_i$. Then all distinct ideals of $\overline{{\mathcal {R}}}_i$ are given by: $\{0\}=\langle {\overline{f}}_i(x)^4\rangle \subset \langle {\overline{f}}_i(x)^3\rangle \subset \langle {\overline{f}}_i(x)^2\rangle \subset \langle {\overline{f}}_i(x)\rangle \subset \langle {\overline{f}}_i(x)^0\rangle =\overline{{\mathcal {R}}}_i$.

• (ii) Every element $\beta$ of $\overline{{\mathcal {R}}}_i$ has a unique ${\overline{f}}_i(x)$-adic expansion:

  $$\begin{aligned} \beta =t_0(x)+t_1(x){\overline{f}}_i(x)+t_2(x){\overline{f}}_i(x)^2+t_3(x){\overline{f}}_i(x)^3, \ t_j(x)\in {\mathcal {F}}_i, \ j=0,1,2,3. \end{aligned}$$

  Then $\beta$ is an invertible element of $\overline{{\mathcal {R}}}_i$, i.e. $\beta \in \overline{{\mathcal {R}}}_i^\times$, if and only if $t_0(x)\ne 0$.

• (iii) $|{\overline{f}}_i(x)^l\overline{{\mathcal {R}}}_i|=|{\mathcal {F}}_i|^{4-l} =2^{m_i(4-l)}$, for $l=0,1,2,3,4$.

As we regard ${\mathbb {F}}_2$ as a subset of ${\mathbb {Z}}_4$, we will regard $\overline{{\mathcal {R}}}_i$ as a subset of ${\mathcal {R}}_i$ hereafter, though $\overline{{\mathcal {R}}}_i$ is not a subring of ${\mathcal {R}}_i$. If needed, the reader is referred back to this identification of $\overline{{\mathcal {R}}}_i$ with a subset of ${\mathcal {R}}_i$. Then we have $2{\mathcal {R}}_i=\overline{{\mathcal {R}}}_i$.

Paralleling to the proof of [6, Lemma 4], one can easily verify the following conclusion. Here, we omit the proof.

Lemma 2.3

Every element a(x) of ${\mathcal {R}}_i$ has a unique 2-adic expansion:

$$\begin{aligned} a(x)=a_0(x)+2a_1(x), \quad a_0(x),a_1(x)\in \overline{{\mathcal {R}}}_i. \end{aligned}$$

Then $a(x)\in {\mathcal {R}}_i^{\times }$ if and only if $a_0(x)\in \overline{{\mathcal {R}}}_i^{\times }$.

Let $\tau$ be the surjective ring homomorphism from ${\mathcal {R}}_i$ onto $\overline{{\mathcal {R}}}_i$ induced by $^{-}: {\mathbb {Z}}_4 \rightarrow {\mathbb {F}}_2$ in the natural way:

$$\begin{aligned} \tau : a(x)=a_0(x)+2a_1(x)\mapsto \tau (a(x))=\overline{a(x)}=a_0(x), \end{aligned}$$

for all $a_0(x),a_1(x)\in \overline{{\mathcal {R}}}_i$. Let C be an ideal of ${\mathcal {R}}_i$. We define

$$\begin{aligned} {\overline{C}}=\tau (C)=\{\overline{a(x)}\mid a(x)\in C\} \quad \text{ and } \quad (C:2)=\{b(x)\in {\mathcal {R}}_i\mid 2b(x)\in C\}. \end{aligned}$$

Then ${\overline{C}}$ is an ideal of $\overline{{\mathcal {R}}}_i$ and (C : 2) is an ideal of ${\mathcal {R}}_i$ satisfying $C\subseteq (C:2)$.

Lemma 2.4

Let l, s be integers satisfying $0\le s\le l\le 4$ and $v(x)\in \overline{{\mathcal {R}}}_i$. Denote the ideal of ${\mathcal {R}}_i$ generated by $f_i(x)^l+2v(x)$ and $2{\overline{f}}_i(x)^s$ as follows

$$\begin{aligned} C_{(l,s;v(x))}:= & {} \langle f_i(x)^l+2v(x), 2{\overline{f}}_i(x)^s\rangle \\= & {} \{a(x)\cdot (f_i(x)^l+2v(x))+b(x)\cdot 2{\overline{f}}_i(x)^s\mid a(x),b(x)\in {\mathcal {R}}_i\}. \end{aligned}$$

Then we have the following conclusions:

• (i) We have ${\overline{C}}_{(l,s;v(x))}=\langle {\overline{f}}_i(x)^l\rangle$ and $\overline{(C_{(l,s;v(x))}:2)}=\langle {\overline{f}}_i(x)^s\rangle$ in $\overline{{\mathcal {R}}}_i$.

• (ii) The number of elements in $C_{(l,s;v(x))}$ is $|C_{(l,s;v(x))}|=2^{m_i(8-(l+s))}$.

• (iii) If $s=l$, we have $C_{(l,s;v(x))} =\langle f_i(x)^l+2v(x)\rangle$.

• (iv) For any $u(x),v(x)\in \overline{{\mathcal {R}}}_i$, we have that

  $$\begin{aligned} C_{(l,s;u(x))}=C_{(l,s;v(x))} \Longleftrightarrow u(x)\equiv v(x) \ (\mathrm{mod} \ {\overline{f}}_i(x)^s) \ \mathrm{in} \ {\mathbb {F}}_2[x]. \end{aligned}$$

  (4)

Proof

(i) Denote $C=C_{(l,s;v(x))}$ in the following. Then by the definition of ${\overline{C}}$ and $\overline{f_i(x)^l+2v(x)}={\overline{f}}_i(x)^l$, it follows that ${\overline{C}}=\langle {\overline{f}}_i(x)^l\rangle$ immediately.

By $2{\overline{f}}_i(x)^s\in C$, we have that ${\overline{f}}_i(x)^s\in (C:2)$. This implies ${\overline{f}}_i(x)^s=\tau ({\overline{f}}_i(x)^s)\in \tau (C:2)=\overline{(C:2)}$. Hence $\langle {\overline{f}}_i(x)^s\rangle \subseteq \overline{(C:2)}$.

Conversely, let $e(x)\in \overline{(C:2)}$. Then $e(x)\in \overline{{\mathcal {R}}}_i$ and $e(x)+2c(x)\in (C:2)$ for some $c(x)\in {\mathcal {R}}_i$. This implies $2e(x)=2(e(x)+2c(x))\in C$. Hence there exist $a(x),b(x)\in {\mathcal {R}}_i$ such that $2e(x)=a(x)\cdot (f_i(x)^l+2v(x))+b(x)\cdot 2{\overline{f}}_i(x)^s.$ Since $f_i(x)^l$ is a monic polynomial in ${\mathbb {Z}}_4[x]$, by comparing the coefficients on both sides of the equation, we know that $a(x)=2a_1(x)$ for some $a_1(x)\in \overline{{\mathcal {R}}}_i$. Therefore, by $2f_i(x)^l=2{\overline{f}}_i(x)^l$, $2b(x)=2{\overline{b}}(x)$, $4=0$ and $s\le l$, it follows that

$$\begin{aligned} 2e(x)= & {} 2a_1(x)\cdot (f_i(x)^l+2v(x))+2{\overline{b}}(x){\overline{f}}_i(x)^s =2a_1(x){\overline{f}}_i(x)^l+2{\overline{b}}(x){\overline{f}}_i(x)^s\\= & {} 2(a_1(x){\overline{f}}_i(x)^{l-s}+{\overline{b}}(x))\cdot {\overline{f}}_i(x)^s. \end{aligned}$$

This implies $e(x)\in \langle {\overline{f}}_i(x)^s\rangle$. Hence $\overline{(C:2)}=\langle {\overline{f}}_i(x)^s\rangle$.

(ii) Let $\tau |_C$ be the restriction of $\tau$ on the ideal $C=C_{(l,s;v(x))}$. Then $\tau |_C$ is a surjective ring homomorphism from C onto $\tau (C)={\overline{C}}=\{\tau (c(x))\mid c(x)\in C\}$. This implies $\tau (C)\cong C/\mathrm{ker}(\tau |_C)$, where $\mathrm{ker}(\tau |_C)=\{c(x)\in C\mid \tau (c(x))={\overline{c}}(x)=0\}$ is the kernel of $\tau |_C$. Therefore, $|C|=|\tau (C)||\mathrm{ker}(\tau |_C)|$. By the definition of $\tau$ and $2\cdot 2=0$, we have

$$\begin{aligned} {\mathrm{ker}}(\tau |_C)= & {} \{2c_1(x)\in C\mid c_1(x)\in \overline{{\mathcal {R}}}_i\}\\= & {} \{2(c_1(x)+2b(x))\in C\mid c_1(x)+2b(x)\in {\mathcal {R}}_i, \ c_1(x),b(x)\in \overline{{\mathcal {R}}}_i\}\\= & {} 2(C:2)=2\tau (C:2). \end{aligned}$$

This implies $|\mathrm{ker}(\tau |_C)|=|\tau (C:2)|$. Then by (i) and Lemma 2.2 (iii), we obtain $|C|=|\tau (C)||\tau (C:2)|=|{\overline{C}}||\overline{(C:2)}|=2^{m_i(4-l)}\cdot 2^{m_i(4-s)}=2^{m_i(8-(l+s))}$.

(iii) Let $s=l$. Then $2{\overline{f}}_i(x)^s=2f_i(x)^l=2(f_i(x)^l+2v(x))\subseteq \langle f_i(x)^l+2v(x)\rangle$. Therefore, $C_{(l,s;v(x))}=\langle f_i(x)^l+2v(x), 2{\overline{f}}_i(x)^s\rangle =\langle f_i(x)^l+2v(x)\rangle$.

(iv) The part “$\Longleftarrow$” can be easily verified. Here, we only prove the part “$\Longrightarrow$”. Now, let $C_{(l,s;u(x))}=C_{(l,s;v(x))}=C$. Then we see that $2(u(x)-v(x))=(f_i(x)^l+2u(x))-(f_i(x)^l+2v(x))\in C$. This implies $u(x)-v(x)\in (C:2)$. As $u(x),v(x)\in \overline{{\mathcal {R}}}_i$, by (i) it follows that $u(x)-v(x)\in \overline{(C:2)}=\langle {\overline{f}}_i(x)^s\rangle$. Hence $u(x)\equiv v(x)$ (mod ${\overline{f}}_i(x)^s$) in ${\mathbb {F}}_2[x]$. $\square$

Now, we illustrate how to determine elements $w_i(x), w_{i,0}(x), w_{i,0}(x)\in {\mathcal {T}}_i$.

Theorem 2.5

Let $1\le i\le r$. Then $2f_i(x)$ is a divisor of $f_i(x)^2-f_i(x^2)$ in ${\mathbb {Z}}_4[x]$ $(\mathrm{cf.}$ [6, Lemma 5 (i)]). Denote $g_i(x)=\frac{f_i(x)^2-f_i(x^2)}{2f_i(x)}\in {\mathbb {Z}}_4[x]$ and set $w_i(x)=g_i(x)$ $(\mathrm{mod} \ 2)$. Then

• (i) (cf. [6, Lemma 5 (i)]) We have $f_i(x)^2=f_i(x^2)+2f_i(x)w_i(x)$ in ${\mathbb {Z}}_4[x]$ and $0\ne w_i(x)\in {\mathcal {T}}_i$.

• (ii) We have $f_i(x)^4=2{\overline{f}}_i(x)^2w_i(x)^2$. Moreover, let

  $$\begin{aligned} w_{i,0}(x)=w_i(x)^2 \ (\mathrm{mod} \ {\overline{f}}_i(x)) \ \mathrm{and} \ w_{i,1}(x)=\frac{w_i(x)^2-w_{i,0}(x)}{{\overline{f}}_i(x)} \ (\mathrm{mod} \ {\overline{f}}_i(x)) \end{aligned}$$

  in ${\mathbb {F}}_2[x]$. Then $w_{i,0}(x), w_{i,1}(x)\in {{{\mathcal {T}}}}_i$ satisfying

  $$\begin{aligned} w_i(x)^2=w_{i,0}(x)+w_{i,1}(x){\overline{f}}_i(x) \ (\text{ mod } \ ({\overline{f}}_i(x))^2) \ \text{ and } \ w_{i,0}(x)\ne 0 \ \text{ in } \ {\mathbb {F}}_2[x]. \end{aligned}$$

Proof

(ii) By $4=0$ and $f_i(x)^2=f_i(x^2)+2f_i(x)g_i(x)$ in ${\mathbb {Z}}_4[x]$, it follows that

$$\begin{aligned} f_i(x)^4= & {} \left( f_i(x^2)+2f_i(x)g_i(x)\right) ^2=f_i(x^2)^2 =f_i((x^2)^2)+2f_i(x^2)g_i(x^2)\\= & {} f_i(x^4)+2\left( f_i(x)^2-2f_i(x)g_i(x)\right) g_i(x^2)\\= & {} f_i(x^4)+2f_i(x)^2g_i(x^2). \end{aligned}$$

From this and by $w_i(x)=g_i(x)$ (mod 2), we deduce that

$$\begin{aligned} f_i(x)^4\equiv & {} 2f_i(x)^2g_i(x^2)\equiv 2{\overline{f}}_i(x)^2{\overline{g}}_i(x^2)\equiv 2{\overline{f}}_i(x)^2w_i(x^2)\\\equiv & {} 2{\overline{f}}_i(x)^2w_i(x)^2 \ (\mathrm{mod} \ \langle f_i(x^4), 4\rangle ), \end{aligned}$$

i.e. $f_i(x)^4=2{\overline{f}}_i(x)^2w_i(x)^2$ in the ring ${\mathcal {R}}_i$. By $0\ne w_i(x)\in {\mathcal {F}}_i$, we see that $w_i(x)$ is an invertible element of $\overline{{\mathcal {R}}}_i$, i.e. $w_i(x)\in \overline{{\mathcal {R}}}_i^\times$, by Lemma 2.2 (ii). This implies $w_i(x)^2\in \overline{{\mathcal {R}}}_i^\times$. Since $\mathrm{deg}(w_i(x)^2)=2\mathrm{deg}(w_i(x))\le 2m_i-2$, by Lemma 2.2 (ii), there exists a unique ordered pair $(w_{i,0}(x),w_{i,1}(x))$ of elements in ${{{\mathcal {F}}}}_i$ such that $w_{i,0}(x)\ne 0$ and $w_i(x)^2=w_{i,0}(x)+w_{i,1}(x){\overline{f}}_i(x)$ in $\overline{{{{\mathcal {R}}}}}_i$. The latter implies $w_i(x)^2\equiv w_{i,0}(x)+w_{i,1}(x){\overline{f}}_i(x)$ (mod ${\overline{f}}_i(x)^4$), and hence $w_i(x)^2\equiv w_{i,0}(x)+w_{i,1}(x){\overline{f}}_i(x)$ (mod ${\overline{f}}_i(x)^2$) in ${\mathbb {F}}_2[x]$. $\square$

In order to determine the subsets ${\mathcal {V}}_i$, ${\mathcal {W}}_i^{(0)}$ and ${\mathcal {W}}_{i,h_0(x)}^{(1)}$ of ${{{\mathcal {F}}}}_i$, for any $h_0(x)\in {\mathcal {W}}_i^{(0)}$ and $2\le i\le \lambda$, we need the following lemma.

Lemma 2.6

(cf. [6, Lemma 5.1 (i)–(iv)]) Let $2\le i\le \lambda$ and set

$$\begin{aligned} {\mathcal {H}}_i=\{a(x)\in {\mathcal {F}}_i={\mathbb {F}}_2[x]/\langle {\overline{f}}_i(x)\rangle \mid a(x)^{2^{\frac{m_i}{2}}}\equiv a(x) \ (\mathrm{mod} \ {\overline{f}}_i(x))\}. \end{aligned}$$

Then we have the following conclusions:

• (i) In the finite field ${\mathcal {F}}_i$, we have $x^{-1}=x^{2^{\frac{m_i}{2}}}$.

• (ii) ${\mathcal {H}}_i$ is a subfield of ${\mathcal {F}}_i$ with $2^{\frac{m_i}{2}}$ elements.

• (iii) Let $\mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}$ be the trace function from ${\mathcal {F}}_i$ onto ${\mathcal {H}}_i$ defined by:

  $$\begin{aligned} {\mathrm{Tr}}_{{\mathcal {F}}_i/{\mathcal {H}}_i}(\xi )=\xi +\xi ^{2^{\frac{m_i}{2}}}, \quad \forall \xi \in {\mathcal {F}}_i. \end{aligned}$$

  Then for any $\alpha \in {\mathcal {H}}_i$, the number of elements $\xi \in {\mathcal {F}}_i$ such that $\mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}(\xi )=\alpha$ is $2^{\frac{m_i}{2}}$, i.e. $|\mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}(\alpha )|=2^{\frac{m_i}{2}}$ (cf. [24, Corollary 7.17]).

• (iv) Let $w_i(x)$ be determined by Lemma 2.5. Then $x^{m_i}w_i(x^{-1})\equiv w_i(x)$ (mod ${\overline{f}}_i(x)$) in ${\mathbb {F}}_2[x]$ and $x^{\frac{m_i}{2}}w_i(x^{-1})\in {\mathcal {H}}_i$.

Finally, the subsets ${\mathcal {V}}_i$, ${\mathcal {W}}_i^{(0)}$ and ${\mathcal {W}}_{i,h_0(x)}^{(1)}$ of ${{{\mathcal {F}}}}_i$ can be calculated by the following theorem.

Theorem 2.7

Using the notations in Lemma 2.6, we have the following:

• (i) ${\mathcal {V}}_i=\left\{ x^{\frac{3m_i}{2}}\xi (x) \ \left( \mathrm{mod} \ {\overline{f}}_i(x)\right) \mid \xi (x)\in \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}(0)\right\}$. Therefore, we have $|{\mathcal {V}}_i|=|\mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}(0)|=2^{\frac{m_i}{2}}$.

• (ii) We have that $x^{m_i}w_i(x^{-1})^2\in {\mathcal {H}}_i$ and

  $$\begin{aligned} {\mathcal {W}}_i^{(0)}=\left\{ x^{m_i}\xi (x) \ \left( \mathrm{mod} \ {\overline{f}}_i(x)\right) \mid \xi (x)\in \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}\left( x^{m_i}w_i(x^{-1})^2\right) \right\} . \end{aligned}$$

  Hence $|{\mathcal {W}}_i^{(0)}|=|\mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}\left( x^{m_i}w_i(x^{-1})^2 \right) |=2^{\frac{m_i}{2}}$.

• (iii) Let $h_0(x)\in {\mathcal {W}}_i^{(0)}$. Then $h_0(x)+x^{2m_i}(w_i(x^{-1})^2+h_0(x^{-1})) \equiv 0$ (mod ${\overline{f}}_i(x)$) in ${\mathbb {F}}_2[x]$. Let $\varsigma _i(x)\in {\mathcal {F}}_i$ be defined by

  $$\begin{aligned} \varsigma _i(x):=\frac{h_0(x)+x^{2m_i}(w_i(x^{-1})^2+h_0(x^{-1}))}{{\overline{f}}_i(x)} \ \left( \mathrm{mod} \ {\overline{f}}_i(x)\right) \ \mathrm{in} \ {\mathbb {F}}_2[x]. \end{aligned}$$

  Then we have that $x^{-\frac{m_i}{2}}\varsigma _i(x)\in {\mathcal {H}}_i$ and

  $$\begin{aligned} {\mathcal {W}}_{i,h_0(x)}^{(1)}=\left\{ x^{\frac{m_i}{2}}\xi (x) \ \left( \mathrm{mod} \ {\overline{f}}_i(x)\right) \mid \xi (x)\in \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}\left( x^{-\frac{m_i}{2}} \varsigma _i(x)\right) \right\} . \end{aligned}$$

  Hence $|{\mathcal {W}}_{i,h_0}^{(1)}| =|\mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}(x^{-\frac{m_i}{2}} \varsigma _i(x))|=2^{\frac{m_i}{2}}$.

Proof

• (i) Let $h(x)\in {\mathcal {T}}_i={\mathcal {F}}_i$. Then $h(x)\in {\mathcal {V}}_i$ if and only if h(x) satisfies the following congruence relation:

  $$\begin{aligned} h(x)+x^{3m_i}h(x^{-1})\equiv 0 \ (\mathrm{mod} \ {\overline{f}}_i(x)) \ \mathrm{in} \ {\mathbb {F}}_2[x]. \end{aligned}$$

  (5)

  By Lemma 2.6 (i), we have $x^{2^{\frac{m_i}{2}}}=x^{-1}$ in ${\mathcal {F}}_i$. This implies that $(h(x))^{2^{\frac{m_i}{2}}}=h(x^{-1})$. Hence Eq. (5) is equivalent to

  $$\begin{aligned} \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}(x^{-\frac{3m_i}{2}}h(x))= & {} x^{-\frac{3m_i}{2}}h(x)+(x^{-\frac{3m_i}{2}}h(x))^{2^{\frac{m_i}{2}}}\\= & {} x^{-\frac{3m_i}{2}}h(x)+x^{\frac{3m_i}{2}}h(x^{-1})\\= & {} x^{-\frac{3m_i}{2}}\left( h(x)+x^{3m_i}h(x^{-1})\right) \\= & {} 0 \ \mathrm{in} \ {\mathcal {F}}_i. \end{aligned}$$

  Now, denote $\xi (x)=x^{-\frac{3m_i}{2}}h(x)$. Then $\xi (x)\in \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}(0)$ and $h(x)=x^{\frac{3m_i}{2}}\xi (x)$. Hence ${\mathcal {V}}_i=x^{\frac{3m_i}{2}}\cdot \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}(0)=\left\{ x^{\frac{3m_i}{2}}\xi (x) \in {\mathcal {F}}_i \mid \xi (x)\in \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}(0)\right\}$.

• (ii) By Lemma 2.6 (iv), we have $x^{\frac{m_i}{2}}w_i(x^{-1})\in {\mathcal {H}}_i$. Since ${\mathcal {H}}_i$ is a subfield of ${\mathcal {F}}_i$, we obtain $x^{m_i}w_i(x^{-1})^2=(x^{\frac{m_i}{2}}w_i(x^{-1}))^2\in {\mathcal {H}}_i$.

  Let $h_0(x)\in {\mathcal {F}}_i$. Then $h_0(x)\in {\mathcal {W}}_i^{(0)}$ if and only if $h_0(x)$ satisfies the following congruence relation:

  $$\begin{aligned} h_0(x)+x^{2m_i}(w_i(x^{-1})^2+h_0(x^{-1}))\equiv 0 \ (\mathrm{mod} \ {\overline{f}}_i(x)) \ \mathrm{in} \ {\mathbb {F}}_2[x]. \end{aligned}$$

  (6)

  By $x^{2^{\frac{m_i}{2}}}=x^{-1}$ and $(h_0(x))^{2^{\frac{m_i}{2}}}=h_0(x^{-1})$, Eq. (6) is equivalent to that $\mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}(x^{-m_i}h_0(x))=x^{-m_i}h_0(x)+x^{m_i}h_0(x^{-1})=x^{m_i}w_i(x^{-1})^2.$ Now, we set $\xi (x)=x^{-m_i}h_0(x)$. Then $\xi (x)\in \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}\left( x^{m_i}w_i(x^{-1})^2\right)$ and $h_0(x)=x^{m_i}\xi (x)$. So ${\mathcal {W}}_i^{(0)}=\{x^{m_i}\xi (x) \in {\mathcal {F}}_i \mid \xi (x)\in \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}(x^{m_i}w_i(x^{-1})^2)\}$.

• (iii) Let $h_0(x)\in {\mathcal {W}}_i^{(0)}$. Then $h_0(x)$ satisfies Eq. (6). This implies that $h_0(x)+x^{2m_i}(w_i(x^{-1})^2+h_0(x^{-1}))$ is a multiple of the polynomial ${\overline{f}}_i(x)$ in ${\mathbb {F}}_2[x]$. Hence the polynomial $\varsigma _i(x)\in {\mathcal {F}}_i$ is well-defined.

As $2\le i\le \lambda$, $f_i(x)$ is self-reciprocal in ${\mathbb {Z}}_4[x]$. This implies that ${\overline{f}}_i(x)$ is a self-reciprocal polynomial in ${\mathbb {F}}_2[x]$, i.e., $\widetilde{{\overline{f}}_i(x)}={\overline{f}}_i(x)$. From this and by $\mathrm{deg}({\overline{f}}_i(x))=m_i$, we deduce $x^{m_i}{\overline{f}}_i(x^{-1})=\widetilde{{\overline{f}}_i(x)}={\overline{f}}_i(x)$. Further, by Lemma 2.6 (iv), it follows that $w_i(x)^2=(x^{m_i}w_i(x^{-1}))=x^{2m_i}w_i(x^{-2})$. Hence

$$\begin{aligned} \left( x^{-\frac{m_i}{2}}\varsigma _i(x)\right) ^{2^{\frac{m_i}{2}}}= & {} x^{\frac{m_i}{2}}\left( \frac{h_0(x)+x^{2m_i}(w_i(x^{-1})^2+h_0(x^{-1}))}{{\overline{f}}_i(x)}\right) ^{2^{\frac{m_i}{2}}}\\= & {} x^{\frac{m_i}{2}}\cdot \frac{h_0(x^{-1})+x^{-2m_i}(w_i(x)^2+h_0(x))}{{\overline{f}}_i(x^{-1})}\\= & {} x^{\frac{m_i}{2}}\cdot \frac{h_0(x^{-1})+x^{-2m_i}(w_i(x)^2+h_0(x))}{x^{-m_i}{\overline{f}}_i(x)}\\= & {} x^{-\frac{m_i}{2}}\cdot \frac{x^{2m_i}h_0(x^{-1})+w_i(x)^2+h_0(x)}{{\overline{f}}_i(x)}\\= & {} x^{-\frac{m_i}{2}}\varsigma _i(x). \end{aligned}$$

This implies $x^{-\frac{m_i}{2}}\varsigma _i(x)\in {\mathcal {H}}_i$.

Now, let $h_1(x)\in {\mathcal {F}}_i$. Then $h_1(x)\in {\mathcal {W}}_{i,h_0(x)}^{(1)}$ if and only if $h_1(x)$ satisfies the following congruence relation:

$$\begin{aligned} h_0(x)+h_1(x){\overline{f}}_i(x)\equiv {\widehat{\delta }}_{i,0}(x)+{\widehat{\delta }}_{i,1}(x){\overline{f}}_i(x) \ (\mathrm{mod} \ {\overline{f}}_i(x)^2) \ \mathrm{in} \ {\mathbb {F}}_2[x], \end{aligned}$$

(7)

where ${\widehat{\delta }}_{i,0}(x)=x^{2m_i}(w_i(x^{-1})^2+h_0(x^{-1}))$ and ${\widehat{\delta }}_{i,1}(x)=x^{m_i}h_1(x^{-1})$. Since

$$\begin{aligned}&h_0(x)+{\widehat{\delta }}_{i,0}(x)+{\widehat{\delta }}_{i,1}(x){\overline{f}}_i(x) \\&\quad =h_0(x)+x^{2m_i}(w_i(x^{-1})^2+h_0(x^{-1}))+x^{m_i}h_1(x^{-1}){\overline{f}}_i(x) \\&\quad =\varsigma _i(x){\overline{f}}_i(x)+x^{m_i}h_1(x^{-1}){\overline{f}}_i(x) \ (\mathrm{mod} \ {\overline{f}}_i(x)^2), \end{aligned}$$

Equation (7) is equivalent to $\varsigma _i(x)+h_1(x)+x^{m_i}h_1(x^{-1}) \equiv 0$ (mod ${\overline{f}}_i(x)$) in ${\mathbb {F}}_2[x]$, i.e., $\mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}(x^{-\frac{m_i}{2}}h_1(x))= x^{-\frac{m_i}{2}}h_1(x)+x^{\frac{m_i}{2}}h_1(x^{-1})=x^{-\frac{m_i}{2}}\varsigma _i(x)$. Therefore, $h_1(x)=x^{\frac{m_i}{2}}\xi (x)\in {\mathcal {F}}_i$, where $\xi (x)\in \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}(x^{-\frac{m_i}{2}}\varsigma _i(x))$. Hence ${\mathcal {W}}_{i,h_0(x)}^{(1)}=\left\{ x^{\frac{m_i}{2}}\xi (x)\in {\mathcal {F}}_i \mid \xi (x)\in \mathrm{Tr}_{{\mathcal {F}}_i/{\mathcal {H}}_i}^{-1}\left( x^{-\frac{m_i}{2}}\varsigma _i(x) \right) \right\}$.

Finally, the conclusions $|{\mathcal {V}}_i|=|{\mathcal {W}}_i^{(0)}|=|{\mathcal {W}}_{i,h_0(x)}^{(1)}| =2^{\frac{m_i}{2}}$ follow from Lemma 2.6 (iii) immediately. $\square$

3 Main results

In this section, we list the main results of this paper by four theorems. Every result here is a significant simplification of the original in that it requires only 10 cases, compared to the 20 cases that the paper [5] requires. In particular, the expression here is much more direct and explicit. Moreover, they are interesting development and a non-trivial extension of the theory in [6].

First, we list all distinct ideals of ${\mathcal {B}}$ by the following theorem.

Theorem 3.1

All distinct cyclic codes over ${\mathbb {Z}}_4$ of length 4n are given by:

$$\begin{aligned} {{{\mathcal {C}}}}=\bigoplus _{i=1}^r{{{\mathcal {C}}}}_i=\sum _{i=1}^r{{{\mathcal {C}}}}_i =\{\xi _1(x)+\cdots +\xi _r(x)\mid \xi _i(x)\in {{{\mathcal {C}}}}_i, \ i=1,\ldots ,r\}, \end{aligned}$$

where

$$\begin{aligned} {{{\mathcal {C}}}}_i=\varepsilon _i(x)C_i=\{\varepsilon _i(x)b(x)\mid b(x)\in C_i\} \ (\mathrm{mod} \ x^{4n}-1), \end{aligned}$$

which is a subcode of ${\mathcal {C}}$ for all i, $1\le i\le r$, and $C_i$ is an ideal of the ring ${{{\mathcal {R}}}}_i$ listed by the following table:

Case	$C_i$	Type of ${\mathcal {C}}_i$	$|{\mathcal {C}}_i|$	L
1.	$\langle 0\rangle$	$4^02^0$	1	1
2.	$\langle 1\rangle$	$4^{4m_i}2^0$	$2^{8m_i}$	1
3.	$\langle 2{\overline{f}}_i(x)^s\rangle$ $(s=0,1,2,3)$	$4^02^{(4-s)m_i}$	$2^{(4-s)m_i}$	4
4.	$\langle f_i(x)^l,2\rangle$ $(l=1,2,3)$	$4^{(4-l)m_i}2^{lm_i}$	$2^{(8-l)m_i}$	3
5.	$\langle f_i(x)+2h(x)\rangle$	$4^{3m_i}2^0$	$2^{6m_i}$	$2^{m_i}$
6.	$\langle f_i(x)^2+2h(x),2{\overline{f}}_i(x)\rangle$	$4^{2m_i}2^{m_i}$	$2^{5m_i}$	$2^{m_i}$
7.	$\langle f_i(x)^2+2(h_0(x)+h_1(x){\overline{f}}_i(x))\rangle$	$4^{2m_i}2^0$	$2^{4m_i}$	$(2^{m_i})^2$
8.	$\langle f_i(x)^3+2h(x),2{\overline{f}}_i(x)\rangle$	$4^{m_i}2^{2m_i}$	$2^{4m_i}$	$2^{m_i}$
9.	$\langle f_i(x)^3+2{\overline{f}}_i(x)h(x),2{\overline{f}}_i(x)^2\rangle$	$4^{m_i}2^{m_i}$	$2^{3m_i}$	$2^{m_i}$
10.	$\langle f_i(x)^3+2{\overline{f}}_i(x)\cdot (w_{i,0}(x)+h(x){\overline{f}}_i(x))\rangle$	$4^{m_i}2^0$	$2^{2m_i}$	$2^{m_i}$

where $h(x),h_0(x),h_1(x)\in {\mathcal {T}}_i$ arbitrary, and L is the number of ideals $C_i$ in the same row.

Further, let $4^{k_{0,i}}2^{k_{1,i}}$ be the type of the subcode ${\mathcal {C}}_i$ listed in the table above for all integers i: $1\le i\le r$. Then the cyclic code ${\mathcal {C}}$ is of type

$$\begin{aligned} 4^{\sum _{i=1}^rk_{0,i}}2^{\sum _{i=1}^rk_{1,i}}. \end{aligned}$$

Hence the number of codewords in ${{{\mathcal {C}}}}$ is $\prod _{i=1}^r|{\mathcal {C}}_i|=2^{2\sum _{i=1}^rk_{0,i}+\sum _{i=1}^rk_{1,i}}$. Moreover, the minimum Hamming distance (Lee distance and Euclidean distance) of ${{{\mathcal {C}}}}$ satisfies $d_{\mathrm{min}}({{{\mathcal {C}}}})\le \mathrm{min}\{d_{\mathrm{min}}(\varepsilon _i(x)C_i)\mid i=1,\ldots ,r \}.$

Therefore, the number of all cyclic codes ${\mathcal {C}}$ over ${\mathbb {Z}}_4$ of length 4n is equal to $\prod _{i=1}^r(9+5\cdot 2^{m_i}+4^{m_i})$.

Using the notation of Theorem 3.1, ${{{\mathcal {C}}}}=\bigoplus _{j=1}^r\varepsilon _j(x)C_j$ is called the canonical form decomposition of the cyclic code ${\mathcal {C}}$ over ${\mathbb {Z}}_{4}$.

Similar to [8, Eq. (9)], in the rest of this paper we identify each polynomial $a(x)=a_0+a_1x+\cdots +a_{4n-1}x^{4n-1}\in {\mathcal {B}}=\frac{{\mathbb {Z}}_{4}[x]}{\langle x^{4n}-1\rangle }$ with $(a_0,a_1,\ldots ,a_{4n-1})\in {\mathbb {Z}}_{4}^{4n}$. Further, for any integer $1\le \rho \le 4(n-1)$, define:

$$\begin{aligned} {[}a(x)]_{\rho ,4n}=\left( \begin{array}{c}a(x)\\ xa(x)\\ \ldots \\ x^{\rho -1}a(x)\end{array}\right) =\left( \begin{array}{ccccc}a_0 &{}\quad a_1 &{}\quad \ldots &{}\quad a_{4n-2} &{}\quad a_{4n-1}\\ a_{4n-1} &{}\quad a_0 &{}\quad \ldots &{}\quad a_{4n-3} &{}\quad a_{4n-2}\\ \ldots &{}\quad \ldots &{}\quad \ldots &{}\quad \ldots &{}\quad \ldots \\ a_{4n-\rho +1} &{}\quad a_{4n-\rho +2} &{}\quad \ldots &{}\quad a_{4n-\rho -1} &{}\quad a_{4n-\rho }\end{array}\right) , \end{aligned}$$

(8)

which is a matrix over ${\mathbb {Z}}_{4}$ of size $\rho \times 4n$. Then we provide an efficient encoder for each cyclic code ${{{\mathcal {C}}}}$ over ${\mathbb {Z}}_{4}$ of length 4n by the following theorem.

Theorem 3.2

Let ${\mathcal {C}}$ be a cyclic code over ${\mathbb {Z}}_4$ of length 4n with canonical form decomposition ${{{\mathcal {C}}}}=\bigoplus _{i=1}^r{{{\mathcal {C}}}}_i$, where ${{{\mathcal {C}}}}_i=\varepsilon _i(x)C_i$ and $C_i$ is an ideal of ${\mathcal {R}}_i$ given by the table in Theorem 3.1. Then a generator matrix $G_i$ for each subcode ${\mathcal {C}}_i$, $1\le i\le r$, is given by the following table:

Case	Generator matrix $G_i$	Subcode ${\mathcal {C}}_i$
1.	0	$\{0\}$
2.	$[\varepsilon _i(x)]_{4m_i,4n}$	$\{{\underline{u}}G_i\mid {\underline{u}}\in {\mathbb {Z}}_4^{4m_i}\}$
3.	$[2{\overline{f}}_i(x)^{s}\varepsilon _i(x)]_{(4-s)m_i,4n}$	$\{{\underline{v}}G_i\mid {\underline{v}}\in {\mathbb {F}}_2^{(4-s)m_i}\}$
4.	$\left( \begin{array}{c}{[}f_i(x)^{l}\varepsilon _i(x){]}_{(4-l)m_i,4n}\\ {[}2\varepsilon _i(x){]}_{lm{_i,4n}}\end{array}\right)$	$\{({\underline{u}},{\underline{v}})G_i \mid {\underline{u}}\in {\mathbb {Z}}_4^{(4-l)m_i}, {\underline{v}}\in {\mathbb {F}}_{2}^{{lm_i}}\}$
5.	${[}(f_i(x)+2h(x))\varepsilon _i(x){]}_{3m_i,4n}$	$\{{\underline{u}}G_i\mid {\underline{u}}\in {\mathbb {Z}}_4^{3m_i}\}$
6.	$\left( \begin{array}{c}{[}(f_i(x)^2+2h(x))\varepsilon _i(x){]}_{2m_i,4n}\\ {[}2{\overline{f}}_i(x)\varepsilon _i(x){]}_{m_i,4n}\end{array}\right)$	$\{({\underline{u}},{\underline{v}})G_i \mid {\underline{u}}\in {\mathbb {Z}}_{4}^{2m_i}, {\underline{v}}\in {\mathbb {F}}_2^{m_i}\}$
7.	${[}(f_i(x)^2 +2(h_0(x)+h_1(x){\overline{f}}_i(x)))\varepsilon _i(x){]}_{2m_i,4n}$	$\{{\underline{u}}G_i\mid {\underline{u}}\in {\mathbb {Z}}_4^{2m_i}\}$
8.	$\left( \begin{array}{c}{[}(f_i(x)^3+2h(x))\varepsilon _i(x){]}_{m_i,4n}\\ {[}2{\overline{f}}_i(x)\varepsilon _i(x){]}_{2m_i,4n}\end{array}\right)$	$\{({\underline{u}},{\underline{v}})G_i \mid {\underline{u}}\in {\mathbb {Z}}_4^{m_i}, {\underline{v}}\in {\mathbb {F}}_2^{2m_i}\}$
9.	$\left( \begin{array}{c}{[}(f_i(x)^3+2{\overline{f}}_i(x)h(x))\varepsilon _i(x){]}_{m_i,4n} \\ {[}2{\overline{f}}_i(x)^2\varepsilon _i(x){]}_{m_i,4n}\end{array}\right)$	$\{({\underline{u}},{\underline{v}})G_i \mid {\underline{u}}\in {\mathbb {Z}}_4^{m_i}, {\underline{v}}\in {\mathbb {F}}_2^{m_i}\}$
10.	${[}(f_i(x)^3 +2{\overline{f}}_i(x)(w_{i,0}(x)+h(x){\overline{f}}_i(x)))\varepsilon _i(x){]}_{m_i,4n}$	$\{{\underline{u}}G_i\mid {\underline{u}}\in {\mathbb {Z}}_4^{m_i}\}$

where $h(x),h_0(x),h_1(x)\in {\mathcal {T}}_i$ arbitrary.

Now, we determine the dual code of each cyclic code.

Theorem 3.3

Let ${{{\mathcal {C}}}}$ be a cyclic code over ${\mathbb {Z}}_4$ of length 4n with canonical form decomposition ${{{\mathcal {C}}}}=\bigoplus _{i=1}^r\varepsilon _i(x)C_i$, where $C_i$ is an ideal of the ring ${{{\mathcal {R}}}}_i$. Then the dual code of ${{{\mathcal {C}}}}$ is given by ${{{\mathcal {C}}}}^{\bot }=\bigoplus _{i=1}^r\varepsilon _{\mu (i)}(x)D_{\mu (i)} =\bigoplus _{j=1}^r\varepsilon _j(x)D_j,$ where $D_{\mu (i)}$ is an ideal of ${{{\mathcal {R}}}}_{\mu (i)}$ determined by the following table:

$C_i$ (mod $f_i(x^4)$)	$D_{\mu (i)}$ (mod $f_{\mu (i)}(x^4)$)
$\diamond$ $\langle 0\rangle$	$\bullet$ $\langle 1\rangle$
$\diamond$ $\langle 1\rangle$	$\bullet$ $\langle 0\rangle$
$\diamond$ $\langle 2{\overline{f}}_i(x)^s\rangle$ $(s=0,1,2,3)$	$\bullet$ $\langle f_{\mu (i)}(x)^{4-s},2\rangle$
$\diamond$ $\langle f_i(x)^l,2\rangle$ $(l=1,2,3)$	$\bullet$ $\langle 2{\overline{f}}_{\mu (i)}(x)^{4-l}\rangle$
$\diamond$ $\langle f_i(x)+2h(x)\rangle$	$\bullet$ $\langle f_{\mu (i)}(x)^3+ 2{\overline{f}}_{\mu (i)}(x)\cdot (x^{2m_i}w_{i,0}(x^{-1})+{\widehat{\vartheta }}_i(x){\overline{f}}_{\mu (i)}(x))\rangle$
	where ${\widehat{\vartheta }}_i(x)=x^{m_i}\left( w_{i,1}(x^{-1})+h(x^{-1})\right)$
$\diamond$ $\langle f_i(x)^2+2h(x),2{\overline{f}}_i(x)\rangle$	$\bullet$ $\langle f_{\mu (i)}(x)^3+2{\overline{f}}_{\mu (i)}(x){\widehat{t}}_i(x),2{\overline{f}}_{\mu (i)}(x)^2\rangle$
	where ${\widehat{t}}_i(x)=x^{2m_i}\left( w_{i,0}(x^{-1})+h(x^{-1})\right)$
$\diamond$ $\langle f_i(x)^2+2(h_0(x)+h_1(x){\overline{f}}_i(x))\rangle$	$\bullet$ $\langle f_{\mu (i)}(x)^2+ 2({\widehat{\delta }}_{i,0}(x)+{\widehat{\delta }}_{i,1}(x){\overline{f}}_{\mu (i)}(x))\rangle$
	where ${\widehat{\delta }}_{i,0}(x)=x^{2m_i}(w_{i}(x^{-1})^2+h_0(x^{-1}))$
	and ${\widehat{\delta }}_{i,1}(x)=x^{m_i}h_1(x^{-1})$
$\diamond$ $\langle f_i(x)^3+2h(x),2{\overline{f}}_i(x)\rangle$	$\bullet$ $\langle f_{\mu (i)}(x)^3+2x^{3m_i}h(x^{-1}),2{\overline{f}}_{\mu (i)}(x)\rangle$
$\diamond$ $\langle f_i(x)^3+2{\overline{f}}_i(x)h(x), 2{\overline{f}}_i(x)^2\rangle$	$\bullet$ $\langle f_{\mu (i)}(x)^2+2{\widehat{t}}_i(x),2{\overline{f}}_{\mu (i)}(x)\rangle$
	where ${\widehat{t}}_i(x)=x^{2m_i}\left( w_{i,0}(x^{-1})+h(x^{-1})\right)$
$\diamond$ $\langle f_i(x)^3 + 2{\overline{f}}_i(x)(w_{i,0}(x)+h(x){\overline{f}}_i(x))\rangle$	$\bullet$ $\langle f_{\mu (i)}(x)+2{\widehat{\vartheta }}_i(x)\rangle$
	where ${\widehat{\vartheta }}_i(x)=x^{m_i}(w_{i,1}(x^{-1})+h(x^{-1}))$

in which $\mu$ is a permutation on the set $\{1,\ldots ,r\}$ defined by

$$\begin{aligned} \mu (i)=i, \ \mathrm{if} \ 1\le i\le \lambda ; \ \mu (\lambda +j)=\lambda +j+\epsilon \ \mathrm{and} \ \mu (\lambda +j+\epsilon )=\lambda +j, \ \forall j=1,\ldots ,\epsilon , \end{aligned}$$

and $h(x),h_0(x),h_1(x)\in {\mathcal {T}}_i$ arbitrary.

Finally, we list all distinct self-dual cyclic codes by the following theorem.

Theorem 3.4

Using the notation of Sect. 2, all distinct self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n are given by:

$$\begin{aligned} {{{\mathcal {C}}}}=\bigoplus _{1\le i\le r}\varepsilon _i(x)C_i, \end{aligned}$$

where for each integer i, $1\le i\le r$, $C_i$ is an ideal of ${{{\mathcal {R}}}}_i$ given by the following three cases:

• (i) $C_1$ is one of the following 3 ideals:

  $$\begin{aligned} C_1=\langle 2\rangle , \quad C_1=\langle (x-1)^3, 2(x-1)\rangle , \ C_1=\langle (x-1)^3+2, 2(x-1)\rangle , \end{aligned}$$

• (ii) If $2\le i\le \lambda$, $C_i$ is one of the following $1+2^{\frac{m_i}{2}}+2^{m_i}$ ideals:

• (ii-1) $C_i=\langle 2\rangle$.

• (ii-2) $C_i=\langle f_i(x)^2+2(h_0(x)+h_1(x){\overline{f}}_i(x))\rangle$, where $h_0(x)\in {\mathcal {W}}_{i}^{(0)}$ and $h_1(x)\in {\mathcal {W}}_{i,h_0(x)}^{(1)}$.

• (ii-3) $C_i=\langle f_i(x)^3+2h(x),2{\overline{f}}_i(x)\rangle$, where $h(x)\in {{{\mathcal {V}}}}_i$.

• (iii) If $i=\lambda +j$, where $1\le j\le \epsilon$, $(C_i,C_{i+\epsilon })$ is one of $4^{m_{\lambda +j}}+5\cdot 2^{m_{\lambda +j}}+9$ pairs of ideals given by the following table:

$C_i$ (mod $f_i(x^4)$)	$C_{i+\epsilon }$ (mod $f_{i+\epsilon }(x^4)$)
$\diamond$ $\langle 0\rangle$	$\bullet$ $\langle 1\rangle$
$\diamond$ $\langle 1\rangle$	$\bullet$ $\langle 0\rangle$
$\diamond$ $\langle 2{\overline{f}}_i(x)^s\rangle$ $(s=0,1,2,3)$	$\bullet$ $\langle f_{i+\epsilon }(x)^{4-s},2\rangle$
$\diamond$ $\langle f_i(x)^l,2\rangle$ $(l=1,2,3)$	$\bullet$ $\langle 2{\overline{f}}_{i+\epsilon }(x)^{4-l}\rangle$
$\diamond$ $\langle f_i(x)+2h(x)\rangle$	$\bullet$ $\langle f_{i+\epsilon }(x)^3+ 2{\overline{f}}_{i+\epsilon }(x)\cdot (x^{2m_i}w_{i,0}(x^{-1})+{\widehat{\vartheta }}_i(x){\overline{f}}_{i+\epsilon }(x))\rangle$
	where ${\widehat{\vartheta }}_i(x)=x^{m_i}\left( w_{i,1}(x^{-1})+h(x^{-1})\right)$
$\diamond$ $\langle f_i(x)^2+2h(x),2{\overline{f}}_i(x)\rangle$	$\bullet$ $\langle f_{i+\epsilon }(x)^3+2{\overline{f}}_{i+\epsilon }(x){\widehat{t}}_i(x),2{\overline{f}}_{i+\epsilon }(x)^2\rangle$
	where ${\widehat{t}}_i(x)=x^{2m_i}\left( w_{i,0}(x^{-1})+h(x^{-1})\right)$
$\diamond$ $\langle f_i(x)^2+2(h_0(x)+h_1(x){\overline{f}}_i(x))\rangle$	$\bullet$ $\langle f_{i+\epsilon }(x)^2+ 2({\widehat{\delta }}_{i,0}(x)+{\widehat{\delta }}_{i,1}(x){\overline{f}}_{i+\epsilon }(x))\rangle$
	where ${\widehat{\delta }}_{i,0}(x)=x^{2m_i}(w_{i}(x^{-1})^2+h_0(x^{-1}))$
	and ${\widehat{\delta }}_{i,1}(x)=x^{m_i}h_1(x^{-1})$
$\diamond$ $\langle f_i(x)^3+2h(x),2{\overline{f}}_i(x)\rangle$	$\bullet$ $\langle f_{i+\epsilon }(x)^3+2x^{3m_i}h(x^{-1}),2{\overline{f}}_{i+\epsilon }(x)\rangle$
$\diamond$ $\langle f_i(x)^3+2{\overline{f}}_i(x)h(x),2{\overline{f}}_i(x)^2\rangle$	$\bullet$ $\langle f_{i+\epsilon }(x)^2+2{\widehat{t}}_i(x),2{\overline{f}}_{i+\epsilon }(x)\rangle$
	where ${\widehat{t}}_i(x)=x^{2m_i}\left( w_{i,0}(x^{-1})+h(x^{-1})\right)$
$\diamond$ $\langle f_i(x)^3 + 2{\overline{f}}_i(x)(w_{i,0}(x)+h(x){\overline{f}}_i(x))\rangle$	$\bullet$ $\langle f_{i+\epsilon }(x)+2{\widehat{\vartheta }}_i(x)\rangle$
	where ${\widehat{\vartheta }}_i(x)=x^{m_i}(w_{i,1}(x^{-1})+h(x^{-1}))$

in which $h(x),h_0(x), h_1(x)\in {{{\mathcal {T}}}}_i$ arbitrary.

Therefore, the number of self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n is

$$\begin{aligned} 3\cdot \prod _{2\le i\le \lambda }(1+2^{\frac{m_i}{2}}+2^{m_i}) \cdot \prod _{j=1}^\epsilon (9+5\cdot 2^{m_{\lambda +j}}+4^{m_{\lambda +j}}). \end{aligned}$$

Remark

• (i) In [5, Corollary 4.6], the number of self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n was proposed as:

  $$\begin{aligned} 3^\lambda \cdot \prod _{1\le j\le \epsilon }(9+5\cdot 2^{m_{\lambda +j}}+4^{m_{\lambda +j}}). \end{aligned}$$

  The mistake in this formula occurs in the cases (ii-2) and (ii-3). Obviously, the conclusion of [5, Corollary 4.6] holds when $\lambda =1$.

• (ii) By the table in p. 302 of [5], the number ${{{\mathcal {N}}}}$ of self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n was proposed by the following table, where n is odd and $12\le 4n\le 100$:

4n	${{{\mathcal {N}}}}$	4n	${{{\mathcal {N}}}}$	4n	${{{\mathcal {N}}}}$
12, 20, 44, 52, 76	9	28	339	84	4500225
36, 68, 100	27	60	9315	92	12613659

There are mistakes in this table. Now, we correct them as follows:

4n	${\mathcal {N}}$	corrected
12	$21=3\cdot (1+2+2^2)$	$\checkmark$
20	$63=3\cdot (1+2^2+2^4)$	$\checkmark$
28	$339=3\cdot (9+5\cdot 2^3+4^3)$
36	$1533=3\cdot (1+2+2^2)\cdot (1+2^3+2^6)$	$\checkmark$
44	$3171=3\cdot (1+2^5+2^{10})$	$\checkmark$
52	$12483=3\cdot (1+2^6+2^{12})$	$\checkmark$
60	$152145=3\cdot (1+2+2^2)\cdot (1+2^2+2^4)\cdot (9+5\cdot 2^4+4^4)$	$\checkmark$
68	$223587=3\cdot (1+2^4+2^8)^2$	$\checkmark$
76	$787971=3\cdot (1+2^9+2^{18})$	$\checkmark$
84	$10500525=3\cdot (1+2+2^2)\cdot (9+5\cdot 2^3+4^3)\cdot (9+5\cdot 2^6+4^6)$	$\checkmark$
92	$12613659=3\cdot (9+5\cdot 2^{11}+4^{11})$
100	$66124863=3\cdot (1+2^2+2^4)\cdot (1+2^{10}+2^{20})$	$\checkmark$

4 Examples

In this section, we show how to use Theorem 3.4 to construct self-dual codes over ${\mathbb {Z}}_{4}$ of length 4n.

Example 4.1

We construct all 21 self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 12.

First, we have $y^3-1=f_1(y)f_2(y)$, where $f_1(y)=y-1$ and $f_2(y)=y^2+y+1$. In this case, $\lambda =r=2$, $\epsilon =0$. ${\widetilde{f}}_1(y)=c_1f_1(y)$, ${\widetilde{f}}_2(y)=c_2f_1(y)$, $c_1=-1$ and $c_2=1$. Obviously, $3\cdot f_2(y)+(y+2)f_1(y)=1$ in ${\mathbb {Z}}_4[y]$.

Using the notation in Sect. 2, we obtain

$$\begin{aligned} \varepsilon _1(x)=3f_2(x^4)=3x^8+3x^4+3 \ \text{ and } \ \varepsilon _2(x)=(x^4+2)f_1(x^4)=x^8+x^4+2. \end{aligned}$$

Further, by Theorems 2.5, 2.7 and Lemma 2.6 we have the following:

$\diamond$:

${\overline{f}}_2(x)=x^2+x+1$, $\overline{{\mathcal {R}}}_2=\frac{{\mathbb {F}}_2[x]}{\langle (x^2+x+1)^4\rangle }$ and ${\mathcal {T}}_2=\{a+bx\mid a,b\in {\mathbb {F}}_2\}$.

$\diamond$:

$g_2(x)=\frac{(x^2+x+1)^2-((x^2)^2+x^2+1)}{2(x^2+x+1)}=x\in {\mathbb {Z}}_4[x]$, $w_2(x)=g_2(x)=x$ (mod 2), and $w_2(x)^2=x^2=1+x+1\cdot (x^2+x+1)$.

Hence $w_{i,0}(x)=1+x$ and $w_{i,1}(x)=1$.

$\diamond$:

${\mathcal {F}}_2=\frac{{\mathbb {F}}_2[x]}{\langle x^2+x+1\rangle }=\{a+bx\mid a,b\in {\mathbb {F}}_2\}$ and ${\mathcal {H}}_2=\{\xi \in {\mathcal {F}}_2\mid \xi ^{2^1}=\xi \}={\mathbb {F}}_2$.

$\diamond$:

$\mathrm{Tr}^{-1}_{{\mathcal {F}}_2/{\mathcal {H}}_2}(0)=\{\xi \in {\mathcal {F}}_2\mid \xi +\xi ^2=0\}=\{0,1\}$; $\mathrm{Tr}^{-1}_{{\mathcal {F}}_2/{\mathcal {H}}_2}(1)=\{\xi \in {\mathcal {F}}_2\mid \xi +\xi ^2=1\}=\{x,1+x\}$.

$\diamond$:

${\mathcal {V}}_2=\{x^{\frac{3\cdot 2}{2}}\xi \mid \xi \in \mathrm{Tr}^{-1}_{{\mathcal {F}}_2/{\mathcal {H}}_2}(0)\} =x^3\cdot \mathrm{Tr}^{-1}_{{\mathcal {F}}_2/{\mathcal {H}}_2}(0)=\{0,1\}$ (mod $x^2+x+1$).

$\diamond$:

${\mathcal {W}}_2^{(0)}=\{x^{2}\xi \mid \xi \in \mathrm{Tr}^{-1}_{{\mathcal {F}}_2/{\mathcal {H}}_2}(1)\}=\{1,x\}$, since $x^{m_2}w_2(x^{-1})^2=x^2x^{-2}=1$.

$\diamond$:

${\mathcal {W}}_{2,1}^{(1)}=\{x^{-\frac{2}{2}}\xi \mid \xi \in \mathrm{Tr}^{-1}_{{\mathcal {F}}_2/{\mathcal {H}}_2}(0)\}=\{0,x\}$. This conclusion is due to $\varsigma _2(x)=\frac{1+x^{4}(x^{-2}+1)}{x^2+x+1}= \frac{1+x^{2}+x^{4}}{x^2+x+1}=x^2+x+1=0\in {\mathcal {F}}_2$ and $x^{-\frac{2}{2}}\varsigma _2(x)=0\in {\mathcal {H}}_2$.

$\diamond$:

${\mathcal {W}}_{2,x}^{(1)}=\{x^{-\frac{2}{2}}\xi \mid \xi \in \mathrm{Tr}^{-1}_{{\mathcal {F}}_2/{\mathcal {H}}_2}(1)\}=\{1,1+x\}$. This conclusion is due to $\varsigma _2(x)=\frac{x+x^{4}(x^{-2}+x^{-1})}{x^2+x+1}= \frac{x+x^{2}+x^{3}}{x^2+x+1}=x\in {\mathcal {F}}_2$ and $x^{-\frac{2}{2}}\varsigma _2(x)=1\in {\mathcal {H}}_2$.

Then by Theorem 3.4 (i) and (ii), all distinct 21 self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 12 are given by ${\mathcal {C}}=\varepsilon _1(x)C_{1}\oplus \varepsilon _2(x)C_{2},$ where $C_1$ is given by Theorem 3.4 (i) and $C_2$ is an ideals of the ring $\frac{{\mathbb {Z}}_4[x]}{\langle x^8+x^4+1\rangle }$ given by:

$C_2=\langle 2\rangle$.

$C_2=\langle f_2(x)^2+2(h_0(x)+h_1(x){\overline{f}}_2(x))\rangle$, $h_1(x)\in {\mathcal {W}}_{2,h_0(x)}^{(1)}$ and $h_0(x)\in {\mathcal {W}}_{2}^{(0)}$.

$C_2=\langle f_2(x)^3+2h(x),2{\overline{f}}_2(x)\rangle$, $h(x)\in {{{\mathcal {V}}}}_2$.

Specifically, all these 21 self-dual cyclic codes over ${\mathbb {Z}}_4$ are the following:

$$\begin{aligned} {\mathcal {C}}_{j,k}=\varepsilon _1(x)C_{1,j}\oplus \varepsilon _1(x)C_{2,k}, \ j=1,2,3 \ \mathrm{and} \ k=1,2,\ldots ,7, \end{aligned}$$

where $C_{1,j}$ is an ideals of the ring $\frac{{\mathbb {Z}}_4[x]}{\langle x^4-1\rangle }$ given by:

$C_{1,1}=\langle 2\rangle , \quad C_{1,2}=\langle (x-1)^3, 2(x-1)\rangle , \ C_{1,3}=\langle (x-1)^3+2, 2(x-1)\rangle ;$

and $C_{2,k}$ is an ideals of the ring $\frac{{\mathbb {Z}}_4[x]}{\langle x^8+x^4+1\rangle }$ given by:

$C_{2,1}=\langle 2\rangle$,     $C_{2,2}=\langle (x^2+x+1)^2+2\rangle$,

$C_{2,3}=\langle (x^2+x+1)^2+2(1+x\cdot (x^2+x+1))\rangle$,

$C_{2,4}=\langle (x^2+x+1)^2+2(x+(x^2+x+1))\rangle$,

$C_{2,5}=\langle (x^2+x+1)^2+2(x+(1+x)(x^2+x+1))\rangle$,

$C_{2,6}=\langle (x^2+x+1)^3, 2(x^2+x+1)\rangle$,

$C_{2,7}=\langle (x^2+x+1)^3+2, 2(x^2+x+1)\rangle$.

Now, let

$G_{1,1}=\left( \begin{array}{cccccccccccc} 2 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0\\ 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 \\ 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 \\ 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2\end{array}\right)$, $G_{1,2}=\left( \begin{array}{cccccccccccc} 1 &{} 1 &{} 3 &{} 3 &{} 1 &{} 1 &{} 3 &{} 3 &{} 1 &{} 1 &{} 3 &{} 3\\ \hline 2 &{} 2 &{} 0 &{} 0 &{} 2 &{} 2 &{} 0 &{} 0 &{} 2 &{} 2 &{} 0 &{} 0 \\ 0 &{} 2 &{} 2 &{} 0 &{} 0 &{} 2 &{} 2 &{} 0 &{} 0 &{} 2 &{} 2 &{} 0 \end{array}\right)$,

$G_{1,3}=\left( \begin{array}{cccccccccccc} 3 &{} 1 &{} 3 &{} 3 &{} 3 &{} 1 &{} 3 &{} 3 &{} 3 &{} 1 &{} 3 &{} 3\\ \hline 2 &{} 2 &{} 0 &{} 0 &{} 2 &{} 2 &{} 0 &{} 0 &{} 2 &{} 2 &{} 0 &{} 0 \\ 0 &{} 2 &{} 2 &{} 0 &{} 0 &{} 2 &{} 2 &{} 0 &{} 0 &{} 2 &{} 2 &{} 0 \end{array}\right)$; $G_{2,1}=\left( \begin{array}{cccccccccccc} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 2 \\ 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 \\ 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 \\ 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 \\ 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 \end{array}\right)$,

$G_{2,2}=\left( \begin{array}{cccccccccccc} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 2 &{} 3 &{} 2 &{} 0 &{} 2 &{} 3 &{} 2\\ 2 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 2 &{} 3 &{} 2 &{} 0 &{} 2 &{} 3\\ 3 &{} 2 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 2 &{} 3 &{} 2 &{} 0 &{} 2\\ 2 &{} 3 &{} 2 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 2 &{} 3 &{} 2 &{} 0\end{array}\right)$, $G_{2,3}=\left( \begin{array}{cccccccccccc} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 \\ 0 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 1 \\ 1 &{} 0 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 0 &{} 1 &{} 0 &{} 0\end{array}\right)$,

$G_{2,4}=\left( \begin{array}{cccccccccccc} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 2 &{} 1 &{} 2 &{} 0 &{} 2 &{} 1 &{} 2\\ 2 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 2 &{} 1 &{} 2 &{} 0 &{} 2 &{} 1\\ 1 &{} 2 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 2 &{} 1 &{} 2 &{} 0 &{} 2\\ 2 &{} 1 &{} 2 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 2 &{} 1 &{} 2 &{} 0\end{array}\right)$, $G_{2,5}=\left( \begin{array}{cccccccccccc} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 0 &{} 3 &{} 0 &{} 0 &{} 0 &{} 3 &{} 0 \\ 0 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 0 &{} 3 &{} 0 &{} 0 &{} 0 &{} 3 \\ 3 &{} 0 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 0 &{} 3 &{} 0 &{} 0 &{} 0 \\ 0 &{} 3 &{} 0 &{} 3 &{} 0 &{} 2 &{} 0 &{} 1 &{} 0 &{} 3 &{} 0 &{} 0\end{array}\right)$

$G_{2,6}=\left( \begin{array}{cccccccccccc} 0 &{} 1 &{} 1 &{} 2 &{} 1 &{} 1 &{} 0 &{} 3 &{} 3 &{} 2 &{} 3 &{} 3 \\ 3 &{} 0 &{} 1 &{} 1 &{} 2 &{} 1 &{} 1 &{} 0 &{} 3 &{} 3 &{} 2 &{} 3 \\ \hline 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 2 &{} 2 &{} 0 &{} 2 &{} 2 &{} 2 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 2 &{} 2 &{} 0 &{} 2 &{} 2 &{} 2 \\ 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 2 &{} 2 &{} 0 &{} 2 &{} 2 \\ 2 &{} 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 2 &{} 2 &{} 0 &{} 2 \end{array}\right)$, $G_{2,7}=\left( \begin{array}{cccccccccccc} 0 &{} 1 &{} 1 &{} 2 &{} 3 &{} 1 &{} 0 &{} 3 &{} 1 &{} 2 &{} 3 &{} 3 \\ 3 &{} 0 &{} 1 &{} 1 &{} 2 &{} 3 &{} 1 &{} 0 &{} 3 &{} 1 &{} 2 &{} 3 \\ \hline 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 2 &{} 2 &{} 0 &{} 2 &{} 2 &{} 2 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 2 &{} 2 &{} 0 &{} 2 &{} 2 &{} 2 \\ 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 2 &{} 2 &{} 0 &{} 2 &{} 2 \\ 2 &{} 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 2 &{} 2 &{} 0 &{} 2 \end{array}\right)$.

Then by Theorem 3.2, the self-dual cyclic code ${\mathcal {C}}_{j,k}$ over ${\mathbb {Z}}_4$ of length 12 is generated by the matrix $\left( \begin{array}{c}G_{1,j} \\ G_{2,k}\end{array}\right)$, and the type of ${\mathcal {C}}_{j,k}$ is given by:

${\mathcal {C}}_{1,1}=2({\mathbb {Z}}_4^{12})$ is the trivial self-dual cyclic code of type $4^02^{12}$;

${\mathcal {C}}_{1,k}$ is of type $4^42^4$ for $k=2,3,4,5$; ${\mathcal {C}}_{1,k}$ is of type $4^22^8$ for $k=6,7$;

${\mathcal {C}}_{j,k}$ is of type $4^52^2$ for $j=2,3$ and $k=2,3,4,5$;

${\mathcal {C}}_{j,k}$ is of type $4^32^6$ for $j=2,3$ and $k=6,7$.

Example 4.2

We construct self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 28.

In this case, $y^7-1=f_1(y)f_2(y)f_3(y)$, where $f_1(y)=y-1$, $f_2(y)=y^3+2y^2+y+3$ and $f_3(y)=y^3+3y^2+2y+3=3{\widetilde{f}}_3(y)$. These imply ${\overline{f}}_1(x)=x+1$, ${\overline{f}}_2(x)=x^3+x+1$ and ${\overline{f}}_3(x)=x^3+x^2+1$ in ${\mathbb {F}}_2[x]$.

Using the notations in Sect. 2, we have $r=3$, $\lambda =1$, $\epsilon =1$, $m_1=1$ and $m_2=m_3=3$. Hence there are $\prod _{i=1}^3(9+5\cdot 2^{m_i}+4^{m_i})=293687$ distinct cyclic codes over ${\mathbb {Z}}_{4}$ of length 28.

For each $i=1,2,3$, let $F_i(y)=\frac{x^7-1}{f_i(y)}$ and find polynomials $u_i(y),v_i(y) \in {\mathbb {Z}}_{4}[y]$ satisfying $u_i(y)F_i(y)+v_i(y)f_i(y)=1$. Then set $\varepsilon _i(x)=u_i(x^4)F_i(x^4)$ (mod $x^{28}-1$). Precisely, we have

$\varepsilon _1(x)=3x^{24}+3x^{20}+3x^{16}+3x^{12}+3x^{8}+3x^4+3$,

$\varepsilon _2(x)=2x^{24}+2x^{20}+3x^{16}+2x^{12}+3x^{8}+3x^4+1$,

$\varepsilon _3(x)=3x^{24}+3x^{20}+2x^{16}+3x^{12}+2x^{8}+2x^4+1$.

$\diamond$:

${\mathcal {R}}_1=\frac{{\mathbb {Z}}_{4}[x]}{\langle f_1(x^4)\rangle }=\frac{{\mathbb {Z}}_{4}[x]}{\langle x^4-1\rangle }$ and ${\mathcal {F}}_1={\mathbb {F}}_{2}[x]/\langle {\overline{f}}_1(x)\rangle =\{0,1\}={\mathcal {T}}_1$.

$\diamond$:

${\mathcal {R}}_i=\frac{{\mathbb {Z}}_{4}[x]}{\langle f_i(x^4)\rangle }$ and ${\mathcal {F}}_i=\frac{{\mathbb {F}}_{2}[x]}{\langle {\overline{f}}_i(x)\rangle }=\{t_0+t_1x+t_2x^2\mid t_0,t_1,t_2\in \{0,1\}\}={\mathcal {T}}_i$, for $i=2,3$.

$\diamond$:

$g_2(x)=\frac{(x^3+2x^2+x+3)^2-(x^6+2x^4+x^2+3)}{2(x^3+2x^2+x+3)}=2x^2+2x+1\in {\mathbb {Z}}_4[x]$,

$w_2(x)=1=g_2(x)$ (mod 2), $w_{2,0}(x)=1$ and $w_{2,1}(x)=0$ which satisfy

$w_2(x)^2=1=w_{2,0}(x)+w_{2,1}(x){\overline{f}}_2(x) \ \text{ in } \ \overline{{\mathcal {R}}}_2={\mathbb {F}}_{2}[x]/\langle {\overline{f}}_2(x)^4\rangle .$

By Theorem 3.4, all 339 self-dual codes over ${\mathbb {Z}}_4$ of length 28 are given by:

$$\begin{aligned} {\mathcal {C}}=\varepsilon _1(x)C_1\oplus \varepsilon _2(x)C_2\oplus \varepsilon _3(x)C_3, \end{aligned}$$

where $C_i$ is an ideal of ${\mathcal {R}}_i$, $1\le i\le 3$, given by the following:

$\diamondsuit$ $C_1$ is one of the following 3 ideals in ${\mathcal {R}}_1$:

$$\begin{aligned} \langle 2\rangle ,\quad \langle (x-1)^3, 2(x-1)\rangle , \quad \langle (x-1)^3+2, 2(x-1)\rangle . \end{aligned}$$

$\diamondsuit$ $(C_2,C_3)$ is one of the following 113 pairs of ideals, where $C_2$ is an ideal of ${\mathcal {R}}_2$ and $C_3$ is an ideal of ${\mathcal {R}}_3$:

• 1. $C_2=\langle 0\rangle$ and $C_3=\langle 1\rangle$;

• 2. $C_2=\langle 1\rangle$ and $C_3=\langle 0\rangle$;

• 3. $C_2=\langle 2{\overline{f}}_2(x)^s\rangle$ and $C_3=\langle f_3(x)^{4-s},2\rangle$, where $s=0,1,2,3$;

• 4. $C_2=\langle f_2(x)^l, 2\rangle$ and $C_3=\langle 2{\overline{f}}_3(x)^{4-l}\rangle$, where $l=1,2,3$;

• 5. $C_2=\langle f_2(x)+2h(x)\rangle$ and $C_3=\langle f_{3}(x)^3 +2{\overline{f}}_{3}(x)(x^{6} +{\widehat{\vartheta }}_2(x){\overline{f}}_{3}(x))\rangle$;

• 6. $C_2=\langle f_2(x)^2+2h(x),2{\overline{f}}_2(x)\rangle$ and $C_{3}=\langle f_{3}(x)^3+2{\overline{f}}_{3}(x){\widehat{t}}_2(x),2{\overline{f}}_{3}(x)^2\rangle$;

• 7. $C_2=\langle f_2(x)^2+2(h_0(x)+h_1(x){\overline{f}}_2(x))\rangle$ and $C_{3}=\langle f_{3}(x)^2+2({\widehat{\delta }}_{2,0}(x)+{\widehat{\delta }}_{2,1}(x){\overline{f}}_{3}(x)) \rangle ;$

• 8. $C_2=\langle f_2(x)^3+2h(x),2{\overline{f}}_2(x)\rangle$ and $C_{3}=\langle f_{3}(x)^3+2(ax^2+bx+c),2{\overline{f}}_{3}(x)\rangle$ in which $ax^2+bx+c\equiv x^{3\cdot 3}h(x^{-1})$ (mod $\langle 2, {\overline{f}}_3(x)\rangle$);

• 9. $C_2=\langle f_2(x)^3+2{\overline{f}}_2(x)h(x),2{\overline{f}}_2(x)^2\rangle$ and $C_{3}=\langle f_{3}(x)^2+2{\widehat{t}}_2(x),2{\overline{f}}_{3}(x)\rangle$;

• 10. $C_2=\langle f_2(x)^3+2{\overline{f}}_2(x)(1+h(x){\overline{f}}_2(x))\rangle$ and $C_{3}=\langle f_{3}(x)+2{\widehat{\vartheta }}_2(x)\rangle$,

where

• $h(x)=h_0(x)=a+bx+cx^2$ and $h_1(x)=g+ux+vx^2$ with $a,b,c,g,u,v\in {\mathbb {F}}_2$;

• ${\widehat{\vartheta }}_2(x) =(a+b)x^2+cx+a\equiv x^{3}(w_{2,1}(x^{-1})+h(x^{-1}))\equiv x^3h(x^{27})$ $\equiv x^3(a+bx^{27}+cx^{26})$ (mod ${\overline{f}}_{3}(x)=x^3+x^2+1$) in ${\mathbb {F}}_2[x]$;

• ${\widehat{t}}_2(x) =(a+c+1)x^2+(a+b+c+1)x+b+c$ $\equiv x^{2\cdot 3}(w_{2,0}(x^{-1})+h(x^{-1}))$ (mod ${\overline{f}}_{3}(x)$) in ${\mathbb {F}}_2[x]$;

• ${\widehat{\delta }}_{2,0}(x) =bx^5+(a+c+1)x^4+1+a$ $\equiv x^{2\cdot 3}(w_{2}(x^{-1})^2+h_0(x^{-1}))$ (mod ${\overline{f}}_3(x)^2$) in ${\mathbb {F}}_2[x]$;

• ${\widehat{\delta }}_{2,1}(x) =(g+u)x^2+vx+g\equiv x^{3}h_1(x^{-1})$ (mod ${\overline{f}}_3(x)$) in ${\mathbb {F}}_2[x]$.

Shi et al. obtained some good cyclic codes over ${\mathbb {Z}}_4$ from $(1 +2u)$-constacyclic codes over the ring ${\mathbb {Z}}_4[u]/\langle u^2-1\rangle$ by [22, Tables 1–3]. From self-dual negacyclic codes over the ring $\frac{{\mathbb {Z}}_4[v]}{\langle v^2+2v\rangle }$ of length 14, 36 new and good self-dual 2-quasi-twisted ${\mathbb {Z}}_4$-codes with parameters $(28, 2^{28}, d_L=8, d_E=12)$ and of type $4^72^{14}$ and parameters $(28, 2^{28}, d_L=6, d_E=12)$ and of type $4^62^{16}$ were given in [9]; From self-dual cyclic codes over $\frac{{\mathbb {Z}}_4[v]}{\langle v^2+2v\rangle }$ of length 15, 70 new and good self-dual 2-quasi-cyclic ${\mathbb {Z}}_4$-codes with parameters $(30, 2^{30}, d_L=12)$ and 92 new and good self-dual 2-quasi-cyclic ${\mathbb {Z}}_4$-codes with parameters $(30, 2^{30}, d_L=10)$ were obtained in [10], where $d_H,d_L$ and $d_E$ be the minimum Hamming distance, Lee distance and Euclidean distance of a ${\mathbb {Z}}_4$-code respectively (cf. [25]).

Now, among the 339 self-dual codes over ${\mathbb {Z}}_4$ of length 28 listed above, we have the following 50 new and good self-dual cyclic ${\mathbb {Z}}_4$-codes with basic parameters $(28,|C|=2^{28},d_H=4,d_L=8,d_E=8)$:

• 8 self-dual cyclic ${\mathbb {Z}}_4$-codes of type $4^{12}2^4$ determined by:

  $C_1=\langle 2\rangle$, $C_2=\langle f_2(x)+2(a+bx+cx^2)\rangle$,

  $C_3=\langle f_{3}(x)^3+2{\overline{f}}_{3}(x)(x^{6}+((a+b)x^2+cx+a) {\overline{f}}_{3}(x))\rangle$,

  where $a,b,c\in {\mathbb {F}}_2=\{0,1\}$.

• 8 self-dual cyclic ${\mathbb {Z}}_4$-codes of type $4^{12}2^4$ determined by:

  $C_1=\langle 2\rangle$, $C_2=\langle f_2(x)^3+2{\overline{f}}_2(x)(1 +(a+bx+cx^2){\overline{f}}_2(x))\rangle$,

  $C_{3}=\left\langle f_{3}(x)+2((a+b)x^2+cx+a)\right\rangle$, where $a,b,c\in {\mathbb {F}}_2$.

• 4 self-dual cyclic ${\mathbb {Z}}_4$-codes of type $4^{10}2^8$ determined by:

  $C_1=\langle (x-1)^3, 2(x-1)\rangle$, $C_2=\langle 2{\overline{f}}_2(x)^3\rangle$ and $C_3=\langle f_3(x),2\rangle$, or $C_2=\langle f_2(x), 2\rangle$ and $C_3=\langle 2{\overline{f}}_3(x)^{3}\rangle$;

  $C_1=\langle (x-1)^3+2, 2(x-1)\rangle$, $C_2=\langle 2{\overline{f}}_2(x)^3\rangle$ and $C_3=\langle f_3(x),2\rangle$, or $C_2=\langle f_2(x), 2\rangle$ and $C_3=\langle 2{\overline{f}}_3(x)^{3}\rangle$.

• 2 self-dual cyclic ${\mathbb {Z}}_4$-codes of type $4^{7}2^{14}$ determined by:

  $C_1=\langle (x-1)^3, 2(x-1)\rangle$, $C_2=\langle f_2(x)^2, 2\rangle$ and $C_3=\langle 2{\overline{f}}_3(x)^{2}\rangle$;

  $C_1=\langle (x-1)^3+2, 2(x-1)\rangle$, $C_2=\langle f_2(x)^2, 2\rangle$ and $C_3=\langle 2{\overline{f}}_3(x)^{2}\rangle$.

• 6 self-dual cyclic ${\mathbb {Z}}_4$-codes of type $4^{13}2^2$ determined by:

  $C_1=\langle (x-1)^3, 2(x-1)\rangle$, $C_2=\langle f_2(x)^2+2h(x),2{\overline{f}}_2(x)\rangle$,

  $C_{3}=\langle f_{3}(x)^3+2{\overline{f}}_{3}(x){\widehat{t}}_2(x),2{\overline{f}}_{3}(x)^2\rangle$,

  where $(h(x),{\widehat{t}}_2(x))\in \{(x^2, 1), (x, x^2 + 1), (x^2 + x, x), (x^2 + 1, x^2 + x + 1)$, $(x + 1, x + 1), (x^2 + x + 1, x^2)\}$.

• 6 self-dual cyclic ${\mathbb {Z}}_4$-codes of type $4^{13}2^2$ determined by:

  $C_1=\langle (x-1)^3, 2(x-1)\rangle$, $C_2=\langle f_2(x)^3+2{\overline{f}}_2(x)(1+h(x){\overline{f}}_2(x))\rangle$,

  $C_{3}=\langle f_{3}(x)+2{\widehat{\vartheta }}_2(x)\rangle$,

  where $(h(x),{\widehat{\vartheta }}_2(x))\in \{(x^2, x), (x, x^2), (x^2 + x, x^2 + x)$, $(x^2 + 1, x^2 + x + 1), (x + 1, 1), (x^2 + x + 1, x + 1)\}$.

• 8 self-dual cyclic ${\mathbb {Z}}_4$-codes of type $4^{13}2^2$ determined by:

  $C_1=\langle (x-1)^3+2, 2(x-1)\rangle$, $C_2=\langle f_2(x)^2+2(a+bx+cx^2),2{\overline{f}}_2(x)\rangle$,

  $C_{3}=\langle f_{3}(x)^3+2{\overline{f}}_{3}(x)((a+c+1)x^2+(a+b+c+1)x+b+c),2{\overline{f}}_{3}(x)^2\rangle$,

  where $a,b,c\in {\mathbb {F}}_2$.

• 8 self-dual cyclic ${\mathbb {Z}}_4$-codes of type $4^{13}2^2$ determined by:

  $C_1=\langle (x-1)^3+2, 2(x-1)\rangle$, $C_2=\langle f_2(x)^3+2{\overline{f}}_2(x)(1+(a+bx+cx^2){\overline{f}}_2(x))\rangle$,

  $C_{3}=\langle f_{3}(x)+2((a+b)x^2+cx+a)\rangle$, where $a,b,c\in {\mathbb {F}}_2$.

5 Proofs of Theorems 3.1–3.4

In this section, we give the detail proofs for Theorems 3.1–3.4. First, we prove that ideals of the ring ${\mathcal {B}}$ are determined by ideals of rings ${\mathcal {R}}_i$, $i=1,\ldots ,r$.

Proposition 5.1

Let ${\mathcal {C}}\subseteq {\mathcal {B}}=\frac{{\mathbb {Z}}_4[x]}{\langle x^{4n}-1\rangle }$. Then ${\mathcal {C}}$ is a cyclic code over ${\mathbb {Z}}_4$ of length 4n if and only if for each integer i, $1\le i\le r$, there is a unique ideal $C_i$ of the ring ${\mathcal {R}}_i$ such that ${\mathcal {C}}=\bigoplus _{i=1}^r{{{\mathcal {C}}}}_i=\sum _{i=1}^r{{{\mathcal {C}}}}_i$, where

$$\begin{aligned} {{{\mathcal {C}}}}_i=\varepsilon _i(x)C_i=\{\varepsilon _i(x)c_i(x) \mid c_i(x)\in C_i\} \ (\mathrm{mod} \ x^{4n}-1). \end{aligned}$$

Moreover, the number of codewords in ${\mathcal {C}}$ is equal to $|{\mathcal {C}}|=\prod _{i=1}^r|C_i|$.

Proof

Let ${\mathcal {C}}$ be a cyclic code over ${\mathbb {Z}}_4$ of length 4n. By Lemma 2.1 and properties of isomorphic rings, there is a unique ideal C of the direct product ring ${\mathcal {R}}_1\times \cdots \times {\mathcal {R}}_r$ such that ${{{\mathcal {C}}}}=\psi (C)$. Hence for each integer i, $1\le i\le r$, there is a unique ideal $C_i$ of ${\mathcal {R}}_i$ such that $C=C_1\times \cdots \times C_r$. This implies

$$\begin{aligned} {\mathcal {C}}= & {} \{\psi (c_1(x),\ldots ,c_r(x))\mid c_i(x)\in C_i, \ i=1,\ldots ,r\}\\= & {} \sum _{i=1}^r\{\varepsilon _i(x)c_i(x)\mid c_i(x)\in C_i\} \ (\mathrm{mod} \ x^{4n}-1). \end{aligned}$$

Then the conclusion follows from Lemma 2.1(i), ${{{\mathcal {C}}}}_i=\varepsilon _i(x)C_i\subseteq {\mathcal {B}}_i$ for all i and $|{\mathcal {C}}|=|C|=|C_1\times \cdots \times C_r|=\prod _{i=1}^r|C_i|$. $\square$

Then we investigate properties of ideals in the ring ${\mathcal {R}}_i=\frac{{\mathbb {Z}}_{4}[x]}{\langle f_i(x^4)\rangle }$.

Proposition 5.2

For any integer i, $1\le i\le r$, let $C_i=\langle f_i(x)^l+2v(x),2{\overline{f}}_i(x)^s\rangle$ be an ideal of ${\mathcal {R}}_i$, where $v(x)\in \overline{{\mathcal {R}}}_i$ and $0\le s\le l\le 4$, and set ${\mathcal {C}}_i=\varepsilon _i(x)C_i\subseteq {\mathcal {B}}$. Then we have the following conclusions:

• (i) The set ${\mathcal {C}}_i$ is a cyclic code of type $4^{(4-l)m_i}2^{(l-s)m_i}$ over ${\mathbb {Z}}_4$ with length 4n.

• (ii) Using the notation in Eq. (8), let $G=\left( \begin{array}{c}G_{i,(1)}\\ G_{i,(2)}\end{array}\right)$, where

  $$\begin{aligned} G_{i,(1)}=\left[ \left( f_i(x)^l+2v(x)\right) \varepsilon _i(x)\right] _{(4-l)m_i,4n}, \ \ G_{i,(2)}=\left[ 2{\overline{f}}_i(x)\varepsilon _i(x)\right] _{(l-s)m_i,4n}. \end{aligned}$$

  Then ${\mathcal {C}}_i=\left\{ ({\underline{a}}, {\underline{b}})G_i\mid {\underline{a}}\in {\mathbb {Z}}_4^{(4-l)m_i}, {\underline{b}}\in {\mathbb {F}}_2^{(l-s)m_i}\right\} .$

Proof

(i) For any vectors ${\underline{a}}=(a_0,a_1,\ldots ,a_{(4-l)m_i-1})\in {\mathbb {Z}}_4^{(4-l)m_i}$ and ${\underline{b}}=(b_0,b_1$, $\ldots ,b_{(l-s)m_i-1})\in {\mathbb {F}}_2^{(l-s)m_i}$, define a map $\varrho$ by

$$\begin{aligned} \varrho ({\underline{a}}, {\underline{b}}) =\sum _{0\le j\le (4-l)m_i-1}a_jx^j\left( f_i(x)^l+2v(x)\right) +\sum _{0\le t\le (l-s)m_i-1}2b_tx^t{\overline{f}}_i(x)^s \end{aligned}$$

(mod $f_i(x^4)$). Now, we claim that $\varrho$ is a bijection from ${\mathbb {Z}}_4^{(4-l)m_i} \times {\mathbb {F}}_2^{(l-s)m_i}$ onto $C_i$. In fact, since $C_i$ is an ideal of ${\mathcal {R}}_i$ generated by $f_i(x)^l+2v(x)$ and $2{\overline{f}}_i(x)^s$, we see that $\varrho ({\underline{a}}, {\underline{b}})\in C_i$. By Lemma 2.4, it follows that

$$\begin{aligned} |C_i|=2^{m_i(8-(l+s))}=2^{2(4-l)m_i+(l-s)m_i}=|{\mathbb {Z}}_4^{(4-l)m_i} \times {\mathbb {F}}_2^{(l-s)m_i}|. \end{aligned}$$

Hence we only need to prove that the map $\varrho$ is injective.

In fact, let ${\underline{c}}=(c_0,c_1,c_2,\ldots ,c_{(4-l)m_i-1})\in {\mathbb {Z}}_4^{(4-l)m_i}$ and ${\underline{d}}=(d_0,d_1, \ldots$, $d_{(l-s)m_i-1})\in {\mathbb {F}}_2^{(l-s)m_i}$ satisfying $\varrho ({\underline{a}}, {\underline{b}})=\varrho ({\underline{c}}, {\underline{d}})$ (mod $f_i(x^4)$). For each integer j, $0\le j\le (4-l)m_i-1$, we write: $a_j=a_{0,j}+2a_{1,j}$ and $c_j=c_{0,j}+2c_{1,j}$, where $a_{0,j},a_{1,j},c_{0,j},c_{1,j}\in {\mathbb {F}}_2$. Then by $\varrho ({\underline{a}}, {\underline{b}})=\varrho ({\underline{c}}, {\underline{d}})$ (mod $f_i(x^4)$), we have that $\varrho ({\underline{a}}, {\underline{b}}) \ (\text{ mod } \ \langle {\overline{f}}_i(x)^4, 2\rangle ) =\varrho ({\underline{c}}, {\underline{d}}) \ (\text{ mod } \ \langle {\overline{f}}_i(x)^4, 2\rangle )$. This implies

$$\begin{aligned} \left( \sum _{0\le j\le (4-l)m_i-1}a_{0,j}x^j\right) {\overline{f}}_i(x)^l=\left(\sum _{0\le j\le (4-l)m_i-1}c_{0,j}x^j\right){\overline{f}}_i(x)^l \ (\text{ mod } \ {\overline{f}}_i(x)^4). \end{aligned}$$

Since both $\sum _{0\le j\le (4-l)m_i-1}a_{0,j}x^j$ and $\sum _{0\le j\le (4-l)m_i-1}c_{0,j}x^j$ are polynomials in ${\mathbb {F}}_2[x]$ of degree $\le (4-l)m_i-1<(4-l)\mathrm{deg}({\overline{f}}_i(x))$, From the equation above, we can deduce $\sum _{0\le j\le (4-l)m_i-1}a_{0,j}x^j=\sum _{0\le j\le (4-l)m_i-1}c_{0,j}x^j$ in ${\mathbb {F}}_2[x]$, by Lemma 2.2 (ii). This implies $a_{0,j}=c_{0,j}, \ \forall j=0,1,\ldots ,(4-l)m_i-1.$

Moreover, by $\varrho ({\underline{a}}, {\underline{b}})=\varrho ({\underline{c}}, {\underline{d}})$ (mod $f_i(x^4)$) and $2(f_i(x)^l+2v(x))=2{\overline{f}}_i(x)^l$, we obtain

$$\begin{aligned}&2\left( \left( \sum _{0\le j\le (4-l)m_i-1}a_{1,j}x^j\right) \left( {\overline{f}}_i(x)\right) ^l +\left( \sum _{0\le t\le (l-s)m_i-1}b_tx^t\right) \left( {\overline{f}}_i(x)\right) ^s\right) \\= & {} 2\left( \left( \sum _{0\le j\le (4-l)m_i-1}c_{1,j}x^j\right) \left( {\overline{f}}_i(x)\right) ^l +\left( \sum _{0\le t\le (l-s)m_i-1}d_tx^t\right) \left( {\overline{f}}_i(x)\right) ^s\right) . \end{aligned}$$

Since $\sum _{0\le j\le (4-l)m_i-1}a_{1,j}x^j$ and $\sum _{0\le j\le (4-l)m_i-1}c_{1,j}x^j$ are polynomials in ${\mathbb {F}}_2[x]$ of degree $\le (4-l)m_i-1<(4-l)\mathrm{deg}({\overline{f}}_i(x))$, $\sum _{0\le t\le (l-s)m_i-1}b_tx^t$ and $\sum _{0\le t\le (l-s)m_i-1}d_tx^t$ are polynomials in ${\mathbb {F}}_2[x]$ of degree $\le (l-s)m_i-1<(l-s)\mathrm{deg}({\overline{f}}_i(x))$, we see that $\sum _{0\le j\le (4-l)m_i-1}a_{1,j}x^j=\sum _{0\le j\le (4-l)m_i-1}c_{1,j}x^j$ and $\sum _{0\le t\le (l-s)m_i-1}b_tx^t=\sum _{0\le t\le (l-s)m_i-1}d_tx^t$ in ${\mathbb {F}}_2[x]$. This implies

$$\begin{aligned} a_{1,j}=c_{1,j}, \ 0\le j\le (4-l)m_i-1; \ \text{ and } \ b_t=d_t, \ 0\le t\le (l-s)m_i-1. \end{aligned}$$

Summing up the results above, we get the following: ${\underline{a}}={\underline{c}}$ and ${\underline{b}}={\underline{d}}$. Therefore, $\varrho$ is injective and hence a bijection. Moreover, it is clear that

$$\begin{aligned} \varrho (({\underline{a}}, {\underline{b}})+({\underline{c}}, {\underline{d}})) =\varrho (({\underline{a}}+{\underline{c}}, {\underline{b}}+{\underline{d}})=\varrho ({\underline{a}}, {\underline{b}})+\varrho ({\underline{c}}, {\underline{d}}). \end{aligned}$$

Hence $\varrho$ is additive group isomorphism from $({\mathbb {Z}}_4^{(4-l)m_i},+) \times ({\mathbb {F}}_2^{(l-s)m_i},+)$ onto $(C_i,+)$. So $C_i$ is an abelian group of type $4^{(4-l)m_i}2^{(l-s)m_i}$.

By Lemma 5.1, ${{{\mathcal {C}}}}_i=\varepsilon _i(x)C_i$ is a cyclic code over ${\mathbb {Z}}_4$ of length 4n. Using Lemma 2.1 (iii), the map $\psi _i$ induces an additive group isomorphism from $(C_i,+)$ onto $({{{\mathcal {C}}}}_i,+)$. Hence, ${\mathcal {C}}_i$ is an abelian group of type $4^{(4-l)m_i}2^{(l-s)m_i}$.

(ii) For any positive integer k, denote $X_k=(1,x,\ldots ,x^{k-1})^{\mathrm{tr}}$ where $M^{\mathrm{tr}}$ represents the transpose of a matrix M. Using the notation of Eq. (8), by the proof of (i), each codeword in ${\mathcal {C}}_i$ can be uniquely expessed as

$$\begin{aligned} \varepsilon _i(x)\varrho ({\underline{a}}, {\underline{b}})= & {} \left( {\underline{a}}X_{(4-l)m_i}\right) \left( f_i(x)^l+2v(x)\right) +({\underline{b}}X_{(l-s)m_i}){\overline{f}}_i(x)^s \\= & {} {\underline{a}}\left( X_{(4-l)m_i}f_i(x)^l+2v(x)\right) +{\underline{b}}(X_{(l-s)m_i}{\overline{f}}_i(x)^s)\\= & {} {\underline{a}}G_{i,(1)}+{\underline{b}}G_{i,(2)}, \end{aligned}$$

i.e., $\varepsilon _i(x)\varrho ({\underline{a}}, {\underline{b}})=({\underline{a}}, {\underline{b}})G_i$, where ${\underline{a}}\in {\mathbb {Z}}_4^{(4-l)m_i}$ and ${\underline{b}}\in {\mathbb {F}}_2^{(l-s)m_i}$.

Therefore, ${\mathcal {C}}_i=\{({\underline{a}}, {\underline{b}})G_i\mid {\underline{a}}\in {\mathbb {Z}}_4^{(4-l)m_i}$, ${\underline{b}}\in {\mathbb {F}}_2^{(l-s)m_i}\}$. $\square$

A. Proof for Theorem 3.1.

By Proposition 5.2, the ideal $C_i$ of ${\mathcal {R}}_i$ listed in the table of Theorem 3.1 is of the type as given by the table, for all Cases 1–10. Further, as ideals in distinct cases are of different types, any two ideals are distinct in different cases. Obviously, the 7 ideals in Cases 3 and 4 are distinct, as they are of different types. Then we need to consider ideals $C_i$ of ${\mathcal {R}}_i$ in Cases 5–10.

Case 5. Let $h(x), q(x)\in {\mathcal {T}}_i=\{\sum _{j=0}^{m_i-1}t_jx^j\mid t_j\in \{0,1\}, \ j=0,1,\ldots ,m_i-1\}$ be such that $\langle f_i(x)+2h(x)\rangle =\langle f_i(x)+2q(x)\rangle$ as ideals of the ring ${\mathcal {R}}_i$. Then by Lemma 2.4 (iii) and Equation (4), we have $h(x)\equiv q(x)$ (mod ${\overline{f}}_i(x)$) in ${\mathbb {F}}_2[x]$. This implies $h(x)=q(x)$. Therefore, the ideals $C_i$ in Case 5 are distinct.

Case 6. Let $h(x), q(x)\in {\mathcal {T}}_i$ be such that $\langle f_i(x)^2+2h(x),2{\overline{f}}_i(x)\rangle =\langle f_i(x)^2+2q(x),2{\overline{f}}_i(x)\rangle$. Then by Eq. (4) in Lemma 2.4, we have that $h(x)\equiv q(x)$ (mod ${\overline{f}}_i(x)$) in ${\mathbb {F}}_2[x]$. This implies $h(x)=q(x)$.

Similarly, one can easily verify that the ideals $C_i$ in Case 8 are distinct.

Case 7. Let $h_0(x), h_1(x), q_0(x), q_1(x)\in {\mathcal {T}}_i$ be such that $\langle f_i(x)^2+2(h_0(x)+h_1(x){\overline{f}}_i(x))\rangle =\langle f_i(x)^2+2(q_0(x)+q_1(x){\overline{f}}_i(x))\rangle$ as ideals of ${\mathcal {R}}_i$. By Lemma 2.4 (iii) and Eq. (4), we have $h_0(x)+h_1(x){\overline{f}}_i(x)\equiv q_0(x)+q_1(x){\overline{f}}_i(x)$ (mod ${\overline{f}}_i(x)^2$) in ${\mathbb {F}}_2[x]$. This implies $h_0(x)=q_0(x)$ and $h_1(x)=q_1(x)$. Therefore, the ideals $C_i$ in Case 7 are distinct.

Case 9. Let $h(x), q(x)\in {\mathcal {T}}_i$ be such that $\langle f_i(x)^3+2h(x){\overline{f}}_i(x),2{\overline{f}}_i(x)^2\rangle$ $=\langle f_i(x)+2q(x){\overline{f}}_i(x),2{\overline{f}}_i(x)^2\rangle$ as ideals of ${\mathcal {R}}_i$. By Lemma 2.4 (iii) and Eq.  (4), we have $h(x){\overline{f}}_i(x)\equiv q(x){\overline{f}}_i(x)$ (mod ${\overline{f}}_i(x)^2$) in ${\mathbb {F}}_2[x]$. This implies that $h(x)=q(x)$. Therefore, the ideals $C_i$ in Case 9 are distinct.

Case 10. Let $h(x), q(x)\in {\mathcal {T}}_i$ be such that $\langle f_i(x)^3+2{\overline{f}}_i(x)\cdot (w_{i,0}(x)+h(x){\overline{f}}_i(x))\rangle =\langle f_i(x)^3+2{\overline{f}}_i(x)\cdot (w_{i,0}(x)+q(x){\overline{f}}_i(x))\rangle$. By Lemma 2.4 (iii) and Eq. (4), we have ${\overline{f}}_i(x)\cdot (w_{i,0}(x)+h(x){\overline{f}}_i(x)) \equiv {\overline{f}}_i(x)\cdot (w_{i,0}(x)+q(x){\overline{f}}_i(x))$ (mod ${\overline{f}}_i(x)^3$) in ${\mathbb {F}}_2[x]$. This implies $w_{i,0}(x)+h(x){\overline{f}}_i(x)) \equiv w_{i,0}(x)+q(x){\overline{f}}_i(x)$ (mod ${\overline{f}}_i(x)^2$), and hence $h(x)=q(x)$. So, the ideals $C_i$ in Case 10 are distinct.

As stated above, we conclude that all ideals of ${\mathcal {R}}_i$ in Cases 1–10 are distinct. Now, let ${\mathcal {L}}_i$ be the number of all ideals in ${\mathcal {R}}_i$. Then we have

$$\begin{aligned} {\mathcal {L}}_i\ge 9+5\cdot 2^{m_i}+(2^{m_i})^2, \quad i=1,2,\ldots ,r. \end{aligned}$$

Further, by Proposition 5.1 the number of cyclic ideals over ${\mathbb {Z}}_4$ of length 4n is equal to $\prod _{i=1}^r{\mathcal {L}}_i\ge \prod _{i=1}^r(9+5\cdot 2^{m_i}+(2^{m_i})^2)$. On the other hand, by [11, Corollary 3.4], we know that the number of cyclic ideals over ${\mathbb {Z}}_4$ of length 4n equals $\prod _{i=1}^r(9+5\cdot 2^{m_i}+(2^{m_i})^2)$. From these, we deduce that

$$\begin{aligned} {\mathcal {L}}_i=9+5\cdot 2^{m_i}+(2^{m_i})^2, \quad i=1,2,\ldots ,r. \end{aligned}$$

Therefore, all distinct ideals of ${\mathcal {R}}_i$ have been listed by the table in Theorem 3.1. The other conclusions follow from Proposition 5.1 immediately.

B. Proof for Theorem 3.2.

The result follows directly from Theorem 3.1 and Proposition 5.2.

C. Proof for Theorem 3.3

As usual, for any ${\underline{a}}=(a_0,a_1,\ldots ,a_{4n-1})\in {\mathbb {Z}}_4^{4n}$, we identify ${\underline{a}}$ with $a(x)=\sum _{j=0}^{4n-1}a_jx^j\in {\mathcal {B}}={\mathbb {Z}}_4[x]/\langle x^{4n}-1\rangle$ in the following. In the ring ${\mathcal {B}}$, we have that $x^{4n}=1$, and hence $x^{-1}=x^{4n-1}$. Moreover, we have $x^{4n}\equiv 1$ (mod $f_i(x^4)$), since $f_i(x^4)$ is a divisor of $x^{4n}-1$ in ${\mathbb {Z}}_4[x]$. This implies $x^{4n}=1$ and $x^{-1}=x^{4n-1}$ in the ring ${\mathcal {R}}_i={\mathbb {Z}}_4[x]/\langle f_i(x^4)\rangle$ for all i. Define

$$\begin{aligned} \mu (a(x))=a(x^{-1})=a_0+\sum _{1\le j\le 4n-1}a_jx^{4n-j}, \quad \forall a(x)\in {\mathcal {B}}. \end{aligned}$$

It is clear that $\mu$ is a ring automorphism of ${\mathcal {B}}$ and satisfies $\mu ^{-1}=\mu$. Then, by a direct calculation, we get the following lemma.

Lemma 5.3

Let ${\underline{a}},{\underline{b}}\in {\mathbb {Z}}_4^{4n}$ where ${\underline{b}}=(b_0,b_1,\ldots ,b_{4n-1})$. Then $[{\underline{a}},{\underline{b}}]=0$ if $a(x)\mu (b(x))=0$ in the ring ${\mathcal {B}}$ where $b(x)=\sum _{j=0}^{4n-1}b_jx^j$.

In Theorem 3.4, we define by $\mu$ the permutation on the set $\{1,\ldots ,r\}$:

$$\begin{aligned} \mu (i)=i, \ \mathrm{if} \ 1\le i\le \lambda ; \ \mu (\lambda +j)=\lambda +j+\epsilon \ \mathrm{and} \ \mu (\lambda +j+\epsilon )=\lambda +j, \ \forall j=1,\ldots ,\epsilon . \end{aligned}$$

Whether $\mu$ denotes the automorphism of ${\mathcal {B}}$ or this map on the set $\{1,\ldots ,r\}$ is determined by the context. By a method paralleling to that above [6, Lemma 7] and its proof, we can prove the following lemma. Here we omit its proof.

Lemma 5.4

Using the notations in Sect. 2, we have the following:

• (i) For each integer i, $1\le i\le r$, there is a unique element $c_i\in {\mathbb {Z}}_4^\times =\{1,-1\}$ such that ${\widetilde{f}}_i(x)=c_i f_{\mu (i)}(x)$.

• (ii) For any integer i, $1\le i\le r$, we have that $\mu (\varepsilon _i(x))=\varepsilon _i(x^{-1})=\varepsilon _{\mu (i)}(x)$ in the ring ${{{\mathcal {B}}}}$ and $\mu ({{{\mathcal {B}}}}_{i})={{{\mathcal {B}}}}_{\mu (i)}$.

• (iii) Let $\mu |_{{{{\mathcal {B}}}}_{i}}: {{{\mathcal {B}}}}_{i}\rightarrow {{{\mathcal {B}}}}_{\mu (i)}$ be the restriction of $\mu$ on ${{{\mathcal {B}}}}_{i}$, and define

  $$\begin{aligned} \mu _i(c(x))=c(x^{-1})=c(x^{4n-1})\ \left( \mathrm{mod} \ f_{\mu (i)}(x^4)\right) , \ \forall c(x)\in {\mathcal {R}}_i. \end{aligned}$$

  Using the notation in Lemma 2.1(iii), we have the following commutative diagram of ring isomorphisms:

  $$\begin{aligned} \begin{array}{ccc} \ \ \ \ {\mathcal {R}}_i={\mathbb {Z}}_4[x]/\langle f_i(x^4)\rangle &{} {\mathop {\longrightarrow }\limits ^{\psi _{\mu (i)}^{-1}(\mu |_{{{{\mathcal {B}}}}_i})\psi _i}} &{} {\mathcal {R}}_{\mu (i)}={\mathbb {Z}}_4[x]/\langle f_{\mu (i)}(x^4)\rangle \\ \psi _i \downarrow &{} &{} \ \ \ \downarrow \psi _{\mu (i)} \\ \ \ \ \ {{{\mathcal {B}}}}_i=\varepsilon _i(x){\mathcal {R}}_i &{} {\mathop {\longrightarrow }\limits ^{\mu |_{{{{\mathcal {B}}}}_i}}} &{} {{{\mathcal {B}}}}_{\mu (i)}=\varepsilon _{\mu (i)}(x){\mathcal {R}}_i \end{array} \end{aligned}$$

  Let $\mu _i=\psi _{\mu (i)}^{-1}(\mu |_{{{{\mathcal {B}}}}_i})\psi _i$. Then $\mu _i$ is a ring isomorphism from ${{{\mathcal {R}}}}_i$ onto ${{{\mathcal {R}}}}_{\mu (i)}$. Moreover, we have $\mu _i^{-1}=\mu _{\mu (i)}$ where $\mu _{\mu (i)}: {{{\mathcal {R}}}}_{\mu (i)}\rightarrow {{{\mathcal {R}}}}_i$ is defined by $\mu _{\mu (i)}(a(x))=a(x^{-1})=a(x^{4n-1})$ $(\mathrm{mod} \ f_i(x^4))$ for all $a(x)\in {{{\mathcal {K}}}}_{\mu (i)}$.

For any ideal $C_i$ of the ring ${\mathcal {R}}_i={\mathbb {Z}}_4[x]/\langle f_i(x^4)\rangle$, recall that the annihilator of $C_i$ is defined as the ideal $\mathrm{Ann}(C_i)=\{\alpha \in {\mathcal {R}}_i\mid \alpha \beta =0, \forall \beta \in C_i \}$ of ${\mathcal {R}}_i$. The annihilator of each ideal in ${\mathcal {R}}_i$ is given by the following proposition.

Proposition 5.5

Using the notation in Sect. 2and Theorem 3.1, for any integer i, $1\le i\le r$, the annihilator $\mathrm{Ann}(C_i)$ of each ideal $C_i$ in ${\mathcal {R}}_i$ is given by the following table:

$C_i$ in Theorem 3.1	Ann$(C_i)$
Case 1	$\langle 1\rangle$
Case 2	$\langle 0\rangle$
Case 3	$\langle f_i(x)^{4-s},2\rangle$
Case 4	$\langle 2{\overline{f}}_i(x)^{4-l}\rangle$
Case 5	$\left\langle f_i(x)^3+2{\overline{f}}_i(x)\cdot \left( w_{i,0}(x)+(w_{i,1}(x)+h(x)){\overline{f}}_i(x)\right) \right\rangle$
Case 6	$\langle f_i(x)^3+2{\overline{f}}_i(x)\cdot (w_{i,0}(x)+h(x)), 2{\overline{f}}_i(x)^2\rangle$
Case 7	$\left\langle f_i(x)^2+2\left( w_i(x)^2+h_{0}(x)+h_{1}(x){\overline{f}}_i(x)\right) \right\rangle$
Case 8	$\langle f_i(x)^3+2h(x),2{\overline{f}}_i(x)\rangle$
Case 9	$\langle f_i(x)^2+2(w_{i,0}(x)+h(x)),2{\overline{f}}_i(x)\rangle$
Case 10	$\langle f_i(x)+2(w_{i,1}(x)+h(x))\rangle$

where $0\le s\le 3$, $1\le l\le 3$, and $h(x),h_0(x),h_1(x)\in {\mathcal {T}}_i$.

Proof

Let ${\mathcal {S}}_i$ be the set of all ideals in ${\mathcal {R}}_i$ listed in the table of Theorem 3.1, and assume $C_i\in {\mathcal {S}}_i$. It is clear that $\mathrm{Ann}(C_i)=D$, where $D\in {{{\mathcal {S}}}}_i$ satisfying the following conditions:

$$\begin{aligned} C_i\cdot D=\{0\} \ \mathrm{and} \ |D|=\mathrm{Max}\{|J|\mid C_i\cdot J=\{0\}, \ J\in {{{\mathcal {S}}}}_i\}. \end{aligned}$$

(9)

By the table in Theorem 3.1, there are 10 cases for all distinct ideals of ${\mathcal {R}}_i$. Hence we have the following four situations:

(i) Let $C_i$ be given by Case 7 in the table of Theorem 3.1. Then $|C_i|=2^{4m_i}$. We set $D=\langle f_i(x)^2+2(w_i(x)^2+h_{0}(x)+h_{1}(x){\overline{f}}_i(x))\rangle$. By Theorem 3.1, we see that D is an ideal of ${\mathcal {R}}_i$ and $|D|=2^{4m_i}$. Now, denote

$\alpha =f_i(x)^2+2(h_0(x)+h_1(x){\overline{f}}_i(x))$,

$\beta =f_i(x)^2+2(w_i(x)^2+h_0(x)+h_1(x){\overline{f}}_i(x))$.

Then $C_i=\{\xi \alpha \mid \xi \in {\mathcal {R}}_i\}$ and $D=\{\eta \beta \mid \eta \in {\mathcal {R}}_i\}$. By Theorem 2.5 (ii), we have $f_i(x)^4=2{\overline{f}}_i(x)^2w_i(x)^2=2f_i(x)^2w_i(x)^2$ in ${\mathcal {R}}_i$. This implies that $\alpha \beta =f_i(x)^4+2f_i(x)^2\cdot (h_0(x)+h_1(x){\overline{f}}_i(x)+w_i(x)^2+h_0(x)+h_1(x){\overline{f}}_i(x))=0,$ and hence $C_i\cdot D=\{\xi \alpha \cdot \eta \beta \mid \xi ,\eta \in {\mathcal {R}}_i\}=\{0\}$. Then by Condition (9), we conclude that $D=\mathrm{Ann}(C_i)$.

(ii) Let $C_i=\langle f_i(x)^3+2{\overline{f}}_i(x)h(x), 2{\overline{f}}_i(x)^2\rangle$ be given by Case 9 in the table of Theorem 3.1. Then $|C_i|=2^{3m_i}$. We set $D=\langle f_i(x)^2+2(w_{i,0}(x)+h(x)),2{\overline{f}}_i(x)\rangle$. Then by Theorem 3.1, D is an ideal of ${\mathcal {R}}_i$ and $|D|=2^{5m_i}$. As ${\overline{f}}_i(x)^4=0$ in $\overline{{\mathcal {R}}}_i$, we have $2{\overline{f}}_i(x)^2\cdot f_i(x)^2=f_i(x)^3\cdot 2{\overline{f}}_i(x)=2{\overline{f}}_i(x)^4=0$. This implies

$$\begin{aligned} 2{\overline{f}}_i(x)^2\cdot (f_i(x)^2+2(w_{i,0}(x)+h(x)))=(f_i(x)^3+2{\overline{f}}_i(x)h(x))\cdot 2{\overline{f}}_i(x)=0. \end{aligned}$$

Since $f_i(x)^4=2{\overline{f}}_i(x)^2w_i(x)^2=2{\overline{f}}_i(x)^2(w_{i,0}(x)+w_{i,1}(x){\overline{f}}_i(x))$ by Theorem 2.5 (ii), it follows that

$$\begin{aligned}&\left( f_i(x)^3+2{\overline{f}}_i(x)h(x)\right) \cdot \left( f_i(x)^2+2(w_{i,0}(x)+h(x))\right) \\&\quad = f_i(x)^5+2{\overline{f}}_i(x)^3(w_{i,0}(x)+2h(x))\\&\quad = 2{\overline{f}}_i(x)^3(w_{i,0}(x)+w_{i,1}(x){\overline{f}}_i(x))+2{\overline{f}}_i(x)^3w_{i,0}(x)\\&\quad = 0. \end{aligned}$$

This implies $C_i\cdot D=\{0\}$. Then by Condition (9), we have $D=\mathrm{Ann}(C_i)$.

(iii) Let $C_i=\langle f_i(x)^3+2{\overline{f}}_i(x)(w_{i,0}(x)+h(x){\overline{f}}_i(x))\rangle$ be given by Case 10 in the table of Theorem 3.1. Then $|C_i|=2^{2m_i}$. Set $D=\langle f_i(x)+2(w_{i,1}(x)+h(x))\rangle$. By Theorem 3.1, we see that D is an ideal of ${\mathcal {R}}_i$ and $|D|=2^{6m_i}$.

By Theorem 2.5 (ii), it follows that

$$\begin{aligned}&\left( f_i(x)^3+2{\overline{f}}_i(x)(w_{i,0}(x)+h(x){\overline{f}}_i(x)) \right) \cdot \left( f_i(x)+2(w_{i,1}(x)+h(x))\right) \\&\quad = f_i(x)^4+2{\overline{f}}_i(x)^3(w_{i,1}(x)+h(x)) +2{\overline{f}}_i(x)^2(w_{i,0}(x)+h(x){\overline{f}}_i(x)) \\&\quad = 2{\overline{f}}_i(x)^2(w_{i,0}(x)+w_{i,1}(x){\overline{f}}_i(x)) +2{\overline{f}}_i(x)^3w_{i,1}(x)+2{\overline{f}}_i(x)^2w_{i,0}(x)\\&\quad = 0. \end{aligned}$$

This implies $C_i\cdot D=\{0\}$. Then by Condition (9), we have $D=\mathrm{Ann}(C_i)$.

(iv) If $C_i$ is an ideal of ${\mathcal {R}}_i$ given by Cases 1–6 and Case 8 in the table of Theorem 3.1, the conclusions for each $\mathrm{Ann}(C_i)$ follow from Theorem 2.5 and direct calculations. Here, we omit these elementary calculations. $\square$

Lemma 5.6

Let $a(x)=\sum _{i=1}^r\varepsilon _i(x)\xi _i, b(x)=\sum _{i=1}^r\varepsilon _i(x)\eta _i\in {{{\mathcal {B}}}}$, where $\xi _i, \eta _i\in {\mathcal {R}}_i$. Then $a(x)\mu (b(x))=\sum _{i=1}^r\varepsilon _i(x)(\xi _i\cdot \mu _i^{-1}(\eta _{\mu (i)}))$.

Proof

By Lemma 5.4 (ii), we have $\mu _i^{-1}(\eta _{\mu (i)})\in \mu _i^{-1}({{{\mathcal {R}}}}_{\mu (i)})={{{\mathcal {R}}}}_i$. This implies $\xi _i\cdot \mu _i^{-1}(\eta _{\mu (i)})\in {{{\mathcal {R}}}}_i$ for all i. If $j\ne \mu (i)$, then $i\ne \mu (j)$ and so $\varepsilon _i(x)\varepsilon _{\mu (j)}(x)=0$ in the ring ${{{\mathcal {B}}}}$, by Lemma 2.1 (i). Therefore, by Lemma 5.4 (iii) it follows that

$$\begin{aligned} a(x)\mu (b(x))= & {} \sum _{i,j=1}^r\varepsilon _i(x)\xi _i\cdot \mu (\varepsilon _j(x)\eta _j) =\sum _{i,j=1}^r\varepsilon _i(x)\xi _i\cdot \mu (\varepsilon _j(x))\mu _j(\eta _j)\\= & {} \sum _{i,j=1}^r\varepsilon _i(x)\xi _i\cdot \varepsilon _{\mu (j)}(x)\mu _j(\eta _j) =\sum _{i=1}^r\varepsilon _i(x)\xi _i\cdot \varepsilon _{i}(x)\mu _{\mu (i)}(\eta _{\mu (i)}). \end{aligned}$$

Hence $a(x)\mu (b(x))=\sum _{i=1}^r\varepsilon _i(x)(\xi _i\cdot \mu _i^{-1}(\eta _{\mu (i)}))$ by Lemma 2.1 (i). $\square$

Now, we prove Theorem 3.3 as follows:

Let ${{{\mathcal {C}}}}=\bigoplus _{i=1}^r\varepsilon _i(x)C_i$ be a cyclic code over ${\mathbb {Z}}_4$ of length 4n, where $C_i$ is an ideal of the ring ${{{\mathcal {R}}}}_i$ given by Theorem 3.1. For each integer $1\le j\le r$, define

$$\begin{aligned} H_j=\mu _{\mu (j)}(\mathrm{Ann}(C_{\mu (j)}))=\mu _{j}^{-1}(\mathrm{Ann}(C_{\mu (j)})), \end{aligned}$$

where $\mathrm{Ann}(C_{\mu (j)})$ is the annihilator of $C_{\mu (j)}$ determine by Proposition 5.5, for all $j=1,\ldots ,r$. Then by Lemma 5.4 (iii), we have $\mu _{\mu (j)}^{-1}(H_j)=\mu _{j}(H_j)=\mathrm{Ann}(C_{\mu (j)})$. Assume $i=\mu (j)$. Then we have $\mu (i)=j$ and $\mu _i^{-1}(H_{\mu (i)})=\mathrm{Ann}(C_{i})$. Now, let ${{{\mathcal {H}}}}=\bigoplus _{i=1}^r\varepsilon _{\mu (i)}(x)H_{\mu (i)}=\sum _{j=1}^r\varepsilon _j(x)H_j \ (\text{ mod } \ x^{4n}-1).$

By Proposition 5.1, we know that ${{{\mathcal {H}}}}$ is a cyclic code over ${\mathbb {Z}}_4$ of length 4n. As $C_i\cdot \mathrm{Ann}(C_i)=\{0\}$, by Lemma 5.6 it follows that

$$\begin{aligned} {{{\mathcal {C}}}}\cdot \mu ({{{\mathcal {H}}}})=\sum _{i=1}^r\varepsilon _i(x)\left( C_i\cdot \mu _i^{-1}(H_{\mu (i)})\right) =\sum _{i=1}^r\varepsilon _i(x)\left( C_i\cdot \mathrm{Ann}(C_i)\right) =\{0\}. \end{aligned}$$

From this and by Lemma 5.3, we deduce that ${{{\mathcal {H}}}}\subseteq {{{\mathcal {C}}}}^{\bot }$. Further, by Theorem 3.1 and Proposition 5.5, we have $|C_i||\mathrm{Ann}(C_i)|=2^{8m_i}$ for all i. This implies

$$\begin{aligned} |{{{\mathcal {C}}}}||{{{\mathcal {H}}}}|=(\prod _{i=1}^r|C_i|)(\prod _{i=1}^r|H_{\mu (i)}|) =\prod _{i=1}^r\left( |C_i||\mathrm{Ann}(C_i)|\right) =2^{8\sum _{i=1}^rm_i}=|{\mathbb {Z}}_4|^{4n}, \end{aligned}$$

by Proposition 5.1 and $\sum _{i=1}^rm_i=n$. Then from the theory of linear codes over ${\mathbb {Z}}_4$ (cf. [23]), we deduce that ${{{\mathcal {C}}}}^{\bot }={{{\mathcal {H}}}}$.

To prove Theorem 3.3, it is sufficient to prove $H_{\mu (i)}=D_{\mu (i)}$, where $D_{\mu (i)}$ is an ideal of the ring ${\mathcal {R}}_{\mu (i)}$ listed in the table of Theorem 3.3, $1\le i\le r$.

Since $x^{-1}=x^{4n-1}$ in ${\mathcal {R}}_i$, we have $x\in {\mathcal {R}}_i^{\times }$, for all i. By Lemma 5.4 (i), we know that ${\widetilde{f}}_i(x)=c_if_{\mu (i)}(x)$ where $c_i\in \{1,-1\}$. This implies $2c_i=2$ in ${\mathbb {Z}}_4$. Then by the definition of $\mu _i$ in Lemma 5.4 (iii), for $k=1,2,3$, we have

$$\begin{aligned} \mu _i(f_i(x)^k)= & {} (\mu _i(f_i(x)))^k=f_i(x^{-1})^k =x^{-km_i}(x^{m_i}f_i(x^{-1}))^k\\= & {} x^{-km_i}({\widetilde{f}}_i(x))^k=c_i^kx^{-km_i}f_{\mu (i)}(x)^k, \end{aligned}$$

and so $\mu _i(2{\overline{f}}_i(x)^k)=\mu _i(2f_i(x)^k)=2c_i^kx^{-km_i}f_{\mu (i)}(x)^k =2x^{-km_i}{\overline{f}}_{\mu (i)}(x)^k.$

(i) Let $\mathrm{Ann}(C_i)=\langle f_i(x)^{4-s}, 2\rangle$ be given in Case 3 of Proposition 5.5. By $x\in {\mathcal {R}}_{\mu (i)}^\times$, it follows that

$$\begin{aligned} H_{\mu (i)}= & {} \langle \mu _i(f_i(x)^{4-s}), \mu _i(2)\rangle =\langle c_{i}^{4-s}x^{-(4-s)m_i}f_{\mu (i)}(x)^{4-s},2\rangle \\= & {} \langle f_{\mu (i)}(x)^{4-s},2\rangle =D_{\mu (i)}. \end{aligned}$$

The equations $H_{\mu (i)}=D_{\mu (i)}$ for Cases 1, 2, 4 can be proved similarly as Case 3.

(ii) Let $\mathrm{Ann}(C_i)=\langle f_i(x)^3+2{\overline{f}}_i(x)(w_{i,0}(x)+\vartheta _i(x){\overline{f}}_i(x))\rangle$ be given in Case 5 of Proposition 5.5, where $\vartheta _i(x)=w_{i,1}(x)+h(x)$. Then we have

$$\begin{aligned} H_{\mu (i)}= & {} \left\langle \mu _i\left( f_i(x)^3+2{\overline{f}}_i(x)(w_{i,0}(x)+\vartheta _i(x){\overline{f}}_i(x))\right) \right\rangle \\= & {} \langle c_i^3x^{-3m_i}f_{\mu (i)}(x)^3\\&+2x^{-m_i}{\overline{f}}_{\mu (i)}(x) \cdot (w_{i,0}(x^{-1})+\vartheta _i(x^{-1})x^{-m_i}{\overline{f}}_{\mu (i)}(x))\rangle \\= & {} \left\langle f_{\mu (i)}(x)^3+2{\overline{f}}_{\mu (i)}(x)\left( x^{2m_i}w_{i,0}(x^{-1}) +{\widehat{\vartheta }}_i(x){\overline{f}}_{\mu (i)}(x)\right) \right\rangle , \end{aligned}$$

where ${\widehat{\vartheta }}_i(x)=x^{m_i}(w_{i,1}(x^{-1})+h(x^{-1}))\in \overline{{{{\mathcal {R}}}}}_{\mu (i)}$. Hence $H_{\mu (i)}=D_{\mu (i)}$.

The equation $H_{\mu (i)}=D_{\mu (i)}$ for Case 10 can be proved similarly as Case 5.

(iii) Let $\mathrm{Ann}(C_i)=\langle f_i(x)^3+2{\overline{f}}_i(x)t_i(x),2{\overline{f}}_i(x)^2\rangle$ be given in case 6 of Proposition 5.5, where $t_i(x)=w_{i,0}(x)+h(x)$. Then we have

$$\begin{aligned} H_{\mu (i)}= & {} \left\langle c_i^3x^{-3m_i}f_{\mu (i)}(x)^3+2x^{-m_i}{\overline{f}}_{\mu (i)}(x)t_i(x^{-1}),2x^{-2m_i}{\overline{f}}_{\mu (i)}(x)^2\right\rangle \\= & {} \langle f_{\mu (i)}(x)^3+2{\overline{f}}_{\mu (i)}(x){\widehat{t}}_i(x),{\overline{f}}_{\mu (i)}(x)^2\rangle , \end{aligned}$$

where ${\widehat{t}}_i(x)=x^{2m_i}(w_{i,0}(x^{-1})+h(x^{-1}))\in \overline{{{{\mathcal {R}}}}}_{\mu (i)}$. Hence $H_{\mu (i)}=D_{\mu (i)}$.

The equation $H_{\mu (i)}=D_{\mu (i)}$ for Case 9 can be proved similarly as Case 6.

(iv) Let $\mathrm{Ann}(C_i)=\langle f_i(x)^2+2(w_i(x)^2+h_0(x)+h_1(x){\overline{f}}_i(x))\rangle$ be given in Case 7 of Proposition 5.5, where $\delta _{i,j}(x)=w_{i,j}(x)+h_j(x)$ for $j=0,1$. Then

$$\begin{aligned} H_{\mu (i)}= & {} \langle c_i^2x^{-2m_i}f_{\mu (i)}(x)^2+2(w_i(x^{-1})^2+h_0(x^{-1})+h_1(x^{-1})x^{-m_i}{\overline{f}}_{\mu (i)}(x))\rangle \\= & {} \left\langle f_{\mu (i)}(x)^2+2\left( {\widehat{\delta }}_{i,0}(x) +{\widehat{\delta }}_{i,1}(x){\overline{f}}_{\mu (i)}(x)\right) \right\rangle = D_{\mu (i)}, \end{aligned}$$

where ${\widehat{\delta }}_{i,0}(x)=x^{2m_i}(w_{i}(x^{-1})^2+h_0(x^{-1}))$ and ${\widehat{\delta }}_{i,1}(x)=x^{m_i}h_1(x^{-1})$ in $\overline{{{{\mathcal {R}}}}}_{\mu (i)}$.

(v) Let $\mathrm{Ann}(C_i)=\langle f_i(x)^3+2h(x),2{\overline{f}}_i(x)\rangle$ be given in Case 8 of Proposition 5.5. Then we have $H_{\mu (i)}=\langle c_i^3x^{-3m_i}f_{\mu (i)}(x)^3+2h(x^{-1}),2x^{-m_i}{\overline{f}}_{\mu (i)}(x)\rangle$ $=\langle f_{\mu (i)}(x)^3+2x^{3m_i}h(x^{-1}),2{\overline{f}}_{\mu (i)}(x)\rangle =D_{\mu (i)}$.

On the basis of the above discussion, we get ${{{\mathcal {C}}}}^{\bot }=\bigoplus _{i=1}^r\varepsilon _{\mu (i)}(x)D_{\mu (i)}$, where $D_{\mu (i)}$ is an ideal of ${\mathcal {R}}_{\mu (i)}$ given in the table of Theorem 3.3 for all $i=1,\ldots ,r$. This proves the theorem 3.3.

D. Proof for Theorem 3.4

Using the notation of Theorem 3.3, by Proposition 5.1, we see that ${\mathcal {C}}$ is self-dual if and only if the ideal $C_i$ of ${\mathcal {R}}_i$ satisfies $C_i=D_{i}$ for all $i=1,\ldots ,r$, where the pair $(C_i, D_{\mu (i)})$ of ideals is listed in the table of Theorem 3.3. Then the latter condition is equivalent to that $C_i$ satisfies the following conditions:

$\diamondsuit$ Let $i=1$. There are 3 ideals of ${\mathcal {R}}_1$ satisfies $C_1=D_{1}=D_{\mu (1)}$:

$C_1=\langle 2\rangle$, $C_1=\langle (x-1)^3, 2(x-1)\rangle$ and $C_1=\langle (x-1)^3+2, 2(x-1)\rangle .$

$\diamondsuit$ Let $2\le i\le \lambda$. In this case, $\mu (i)=i$. Then by Theorems 3.1 and 3.3, we see that $C_i$ is one of the following three subcases:

$\triangleright$ $C_i=\langle 2\rangle$.

$\triangleright$ $C_i=\langle f_i(x)^3+2h(x), 2{\overline{f}}_i(x)\rangle$, where $h(x)\in {\mathcal {T}}_i$ satisfying Eq. (5) in the proof of Theorem 2.7, i.e., $h(x)\in {\mathcal {V}}_i$ by Theorem 2.7 (i).

$\triangleright$ $C_i=\langle f_i(x)^2+2(h_0(x)+h_1(x){\overline{f}}_i(x))\rangle$, where $h_0(x),h_1(x)\in {\mathcal {T}}_i$ satisfying Equation (7) in the proof of Theorem 2.7, i.e., $h_0(x)\in {\mathcal {W}}_i^{(0)}$ and $h_1(x)\in {\mathcal {W}}_{i,h_0(x)}^{(1)}$ by Theorem 2.7 (ii) and (iii).

By Theorem 2.7, there are $1+2^{\frac{m_i}{2}}+(2^{\frac{m_i}{2}})^2$ ideals $C_i$ of ${\mathcal {R}}_i$ satisfies $C_i=D_{i}=D_{\mu (i)}$, for all $i=2,\ldots ,\lambda$.

$\diamondsuit$ Let $i=\lambda +j$ where $1\le j\le \epsilon$. Then $\mu (i)=i+\epsilon$ and $\mu (i+\epsilon )=i$. In this case, $C_t=D_t$ for all $t=\lambda +1,\ldots , \lambda +2\epsilon$ if and only if: $C_i$ is any ideal of ${\mathcal {R}}_i$ listed in the table of Theorem 3.1 and $C_{i+\epsilon }=D_{\mu (i)}$, where $D_{\mu (i)}=D_{i+\epsilon }$ is given by the table in Theorem 3.3, for all $i=\lambda +j$ and $1\le j\le \epsilon$.

As stated above, we see that the number of self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n is $3\cdot \prod _{2\le i\le \lambda }(1+2^{\frac{m_i}{2}}+2^{m_i}) \cdot \prod _{j=1}^\epsilon (9+5\cdot 2^{m_{\lambda +j}}+4^{m_{\lambda +j}})$.

6 Conclusions

We give an explicit representation and enumeration for all distinct cyclic codes over ${\mathbb {Z}}_{4}$ of length 4n where n is odd. Using this representation, we provide an efficient encoder for each code and determine its type explicitly. Then we give a precise description for the dual codes and listed explicitly all distinct self-dual cyclic codes over ${\mathbb {Z}}_{4}$ of length 4n. Compared with the results in the literature, the results in this paper are more simple and practical for constructing self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 4n, for a given odd positive integer n.

Obtaining some bounds for the minimal distance such as BCH-like of a self-dual cyclic code over ${\mathbb {Z}}_{4}$ of length 4n by just looking at the representation of such codes are future topics of interest.