Cyclic codes of odd length over ℤ₄ [u] / 〈u ^k〉

Abstract

Let $R=\mathbb{Z}_{4}[u]/ \langle u^k \rangle=\mathbb{Z}_{4}+u \mathbb{Z}_{4}+\ldots+u^{k-1}\mathbb{Z}_{4}$ ($u^{k}=0$), where k ≥ 2 is an positive integer. For any odd positive integer n, it is known that cyclic codes of length n over R are identified with ideals of the ring $R[x]/\langle x^{n}-1\rangle$. In this paper, an explicit representation for each cyclic code over R of length n is provided and a formula to count the number of codewords in each code is given. Then a formula to calculate the number of cyclic codes of length n over R is obtained. Precisely, the dual code of each cyclic code and self-dual cyclic codes of length n over R are investigated. As an application, some good quasi-cyclic codes of length 7k and index k over ℤ₄ are obtained from cyclic codes over R = ℤ₄ [u] / 〈u ^k〉 when k = 2, 3, 4.

Appendices

Appendix A: Proof of Theorem 3.7

In order to simplify the notations, we denote $\mathcal {Q}_{j}={\mathcal {K}}_{j}[u]/\langle u^{k}\rangle$ and ${\mathcal {R}}_{j}={\mathcal {F}}_{j}[u]/\langle u^{k}\rangle$. By Lemma 3.5(i), ${\mathcal {R}}_{j}$ is a finite chain ring with maximal ideal $u{\mathcal {R}}_{j}$, the nilpotency index of u is equal to k and the residue field of ${\mathcal {R}}_{j}$ is ${\mathcal {R}}_{j}/(u{\mathcal {R}}_{j})\cong {\mathcal {F}}_{j}$. Then by finite chain ring theory (cf. [9]) and $|{\mathcal {F}}_{j}|=2^{d_{j}}$, we deduce that all ideals of ${\mathcal {R}}_{j}$ are given by $\{0\}=u^{k}{\mathcal {R}}_{j}\subset u^{k-1}{\mathcal {R}}_{j}\subset \ldots \subset u{\mathcal {R}}_{j}\subset u^{0}{\mathcal {R}}_{j}={\mathcal {R}}_{j}$ and $|u^{l}{\mathcal {R}}_{j}|=|{\mathcal {F}}_{j}|^{k-l}=2^{d_{j}(k-l)}$. Moreover, we have $2\mathcal {Q}_{j}=2{\mathcal {R}}_{j}$ by Lemma 3.6 where we regard ${\mathcal {R}}_{j}$ as a subset of $\mathcal {Q}_{j}$.

Let C be any ideal of $\mathcal {Q}_{j}$. We define $\text {Tor}_{0}(C)=\tau (C)$ snd $\text {Tor}_{1}(C)=\tau (\{\alpha \in \mathcal {Q}_{j}\mid 2\alpha \in C\})$. Since $\tau$ is a surjective ring homomorphism from $\mathcal {Q}_{j}$ onto ${\mathcal {R}}_{j}$, both $\text {Tor}_{0}(C)$ and $\text {Tor}_{1}(C)$ are ideals of ${\mathcal {R}}_{j}$ satisfying $\text {Tor}_{0}(C)\subseteq \text {Tor}_{1}(C)$. Hence there is a unique pair $(i,s)$ of integers, $0\leq s\leq i\leq k$, such that

$$\text{Tor}_{0}(C)=u^{i}{\mathcal{R}}_{j} \ \text{and} \ \text{Tor}_{1}(C)=u^{s}{\mathcal{R}}_{j}.$$

Let $\tau |_{C}$ be the restriction of $\tau$ to C. Then $\tau |_{C}$ is a surjective ring homomorphism from C onto $\text {Tor}_{0}(C)$ with kernel Ker(τ| _C )=2Tor₁(C) , which implies |Ker(τ| _C )|=|Tor₁(C)|. Hence we deduce that

$$ |C|=|\text{Tor}_{0}(C)||\text{Tor}_{1}(C)|=|u^{i}{\mathcal{R}}_{j}||u^{s}{\mathcal{R}}_{j}|=2^{d_{j}(2k-(i+s))} $$

(10)

by the ring isomorphism theorems and Lemma 3.5(iii). Then we have the following cases.

• Case (i) $s=i$. In this case, we have $|C|=2^{2d_{j}(k-i)}$ by (10).

• (i-1) When $s=k$, then $i=k$ and $C=\{0\}=u^{k}\mathcal {Q}_{j}=\langle u^{k}\rangle$.

• (i-2) Let $0\leq i\leq k-1$. By $u^{i}\in u^{i}{\mathcal {R}}_{j}=\text {Tor}_{0}(C)$, there exists $\alpha \in \mathcal {Q}_{j}$ such that $u^{i}+2\alpha \in C$. Then by Lemma 3.6 we can assume that $\alpha \in {\mathcal {R}}_{j}$.

  It is obvious that $\langle u^{i}+2\alpha \rangle =\mathcal {Q}_{j}(u^{i}+2\alpha )\subseteq C$. Conversely, let $\xi \in C$. Since $\text {Tor}_{0}(C)=u^{i}{\mathcal {R}}_{j}$, by (6) and Lemma 3.6 there exist $\beta ,\gamma \in {\mathcal {R}}_{j}$ such that $\xi =u^{i}\beta +2\gamma$, which implies $2(\gamma -\alpha \beta )=\xi -(u^{i}+2\alpha )\beta \in C$ and so $\tau (\gamma -\alpha \beta )\in \text {Tor}_{1}(C)=u^{i}{\mathcal {R}}_{j}$. Hence $\tau (\gamma -\alpha \beta )=u^{i}\delta$ for some $\delta \in {\mathcal {R}}_{j}$. From this and by $\text {Ker}(\tau )=2{\mathcal {Q}}_{j}$, we deduce that $\gamma -\alpha \beta =u^{i}\delta +2w$ for some $w\in {\mathcal {Q}}_{j}$, which implies $2\gamma =2(\alpha \beta +u^{i}\delta +2w)=2\alpha \beta +2u^{i}\delta$. Therefore, ξ=(u ⁱ+2α)β+(u ⁱ+2α)⋅2δ=(u ⁱ+2α)(β+2δ)∈〈u ⁱ+2α〉. Hence $C=\langle u^{i}+2\alpha \rangle$.

• (i-2-1) When $i=0$, $1+2\alpha \in C$. By $(1+2\alpha )(1+2\alpha )=1$ in $\mathcal {Q}_{j}$, we see that $1+2\alpha$ is an invertible element of $\mathcal {Q}_{j}$, which implies $C=\mathcal {Q}_{j}=\langle u^{0}\rangle$.

• (i-2-2) Let $1\leq i\leq k-1$.

Since $\text {Tor}_{1}(C)=u^{i}{\mathcal {R}}_{j}$, we have $2u^{i}=2(u^{i}+2\alpha )\in C$. As $\alpha \in {\mathcal {R}}_{j}$, by Lemma 3.5(v) we can write $\alpha$ as $\alpha =u^{i}\beta$ or $\alpha =u^{t}h+u^{i}\beta$ for some $\beta \in {\mathcal {R}}_{j}$, $0\leq t\leq i-1$ and $h\in {\mathcal {R}}_{j}^{\times }$.

• ♢ When $\alpha =u^{i}\beta$, we have $u^{i}=u^{i}+2\alpha -(2u^{i})\beta \in C$, and so $C=\langle u^{i}\rangle$.

• ♢ Let $\alpha =u^{t}h+u^{i}\beta$. Then $u^{i}+2u^{t}h=u^{i}+2\alpha -(2u^{i})\beta \in C,$and hence $C=\langle u^{i}+2u^{t}h\rangle$. In this case, by $u^{k}=0$ we have $2u^{k-i+t}= u^{k-i}h^{-1}(u^{i}+2u^{t}h)\in C$ where $h^{-1}$ is the inverse of h in ${\mathcal {R}}_{j}$. Therefore, $u^{k-i+t}\in \text {Tor}_{1}(C)=u^{i}{\mathcal {R}}_{j}$, which implies $k-i+t\geq i$, and hence $t\geq 2i-k$.

Now, let $h_{1},h_{2}\in {\mathcal {R}}_{j}^{\times }$ and $0\leq t_{1},t_{2}\leq k-1$ satisfying $C=\langle u^{i}+2u^{t_{1}}h_{1}\rangle =\langle u^{i}+2u^{t_{2}}h_{2}\rangle$. Then $2(u^{t_{1}}h_{1}-u^{t_{2}}h_{2})=(u^{i}+2u^{t_{1}}h_{1})-(u^{i}+2u^{t_{2}}h_{2})\in C$, which implies $u^{t_{1}}h_{1}-u^{t_{2}}h_{2}\in \text {Tor}_{1}(C)=u^{i}{\mathcal {R}}_{j}$. From this we deduce that $t_{1}=t_{2}=t$ and $u^{t}(h_{1}-h_{2})\in u^{i}{\mathcal {R}}_{j}$. Then by finite chain ring theory (cf. [9]) and $u^{t}(h_{1}-h_{2})\in u^{i}{\mathcal {R}}_{j}$, it follows that $h_{1}\equiv h_{2}$ (mod $u^{i-t}$), i.e., $h_{1}=h_{2}$ as elements of $({\mathcal {R}}_{j}/(u^{i-t}{\mathcal {R}}_{j})^{\times }$. Finally, by ${\mathcal {R}}_{j}={\mathcal {F}}_{j}[u]/\langle u^{k}\rangle$ we have $({\mathcal {R}}_{j}/(u^{i-t}{\mathcal {R}}_{j})^{\times }={\mathcal {F}}_{j}[u]/\langle u^{i-t}\rangle$ up to a natural ring isomorphism.

As stated above, we conclude that all distinct ideals C of $\mathcal {Q}_{j}$ satisfying $\text {Tor}_{0}(C)=\text {Tor}_{1}(C)=u^{i}{\mathcal {R}}_{j}$ are given by (I) and (III) in the table.

• Case (ii) $i=k$ and $0\leq s\leq k-1$.

In this case, we have $\text {Tor}_{0}(C)=\{0\}$ and $|C|=2^{d_{j}(2k-(k+s))}=2^{d_{j}(k-s)}$ by (10). As $\text {Tor}_{1}(C)=u^{s} {\mathcal {R}}_{j}$, one can easily verify that all distinct ideals of $\mathcal {Q}_{j}$ in this case are given by: (II) $C=\langle 2u^{s}\rangle$ where $0\leq s\leq k-1$.

• Case (iii) $s=0$ and $1\leq i\leq k-1$.

  In this case, we have $|C|=2^{d_{j}(2k-i)}$ by (10). Moreover, by $1\in \text {Tor}_{1}(C)=u^{0} {\mathcal {R}}_{j}$ we conclude that $2=2u^{0}\in C$. Then by $\text {Tor}_{0}(C)=u^{i} {\mathcal {R}}_{j}$ it follows that $C=\langle u^{i},2\rangle$ immediately.

• Case (iv) $1\leq s<i\leq k-1$. In this case, $|C|=2^{d_{j}(2k-(i+s))}$ by (10).

  By $\text {Tor}_{0}(C)=u^{i} {\mathcal {R}}_{j}$ and $\text {Tor}_{1}(C)=u^{s} {\mathcal {R}}_{j}$ we have $2u^{s}\in C$ and there exists $\alpha \in {\mathcal {R}}_{j}$ such that $u^{i}+2\alpha \in C$, which implies $\langle u^{i}+2\alpha , 2u^{s}\rangle \subseteq C$. Conversely, let $\xi \in C$. By $\text {Tor}_{0}(C)=u^{i} {\mathcal {R}}_{j}$ there exist $\beta ,\gamma \in {\mathcal {R}}_{j}$ such that $\xi =u^{i}\beta +2\gamma$. Then from $2(\gamma -\alpha \beta )=\xi -(u^{i}+2\alpha )\beta \in C$, we deduce $\tau (\gamma -\alpha \beta )\in \text {Tor}_{1}(C)=u^{s} {\mathcal {R}}_{j}$, which implies $\gamma -\alpha \beta =u^{s} \delta +2w$ for some $\delta ,w\in {\mathcal {R}}_{j}$. By $2\gamma =2\alpha \beta +2u^{s} \delta$, it follows that $\xi =(u^{i}+2\alpha )\beta +(2u^{s})\delta \in \langle u^{i}+2\alpha , 2u^{s}\rangle$. Hence $C=\langle u^{i}+2\alpha , 2u^{s}\rangle$.

Furthermore, by $2u^{s}\in C$ and an argument similar to the proof of Case (i) we can assume that $\alpha =0$ or $\alpha =u^{t}h$, where $h\in {\mathcal {R}}_{j}^{\times }$ and $0\leq t\leq s-1$.

• (iv-1) When $\alpha =0$, $C=\langle u^{i}, 2u^{s}\rangle$ which is given by (V) in the table.

• (iv-2) Let $C=\langle u^{i}+2u^{t}h, 2u^{s}\rangle$ where $h\in {\mathcal {R}}_{j}^{\times }$ and $0\leq t\leq s-1$.

Assume $C=\langle u^{i}+2u^{t_{1}}h_{1}, 2u^{s}\rangle =\langle u^{i}+2u^{t_{2}}h_{2}, 2u^{s}\rangle$ where $0\leq t_{1},t_{2}\leq s-1$ and $h_{1},h_{2}\in {\mathcal {R}}^{\times }$. Then $2(u^{t_{1}}h_{1}-u^{t_{2}}h_{2})=(u^{i}+2u^{t_{1}}h_{1})-(u^{i}+2u^{t_{2}}h_{2})\in C$, which implies $u^{t_{1}}h_{1}-u^{t_{2}}h_{2}\in \text {Tor}_{1}(C)=u^{s} {\mathcal {R}}_{j}$. From this and by an argument similar to the proof of Case (i), we deduce that $t_{1}=t_{2}=t$ and $h_{1}\equiv h_{2}$ (mod $u^{s-t} {\mathcal {R}}_{j}$). By the latter condition, we have $h=h_{1}=h_{2}\in ({\mathcal {R}}_{j}/(u^{s-t}{\mathcal {R}}_{j}))^{\times }$. From this and by ${\mathcal {R}}_{j}/(u^{s-t}{\mathcal {R}}_{j})={\mathcal {F}}_{j}[u]/\langle u^{s-t}\rangle$, we deduce that all distinct ideals of $\mathcal {Q}_{j}$ in this case are given by: $C=\langle u^{i}+2u^{t}h, 2u^{s}\rangle$ where $h\in ({\mathcal {F}}_{j}[u]/\langle u^{s-t}\rangle )^{\times }$.

As $2u^{k-i+t}=u^{k-i}h^{-1}(u^{i}+2u^{t}h)\in C$, we have $u^{k-i+t}\in \text {Tor}_{1}(C)=u^{s} {\mathcal {R}}_{j}$, which implies $k-i+t\geq s$. So we have one of the following two cases:

• (♢-1) $k-i+t=s$, i.e., $s-t=k-i$.

  In this case, by $2u^{s}=u^{k-i}h^{-1}(u^{i}+2u^{t}h)$ we have $C=\langle u^{i}+2u^{t}h\rangle$ and $2i>i+s=k+t$, i.e., $t<2i-k$.

  Furthermore, we have $h\in ({\mathcal {F}}_{j}[u]/\langle u^{k-i}\rangle )^{\times }$ and $|C|=2^{d_{j}(2k-(i+s))} =2^{d_{j}(2k-(k+t))}=2^{d_{j}(k-t)}$. Hence C is given by (IV) in the table.

• (♢-2) $k-i+t>s$, i.e., $i+s\leq k+t-1$. In this case, C is given by (VI) in the table.

  As stated above, we conclude that all distinct ideals and the number of elements in each ideal of $\mathcal {Q}_{j}$ are given by (I)–(VI) of the table.

It is obvious that the number of ideals in (I), (II) and (V) is equal to $k+1$, k and $\frac {1}{2}k(k-1)$ respectively. Then we count the number of ideals in (III), (IV) and (V) respectively. In order to simplify the notation, we denote $q=2^{d_{j}}$ in the following.

First, we count the number of ideals in (III). Let $0\leq t<i\leq k-1$ and $t\geq 2i-k$. By Lemma 3.5(iv), it follows that $|(\mathcal {F}_{j}[u]/\langle u^{i-t}\rangle )^{\times }|=(2^{d_{j}}-1)(2^{d_{j}})^{(i-t)-1}=(q-1)q^{i-t-1}$. Then we have one of the following two cases:

• (♢-1) k is even. In this case, $t\geq 2i-k$ if and only if i and t satisfy one of the following two conditions:

• (♢-1-1) $i\leq \frac {k}{2}$, i.e., $2i\leq k$, and $0\leq t\leq i-1$. In this case, the number of ideals is equal to

  $$N_{1}=\sum\limits_{i=1}^{\frac{k}{2}}\sum\limits_{t=0}^{i-1}|(\mathcal{F}_{j}[u]/\langle u^{i-t}\rangle)^{\times}|=\sum\limits_{i=1}^{\frac{k}{2}}\sum\limits_{t=0}^{i-1}(q-1)q^{i-t-1}=\frac{q^{\frac{k}{2}+1}-1}{q-1}-(\frac{k}{2}+1).$$

• (♢-1-2) $i\geq \frac {k}{2}+1$, i.e., $2i> k$, and $2i-k\leq t\leq i-1$. In this case, the number of ideals is equal to

  $$N_{2}=\sum\limits_{i=1}^{\frac{k}{2}}\sum\limits_{t=0}^{i-1}|(\mathcal{F}_{j}[u]/\langle u^{i-t}\rangle)^{\times}|=\sum\limits_{i=\frac{k}{2}+1}^{k-1}\sum\limits_{t=2i-k}^{i-1}(q-1)q^{i-t-1} =\frac{q^{\frac{k}{2}}-1}{q-1}-\frac{k}{2}.$$

  Therefore, the number of ideals in (III) is equal to ${\Omega }_{1}(q,k)=N_{1}+N_{2}=\frac {q^{\frac {k}{2}+1}+ q^{\frac {k}{2}}-2}{q-1}-(k+1)$, where k is even.

• (♢-2) k is odd. In this case, $t\geq 2i-k$ if and only if i and t satisfy one of the following two conditions:

• (♢-2-1) $i\leq \frac {k-1}{2}$, i.e., $2i\leq k$, and $0\leq t\leq i-1$. Then the number of ideals is equal to $N_{1}={\sum }_{i=1}^{\frac {k-1}{2}}{\sum }_{t=0}^{i-1}(q-1)q^{i-t-1}=\frac {q^{\frac {k+1}{2}}-1}{q-1}-\frac {k+1}{2}$.

• (♢-2-2) $i\geq \frac {k-1}{2}+1$, i.e., $2i> k$, and $2i-k\leq t\leq i-1$. Then the number of ideals is equal to $N_{2}={\sum }_{i=\frac {k-1}{2}+1}^{k-1}{\sum }_{t=2i-k}^{i-1}(q-1)q^{i-t-1} =\frac {q^{\frac {k+1}{2}}-1}{q-1}-\frac {k+1}{2}$.

  Therefore, the number of ideals in (III) is equal to ${\Omega }_{1}(q,k)=N_{1}+N_{2}=\frac {2q^{\frac {k+1}{2}}-2}{q-1}-(k+1)$, where k is odd.

  Next we count the number of ideals in (IV). Let $0\leq t<i\leq k-1$ and $t<2i-k$. By Lemma 3.5(iv), it follows that $|(\mathcal {F}_{j}[u]/\langle u^{k-i}\rangle )^{\times }|=(q-1)q^{k-i-1}$. Then we have one of the following two cases:

• (♢-1) k is even. Then $t<2i-k$ if and only if $2i>k$, i.e., $i\geq \frac {k}{2}+1$, and $0\leq t\leq 2i-k-1$. Hence the number of ideals in (IV) is equal to

  $$\begin{array}{@{}rcl@{}} {\Omega}_{2}(q,k)&=&\sum\limits_{i=\frac{k}{2}+1}^{k-1}\sum\limits_{t=0}^{2i-k-1}|(\mathcal{F}_{j}[u]/\langle u^{k-i}\rangle)^{\times}|=\sum\limits_{i=\frac{k}{2}+1}^{k-1}\sum\limits_{t=0}^{2i-k-1} (q-1)q^{k-i-1}\\ &=&(q-1)\sum\limits_{i=\frac{k}{2}+1}^{k-1}(2i-k)q^{k-i-1}. \end{array} $$

• (♢-2) k is odd. Then $t<2i-k$ if and only if $2i>k$, i.e., $i\geq \frac {k+1}{2}$, and $0\leq t\leq 2i-k-1$. Hence the number of ideals in (IV) is equal to

  $${\Omega}_{2}(q,k)=\sum\limits_{i=\frac{k+1}{2}}^{k-1}\sum\limits_{t=0}^{2i-k-1} (q-1)q^{k-i-1} =(q-1)\sum\limits_{i=\frac{k+1}{2}}^{k-1}(2i-k)q^{k-i-1}.$$

Finally, we count the number of ideals in (VI). By Lemma 3.5(iv), we have $|(\mathcal {F}_{j}[u]/\langle u^{s-t}\rangle )^{\times }|=(q-1)q^{s-t-1}$. So the number of ideals in (VI) is equal to

$$\sum\limits_{t=0}^{k-4}\sum\limits_{t+1\leq s\leq \frac{k+t}{2}-1}\sum\limits_{s+1\leq i\leq k+t-1-s}|(\mathcal{F}_{j}[u]/\langle u^{s-t}\rangle)^{\times}| =(q-1){\Gamma}(q,k),$$

where ${\Gamma }(q,k)={\sum }_{t=0}^{k-4}{\sum }_{t+1\leq s\leq \frac {k+t}{2}-1}{\sum }_{s+1\leq i\leq k+t-1-s}q^{s-t-1}$. When $k\geq 5$,

$$\begin{array}{@{}rcl@{}} {\Gamma}(q,k)&=&\sum\limits_{t=1}^{k-4}\sum\limits_{t+1\leq s\leq \frac{k+t}{2}-1}{\sum}_{s+1\leq i\leq k+t-1-s}q^{s-t-1} +\sum\limits_{1\leq s\leq \frac{k}{2}-1}\sum\limits_{s+1\leq i\leq k-1-s}q^{s-1}\\ &=&\sum\limits_{t^{\prime}=0}^{(k-1)-4}\sum\limits_{t^{\prime}+1\leq s^{\prime}\leq \frac{(k-1)+t^{\prime}}{2}-1}\sum\limits_{s^{\prime}+1\leq i^{\prime}\leq (k-1)+t^{\prime}-1-s^{\prime}}q^{s^{\prime}-t^{\prime}-1} \ \ \ \ \ \ \ \\ &&+\sum\limits_{s=1}^{\lfloor \frac{k}{2}\rfloor-1}(k-2s-1)q^{s-1}\\ &=&{\Gamma}(q,k-1)+\sum\limits_{s=1}^{\lfloor \frac{k}{2}\rfloor-1}(k-2s-1)q^{s-1}. \end{array} $$

If $1\leq k\leq 3$, there is no triple $(t,s,i)$ of integers satisfying $0\leq t<s<i\leq k-1$ and $i+s\leq k+t-1$. In this case, the number of ideals in (VI) is equal to 0. Then we set ${\Gamma }(q,k)=0$ for $k=1,2,3$.

If $k=4$, there is a unique triple $(t,s,i)=(0,1,2)$ of integers satisfying $0\leq t<s<i\leq k-1$ and $i+s\leq k+t-1$. In this case, all distinct ideals in (VI) are given by $\langle u^{2}+2h,2u\rangle$, where $h\in ({\mathcal {F}}_{j}[u]/\langle u\rangle )^{\times }={\mathcal {F}}_{j}^{\times }$ and $|{\mathcal {F}}_{j}^{\times }|=q-1$. Then we set ${\Gamma }(q,4)=1$.

Therefore, the number $N_{(2,d_{j},k)}$ of ideals of $\mathcal {Q}_{j}$ is equal to

$$\begin{array}{@{}rcl@{}} N_{(2,d_{j},k)}&=&k+1+k+{\Omega}_{1}(q,k)+{\Omega}_{2}(q,k)+\frac{1}{2}k(k-1)+(q-1){\Gamma}(q,k)\\ &=&1+\frac{1}{2}k(k+3)+{\Omega}_{1}(q,k)+{\Omega}_{2}(q,k)+(q_{j}-1){\Gamma}(q,k). \end{array} $$

Then by a direct calculation, we deduce that $N_{(2,d_{j},2\rho )}={\sum }_{i=0}^{\rho }(1+4i)2^{(\rho -i)d_{j}}$ and $N_{(2,d_{j},2\rho +1)}={\sum }_{i=0}^{\rho }(3+4i)2^{(\rho -i)d_{j}}$ for any positive integer $\rho$.

Appendix B: Proof of Theorem 4.5

Let $1\leq j\leq r$, and $E_{j}$ be an ideal of ${\mathcal {K}}_{j}[u]/\langle u^{k}\rangle$ given by one of the following eight cases:

• (i) $E_{j}=\langle u^{k-i}\rangle$, if $C_{j}=\langle u^{i}\rangle$, where $0\leq i\leq k$.

• (ii) $E_{j}=\langle u^{k-s},2\rangle$, if $C_{j}=\langle 2u^{s}\rangle$, where $0\leq s\leq k-1$.

• (iii) $E_{j}=\langle u^{k-i}+2u^{k+t-2i}h(x)\rangle$, if $C_{j}=\langle u^{i}+2u^{t}h(x)\rangle$ where $h(x)\in ({\mathcal {F}}_{j}[u]/\langle u^{i-t}\rangle )^{\times }$, $0\leq t<i\leq k-1$ and $t\geq 2i-k$.

• (iv) $E_{j}=\langle u^{i}+2h(x)\rangle$, if $C_{j}=\langle u^{i}+2h(x)\rangle$ where $h(x)\in ({\mathcal {F}}_{j}[u]/\langle u^{k-i}\rangle )^{\times }$, $0<i\leq k-1$ and $2i>k$.

• (v) $E_{j}=\langle u^{i-t}+2h(x),2u^{k-i}\rangle$, if $C_{i}=\langle u^{i}+2u^{t}h(x)\rangle$ where $h(x)\in ({\mathcal {F}}_{j}[u]/\langle u^{k-i}\rangle )^{\times }$, $1\leq t<i\leq k-1$ and $t<2i-k$.

• (vi) $E_{j}=\langle u^{k-s},2u^{k-i}\rangle$, if $C_{j}=\langle u^{i},2u^{s}\rangle$, where $0\leq s<i\leq k-1$.

• (vii) $E_{j}=\langle u^{k-s}+2u^{k-i-s}h(x)\rangle$, if $C_{j}=\langle u^{i}+2h(x), 2u^{s}\rangle$, where $h(x)\in ({\mathcal {F}}_{j}[u]/\langle u^{s}\rangle )^{\times }$, $1\leq s<i\leq k-1$, $i+s\leq k-1$.

• (viii) $E_{j}=\langle u^{k-s}+2u^{k+t-i-s}h(x), 2u^{k-i}\rangle$, when $C_{j}=\langle u^{i}+2u^{t}h(x), 2u^{s}\rangle$, where $h(x)\in ({\mathcal {F}}_{j}[u]/\langle u^{s-t}\rangle )^{\times }$, $1\leq t<s<i\leq k-1$, $i+s\leq k+t-1$.

Then by a direct calculation we deduce that $C_{j}E_{j}=\{a(x)b(x)\mid a(x)\in C_{j}, \ b(x)\in E_{j}\}=\{0\}$. Further, we have $|C_{j}||E_{j}|=2^{2d_{j}k}$ by Theorem 3.7.

Since $\sigma _{j}$ is a ring isomorphism from ${\mathcal {K}}_{j}[u]/\langle u^{k}\rangle$ onto ${\mathcal {K}}_{\sigma (j)}[u]/\langle u^{k}\rangle$ define by Lemma 4.4(ii), $\sigma _{j}(E_{j})$ is an ideal of ${\mathcal {K}}_{\sigma (j)}[u]/\langle u^{k}\rangle$ and $|\sigma _{j}(E_{j})|=|E_{j}|$. We denote $D_{\sigma (j)}=\sigma _{j}(E_{j})$. Let $l=\sigma (j)$. Then $j=\sigma (l)$ as $\sigma ^{-1}=\sigma$.

Now, set ${\mathcal {D}}={\sum }_{j=1}^{r}{\Psi }(e_{\sigma (j)}(x) D_{\sigma (j)}) =\bigoplus _{l=1}^{r}{\Psi }(e_{l}(x)D_{l})$ , where D _l = σ _σ(l)(E _σ(l)) is an ideal of ${\mathcal {K}}_{l}[u]/\langle u^{k}\rangle$ since $\sigma _{\sigma (l)}=\sigma _{l}^{-1}: \mathcal {K}_{\sigma (l)}[u]/\langle u^{k}\rangle \rightarrow \mathcal {K}_{l}[u]/\langle u^{k}\rangle$ . By Theorem 3.4, we see that ${\mathcal {D}}$ is a cyclic code of length n over R.

For any $1\leq j\leq r$, by Lemma 3.2(iii) we deduce that $e_{\sigma (j)}(x) D_{\sigma (j)}$ is an ideal of ${\mathcal {A}}_{\sigma (j)}[u]/\langle u^{k}\rangle$. Then we have $\sigma (D_{\sigma (j)})=\sigma ^{2}(E_{j})=E_{j}$ and $\sigma (e_{\sigma (j)}(x))=e_{j}(x)$ by Lemma 4.2(iv). Using Lemma 4.3 we deduce

$$\begin{array}{@{}rcl@{}} {\mathcal{C}}\cdot \sigma({\mathcal{D}})&=&(\sum\limits_{j=1}^{r}{\Psi}(e_{j}(x)C_{j}))\cdot \sigma({\sum}_{l=1}^{r}{\Psi}(e_{l}(x)D_{l}))\\ &=&{\Psi}(\sum\limits_{j=1}^{r}(e_{j}(x)C_{j})\cdot \sigma(e_{\sigma(j)}(x)D_{\sigma(j)})) ={\Psi}(\sum\limits_{j=1}^{r}(e_{j}(x)C_{j}) (e_{j}(x)E_{j}))\\ &=&{\Psi}(\sum\limits_{j=1}^{r}e_{j}(x)(C_{j} E_{j})) =\{0\}, \end{array} $$

which implies ${\mathcal {D}}\subseteq {\mathcal {C}}^{\bot _{E}}$ by Lemma 4.1. On the other hand, by Theorems 3.4 and 3.7, we have that $|{\mathcal {C}}||{\mathcal {D}}|=({\prod }_{j=1}^{r}|C_{j}|)({\prod }_{j=1}^{r}|E_{j}|)={\prod }_{j=1}^{r}(|C_{j}||E_{j}|) ={\prod }_{j=1}^{r}2^{2d_{j}k} =2^{2k{\sum }_{j=1}^{r}d_{j}}=(4^{k})^{n}=|R^{n}|$. Since $R=\mathbb {Z}_{4}[u]/\langle u^{k}\rangle$ is a finite Frobenius ring, by the theory of linear codes over Frobenius rings (See Dougherty et al. [7]) we deduce that ${\mathcal {C}}^{\bot _{E}}={\mathcal {D}}$.

Finally, we give the generator set of $D_{\sigma (j)}=\sigma _{j}(E_{j})$, $1\leq j\leq r$. By Lemma 4.4, we have $\sigma _{j}(u^{l})=u^{l}$ for all $0\leq l\leq k$, and for any $h(x)\in {\mathcal {F}}_{j}[u]/\langle u^{k}\rangle$

$$\sigma_{j}(h(x))=h(x^{-1})\equiv h(x^{n-1}) \ (\text{mod} \ \overline{f}_{\sigma(j)}(x)),$$

which implies

$$\sigma_{j}(2u^{l}h(x))=2u^{l}h(x^{-1}) \ \text{with} \ h(x^{-1})\in {\mathcal{F}}_{\sigma(j)}[u]/\langle u^{k}\rangle.$$

Then the conclusions follows from (i)–(viii) immediately.