Symplectic QSD, LCD, and ACD Codes over a Non-Commutative Non-Unitary Ring of Order Nine

Abstract

We introduce quasi self-dual (QSD), linear complementary dual (LCD), and additive complementary dual (ACD) codes for the symplectic inner product over a non-commutative non-unitary ring of order 9. We establish connections with symplectic–self-orthogonal and LCD ternary codes. We characterize right-symplectic ACD codes.

1. Introduction

Codes over non-unitary rings have received significant attention in the last five years [1,2] especially in the context of duality [3,4,5,6]. The notion of quasi self-dual codes (QSD) plays here the role of self-dual codes over unitary rings and finite fields.

Recently, Alahmadi et al. [7] derived a mass formula for self-orthogonal, QSD and self-dual codes of non-unitary rings of four elements. Later, this research was extended to non-unitary rings of the size of the square of an odd prime [5,6]. Furthermore, in [3], the concept of linear complementary dual (LCD) codes was characterized over a non-unitary ring of order four. Additive complementary dual (ACD) codes were introduced in [4]. Kushwaha et al. [8] pushed further, with a study in the characteristic three. These studies were conducted using the Euclidean inner product.

More recently, in a series of papers [9,10,11], we introduced and characterized self-orthogonal and QSD codes over non-unitary rings of the order of four under the symplectic inner product. Later, we classified all distinct symplectic–self-orthogonal and QSD codes over commutative non-unitary rings with four elements [9,11]. Additionally, [10] introduced symplectic LCD and ACD codes over a non-commutative non-unitary ring of size four, in which the authors focused on the study of symplectic left ACD and LCD codes. This motivated us to extend the study of left- and right-symplectic LCD and ACD codes to the non-commutative non-unitary ring of order nine. Therefore, in the present manuscript, we consider the case of a ring of order 9 and E in the notation of Fine [12]. By forgetting the module structure, our codes can be considered as additive codes over ${\mathbb{F}}_9,$ a viewpoint that allows us, for numerical examples, to use the additive codes Magma package [13]. Moreover, we introduce QSD, right (respectively, left) ACD, nice, and LCD codes for the symplectic inner product over that ring, here denoted by ${\mathsf{E}}_3$. We establish connections with ternary symplectic–self-orthogonal and symplectic LCD codes. In particular, we use a multilevel construction from ternary symplectic–self-orthogonal codes to construct symplectic QSD codes. We investigate the right-symplectic LCD codes. Furthermore, we characterize free right-symplectic LCD codes over ${\mathsf{E}}_3$ in terms of a ternary generator matrix. Since non-trivial left-symplectic ACD codes do not exist, we introduce the concept of a right-symplectic ACD code over ${\mathsf{E}}_3$. Additionally, we establish several results for the existence of these codes.

This manuscript is organized as follows: The basic notions and notations used in the remainder of this paper are presented in the next section. Section 3 studies the symplectic QSD and LCD codes over ${\mathsf{E}}_3$. Section 4 treats right-symplectic ACD ${\mathsf{E}}_3$ codes. Section 5 provides the conclusion.

2. Preliminaries

2.1. Ternary Linear Codes

A linear code C having a dimension, k, and length, n, over the ternary field ${\mathbb{F}}_3$ is an ${\mathbb{F}}_3$ subspace of ${\mathbb{F}}_3^n$ of the k dimension. Briefly, we say C is an $[n,k]$–code and its elements are known by codewords.

Denote the vector $\left({\mathbf{x}}_{\mathbf{1}}\right|{\mathbf{x}}_{\mathbf{2}})$ as the juxtaposition of two vectors, ${\mathbf{x}}_{\mathbf{1}},{\mathbf{x}}_{\mathbf{2}}\in {\mathbb{F}}_3^{2n}$. Let $\mathbf{x}=\left({\mathbf{x}}_1\right|{\mathbf{x}}_2)$ and $\mathbf{y}=\left({\mathbf{y}}_1\right|{\mathbf{y}}_2)$, where ${\mathbf{x}}_{\mathbf{1}}=(x_1,x_2,\cdots ,x_n),{\mathbf{x}}_{\mathbf{2}}=(x_{n+1},x_{n+2},\cdots ,x_{2n})\in {\mathbb{F}}_3^{2n},{\mathbf{y}}_{\mathbf{1}}=(y_1,y_2,\cdots ,y_n),{\mathbf{y}}_{\mathbf{2}}=(y_{n+1},y_{n+2},\cdots ,y_{2n})\in {\mathbb{F}}_3^{2n}$ are the vectors of ${\mathbb{F}}_3^{2n}$. The Euclidean and symplectic inner products of $\mathbf{x}$ and $\mathbf{y}$ are, respectively, defined by

$${(\mathbf{x},\mathbf{y})}_e={\displaystyle \sum _{i=1}^{2n}}{x}_i{y}_i\mathrm{and}{(\mathbf{x},\mathbf{y})}_s={(\mathbf{x},\mathbf{y}{\Omega }_n^T)}_e={({\mathbf{x}}_{\mathbf{1}},{\mathbf{y}}_{\mathbf{2}})}_e-{({\mathbf{x}}_{\mathbf{2}},{\mathbf{y}}_{\mathbf{1}})}_e={\sum }_{i=1}^n(x_i{y}_{n+i}-x_{n+i}{y}_i),$$

where ${\Omega }_n=\left[\begin{array}{cc}{0}_n& I_n\\ -I_n& 0_n\end{array}\right]$ with $0_n$ is the $n\times n$ zero matrix and $I_n$ is the identity matrix.

Two vectors $\mathbf{x}$ and $\mathbf{y}$ are symplectic–orthogonals if they satisfy ${(\mathbf{x},\mathbf{y})}_s=0$. Moreover, the vector space consists of all vectors of ${\mathbb{F}}_3^{2n}$ that are symplectic–orthogonal to each codeword in C, which is known by the symplectic dual of C, denoted by $C^{{\perp }_s}$.

The code C is said to be symplectic–self-orthogonal (resp. symplectic self-dual) if $C\subseteq C^{{\perp }_s}$ (resp. $C=C^{{\perp }_s}$).

2.2. Non-Commutative Non-Unitary Ring of Nine Elements

Define the ring ${\mathsf{E}}_3$ on two generators, a and b, under certain relations, such as:

$${\mathsf{E}}_3=\langle a,b|3a=3b=0,ab=a,ba=b,a^2=a,b^2=b\rangle .$$

Assume that $c=a+2b$. According to [8], the ring ${\mathsf{E}}_3$ has characteristic 3 and 9 elements, given by

$${\mathsf{E}}_3=\langle 0,a,b,c,2a,2b,2c,a+b,a+c\rangle .$$

It is easy to derive the addition table of ${\mathsf{E}}_3$ and its multiplication table is in

Table 1. Multiplication table for ${\mathsf{E}}_3$.

×	0	a	b	c	$2\mathit{a}$	$2\mathit{b}$	$2\mathit{c}$	$\mathit{a}+\mathit{b}$	$\mathit{a}+\mathit{c}$
0	0	0	0	0	0	0	0	0	0
a	0	a	b	c	$2a$	$2b$	$2c$	$a+b$	$a+c$
b	0	a	b	c	$2a$	$2b$	$2c$	$a+b$	$a+c$
c	0	0	0	0	0	0	0	0	0
$2a$	0	$2a$	$2b$	$2c$	a	b	c	$a+c$	$a+b$
$2b$	0	$2a$	$2b$	$2c$	a	b	c	$a+c$	$a+b$
$2c$	0	0	0	0	0	0	0	0	0
$a+b$	0	$2a$	$2b$	$2c$	a	b	c	$a+c$	$a+b$
$a+c$	0	a	b	c	$2a$	$2b$	$2c$	$a+b$	$a+c$

.

Thus, ${\mathsf{E}}_3$ is a non-commutative ring without identity element with respect to the given multiplication and has a unique maximal ideal $\mathcal{J}=\{0,c,2c\}$. So, the ring ${\mathsf{E}}_3$ is local having a residue class field ${\mathsf{E}}_3/\mathcal{J}\cong {\mathbb{F}}_3$.

Every element, $r\in {\mathsf{E}}_3$, has a c-adic decomposition, expressed as $r=au+cv$, for unique scalars $u,v$ of ${\mathbb{F}}_3$. Furthermore, $ax=x$, $bx=x$, and $cx=0$ for all $x\in {\mathsf{E}}_3$.

Define a natural action of ${\mathbb{F}}_3$ on the ring ${\mathsf{E}}_3$ as

$$r0=0r=0,r1=1r=r,\mathrm{and}r2=2r=2r.$$

This action is clearly distributive in the sense that, for any $r\in {\mathsf{E}}_3$ and $u,vs.\in {\mathbb{F}}_3$, we have $r(u\oplus v)=ru+rv$, where ⊕ is denoted in the addition in ${\mathbb{F}}_3$.

The symplectic inner product of the two elements $\mathbf{u}\in {\mathbb{F}}_3^{2n}$ and $\mathbf{r}\in {\mathsf{E}}_3^{2n}$, denoted by ${(\mathbf{u},\mathbf{r})}_s$ is defined as

$${(\mathbf{u},\mathbf{r})}_s=\sum _{i=1}^n(u_i{r}_{n+i}-u_{n+i}{r}_i).$$

We define the reduction map modulo $\mathcal{J}$ as: ${\alpha }_{{\mathsf{E}}_3}:{\mathsf{E}}_3\to {\mathsf{E}}_3/\mathcal{J}\cong {\mathbb{F}}_3$ by

$$\begin{array}{c}{\alpha }_{{\mathsf{E}}_3}\left(0\right)={\alpha }_{{\mathsf{E}}_3}\left(c\right)={\alpha }_{{\mathsf{E}}_3}\left(2c\right)=0,\\ {\alpha }_{{\mathsf{E}}_3}\left(a\right)={\alpha }_{{\mathsf{E}}_3}\left(b\right)={\alpha }_{{\mathsf{E}}_3}(a+c)=1,\\ {\alpha }_{{\mathsf{E}}_3}\left(2a\right)={\alpha }_{{\mathsf{E}}_3}\left(2b\right)={\alpha }_{{\mathsf{E}}_3}(a+b)=2.\end{array}$$

This map can be extended naturally to a map from ${\mathsf{E}}_3^n$ to ${\mathbb{F}}_3^n$.

2.3. Linear ${\mathsf{E}}_3$ Codes

A linear code C over ${\mathsf{E}}_3$ or a linear ${\mathsf{E}}_3$ –code of length n is a right ${\mathsf{E}}_3$ submodule of ${\mathsf{E}}_3^n$. A linear ${\mathsf{E}}_3$ –code C is free if it is free as a right ${\mathsf{E}}_3$–module.

An additive code over ${\mathsf{E}}_3$ or an additive ${\mathsf{E}}_3$–code of length n is an additive subgroup of ${\mathsf{E}}_3^n$.

For any linear code C over ${\mathsf{E}}_3$, there are the two ternary codes of length n associated with C.

• The residue code of C is defined as

  $$\mathrm{res}\left(C\right)=\left\{{\alpha }_{{\mathsf{E}}_3}\left(\mathbf{r}\right)\right|\mathbf{r}\in C\}.$$

The torsion code of C is

$$\mathrm{tor}\left(C\right)=\{\mathbf{u}\in {\mathbb{F}}_3^n|\mathbf{u}c\in C\}.$$

From [2], we have $\mathrm{res}\left(C\right)\subseteq \mathrm{tor}\left(C\right)$. We denote $k_1$ as the dimension of $\mathrm{res}\left(C\right)$, and $k_1+k_2$ as the dimension of $\mathrm{tor}\left(C\right)$. We say that a linear ${\mathsf{E}}_3$ code, C, is of type $\{k_1,k_2\}$ and $\left|C\right|=\left|\mathrm{res}\left(C\right)\right|\left|\mathrm{tor}\left(C\right)\right|=3^{2k_1+k_2}$. From [8], C is free if, and only if, $k_2=0$.

The symplectic inner product between two codewords, $\mathbf{x}=(x_1,x_2,\cdots ,x_{2n}),\mathbf{y}=(y_1,y_2,\cdots ,y_{2n})\in C$, is defined as

$${(\mathbf{x},\mathbf{y})}_s=\sum _{i=1}^n(x_i{y}_{n+i}-x_{n+i}{y}_i).$$

Under this symplectic inner product, the duals of an ${\mathsf{E}}_3$ code, C, are defined as:

(1).

The left-symplectic dual is defined as

$$C^{{\perp }_{L_s}}=\{\mathbf{y}\in {\mathsf{E}}_3^{2n}\mid {(\mathbf{y},\mathbf{x})}_s=0,\forall \mathbf{x}\in C\}.$$

(2).

The right-symplectic dual of C is defined by

$$C^{{\perp }_{R_s}}=\{\mathbf{y}\in {\mathsf{E}}_3^{2n}\mid {(\mathbf{x},\mathbf{y})}_s=0,\forall \mathbf{x}\in C\}.$$

An ${\mathsf{E}}_3$ code C, is called left-symplectic self-dual if $C^{{\perp }_{L_s}}=C$. Likewise, C is called right-symplectic self-dual if $C^{{\perp }_{R_s}}=C$. Then, we set $C^{{\perp }_s}=C^{{\perp }_{L_s}}\cap C^{{\perp }_{R_s}}$.

If any two codewords, $\mathbf{x},\mathbf{y}\in C$, satisfy ${(\mathbf{x},\mathbf{y})}_s=0$, the code, C, is called a symplectic–self-orthogonal code. It is evident that C is symplectic–self-orthogonal if $C\subseteq C^{{\perp }_s}$.

A linear code, C, of length $2n$ is called left-symplectic-nice (resp. right-symplectic-nice) if it verifies $\left|C\right||C^{{\perp }_{L_s}}| =9^{2n}$ (respectively, if $\left|C\right||C^{{\perp }_{R_s}}| =9^{2n}$).

Remark 1. 

In general, the relation “$\left|C\right||C^{{\perp }_{R_s}}| =9^{2n}$” does not hold. For example, let C be an additive ${\mathsf{E}}_3$ code of length 4 with (additive) generator matrix, G, where

$$G=\left[\begin{array}{cccc}a& 2b& 0& 0\\ 0& 0& a& 2b\end{array}\right]$$

Let $(y_1,y_2,y_3,y_4)$ be an arbitrary vector of $C^{{\perp }_{R_s}}$, then

$${((a,2b,0,0),(y_1,y_2,y_3,y_4))}_s=0and{((0,0,a,2b),(y_1,y_2,y_3,y_4))}_s=0.$$

These imply that, $a{y}_3+2b{y}_4=0$ and $2a{y}_1+b{y}_2=0$. Since $ax=x$ and $bx=x$ for any $x\in {\mathsf{E}}_3$, $y_3+2y_4=0$ and $2y_2+y_1=0$. So, $y_4=y_3$ and $y_2=y_1$. Hence, $C^{{\perp }_{R_s}}=\left\{(y_1,y_1,y_3,y_3)\right|y_1,y_3\in {\mathsf{E}}_3\}.$

• It is easy to derive, $\left|C\right|=9$ and $|C^{{\perp }_{R_s}}| =9^2$. Thus, $\left|C\right||C^{{\perp }_{R_s}}| =9^3<9^4$.

3. Symplectic QSD and LCD ${\mathsf{E}}_3$ Codes

In this section, first we define the notion symplectic QSD codes over the ring ${\mathsf{E}}_3$ and then prove their multilevel construction from ternary linear codes. Further, we investigate the symplectic LCD codes and characterize free symplectic LCD code over ${\mathsf{E}}_3$ in term of ternary generator matrix.

A symplectic–self-orthogonal ${\mathsf{E}}_3$ code with length $2n$ and $9^n$ elements is called symplectic quasi self-dual (symplectic QSD).

In the following result, we establish a necessary and sufficient condition for a linear ${\mathsf{E}}_3$ code to be self-orthogonal under the symplectic inner product.

Lemma 1. 

A linear ${\mathsf{E}}_3$ code C is symplectic–self-orthogonal if $\mathrm{res}\left(C\right)\subseteq \mathrm{tor}{\left(C\right)}^{{\perp }_s}$.

Proof. 

Suppose C is symplectic–self-orthogonal. Let $\mathbf{x}$ be an arbitrary codeword of $\mathrm{res}\left(C\right)$, then there exists $\mathbf{x}a+\mathbf{y}c\in C$, such that ${\alpha }_{{\mathsf{E}}_3}(\mathbf{x}a+\mathbf{y}c)=\mathbf{x}$. By definition, for any element $\mathbf{z}\in \mathrm{tor}\left(C\right)$, $\mathbf{z}c\in C$. From the symplectic–self-orthogonality of C, we have ${(\mathbf{x}a+\mathbf{y}c,\mathbf{z}c)}_s={(\mathbf{x},\mathbf{z})}_{s}c=0$. This implies ${(\mathbf{x},\mathbf{z})}_s=0$. This means $\mathbf{z}\in \mathrm{res}{\left(C\right)}^{{\perp }_s}$. Hence, $\mathrm{tor}\left(C\right)\subseteq \mathrm{res}{\left(C\right)}^{{\perp }_s}$. So, $\mathrm{res}\left(C\right)\subseteq \mathrm{tor}{\left(C\right)}^{{\perp }_s}$.

For the inverse assertion. Let $\mathbf{c},{\mathbf{c}}^{\prime }\in C$. By Theorem 3.5 in [8], there exist $\mathbf{x},{\mathbf{x}}^{\prime }\in \mathrm{res}\left(C\right)$ and $\mathbf{y},{\mathbf{y}}^{\prime }\in \mathrm{tor}\left(C\right)$ such that $\mathbf{c}=\mathbf{x}a+\mathbf{y}c$ and ${\mathbf{c}}^{\prime }={\mathbf{x}}^{\prime }a+{\mathbf{y}}^{\prime }c$. For an arbitrary linear code, C, over ${\mathsf{E}}_3$. Using a procedure similar to that given in [10], obtain $\mathrm{tor}{\left(C\right)}^{{\perp }_s}\subseteq \mathrm{res}{\left(C\right)}^{{\perp }_s}$ as $\mathrm{res}\left(C\right)\subseteq \mathrm{tor}\left(C\right)$. Moreover, $\mathrm{res}\left(C\right)\subseteq \mathrm{tor}{\left(C\right)}^{{\perp }_s}\subseteq \mathrm{res}{\left(C\right)}^{{\perp }_s}$. This yields

$${(\mathbf{c},{\mathbf{c}}^{\prime })}_s={(\mathbf{x}a+\mathbf{y}c,{\mathbf{x}}^{\prime }a+{\mathbf{y}}^{\prime }c)}_s={(\mathbf{x},{\mathbf{x}}^{\prime })}_{s}a+{(\mathbf{x},{\mathbf{y}}^{\prime })}_{s}c=0.$$

Hence, C is symplectic–self-orthogonal over ${\mathsf{E}}_3$. □

Next, we state the multilevel construction of symplectic QSD ${\mathsf{E}}_3$ codes over from ternary linear codes.

Theorem 1. 

Let $\mathsf{B}$ be a ternary linear code of length $2n$. If $\mathsf{B}$ is a symplectic–self-orthogonal code, then the ${\mathsf{E}}_3$ code, C, constructed through the relationship $C=\mathsf{B}a+{\mathsf{B}}^{{\perp }_s}c$ is symplectic QSD. Furthermore, $\mathrm{res}\left(C\right)=\mathsf{B}$ and $\mathrm{tor}\left(C\right)={\mathsf{B}}^{{\perp }_s}$.

Proof. 

We begin by proving that C is a linear ${\mathsf{E}}_3$ code. Using the linearity of $\mathsf{B}$ and ${\mathsf{B}}^{{\perp }_s}$, we note that C is closed under the addition operation. Let $\mathbf{c}\in C$ and $e\in {\mathsf{E}}_3$, then $\mathbf{c}=\mathbf{b}a+{\mathbf{b}}^{\prime }c$, where $\mathbf{b}\in \mathsf{B}$, ${\mathbf{b}}^{\prime }\in {\mathsf{B}}^{{\perp }_s}$ and by c-adic decomposition of e, we have $e=ra+r^{\prime }c$ where $r,r^{\prime }\in {\mathbb{F}}_3$. So, we get

$$\begin{array}{cc}\hfill \mathbf{c}e& =(\mathbf{b}a+{\mathbf{b}}^{\prime }c)(ra+r^{\prime }c)=\mathbf{b}ara+\mathbf{b}a{r}^{\prime }c+{\mathbf{b}}^{\prime }cra+{\mathbf{b}}^{\prime }c{r}^{\prime }c\hfill \\ & =\mathbf{b}r{a}^2+\mathbf{b}{r}^{\prime }ac+{\mathbf{b}}^{\prime }rca+{\mathbf{b}}^{\prime }{r}^{\prime }{c}^2\hfill \\ & =\mathbf{b}ra+\mathbf{b}{r}^{\prime }c.\hfill \end{array}$$

Therefore, $\mathbf{b}r,\mathbf{b}{r}^{\prime }\in \mathsf{B}$ since the code $\mathsf{B}$ is linear, $\mathbf{d}\in \mathsf{B}$ and $r,r^{\prime }\in {\mathbb{F}}_3$. From the symplectic–orthogonality of $\mathsf{B}$, we see that $\mathbf{b}{r}^{\prime }\in {\mathsf{B}}^{{\perp }_s}$. We conclude that, $\mathbf{c}e\in C$. Hence, C is a linear ${\mathsf{E}}_3$ code.

Let $\mathbf{c}=\mathbf{b}a+{\mathbf{b}}^{\prime }c,{\mathbf{c}}^{\prime }=\mathbf{d}a+{\mathbf{d}}^{\prime }c\in C$ for some $\mathbf{b},\mathbf{d}\in \mathsf{B}$ and ${\mathbf{b}}^{\prime },{\mathbf{d}}^{\prime }\in {\mathsf{B}}^{{\perp }_s}$. Then,

$$\begin{array}{cc}\hfill {(\mathbf{c},{\mathbf{c}}^{\prime })}_s& ={(\mathbf{b}a+{\mathbf{b}}^{\prime }c,\mathbf{d}a+{\mathbf{d}}^{\prime }c)}_s\hfill \\ & ={(\mathbf{b},\mathbf{d})}_s{a}^2+{(\mathbf{b},{\mathbf{d}}^{\prime })}_{s}ac+{({\mathbf{b}}^{\prime },\mathbf{d})}_{s}ca+{({\mathbf{b}}^{\prime },{\mathbf{d}}^{\prime })}_s{c}^2\hfill \\ & ={(\mathbf{b},\mathbf{d})}_{s}a+{(\mathbf{b},{\mathbf{d}}^{\prime })}_{s}c=0.\hfill \end{array}$$

The last equality is derived because $\mathsf{B}$ is a symplectic–self-orthogonal code. Therefore, we have $C=\mathsf{B}a+{\mathsf{B}}^{{\perp }_s}c$. This implies that $\left|C\right|=\left|\mathsf{B}\right||{\mathsf{B}}^{{\perp }_s}| =9^n$. Consequently, C is a symplectic QSD code. Thus, the torsion and residue codes can be obtained by applying their definitions. □

The following result demonstrates that Theorem 1 does not hold if $\mathsf{B}$ is not a symplectic–self-orthogonal code.

Theorem 2. 

For every ${\mathsf{E}}_3$ code, C, given by $C=\mathsf{B}a+{\mathsf{B}}^{{\perp }_s}c$ where $\mathsf{B}$ is a ternary linear code. Then, C is always an additive code over ${\mathsf{E}}_3$, but never a linear code over ${\mathsf{E}}_3$, unless $\mathsf{B}$ is symplectic–self-orthogonal.

Proof. 

First, we have to prove that C is an additive code over ${\mathsf{E}}_3$. Let ${\mathbf{c}}_{\mathbf{1}},{\mathbf{c}}_{\mathbf{2}}\in C$, then ${\mathbf{c}}_{\mathbf{1}}={\mathbf{b}}_{\mathbf{1}}a+{\mathbf{b}}_{\mathbf{1}}^{\prime }c$ and ${\mathbf{c}}_{\mathbf{2}}={\mathbf{b}}_{\mathbf{2}}a+{\mathbf{b}}_{\mathbf{2}}^{\prime }c$, where ${\mathbf{b}}_{\mathbf{1}},{\mathbf{b}}_{\mathbf{2}}\in \mathsf{B}$ and ${\mathbf{b}}_{\mathbf{1}}^{\prime },{\mathbf{b}}_{\mathbf{2}}^{\prime }\in {\mathsf{B}}^{{\perp }_s}$. Thus, $\mathbf{c}+{\mathbf{c}}^{\prime }=({\mathbf{b}}_{\mathbf{1}}+{\mathbf{b}}_{\mathbf{2}})a+({\mathbf{b}}_{\mathbf{1}}^{\prime }+{\mathbf{b}}_{\mathbf{2}}^{\prime })c$. From the linearity of $\mathsf{B}$ and ${\mathsf{B}}^{{\perp }_s}$ and the definition of C, it is easy to see that $\mathbf{c}+{\mathbf{c}}^{\prime }\in C$. Hence, C is an additive code over ${\mathsf{E}}_3$.

Now, suppose that C is a linear code over ${\mathsf{E}}_3$ and there exists a codeword $\mathbf{b}\in \mathsf{B}$. Then, a vector ${\mathbf{b}}^{\prime }$ exists in ${\mathsf{B}}^{{\perp }_s}$, such that $\mathbf{c}=\mathbf{b}a+{\mathbf{b}}^{\prime }c\in C$. Thus, let $e\in {\mathsf{E}}_3$; then, e has a c-adic decomposition form $e=ra+r^{\prime }c$ with $r,r^{\prime }\in {\mathbb{F}}_3$. Further,

$$\mathbf{c}e=(\mathbf{b}a+{\mathbf{b}}^{\prime }c)(ra+r^{\prime }c)=\mathbf{b}r{a}^2+\mathbf{b}{r}^{\prime }ac+{\mathbf{b}}^{\prime }rca+{\mathbf{b}}^{\prime }{r}^{\prime }{c}^2=\left(\mathbf{b}r\right)a+\left(\mathbf{b}{r}^{\prime }\right)c.$$

The last equality derived from $ca=c^2=0,a^2=a,ac=c$. Using the linearity of C, we have $\mathbf{c}e\in C$ and $\mathbf{b}{r}^{\prime }\in {\mathsf{B}}^{{\perp }_s}$. Since ${\mathsf{B}}^{{\perp }_s}$ is a ternary linear code, we have $\mathbf{b}{r}^{\prime }{r}^{\prime -1}=\mathbf{b}\in {\mathsf{B}}^{{\perp }_s}$, where $r^{\prime -1}$ denotes the multiplicative inverse of $r^{\prime }\in {\mathbb{F}}_3$. This yields $\mathsf{B}\subseteq {\mathsf{B}}^{{\perp }_s}$. □

Next, we characterize the symplectic QSD codes over ${\mathsf{E}}_3$ in terms of their associated ternary codes.

Corollary 1. 

A linear ${\mathsf{E}}_3$ code, C, of length $2n$ is symplectic QSD if $\mathrm{res}\left(C\right)=\mathrm{tor}{\left(C\right)}^{{\perp }_s}$.

Proof. 

For the direct statement, suppose C is symplectic QSD; then, $\left|C\right|=3^{2k_1+k_2}=3^{2n}$. This means $k_1=2n-(k_1+k_2)$. So, $\mathrm{res}\left(C\right)$ and $\mathrm{tor}{\left(C\right)}^{{\perp }_s}$ have the same dimension. Since C is symplectic–self-orthogonal, then by Lemma 1, $\mathrm{res}\left(C\right)\subseteq \mathrm{tor}{\left(C\right)}^{{\perp }_s}$. Hence, $\mathrm{res}\left(C\right)=\mathrm{tor}{\left(C\right)}^{{\perp }_s}$.

Conversely, since $\mathrm{res}\left(C\right)\subseteq \mathrm{tor}\left(C\right)$, then from the given hypotheses, $\mathrm{res}\left(C\right)=\mathrm{tor}{\left(C\right)}^{{\perp }_s}$ and $\mathrm{tor}{\left(C\right)}^{{\perp }_s}\subseteq \mathrm{res}{\left(C\right)}^{{\perp }_s}$. These imply that, $\mathrm{res}\left(C\right)$ is ternary symplectic–self-orthogonal. By Theorem 2 and Theorem 3.5 in [8], the code, C, is symplectic QSD. □

The study of LCD codes with Euclidean inner product over non-unitary ring of order four first appeared in [4]. Further, the authors in [8] established right LCD ${\mathsf{E}}_3$ codes. In addition, LCD codes under the symplectic inner product were introduced in [10]. In the following results, we extend this notion to codes over ${\mathsf{E}}_3$.

A right-symplectic-nice code is said to be right-symplectic linear with complementary dual (right-symplectic LCD) if satisfies: $C\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$; C is right-symplectic LCD if $C\oplus C^{{\perp }_{R_s}}={\mathsf{E}}_3^{2n}$.

Likewise, a left-symplectic-nice code is called left-symplectic linear with complementary dual (left-symplectic LCD) if satisfies $C\cap C^{{\perp }_{L_s}}=\left\{\mathbf{0}\right\}$. Moreover, C is a left-symplectic LCD if $C\oplus C^{{\perp }_{L_s}}={\mathsf{E}}_3^{2n}$.

Remark 2. 

Note that non-trivial left-symplectic LCD code does not exist over ${\mathsf{E}}_3$. If $C\ne \left\{\mathbf{0}\right\}$ is a nonzero linear ${\mathsf{E}}_3$ code of length $2n$. Then, for any $\mathbf{y}\in C$, $\mathbf{y}c\in C$. We can see that $\mathbf{y}c=\mathbf{v}c$, where $\mathbf{v}\in {\mathbb{F}}_3^{2n}$. Since c is a left zero divisor, we have ${(\mathbf{y}c,\mathbf{z})}_s=0$ for all $\mathbf{z}\in C$. This shows $\mathbf{y}c\in C^{{\perp }_{L_s}}$. Thus, $C\cap C^{{\perp }_{L_s}}\ne \left\{\mathbf{0}\right\}$. Hence, the desired result achieved.

Henceforward, we consider the right-symplectic LCD codes as symplectic LCD and therefore only the right (symplectic) dual.

Using part (i) of Theorem 4.10 in [8], Proposition 2.1 in [14] and by following Lemma 5 in [10], we state the next Lemma.

Lemma 2. 

For any free linear ${\mathsf{E}}_3$ code, C, of generator matrix, G, we have the following:

(1). 

$C={\langle G_{3}a\rangle }_{{\mathsf{E}}_3}$ where $G_3$ is a ternary matrix.

(2). 

$C^{{\perp }_{R_s}}={\langle H_3\Omega a\rangle }_{{\mathsf{E}}_3}$ where $H_3$ is a ternary parity check matrix related to $G_3$.

Proof. 

The proof of (1) is achieved in the proof of Theorem 4.10 in [8].

To prove (2), we have $C={\langle G_{3}a\rangle }_{{\mathsf{E}}_3}$ for some ternary matrix $G_3$. Then, let $H_3$ be a parity check matrix of the linear code of generator matrix $G_3$. By Proposition 2.1 in [14], $H_3\Omega$ is a generator matrix of the symplectic dual of the ternary code whose generator matrix is $G_3$. Note that $\left(G_{3}a\right)\Omega {\left(H_3\Omega a\right)}^T=a{G}_3\Omega H_3^{T}a=\mathbf{0}$. Hence, ${\langle H_3\Omega a\rangle }_{{\mathsf{E}}_3}\subseteq C^{{\perp }_{R_s}}$.

For the converse inclusion, assume $\mathbf{z}=(z_1,z_2,\cdots ,z_{2n})\in C^{{\perp }_{R_s}}$ and $G_3$ is in standard form, i.e., $G_3=\left[I_k\right|M]$. Therefore, $G_{3}a=\left[I_k\right|M]a=\left[\begin{array}{c}{\mathbf{r}}_{\mathbf{1}}a\\ {\mathbf{r}}_{\mathbf{2}}a\\ ⋮\\ {\mathbf{r}}_{\mathbf{k}}a\end{array}\right]$ where ${\mathbf{r}}_{\mathbf{1}},\cdots ,{\mathbf{r}}_{\mathbf{k}}$ are the rows of $G_3$.

Furthermore, for all $0\le i\le k$, we have ${({\mathbf{r}}_{\mathbf{i}}a,\mathbf{z})}_s=0$. So, for any $1\le i\le k$,

$$x_{i_1}a{z}_{i_1}+x_{i_2}a{z}_{i_2}+\cdots +x_{i_t}a{z}_{i_t}=0,$$

for some $1\le t\le k$ and $x_{i_j}$ are nonzero elements of ${\mathbb{F}}_3$. Since $ax=x$ for all $x\in {\mathsf{E}}_3$, we get

$$x_{i_1}{z}_{i_1}+x_{i_2}{z}_{i_2}+\cdots +x_{i_t}{z}_{i_t}=0.$$

Further, we obtain the next k equations:

$$\begin{array}{c}{\beta }_1{z}_{n+1}+0+0+\cdots +0+x_{k+i_1}{z}_{k+i_1}+\cdots +x_{i_{l_1}}{z}_{i_{l_1}}=0,\\ 0+{\beta }_2{z}_{n+2}+0+\cdots +0+x_{k+i_2}{z}_{k+i_2}+\cdots +x_{i_{l_2}}{z}_{i_{l_2}}=0,\\ ⋮\\ 0+0+\cdots +0+{\beta }_k{z}_k+x_{k+i_k}{z}_{k+i_k}+\cdots +x_{k+i_k}{z}_{i_k}=0\end{array}$$

where for each $0\le t\le k,$${{\beta }_t}^{’}\mathrm{s}$ are 1 or 2 and the indices $k+i_t,\cdots i_{l_t}$ of z correspond to the nonzero positions of ${\mathsf{r}}_{t}a$. Moreover, there are at most $2n-k$ free variables in the previous k equations, then there are $9^{2n-k}$ solutions of z. This implies that $\left|{\mathbf{C}}^{{\perp }_{R_s}}\right|\le 9^{2n-k}.$ Since ${\langle H_3\Omega a\rangle }_{{\mathsf{E}}_3}=9^{2n-k}$ and ${\langle H_3\Omega a\rangle }_{{\mathsf{E}}_3}\subseteq {\mathbf{C}}^{{\perp }_{R_s}}$, we conclude that ${\mathbf{C}}^{{\perp }_{R_s}}={\langle H_3\Omega a\rangle }_{{\mathsf{E}}_3}$. □

Proposition 1. 

Let C be a linear ${\mathsf{E}}_3$ code having length $2n$. Then, C is a free code if it is right-symplectic-nice.

Proof. 

We have C is a linear code over ${\mathsf{E}}_3$, then $\left|C\right||C^{{\perp }_{R_s}}| =3^{2k_1+k_2}\times 9^{2n-k_1}=3^{4n+k_2}$. Hence, C is right-symplectic-nice if $k_2=0$. □

Corollary 2. 

Every symplectic LCD code, C, over ${\mathsf{E}}_3$ is free.

Proof. 

Using the definition of symplectic LCD code, we found that C is right-symplectic-nice. By Proposition 1 the code, C, is free. □

Now, we construct symplectic LCD codes over ${\mathsf{E}}_3$ from ternary symplectic LCD codes.

Proposition 2. 

Let $\mathsf{B}$ be a symplectic LCD ternary $[2n,k]$ code having a generator matrix $G_3$. Then, the ${\mathsf{E}}_3$ code $C={\langle G_{3}a\rangle }_{{\mathsf{E}}_3}$ is a symplectic LCD ${\mathsf{E}}_3$ code.

Proof. 

We have $C={\langle G_{3}a\rangle }_{{\mathsf{E}}_3}$; then, by statement (2) of Lemma 2, $C^{{\perp }_{R_s}}={\langle H_3\Omega a\rangle }_{{\mathsf{E}}_3}$ where $H_3$ is a parity check matrix of the matrix $G_3$ and $H_3\Omega$ is a generator matrix of the ternary symplectic dual code related to $G_3$.

Further, $\left|C\right||C^{{\perp }_{R_s}}| =3^{2k}\times 9^{2n-k}=9^{2n}$. This implies that C is right-symplectic-nice.

To prove that $C\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$, assume (if possible) $\mathbf{0}\ne \mathbf{z}$ in $C\cap C^{{\perp }_{R_s}}$. Then, there are two possibilities:

(1).

Let $\mathbf{z}=\mathbf{y}c$ where $\mathbf{0}\ne \mathbf{y}\in {\mathbb{F}}_3^{2n}$ and $G_3$ is in standard form. By investigation of the matrix $G_{3}a$, we conclude that $\mathbf{y}$ can be expressed as a sum of scalar multiples of some rows of $G_3$. Hence, $\mathbf{y}\in \mathsf{B}$. Likewise, we can prove that $\mathbf{y}\in {\mathsf{B}}^{{\perp }_s}$.

Consequently, $\mathbf{0}\ne \mathbf{y}\in \mathsf{B}\cap {\mathsf{B}}^{{\perp }_s}$, this contradicts with $\mathsf{B}$ is a ternary symplectic LCD code.

(2).

Suppose $\mathbf{z}$ is not a product of a ternary vector by c, then

$$\mathbf{z}=\sum {\mathbf{g}}_{i}a{u}_i\mathrm{and}\mathbf{z}=\sum {\mathbf{g}}_{j}a{v}_j.$$

where ${\mathbf{g}}_{j}a$ is a row of $G_{3}a$ for some (distinct) i and $u_i\in {\mathsf{E}}_3$ and ${\mathbf{g}}_{j}a$ is a row of $H_3\Omega a$ for some (distinct) j and $v_i\in {\mathsf{E}}_3$. Therefore,

$$\mathbf{z}=\sum {\mathbf{g}}_i{u}_i\mathrm{also}\mathbf{z}=\sum {\mathbf{g}}_j{v}_j$$

where ${\mathbf{g}}_i$ is a row of $G_3$ for some (distinct) i, j and $u_i,v_i\in {\mathsf{E}}_3$.

From the multiplication table of ${\mathsf{E}}_3$, we have $\mathbf{z}a=\sum {\mathbf{g}}_{i_t}$ and $\mathbf{z}=\sum {\mathbf{g}}_{j_l}$, where ${\mathbf{g}}_{i_t}$ is a row of $G_3$ for some distinct $i_t$ and ${\mathbf{g}}_{j_l}$ is a row of $H_3\Omega$ for some distinct $j_l$.

Therefore, $\sum {\mathbf{g}}_{i_t}a-\sum {\mathbf{g}}_{j_l}a=(\sum {\mathbf{g}}_{i_t}-\sum {\mathbf{g}}_{j_l})a=\mathbf{0}$. This implies that, $\sum {\mathbf{g}}_{i_t}-\sum {\mathbf{g}}_{j_l}=\mathbf{0}$, so $\sum {\mathbf{g}}_{i_t}=\sum {\mathbf{g}}_{j_l}\in \mathsf{B}\cap {\mathsf{B}}^{{\perp }_s}$. This implies that $\mathsf{B}\cap {\mathsf{B}}^{{\perp }_s}\ne \left\{\mathbf{0}\right\},$ which contradicts the fact that $\mathsf{B}$ is a symplectic LCD code.

□

The example below presents a symplectic LCD ${\mathsf{E}}_3$ code constructed according to Proposition 2.

Example 1. 

Let $\mathsf{B}$ be the ternary $[4, 2]$ code of generator matrix $G_3=\left[\begin{array}{cccc}1& 0& 1& 1\\ 0& 1& 2& 1\end{array}\right]$.

• Since $G_3\Omega G_3^T=\left[\begin{array}{cc}0& 1\\ 2& 0\end{array}\right]$ is a non-singular matrix over ${\mathbb{F}}_3$; then, by Theorem 1 in [15], $\mathsf{B}$ is a ternary symplectic LCD code of length 4. Then, by Proposition 2, ${\langle G_{3}a\rangle }_{{\mathsf{E}}_3}$ is a symplectic LCD ${\mathsf{E}}_3$ code.

Now, we verify the conditions of the definition of symplectic LCD ${\mathsf{E}}_3$ code to prove that the code ${\langle G_{3}a\rangle }_{{\mathsf{E}}_3}$ is symplectic LCD over ${\mathsf{E}}_3$.

Assume, $C={\langle G_{3}a\rangle }_{{\mathsf{E}}_3}={〈\left[\begin{array}{cccc}a& 0& a& a\\ 0& a& 2a& a\end{array}\right]〉}_{{\mathsf{E}}_3}$. Then, C can be represented by the set

$$C=\left\{(ax,a{x}^{\prime },ax+2a{x}^{\prime },ax+a{x}^{\prime })\right|x,x^{\prime }\in {\mathsf{E}}_3\}=\left\{(x,x^{\prime },x+2x^{\prime },x+x^{\prime })\right|x,x^{\prime }\in {\mathsf{E}}_3\}.$$

For any $(y,y^{\prime },z,z^{\prime })\in C$, we have

$${((a,0,a,a),(y,y^{\prime },z,z^{\prime }))}_s=0and{((0,a,2a,a),(y,y^{\prime },z,z^{\prime }))}_s=0.$$

These imply, $az+2ay+2a{y}^{\prime }=0$ and $a{z}^{\prime }+ay+2a{y}^{\prime }=0$. Thus, $z+2y+2y^{\prime }$ and $z^{\prime }+y+2y^{\prime }=0$. So, $z=y+y^{\prime }$ and $z^{\prime }=2y+y^{\prime }$. Hence,

$$C^{{\perp }_{R_s}}=\left\{(y,y^{\prime },y+y^{\prime },2y+y^{\prime })\right|y,y^{\prime }\in {\mathsf{E}}_3\}.$$

Clearly, $|C^{{\perp }_{R_s}}| =9^2$, $\left|C\right|=9^2$. Then, $\left|C\right||C^{{\perp }_{R_s}}| =9^4$. Furthermore, $C\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$. Therefore, C is symplectic LCD ${\mathsf{E}}_3$ code.

The inverse statement of Proposition 2 is partially true, as the following proposition states.

Proposition 3. 

A free symplectic LCD ${\mathsf{E}}_3$ code, C, is the ${\mathsf{E}}_3$-span by the matrix $Ga$ with G is a generator matrix of a ternary symplectic LCD code $\mathsf{B}$.

Proof. 

Since C is free, then by part (1) of Lemma 2, C is the ${\mathsf{E}}_3$-span by the matrix $Ga$ where G is a generator matrix of the ternary code $\mathsf{B}$. Moreover, from part (2) of Lemma 2, $C^{{\perp }_{R_s}}$ is generated by the matrix $H\Omega a$, where H is a parity check matrix of $\mathsf{B}$. Since C is symplectic LCD, $C\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$. Further, suppose $(\mathbf{0}\ne \mathbf{x})\in \mathsf{B}\cap {\mathsf{B}}^{{\perp }_{R_s}}$, then $\mathbf{x}a\in C\cap C^{{\perp }_{R_s}}$; this contradicts $C\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$. Therefore, $\mathsf{B}$ is a ternary symplectic LCD code. □

Corollary 3. 

For any symplectic LCD code, C, over ${\mathsf{E}}_3$, the ternary codes $\mathrm{tor}\left(C\right)$ and $\mathrm{res}\left(C\right)$ are symplectic LCD codes.

Proof. 

Since C is symplectic LCD code, then by Corollary 2, C is free. Then, by Proposition 3, C is the ${\mathsf{E}}_3$-span by the matrix $Ga$ where G is a generator matrix of ternary symplectic LCD code. From the freeness of C, we have $\mathrm{res}\left(C\right)=\mathrm{tor}\left(C\right)$ and G will be a generator matrix for both $\mathrm{res}\left(C\right)$ and $\mathrm{tor}\left(C\right)$. □

4. Symplectic ACD Codes

In this section, we introduce the right-symplectic-nice and left-symplectic-nice additive codes, as well as right-symplectic ACD and left-symplectic ACD ${\mathsf{E}}_3$ codes. We provide some criteria for the existence of the right-symplectic ACD ${\mathsf{E}}_3$ codes. Moreover, for additive, linear and free ${\mathsf{E}}_3$ codes, in terms of their torsion and residue codes, we also determine the residue and torsion codes of the right-symplectic dual of such codes. At the end, we show that any linear ${\mathsf{E}}_3$ code is free if, and only if, its right and left-symplectic duals have the same residue code.

An additive ${\mathsf{E}}_3$ code of length $2n$ is called right-symplectic-nice (respectively, left-symplectic-nice) code if $\left|C\right||C^{{\perp }_{R_s}}| =9^{2n}$ (respectively, $\left|C\right||C^{{\perp }_{L_s}}| =9^{2n}$).

A right-symplectic-nice additive ${\mathsf{E}}_3$ code, C, is called right-symplectic additive complementary dual (right-symplectic ACD) if $C\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$.

Likewise, a left-symplectic-nice additive ${\mathsf{E}}_3$ code, C, is called left-symplectic additive complementary dual (left-symplectic ACD) if $C\cap C^{{\perp }_{L_s}}=\left\{\mathbf{0}\right\}$.

Lemma 3. 

Let C be an additive ${\mathsf{E}}_3$ code. Then, the right-symplectic dual of C is a free linear ${\mathsf{E}}_3$ code and its left-symplectic dual is a non-free linear ${\mathsf{E}}_3$ code.

Proof. 

Firstly, we have to prove that $C^{{\perp }_{R_s}}$ and $C^{{\perp }_{L_s}}$ are linear codes over ${\mathsf{E}}_3$. Let $\mathbf{z},{\mathbf{z}}^{\prime }\in C^{{\perp }_{R_s}}$ and $\mathbf{y}\in C$. Then,

$${(\mathbf{y},\mathbf{z}+{\mathbf{z}}^{\prime })}_s={(\mathbf{y},\mathbf{z})}_s+{(\mathbf{y},{\mathbf{z}}^{\prime })}_s=0+0=0.$$

Hence, $\mathbf{z}+{\mathbf{z}}^{\prime }\in C^{{\perp }_{R_s}}$. Thus, for any $\mathbf{z}\in C^{{\perp }_{R_s}}$, $u\in {\mathsf{E}}_3$ and $\mathbf{y}\in C$, we have

$${(\mathbf{y},\mathbf{z}r)}_s={(\mathbf{y},\mathbf{z})}_{s}r=0.$$

So, $\mathbf{z}r\in C^{{\perp }_{R_s}}$. Further, $C^{{\perp }_{R_s}}$ is a right submodule of ${\mathsf{E}}_3^{2n}$. This shows that $C^{{\perp }_{R_s}}$ is a linear ${\mathsf{E}}_3$ code.

Next, we prove that $C^{{\perp }_{R_s}}$ is free code. Let $\mathbf{z}=\mathbf{u}c\in C^{{\perp }_{R_s}}$ where $\mathbf{u}\in {\mathbb{F}}_3^{2n}$. Then according to multiplication table of ${\mathsf{E}}_3$, for any $\mathbf{y}\in C$, we get ${(\mathbf{y},\mathbf{z})}_s={(\mathbf{y},\mathbf{u}c)}_s=tc=0$. This implies, $\mathbf{t}$ is multiple of 3. Therefore, ${(\mathbf{y},\mathbf{u}a)}_s={(\mathbf{y},\mathbf{u}a)}_s=ta=0$. This means $\mathbf{u}a\in C^{{\perp }_{R_s}}$. By Theorem 4.12 in [8], consequently $C^{{\perp }_{R_s}}$ is free.

Now, we have to prove $C^{{\perp }_{L_s}}$ is also linear over ${\mathsf{E}}_3$. Let $\mathbf{x},{\mathbf{x}}^{\prime }\in C^{{\perp }_{L_s}}$ and $\mathbf{y}\in C$. Then,

$${(\mathbf{x}+{\mathbf{x}}^{\prime },\mathbf{y})}_s={(\mathbf{z},\mathbf{y})}_s+{({\mathbf{z}}^{\prime },\mathbf{y})}_s=0+0=0.$$

So, $\mathbf{x}+{\mathbf{x}}^{\prime }\in C^{{\perp }_{L_s}}$. Therefore, let $\mathbf{x}\in C^{{\perp }_{R_s}}$, $\mathbf{y}\in C$ and let $r\in {\mathsf{E}}_3$, then $r=ea+e^{\prime }c$ where $e,e^{\prime }\in {\mathbb{F}}_3$. Thus,

$${(\mathbf{x}r,\mathbf{y})}_s={(\mathbf{x}(ea+e^{\prime }c),\mathbf{y})}_s={(\mathbf{x}ea,\mathbf{y})}_s+{(\mathbf{x}{e}^{\prime }c,\mathbf{y})}_s.$$

Since ${(\mathbf{x}a,\mathbf{y})}_s={(\mathbf{x}b,\mathbf{y})}_s={(\mathbf{x},\mathbf{y})}_s=0$, we get ${(\mathbf{x}ea,\mathbf{y})}_s+{(\mathbf{x}{e}^{\prime }c,\mathbf{y})}_s={(\mathbf{x}e,\mathbf{y})}_s+0={(\mathbf{x},\mathbf{y})}_{s}e=0$. So, ${(\mathbf{x}r,\mathbf{y})}_s=0$. Hence, $\mathbf{x}r\in C^{{\perp }_{L_s}}$. Consequently, $C^{{\perp }_{L_s}}$ is a linear ${\mathsf{E}}_3^{2n}$ code.

It remains to prove $C^{{\perp }_{L_s}}$ is not a free code. Suppose $C^{{\perp }_{L_s}}$ is free over ${\mathsf{E}}_3$. Since c is left zero divisor, we get ${\langle c{I}_{2n}\rangle }_{{\mathbb{F}}_3}\subseteq C^{{\perp }_{L_s}}$. Thus, according to Theorem 4.12 in [8], ${\langle a{I}_{2n}\rangle }_{{\mathbb{F}}_3}\subseteq C^{{\perp }_{L_s}}$. These yield $|C^{{\perp }_{L_s}}|\ge 9^n\times 9^n=9^{2n}$. Since $C^{{\perp }_{L_s}}\subseteq {\mathsf{E}}_3^{2n}$, we have $|C^{{\perp }_{L_s}}|\le 9^{2n}$. Hence, $C^{{\perp }_{L_s}}={\mathsf{E}}_3^{2n}$. This implies that $C=\left\{\mathbf{0}\right\}$, which is impossible. Hence, $C^{{\perp }_{L_s}}$ is not a free ${\mathsf{E}}_3$ code. □

Theorem 3. 

Let C be a linear code over ${\mathsf{E}}_3$ of length $2n$. Then, $C={\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}$ if C is right-symplectic-nice.

Proof. 

Assume, ${\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}=C$. By Lemma 3, $C^{{\perp }_{R_s}}$ is a free linear ${\mathsf{E}}_3$ code. Thus, from Proposition 1, $C^{{\perp }_{R_s}}$ is right-symplectic-nice. This yields $|C^{{\perp }_{R_s}}\left|\right|{\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}| = |C\left|\right|C^{{\perp }_{R_s}}| =9^{2n}$. Hence, C is right-symplectic-nice.

Conversely, assume C is right-symplectic-nice. From Proposition 1, we have C is free. Furthermore, by Lemma 3, $C^{{\perp }_{R_s}}$ is free. Thus, by Lemma 2, for any $\mathbf{y}\in C$ and $\mathbf{z}\in C^{{\perp }_{R_s}}$, we assume $\mathbf{y}=\mathbf{l}a$ and $\mathbf{z}=\mathbf{t}a$ for some $\mathbf{l},{\mathbf{t}}^{\prime }\in {\mathbb{F}}_3^{2n}$. Then, $0={(\mathbf{y},\mathbf{z})}_s={(\mathbf{l}a,\mathbf{t}a)}_s={(\mathbf{l},\mathbf{t})}_s{a}^2={(\mathbf{l},\mathbf{t})}_{s}a=2{(\mathbf{t},\mathbf{l})}_{s}a=2{(\mathbf{z},\mathbf{y})}_s$. So, ${(\mathbf{z},\mathbf{y})}_s=0$, for any $\mathbf{z}\in C^{{\perp }_{R_s}}$. Then, $\mathbf{y}\in {\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}$. This shows that $C\subseteq {\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}$. Furthermore, C, $C^{{\perp }_{R_s}}$ are both right-symplectic-nice, which means $\left|C\right||C^{{\perp }_{R_s}}| = |{\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}\left|\right|C^{{\perp }_{R_s}}|$. So, ${\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}| = |C\left|\right|$. This together with $C\subseteq {\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}$ imply that $C={\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}$. □

The next result characterizes the one-sided duals of a symplectic LCD ${\mathsf{E}}_3$ code.

Proposition 4. 

Let C be a symplectic LCD ${\mathsf{E}}_3$ code, then $C^{{\perp }_{R_s}}$ is symplectic LCD but $C^{{\perp }_{L_s}}$ is not.

Proof. 

Suppose C is symplectic LCD code. Then, C is right-symplectic-nice with $C\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$. By Theorem 3, ${\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}=C$. Thus, $|C^{{\perp }_{R_s}}\left|\right|{\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}| = |C^{{\perp }_{R_s}}\left|\right|C|=9^{2n}$. Furthermore, $C^{{\perp }_{R_s}}\cap {\left(C^{{\perp }_{R_s}}\right)}^{{\perp }_{R_s}}=C^{{\perp }_{R_s}}\cap C=\mathbf{0}$. These imply that $C^{{\perp }_{R_s}}$ is a symplectic LCD ${\mathsf{E}}_3$ code.

Further, from Lemma 3, $C^{{\perp }_{L_s}}$ is non-free linear ${\mathsf{E}}_3$ code. Hence, by Proposition 1, $C^{{\perp }_{L_s}}$ is not right-symplectic-nice. So, $C^{{\perp }_{L_s}}$ is not symplectic LCD. □

Proposition 5. 

If C is a right-symplectic ACD ${\mathsf{E}}_3$ code with $3^t$ elements then t is even.

Proof. 

By Lemma 3, $C^{{\perp }_{R_s}}$ is free. This yields, $C^{{\perp }_{R_s}}=9^{k_1}=3^{2k_1}$. Since C is right-symplectic ACD, then $\left|C\right||C^{{\perp }_{R_s}}| =9^{2n}$. Thus, $\left|C\right|=3^{4n-2k_1}$. Hence, $t=4n-2k_1$; this means t is even. □

Next, we construct the right-symplectic ACD codes over ${\mathsf{E}}_3$.

Proposition 6. 

Let C be an additive ${\mathsf{E}}_3$ code of length $2n$ with $C\cap C^{{\perp }_{R_s}}=Cc\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$ and $|C^{{\perp }_{R_s}}\left|\right|C|<9^{2n}$. Then, for any nonzero element $\mathbf{y}=\mathbf{z}+{\mathbf{z}}^{\prime }c\in C+Cc$, if $\mathbf{y}\in C^{{\perp }_{R_s}}$, then $\mathbf{z}$ is a multiple of a ternary vector by c, $\mathbf{z}\ne \mathbf{0}$, ${\mathbf{z}}^{\prime }c\ne \mathbf{0}$ and $\mathbf{z}\in C\setminus Cc$.

Proof. 

Let $\mathbf{0}\ne \mathbf{y}=\mathbf{z}+{\mathbf{z}}^{\prime }c\in C+Cc$, where $\mathbf{z},{\mathbf{z}}^{\prime }\in C$. We suppose that $\mathbf{y}\in C^{{\perp }_{R_s}}$. Then, we get the following cases:

(1).

If $\mathbf{z}=\mathbf{0}$ and ${\mathbf{z}}^{\prime }c=\mathbf{0}$, then $\mathbf{y}=\mathbf{0}$ which is not possible.

(2).

If $\mathbf{z}=\mathbf{0}$ and ${\mathbf{z}}^{\prime }c\ne \mathbf{0}$, then $\mathbf{y}={\mathbf{z}}^{\prime }c\in C^{{\perp }_{R_s}}$. This implies that, $\mathbf{0}\ne \mathbf{y}\in Cc\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$ which is not possible.

(3).

If $\mathbf{z}\ne \mathbf{0}$ and ${\mathbf{z}}^{\prime }c=\mathbf{0}$, then $\mathbf{y}=\mathbf{z}\in C^{{\perp }_{R_s}}$. This shows that $\mathbf{0}\ne \mathbf{y}\in C\cap C^{{\perp }_{R_s}}=\mathbf{0}$ which leads to a contradiction.

(4).

If $\mathbf{z}\ne \mathbf{0}$ and ${\mathbf{z}}^{\prime }c\ne \mathbf{0}$, then $\mathbf{y}=\mathbf{z}+{\mathbf{z}}^{\prime }c$. Suppose, $\mathbf{z}$ is not a multiple of a ternary vector by c. So, $\mathbf{z}c\ne \mathbf{0}$. By Lemma 3, $C^{{\perp }_{R_s}}$ is a free linear ${\mathsf{E}}_3$ code. This together with $\mathbf{y}\in C^{{\perp }_{R_s}}$, we obtain $\mathbf{y}c=(\mathbf{z}+{\mathbf{z}}^{\prime }c)c=\mathbf{z}c\in C^{{\perp }_{R_s}}$. Thus, $\mathbf{z}c\in Cc\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$ which is not possible. Hence, $\mathbf{z}$ is a multiple of a ternary vector by c. Further, if $\mathbf{z}\in Cc$, then $\mathbf{y}\in Cc$, also $\mathbf{y}\in Cc\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$. It again leads to a contradiction. Hence, $\mathbf{z}\in C\setminus Cc$.

By (1), (2), (3), and (4), if $\mathbf{0}\ne \mathbf{y}=\mathbf{z}+{\mathbf{z}}^{\prime }c\in C^{{\perp }_{R_s}}$, then $\mathbf{z}\ne \mathbf{0}$, ${\mathbf{z}}^{\prime }c\ne \mathbf{0}$, $\mathbf{z}$ is a multiple of a ternary vector by c, and $\mathbf{z}\in C\setminus Cc$. This completes the proof. □

Theorem 4. 

Let C be an additive ${\mathsf{E}}_3$ code of length $2n$. If $\left|C\right||C^{{\perp }_{R_s}}|<9^{2n}$ and $C\cap C^{{\perp }_{R_s}}=Cc\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$, then C can be extended to a right-symplectic ACD code over ${\mathsf{E}}_3$ by adding some elements of $Cc$.

Proof. (1). First, we show that $Cc⊈C$. Suppose $\mathsf{B}={\langle C\rangle }_{{\mathsf{E}}_3}=Ca+Cc=\{\mathbf{z}a+{\mathbf{z}}^{\prime }c|\mathbf{z},{\mathbf{z}}^{\prime }\in C\setminus Cc\}$.

• Since c is a left zero-divisor, we suppose that $\mathbf{z}$ and ${\mathbf{z}}^{\prime }$ are not multiples of a ternary vectors by c. Clearly, $\mathsf{B}$ is a free linear ${\mathsf{E}}_3$ code and ${\mathsf{B}}^{{\perp }_{R_s}}=C^{{\perp }_{R_s}}$. From Lemma 2 and Proposition 2, $\mathsf{B}$ is right-symplectic-nice.
  If $Cc\subseteq C$, according to [8], we define a map ${\pi }_1:C\to \mathsf{B}$ by

  $${\pi }_1\left(\mathbf{0}\right)=\mathbf{0},{\pi }_1\left(\mathbf{z}\right)=\mathbf{z}a\mathrm{and}{\pi }_1\left(\mathbf{z}c\right)=\mathbf{z}c,\mathrm{for}\mathrm{any}\mathbf{z}\in C\setminus Cc.$$
  Note that, for any $\mathbf{z},{\mathbf{z}}^{\prime }\in C$, ${\pi }_1(\mathbf{z}+{\mathbf{z}}^{\prime })={\pi }_1\left(\mathbf{z}\right)+{\pi }_1\left({\mathbf{z}}^{\prime }\right)$. It is clear that ${\pi }_1$ is a surjective linear mapping over ${\mathbb{F}}_3$. Thus, $\left|C\right|\ge \left|\mathsf{B}\right|$ since ${\pi }_1$ is surjective. Thus, $\left|C\right||C^{{\perp }_{R_s}}| \ge |\mathsf{B}\left|\right|C^{{\perp }_{R_s}}| =|\mathsf{B}\left|\right|{\mathsf{B}}^{{\perp }_{R_s}}| =9^{2n}$. So, $\left|C\right||C^{{\perp }_{R_s}}|\ge 9^{2n}$ which contradicts the hypothesis. Hence, $Cc⊈C$. Moreover, C can be extended to a right-symplectic-nice code by adding some codewords of $Cc$.

(2).

Now, we prove that a vector ${\mathbf{z}}^{\prime }c$ exists in $Cc\setminus C$ such that $\mathbf{z}+{\mathbf{z}}^{\prime }c\notin C^{{\perp }_{R_s}}$ for any $\mathbf{z}\in C$. Assume this is not true, i.e., there is a codeword $\mathbf{z}$ of C such that $\mathbf{z}+{\mathbf{z}}^{\prime }c\in C^{{\perp }_{R_s}}$ for any ${\mathbf{z}}^{\prime }c\in Cc\setminus C$. Then, by Proposition 6, a ternary vector $\mathbf{y}$ exists such that $\mathbf{z}=\mathbf{y}c$ and $\mathbf{z}$ in $C\setminus Cc$.

Suppose for any ${\mathbf{z}}_{\mathbf{1}}^{\prime }c,{\mathbf{z}}_{\mathbf{2}}^{\prime }c\in C\setminus Cc$ with ${\mathbf{z}}_{\mathbf{1}}^{\prime }c\ne {\mathbf{z}}_{\mathbf{2}}^{\prime }c$, there are ${\mathbf{z}}_{\mathbf{1}},{\mathbf{z}}_{\mathbf{2}}\in C$ such that ${\mathbf{z}}_{\mathbf{1}}+{\mathbf{z}}_{\mathbf{1}}^{\prime }c,{\mathbf{z}}_{\mathbf{2}}+{\mathbf{z}}_{\mathbf{2}}^{\prime }c\in C^{{\perp }_{R_s}}$.

If possible, let ${\mathbf{z}}_{\mathbf{1}}={\mathbf{z}}_{\mathbf{2}}$, then $({\mathbf{z}}_{\mathbf{1}}^{\prime }+2{\mathbf{z}}_{\mathbf{2}}^{\prime })c=({\mathbf{z}}_{\mathbf{1}}+{\mathbf{z}}_{\mathbf{1}}^{\prime }c)+2({\mathbf{z}}_{\mathbf{2}}+{\mathbf{z}}_{\mathbf{2}}^{\prime }c)\in C^{{\perp }_{R_s}}$ (since $C^{{\perp }_{R_s}}$ is a linear code over ${\mathsf{E}}_3$ and ${\mathbf{z}}_{\mathbf{1}}+{\mathbf{z}}_{\mathbf{1}}^{\prime }c,{\mathbf{z}}_{\mathbf{2}}+{\mathbf{z}}_{\mathbf{2}}^{\prime }c\in C^{{\perp }_{R_s}}$).

Thus, ${\mathbf{z}}_{\mathbf{1}}^{\prime }+2{\mathbf{z}}_{\mathbf{2}}^{\prime }\in C$ as ${\mathbf{z}}_{\mathbf{1}}^{\prime },{\mathbf{z}}_{\mathbf{2}}^{\prime }\in C$. This yields $({\mathbf{z}}_{\mathbf{1}}^{\prime }+2{\mathbf{z}}_{\mathbf{2}}^{\prime })c\in Cc$. So, $({\mathbf{z}}_{\mathbf{1}}^{\prime }+2{\mathbf{z}}_{\mathbf{2}}^{\prime })c\in Cc\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$. Therefore, ${\mathbf{z}}_{\mathbf{1}}^{\prime }c={\mathbf{z}}_{\mathbf{2}}^{\prime }c$ which is not possible.

Hence, we have ${\mathbf{z}}_{\mathbf{1}}\ne {\mathbf{z}}_{\mathbf{2}}$. Let us define ${\pi }_2:\mathsf{B}\to C$. We will discuss the following cases:

(2.1).

For any $\mathbf{z}\in C\setminus Cc$, if $\mathbf{z}c\in Cc\setminus C$, we define ${\pi }_2$ as

$${\pi }_2\left(\mathbf{0}\right)=\mathbf{0},{\pi }_2\left(\mathbf{za}\right)=\mathbf{z}\mathrm{and}{\pi }_2\left(\mathbf{z}c\right)=\mathbf{y}.$$

where, $\mathbf{y}$ is the given vector satisfies $\mathbf{y}+\mathbf{z}a\in C^{{\perp }_{R_s}}$. For all $\mathbf{z},{\mathbf{z}}^{\prime }\in \mathsf{B}$, ${\pi }_2(\mathbf{z}+{\mathbf{z}}^{\prime })={\pi }_2\left(\mathbf{z}\right)+{\pi }_2\left({\mathbf{z}}^{\prime }\right)$. Thus, it is easy to see that ${\pi }_2$ is injective linear map over ${\mathbb{F}}_3$. Therefore, if ${\pi }_2\left(\mathbf{z}c\right)\ne {\pi }_2\left({\mathbf{z}}^{\prime }c\right)$, then $\mathbf{y}\ne {\mathbf{y}}^{\prime }$, as for $\mathbf{z}c,{\mathbf{z}}^{\prime }c\in Cc$, there exist $\mathbf{y},{\mathbf{y}}^{\prime }\in C$ such that $\mathbf{y}+\mathbf{z}c,\mathbf{y}+\mathbf{z}c\in C^{{\perp }_{R_s}}$ with $\mathbf{y}\ne {\mathbf{y}}^{\prime }$.

(2.2).

For any $\mathbf{z}\in C\setminus Cc$, if $\mathbf{z}c\in Cc\setminus C$, we define ${\pi }_2$ by

$${\pi }_2\left(\mathbf{0}\right)=\mathbf{0},{\pi }_2\left(\mathbf{za}\right)=\mathbf{z}\mathrm{and}{\pi }_2\left(\mathbf{z}c\right)=\mathbf{z}c.$$

For all $\mathbf{z},{\mathbf{z}}^{\prime }\in \mathsf{B}$, ${\pi }_2(\mathbf{z}+{\mathbf{z}}^{\prime })={\pi }_2\left(\mathbf{z}\right)+{\pi }_2\left({\mathbf{z}}^{\prime }\right)$. Furthermore, ${\pi }_2$ is injective ${\mathbb{F}}_3$-linear map.

For both cases, since ${\pi }_2$ is injective, we have $\left|C\right|\ge \left|\mathsf{B}\right|$. This yields $\left|C\right||C^{{\perp }_{R_s}}|\ge |\mathsf{B}\left|\right|{\mathsf{B}}^{{\perp }_{R_s}}|=9^{2n}$. So, $\left|C\right||C^{{\perp }_{R_s}}|\ge 9^{2n}$, this is a contradiction.

Hence, ${\mathbf{z}}^{\prime }c\in Cc\setminus C$ such that $\mathbf{z}+{\mathbf{z}}^{\prime }c\notin C^{{\perp }_{R_s}}$ for any $\mathbf{z}\in C$.

At this time, we add $\mathbf{z}c$ to C to get a new code D. We can check that, $D\cap D^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$. If D is right-symplectic-nice, then D is right-symplectic ACD. Otherwise, repeat steps $\left(1\right)$ and $\left(2\right)$ for D. □

The definitions of the associated ternary codes (i.e., residue and torsion codes) of a linear ${\mathsf{E}}_3$ code can be extended into an additive ${\mathsf{E}}_3$ code C. Clearly, these codes are linear codes. Consider, $\dim \left(\mathrm{res}\left(C\right)\right)=d_1$ and $\dim \left(\mathrm{tor}\left(C\right)\right)=d_2$.

For a linear ${\mathsf{E}}_3$ code, C, of generator matrix G in the form given in Corollary 4.11 in [8]. Then, $\dim (\mathrm{res}\left(C\right)=k_1=d_1$ and $\dim \left(\mathrm{tor}\left(C\right)\right)=k_1+k_2=d_1+k_2=d_2$, where $k_2$ is the ${\mathbb{F}}_3$-dimension of the set of all codewords of C which are scalar multiples of $c\in {\mathsf{E}}_3$ and which can not be obtained from the upper two blocks of G.

For any codeword $\mathbf{r}$ of the code, C, we can write it in c-adic decomposition form as $\mathbf{r}=\mathbf{u}a+\mathbf{v}c$ with $\mathbf{u},\mathbf{v}$ are ternary vectors, so that ${\alpha }_{{\mathsf{E}}_3}\left(\mathbf{r}\right)=\mathbf{u}$.

Theorem 5.

For any additive ${\mathsf{E}}_3$ code, C, of length $2n$. We have

$$C^{{\perp }_{R_s}}={\langle \mathrm{res}{\left(C\right)}^{{\perp }_s}\rangle }_{{\mathsf{E}}_3}and\left|C^{{\perp }_{R_s}}\right|=9^{2n-\dim \left(\mathrm{res}\right(C\left)\right)}=9^{2n-d_1}.$$

Proof. 

For any $\mathbf{z}\in C$, there exists $\mathbf{v}\in \mathrm{res}\left(C\right)$ such that $\mathbf{z}a=\mathbf{v}a$. Let $\mathbf{u}\in \mathrm{res}{\left(C\right)}^{{\perp }_{R_s}}$. Then,

$${(\mathbf{z},\mathbf{u}a)}_s={(\mathbf{z}a,\mathbf{u}a)}_s={(\mathbf{v}a,\mathbf{u}a)}_s={(\mathbf{v},\mathbf{u})}_s{a}^2={(\mathbf{v},\mathbf{u})}_{s}a=0.$$

$${(\mathbf{z},\mathbf{u}b)}_s={(\mathbf{z}a,\mathbf{u}b)}_s={(\mathbf{v}a,\mathbf{u}b)}_s={(\mathbf{v}b,\mathbf{u}b)}_s={(\mathbf{v},\mathbf{u})}_{s}b=0.$$

These show that $\mathbf{u}a,\mathbf{u}b\in C^{{\perp }_{R_s}}$. Hence, ${\langle \mathrm{res}{\left(C\right)}^{{\perp }_s}\rangle }_{{\mathsf{E}}_3}\subseteq C^{{\perp }_{R_s}}$ since a and b are generators of ${\mathsf{E}}_3$ and $\mathbf{u}a,\mathbf{u}a\in C^{{\perp }_{R_s}}$. Clearly, ${\langle \mathrm{res}{\left(C\right)}^{{\perp }_s}\rangle }_{{\mathsf{E}}_3}={\langle \mathrm{res}\left(C\right)\rangle }_{{\mathsf{E}}_3}^{{\perp }_{Rs}}$.

Now, we have to prove $C^{{\perp }_{R_s}}\subseteq {\langle \mathrm{res}{\left(C\right)}^{{\perp }_s}\rangle }_{{\mathsf{E}}_3}$. Let $\mathbf{x}\in \mathrm{res}\left(C\right)$, then there exists $\mathbf{x}a+\mathbf{y}c\in C^{{\perp }_{R_s}}$ such that ${\alpha }_{{\mathsf{E}}_3}(\mathbf{x}a+\mathbf{y}c)=\mathbf{x}$.

Since c is a left-zero divisor, then for any $\mathbf{z}\in C^{{\perp }_{R_s}}$, we get

$${(\mathbf{x}a,\mathbf{z})}_s={(\mathbf{x}a+\mathbf{y}c,\mathbf{z})}_s=0\mathrm{and}{(\mathbf{x}b,\mathbf{z})}_s={(\mathbf{x}a,\mathbf{z})}_s=0.$$

This yields $\mathbf{z}\in {\langle \mathrm{res}\left(C\right)\rangle }_{{\mathsf{E}}_3}^{{\perp }_{R_s}}$. So, $C^{{\perp }_{R_s}}\subseteq {\langle \mathrm{res}\left(C\right)\rangle }_{{\mathsf{E}}_3}^{{\perp }_{R_s}}$.

Consequently, $C^{{\perp }_{R_s}}={\langle \mathrm{res}{\left(C\right)}^{{\perp }_s}\rangle }_{{\mathsf{E}}_3}$. Further, $|C^{{\perp }_{R_s}}| = |{\langle \mathrm{res}{\left(C\right)}^{{\perp }_s}\rangle }_{{\mathsf{E}}_3}| =9^{2n-d_1}$. □

The following result derives a sufficient condition for an additive ${\mathsf{E}}_3$ code to be right-symplectic ACD.

Proposition 7. 

Let C be an additive ${\mathsf{E}}_3$ code of length $2n$. Then, C is a right-symplectic ACD code if the next statements hold:

(1).

The ternary code $\mathrm{res}\left(C\right)$ is symplectic LCD.

(2).

$\mathrm{tor}\left(C\right)c\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$.

(3).

$3^{d_1+d_2}\times 9^{2n-d_1}=9^{2n}$ (i.e., $d_1=d_2$).

Proof. 

First, it is clear that, $\left|C\right|=\left|\mathrm{res}\left(C\right)\right|\left|\mathrm{tor}\left(C\right)\right|=3^{d_1+d_2}$ and by Theorem 5, $|C^{{\perp }_{R_s}}| =9^{2n-d_1}$. So, $\left|C\right||C^{{\perp }_{R_s}}| =9^{2n}$. This demonstrates that C is a right-symplectic-nice ${\mathsf{E}}_3$ code. The final step is to show that $C\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$. If it is possible, let $\mathbf{0}\ne \mathbf{z}=(z_1,z_2,\cdots ,z_{2n})\in C\cap C^{{\perp }_{R_s}}$. As $\mathrm{tor}\left(C\right)c\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$, $\mathbf{z}$ is not a multiple of ternary vector by c. This yields ${\alpha }_{{\mathsf{E}}_3}\left(\mathbf{z}\right)\ne \mathbf{0}$. Thus, for any $\mathbf{y}=(y_1,y_2,\cdots ,y_{2n})\in C$, we have ${(\mathbf{y},\mathbf{z})}_s=0$. As ${\alpha }_{{\mathsf{E}}_3}$ is a (ring) morphism, we get

$$\begin{array}{cc}\hfill {({\alpha }_{{\mathsf{E}}_3}\left(\mathbf{y}\right),{\alpha }_{{\mathsf{E}}_3}\left(\mathbf{z}\right))}_s& ={\sum }_{i=1}^n({\alpha }_{{\mathsf{E}}_3}\left(y_i\right){\alpha }_{{\mathsf{E}}_3}\left(z_{n+i}\right)-{\alpha }_{{\mathsf{E}}_3}\left(y_{n+i}\right){\alpha }_{{\mathsf{E}}_3}\left(z_i\right))\hfill \\ & ={\alpha }_{{\mathsf{E}}_3}\left({\sum }_{i=1}^n(y_i{z}_{n+i}-y_{n+i}{z}_i)\right)={\alpha }_{{\mathsf{E}}_3}\left({(\mathbf{y},\mathbf{z})}_s\right)\hfill \\ & ={\alpha }_{{\mathsf{E}}_3}\left(0\right)=0.\hfill \end{array}$$

This derived that ${\alpha }_{{\mathsf{E}}_3}\left(\mathbf{z}\right)\in \mathrm{res}{\left(C\right)}^{{\perp }_s}$, which contradicts that $\mathrm{res}\left(C\right)$ is symplectic LCD. Hence, $\mathbf{z}=\mathbf{0}$. So, $C\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$.

Hence, C is a right-symplectic ACD code over ${\mathsf{E}}_3$. □

The inverse statement of Corollary 3 is also true, as the corollary below establishes.

Corollary 4. 

Let C be a free linear ${\mathsf{E}}_3$ code with residue code is ternary symplectic LCD, then C is symplectic LCD.

Proof. 

Since $\mathrm{res}\left(C\right)$ is a symplectic LCD code, then to prove C is symplectic LCD is sufficient to check the conditions $\left(2\right)$ and $\left(3\right)$ of Proposition 7. We have C is an additive ${\mathsf{E}}_3$ code, then by Theorem 5,

$$\begin{array}{cc}\hfill \mathrm{tor}\left(C\right)c\cap C^{{\perp }_{R_s}}& =\mathrm{tor}\left(C\right)c\cap {\langle \mathrm{res}{\left(C\right)}^{{\perp }_s}\rangle }_{{\mathsf{E}}_3}=\mathrm{res}\left(C\right)c\cap {\langle \mathrm{res}{\left(C\right)}^{{\perp }_s}\rangle }_{{\mathsf{E}}_3}\hfill \\ & =\mathrm{res}\left(C\right)c\cap \mathrm{res}{\left(C\right)}^{{\perp }_s}c=(\mathrm{res}\left(C\right)\cap \mathrm{res}{\left(C\right)}^{{\perp }_s})c\hfill \\ & =\left\{\mathbf{0}\right\}.\hfill \end{array}$$

This shows that condition (2) is satisfied. Therefore, from the freeness of C, we get $d_1=d_2$, so $3^{d_1+d_2}\times 9^{2n-d_1}=9^{d_1}\times 9^{2n-d_1}=9^{2n}$. This yields condition (3) is satisfied.

Hence, C is right-symplectic ACD. Moreover, C is a symplectic LCD code over ${\mathsf{E}}_3$. □

We determine the residue and torsion codes of the right-symplectic dual codes of additive ${\mathsf{E}}_3$ codes as the following theorem shows.

Theorem 6. 

Let C be an additive code over ${\mathsf{E}}_3$ of length $2n$. Then,

$$\mathrm{res}\left(C^{{\perp }_{R_s}}\right)=\mathrm{tor}\left(C^{{\perp }_{R_s}}\right)=\mathrm{res}{\left(C\right)}^{{\perp }_s}.$$

Proof. 

By Lemma 3, $C^{{\perp }_{R_s}}$ is a free linear code, then $\mathrm{res}\left(C^{{\perp }_{R_s}}\right)=\mathrm{tor}\left(C^{{\perp }_{R_s}}\right)$. Then, we have to check that $\mathrm{res}\left(C^{{\perp }_{R_s}}\right)\subseteq \mathrm{res}{\left(C\right)}^{{\perp }_s}$. Let $\mathbf{x}a+\mathbf{y}c\in C^{{\perp }_{R_s}}$ and ${\mathbf{x}}^{\prime }a+{\mathbf{y}}^{\prime }c\in C$ where $\mathbf{x},{\mathbf{x}}^{\prime },\mathbf{y},{\mathbf{y}}^{\prime }\in {\mathbb{F}}_3^{2n}$. Then, ${\alpha }_{{\mathsf{E}}_3}(\mathbf{x}a+\mathbf{y}c)=\mathbf{x}\in \mathrm{res}\left(C^{{\perp }_{R_s}}\right)$ and ${\alpha }_{{\mathsf{E}}_3}({\mathbf{x}}^{\prime }a+{\mathbf{y}}^{\prime }c)={\mathbf{x}}^{\prime }\in \mathrm{res}\left(C\right)$. Since ${\alpha }_{{\mathsf{E}}_3}$ is a ring morphism and ${({\mathbf{x}}^{\prime }a+{\mathbf{y}}^{\prime }c,\mathbf{x}a+\mathbf{y}c)}_s=0$, we get

$${\alpha }_{{\mathsf{E}}_3}\left({({\mathbf{x}}^{\prime }a+{\mathbf{y}}^{\prime }c,\mathbf{x}a+\mathbf{y}c)}_s\right)={({\alpha }_{{\mathsf{E}}_3}({\mathbf{x}}^{\prime }a+{\mathbf{y}}^{\prime }c),{\alpha }_{{\mathsf{E}}_3}(\mathbf{x}a+\mathbf{y}c))}_s={({\mathbf{x}}^{\prime },\mathbf{x})}_s=0.$$

This implies that $\mathbf{x}\in \mathrm{res}{\left(C\right)}^{{\perp }_s}$. So, $\mathrm{res}\left(C^{{\perp }_{R_s}}\right)\subseteq \mathrm{res}{\left(C\right)}^{{\perp }_s}$.

To prove that $\mathrm{res}{\left(C\right)}^{{\perp }_s}\subseteq \mathrm{res}\left(C^{{\perp }_{R_s}}\right)$, suppose $\mathbf{x}$ is an arbitrary element of $\mathrm{res}{\left(C\right)}^{{\perp }_s}$. Let $\mathbf{z}=\mathbf{y}a+{\mathbf{y}}^{\prime }c\in C$ where $\mathbf{y},{\mathbf{y}}^{\prime }\in {\mathbb{F}}_3^{2n}$. Then, ${\alpha }_{{\mathsf{E}}_3}\left(\mathbf{z}\right)=\mathbf{y}\in \mathrm{res}\left(C\right)$. So,

$${(\mathbf{z},\mathbf{x}a)}_s={(\mathbf{y}a+{\mathbf{y}}^{\prime }c,\mathbf{x}a)}_s={(\mathbf{y}a,\mathbf{x}a)}_s+{({\mathbf{y}}^{\prime }c,\mathbf{x}a)}_s={(\mathbf{y}a,\mathbf{x}a)}_s={(\mathbf{y},\mathbf{x})}_{s}a+0=0.$$

This shows that $\mathbf{x}a\in C^{{\perp }_{R_s}}$. Thus, $\mathbf{x}\in \mathrm{res}\left(C^{{\perp }_{R_s}}\right)$ as ${\alpha }_{{\mathsf{E}}_3}\left(\mathbf{x}a\right)=\mathbf{x}$. Hence, $\mathrm{res}{\left(C\right)}^{{\perp }_s}\subseteq \mathrm{res}\left(C^{{\perp }_{R_s}}\right)$. Therefore, $\mathrm{res}\left(C^{{\perp }_{R_s}}\right)=\mathrm{res}{\left(C\right)}^{{\perp }_s}$. □

Corollary 5. 

For any linear ${\mathsf{E}}_3$ code, C, we have $\mathrm{tor}{\left(C\right)}^{{\perp }_s}\subseteq \mathrm{tor}\left(C^{{\perp }_{R_s}}\right)$. Particularly, if C is free, then $\mathrm{tor}\left(C^{{\perp }_{R_s}}\right)=\mathrm{tor}{\left(C\right)}^{{\perp }_s}$.

Proof. 

First we have, $\mathrm{res}\left(C\right)\subseteq \mathrm{tor}\left(C\right)$ as C is a linear ${\mathsf{E}}_3$ code, then $\mathrm{tor}{\left(C\right)}^{{\perp }_s}\subseteq \mathrm{res}{\left(C\right)}^{{\perp }_s}$. By Theorem 6, $\mathrm{res}\left(C^{{\perp }_{R_s}}\right)=\mathrm{res}{\left(C\right)}^{{\perp }_s}\supseteq \mathrm{tor}{\left(C\right)}^{{\perp }_s}$. According to Lemma 3, $C^{{\perp }_{R_s}}$ is free, so $\mathrm{tor}\left(C^{{\perp }_{R_s}}\right)=\mathrm{res}\left(C^{{\perp }_{R_s}}\right)$. Hence, $\mathrm{tor}{\left(C\right)}^{{\perp }_s}\subseteq \mathrm{tor}\left(C^{{\perp }_{R_s}}\right)$. Therefore, if C is free, then $\mathrm{tor}\left(C\right)=\mathrm{res}\left(C\right)$. Thus, $\mathrm{tor}{\left(C\right)}^{{\perp }_s}=\mathrm{res}{\left(C\right)}^{{\perp }_s}$. By Lemma 3 and Theorem 6, $\mathrm{tor}\left(C^{{\perp }_{R_s}}\right)=\mathrm{res}\left(C^{{\perp }_{R_s}}\right)=\mathrm{res}{\left(C\right)}^{{\perp }_s}$. Hence, $\mathrm{tor}\left(C^{{\perp }_{R_s}}\right)=\mathrm{tor}{\left(C\right)}^{{\perp }_s}$. □

The converse statement of Proposition 7 partially holds.

Proposition 8. 

For an additive ${\mathsf{E}}_3$ code, C, of length $2n$. We have, $\mathrm{tor}\left(C\right)c\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$, $\mathrm{res}\left(C\right)$ is a ternary symplectic LCD code and $3^{d_1+d_2}\times 9^{2n-d_1}=9^{2n}$ (i.e., $d_1=d_2$) if C is a right-symplectic ACD code over ${\mathsf{E}}_3$ with $Cc\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$.

Proof. 

Let $\mathbf{z}\in \mathrm{tor}\left(C\right)c\cap C^{{\perp }_{R_s}}$. Then, $\mathbf{z}\in \mathrm{tor}\left(C\right)c$; this means $\mathbf{z}=\mathbf{t}c$ with $\mathbf{t}\in \mathrm{tor}\left(C\right)$. By the definition of the torsion code, $\mathbf{z}\in C$. This together with $\mathbf{z}\in C^{{\perp }_{R_s}}$ imply that $\mathbf{z}\in C\cap C^{{\perp }_{R_s}}$. Since C is right-symplectic ACD code, $\mathbf{z}=\mathbf{0}$. Hence, $\mathrm{tor}\left(C\right)c\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$.

We have $\left|C\right|=\left|\mathrm{res}\left(C\right)\right|\left|\mathrm{tor}\left(C\right)\right|=3^{d_1+d_2}$ and by Theorem 5, $|C^{{\perp }_{R_s}}| =9^{2n-d_1}$. Therefore, $\left|C\right||C^{{\perp }_{R_s}}| =3^{d_1+d_2}\times 9^{2n-d_1}$. Since C is right-symplectic-nice, $3^{d_1+d_2}\times 9^{2n-d_1}=9^{2n}$. Hence, $d_1=d_2$. Now, let $\mathbf{r}\in \mathrm{res}\left(C\right)\cap \mathrm{res}{\left(C\right)}^{{\perp }_s}$. By Theorem 6, $\mathbf{r}\in \mathrm{res}{\left(C\right)}^{{\perp }_s}=\mathrm{res}\left(C^{{\perp }_{R_s}}\right)$. Thus, there are $\mathbf{r}a+\mathbf{t}a\in C$ and $\mathbf{r}a+{\mathbf{t}}^{\prime }a\in C^{{\perp }_{R_s}}$ such that ${\alpha }_{{\mathsf{E}}_3}(\mathbf{r}a+\mathbf{t}a)={\alpha }_{{\mathsf{E}}_3}(\mathbf{r}a+{\mathbf{t}}^{\prime }a)=\mathbf{r}$. From the linearity of $C^{{\perp }_{R_s}}$ over ${\mathsf{E}}_3$, we get $(\mathbf{r}a+{\mathbf{t}}^{\prime }a)c=\mathbf{r}c\in C^{{\perp }_{R_s}}$. Therefore, $(\mathbf{r}a+{\mathbf{t}}^{\prime }a)c=\mathbf{r}c\in Cc$. So, $(\mathbf{r}a+{\mathbf{t}}^{\prime }a)c\in Cc\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$. Thus, $\mathbf{r}c=\mathbf{0}$, also $\mathbf{r}=\mathbf{0}$. Hence, $\mathrm{res}\left(C\right)$ is a ternary symplectic LCD code. □

Proposition 9. 

Let C be an additive code of length $2n$ over ${\mathsf{E}}_3$. Then, $\mathrm{tor}\left(C^{{\perp }_{L_s}}\right)\subseteq \mathrm{res}{\left(C\right)}^{{\perp }_s}$. Equality holds if C is linear.

Proof. 

Let $\mathbf{z}=\mathbf{r}a+\mathbf{t}c\in C^{{\perp }_{L_s}}$ where $\mathbf{r},\mathbf{t}\in {\mathbb{F}}_3^{2n}$. Then, ${\alpha }_{{\mathsf{E}}_3}\left(\mathbf{z}\right)={\alpha }_{{\mathsf{E}}_3}(\mathbf{r}a+\mathbf{t}c)=\mathbf{r}$ is an arbitrary codeword of $\mathrm{res}\left(C^{{\perp }_{L_s}}\right)$. Therefore, let ${\mathbf{t}}^{\prime }\in \mathrm{tor}\left(C\right)$, then ${\mathbf{t}}^{\prime }c\in C$. This together with $\mathbf{z}\in C^{{\perp }_{L_s}}$, we get

$$0={(\mathbf{z},{\mathbf{t}}^{\prime }c)}_s={(\mathbf{r}a+\mathbf{t}c,{\mathbf{t}}^{\prime }c)}_s={(\mathbf{r}a,{\mathbf{t}}^{\prime }c)}_s+{(\mathbf{t}c,{\mathbf{t}}^{\prime }c)}_s={(\mathbf{r},{\mathbf{t}}^{\prime })}_{s}c.$$

This yields ${(\mathbf{r},{\mathbf{t}}^{\prime })}_s=0$. Hence, $\mathbf{r}\in \mathrm{tor}\left(C^{{\perp }_{L_s}}\right)$. So, $\mathrm{tor}\left(C^{{\perp }_{L_s}}\right)\subseteq \mathrm{res}{\left(C\right)}^{{\perp }_s}$.

Now, suppose that C is a linear code. Then, let $\mathbf{z}\in C$. By Theorem 3.5 in [8], $\mathbf{z}=\mathbf{x}a+\mathbf{y}c$ for some $\mathbf{x}\in \mathrm{res}\left(C\right)$ and $\mathbf{y}\in \mathrm{tor}\left(C\right)$. Let ${\mathbf{y}}^{\prime }\in \mathrm{tor}{\left(C\right)}^{{\perp }_s}$, we have $\mathrm{res}\left(C\right)\subseteq \mathrm{tor}\left(C\right)$. Then, we obtain

$${({\mathbf{y}}^{\prime }a,\mathbf{z})}_s={({\mathbf{y}}^{\prime }a,{\mathbf{y}}^{\prime }a+\mathbf{y}c)}_s={({\mathbf{y}}^{\prime }a,\mathbf{x}a)}_s+{({\mathbf{y}}^{\prime }a,\mathbf{y}c)}_s={({\mathbf{y}}^{\prime },\mathbf{x})}_{s}a+{({\mathbf{y}}^{\prime },\mathbf{y})}_{s}c=0+0=0.$$

Hence, ${\mathbf{y}}^{\prime }a\in C^{{\perp }_{L_s}}$. This implies that, ${\alpha }_{{\mathsf{E}}_3}\left({\mathbf{y}}^{\prime }a\right)={\mathbf{y}}^{\prime }\in \mathrm{res}\left(C^{{\perp }_{L_s}}\right)$. So, $\mathrm{tor}{\left(C\right)}^{{\perp }_s}\subseteq \mathrm{res}\left(C^{{\perp }_{L_s}}\right)$. Therefore, $\mathrm{res}\left(C^{{\perp }_{L_s}}\right)=\mathrm{tor}{\left(C\right)}^{{\perp }_s}$. □

Proposition 10. 

For a linear ${\mathsf{E}}_3$ code, C, of length $2n$. We have,

$$C\mathit{is}\mathit{free}\mathit{if}, \mathit{and}\mathit{only}\mathit{if}, \mathrm{res}\left(C^{{\perp }_{R_s}}\right)=\mathrm{res}\left(C^{{\perp }_{L_s}}\right)$$

Proof. 

Suppose C is a linear ${\mathsf{E}}_3$ code. By Theorem 6, $\mathrm{res}\left(C^{{\perp }_{R_s}}\right)=\mathrm{res}{\left(C\right)}^{{\perp }_s})$. According to Proposition 9, $\mathrm{res}\left(C^{{\perp }_{L_s}}\right)=\mathrm{tor}{\left(C\right)}^{{\perp }_s}$. Since C is free, $\mathrm{res}\left(C\right)=\mathrm{tor}\left(C\right)$. Thus, $\mathrm{res}{\left(C\right)}^{{\perp }_s}=\mathrm{tor}{\left(C\right)}^{{\perp }_s}$. Hence, $\mathrm{res}\left(C^{{\perp }_{R_s}}\right)=\mathrm{res}\left(C^{{\perp }_{L_s}}\right)$.

For the inverse assertion. Assume, C is a linear ${\mathsf{E}}_3$ code with $\mathrm{res}\left(C^{{\perp }_{R_s}}\right)=\mathrm{res}\left(C^{{\perp }_{L_s}}\right)$. By Theorem 6 and Proposition 9, $\mathrm{tor}{\left(C\right)}^{{\perp }_s}=\mathrm{res}{\left(C\right)}^{{\perp }_s}$. So, $\mathrm{tor}\left(C\right)=\mathrm{res}\left(C\right)$. Hence, C is free. □

In what follows, we give an example to describe Theorem 4 and Proposition 7.

Example 2. 

Let C be an additive ${\mathsf{E}}_3$ code of length 4 and additive generator matrix

$$\mathsf{G}=\left[\begin{array}{cccc}a& 2b& 0& 0\\ 0& 0& a& 2b\end{array}\right]$$

Let $(y,y^{\prime },z,z^{\prime })\in C^{{\perp }_{R_s}}$. Then,

$${((a,2b,0,0),(y,y^{\prime },z,z^{\prime }))}_s=0and{((0,0,a,2b),(y,y^{\prime },z,z^{\prime }))}_s=0.$$

These imply that $az+2b{z}^{\prime }=0$ and $2ay+b{y}^{\prime }=0$. Since $ax=bx=x$ for any $x\in {\mathsf{E}}_3$, we get $z+2z^{\prime }=0$ and $2y+y^{\prime }=0$. So, $z^{\prime }=z$ and $y^{\prime }=y$. Hence, $C^{{\perp }_{R_s}}=\left\{(y,y,z,z)\right|y,z\in {\mathsf{E}}_3\}$.

It is easy to derive, $C\cap C^{{\perp }_{R_s}}=Cc\cap C^{{\perp }_{R_s}}=\left\{\mathbf{0}\right\}$. Furthermore, $\left|C\right| =9$ and $|C^{{\perp }_{R_s}}| =9^2$. So, $\left|C\right||C^{{\perp }_{R_s}}| =9^3<9^4$. We add $(a,2b,0,0)c=(c,2c,0,0)$ and $(0,0,a,2b)c=(0,0,2c,c)$ to $\mathsf{G}$, then we get ${\mathsf{G}}^{\prime }$ where

$${\mathsf{G}}^{\prime }=\left[\begin{array}{cccc}a& 2b& 0& 0\\ 0& 0& a& 2b\\ c& 2c& 0& 0\\ 0& 0& c& 2c\end{array}\right]$$

Let $C^{\prime }$ be the additive ${\mathsf{E}}_3$–code of length 4 with an additive generator matrix ${\mathsf{G}}^{\prime }$. Then, according to Theorem 4, $C^{\prime }$ is right-symplectic ACD over ${\mathsf{E}}_3$.

To prove $C^{\prime }$ is right-symplectic ACD, we check is this code satisfies the conditions of Proposition 7. First, we prove that $\mathrm{res}\left(C^{\prime }\right)$ is ternary symplectic LCD code. Let $(z_1,z_2,z_3,z_4)\in \mathrm{res}{\left(C^{\prime }\right)}^{{\perp }_s}$. Then,

$${((y,y^{\prime },z,z^{\prime }),(1,2,0,0))}_s=0and{((y,y^{\prime },z,z^{\prime }),(0,0,1,2))}_s=0.$$

Thus, $2z+z^{\prime }=0$ and $2y+y^{\prime }=0$. So, $z^{\prime }=z_3$ and $y^{\prime }=y$. Hence, $\mathrm{res}{\left(C^{\prime }\right)}^{{\perp }_s}=\left\{(y,y,z,z)\right|;z\in {\mathbb{F}}_3\}$. This gives $\mathrm{res}\left(C^{\prime }\right)\cap \mathrm{res}{\left(C^{\prime }\right)}^{{\perp }_s}=\left\{\mathbf{0}\right\}$. Hence, $\mathrm{res}\left(C^{\prime }\right)$ is a ternary symplectic LCD.

Clearly, $\mathrm{res}\left(C^{\prime }\right)$ has the generator matrix ${\mathsf{G}}_1^{\prime }=\left[\begin{array}{cccc}1& 2& 0& 0\\ 0& 0& 1& 2\end{array}\right]$ and the generator matrix of $\mathrm{tor}\left(C^{\prime }\right)$ is ${\mathsf{G}}_2^{\prime }=\left[\begin{array}{cccc}1& 2& 0& 0\\ 0& 0& 1& 2\end{array}\right]$. These show that $d_1=d_2=2$.

Therefore, $\mathrm{tor}\left(C^{\prime }\right)c$ has generator matrix $\left[\begin{array}{cccc}c& 2c& 0& 0\\ 0& 0& c& 2c\end{array}\right]$.

By Theorem 5, $C^{\prime {\perp }_{R_s}}={\langle \mathrm{res}{\left(C^{\prime }\right)}^{{\perp }_s}\rangle }_{{\mathsf{E}}_3}$. It is easy to see that, $\mathrm{tor}\left(C^{\prime }\right)c\cap C^{\prime {\perp }_{R_s}}=\left\{\mathbf{0}\right\}$.

Hence, $C^{\prime }$ is right-symplectic ACD code over ${\mathsf{E}}_3$ according to Proposition 7.

5. Conclusions

We have introduced symplectic QSD, LCD, and ACD codes over non-commutative non-unitary local ring ${\mathsf{E}}_3$ with nine elements. Using symplectic–self-orthogonal ternary linear codes, we have given the multi-level construction of symplectic QSD codes. We have shown the relationship between symplectic LCD ${\mathsf{E}}_3$ codes and ternary symplectic LCD codes. Further, the right-symplectic ACD ${\mathsf{E}}_3$ codes were introduced and for the existence of such codes, and several conditions have been stated. We have also characterized the right-symplectic ACD codes over ${\mathsf{E}}_3$.

The construction of quantum codes can be achieved by deriving classical codes over finite fields with the symplectic–self-orthogonal property [14,15]. Therefore, these codes could be applied to construct quantum codes over this ring.

Additionally, it will be interesting to derive a mass formula for symplectic QSD and self-dual ${\mathsf{E}}_3$ codes and classify them in modest lengths. Moreover, the absence of non-trivial symplectic left LCD codes over ${\mathsf{E}}_3$ makes the study of symplectic left hull codes over this ring a natural direction for future work.