A note on complete classification of $(\delta+\alpha u^2)$-constacyclic codes of length $p^k$ over $\F_{p^m}+u\F_{p^m}+u^2\F_{p^m}$

TL;DR

This paper classifies and analyzes the structure of a specific class of constacyclic codes over a finite ring, focusing on their self-duality properties, for codes of length a power of a prime.

Contribution

It provides a complete classification of $( ext{delta}+ ext{alpha} u^2)$-constacyclic codes over a finite ring of length $p^k$, including their self-dual variants.

Findings

Complete classification of the codes

Characterization of self-dual codes

Structural insights into the code algebra

Abstract

For units $\delta$ and $\alpha$ in $\F_{p^m}$, the structure of $(\delta+\alpha u^2)$-constacyclic codes of length $p^k$ over $\mathbb{F}_{p^m}+u\mathbb{F}_{p^m}+u^2\mathbb{F}_{p^m}$ is studied and self-dual $(\delta+\alpha u^2)$-constacyclic codes are analyzed.

Equations

$$f_0(x)=(x-1{)}^a+u(x-1{)}^{t}g(x)+u^{2}h(x),$$

$$f_1(x)=u(x-1{)}^b+u^{2}r(x),$$

$$f_2(x)=u^2(x-1{)}^c,$$

$$\mathcal{A}:=\{C\in \mathcal{I}|c=0\mathrm{and}a+b\le p^k\},$$

$${\mathcal{A}}^{\prime }:=\{C\in \mathcal{I}|c=0\mathrm{and}a+b\ge p^k\},$$

$$\mathcal{B}:=\{C\in \mathcal{I}|a=p^k\mathrm{and}b+c\le p^k\},$$

$${\mathcal{B}}^{\prime }:=\{C\in \mathcal{I}|a=p^k\mathrm{and}b+c\ge p^k\},$$

$$\mathcal{C}:=\{C\in \mathcal{I}|a<p^k,c>0\mathrm{and}a+c\le p^k\},$$

$${\mathcal{C}}^{\prime }:=\{C\in \mathcal{I}|a<p^k,c>0\mathrm{and}a+c\ge p^k\}.$$

$$C=⟨⟨f_0(x),f_1(x),f_2(x)⟩⟩,$$

$$f_0(x)=(x-1{)}^7+u(x-1{)}^2(1+(x-1))$$

$$f_1(x)=u(x-1{)}^4+u^2(1-(x-1))$$

$$f_2(x)=u^2(x-1{)}^2.$$

$$Ann(C)=⟨⟨f_0^{\prime }(x),f_1^{\prime }(x),f_2^{\prime }(x)⟩⟩,$$

$$f_0^{\prime }(x)=(x-1{)}^{p^k-c}+u(x-1{)}^{a-c-t}[-r(x)]+u^2[-l(x)-(x-1{)}^{p^k-a-c}h(x)],$$

$$f_1^{\prime }(x)=u(x-1{)}^{a-t}+u^2[-g(x)],$$

$$f_2^{\prime }(x)=u^2(x-1{)}^{p^k-a}.$$

$$F_0^{\prime }(x):=(x-1{)}^{2^k-c}+u(x-1{)}^{2^{k-1}-c}{g}^{\prime }(x)+u^2{h}^{\prime }(x),$$

$$F_1^{\prime }(x):=u(x-1{)}^{2^{k-1}}+u^2{r}^{\prime }(x),$$

$$F_2^{\prime }(x):=u^2(x-1{)}^{2^k-a}.$$

$$f(x)\cdot g(x)=\sum _{i=0}^{p^k-1}{f}_i{g}_i.$$

$$C^{\perp }=\{f(x)\in S|f(x)\cdot g(x)=0\mathrm{for}\mathrm{all}g(x)\in C\}.$$

$$2^k-a+t=T_1(C)=T_1(C^{\perp })=T_1(\mathrm{Ann}(C))=a-t,$$

$$a+c=T_0(C)+T_2(C)=T_0(C^{\perp })+T_2(C^{\perp })=T_2(\mathrm{Ann}(C))+T_0(\mathrm{Ann}(C))=2^k-a+2^k-c,$$

$$f_0(x)=(x+1{)}^{2^{k-1}+t}+u(x+1{)}^{t}g(x)+u^{2}h(x),$$

$$f_1(x)=u(x+1{)}^{2^{k-1}}+u^{2}r(x),$$

$$f_2(x)=u^2(x+1{)}^{2^{k-1}-t},$$

$$G(x)=(\alpha g^{-1}(x){)}^{\ast }\mathrm{mod}(x+1{)}^{2^{k-1}-t},$$

$$H(x)=[\alpha (x+1{)}^t+x^{2^{k-1}+t}{h}^{\ast }(x)+q(x)g^{\ast }(x)]\mathrm{mod}(x+1{)}^{2^{k-1}-t},$$

$$g^{\ast }(x)$$

$$\displaystyle\left\{\begin{array}[]{ll}\sigma_{0}=g^{2}_{0}=\alpha\\ \sigma_{1}=0\\ \sigma_{2}=\sigma_{3}=g_{1}(g_{0}+g_{1})=0\\ \sigma_{4}=(g_{0}+g_{1})(g_{1}+g_{2})+g_{2}^{2}=0\\ \ \ \ \ \ \ \ \vdots\end{array}\right.$$

$$\displaystyle\begin{array}[]{lll}2^{k-1}-t=2&\Longrightarrow&g(x)=\sqrt{\alpha}+\beta(x+1)\\ 2^{k-1}-t\in\{3,4\}&\Longrightarrow&g(x)=\sqrt{\alpha}+\beta(x+1)^{2}\ {\rm or}\ g(x)=\sqrt{\alpha}x+\beta(x+1)^{2},\end{array}$$

$$h(x)=h_0+h_1(x+1)+\cdots +h_{2^{k-1}-t-1}(x+1{)}^{2^{k-1}-t-1}$$

$$\displaystyle M(2^{k},t):=\left(\begin{array}[]{cccccc}0&0&\dots&\dots&0\\ {{2^{k-1}+t}\choose 1}&0&\ddots&\ddots&0\\ {{2^{k-1}+t}\choose 2}&{{2^{k-1}+t-1}\choose 1}&\ddots&\ddots&0\\ \vdots&\vdots&\mbox{}&\ddots&\vdots\\ {2^{k-1}+t}\choose{2^{k-1}-t-1}&{2^{k-1}+t-1}\choose{2^{k-1}-t-2}&\dots&\dots&0\\ \end{array}\right).$$

$$u^{2}M(2^k,t)(h_0,h_1,\cdots ,h_{2^{k-1}-t-1}{)}^{tr}=u^2(d_0,d_1,\cdots ,d_{2^{k-1}-t-1}{)}^{tr},$$

$$M(2^k,t)(h_0,h_1,\cdots ,h_{2^{k-1}-t-1}{)}^{tr}=(d_0,d_1,\cdots ,d_{2^{k-1}-t-1}{)}^{tr},$$

Full text

1: Department of Mathematics, University of Isfahan, 81746-73441 Isfahan, Iran.

2: School of Mathematics, Institute for Research in Fundamental Sciences (IPM), 19395-5746 Tehran, Iran.

3: School of Mathematics, Hefei University of Technology, Hefei 230009, Anhui, P.R.China.

111E-mail addresses: r.sobhani@sci.ui.ac.ir (R. Sobhani), sunzhonghuas@163.com (Z. Sun), liqiwangg@163.com (L. Wang), zhushixin@hfut.edu.cn (S. Zhu).

Keywords: Constacyclic codes, Self-dual codes, Torsion codes.

1 Introduction

The class of constacyclic codes is an important class of linear codes in coding theory, which can be viewed as a generalization of cyclic codes. Let $R$ be a finite chain ring, a nonempty subset $C$ of $R^{n}$ is called a linear code of length $n$ over $R$ if $C$ is an $R$-submodule of $R^{n}$. For any unit $\lambda$ in the ring $R$, $C$ is said to be $\lambda$-constacyclic if $(\lambda c_{n-1},c_{0},\dots,c_{n-2})\in C$ for all $(c_{0},c_{1},\dots,c_{n-1})\in C$. If an $n$-tuple $(c_{0},c_{1},\dots,c_{n-1})$ is identified with the polynomial $c_{0}+c_{1}x+\dots+c_{n-1}x^{n-1}$ in the ring $R[x]/{\left<x^{n}-\lambda\right>}$, then $\lambda$-constacyclic codes of a given length $n$ over $R$ are in correspondence with ideals of the ring $R[x]/{\left<x^{n}-\lambda\right>}$. Special classes of repeated-root constacyclic codes over certain finite chain rings have been studied by numerous authors.

For any $s\geq 2$, let $R$ be the ring ${\mathbb{F}}_{p^{m}}[u]/{\left<u^{s}\right>}$. The ring $R$ has been widely used as alphabets in certain constacyclic codes (See, for example, [1, 2, 7, 9, 10, 13, 15] and the references therein). In general, it seems to be difficult to classify all constacyclic codes over $R$ and only some constacyclic codes of certain lengths are classified yet. Dinh [4] classified all constacyclic codes of length $2^{s}$ over the Galois extension rings of ${\mathbb{F}}_{2}[u]/{\left<u^{2}\right>}$ and gave their detailed structure. Later, he classified and gave all constacyclic codes of length $p^{s}$ over ${\mathbb{F}}_{p^{m}}[u]/{\left<u^{2}\right>}$ in [5]. Recently, Dinh et al. [8] classified negacyclic codes of $2p^{s}$ over ${\mathbb{F}}_{p^{m}}[u]/{\left<u^{2}\right>}$, analyzed the form of dual codes due to each type and identified self-dual constacyclic codes. Chen et al. [3] classified all constacyclic codes of length $2p^{s}$ over ${\mathbb{F}}_{p^{m}}[u]/{\left<u^{2}\right>}$. Also $(1+\alpha u)$-constacyclic codes of arbitrary length $n$ over ${\mathbb{F}}_{p}[u]/{\left<u^{s}\right>}$ have been studied in details in [9], where $\alpha$ is a unit element in ${\mathbb{F}}_{p}[u]/{\left<u^{s}\right>}$.

From the above studies, we could see that little work had been done on repeated-root $(\delta+\alpha u^{i})$-constacyclic codes over $R$, where $s\geq 3$ and $i\geq 2$. Recently, the first author considered $(\delta+\alpha u^{2})$-constacyclic codes of length $p^{k}$ over ${\mathbb{F}}_{p^{m}}[u]/{\left<u^{3}\right>}$ and classified all $(\delta+\alpha u^{2})$-constacyclic codes of length $p^{k}$ over such ring in [14]. However, we have discovered a wrong conclusion claiming that a certain polynomial $g(x)$ is constant, in the proof of Theorem 8 in [14]. Even though, that polynomial may not be constant, the results before that theorem in [14] are all correct and those in the continuation of that theorem are correct only for all constant such polynomials. Hence the classification of the corresponding codes (for the non-constant case) and specially self-dual codes, would be still incomplete. In this paper we review some of the results presented in [14] and provide a complete classification for $(\delta+\alpha u^{2})$-constacyclic codes of length $p^{k}$ over ${\mathbb{F}}_{p^{m}}[u]/{\left<u^{3}\right>}$. We also try to classify self-dual such codes and obtain some results in this respect.

2 $(1+\alpha u^{2})$-Constacyclic codes

In this section we take a review to some structural results presented in [14]. Let $R$ be the ring ${\mathbb{F}}_{p^{m}}[u]/{\left<u^{3}\right>}$, $\alpha$ be a nonzero element in ${\mathbb{F}}_{p^{m}}$ and $S$ be the ring $R[x]/\langle x^{p^{k}}-(1+\alpha u^{2})\rangle$. Let $\overline{S}$ be the ring ${\mathbb{F}}_{p^{m}}[x]/\langle x^{p^{k}}-1\rangle$ and $\mu:S\longrightarrow\overline{S}$ be the map which sends $f(x)$ to $f(x)\bmod u$. For an ideal $C$ in $S$ and $0\leq i\leq 2$, define ${\rm Tor}_{i}(C)$ to be $\mu(\{f(x)\in S|u^{i}f(x)\in C\})$ which is an ideal of the ring $\overline{S}$ and we call it the $i$-th torsion code of $C$. Clearly we have ${\rm Tor}_{0}(C)\subseteq{\rm Tor}_{1}(C)\subseteq{\rm Tor}_{2}(C)$. Hence, related to $C$, there are integers $a\geq b\geq c$ such that ${\rm Tor}_{0}(C)={\left<(x-1)^{a}\right>}$, ${\rm Tor}_{1}(C)={\left<(x-1)^{b}\right>}$ and ${\rm Tor}_{2}(C)={\left<(x-1)^{c}\right>}$. The following theorem is a variation of Theorem 2 in [14].

Theorem 1

Let $C$ be an ideal of $S$ and $a,b,c$ are as above. Then $C$ is uniquely generated by the polynomials

[TABLE]

in $C$, where $g(x),h(x)$ and $r(x)$ are polynomials in ${\mathbb{F}}_{p^{m}}[x]$ with the property that $g(x)=0$ or $g(x)$ is a unit in $\overline{S}$ with $t+{\rm deg}(g(x))<b$, $r(x)=0$ or ${\rm deg}(r(x))<c$ and $h(x)=0$ or ${\rm deg}(h(x))<c$. $\square$

If $C$ is an ideal with the unique generators described in Theorem 1, we write $C={\left<{\left<f_{0}(x),f_{1}(x),f_{2}(x)\right>}\right>}$. Let ${\mathcal{I}}$ be the set of all ideals of $S$. In [14], the set ${\mathcal{I}}$ has been divided to the following subsets.

[TABLE]

We refer to [14] for the structure of ideals in ${\mathcal{A}}\cup{\mathcal{A}}^{\prime}\cup{\mathcal{B}}\cup{\mathcal{B}}^{\prime}$. Here we only review the structure of ideals in ${\mathcal{C}}\cup{\mathcal{C}}^{\prime}$. Let $C$ be an ideal of $S$. The annihilator of $C$, denoted by $Ann(C)$, is defined to be the set $\{f(x)|f(x)g(x)=0,\ for\ all\ g(x)\in C\}$. Clearly, $Ann(C)$ is also an ideal of $S$. As declared in [14], the map $\eta:{\mathcal{C}}\longrightarrow{\mathcal{C}}^{\prime}$ sending $C$ to $Ann(C)$ is a bijection and hence we only need to determine ideals in ${\mathcal{C}}$. The following theorem is indeed [14, Theorem 7].

Theorem 2

Let $C={\left<{\left<f_{0}(x),f_{1}(x),f_{2}(x)\right>}\right>}$ be an element of ${\mathcal{C}}$. Then we must have the following:

• (I)

$b=p^{k}-a+t$,

• (II)

$g(x)$ is a unit and $r(x)=\alpha g^{-1}(x)\bmod{\left<u,(x-1)^{c}\right>}$,

• (III)

when $t\neq 2a-p^{k}-t$ we have $c\leq{\rm min}\{t,2a-p^{k}-t\}$. $\square$

In Theorem 8 of [14] it is claimed that for ideals in ${\mathcal{C}}$, the polynomial $g(x)$ is a constant polynomial. This claim is not true in general and $g(x)$ may be non-constant. In fact, the conclusion made at lines 4 and 5 at page 131 in [14] is wrong. To be more accurate, in the following, we present a counter-example.

Example 1

Let $R={\mathbb{F}}_{3}[u]/{\left<u^{3}\right>}$, $S=R[x]/{\left<x^{9}-(1+u^{2})\right>}$ and

[TABLE]

where

[TABLE]

It is easy to verify that $C$ is in the unique form and we are done.

In the following theorem we provide a correct version of Theorems 8, 9 and 10 in [14].

Theorem 3

Let $C={\left<{\left<f_{0}(x),f_{1}(x),f_{2}(x)\right>}\right>}$ be an element of ${\mathcal{C}}$. Then

If $1\leq c\leq{\rm min}\{t,2a-p^{k}-t\}$ then we must have $\lceil(p^{k}+2)/2\rceil\leq a\leq p^{k}-1$, $1\leq t\leq 2a-p^{k}-1$, $a+c\leq p^{k}$, $g(x)\in\overline{S}$ is a unit with ${\rm deg}(g(x))<p^{k}-a$, $r(x)=\alpha g^{-1}(x)\bmod(x-1)^{c}$ and $h(x)\in\overline{S}$ is a polynomial with ${\rm deg}(h(x))<c$.

• 2)

If $c>{\rm min}\{t,2a-p^{k}-t\}$ then we must have $p=2$, $0\leq t<2^{k-2}$, $a=t+2^{k-1}$, $t<c\leq 2^{k-1}-t$, $g(x)\in\overline{S}$ is a unit with ${\rm deg}(g(x))<2^{k-1}-t$, $r(x)=\alpha g^{-1}(x)\bmod(x-1)^{c}$, $r(x)=g(x)\bmod(x-1)^{c-t}$ and $h(x)\in\overline{S}$ is a polynomial with ${\rm deg}(h(x))<c$.

Moreover, in both cases, if $g(x)r(x)=\alpha+(x-1)^{c}l(x)$ then we have

[TABLE]

where

[TABLE]

[TABLE]

[TABLE]

**Proof.  **First, we assume that $c\leq{\rm min}\{t,2a-p^{k}-t\}$. Note that, since $c\geq 1$, we can not have $t=0$ and also we can not have $2a=p^{k}+t$. Hence $t\geq 1$ and $2a\geq p^{k}+t+1\geq p^{k}+2$ and consequently $a\geq\lceil\frac{p^{k}+2}{2}\rceil$. Also we have $1\leq t\leq 2a-p^{k}-1$. Other conditions comes from Theorems 1 and 2. Also, if $D:={\left<f^{\prime}_{0}(x),f^{\prime}_{1}(x),f^{\prime}_{2}(x)\right>}$ one can easily conclude that $CD=0$ and hence similar arguments as those used in [14] shows that $D=Ann(C)$ and both $C$ and $D$ are in the unique form. Now, if $c>{\rm min}\{t,2a-p^{k}-t\}$ then according to Theorem 2 we must have $t=2a-p^{k}-t$ and hence $p=2$ and $a=2^{k-1}+t$. Clearly, in this case, the representation of ${\rm Ann}(C)$ must has the form ${\left<{\left<F^{\prime}_{0}(x),F^{\prime}_{1}(x),F^{\prime}_{2}(x)\right>}\right>}$, where

[TABLE]

From $0=F_{0}(x)F^{\prime}_{1}(x)$ we can conclude that $g^{\prime}(x)=r(x)$ and from $0=F^{\prime}_{0}(x)F_{1}(x)$ we can deduce that $r^{\prime}(x)=g(x)$. Now, $0=F_{0}(x)F^{\prime}_{0}(x)$ implies that $r(x)=g(x)\bmod(x-1)^{c-t}$. Now, setting $D:={\left<f^{\prime}_{0}(x),f^{\prime}_{1}(x),f^{\prime}_{2}(x)\right>}$, the remaining of the proof is similar to that for the first part. $\blacksquare$

Remark 1

As it was proved in [13], the above obtained results can be extended to $(\delta+\alpha u^{2})$-constacyclic codes of length $p^{k}$ over $R$.

3 Self-dual $(1+\alpha u^{2})$-constacyclic codes

In this section we try to find Euclidean self-dual $(1+\alpha u^{2})$-constacyclic codes of length $p^{k}$. For two polynomials $f(x)=\sum_{i=0}^{p^{k}-1}f_{i}x^{i}$ and $g(x)=\sum_{i=0}^{p^{k}-1}g_{i}x^{i}$ in $S$ we define the inner product of $f(x)$ and $g(x)$ to be

[TABLE]

With respect to this inner product, the dual of an ideal $C$ of $S$, denoted by $C^{\perp}$, is defined as

[TABLE]

The ideal $C$ is said to be self-dual, if $C=C^{\perp}$. To determine self-dual codes we need the following definition.

Definition 1

Let $f(x)=\sum_{i=0}^{p^{k}-1}{a_{i}x^{i}}$ be an element of $S$. The reciprocal of $f(x)$, denoted $f^{*}(x)$, is the element $\sum_{i=0}^{p^{k}-1}{a_{i}x^{p^{k}-i}}$ of $S$. For any subset $E$ of $S$, the set $\{e^{*}\ |\ e\in E\}$ is denoted by $E^{*}$.

We know from [14, Lemma 1] that $C^{\perp}={\rm Ann}(C)^{*}$. Also, we know from [14] that, self-dual $(1+\alpha u^{2})$-constacyclic codes exist only in the case $p=2$ and the only self-dual ideal in ${\mathcal{A}}\cup{\mathcal{A}}^{\prime}\cup{\mathcal{B}}\cup{\mathcal{B}}^{\prime}$ is the ideal ${\left<{\left<u(x-1)^{2^{k-1}},u^{2}\right>}\right>}$.

Now, let $C$ be a self-dual ideal in ${\mathcal{C}}$. Since we must have

[TABLE]

hence we have $t=a-2^{k-1}$. Also we have

[TABLE]

and hence $a+c=2^{k}$. Therefore a self-dual ideal $C$ must belong to ${\mathcal{C}}\cap{\mathcal{C}^{\prime}}$. By now, a self-dual ideal $C$ in $S$ must has the representation $C={\left<{\left<f_{0}(x),f_{1}(x),f_{2}(x)\right>}\right>}$, where

[TABLE]

[TABLE]

[TABLE]

$0\leq t<2^{k-1}$, $g(x)\in\overline{S}$ is invertible with ${\rm deg}(g(x))<2^{k-1}-t$, $r(x)=\alpha g^{-1}(x)\bmod(x+1)^{2^{k-1}-t}$ and $h(x)\in\overline{S}$ is a polynomial with ${\rm deg}(h(x))<2^{k-1}-t$. Moreover if $t<2^{k-2}$ then, additionally, we must have $g(x)=r(x)\bmod(x+1)^{2^{k-1}-t}$. From now on, we denote such an ideal $C$ with $Ideal(k,t,g(x),h(x))$. It can be deduced from $C^{\perp}={\rm Ann}(C)^{*}$ and some simple calculations that $C^{\perp}=Ideal(k,t,G(x),H(x))$, where

[TABLE]

[TABLE]

and $q(x)$ is such that $(\alpha g^{-1}(x))^{*}=(x+1)^{2^{k-1}-t}q(x)+G(x)$. As a consequence of the above discussions and with the same notation, we have the following theorem.

Theorem 4

Let $k$ be a positive integer, $0\leq t<2^{k-1}$ and $C=Ideal(k,t,g(x),h(x))$ be an element of $S$. Then $C$ is a self-dual ideal, if and only if, all the following conditions hold.

$g(x)g^{*}(x)=\alpha\bmod(x+1)^{2^{k-1}-t}$.

• 2)

if $t<2^{k-2}$, then $g^{2}(x)=\alpha\bmod(x+1)^{2^{k-1}-2t}$.

• 3)

$H(x)=h(x)$. $\square$

The first step to find self-dual ideals of $S$ is to find polynomials $g(x)\in{\mathbb{F}}_{p^{m}}[x]$ for which we have ${\rm deg}(g(x))<2^{k-1}-t$ and $g(x)g^{*}(x)=\alpha\bmod(x+1)^{2^{k-1}-t}$. Clearly, the constant polynomial $g(x)=\sqrt{\alpha}$, considered in [14], is one of such polynomials. Write $g(x)=\sum_{i=0}^{2^{k-1}-t-1}g_{i}(x+1)^{i}$. Then we have

[TABLE]

For $0\leq s\leq 2^{k-1}-t-1$, set $\sigma_{s}:=\sum_{i=0}^{s}\sum_{j=0}^{s-i}{{2^{k}-j}\choose s-i-j}g_{i}g_{j}$. The first condition in Theorem 4 now becomes $\sigma_{0}=\alpha$, i.e. $g_{0}=\sqrt{\alpha}$, and for $1\leq s\leq 2^{k-1}-t-1$, $\sigma_{s}=0$. This is a nonlinear system of equations and solving it is not an easy task! To estimate the form of solutions, let us assume that $g(x)$ has the form $g_{0}+g_{1}(x+1)+g_{2}(x+1)^{2}$, i.e. $g_{i}=0$ for $i\geq 3$. In this case, the equations become

[TABLE]

Therefore, if $2^{k-1}-t\geq 5$ then $g(x)=\sqrt{\alpha}$, $g(x)=\sqrt{\alpha}x$ and $g(x)=\sqrt{\alpha}x^{2}$ are the only solutions for the system of equations. Other solutions corresponding to the case $2^{k-1}-t\leq 4$, rather than those of the form given above, are as follows:

[TABLE]

where, $\beta\in{\mathbb{F}}_{2^{m}}$. As we mentioned, obtaining all solutions are difficult but we can now conjecture that, if we restrict the degree of $g(x)$ to the positive integer $d$ then there exist a positive integer $D$ depending on $d$ such that for all $k,t$ for which $2^{k-1}-t>D$, the polynomials $g(x)=\sqrt{\alpha}x^{s}$, are the only solutions for the system of equations. From this motivation, in what follows, we classify self-dual codes for which the polynomial $g(x)$ has the form $g(x)=\sqrt{\alpha}x^{s}$, for some $0\leq s\leq 2^{k-1}-t$. Let $w=w(s)$ be such that when $s\neq 0$, $2^{w}\|s$ and when $s=0$, $w=k-2$. We now have the following theorem.

Theorem 5

Let $k$ be a positive integer, $0\leq t<2^{k-1}$, $0\leq s<2^{k-1}-t$, $h(x)\in\overline{S}$ with ${\rm deg}(h(x))<2^{k-1}-t$ and $C=Ideal(k,t,\sqrt{\alpha}x^{s},h(x))$ be an element of $S$. Then $C$ is a self-dual ideal, if and only if, one of the following conditions hold.

$2^{k-2}\leq t<2^{k-1}$ and $x^{2^{k-1}+t}h^{*}(x)=h(x)\bmod(x+1)^{2^{k-1}-t}$.

• 2)

$2^{k-2}-2^{w}\leq t<2^{k-2}$ and $[\alpha(x+1)^{t}+x^{2^{k-1}+t}h^{*}(x)]=h(x)\bmod(x+1)^{2^{k-1}-t}$.

**Proof.  **The first part follows from Theorem 4 and the facts that $\alpha(x+1)^{t}=0\bmod(x+1)^{2^{k-1}-t}$ and $q(x)=0$. For the second part, note that if $t<2^{k-2}$ then according to the second part of Theorem 4 we must have $g^{2}(x)=\alpha\bmod(x+1)^{2^{k-1}-2t}$ and hence we must have $(x+1)^{2^{k-1}-2t}\mid(x^{2s}+1)$. But, if we write $s=2^{w}s^{\prime}$ with $s^{\prime}$ odd, then we have $x^{2s}+1=(x^{s^{\prime}}+1)^{2^{w+1}}$. Hence $(x+1)^{2^{k-1}-2t}\mid(x^{2s}+1)$ if and only if $2^{k-1}-2t\leq 2^{w+1}$ or equivalently, $t\geq 2^{k-2}-2^{w}$. The remaining of the proof follows from the third part of Theorem 4 and the fact that $q(x)=0$. $\blacksquare$

Corollary 1

Let $k$ be a positive integer, $0\leq t<2^{k-1}$ and $1\leq s<2^{k-1}-t$. If $C=Ideal(k,t,\sqrt{\alpha}x^{s},h(x))$ is a self-dual ideal then we must have $s\leq 2^{k-2}$.

**Proof.  **If $t\geq 2^{k-2}$ then clearly we must have $s<2^{k-2}$. If $t<2^{k-1}$ then according to the second part of Theorem 5, we must have $t\geq 2^{k-2}-2^{w}$, where $2^{w}\|s$. But if $s>2^{k-2}$ then we have $2^{k-2}-2^{w}>2^{k-1}-s-1$ implying $s+t>2^{k-1}-1$ which is a contradiction. Hence we always must have $s\leq 2^{k-2}$ and the proof is completed. $\blacksquare$

Let us denote by $N(2^{k},t,s)$ the number of self-dual ideals of $S$ of the form $C=Ideal(k,t,\sqrt{\alpha}x^{s},h(x))$, where

[TABLE]

is a polynomial in $\overline{S}$. Recall that $M(2^{k},t)$ is the following matrix defined in [11, 14]:

[TABLE]

Now, from Theorem 5, we have $C=Ideal(k,t,\sqrt{\alpha}x^{s},h(x))$ is self-dual, if and only if

[TABLE]

where $d_{i}=\alpha$ if $i=t$ and $d_{i}=0$ otherwise. But Equation (4) has solutions for $(h_{0},h_{1},\cdots,h_{2^{k-1}-t-1})$, if and only if

[TABLE]

has solutions for $(h_{0},h_{1},\cdots,h_{2^{k-1}-t-1})$ in ${\mathbb{F}}_{2^{m}}$. Now we have the following theorem:

Theorem 6

Let ${\mathcal{K}}$ be the nullity of $M(2^{k},2^{k-1}-t)$ over ${\mathbb{F}}_{2^{m}}$ Then we have

[TABLE]

$\square$

On the other hand, it has been proved in [12] that Equation (5) always has solutions for $(h_{0},h_{1},\cdots,h_{2^{k-1}-t-1})$ in ${\mathbb{F}}_{2^{m}}$ except when $t=0$. Also it has been shown there that ${\mathcal{K}}=\lceil\frac{2^{k-1}-t+1}{2}\rceil$. Therefore we have

[TABLE]

Hence we have the following corollary.

Corollary 2

Let $N(2^{k})$ denote the number of self-dual ideals of $S$ having the form $Ideal(k,t,\sqrt{\alpha}x^{s},h(x))$. We have

[TABLE]

$\square$

As described before, when we restrict the degree of $g(x)$ to a non-negative integer $d$ then, for some limited values of $t$, there might exist some choises for $g(x)$ rather than those of the form $\sqrt{\alpha}x^{s}$. Let us complete the classification of self-dual ideals of $S$ for which $g(x)=g_{0}+g_{1}(x+1)+g_{2}(x+1)^{2}$ has degree at most $2$. We now only need to consider the following three cases

[TABLE]

But, conditions 2 and 3 in Theorem 4 imply that there is no self-dual ideal in the cases $(I)$ and $(III)$ . Also, in the case $(II)$, an easy computation gives us the following self-dual ideals:

[TABLE]

where $g(x)=\sqrt{\alpha}+\beta(x+1)^{2}$, $h(x)=h_{0}+h_{1}(x+1)+h_{2}(x+1)^{2}$, $h_{0}=\alpha+\sqrt{\alpha}\beta+\beta^{2}$, and $\beta,h_{1},h_{2}\in{\mathbb{F}}_{2^{m}}$ with $\beta\notin\{0,\sqrt{\alpha}\}$.

[TABLE]

where $g(x)=\sqrt{\alpha}x+\beta(x+1)^{2}$, $r(x)=g(x)+\sqrt{\alpha}(x+1)^{2}$, $h(x)=h_{0}+h_{1}(x+1)+h_{2}(x+1)^{2}$, $h_{0}=\beta^{2}+\alpha$ and $\beta,h_{1},h_{2}\in{\mathbb{F}}_{2^{m}}$ with $\beta\neq 0$.

[TABLE]

where $g(x)=\sqrt{\alpha}+\beta(x+1)^{2}$, $h(x)=h_{0}+h_{1}(x+1)+h_{2}(x+1)^{2}$, $h_{0}=\sqrt{\alpha}\beta+\beta^{2}$, $k\geq 4$ and $\beta,h_{1},h_{2}\in\mathbb{F}_{2^{m}}$ with $\beta\notin\{0,\sqrt{\alpha}\}$.

[TABLE]

where $g(x)=\sqrt{\alpha}x+\beta(x+1)^{2}$, $h(x)=h_{0}+h_{1}(x+1)+h_{2}(x+1)^{2}$, $h_{0}=\beta^{2}$, $k\geq 4$ and $\beta,h_{1},h_{2}\in{\mathbb{F}}_{2^{m}}$ with $\beta\neq 0$.

4 Acknowledgments

The research of the first author was in part supported by a grant from IPM (No. 94050080). The research of the third author was in part supported by Fundamental Research Funds through the Central Universities (No.JZ2015HGBZ0499, JZ2016HGXJ0089), and the research of the fourth author was in part supported by the National Natural Science Foundation of China(No.61370089).