A New Method for Constructing Self-Dual Codes over Finite Commutative Rings with Characteristic 2

Abstract

In this work, we present a new method for constructing self-dual codes over finite commutative rings R with characteristic 2. Our method involves searching for $k\times 2k$ matrices M over R satisfying the conditions that its rows are linearly independent over R and $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$ for an R-linearly independent vector $\mathit{\alpha }\in R^k.$ Let $\mathcal{C}$ be a linear code generated by such a matrix $M.$ We prove that the dual code ${\mathcal{C}}^{\perp }$ of $\mathcal{C}$ is also a free linear code with dimension k, as well as $\mathcal{C}/Hull\left(\mathcal{C}\right)$ and ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)$ are one-dimensional free R-modules, where $Hull\left(\mathcal{C}\right)$ represents the hull of $\mathcal{C}$. Based on these facts, an isometry from $R\mathit{x}+R\mathit{y}$ onto $R^2$ is established, assuming that $\mathit{x}+Hull\left(\mathcal{C}\right)$ and $\mathit{y}+Hull\left(\mathcal{C}\right)$ are bases for $\mathcal{C}/Hull\left(\mathcal{C}\right)$ and ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)$ over R, respectively. By utilizing this isometry, we introduce a new method for constructing self-dual codes from self-dual codes of length 2 over finite commutative rings with characteristic 2. To determine whether the matrix $M{M}^{\top }$ takes the form of ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ with $\mathit{\alpha }$ being a linearly independent vector in $R^k$, a necessary and sufficient condition is provided. Our method differs from the conventional approach, which requires the matrix M to satisfy $M{M}^{\top }=0$. The main advantage of our method is the ability to construct nonfree self-dual codes over finite commutative rings, a task that is typically unachievable using the conventional approach. Therefore, by combining our method with the conventional approach and selecting an appropriate matrix construction, it is possible to produce more self-dual codes, in contrast to using solely the conventional approach.

1. Introduction

Self-dual codes are a significant class of codes that possess both algebraic and combinatorial properties, connecting them to various other areas of mathematics such as design theory, combinatorics, number theory, invariant theory, unimodular lattices and quantum codes. In particular, binary self-dual codes have garnered extensive attention, with considerable research effort dedicated to developing techniques for constructing new extremal and optimal binary self-dual codes. These known construction techniques include the double-circulant and bordered double-circulant constructions [1,2,3,4], the four-circulant construction and its variations [5,6,7,8,9,10,11], group rings and their connection to self-dual codes [12,13,14,15,16,17,18,19,20], bordered matrix constructions [12,13,14,17,19,21,22,23,24], neighbours of binary self-dual codes [25,26], the widely employed building-up construction [27,28,29], the production of binary self-dual codes as Gray images of self-dual codes over finite commutative Frobenius rings of characteristic 2 [28,30,31,32], and the well-known lifting method [10,11,33,34,35,36,37].

All of the matrix constructions described in the literature mentioned above are defined over finite commutative Frobenius rings with characteristic 2. It is worth noting that all these matrices have the property that the number of rows is half of the number of columns. As a result, all the self-dual codes produced from these matrix constructions are free.

After conducting a comprehensive review of the literature mentioned above, we can outline the conventional method for the construction of extremal binary self-dual codes as follows: One needs to present a new matrix construction, M, over a certain finite commutative Frobenius ring of characteristic 2 equipped with a Gray map, and establish the necessary conditions for M to satisfy $M{M}^{\top }=0.$ In some cases, it may be necessary to provide additional conditions to ensure the linear independence of the rows of $M.$ If these conditions are met, then M may be used to generate a self-dual code. By considering the Gray image of this self-dual code and employing well-known techniques such as the neighbor method and building-up construction, it may be possible to obtain a new extremal binary self-dual code. This conventional approach has been extensively utilized in the literature, leading to the construction of numerous new extremal binary self-dual codes. For further details, please refer to the aforementioned literature.

For the identical task, namely finding new extremal binary self-dual codes, we introduce a new method in this paper. This method, based on an isometry defined in Theorem 1, can be used to construct self-dual codes over finite commutative rings of characteristic 2, regardless of whether they are Frobenius rings or not. We now outline it as follows: Let $\mathcal{C}$ be a linear code generated by M over R, where R is a finite commutative ring with characteristic 2, M is a $k\times 2k$ matrix over R with its rows being linearly independent and satisfies $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$ with a linearly independent vector $\mathit{\alpha }$ in $R^k.$ We can prove that the dual code ${\mathcal{C}}^{\perp }$ of $\mathcal{C}$ is a free linear code with dimension k (see Proposition 2). Furthermore, we can demonstrate that both $\mathcal{C}/Hull\left(\mathcal{C}\right)$ and ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)$ are one-dimensional free R-modules (see Propositions 4 and 5), where $Hull\left(\mathcal{C}\right)=\mathcal{C}\cap {\mathcal{C}}^{\perp }.$ Based on these facts, we can establish an isometry from $R\mathit{x}+R\mathit{y}$ onto $R^2$, where $\mathit{x}+Hull\left(\mathcal{C}\right)$ and $\mathit{y}+Hull\left(\mathcal{C}\right)$ serve as bases for $\mathcal{C}/Hull\left(\mathcal{C}\right)$ and ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)$ over R, respectively. By utilizing this isometry, we can construct self-dual codes of length $2k$ from the matrix M and self-dual codes in $R^2$, and prove that they are exactly those self-dual codes containing $Hull\left(\mathcal{C}\right)$ as a subcode (see Theorem 1).

An apparent difference between our method and the conventional approach is that the matrix M plays different roles in both methods. In the conventional method, M is directly used as a generator matrix to generate a self-dual code. Therefore, one must search for matrices M that satisfy the condition $M{M}^{\top }=0$ concerning a chosen matrix construction. However, as illustrated above, our method imposes distinct requirements on the conditions that M must adhere to, which leads to the conclusion that M is unable to generate any self-dual code constructed using our method. More specially, let M be a $k\times 2k$ matrix over R satisfying the conditions mentioned in the previous paragraph, and let $\mathcal{D}$ be a self-dual code constructed from the matrix M and ${\mathcal{D}}_0$ using our method, where R remains a finite commutative ring with characteristic 2 and ${\mathcal{D}}_0$ is a self-dual code in $R^2$ with generator matrix $M_0.$ Due to the involvement of ${\mathcal{D}}_0$, the generator matrix for $\mathcal{D}$ will be a matrix derived from M and $M_0$ rather than M itself. This means that the matrix M no longer acts directly as a generator matrix for any self-dual code created using our method. Therefore, our method diverges from the conventional approach, providing a new perspective on constructing self-dual codes over finite commutative rings of characteristic 2. In this sense, our method can be seen as a complement to the conventional approach rather than simply a continuation (see Remark 8 or the last three examples in Section 5).

In addition to the distinction mentioned above, another significant difference between these two construction methods is that our method can construct nonfree self-dual codes over finite commutative rings of characteristic 2, in contrast to the conventional approach. To further clarify this point, let us revisit the aforementioned self-dual code $\mathcal{D}$. We can demonstrate that $\mathcal{D}$ will be nonfree if ${\mathcal{D}}_0$ is nonfree. This implies that our method can construct nonfree self-dual codes over finite commutative rings of characteristic 2, as long as there exists a nonfree self-dual code of length 2 over the corresponding rings. However, the conventional approach frequently fails to achieve this objective. A possible explanation for this is the difficulty of determining the code size directly from its generator matrix, as highlighted in [38]. For this difficulty, it is customary to demand that the rows of the generator matrix are linearly independent when constructing self-dual codes over a finite commutative ring (like a non-chain ring) using the conventional approach. This simplifies the task of determining the size of the resulting self-orthogonal code and confirming whether it is indeed a self-dual code. As a side effect, the self-dual codes produced by the conventional method are typically free. Therefore, our method has the potential to construct self-dual codes that would be missed by the conventional approach. Additionally, once a matrix M that satisfies the aforementioned corresponding conditions has been identified, it can be utilized to construct multiple self-dual codes through our method. This is because, in general, there is more than one self-dual code of length 2 over a finite commutative ring that is not a field.

For our method to be considered a viable option for producing self-dual codes, we must address the question of determining whether the matrix $M{M}^{\top }$ has the form of ${\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }$ is a linearly independent vector. To address this question, we present a necessary and sufficient condition for $M{M}^{\top }$ to have this form (see Theorem 2). We also demonstrate that some known matrix constructions, like the four-circulant construction and its certain variations, are not suitable for our method to produce self-dual codes over finite commutative rings of characteristic 2 (see Theorems 3 and 4).

The rest of the paper is organized as follows: Section 2 provides preliminary definitions and results on codes and group rings, as well as discusses certain properties concerning matrices over finite commutative rings. In Section 3, we present several key results on free linear codes over finite commutative rings, which are necessary for proving our main theorem. Section 4 contains the proof of our main theorem (Theorem 1), which establishes an isometry. Based on this theorem, we introduce a new method for constructing self-dual codes over finite commutative rings with characteristic 2. Additionally, this section includes two results on constructing self-dual codes of length 2 over finite commutative rings of characteristic 2 (Propositions 8 and 9). In Section 5, we provide a necessary and sufficient condition for a symmetric matrix to have the form ${\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }$ is a linearly independent vector (Theorem 2). This result is crucial for our construction method. We also prove some results in this section that can be used to exclude certain matrix constructions as candidates for producing self-dual codes using our construction method (Theorems 3 and 4). Several examples are provided to propose suitable matrix constructions for our method. Finally, Section 6 ends the paper with concluding remarks and suggestions for possible future directions.

2. Preliminaries

2.1. Codes over Finite Commutative Rings

We begin by recalling the standard definitions from coding theory. In this paper, all rings are assumed to be finite commutative rings with an identity element $1\ne 0$.

Let R be a finite commutative ring. A code $\mathcal{C}$ over R of length n is a non-empty subset of $R^n$, whose elements are called codewords. If $\mathcal{C}$ is an R-submodule of $R^n$, then we say that $\mathcal{C}$ is linear. A linear code $\mathcal{C}$ is said to be free if it is a free submodule over R, with its dimension denoted as ${dim}_R\mathcal{C}$. Let $\mathit{x}\in R^n$. The hamming weight of $\mathit{x}$, denoted by $w{t}_H\left(\mathit{x}\right)$, is defined to be the number of non-zero coordinates in $\mathit{x}$. The Euclidean inner product on $R^n$ is defined by

$$[\mathit{x},\mathit{y}]=\sum _{i=1}^n{x}_i{y}_i,$$

where $\mathit{x},\mathit{y}\in R^n$ such that $\mathit{x}=(x_1,x_2,\dots ,x_n)$ and $\mathit{y}=(y_1,y_2,\dots ,y_n)$. The dual code of $\mathcal{C}$ with respect to the Euclidean inner product is denoted by ${\mathcal{C}}^{\perp }$ and defined as

$${\mathcal{C}}^{\perp }=\{\mathit{x}\in R^n\mid [\mathit{x},\mathit{y}]=0,\forall \mathit{y}\in \mathcal{C}\},$$

and we say that $\mathcal{C}$ is self-orthogonal if $\mathcal{C}\subseteq {\mathcal{C}}^{\perp }$ and self-dual if $\mathcal{C}={\mathcal{C}}^{\perp }$. Additionally, the hull of a code $\mathcal{C}$ is defined as $Hull\left(\mathcal{C}\right)=\mathcal{C}\cap {\mathcal{C}}^{\perp }.$ By definition, $Hull\left(\mathcal{C}\right)$ is self-orthogonal.

Given a $k\times n$ matrix M over R, we define $\mathcal{C}\left(M\right)$ to be the R-submodule $R\langle M_1,M_2,\dots ,M_k\rangle$ of $R^n$ generated by $M_1,M_2,\dots ,M_k$, where $M_i$$(1\le i\le k)$ denotes the i-th row of M. For a linear code $\mathcal{C}$ over R, if $\mathcal{C}=\mathcal{C}\left(M\right)$, we call M a generator matrix for $\mathcal{C}$.

Two codes $\mathcal{C}$ and ${\mathcal{C}}^{\prime }$ in $R^n$ are said to be equivalent provided there exists a monomial matrix $\mathcal{M}$ such that ${\mathcal{C}}^{\prime }=\mathcal{C}\mathcal{M}:=\left\{\mathit{x}\mathcal{M}\right|\mathit{x}\in \mathcal{C}\}.$ In particular, if $\mathcal{M}$ is a permutation matrix, then we say that $\mathcal{C}$ and ${\mathcal{C}}^{\prime }$ are permutative equivalent codes.

2.2. Matrices over Finite Commutative Rings

Let us denote the set of $k\times n$ matrices in R by $Ma{t}_{k\times n}\left(R\right)$, where R is a finite commutative ring. When $k=n$, we write it as $Ma{t}_k\left(R\right)$ for simplicity. Given $M\in Ma{t}_{k\times n}\left(R\right)$, we use $M_i$ (resp. $M_{\left(j\right)}$) to denote the i-th row (resp. j-th column) of M. Additionally, the transpose of matrix M is denoted as $M^{\top }.$

We shall now discuss the properties of matrices over finite commutative rings, some of which are derived from the Chinese Remainder Theorem.

Let ${\mathfrak{m}}_l(1\le l\le s)$ be the distinct maximal ideals of R. The Jacobson radical $J\left(R\right)$ of R is defined as the intersection of all the maximal ideals of R, i.e., $J\left(R\right)={\cap }_{l=1}^s{\mathfrak{m}}_l$, which in our case is a nilpotent ideal. Let e be a positive integer such that $J{\left(R\right)}^e=\left\{0\right\}$. By the Chinese Reminder Theorem, we can obtain the canonical ring-isomorphism

$$\Psi :R\stackrel{\approx }{⟶}{R}_1\times R_2\times \cdots \times R_s,$$

$$r\mapsto (r^{\left(1\right)},r^{\left(2\right)},\dots ,r^{\left(s\right)}),$$

where $R_l(l\le l\le s)$ is the local ring $R/{\mathfrak{m}}_l^e$ and $r^{\left(l\right)}:=r+{\mathfrak{m}}_l^e$ is the canonical image of r in $R/{\mathfrak{m}}_l^e$. We can further extend $\Psi$ to the following R-algebra isomorphism

$$\overline{\Psi }:R^n\stackrel{\approx }{⟶}{R}_1^n\times R_2^n\times \cdots \times R_s^n,$$

which is given by

$$(r_1,r_2,\dots ,r_n)\mapsto \left((r_1^{\left(1\right)},r_2^{\left(1\right)},\dots ,r_n^{\left(1\right)}),(r_1^{\left(2\right)},r_2^{\left(2\right)},\dots ,r_n^{\left(2\right)}),\dots ,(r_1^{\left(s\right)},r_2^{\left(s\right)},\dots ,r_n^{\left(s\right)})\right),$$

where the R-algebra structure on $R_1^n\times R_2^n\times \cdots \times R_s^n$ is defined by the ring-homomorphism

$$R\to R_1^n\times R_2^n\times \cdots \times R_s^n,$$

$$r\mapsto \left((r^{\left(1\right)},r^{\left(1\right)},\dots ,r^{\left(1\right)}),(r^{\left(2\right)},r^{\left(2\right)},\dots ,r^{\left(2\right)}),\dots ,(r^{\left(s\right)},r^{\left(s\right)},\dots ,r^{\left(s\right)})\right).$$

For a matrix $M=\left(r_{ij}\right)\in Ma{t}_{k\times n}\left(R\right)$, we set $M^{\left(l\right)}=\left(r_{ij}^{\left(l\right)}\right)(1\le l\le s)$, which is obviously a $k\times n$ matrix over $R_l$. Let A and B be two matrices over R, it is clear that ${(A+B)}^{\left(l\right)}=A^{\left(l\right)}+B^{\left(l\right)}$ and ${\left(AB\right)}^{\left(l\right)}=A^{\left(l\right)}{B}^{\left(l\right)}$, provided that these operations are defined for the given matrices. In addition, if $(R,\mathfrak{m})$ is a local ring, then we denote by $\overline{r}$ the residue class of $r\in R$ modulo the unique maximal ideal $\mathfrak{m}$, and define $\overline{M}=\left(\overline{r_{ij}}\right)$ being the $k\times n$ matrix obtained by taking modulo $\mathfrak{m}$.

Let S be a subset of $R^n$ that is not empty. By an R-linear combination of elements of S, one means a sum

$$\sum _{\mathit{v}\in S}{r}_{\mathit{v}}\mathit{v}$$

where $\left\{r_{\mathit{v}}\right\}$ is a set of elements of R, almost all of which are equal to 0. We take the standard definition of linear independence. Namely, S is said to be linearly independent over R whenever we have an R-linear combination

$$\sum _{\mathit{v}\in S}{r}_{\mathit{v}}\mathit{v}$$

which is equal to 0, then $r_{\mathit{v}}=0$ for all $\mathit{v}\in S.$ If S is linearly independent over R, we shall also say that its elements are linearly independent over R. In particular, we refer to $\mathit{v}$ as an R-linearly independent vector if the set $\left\{\mathit{v}\right\}$, which contains only the vector $\mathit{v}\in R^n$, is linearly independent over $R.$ To simplify matters, we sometimes omit the prefix $R-$.

Using the notation given above, and keeping in mind that $\overline{\Psi }$ is an R-algebra isomorphism, the following result can be easily derived.

Lemma 1.

Let R be a finite commutative ring with s distinct maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_s.$ Let M be a $k\times n$ matrix over R. Then, the rows of M are linearly independent over R if and only if the rows of $M^{\left(l\right)}$ are linearly independent over $R_l$ for each $l=1,2,\dots ,s.$

We can utilize the following lemma to determine if the rows of $M^{\left(l\right)}$ are linearly independent over the finite commutative local ring $R_l$.

Lemma 2.

Let $(R,\mathfrak{m})$ be a finite commutative local ring, and M a $k\times n$ matrix over R. Then, the following are equivalent:

(1)

the rows of M are linearly independent over R;

(2)

the rows of $\overline{M}$ are linearly independent over $R/\mathfrak{m}$;

(3)

there exist A and $\mathcal{P}$, where A is an invertible $k\times k$ matrix and $\mathcal{P}$ is an $n\times n$ permutation matrix over R, such that $AM\mathcal{P}=\left(I_k\right|H)$ for some matrix $H\in Ma{t}_{k\times (n-k)}\left(R\right)$, where $I_k$ denotes the $k\times k$ identity matrix over R.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ follows from the fact that $Ann\left(\mathfrak{m}\right):=\{r\in R\mid r\pi =0,\forall \pi \in \mathfrak{m}\}$ is a non-zero ideal.

$\left(2\right)\Rightarrow \left(3\right).$ Since the rows of $\overline{M}$ are linearly independent over $R/\mathfrak{m}$, we can select k columns from $\overline{M}$, say ${\overline{M}}_{\left(j_1\right)},\dots ,{\overline{M}}_{\left(j_k\right)}$, that are also linearly independent over $R/\mathfrak{m}$. Let $B=(M_{\left(j_1\right)},\dots ,M_{\left(j_k\right)})$, which is obviously a $k\times k$ matrix over R. Since $\overline{B}=({\overline{M}}_{\left(j_1\right)},\dots ,{\overline{M}}_{\left(j_k\right)})$ is invertible, we can find a matrix $A^{\prime }\in Ma{t}_k\left(R\right)$ such that $\overline{A^{\prime }}\ast \overline{B}=\overline{I_k}.$ From this, we can conclude that all the diagonal elements of $A^{\prime }B$ belong to $R^{*}$ (here $R^{*}$ denotes the group of units of R). Therefore, we can transform $A^{\prime }B$ into the identity matrix $I_k\in Ma{t}_k\left(R\right)$ by performing a finite number of elementary row operations. This process also yields a suitable invertible $k\times k$ matrix $A^{″}$ over R. Let $A=A^{″}{A}^{\prime }$, and let $\mathcal{P}$ be an $n\times n$ permutation matrix obtained from identity matrix $I_n$ by rearranging the column vectors as follows:

The $j_1-th$, $j_2-th$, …, $j_k-th$ columns are moved to the $1st$, $2nd$, …, $k-th$ columns, respectively. Thus, $AM\mathcal{P}$ is a matrix with the desired form $\left(I_k\right|H)$, where H is a $k\times (n-k)$ matrix over R.

$\left(3\right)\Rightarrow \left(1\right)$ is clear, since A is an invertible matrix. □

From Lemmas 1 and 2, the following is an immediate consequence.

Proposition 1.

Let R be a finite commutative ring, and M a $k\times n$ matrix over R. If the rows of M are linearly independent over R, then there exists an invertible $k\times k$ matrix A and an orthogonal $n\times n$ matrix Q(i.e., $Q{Q}^{\top }=I_n$) over R, such that $AMQ$ is of type $\left(I_k\right|H)$, where H is a $k\times (n-k)$ matrix over R.

Remark 1.

Note, that the matrix Q in Proposition 1 may not necessarily be a permutation matrix, even though all the matrix $Q^{\left(1\right)},Q^{\left(2\right)},\dots ,Q^{\left(s\right)}$ are permutation matrices. We would like to emphasize that the matrices A and Q, which are necessary for our construction method, can be determined by using the proof of Lemma 2 in combination with the algebra-isomorphism $\overline{\Psi }$ given above.

We continue by recalling the definition of $\lambda$-circulant matrices. Let $\mathit{a}=(a_0,a_1,\dots ,a_{n-1})\in R^n$ and let

$$A=\left(\begin{array}{ccccc}{a}_0& a_1& a_2& \cdots & a_{n-1}\\ \lambda a_{n-1}& a_0& a_1& \cdots & a_{n-2}\\ \lambda a_{n-2}& \lambda a_{n-1}& a_0& \cdots & a_{n-3}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ \lambda a_1& \lambda a_2& \lambda a_3& \cdots & a_0\end{array}\right)\in Ma{t}_n\left(R\right),$$

where R is a commutative ring and $\lambda \in R$. Then, A is called the λ-circulant matrix generated by $\mathit{a}$, denoted $A=cir{c}_{\lambda }\left(\mathit{a}\right)$. If we let

$$P_{\lambda }=\left(\begin{array}{cc}\mathbf{0}& I_{n-1}\\ \lambda & \mathbf{0}\end{array}\right)\in Ma{t}_n\left(R\right),$$

then it follows that $A={\sum }_{i=0}^{n-1}{a}_i{P_{\lambda }}^i.$ Since ${P_{\lambda }}^n=\lambda I_n$, the set of all $n\times n$$\lambda$-circulant matrices over R forms a commutative subring of the matrix ring $Ma{t}_n\left(R\right).$ Moreover, if $\lambda \in R^{*}$, where $R^{*}$ denotes the group of units of R, then $A^{\top }$ is a ${\lambda }^{-1}$-circulant matrix. Additionally, if $\lambda =1$, then A is called the circulant matrix generated by $\mathit{a}$ and is more simply denoted by $A=circ\left(\mathit{a}\right)$.

2.3. Group Rings

We will proceed to define group rings and review the common methods for constructing matrices using elements from group rings. This approach was first introduced by Hurley in [20] for codes over finite fields and was later extended to finite commutative rings in [15].

Let R be a commutative ring with identity, and let G be a multiplicative group. Denote by $R\langle G\rangle$ the free R-module with G as its basis. Let r, $r^{\prime }$$\in R$ and g, $g^{\prime }$$\in G.$ We define

$$\left(rg\right)\left(r^{\prime }{g}^{\prime }\right)=\left(r{r}^{\prime }\right)\left(g{g}^{\prime }\right).$$

By means of the above equation, we can define a multiplication operation on $R\langle G\rangle$ according to the distributive law of multiplication over addition. This multiplication operation turns $R\langle G\rangle$ into a ring with identity, denoted by $R\left[G\right]$, which is known as the group ring of G over $R.$

Note, that the unit element of $R\left[G\right]$ is simply $e_G$, where $e_G$ represents the unit element of $G.$ Additionally, R can be viewed as a subring of $R\left[G\right]$ through the natural embedding

$$f_0:R\to R\left[G\right](r\mapsto r{e}_G),$$

which further makes $R\left[G\right]$ an algebra over $R.$

Now, we will restrict G to a finite multiplicative group with order $k.$ Let $G=\{g_1=e_G,g_2,\dots ,g_k\}$ be a predetermined ordering of the elements in the group $G.$ There exists a well-known injective ring homomorphism, denoted by $\sigma$, from $R\left[G\right]$ into $Ma{t}_k\left(R\right)$ (relative to this ordering), which maps an element $v=r_{g_1}{g}_1+r_{g_2}{g}_2+\cdots +r_{g_k}{g}_k\in R\left[G\right]$ to a matrix $\sigma \left(v\right)\in Ma{t}_k\left(R\right)$, defined as

$$\sigma \left(v\right)=\left(\begin{array}{ccccc}{r}_{g_1^{-1}{g}_1}& r_{g_1^{-1}{g}_2}& r_{g_1^{-1}{g}_3}& \cdots & r_{g_1^{-1}{g}_k}\\ r_{g_2^{-1}{g}_1}& r_{g_2^{-1}{g}_2}& r_{g_2^{-1}{g}_3}& \cdots & r_{g_2^{-1}{g}_k}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ r_{g_k^{-1}{g}_1}& r_{g_k^{-1}{g}_2}& r_{g_k^{-1}{g}_3}& \cdots & r_{g_k^{-1}{g}_k}\end{array}\right).$$

Additionally, $\sigma$ is also an R-algebra homomorphism and can be induced from the so-called regular representation of $G.$

For a vector $\mathit{v}$ over R, we denote by $\left|\mathit{v}\right|$ the sum of all coordinates of $\mathit{v}$.

Since G is a group, it follows that

$${|\sigma \left(v\right)}_1{|=\cdots =|\sigma \left(v\right)}_k{|=|\sigma \left(v\right)}_{\left(1\right)}{|=\cdots =|\sigma \left(v\right)}_{\left(k\right)}|=\sum _{i=1}^k{r}_{g_i}.$$

For an element $v=r_{g_1}{g}_1+r_{g_2}{g}_2+\cdots +r_{g_k}{g}_k\in R\left[G\right]$, define the element $v^{*}\in R\left[G\right]$ as

$$v^{\ast }=\sum _{i=1}^k{r}_{g_i}{g}_i^{-1}=\sum _{i=1}^k{r}_{g_i^{-1}}{g}_i.$$

This is commonly referred to as the canonical involution for this group ring. This involution, in effect, is an anti-automorphism of $R\left[G\right]$ of order 2 and satisfies $\sigma \left(v^{\ast }\right)=\sigma {\left(v\right)}^{\top }$ for each $v\in R\left[G\right].$

3. Free Linear Codes over Finite Commutative Rings

In this section, we will present several key results on free linear codes over finite commutative rings. These results will serve as the basis for our main theorem (Theorem 1). Let the notation be as in the previous section. By utilizing Lemma 1 and Proposition 1, we can obtain the following result.

Proposition 2.

Let R be a finite commutative ring with s distinct maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_s.$ Let M be a $k\times n$ matrix over R. Denote the linear code generated by M (resp. $M^{\left(l\right)}$) over R (resp. $R_l$) as $\mathcal{C}$ (resp. ${\mathcal{C}}^{\left(l\right)}$), where $1\le l\le s.$ Then, the following hold:

(1)

$\overline{\Psi }\left(\mathcal{C}\right)={\mathcal{C}}^{\left(1\right)}\times \cdots \times {\mathcal{C}}^{\left(s\right)}.$

(2)

$\overline{\Psi }\left({\mathcal{C}}^{\perp }\right)={{\mathcal{C}}^{\left(1\right)}}^{\perp }\times \cdots \times {{\mathcal{C}}^{\left(s\right)}}^{\perp }.$

(3)

$\overline{\Psi }\left(Hull\left(\mathcal{C}\right)\right)=Hull\left({\mathcal{C}}^{\left(1\right)}\right)\times \cdots \times Hull\left({\mathcal{C}}^{\left(s\right)}\right).$

(4)

$\mathcal{C}$ is a free linear code with ${dim}_R\mathcal{C}=t$ if and only if ${\mathcal{C}}^{\left(l\right)}$ is free over $R_l$ with ${dim}_{R_l}{\mathcal{C}}^{\left(l\right)}=t$ for each $l=1,2,\dots ,s.$

(5)

If the rows of M are linearly independent over R, then ${\mathcal{C}}^{\perp }$ is free, $\left|\mathcal{C}\right||{\mathcal{C}}^{\perp }{|=|R|}^n$, and ${\left({\mathcal{C}}^{\perp }\right)}^{\perp }=\mathcal{C}.$ Here, $\left|\mathcal{C}\right|$ represents the cardinality of the set $\mathcal{C}.$

Proof. 

$\left(1\right)$, $\left(2\right)$, and $\left(3\right)$ follow directly from the R-algebra isomorphism $\overline{\Psi }$ and the definition of hull. $\left(4\right)$ is an immediate consequence of Lemma 1. Alternatively, please refer to Section 2.3 in [39].

To prove $\left(5\right)$, we will use Proposition 1 which states that there exists an invertible $k\times k$ matrix A and an orthogonal $n\times n$ matrix Q over R such that $AMQ=\left(I_k\right|H)$, where $H\in Ma{t}_{k\times (n-k)}\left(R\right)$. This means that $\mathcal{C}=\mathcal{C}\left(\left(I_k\right|H)Q^{\top }\right)$. Therefore,

$${\mathcal{C}}^{\perp }=\mathcal{C}\left((-H^{\top }\mid I_{n-k})\right)Q^{\top }.$$

This shows that ${\mathcal{C}}^{\perp }$ is free with ${dim}_R{\mathcal{C}}^{\perp }=n-{dim}_R\mathcal{C}$, and thus $\left|\mathcal{C}\right||{\mathcal{C}}^{\perp }{|=|R|}^n$. Furthermore, since ${\mathcal{C}}^{\perp }$ is free, we can similarly deduce that ${\left({\mathcal{C}}^{\perp }\right)}^{\perp }$ is also free with ${dim}_R{\left({\mathcal{C}}^{\perp }\right)}^{\perp }=k$. Therefore, considering that $\mathcal{C}$ is a subset of ${\left({\mathcal{C}}^{\perp }\right)}^{\perp }$, we can conclude that $\mathcal{C}={\left({\mathcal{C}}^{\perp }\right)}^{\perp }.$ □

Remark 2.

If R is a finite commutative Frobenius ring, it is well-known that the last two conclusions of $\left(5\right)$ in Proposition 2 remain valid even without any assumptions on the linear code $\mathcal{C}$. However, R is not required to be a Frobenius ring here. To ensure the validity of these two results, we instead require $\mathcal{C}$ to be a free linear code. (see [39] for a full description of Frobenius rings and codes over Frobenius rings).

Proposition 2 has an important implication: it enables us to reduce much of the theory of codes by considering the case where the ring is local.

We will now recall a result on projective modules over local rings from [40]. A module is projective if it is (isomorphic to) a direct summand of a free module.

Proposition 3

([40]). Let $(R,\mathfrak{m})$ be a commutative local ring and E a finitely generated projective R-module. Then, E is free. In fact, if $x_1,\dots ,x_n$ are elements of E whose residue classes $\overline{x_1},\dots ,\overline{x_n}$ are a basis of $E/\mathfrak{m}E$ over $R/\mathfrak{m}$, then $x_1,\dots ,x_n$ are a basis of E over R. If $x_1,\dots ,x_r$ are such that $\overline{x_1},\dots ,\overline{x_r}$ are linearly independent over $R/\mathfrak{m}$, then they can be completed to a basis of E over R.

Corollary 1.

Let $(R,\mathfrak{m})$ be a finite commutative local ring with $\mathfrak{m}\ne 0$. Let $\mathcal{C}\subseteq {\mathcal{C}}^{\prime }$ be two free linear codes in $R^n$. Then, ${\mathcal{C}}^{\prime }/\mathcal{C}$ is a free R-module.

Proof. 

Let $M_1,\dots ,M_k$ be a basis of $\mathcal{C}$ over R, where $k={dim}_R\mathcal{C}$. We will show that this basis can be completed to a basis for ${\mathcal{C}}^{\prime }$ over R by applying Proposition 3. For this, it will suffice to prove that $\overline{M_1},\dots ,\overline{M_k}\in {\mathcal{C}}^{\prime }/\mathfrak{m}{\mathcal{C}}^{\prime }$ are linearly independent over $R/\mathfrak{m}$, where $\overline{M_i}=M_i+\mathfrak{m}{\mathcal{C}}^{\prime }$ for each $i=1,2,\dots ,k.$ Now, let us assume that $r_1,\dots ,r_k\in R$ satisfy the equation $\overline{r_1}\ast \overline{M_1}+\cdots +\overline{r_k}\ast \overline{M_k}=0.$ This implies that

$$r_1{M}_1+\cdots +r_k{M}_k\in \mathfrak{m}{\mathcal{C}}^{\prime }.$$

Take $\pi \in R$ such that $\pi \mathfrak{m}=0$ and $\pi \ne 0.$ (The existence of such an element $\pi$ follows from the fact that $\mathfrak{m}$ is a non-zero nilpotent ideal.) Thus, we can conclude that

$$\pi (r_1{M}_1+\cdots +r_k{M}_k)=0.$$

But $M_1,\dots ,M_k$ are linearly independent over R. Hence,

$$\pi r_1=\cdots =\pi r_k=0,$$

which implies that

$$\overline{r_1}=\cdots =\overline{r_k}=\overline{0}.$$

This shows that $\overline{M_1},\dots ,\overline{M_k}\in {\mathcal{C}}^{\prime }/\mathfrak{m}{\mathcal{C}}^{\prime }$ are linearly independent over $R/\mathfrak{m}$. By Proposition 3, we can extend $M_1,\dots ,M_k$ to form a basis $M_1,\dots ,M_k,\dots ,M_{k^{\prime }}$ of ${\mathcal{C}}^{\prime }$ over R, where $k^{\prime }={dim}_R{\mathcal{C}}^{\prime }$. Setting ${\mathcal{C}}^{″}=R\langle M_{k+1},\dots ,M_{k^{\prime }}\rangle$, which is clearly free over R, we obtain

$${\mathcal{C}}^{\prime }=\mathcal{C}\oplus {\mathcal{C}}^{″},$$

so

$${\mathcal{C}}^{\prime }/\mathcal{C}(\cong {\mathcal{C}}^{″})$$

is a free R-module. □

Using Proposition 2 and Corollary 1, we can deduce the following proposition.

Proposition 4.

Let R be a finite commutative ring, and $\mathcal{C}$ be a free linear code of length n over R. If $\mathcal{C}/Hull\left(\mathcal{C}\right)$ is also free over R, then so are $Hull\left(\mathcal{C}\right)$ and ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right).$ Furthermore, we have

$${dim}_R\mathcal{C}+{dim}_R{\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)={dim}_R{\mathcal{C}}^{\perp }+{dim}_R\mathcal{C}/Hull\left(\mathcal{C}\right),$$

and

$$Hull{\left(\mathcal{C}\right)}^{\perp }=\mathcal{C}+{\mathcal{C}}^{\perp }.$$

Proof. 

We first establish the following claim:

Suppose that $\mathcal{C}$ and ${\mathcal{C}}^{\prime }$ are free linear codes in $R^n$, and $\mathcal{C}\subseteq {\mathcal{C}}^{\prime }$. Then, ${\mathcal{C}}^{\prime }/\mathcal{C}$ is also free as an R-module.

To prove this claim, let $M_1,\dots ,M_k$ be a basis for $\mathcal{C}$ over R, where k is the dimension of $\mathcal{C}$. We can then construct a matrix M with rows consisting of these basis elements. Let $M^{\prime }$ be a generator matrix for ${\mathcal{C}}^{\prime }$ and $k^{\prime }$ the dimension of ${\mathcal{C}}^{\prime }$. By Proposition 2, we have

$$\overline{\Psi }\left(\mathcal{C}\left(M\right)\right)=\mathcal{C}\left(M^{\left(1\right)}\right)\times \cdots \times \mathcal{C}\left(M^{\left(s\right)}\right),$$

and

$$\overline{\Psi }\left(\mathcal{C}\left(M^{\prime }\right)\right)=\mathcal{C}\left({M^{\prime }}^{\left(1\right)}\right)\times \cdots \times \mathcal{C}\left({M^{\prime }}^{\left(s\right)}\right).$$

where s is the number of maximal ideals in $R.$ Since $\mathcal{C}\subseteq {\mathcal{C}}^{\prime }$, it is clear that $\mathcal{C}\left(M^{\left(l\right)}\right)\subseteq \mathcal{C}\left({M^{\prime }}^{\left(l\right)}\right)$ for all l. Furthermore, since the rows of M are linearly independent over R, by Lemma 1, we can conclude that each $\mathcal{C}\left(M^{\left(l\right)}\right)$ is a free linear code over the corresponding local ring $R_l$ with dimension k. Thus, from the proof of the Corollary 1, for each $l=1,2,\dots ,s$, we can add $k^{\prime }-k$ rows below the last row of the matrix $M^{\left(l\right)}$ to obtain a new matrix $M^{″\left(l\right)}$, whose rows exactly form a basis for $\mathcal{C}\left(M^{\prime \left(l\right)}\right)$. Construct a matrix $M^{″}\in Ma{t}_{k^{\prime }\times n}\left(R\right)$ from these matrices $M^{″\left(l\right)}$ using the ring-isomorphism $\Psi$. We can see that the rows of $M^{″}$ are linearly independent over R and form a basis for ${\mathcal{C}}^{\prime }$. Note, that the first k rows of $M^{″}$ are the same as the rows of M. This means we can extend a basis of $\mathcal{C}$ to a basis for ${\mathcal{C}}^{\prime }$, thereby proving that the quotient module ${\mathcal{C}}^{\prime }/\mathcal{C}$ is also free over R.

To prove this proposition, we consider the exact sequence

$$0⟶Hull\left(\mathcal{C}\right)\stackrel{\subseteq }{⟶}\mathcal{C}\stackrel{\phi }{⟶}\mathcal{C}/Hull\left(\mathcal{C}\right)⟶0,$$

where $\phi$ is the canonical map. Since $\mathcal{C}/Hull\left(\mathcal{C}\right)$ is assumed to be free over R, the exact sequence splits, namely, there is an R-module homomorphism

$$\psi :\mathcal{C}/Hull\left(\mathcal{C}\right)\to \mathcal{C}$$

such that $\phi \circ \psi =i{d}_{\mathcal{C}/Hull\left(\mathcal{C}\right)}.$ This, combined with our previous claim and Proposition 2, confirms that both $Hull\left(\mathcal{C}\right)$ and ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)$ are also free over R.

The equation

$${dim}_R\mathcal{C}+{dim}_R{\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)={dim}_R{\mathcal{C}}^{\perp }+{dim}_R\mathcal{C}/Hull\left(\mathcal{C}\right)$$

is derived directly from the identity equation $\left|\mathcal{C}\right||{\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)|=|{\mathcal{C}}^{\perp }\left|\right|\mathcal{C}/Hull\left(\mathcal{C}\right)|.$

As for the last conclusion, by Proposition 2, we know that ${\left({\mathcal{C}}^{\perp }\right)}^{\perp }=\mathcal{C}$, and hence we have

$${(\mathcal{C}+{\mathcal{C}}^{\perp })}^{\perp }={\mathcal{C}}^{\perp }\cap \mathcal{C}=Hull\left(\mathcal{C}\right),$$

which implies that

$$\mathcal{C}+{\mathcal{C}}^{\perp }\subseteq Hull{\left(\mathcal{C}\right)}^{\perp }.$$

Furthermore, considering the exact sequence

$$0⟶Hull\left(\mathcal{C}\right)⟶\mathcal{C}\oplus {\mathcal{C}}^{\perp }⟶\mathcal{C}+{\mathcal{C}}^{\perp }⟶0,$$

where the map

$$Hull\left(\mathcal{C}\right)\to \mathcal{C}\oplus {\mathcal{C}}^{\perp }\mathrm{and}\mathcal{C}\oplus {\mathcal{C}}^{\perp }\to \mathcal{C}+{\mathcal{C}}^{\perp }$$

are defined as

$$\mathit{z}\mapsto (\mathit{z},-\mathit{z})\mathrm{and}(\mathit{x},\mathit{y})\mapsto \mathit{x}+\mathit{y},$$

respectively, we can see that

$${\left|R\right|}^n=|\mathcal{C}\oplus {\mathcal{C}}^{\perp }|=|Hull\left(\mathcal{C}\right)\left|\right|\mathcal{C}+{\mathcal{C}}^{\perp }|.$$

We can also see that ${\left|R\right|}^n={\left|Hull\left(\mathcal{C}\right)\right||Hull\left(\mathcal{C}\right)}^{\perp }|$, as $Hull\left(\mathcal{C}\right)$ is free over $R.$ Therefore, we can conclude that

$$Hull{\left(\mathcal{C}\right)}^{\perp }=\mathcal{C}+{\mathcal{C}}^{\perp }.$$

□

4. The Construction Method

We will hereafter always assume that R is a finite commutative ring of characteristic 2, and it is not necessarily a Frobenius ring.

In this section, armed with the preparation from the preceding section, we will introduce a new method for constructing self-dual codes over R by utilizing self-dual codes of length 2. This method is based on our main theorem, which is stated as follows:

Theorem 1.

Let $\mathcal{C}$ be a free linear code of length n over R with dimension k, where $n=2k$. Assume that $\mathcal{C}/Hull\left(\mathcal{C}\right)$ is a free R-module with dimension 1. Then:

(1)

Both $Hull\left(\mathcal{C}\right)$ and ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)$ are free over R with dimensions $k-1$ and 1, respectively.

(2)

Let $\mathit{x}\in \mathcal{C}$ and $\mathit{y}\in {\mathcal{C}}^{\perp }$. Then, $\mathit{x}+Hull\left(\mathcal{C}\right)$ (resp. $\mathit{y}+Hull\left(\mathcal{C}\right)$) is a basis of $\mathcal{C}/Hull\left(\mathcal{C}\right)$ (resp. ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)$) over R if and only if $[\mathit{x},\mathit{x}]$ (resp. $[\mathit{y},\mathit{y}]$) lies in $R^{*}$. Furthermore, if $\mathit{x}$ and $\mathit{y}$ satisfy these equivalent conditions, then they are linearly independent over R and $Hull{\left(\mathcal{C}\right)}^{\perp }=Hull\left(\mathcal{C}\right)\oplus R\mathit{x}\oplus R\mathit{y}.$

(3)

Let $\mathit{x}\in \mathcal{C}$ and $\mathit{y}\in {\mathcal{C}}^{\perp }$ be two elements satisfying the equivalent conditions in (2), respectively. Then, the map

$$\eta :R\mathit{x}+R\mathit{y}\to R^2$$

defined by

$$\eta (r_1\mathit{x}+r_2\mathit{y})=(r_1\left|\mathit{x}\right|,r_2\left|\mathit{y}\right|)$$

is an isometry, meaning that η is an R-module isomorphism with $\left[\eta \right(\mathit{v}),\eta (\mathit{w}\left)\right]=[\mathit{v},\mathit{w}]$ for all $\mathit{v},\mathit{w}\in R\mathit{x}+R\mathit{y}.$ Define $\mathfrak{D}\left(\mathcal{C}\right)$ to be the set of all self-dual codes of length n over R that contain $Hull\left(\mathcal{C}\right)$ as a subcode. Then,

$$\mathfrak{D}\left(\mathcal{C}\right)=\{Hull\left(\mathcal{C}\right)+{\eta }^{-1}\left({\mathcal{D}}_0\right)|{\mathcal{D}}_0\mathrm{is}\mathrm{a}\mathrm{self-dual}\mathrm{code}\mathrm{in}{R}^2\}.$$

Additionally, the cardinality of $\mathfrak{D}\left(\mathcal{C}\right)$, denoted as $\left|\mathfrak{D}\right(\mathcal{C}\left)\right|$, is equal to the number of self-dual codes in $R^2.$

Proof. 

$\left(1\right)$ follows directly from Propositions 2 and 4 in the previous section.

$\left(2\right)$ We will only provide the proof for $\mathit{x}$, as the proof for $\mathit{y}$ is similar. This is because, by (5) in Proposition 2, we observe that ${\mathcal{C}}^{\perp }$ is also a free linear code with the same dimension as $\mathcal{C}$ and $Hull\left(\mathcal{C}\right)=Hull\left({\mathcal{C}}^{\perp }\right)$. We now let $r\in R$ and $\mathit{x}\in \mathcal{C}$. Then, we have the obvious implications:

$$r(\mathit{x}+Hull(\mathcal{C}\left)\right)=Hull\left(\mathcal{C}\right)\iff [r\mathit{x},\mathcal{C}]=\left\{0\right\}\Rightarrow [r\mathit{x},R\mathit{x}+Hull(\mathcal{C}\left)\right]=0\iff r[\mathit{x},\mathit{x}]=0,$$

where $[r\mathit{x},\mathcal{C}]:=\left\{[r\mathit{x},{\mathit{x}}^{\prime }]\right|{\mathit{x}}^{\prime }\in \mathcal{C}\}.$ From this, if $[\mathit{x},\mathit{x}]\in R^{\ast }$, then $\mathit{x}+Hull\left(\mathcal{C}\right)$ is linearly independent over R, and hence forms a basis for $\mathcal{C}/Hull\left(\mathcal{C}\right)$ over $R.$ Conversely, if $\mathit{x}+Hull\left(\mathcal{C}\right)$ is a basis of $\mathcal{C}/Hull\left(\mathcal{C}\right)$ over R, then

$$\mathcal{C}=R\mathit{x}+Hull\left(\mathcal{C}\right),$$

whence the penultimate implication arrow can be replaced by “$\iff$”. Thus, we conclude

$$[\mathit{x},\mathit{x}]\in R^{*}.$$

Furthermore, let $\mathit{x}\in \mathcal{C},\mathit{y}\in {\mathcal{C}}^{\perp }$, and $r_1,r_2\in R$ be such that $[\mathit{x},\mathit{x}],[\mathit{y},\mathit{y}]\in R^{*}$, and $r_1\mathit{x}+r_2\mathit{y}=0.$ Since

$$[r_1\mathit{x}+r_2\mathit{y},\mathit{x}]=[r_1\mathit{x}+r_2\mathit{y},\mathit{y}]=[\mathit{x},\mathit{y}]=0,$$

we can conclude that

$$r_1[\mathit{x},\mathit{x}]=r_2[\mathit{y},\mathit{y}]=0.$$

But both $[\mathit{x},\mathit{x}]$ and $[\mathit{y},\mathit{y}]$ are units in R, it follows that $r_1=r_2=0.$ Therefore, $\mathit{x}$, $\mathit{y}$ are linearly independent over R.

For the direct sum

$$Hull{\left(\mathcal{C}\right)}^{\perp }=Hull\left(\mathcal{C}\right)\oplus R\mathit{x}\oplus R\mathit{y},$$

it can be derived using Proposition 4.

$\left(3\right)$ We have proven that $\mathit{x}$, $\mathit{y}$ are linearly independent over R. This demonstrates that the map

$$\eta :R\mathit{x}+R\mathit{y}\to R^2$$

defined by

$$\eta (r_1\mathit{x}+r_2\mathit{y})=(r_1\left|\mathit{x}\right|,r_2\left|\mathit{y}\right|)$$

is well-defined. It can be verified that $\eta$ is an isometry, given that $[\mathit{x},\mathit{x}]$ and $[\mathit{y},\mathit{y}]$ are in $R^{*}$, $[\mathit{x},\mathit{y}]=0$, and the characteristic of R is $2.$

Our remaining task is to prove that

$$\mathfrak{D}\left(\mathcal{C}\right)=\{Hull\left(\mathcal{C}\right)+{\eta }^{-1}\left({\mathcal{D}}_0\right)|{\mathcal{D}}_0\mathrm{is}\mathrm{a}\mathrm{self}-\mathrm{dual}\mathrm{code}\mathrm{in}{R}^2\}.$$

To conduct this, we claim that

$$\eta \left({\mathcal{D}}^{\perp }\cap (R\mathit{x}+R\mathit{y})\right)=\eta {\left(\mathcal{D}\cap (R\mathit{x}+R\mathit{y})\right)}^{\perp },$$

if $\mathcal{D}$ is a linear code in $R^n$ satisfying $Hull\left(\mathcal{C}\right)\subset \mathcal{D}\subset Hull{\left(\mathcal{C}\right)}^{\perp }$.

In fact, from $Hull\left(\mathcal{C}\right)\subset \mathcal{D}\subset Hull{\left(\mathcal{C}\right)}^{\perp }$, we can deduce that

$$\mathcal{D}=Hull\left(\mathcal{C}\right)\oplus \left(\mathcal{D}\cap (R\mathit{x}+R\mathit{y})\right).$$

Additionally, since $R\mathit{x}+R\mathit{y}\subset Hull{\left(\mathcal{C}\right)}^{\perp }$ and $\eta$ is an isometry, we obtain that

$$\eta \left({\mathcal{D}}^{\perp }\cap (R\mathit{x}+R\mathit{y})\right)=\eta \left({(\mathcal{D}\cap (R\mathit{x}+R\mathit{y}))}^{\perp }\cap (R\mathit{x}+R\mathit{y})\right)=\eta {\left(\mathcal{D}\cap (R\mathit{x}+R\mathit{y})\right)}^{\perp },$$

and hence

$$\eta \left({\mathcal{D}}^{\perp }\cap (R\mathit{x}+R\mathit{y})\right)=\eta {\left(\mathcal{D}\cap (R\mathit{x}+R\mathit{y})\right)}^{\perp }.$$

Now, let $\mathcal{D}\in \mathfrak{D}\left(\mathcal{C}\right)$. It is clear that

$$Hull\left(\mathcal{C}\right)\subset \mathcal{D}\subset Hull{\left(\mathcal{C}\right)}^{\perp }$$

and

$$\mathcal{D}=Hull\left(\mathcal{C}\right)\oplus \left(\mathcal{D}\cap (R\mathit{x}+R\mathit{y})\right).$$

Set ${\mathcal{D}}_0=\eta \left(\mathcal{D}\cap (R\mathit{x}+R\mathit{y})\right)$, which is a linear code in $R^2$. By the claim, we have

$${\mathcal{D}}_0^{\perp }=\eta {\left(\mathcal{D}\cap (R\mathit{x}+R\mathit{y})\right)}^{\perp }=\eta \left({\mathcal{D}}^{\perp }\cap (R\mathit{x}+R\mathit{y})\right)=\eta \left(\mathcal{D}\cap (R\mathit{x}+R\mathit{y})\right)={\mathcal{D}}_0,$$

hence ${\mathcal{D}}_0$ is self-dual. Thus, we obtain

$$\mathcal{D}=Hull\left(\mathcal{C}\right)\oplus {\eta }^{-1}\left({\mathcal{D}}_0\right).$$

On the other hand, let ${\mathcal{D}}_0$ be a self-dual code in $R^2$ and set

$$\mathcal{D}=Hull\left(\mathcal{C}\right)+{\eta }^{-1}\left({\mathcal{D}}_0\right).$$

Considering that $\mathcal{D}$ is a linear code in $R^n$ and $Hull\left(\mathcal{C}\right)\subset \mathcal{D}\subset Hull{\left(\mathcal{C}\right)}^{\perp }$, we have

$$\mathcal{D}=Hull\left(\mathcal{C}\right)\oplus \left(\mathcal{D}\cap (R\mathit{x}+R\mathit{y})\right)=Hull\left(\mathcal{C}\right)\oplus {\eta }^{-1}\left({\mathcal{D}}_0\right),$$

and hence

$${\eta }^{-1}\left({\mathcal{D}}_0\right)=\mathcal{D}\cap (R\mathit{x}+R\mathit{y}).$$

Again by the above claim, we can see that

$$\eta \left({\mathcal{D}}^{\perp }\cap (R\mathit{x}+R\mathit{y})\right)=\eta {\left(\mathcal{D}\cap (R\mathit{x}+R\mathit{y})\right)}^{\perp }={\mathcal{D}}_0^{\perp }={\mathcal{D}}_0.$$

This shows that

$$\mathcal{D}=Hull\left(\mathcal{C}\right)+{\eta }^{-1}\left({\mathcal{D}}_0\right)=Hull\left(\mathcal{C}\right)+\left({\mathcal{D}}^{\perp }\cap (R\mathit{x}+R\mathit{y})\right)={\mathcal{D}}^{\perp },$$

and, therefore, $\mathcal{D}\in \mathfrak{D}\left(\mathcal{C}\right)$. Thus, we have shown that

$$\mathfrak{D}\left(\mathcal{C}\right)=\{Hull\left(\mathcal{C}\right)+{\eta }^{-1}\left({\mathcal{D}}_0\right)|{\mathcal{D}}_0\mathrm{is}\mathrm{a}\mathrm{self}-\mathrm{dual}\mathrm{code}\mathrm{in}{R}^2\}.$$

As for the conclusion regarding the cardinality of $\mathfrak{D}\left(\mathcal{C}\right)$, it follows directly from the results we have proven. □

As previously mentioned, Theorem 1 outlines a method for constructing self-dual codes over finite commutative rings of characteristic 2. However, before delving into the details, we must address the question of when $\mathcal{C}/Hull\left(\mathcal{C}\right)$ can be a one-dimensional free R-module. To solve this problem, we will first recall two small results.

Lemma 3.

Let R be a finite commutative ring with an identity element $1\ne 0.$ Let $r_1,\dots ,r_k\in R.$ Then, the vector $(r_1,\dots ,r_k)$ is an R-linearly independent vector if and only if the ideal $\langle r_1,\dots ,r_k\rangle$ generated by $r_1,\dots ,r_k$ in R is equal to R itself.

Proof. 

This can be shown by noting that the Jacobson radical $J\left(R\right)$ is a nilpotent ideal, and $\langle r_1,\dots ,r_k\rangle$ would be contained in some maximal ideal if it is a proper ideal. □

Lemma 4.

Let R be a commutative ring, and $M\in Ma{t}_{k\times n}\left(R\right).$ Then, $\mathcal{C}\left(M\right)/Hull\left(\mathcal{C}\left(M\right)\right)\cong \mathcal{C}\left(M{M}^{\top }\right)$ as R-modules.

Proof. 

Consider the map

$$\varphi :\mathcal{C}\left(M\right)\to \mathcal{C}\left(M{M}^{\top }\right)$$

defined by $\varphi \left(\mathit{x}\right)=\mathit{x}{M}^{\top }$ for every $\mathit{x}\in \mathcal{C}\left(M\right)$, and verify $ker\left(\varphi \right)=Hull\left(\mathcal{C}\left(M\right)\right).$ □

Now, we will provide a necessary and sufficient condition for linear code $\mathcal{C}$ to satisfy the property that $\mathcal{C}/Hull\left(\mathcal{C}\right)$ is a one-dimensional free R-module.

Proposition 5.

Let $\mathcal{C}$ be a linear code of length n over $R.$ The following statements are equivalent:

(1)

$\mathcal{C}/Hull\left(\mathcal{C}\right)$ is a one-dimensional free R-module;

(2)

There exists a generator matrix M for $\mathcal{C}$ such that $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$ for some linearly independent vector α over $R;$

(3)

For any generator matrix M for $\mathcal{C}$, we have $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$ for some linearly independent vector α over $R.$

Furthermore, if $M\in Ma{t}_{k\times n}\left(R\right)$ is a matrix that generates $\mathcal{C}$ and $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where α is a linearly independent vector in $R^k$, then the map

$$\mathcal{C}/Hull\left(\mathcal{C}\right)\to R,$$

defined by

$$\mathit{v}M+Hull\left(\mathcal{C}\right)\mapsto \mathit{v}{\mathit{\alpha }}^{\top }$$

for any $\mathit{v}\in R^k$, is an R-module isomorphism.

Proof. 

$\left(1\right)\Rightarrow \left(2\right).$ Let $M_0$ be a generator matrix for $Hull\left(\mathcal{C}\right)$, and let $\mathit{x}\in \mathcal{C}$ be a vector such that $\mathit{x}+Hull\left(\mathcal{C}\right)$ forms a basis for $\mathcal{C}/Hull\left(\mathcal{C}\right)$ over R. Construct a new matrix M by appending the vector $\mathit{x}$ to the bottom of $M_0.$ This results in M being a generator matrix for $\mathcal{C}.$ Set $\mathit{\alpha }=(0,\dots ,0,|\mathit{x}\left|\right).$ It is easy to verify that $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, and hence

$$\mathcal{C}\left(M{M}^{\top }\right)={R\langle (0,\dots ,0,|\mathit{x}|}^2)\rangle .$$

By Lemma 4, we have

$${|R\langle (0,\dots ,0,\left|\mathit{x}\right|}^2)\rangle |=|\mathcal{C}\left(M{M}^{\top }\right)|=|\mathcal{C}/Hull\left(\mathcal{C}\right)|=|R|.$$

This shows that $(0,\dots ,0,|\mathit{x}{|}^2)$, and hence $\mathit{\alpha }$, is an R-linearly independent vector.

$\left(2\right)\Rightarrow \left(3\right).$ Consider M as a generator matrix for $\mathcal{C}$ that satisfies $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }=({\alpha }_1,\dots ,{\alpha }_k)$ is a linearly independent vector in $R^k.$ Thus, for any generator matrix $M^{\prime }$ of $\mathcal{C}$, there exists a suitable matrix A, such that

$$AM=M^{\prime }.$$

Set ${\mathit{\alpha }}^{\prime }=({\alpha }_1^{\prime },\dots ,{\alpha }_{k^{\prime }}^{\prime })=\mathit{\alpha }{A}^{\top }$. Direct calculation shows that

$$M^{\prime }{M}^{\prime \top }={\mathit{\alpha }}^{\prime \top }{\mathit{\alpha }}^{\prime }.$$

We also need to prove that ${\mathit{\alpha }}^{\prime }$ is an R-linearly independent vector. In fact, according to Lemma 4,

$$|\mathcal{C}\left({\mathit{\alpha }}^{\prime \top }{\mathit{\alpha }}^{\prime }\right)\left|\right|Hull\left(\mathcal{C}\right)|=|\mathcal{C}|=|\mathcal{C}\left({\mathit{\alpha }}^{\top }\mathit{\alpha }\right)\left|\right|Hull\left(\mathcal{C}\right)|.$$

Furthermore, we can see that

$$\mathcal{C}\left({\mathit{\alpha }}^{\top }\mathit{\alpha }\right)=\langle {\alpha }_1,\dots ,{\alpha }_k\rangle \mathit{\alpha }$$

and

$$\mathcal{C}\left({\mathit{\alpha }}^{\prime \top }{\mathit{\alpha }}^{\prime }\right)=\langle {\alpha }_1^{\prime },\dots ,{\alpha }_{k^{\prime }}^{\prime }\rangle {\mathit{\alpha }}^{\prime }.$$

Since $\mathit{\alpha }$ is an R-linearly independent vector, Lemma 3 states that $\langle {\alpha }_1,\dots ,{\alpha }_k\rangle =R.$ Thus,

$$|\langle {\alpha }_1^{\prime },\dots ,{\alpha }_{k^{\prime }}^{\prime }\rangle {\mathit{\alpha }}^{\prime }|=|\mathcal{C}\left({\mathit{\alpha }}^{\prime \top }{\mathit{\alpha }}^{\prime }\right)|=|\mathcal{C}\left({\mathit{\alpha }}^{\top }\mathit{\alpha }\right)|=|R|.$$

From this, we can conclude that ${\mathit{\alpha }}^{\prime }$ is an R-linearly independent vector.

$\left(3\right)\Rightarrow \left(1\right).$ Let M be a generator matrix for $\mathcal{C}$ such that $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }=({\alpha }_1,\dots ,{\alpha }_k)$ is a linearly independent vector in $R^k.$ Using Lemma 4 again, we have

$$\mathcal{C}/Hull\left(\mathcal{C}\right)\cong \mathcal{C}\left(M{M}^{\top }\right)=\mathcal{C}\left({\mathit{\alpha }}^{\top }\mathit{\alpha }\right)=\langle {\mathit{\alpha }}_1,\dots ,{\mathit{\alpha }}_k\rangle \mathit{\alpha }.$$

Since $\mathit{\alpha }$ is a linearly independent vector, $\langle {\alpha }_1,\dots ,{\alpha }_k\rangle =R$ by Lemma 3. Thus,

$$\mathcal{C}/Hull\left(\mathcal{C}\right)(\cong R\mathit{\alpha })$$

is a one-dimensional free R-module.

To prove the last statement, consider the map

$$ϵ:\mathcal{C}\left(M\right)\to R$$

defined by

$$ϵ\left(\mathit{v}M\right)=\mathit{v}{\mathit{\alpha }}^{\top }$$

for every $\mathit{v}\in R^k$, where k is the number of rows in M. It is easy to verify that $ϵ$ is well-defined and an epimorphism. Next, let $\mathit{v}\in R^k.$ Since $\mathit{\alpha }$ is a linearly independent vector, we have

$$\mathit{v}M\in ker\left(ϵ\right)\iff \mathit{v}{\mathit{\alpha }}^{\top }=0\iff \mathit{v}{\mathit{\alpha }}^{\top }\mathit{\alpha }=0\iff \mathit{v}M{M}^{\top }=0\iff \mathit{v}M\in Hull\left(\mathcal{C}\left(M\right)\right).$$

Thus, $ker\left(ϵ\right)=Hull\left(\mathcal{C}\right)$, so that

$$\mathcal{C}/Hull\left(\mathcal{C}\right)\to R\left(\mathit{v}M+Hull\left(\mathcal{C}\right)\mapsto \mathit{v}{\mathit{\alpha }}^{\top }\right)$$

is an R-module isomorphism, as desired. □

Remark 3

(The proof for the implication $\left(2\right)\Rightarrow \left(3\right)$ also proves the following statement).

• Let A be a $k\times k$ matrix over R such that A can be written as ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for an R-linearly independent vector α in $R^k.$ Then, for any vector ${\mathit{\alpha }}^{\prime }$ in $R^k$, if ${\mathit{\alpha }}^{\prime \top }{\mathit{\alpha }}^{\prime }$ equals A, then ${\mathit{\alpha }}^{\prime }$ is also an R-linearly independent vector.

Proposition 5 raises the question of how to determine if the square matrix $M{M}^{\top }$ can be expressed as ${\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }$ is a linearly independent vector over $R.$ In order to address this problem, we introduce a method in Section 5, which provides a necessary and sufficient condition for determining if a square matrix has this form.

Example 1.

Let $R={\mathbb{F}}_2\left[x\right]/\langle x^4\rangle$, which is a local Frobenius ring of order 16 with characteristic 2. To simplify notation, we define $\overline{x}=x+\langle x^4\rangle$. Next, we define three $2\times 4$ matrices over $R:$

$$M_1=\left(\begin{array}{cccc}1& 0& 1& \overline{x}\\ 0& 1& \overline{x}& 1\end{array}\right),$$

$$M_2=\left(\begin{array}{cccc}1& 0& \overline{x}+{\overline{x}}^2+{\overline{x}}^3& 1+\overline{x}+{\overline{x}}^2+{\overline{x}}^3\\ 0& 1& 1& \overline{x}\end{array}\right),$$

and

$$M_3=\left(\begin{array}{cccc}1& 0& \overline{x}+{\overline{x}}^2+{\overline{x}}^3& \overline{x}+{\overline{x}}^2+{\overline{x}}^3\\ 0& 1& 1& \overline{x}\end{array}\right).$$

It can be calculated that

$$M_1{M}_1^{\top }=\left(\begin{array}{cc}{\overline{x}}^2& 0\\ 0& {\overline{x}}^2\end{array}\right),M_2{M}_2^{\top }=\left(\begin{array}{cc}0& 0\\ 0& {\overline{x}}^2\end{array}\right),\mathit{and}{M}_3{M}_3^{\top }=\left(\begin{array}{cc}1& \overline{x}\\ \overline{x}& {\overline{x}}^2\end{array}\right).$$

It is easy to verify that $M_2{M}_2^{\top }$ and $M_3{M}_3^{\top }$ can be written as ${(0,\overline{x})}^{\top }(0,\overline{x})$ and ${(1,\overline{x})}^{\top }(1,\overline{x})$, respectively. However, we will obtain a contradiction if we assume that $M_1{M}_1^{\top }$ can be expressed in this form. Let ${\mathcal{C}}_i$ ($i=1,2,3$) denote the linear code generated by $M_i$ over $R.$ It is evident that all of them are two-dimensional free linear codes with length 4 over $R.$ Furthermore, by Proposition 5, it follows that ${\mathcal{C}}_3/Hull\left({\mathcal{C}}_3\right)$ is a one-dimensional free R-module, while both ${\mathcal{C}}_1/Hull\left({\mathcal{C}}_1\right)$ and ${\mathcal{C}}_2/Hull\left({\mathcal{C}}_2\right)$ are not. Additionally, we can obtain that the order of $Hull\left({\mathcal{C}}_1\right)$ is 16 by computer, and hence $|{\mathcal{C}}_1/Hull\left({\mathcal{C}}_1\right)|=|R|.$ Even so, ${\mathcal{C}}_1/Hull\left({\mathcal{C}}_1\right)$ is not a free module with dimension 1 over $R.$

The occurrence of these three cases in Example 1 is not surprising, after all, the alphabets we consider are finite rings rather than finite fields.

-

The construction method

We have now reached the point where we will present our method for constructing self-dual codes of length n over finite commutative rings R with characteristic 2, where $n=2k$ is an even number. This method, founded on Theorem 1 and Proposition 5, consists of six steps.

Step 1: 

Search for matrices M in $Ma{t}_{k\times n}\left(R\right)$ such that the rows of M are linearly independent over R and $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, for an R-linearly independent vector $\mathit{\alpha }=({\alpha }_1,\dots ,{\alpha }_k)$ in $R^k$. The linear code generated by M will be denoted as $\mathcal{C}.$

Step 2: 

Select a codeword $\mathit{x}$ from $\mathcal{C}$ such that $\mathit{x}+Hull\left(\mathcal{C}\right)$ forms a basis for $\mathcal{C}/Hull\left(\mathcal{C}\right)$ over R.

This can be conducted by finding a vector $(r_1,\dots ,r_k)$ in $R^k$ that satisfies $r_1{\alpha }_1+\cdots +r_k{\alpha }_k\in R^{*}.$ Then, set

$$\mathit{x}=(r_1,\dots ,r_k)M.$$

According to the last conclusion of Proposition 5, $\mathit{x}+Hull\left(\mathcal{C}\right)$ forms a basis for $\mathcal{C}/Hull\left(\mathcal{C}\right)$ over $R.$

Step 3: 

Select a codeword $\mathit{y}$ from ${\mathcal{C}}^{\perp }$ such that $\mathit{y}+Hull\left(\mathcal{C}\right)$ forms a basis for ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)$ over $R.$

This can be conducted by finding a generator matrix for ${\mathcal{C}}^{\perp }.$ To conduct this, we can use the proof of Lemma 2, as demonstrated in Remark 1, to determine two suitable matrices A and Q such that $AMQ$ is of the form $\left(I_k\right|H)$, where A is a $k\times k$ invertible matrix, Q is an $n\times n$ orthogonal matrix over R, and H is a $k\times k$ matrix over $R.$ One can verify immediately that $\left(H^{\top }\right|I_k)Q^{\top }$ is a generator matrix for ${\mathcal{C}}^{\perp }.$ Next, according to (2) in Theorem 1, we need to find a vector $(r_1^{\prime },\dots ,r_k^{\prime })\in R^k$ that satisfies the condition

$$r_1^{\prime }\left(\right|H_{\left(1\right)}|+1)+\cdots +r_k^{\prime }\left(\right|H_{\left(k\right)}|+1)\in R^{*},$$

and use this vector to construct the vector $\mathit{y}$ as

$$\mathit{y}=(r_1^{\prime },\dots ,r_k^{\prime })\left(H^{\top }\right|I_k)Q^{\top }.$$

This ensures that $\mathit{y}+Hull\left(\mathcal{C}\right)$ is a basis for ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)$ over $R.$ If M is initially in the form of $\left(I_k\right|H)$, the process of finding $\mathit{y}$ can be simplified significantly.

Step 4: 

Determine a generator matrix for $Hull\left(\mathcal{C}\right).$

From the proof of the last statement in Proposition 5, it follows that

$$Hull\left(\mathcal{C}\right)=\left\{\mathit{v}M\right|\mathit{v}\in R^k\mathrm{and}[\mathit{v},\mathit{\alpha }]=0\}.$$

This means that we only need to find a basis for $\mathcal{C}{\left(\mathit{\alpha }\right)}^{\perp }$ (remember that $\mathit{\alpha }$ is an R-linearly independent vector in $R^k$). Once we have found this basis, denoted as $\{{\mathit{v}}_1,\dots ,{\mathit{v}}_{k-1}\}$, we can use it to construct a basis for $Hull\left(\mathcal{C}\right)$ over R, namely,

$$\{{\mathit{v}}_{1}M,\dots ,{\mathit{v}}_{k-1}M\}.$$

These basis elements can then be arranged as rows to form a $(k-1)\times n$ matrix $M^{\prime }$, which will serve as a generator matrix for $Hull\left(\mathcal{C}\right).$

Step 5: 

Identify the corresponding isometry η as defined in Theorem 1.

Step 6: 

Construct self-dual codes of length n over R from self-dual codes in $R^2.$

Let ${\mathcal{D}}_0$ be a self-dual code in $R^2$ with a generator matrix $M_0$. We can then construct the resulting self-dual code $\mathcal{D}$ as $Hull\left(\mathcal{C}\right)\oplus {\eta }^{-1}\left({\mathcal{D}}_0\right).$ The generator matrix for $\mathcal{D}$ can be easily obtained from the generator matrices $M^{\prime }$ and $M_0.$

Remark 4.

Regarding this construction method, we would like to address two additional points.

Firstly, once a matrix M that satisfies the conditions outlined in Step 1 has been found, it can be utilized to construct multiple self-dual codes through the aforementioned method. This is because there is often more than one self-dual code of length 2 over $R.$ Additionally, the self-dual code $\mathcal{D}$ produced by this method may or may not be free, depending on whether the corresponding self-dual code ${\mathcal{D}}_0$ of length 2 is also a free linear code. Therefore, our method allows for the construction of nonfree self-dual codes, as long as there is a nonfree self-dual code in $R^2.$ This objective is typically unachievable using the conventional approach. These advantages can be seen as benefits of our construction method.

Secondly, when the ring R is a finite commutative local ring, Steps 2 and 4 can be completed effectively. This is because α contains a coordinate component that belongs to $R^{*}$ in this case. Additionally, Step 3 can be easily accomplished by utilizing the proof of Lemma 2. However, even if R is not a local ring, these steps can still be completed with the help of the Chinese Remainder Theorem or the algebra-isomorphism $\overline{\Psi }$ defined in Section 2. Nevertheless, if R has multiple maximal ideals, Step 3 may be time-consuming. Therefore, a more efficient method for finding $\mathit{y}$ is needed.

Examples of constructing self-dual codes using this method will be provided in Section 5.

Next, we will present two propositions: Propositions 6 and 7. One can reduce computation time, while the other, a slight generalization of Theorem 3.2 in [41], can be used to construct more matrices M that meet the requirements outlined in Step 1.

Proposition 6.

Let $\mathcal{C}$ be a linear code over R with the same properties as stated in Theorem 1. Let $\mathcal{P}$ be an $n\times n$ permutation matrix over $R.$ Set ${\mathcal{C}}^{\prime }=\mathcal{C}\mathcal{P}:=\left\{\mathit{x}\mathcal{P}\right|\mathit{x}\in \mathcal{C}\}.$ Then, ${\mathcal{C}}^{\prime }$ is also a free linear code over R with length n and dimension $n/2.$ Additionally, ${\mathcal{C}}^{\prime }/Hull\left({\mathcal{C}}^{\prime }\right)$ is also a one-dimensional free R-module. Furthermore, there is a bijection between $\mathfrak{D}\left(\mathcal{C}\right)$ and $\mathfrak{D}\left({\mathcal{C}}^{\prime }\right)$, given by $\mathcal{D}\mapsto \mathcal{D}\mathcal{P}$.

Proof. 

More generally, let us assume that $\mathcal{P}$ is an orthogonal matrix. It follows from this assumption that the map

$$R^n\to R^n$$

defined by

$$\mathit{v}\mapsto \mathit{v}\mathcal{P}$$

is an isometry. As a result, ${\mathcal{C}}^{\prime }$ is also a $n/2$-dimensional free linear code with length n, and ${\mathcal{C}}^{\prime }/Hull\left({\mathcal{C}}^{\prime }\right)$ is also a one-dimensional free R-module. Additionally, we can see that

$$Hull\left({\mathcal{C}}^{\prime }\right)=Hull\left(\mathcal{C}\right)\mathcal{P}.$$

Thus, we can define a map from $\mathfrak{D}\left(\mathcal{C}\right)$ into $\mathfrak{D}\left({\mathcal{C}}^{\prime }\right)$ by

$$\mathcal{D}\mapsto \mathcal{D}\mathcal{P}$$

for any $\mathcal{D}$ in $\mathfrak{D}\left(\mathcal{C}\right).$ Since $\mathcal{P}$ is an invertible matrix and $|\mathfrak{D}\left({\mathcal{C}}^{\prime }\right)|=|\mathfrak{D}\left(\mathcal{C}\right)|$, which equals the number of self-dual codes in $R^2$, we can conclude that this map is a bijection, as desired. □

Proposition 7.

Let M be a $k\times n$ matrix over R, and let $\mathit{v}$ and $\mathit{w}$ be two vectors in $R^n$. Define $M(\mathit{v},\mathit{w})$ to be the $k\times n$ matrix whose i-th row is $M_i+[\mathit{v},M_i]\mathit{w}+[\mathit{w},M_i]\mathit{v}$, where $M_i$ is the i-th row of M, and $i=1,2,\dots ,k.$ If $[\mathit{v},\mathit{v}]=[\mathit{v},\mathit{w}]=[\mathit{w},\mathit{w}]=0$, then $M(\mathit{v},\mathit{w})M{(\mathit{v},\mathit{w})}^{\top }=M{M}^{\top }.$ Additionally, if the rows of M are linearly independent over R, then the same is true for $M(\mathit{v},\mathit{w}).$

Proof. 

Set $Q(\mathit{v},\mathit{w})=I_n+{\mathit{v}}^{\top }\mathit{w}+{\mathit{w}}^{\top }\mathit{v}$. Given the conditions, it can be deduced that $Q(\mathit{v},\mathit{w})$ is an orthogonal symmetric $n\times n$ matrix over $R.$ Consider the map

$$\lambda :R^n\to R^n$$

defined by

$$\lambda \left({\mathit{v}}^{\prime }\right)={\mathit{v}}^{\prime }+[\mathit{v},{\mathit{v}}^{\prime }]\mathit{w}+[\mathit{w},{\mathit{v}}^{\prime }]\mathit{v}$$

for any ${\mathit{v}}^{\prime }$ in $R^n.$ This map can be easily verified as an isometry, and it can be seen that

$$\lambda \left({\mathit{v}}^{\prime }\right)={\mathit{v}}^{\prime }Q(\mathit{v},\mathit{w}).$$

This implies that the i-th row of $M(\mathit{v},\mathit{w})$ is equal to $\lambda \left(M_i\right)=M_{i}Q(\mathit{v},\mathit{w}).$ Therefore, it follows that

$$M(\mathit{v},\mathit{w})=MQ(\mathit{v},\mathit{w}).$$

From this, we can then deduce that

$$M(\mathit{v},\mathit{w})M{(\mathit{v},\mathit{w})}^{\top }=M{M}^{\top }.$$

Additionally, since $Q(\mathit{v},\mathit{w})$ is an invertible matrix, the condition that the rows of M are linearly independent over R also implies that the rows of $M(\mathit{v},\mathit{w})$ are linearly independent over R. □

Remark 5.

Maintaining the notation used in Proposition 7, we set $n=2k$ and assume that the rows of M are linearly independent over $R.$ Furthermore, we assume that $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where α is an R-linearly independent vector in $R^k.$ By applying Proposition 7, we can see that $M(\mathit{v},\mathit{w})M{(\mathit{v},\mathit{w})}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }.$ Thus, we construct a new matrix $M(\mathit{v},\mathit{w})$ that satisfies the requirements in Step 1. Now, let $\mathcal{C}$ and ${\mathcal{C}}^{\prime }$ be two linear codes generated by M and $M(\mathit{v},\mathit{w})$, respectively. These codes are k-dimensional free linear codes of length n over $R.$ Furthermore, Proposition 5 implies that both $\mathcal{C}/Hull\left(\mathcal{C}\right)$ and ${\mathcal{C}}^{\prime }/Hull\left({\mathcal{C}}^{\prime }\right)$ are one-dimensional free modules over $R.$ This means that both $\mathcal{C}$ and ${\mathcal{C}}^{\prime }$ satisfy the conditions in Theorem 1. Additionally, we can see that

$${\mathcal{C}}^{\prime }=\mathcal{C}\left(M(\mathit{v},\mathit{w})\right)=\mathcal{C}\left(MQ(\mathit{v},\mathit{w})\right)=\mathcal{C}\left(M\right)Q(\mathit{v},\mathit{w})=\mathcal{C}Q(\mathit{v},\mathit{w}).$$

Since $Q(\mathit{v},\mathit{w})$ is orthogonal, it follows from the proof of Proposition 6 that the map

$$\mathfrak{D}\left(\mathcal{C}\right)\to \mathfrak{D}\left({\mathcal{C}}^{\prime }\right)$$

given by

$$\mathcal{D}\mapsto \mathcal{D}Q(\mathit{v},\mathit{w})$$

is a bijection.

To construct self-dual codes using our method, one must first obtain self-dual codes of length 2 over the corresponding ring. In response to this necessity, we will present two results regarding the construction of self-dual codes of length 2 to conclude this section. In the first result, for finite commutative rings of characteristic 2, a comprehensive list of all free self-dual codes of length 2 is provided. In the second result, concerning a finite commutative local ring of characteristic 2, we obtain a nonfree self-dual code of length 2. Moreover, for a finite commutative chain ring with characteristic 2, we construct all nonfree self-dual codes of length 2 over it.

A finite commutative ring which is both principal and local is called a finite chain ring. For an alternative and equivalent definition of chainrings, as well as a discussion of their properties, please refer to [39].

Proposition 8.

The set of all free self-dual codes of length 2 over R is equal to

$$\left\{R\langle (1,1+r)\rangle \right|r\in R\mathrm{and}{r}^2=0\}.$$

Proof. 

On one hand, let $\mathcal{C}=\mathcal{C}\left(\right(1,1+r\left)\right)$, where $r\in R$ and $r^2=0.$ It is clear that $\mathcal{C}$ is self-orthogonal. By Proposition 2, since $\mathcal{C}$ is a one-dimensional free linear code over R, we can conclude that the size of ${\mathcal{C}}^{\perp }$ is also $\left|R\right|$, and hence $\mathcal{C}={\mathcal{C}}^{\perp }$. Thus, $\mathcal{C}$ is a free self-dual code of length 2 over $R.$

On the other hand, let $\mathcal{C}$ be a free self-dual code of length 2 over $R.$ Again from Proposition 2, it follows that the dimension of $\mathcal{C}$ is $1.$ Let $(r_1,r_2)\in R^2$ be a basis for $\mathcal{C}.$ Then, $\mathcal{C}=R\langle (r_1,r_2)\rangle .$ Set $r=r_1+r_2.$ Since $\mathcal{C}$ is self-dual, $r^2=0.$ Furthermore, since $(r_1,r_2)=(r_1,r_1+r)$ is linearly independent over R, according to Lemma 3, there exist $r_3$ and $r_4$ in R such that $r_3{r}_1+r_4(r_1+r)=1$, which implies that $r_3{r}_1+r_4{r}_1=1+r_{4}r.$ Since $r_{4}r$ is a nilpotent element, it follows that $r_1\in R^{*}.$ Thus,

$$\mathcal{C}=R\langle (r_1,r_2)\rangle =R\langle (r_1,r_1+r)\rangle =R\langle (1,1+r_1^{-1}r)\rangle ,$$

where ${\left(r_1^{-1}r\right)}^2=0.$ □

Proposition 9.

Let $(R,\mathfrak{m})$ be a finite commutative local ring with the unique maximal ideal $\mathfrak{m}\ne 0.$ Let e be the minimum positive integer number satisfying ${\mathfrak{m}}^e=0.$ Define $Ann\left(\mathfrak{m}\right)=\{r\in R|r\pi =0,\forall \pi \in \mathfrak{m}\}$ as usual. Then:

$\left(1\right)$ Let $\mathcal{C}=\left\{\right(\pi ,\pi +r\left)\right|\pi \in \mathfrak{m},r\in Ann\left(\mathfrak{m}\right)\}.$ Then, $\mathcal{C}$ is a nonfree self-dual code of length 2 over $R.$ Furthermore, if $r\in Ann\left(\mathfrak{m}\right)$ for each $r\in R$ satisfying $r^2=0$, then code $\mathcal{C}$ is the unique nonfree self-dual code of length 2 over $R.$

$\left(2\right)$ If R is a chain ring with $\mathfrak{m}=\langle \pi \rangle$, then the set of all nonfree self-dual codes of length 2 over R is equal to

$$\left\{\mathcal{C}\right(M\left)\right|M\in \mathfrak{T}\},$$

where

$$\mathfrak{T}=\{\left(\begin{array}{cc}{\pi }^{{\displaystyle \frac{e}{2}}}& 0\\ 0& {\pi }^{{\displaystyle \frac{e}{2}}}\end{array}\right),\left(\begin{array}{cc}{\pi }^{e_1}& r{\pi }^{e_1}\\ 0& {\pi }^{e_2}\end{array}\right)|e_1,e_2\in {\mathbb{Z}}^{+},e_1<e_2,\mathrm{and}{e}_1+e_2=e;r\in 1+{\mathfrak{m}}^{\lceil {\displaystyle \frac{e}{2}}\rceil -e_1}\}.$$

(Note: If e is an odd number, then ${\pi }^{{\displaystyle \frac{e}{2}}}$ is not defined. Therefore, in this case, the former type of matrix in the set $\mathfrak{T}$ will not occur.)

Furthermore, let $\mathcal{C}$ and ${\mathcal{C}}^{\prime }$ be two linear codes generated by

$$\left(\begin{array}{cc}{\pi }^{e_1}& r{\pi }^{e_1}\\ 0& {\pi }^{e_2}\end{array}\right)\mathit{and}\left(\begin{array}{cc}{\pi }^{e_3}& r^{\prime }{\pi }^{e_3}\\ 0& {\pi }^{e_4}\end{array}\right)\in \mathfrak{T},$$

respectively, where $r,r^{\prime }\in 1+\mathfrak{m}.$ Then, $\mathcal{C}={\mathcal{C}}^{\prime }$ if and only if $e_1=e_3$, $e_2=e_4$, and $r+r^{\prime }\in {\mathfrak{m}}^{e_2-e_1}.$

Proof. 

$\left(1\right)$ By definition, $\mathcal{C}$ is self-orthogonal. However, we can also show that $\mathcal{C}$ is self-dual. To do so, let $(r_1,r_2)\in {\mathcal{C}}^{\perp }.$ Since $[(r_1,r_2),(0,r)]=[(r_1,r_2),(\pi ,\pi )]=0$ for each $r\in Ann\left(\mathfrak{m}\right)$ and $\pi \in \mathfrak{m}$, we can conclude that $r_1+r_2\in Ann\left(\mathfrak{m}\right)$ and $r_2\in \mathfrak{m}.$ Let us set $r=r_1+r_2.$ Then, $(r_1,r_2)=(r_1,r_1+r)$, which clearly belongs to $\mathcal{C}.$ Therefore, $\mathcal{C}$ is a self-dual code with length 2 over $R.$ It is clear that $\mathcal{C}$ is not a free linear code.

Next, we will prove the uniqueness of $\mathcal{C}$ under the hypothesis that for every $r\in R$ satisfying $r^2=0$, $r\in Ann\left(\mathfrak{m}\right).$ For this, let us assume ${\mathcal{C}}^{\prime }$ is also a nonfree self-dual code of length 2 over $R.$ It is easy to verify that ${\mathcal{C}}^{\prime }\subset \mathfrak{m}\times \mathfrak{m}$, for otherwise ${\mathcal{C}}^{\prime }$ would contain an R-linearly independent vector and hence be a free linear code, contradicting the fact that ${\mathcal{C}}^{\prime }$ is not free. For each $(r_1,r_2)\in {\mathcal{C}}^{\prime }$, we have $r_1+r_2\in Ann\left(\mathfrak{m}\right)$, since ${(r_1+r_2)}^2=0.$ Therefore, $(r_1,r_2)=\left(r_1,r_1+(r_1+r_2)\right)$ belongs to $\mathcal{C}.$ Thus, we obtain that ${\mathcal{C}}^{\prime }\subset \mathcal{C}$ and hence ${\mathcal{C}}^{\prime }=\mathcal{C}.$ This implies that $\mathcal{C}$ is the unique nonfree self-dual code with length 2 over $R.$

$\left(2\right)$ Consider a matrix M over the chain ring R that can generate a nonfree self-dual code of length 2. It is also easy to verify that all the components of M are contained in $\mathfrak{m}.$ Since R is a chain ring, we can transform the matrix M into a new matrix $M^{\prime }$, which can take one of two forms:

$$\left(\begin{array}{cc}{\pi }^{e_1}& 0\\ 0& {\pi }^{e_2}\end{array}\right)\mathrm{or}\left(\begin{array}{cc}{\pi }^{e_3}& r{\pi }^{e_3}\\ 0& {\pi }^{e_4}\end{array}\right).$$

Here, $e_1$, $e_2$, $e_3$, and $e_4$ are all positive integers that satisfy the conditions $e_1\le e_2$, $e_3<e_4$, respectively, and $r\in R\setminus {\mathfrak{m}}^{e_4-e_3}.$ This can be achieved through a finite sequence of elementary row operations and column permutations and by deleting any rows that consist entirely of zeros. Since $\mathcal{C}\left(M\right)$ is a self-dual code and is permutation equivalent to $\mathcal{C}\left(M^{\prime }\right)$, $\mathcal{C}\left(M^{\prime }\right)$ is also a self-dual code.

If $M^{\prime }$ has the form on the left side, we can then conclude that

$$e_1=e_2=e/2.$$

This is because the surjection

$$R^2\to \mathcal{C}\left(M^{\prime }\right)$$

given by

$$(r_1,r_2)\mapsto r_1({\pi }^{e_1},0)+r_2(0,{\pi }^{e_2})$$

has a kernel of $Ann\left({\mathfrak{m}}^{e_1}\right)\times Ann\left({\mathfrak{m}}^{e_2}\right)$ and $\left|\mathcal{C}\right(M^{\prime }\left)\right|=\left|R\right|.$

On the other hand, If $M^{\prime }$ has the form on the right side, we can then obtain that

$$e_3+e_4=e\mathrm{and}r\in 1+{\mathfrak{m}}^{\lceil {\displaystyle \frac{e}{2}}\rceil -e_3}.$$

This can be derived using a similar approach as in the previous case.

Additionally, it is important to note that if $e_3<e_4$ and $r\in 1+{\mathfrak{m}}^{\lceil {\displaystyle \frac{e}{2}}\rceil -e_3}$, then

$$\left(\begin{array}{cc}r{\pi }^{e_3}& {\pi }^{e_3}\\ {\pi }^{e_4}& 0\end{array}\right)=\left(\begin{array}{cc}r& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ {\pi }^{e_4-e_3}& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& r^{-1}\end{array}\right)\left(\begin{array}{cc}{\pi }^{e_3}& r^{-1}{\pi }^{e_3}\\ 0& {\pi }^{e_4}\end{array}\right),$$

and $r^{-1}$ is also in $1+{\mathfrak{m}}^{\lceil {\displaystyle \frac{e}{2}}\rceil -e_3}.$ Thus, from the above analysis, it follows that every nonfree self-dual code of length 2 over R belongs to the set $\left\{\mathcal{C}\right(M\left)\right|M\in \mathfrak{T}\}.$

As for the reverse inclusion, one can easily verify that all the linear codes generated by matrices in the set $\mathfrak{T}$ are nonfree self-dual codes of length 2 over $R.$

To prove the last statement in the proposition, we first assume $\mathcal{C}={\mathcal{C}}^{\prime }.$ This implies that there exist $r_1$, $r_2$, $r_3$, and $r_4$ in R such that

$$(0,{\pi }^{e_2})=r_1({\pi }^{e_3},r^{\prime }{\pi }^{e_3})+r_2(0,{\pi }^{e_4})$$

and

$$({\pi }^{e_1},r{\pi }^{e_1})=r_3({\pi }^{e_3},r^{\prime }{\pi }^{e_3})+r_4(0,{\pi }^{e_4}).$$

From this, it follows that $\langle {\pi }^{e_2}\rangle \subset \langle {\pi }^{e_4}\rangle .$ Similarly, we can see that the reverse inclusion is also true. Thus, $e_2=e_4$, and then $e_1=e_3$, as $e_1+e_2=e=e_3+e_4.$ Furthermore, from the equation

$$({\pi }^{e_1},r{\pi }^{e_1})=r_3({\pi }^{e_1},r^{\prime }{\pi }^{e_1})+r_4(0,{\pi }^{e_2}),$$

it follows that

$$r{\pi }^{e_1}+r^{\prime }{\pi }^{e_1}=r_4{\pi }^{e_2},$$

and hence $r+r^{\prime }\in {\mathfrak{m}}^{e_2-e_1}.$ Conversely, since $r+r^{\prime }\in {\mathfrak{m}}^{e_2-e_1}$, there exists an element $r_0$ in R such that $r+r^{\prime }=r_0{\pi }^{e_2-e_1}.$ From this, we can deduce that

$$r{\pi }^{e_1}=r^{\prime }{\pi }^{e_1}+r_0{\pi }^{e_2}\mathrm{and}{r}^{\prime }{\pi }^{e_1}=r{\pi }^{e_1}+r_0{\pi }^{e_2}.$$

Therefore,

$$({\pi }^{e_1},r{\pi }^{e_1})=({\pi }^{e_1},r^{\prime }{\pi }^{e_1})+r_0(0,{\pi }^{e_2})$$

and

$$({\pi }^{e_1},r^{\prime }{\pi }^{e_1})=({\pi }^{e_1},r{\pi }^{e_1})+r_0(0,{\pi }^{e_2}).$$

Combining the conditions $e_1=e_3$ and $e_2=e_4$, this shows that $\mathcal{C}={\mathcal{C}}^{\prime }.$ □

Remark 6.

In Proposition 9, when R is a finite commutative chain ring with characteristic 2 and e is an even number, it is easy to prove that the self-dual codes generated by the two types of matrices in the set $\mathfrak{T}$ are not equal. Additionally, assuming that R is a finite commutative principal ring with characteristic 2, we can theoretically construct all self-dual codes of length 2 over R by applying Propositions 8 and 9. This is because R is isomorphic, via the Chinese Remainder Theorem, to a finite direct product of chainrings.

5. Conditions for a Matrix to Take the Form of α^⊤α

In the previous section, we posed a question that still needs to be answered: How can we determine if a symmetric matrix $A\in Ma{t}_k\left(R\right)$ (i.e., $A=A^{\top }$) can be written in the form of ${\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }$ is an R-linearly independent vector in $R^k?$ To address this problem, we introduce the following theorem, which is essential for the construction method discussed in the previous section.

Theorem 2.

Let A be a $k\times k$ symmetric matrix over R, with all its diagonal elements being square elements in $R.$ Denote the $(i,j)$-component of A as $a_{ij}$. Then, the following are equivalent:

(1)

A can be written in the form of ${\mathit{\alpha }}^{\top }\mathit{\alpha }$, where α is an R-linearly independent vector in $R^k.$

(2)

$(a_{11},\dots ,a_{kk})$ is linearly independent over R, and there exist elements $r_1,\dots ,r_k,r_1^{\prime },\dots ,r_k^{\prime }$ in R satisfying $r_i^2=a_{ii}$ for each $i=1,2,\dots ,k$ and $r_1^{\prime }{r}_1+\cdots +r_k^{\prime }{r}_k=1$, such that $A={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }=(r_1^{\prime },\dots ,r_k^{\prime })A.$

(3)

$(a_{11},\dots ,a_{kk})$ is linearly independent over $R.$ Additionally, for any elements $r_1,\dots ,r_k,r_1^{\prime },$$\dots ,r_k^{\prime }$ in R, if $r_i^2=a_{ii}$ for each $i=1,2,\dots ,k$ and $r_1^{\prime }{r}_1+\cdots +r_k^{\prime }{r}_k=1$, then $A={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }=(r_1^{\prime },\dots ,r_k^{\prime })A.$

Proof. 

$\left(1\right)\Rightarrow \left(2\right).$ Let $A={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }=(r_1,\dots ,r_k)$ is an R-linearly independent vector in $R^k.$ It is clear that $a_{ii}=r_i^2$ and $A_i=r_i\mathit{\alpha }$ for each $i=1,2,\dots ,k$, where $A_i$ is the i-th row of $A.$ Since $\mathit{\alpha }$ is linearly independent over R, it follows from Lemma 3 that

$$\langle r_1,\dots ,r_k\rangle =R,$$

and hence

$$\langle a_{11},\dots ,a_{kk}\rangle =R.$$

By Lemma 3, this means that the vector $(a_{11},\dots ,a_{kk})$ is also linearly independent over $R.$ Next, let $r_1^{\prime },\dots ,r_k^{\prime }$ be elements in R such that $r_1^{\prime }{r}_1+\cdots +r_k^{\prime }{r}_k=1.$ We want to show that

$$\mathit{\alpha }=(r_1^{\prime },\dots ,r_k^{\prime })A.$$

Indeed,

$$(r_1^{\prime },\dots ,r_k^{\prime })A=r_1^{\prime }{A}_1+\cdots +r_k^{\prime }{A}_k=r_1^{\prime }{r}_1\mathit{\alpha }+\cdots +r_k^{\prime }{r}_k\mathit{\alpha }=(r_1^{\prime }{r}_1+\cdots +r_k^{\prime }{r}_k)\mathit{\alpha }=\mathit{\alpha }.$$

Thus, we have completed the proof of this part.

$\left(2\right)\Rightarrow \left(3\right).$ Given the conditions in $\left(2\right)$, there exist elements $r_1^{″},\dots ,r_k^{″},r_1^{‴},\dots ,r_k^{‴}$ in R satisfying ${r_i^{″}}^2=a_{ii}$ for each $i=1,2,\dots ,k$ and $r_1^{‴}{r}_1^{″}+\cdots +r_k^{‴}{r}_k^{″}=1$, such that $A={\mathit{\beta }}^{\top }\mathit{\beta }$, where $\mathit{\beta }=(r_1^{‴},\dots ,r_k^{‴})A.$ Now, let $r_1,\dots ,r_k,r_1^{\prime },\dots ,r_k^{\prime }$ be a family of elements in R satisfying $r_i^2=a_{ii}$ for each $i=1,2,\dots ,k$ and $r_1^{\prime }{r}_1+\cdots +r_k^{\prime }{r}_k=1.$ Set $\mathit{\alpha }=(r_1^{\prime },\dots ,r_k^{\prime })A.$ We need to show that

$$A={\mathit{\alpha }}^{\top }\mathit{\alpha }.$$

First, we can see that $r_i^2=a_{ii}={\beta }_i^2$ for any i, where ${\beta }_i$ is the i-th coordinate of $\beta$. Therefore, for each $i=1,2,\dots ,k$, there is an element ${\Delta }_i$ in R satisfying ${\Delta }_i^2=0$ and $r_i={\beta }_i+{\Delta }_i.$ Since $r_1^{\prime }{r}_1+\cdots +r_k^{\prime }{r}_k=1$, it follows that

$$r_1^{\prime }{\beta }_1+\cdots +r_k^{\prime }{\beta }_k=1+{\Delta }_0,$$

where ${\Delta }_0=r_1^{\prime }{\Delta }_1+\cdots +r_k^{\prime }{\Delta }_k.$ Furthermore, since $A={\mathit{\beta }}^{\top }\mathit{\beta }$, we have

$$\mathit{\alpha }=(r_1^{\prime },\dots ,r_k^{\prime })A=(r_1^{\prime }{\beta }_1+\cdots +r_k^{\prime }{\beta }_k)\mathit{\beta }=(1+{\Delta }_0)\mathit{\beta }.$$

Noting that ${\Delta }_0^2=0$, we can conclude that

$${\mathit{\alpha }}^{\top }\mathit{\alpha }={\mathit{\beta }}^{\top }(1+{\Delta }_0)(1+{\Delta }_0)\mathit{\beta }={\mathit{\beta }}^{\top }\mathit{\beta }=A,$$

as desired.

$\left(3\right)\Rightarrow \left(1\right).$ Since $a_{ii}$ is a square element in R, there exists an element $r_i$ in R satisfying $r_i^2=a_{ii}$, for each $i=1,2,\dots ,k.$ By the linear independence of $(a_{11},\dots ,a_{kk})$ over R, it follows that $\langle r_1,\dots ,r_k\rangle =R$, meaning that there are elements $r_1^{\prime },\dots ,r_k^{\prime }$ in R such that $r_1^{\prime }{r}_1+\cdots +r_k^{\prime }{r}_k=1.$ Thus, from (3), we can obtain that

$$A={\mathit{\alpha }}^{\top }\mathit{\alpha },$$

where $\mathit{\alpha }=(r_1^{\prime },\dots ,r_k^{\prime })A.$ Furthermore, the linear independence of $\mathit{\alpha }$ follows directly from Lemma 3 and the fact that the vector $(a_{11},\dots ,a_{kk})$ is linearly independent over R. □

Remark 7.

Let $r_1,\dots ,r_k$ be elements in R such that $\langle r_1,\dots ,r_k\rangle =R.$ In Theorem 2, a family of elements $r_1^{\prime },\dots ,r_k^{\prime }$ in R is needed to satisfy the equation $r_1^{\prime }{r}_1+\cdots +r_k^{\prime }{r}_k=1$ in order to determine the corresponding vector α. The Chinese Remainder Theorem will once again play a crucial role in finding these elements. Additionally, if any of the $r_i$ is a unit in R, the process will be effortless. It is important to note that the vector α is necessary for our construction method to determine the corresponding $\mathit{x}$ and $Hull\left(\mathcal{C}\right)$ in Steps 2 and 4. Fortunately, Theorem 2 also provides a method for constructing the required vector α, as described above.

Let R be a commutative ring. If R is an Artinian ring with $J\left(R\right)=0$, then R is a semisimple ring. All semisimple commutative rings are also Frobenius rings. In the case of semisimple rings with characteristic 2, we have the following proposition:

Proposition 10.

Let R be a finite commutative semisimple ring with characteristic 2. Let A be a $k\times k$ matrix over R. Then, A takes the form ${\mathit{\alpha }}^{\top }\mathit{\alpha }$, where α is a vector in $R^k$, if and only if all the diagonal elements of A are square elements of R and $A={(r_1,\dots ,r_k)}^{\top }(r_1,\dots ,r_k)$, as long as $r_1,\dots ,r_k\in R$ satisfying $r_1^2,\dots ,r_k^2$ are exactly the diagonal elements of A.

Proof. 

It suffices to note that there are no non-zero nilpotent elements in R. □

Remark 8.

In addition to the advantage discussed in Remark 4, we would like to highlight another aspect. Our construction method can be seen as a supplement to the conventional method for producing self-dual codes. As mentioned in Section 1, different types of generation matrices M have been utilized in producing self-dual codes over finite commutative rings of characteristic 2. These matrices typically have half as many rows as columns and have rows that are linearly independent over R. Thus, the linear code $\mathcal{C}$ generated by M is self-dual only when $M{M}^{\top }=0.$ Hence, under the conventional approach, if matrix M fails to satisfy the condition $M{M}^{\top }=0$, it is commonly disregarded and not taken into account for the construction of self-dual codes. Nevertheless, our method offers a new idea. Specifically, if one can confirm through the aforementioned two results that the matrix $M{M}^{\top }$ takes the form ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for some R-linearly independent vector α, then the matrix M can still be utilized for the creation of self-dual codes. This positions our method as a significant supplement to the conventional approach, as demonstrated by the last three examples in this section.

However, if M takes the form $\left(I_k|\sigma \left(v\right)\right)$, where $\sigma$ has been defined in Section 2, v is an element in $R\left[G\right]$, and G is an even-order group (meaning k is even), then it is impossible to express $M{M}^{\top }$ as ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for any vector $\mathit{\alpha }$ that is linearly independent over R. This is due to the following more general result.

Theorem 3.

Let R be a finite commutative ring with characteristic 2. Consider matrix M in the form of $\left(I_k\right|A)$, where A is a $k\times k$ matrix over R. Let $\mathit{a}$ (resp. $\mathit{b}$) denote the vector that is the sum of all rows (resp. columns) of $A.$ Let $G=\{g_1=e_G,g_2,\dots ,g_k\}$ be a finite group with order k, and σ be the corresponding injection from the group ring $R\left[G\right]$ into $Ma{t}_k\left(R\right).$ Denote by $\mathfrak{S}$ the set of all elements $v\in R\left[G\right]$ such that the matrix $M=\left(I_k|\sigma \left(v\right)\right)$ satisfies the condition that $M{M}^{\top }$ takes the form ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for some R-linearly independent vector α in $R^k.$ The following statements hold:

(1)

If there exists a maximal ideal $\mathfrak{m}$ in R such that $\overline{\mathit{a}}={\overline{\mathit{b}}}^{\top }=(\overline{r},\dots ,\overline{r})$ for some element r in R, and k is an even number, then $M{M}^{\top }$ cannot be written in the form ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for any R-linearly independent vector α in $R^k.$ Here, the “bars” always refer to modulo $\mathfrak{m}$.

(2)

If there exists a maximal ideal $\mathfrak{m}$ in R such that $R/\mathfrak{m}$ is a binary field, and either $\overline{\mathit{a}}$ or ${\overline{\mathit{b}}}^{\top }$ is equal to $(\overline{r},\dots ,\overline{r})$ (relative to taking modulo $\mathfrak{m}$) for some element r in R, and k is an even number, then $M{M}^{\top }$ cannot be expressed as ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for any R-linearly independent vector α in $R^k.$

(3)

$\mathfrak{S}$ is a nonempty set if and only if k is an odd number. Furthermore, if k is odd, then

$$\mathfrak{S}=\{v\in R\left[G\right]|e_G+v{v}^{*}=u^2{\pi }_G\mathrm{for}\mathrm{some}u\in R^{*}\},$$

where ${\pi }_G:=g_1+g_2+\cdots +g_k\in R\left[G\right].$

Proof. 

$\left(1\right)$ Assume that $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$ for some R-linearly independent vector $\mathit{\alpha }$ in $R^k.$ By the conditions in $\left(1\right)$, there is a maximal ideal $\mathfrak{m}$ in R such that $\overline{\mathit{a}}={\overline{\mathit{b}}}^{\top }=(\overline{r},\dots ,\overline{r})$ for some element r in R. This implies that all the diagonal elements of $\overline{M}\ast {\overline{M}}^{\top }$ are equal to $\overline{1}+{\overline{r}}^2$, and therefore, the components of $\overline{\mathit{\alpha }}$ are equal to $\overline{1}+\overline{r}.$ However, according to Lemma 3, we know that $\overline{1}+\overline{r}\ne \overline{0}$, since $\mathit{\alpha }$ is linearly independent over $R.$ Now, let us consider the sum of all components of the first row of $\overline{A}\ast {\overline{A}}^{\top }$, denoted as $\Delta .$ Since k is even and all components of $\overline{\mathit{\alpha }}$ are equal to each other, we can conclude that $\Delta =\overline{1}.$ However, we can also express $\Delta$ as ${\overline{r}}^2$, since $\overline{\mathit{a}}={\overline{\mathit{b}}}^{\top }=(\overline{r},\dots ,\overline{r}).$ This leads to the equation ${\overline{r}}^2=\overline{1}$, which in turn, implies that $\overline{1}+\overline{r}=\overline{0}.$ This contradicts our previous statement that $\overline{1}+\overline{r}\ne \overline{0}.$ Therefore, we can conclude that $M{M}^{\top }$ cannot be written in the form ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for any R-linearly independent vector $\mathit{\alpha }$ in $R^k.$

$\left(2\right)$ Assuming $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }$ is an R-linearly independent vector in $R^k$, let $\mathcal{C}$ be the linear code generated by $M.$ It is clear that $\mathcal{C}$ is a free linear code with dimension $k.$ According to Proposition 2, we can show that ${\mathcal{C}}^{\perp }$ is also a free linear code with dimension $k.$ By applying Propositions 4 and 5, we can conclude that both $\mathcal{C}/Hull\left(\mathcal{C}\right)$ and ${\mathcal{C}}^{\perp }/Hull\left(\mathcal{C}\right)={\mathcal{C}}^{\perp }/Hull\left({\mathcal{C}}^{\perp }\right)$ are one-dimensional free R-modules. Noting that $\left(A^{\top }\right|I_k)$ is a generator matrix for ${\mathcal{C}}^{\perp }$, we can use Proposition 5 again to obtain

$$\left(I_k\right|A^{\top }){\left(I_k\right|A^{\top })}^{\top }=I_k+A^{\top }A={\mathit{\beta }}^{\top }\mathit{\beta },$$

where $\mathit{\beta }$ is an R-linearly independent vector in $R^k.$ This implies that we only need to consider the case where ${\overline{\mathit{b}}}^{\top }=(\overline{r},\dots ,\overline{r})$, where $r\in R.$ We will proceed with our proof based on this case. Since $I_k+A{A}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, taking modulo $\mathfrak{m}$, we can deduce that

$$\overline{I_k}+\overline{A}\ast {\overline{A}}^{\top }={\overline{\mathit{\alpha }}}^{\top }\overline{\mathit{\alpha }}.$$

Since ${\overline{\mathit{b}}}^{\top }=(\overline{r},\dots ,\overline{r})$, all the diagonal elements of $\overline{A}\ast {\overline{A}}^{\top }$ are equal to ${(\overline{1}+\overline{r})}^2.$ From this, we conclude that all the components of $\mathit{\alpha }$ are equal to $\overline{1}+\overline{r}.$ Furthermore, since $\mathit{\alpha }$ is linearly independent over R, we can deduce from Lemma 3 that $\overline{r}=\overline{0}.$ This leads us to the conclusion that the diagonal elements of matrix $\overline{A}\ast {\overline{A}}^{\top }$ are all $\overline{0}$, while all other components are $\overline{1}.$ Considering that k is an even number, we can obtain that the determinant $det\left(\overline{A}\right)=\overline{1}.$ However, by adding the last $k-1$ columns of $\overline{A}$ to the first column and taking into account the fact that ${\overline{\mathit{b}}}^{\top }=(\overline{r},\dots ,\overline{r})$ and $\overline{r}=\overline{0}$, we can determine that $det\left(\overline{A}\right)=\overline{0}.$ This is a contradiction, which proves that $M{M}^{\top }$ cannot be expressed as ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for any R-linearly independent vector $\mathit{\alpha }$ in $R^k.$

$\left(3\right)$ We will first prove the necessary and sufficient condition in $\left(3\right).$ On one hand, if $\mathfrak{S}$ is a nonempty set, then k must be an odd number according to $\left(1\right).$ On the other hand, we can take $A=\sigma \left(v\right)$, where $v=g_2+\cdots +g_k\in R\left[G\right].$ It is easy to verify that the matrix $M=\left(I_k\right|A)$ has the property that all the components of matrix $M{M}^{\top }$ are equal to 1 since k is an odd number. This means that $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }$ is a vector that has all its components equal to 1. Therefore, v belongs to $\mathfrak{S}$ by the definition of $\mathfrak{S}$, proving that $\mathfrak{S}$ is indeed a nonempty set.

Next, we will show that

$$\mathfrak{S}=\{v\in R\left[G\right]|e_G+v{v}^{*}=u^2{\pi }_G\mathrm{for}\mathrm{some}u\in R^{*}\}.$$

To conduct this, we only need to prove the inclusion “⊂”. Let $v\in \mathfrak{S}.$ By the definition of $\mathfrak{S}$, we know that the matrix $M=\left(I_k|\sigma \left(v\right)\right)$ satisfies the property that $M{M}^{\top }$ has the form ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for some R-linearly independent vector $\mathit{\alpha }=({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_k)$ in $R^k.$ Since

$$M{M}^{\top }=\sigma (e_G+v{v}^{*}),$$

it follows that the diagonal elements of $M{M}^{\top }$ are equal, and the sums of the components in each column of $M{M}^{\top }$ are also equal. This, combined with the facts that $\mathit{\alpha }$ is linearly independent over R and k is an odd number, allows us to conclude that

$$\left|\mathit{\alpha }\right|{\alpha }_1=\left|\mathit{\alpha }\right|{\alpha }_2=\cdots =\left|\mathit{\alpha }\right|{\alpha }_k$$

and $\left|\mathit{\alpha }\right|\in R^{*}$, and we can then obtain that ${\alpha }_1={\alpha }_2=\cdots ={\alpha }_k\in R^{*}.$ Set ${\alpha }_1=u$, where $u\in R^{*}.$ Since

$$\sigma (e_G+v{v}^{*})=M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$$

is a matrix with all its components being equal to $u^2$, it follows that

$$e_G+v{v}^{\ast }=u^2{\pi }_G,$$

and hence

$$\mathfrak{S}\subset \{v\in R\left[G\right]|e_G+v{v}^{\ast }=u^2{\pi }_G\mathrm{for}\mathrm{some}u\in R^{\ast }\}.$$

□

Example 2.

Let the notation be as in Theorem 3. Now, we consider the matrix $M=\left(I_2\right|A)$ over ${\mathbb{F}}_4$, where A is defined as

$$A=\left(\begin{array}{cc}\omega +1& 0\\ 1& \omega \end{array}\right).$$

Here, ${\mathbb{F}}_4$ denotes a quaternary field and ω is an element of ${\mathbb{F}}_4$ that satisfies the equation ${\omega }^2+\omega +1=0.$ It is easy to verify that $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }=(\omega ,\omega ).$ Clearly, α is linearly independent over ${\mathbb{F}}_4$, with $\mathit{a}=(\omega ,\omega )$ and $\mathit{b}=(\omega +1,\omega +1).$ Despite these conditions, it remains undeniable that $M{M}^{\top }$ takes the form ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for some linearly independent vector α in ${{\mathbb{F}}_4}^2.$ This example highlights the necessity of the condition $\overline{\mathit{a}}={\overline{\mathit{b}}}^{\top }$ in the first statement of Theorem 3. It also emphasizes the significance of $R/\mathfrak{m}$ being a binary field in the second statement of the theorem.

If the matrices A in Theorem 3 are block matrices containing four blocks, then we have the following result.

Theorem 4.

Let R be a finite commutative ring with characteristic $2.$ Let M be a $k\times 2k$ matrix in the form of $\left(I_k\right|A)$, where A is a block matrix of the form

$$A=\left(\begin{array}{cc}{A}_1& A_2\\ A_3& A_4\end{array}\right),$$

with $A_i$$(1\le i\le 4)$ representing $t\times t$ matrices over $R.$ Denote the sum of all rows (resp. columns) of $A_i$ as vector ${\mathit{a}}_i$ (resp. ${\mathit{b}}_i$) for each i. If there exists a maximal ideal $\mathfrak{m}$ in R such that $\overline{{\mathit{a}}_i}={\overline{{\mathit{b}}_i}}^{\top }=(\overline{r_i},\dots ,\overline{r_i})$ for each i, where $r_i$ is an element in R, and if t is an even number, then $M{M}^{\top }$ cannot be expressed as ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for any R-linearly independent vector α in $R^k.$ Here, the “bars” mean modulo $\mathfrak{m}.$

Proof. 

The proof follows a similar approach to the proof of $\left(1\right)$ in Theorem 3. Let us assume that $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$, where $\mathit{\alpha }=({\alpha }_1,\dots ,{\alpha }_t,{\alpha }_{t+1},\dots ,{\alpha }_k)$ is an R-linearly independent vector in $R^k.$ We know that there is a maximal ideal $\mathfrak{m}$ in R such that $\overline{{\mathit{a}}_i}={\overline{{\mathit{b}}_i}}^{\top }=(\overline{r_i},\dots ,\overline{r_i})$ for each $i.$ This means that all the diagonal elements of $\overline{A_i}\ast {\overline{A_i}}^{\top }$ are equal to ${\overline{r_i}}^2.$ Therefore, the first t diagonal elements of $\overline{M}\ast {\overline{M}}^{\top }$ are all $\overline{1}+{\overline{r_1}}^2+{\overline{r_2}}^2$, while the last t are all $\overline{1}+{\overline{r_3}}^2+{\overline{r_4}}^2.$ Using this, along with the equation $\overline{M}\ast {\overline{M}}^{\top }={\overline{\mathit{\alpha }}}^{\top }\overline{\mathit{\alpha }}$, we can conclude that

$$\overline{{\alpha }_1}=\cdots =\overline{{\alpha }_t}=\overline{1+r_1+r_2},$$

and

$$\overline{{\alpha }_{t+1}}=\cdots =\overline{{\alpha }_k}=\overline{1+r_3+r_4}.$$

Next, let us consider the sum of the first t components in the first row of $\overline{A}\ast {\overline{A}}^{\top }$, denoted by ${\Sigma }_1.$ Since t is an even number, we know that ${\Sigma }_1=\overline{1}.$ Additionally, ${\Sigma }_1$ is also equal to the inner product of the first row of $\overline{A}$ and the sum of the first t rows of $\overline{A}$, which is ${(\overline{r_1}+\overline{r_2})}^2.$ Thus, we can obtain that

$$\overline{{\alpha }_1}=\cdots =\overline{{\alpha }_t}=\overline{0}.$$

Moving on, we can also consider the sum ${\Sigma }_2$ of the last t components in the $(t+1)$-st row of $\overline{A}\ast {\overline{A}}^{\top }.$ Similarly, we can see that

$$\overline{{\alpha }_{t+1}}=\cdots =\overline{{\alpha }_k}=\overline{0}.$$

However, this leads to a contradiction due to the linear independence of the vector $\mathit{\alpha }.$ Specifically, according to Lemma 3, the ideal generated by all the components of $\mathit{\alpha }$ is R itself. This implies that not all of $\overline{{\alpha }_1},\dots ,\overline{{\alpha }_k}$ are equal to $\overline{0}$, which contradicts our previous conclusion. Hence, it is not possible for $M{M}^{\top }$ to be expressed as ${\mathit{\alpha }}^{\top }\mathit{\alpha }$ for an R-linearly independent vector $\mathit{\alpha }$ in $R^k.$ □

It is obvious that Theorem 4 can be naturally generalized to the case that the block matrix A is composed of more blocks.

According to Theorems 3 and 4, certain matrix constructions commonly used in the literature to generate self-dual codes over finite commutative rings with characteristic 2, such as those in [5,6], are not compatible with our construction method. This means that the $k\times k$ matrices A, which make up the right half of the matrices $M=\left(I_k\right|A)$, must undergo initial screening if our method is being considered for constructing self-dual codes over finite commutative rings with characteristic 2.

In the following examples, we will preliminarily examine which specific types of matrices A and M are suitable for our construction method.

Example 3.

Let $R={\mathbb{F}}_4+u{\mathbb{F}}_4$, where ${\mathbb{F}}_4$ is the quaternary field and $u^2=0.$ The ring R is a finite commutative chain ring with order 16 and characteristic $2.$ Its unique maximal ideal $\mathfrak{m}=\langle u\rangle$ satisfies $\mathfrak{m}\ne 0$ but ${\mathfrak{m}}^2=0.$ Therefore, by Propositions 8 and 9, we know that there are five self-dual codes of length 2 over $R:$

$$R\langle (1,1)\rangle ,R\langle (1,1+u)\rangle ,R\langle (1,1+u\omega )\rangle ,R\langle (1,1+u(1+\omega \left)\right)\rangle ,\mathrm{and}R\langle (u,0),(0,u)\rangle ,$$

where $\omega \in {\mathbb{F}}_4$ satisfies ${\omega }^2+\omega +1=0.$ Note, that the first four codes are free, while the last one is the unique nonfree self-dual code of length 2 over $R.$

Consider matrices M that have the form $\left(I_k\right|A)$, where A is a $k\times k$λ-circulant matrix over R and $k=6.$ Here, we restrict λ to be a member of $R^{*}.$ We will use this type of matrix M to produce self-dual codes of length 12 over $R.$

When using the conventional method, we need to find matrices M that satisfy the equation $M{M}^{\top }=0.$ As a result, we can construct 29,952 distinct self-dual codes, all of which are free linear codes.

When utilizing our construction approach, the first step involves searching for matrices M that satisfy the condition $M{M}^{\top }={\mathit{\alpha }}^{\top }\mathit{\alpha }$ for a vector α in $R^k$ that is R-linearly independent. By employing these matrices M, our method can construct a total of 2560 self-dual codes. Among these codes, 512 are nonfree, and out of the remaining 2048 free self-dual codes, 1568 cannot be produced with matrices of this type using the conventional method.

In total, by using both methods, we can construct a total of 32,032 self-dual codes of length 12 over $R.$

Example 4.

Let $R_1={\mathbb{F}}_2+u{\mathbb{F}}_2$, where ${\mathbb{F}}_2$ is the binary field and $u^2=0.$ We know that $R_1$ is a finite commutative chain ring of order 4 and characteristic 2, with a unique maximal ideal $\mathfrak{m}=\langle u\rangle$ such that $\mathfrak{m}\ne 0$ but ${\mathfrak{m}}^2=0.$ Similarly, by Propositions 8 and 9, there exist three self-dual codes of length 2 over $R_1:$

$$R_1\langle (1,1)\rangle ,R_1\langle (1,1+u)\rangle ,\mathrm{and}{R}_1\langle (u,0),(0,u)\rangle .$$

Among these self-dual codes, the first two are free, while the last one is the sole nonfree self-dual code in ${R_1}^2.$

We now use matrices M in the form of $\left(I_k\right|A)$ to construct self-dual codes with length 12 over $R_1.$ Here, $k=6$ and A is defined as

$$A=\left(\begin{array}{cccccc}{a}_0+b_0& a_1+b_1& a_2+b_2& a_3+b_3& a_4+b_4& a_5+b_5\\ a_5& a_0& a_1& a_2& a_3& a_4\\ a_4& a_5& a_0& a_1& a_2& a_3\\ a_3& a_4& a_5& a_0& a_1& a_2\\ a_2& a_3& a_4& a_5& a_0& a_1\\ a_1& a_2& a_3& a_4& a_5& a_0\end{array}\right),$$

which can be viewed as a variation of the circulant matrix.

Following the procedure outlined in the previous example, the conventional approach facilitates the generation of 384 self-dual codes. However, our method enables the production of 12,480 self-dual codes, with 192 being nonfree. It is important to stress that all the 384 self-dual codes produced by the conventional approach are free and can also be constructed using our proposed technique.

Therefore, by utilizing both methods simultaneously, a total of 12,480 self-dual codes with length 12 over $R_1$ can be constructed using matrices of the form $\left(I_k\right|A).$

Example 5.

Let $R_1$ be the ring described in the previous example. Next, we will consider the bordered matrix construction, which is defined as

$$M=\left(\begin{array}{cccc}{\gamma }_1& {\beta }_2\cdots {\beta }_2& {\gamma }_2& {\beta }_4\cdots {\beta }_4\\ {\beta }_1& & {\beta }_3& \\ ⋮& I_k& ⋮& B\\ {\beta }_1& & {\beta }_3\end{array}\right).$$

Here, ${\gamma }_1$, ${\gamma }_2$, ${\beta }_1$, ${\beta }_2$, ${\beta }_3$, and ${\beta }_4$ are elements of $R_1$, $k=5$, and B is a $k\times k$ circulant matrix over $R_1.$

By employing matrices M of this type, we can construct 720 self-dual codes utilizing our approach. Out of these, 480 are free linear codes, all of which can also be generated through conventional methods. However, it is noteworthy that the remaining 240 nonfree self-dual codes cannot be acquired via the conventional method.

All of the computational results mentioned above were obtained by conducting searches using PARI/GP (Version 2.15.3) [42].

These examples demonstrate that, in contrast to using solely the conventional approach, utilizing our method alongside the conventional approach can lead to the construction of a greater number of self-dual codes, as long as a suitable matrix construction is taken into account. This is because the chosen matrix construction could be more fully utilized by applying both methods simultaneously, as well as our method can construct nonfree self-dual codes over a finite commutative ring of characteristic 2 (provided a nonfree self-dual code of length 2 exists over the corresponding ring). As a result, this enhances the probability of discovering new extremal binary self-dual codes. Therefore, our technology serves as a valuable complement to the conventional approach.

Finally, we wish to emphasize that, besides the matrix types mentioned in the previous examples, all bordered matrix constructions presented in [12,13,14,17,19,21,22,23,24] are potentially applicable to our method. Nonetheless, further exploration is necessary to determine which other matrix types would also be suitable for our construction method to produce self-dual codes over finite commutative rings with characteristic 2.

6. Conclusions

In this paper, we presented a new method for constructing self-dual codes over finite commutative rings with characteristic 2, not necessarily Frobenius rings. Our approach differs from the conventional method and can be seen as a complementary approach. The main advantage of our method is its ability to construct nonfree self-dual codes over finite commutative rings, a task that is typically unachievable using the conventional approach. Therefore, it is possible to produce more self-dual codes by combining our method with the conventional approach, when a specific type of matrix is chosen to construct self-dual codes over a finite commutative ring of characteristic 2.

A natural direction for future work could be to employ specific matrix constructions from existing literature, particularly bordered matrix constructions, in conjunction with our methods and well-established construction techniques, such as the neighbor method, the building-up construction, and the lifting method, to discover new extremal binary self-dual codes.