An explicit expression for all distinct self-dual cyclic codes of length $p^k$ over Galois ring $\mathrm{GR}(p^2,m)$

Abstract

Let p be any odd prime number and let m, k be arbitrary positive integers. The construction for self-dual cyclic codes of length $p^k$ over the Galois ring $\mathrm{GR}(p^2,m)$ is the key to construct self-dual cyclic codes of length $p^kn$ over the integer residue class ring ${\mathbb {Z}}_{p^2}$ for any positive integer n satisfying $\mathrm{gcd}(p,n)=1$. So far, existing literature has only determined the number of these self-dual cyclic codes (Des Codes Cryptogr 63:105–112, 2012). In this paper, we give an efficient construction for all distinct self-dual cyclic codes of length $p^k$ over $\mathrm{GR}(p^2,m)$ by using column vectors of Kronecker products of matrices with specific types. On this basis, we further obtain an explicit expression for all these self-dual cyclic codes by using binomial coefficients.

Appendix: The construction of self-dual codes in Example 1

In this appendix, we illustrate how to construct all distinct self-dual cyclic codes length $3^3$ over $\mathrm{GR}(3^2,m)$ by use of Theorem 2. First, we have

$$\begin{aligned}&G_3=\left( \begin{array}{ccc} 1 &{} 0 &{} 0\\ -1 &{} -1 &{} 0 \\ 1 &{} -1 &{} 1\end{array}\right) , \quad G_3-I_3=\left( \begin{array}{ccc} 0 &{} 0 &{} 0\\ -1 &{} \mathbf{1} &{} 0 \\ 1 &{} \mathbf{2} &{} 0\end{array}\right) ,\quad G_9=\left( \begin{array}{ccc} G_3 &{} 0 &{} 0\\ -G_3 &{} -G_3 &{} 0 \\ G_3 &{} -G_3 &{} G_3\end{array}\right) ;\\&G_9-I_9 = \left( \begin{array}{ccc} G_3-I_3 &{} 0 &{} 0\\ -G_3 &{} -G_3-I_3 &{} 0 \\ G_3 &{} -G_3 &{} G_3-I_3\end{array}\right) =\left( \begin{array}{ccc|ccc|ccc} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ -1 &{} \mathbf{1} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 1 &{} 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ \hline -1 &{} 0 &{} 0 &{} \mathbf{1} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 1 &{} 1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ -1 &{} 1 &{} -1 &{} 2 &{} 1 &{} \mathbf{1} &{} 0 &{} 0 &{} 0 \\ \hline 1 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ -1 &{} -1 &{} 0 &{} 1 &{} 1 &{} 0 &{} -1 &{} \mathbf{1} &{} 0 \\ 1 &{} -1 &{} 1 &{} -1 &{} 1 &{} 2 &{} 1 &{} 2 &{} 0\end{array}\right) ;\\&G_{27} = \left( \begin{array}{ccc} G_9 &{} 0 &{} 0\\ -G_9 &{} -G_9 &{} 0 \\ G_9 &{} -G_9 &{} G_9\end{array}\right) \quad \hbox {and}\quad G_{18}-I_{18}=\left( \begin{array}{cc} G_9-I_9 &{} 0 \\ -G_9 &{} -G_9-I_9 \end{array}\right) , \end{aligned}$$

where $-1=2$ and $-2=1$ (mod 3). Then by $2^{-1}=2$ (mod 3), we have the following conclusions:

$\bullet$ The codes is Case II of Example 1 are constructed as follows:

• $3^m$ codes: $\langle (u-1)^{25}+3b(u),3(u-1)^2\rangle$, where

  • $\bullet$ $b(u)=2(u-1)^7+b^{[1,3)}(u)$,

  • $\bullet$ $b^{[1,3)}(u)=b_1+b_2(u-1)$, where $\left( \begin{array}{c}b_1 \\ b_2 \end{array}\right) =a_2\xi _2^{[1,3)} =a_2\left( \begin{array}{c}{} \mathbf{1} \\ 2 \end{array}\right) =\left( \begin{array}{c}a_2 \\ 2a_2 \end{array}\right)$ and $a_2\in {\mathbb {F}}_{3^m}$ arbitrary.

• $(3^m)^2$ codes: $\langle (u-1)^{23}+3b(u),3(u-1)^4\rangle$, where

  • $\bullet$ $b(u)=2(u-1)^5+b^{[3,7)}(u)$,

  • $\bullet$ $b^{[3,7)}(u)=b_3+b_4(u-1)+b_5(u-1)^2+b_6(u-1)^3$, where $\left( \begin{array}{c}b_3 \\ b_4 \\ b_5 \\ b_6\end{array}\right) =a_4\xi _4^{[3,7)}+a_6\xi _6^{[3,7)} =a_4\left( \begin{array}{c}{} \mathbf{1} \\ 1 \\ 2 \\ 2\end{array}\right) +a_6\left( \begin{array}{c}0 \\ 0 \\ \mathbf{1} \\ 0\end{array}\right) =\left( \begin{array}{c} a_4\\ a_4\\ 2a_4+a_6\\ 2a_4 \end{array}\right)$ and $a_4,a_6\in {\mathbb {F}}_{3^m}$ arbitrary.

• $(3^m)^3$ codes: $\langle (u-1)^{21}+3b(u),3(u-1)^6\rangle$, where

  • $\bullet$ $b(u)=2(u-1)^3+b^{[5,11)}(u)$,

  • $\bullet$ $b^{[5,11)}(u)=\sum _{j=5}^{10}b_j(u-1)^{j-5}$, where

    $$\begin{aligned} \left( \begin{array}{c} b_5 \\ b_6 \\ b_7 \\ b_8 \\ b_9 \\ b_{10}\end{array}\right)= & {} a_6\xi _6^{[5,11)}+a_8\xi _8^{[5,11)}+a_{10}\xi _{10}^{[5,11)} = \left( \begin{array}{c|c|c} \mathbf{1} &{} 0 &{} 0 \\ 0 &{} 0 &{} 0\\ 0 &{} \mathbf{1} &{} 0\\ 2 &{} 2 &{} 0 \\ 0 &{} 0 &{} \mathbf{1} \\ 0 &{} 0 &{} 1\end{array}\right) \left( \begin{array}{c} a_6 \\ a_8 \\ a_{10}\end{array}\right) \\= & {} (a_6, 0, a_8, 2a_6+2a_8, a_{10}, a_{10})^{\mathrm{tr}} \end{aligned}$$

    and $a_6,a_8,a_{10}\in {\mathbb {F}}_{3^m}$ arbitrary.

• $(3^m)^4$ codes: $\langle (u-1)^{19}+3b(u),3(u-1)^8\rangle$, where

  • $\bullet$ $b(u)=2(u-1)+b^{[7,15)}(u)$,

  • $\bullet$ $b^{[7,15)}(u)=\sum _{j=7}^{14}b_j(u-1)^{j-7}$, where

    $$\begin{aligned}&(b_7,b_8,b_9,b_{10},b_{11},b_{12},b_{13},b_{14})^{\mathrm{tr}} \\= & {} \left( \xi _8^{[7,15)},\xi _{10}^{[7,15)},\xi _{12}^{[7,15)},\xi _{14}^{[7,15)}\right) \left( \begin{array}{c} a_8 \\ a_{10} \\ a_{12}\\ a_{14}\end{array}\right) =\left( \begin{array}{c|c|c|c} \mathbf{1} &{} 0&{} 0&{} 0\\ 2 &{} 0 &{} 0 &{} 0 \\ 0 &{} \mathbf{1} &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 &{} 0 \\ 0 &{} 2 &{} \mathbf{1} &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 2 &{} 0 &{} \mathbf{1} \\ 0 &{} 1 &{} 1 &{} 2\end{array}\right) \left( \begin{array}{c} a_8 \\ a_{10} \\ a_{12}\\ a_{14}\end{array}\right) \\= & {} (a_8, 2a_8, a_{10}, a_{10}, 2a_{10}+a_{12}, a_{10}, 2a_{10}+a_{14}, a_{10}+a_{12}+2a_{14})^{\mathrm{tr}} \end{aligned}$$

    and $a_8,a_{10},a_{12},a_{14}\in {\mathbb {F}}_{3^m}$ arbitrary.

$\bullet$ The codes is Case III of Example 1 are constructed as follows:

• $3^m$ codes: $\langle (u-1)^{24}+3b(u),3(u-1)^3\rangle$, where

  • $\bullet$ $b(u)=2(u-1)^6+b^{[3,5)}(u)$,

  • $\bullet$ $b^{[3,5)}(u)=b_3(u-1)+b_4(u-1)^2$, where $\left( \begin{array}{c}b_3 \\ b_4 \end{array}\right) =a_4\xi _4^{[3,5)} =a_4\left( \begin{array}{c}{} \mathbf{1} \\ 1 \end{array}\right) =\left( \begin{array}{c}a_4 \\ a_4 \end{array}\right)$ and $a_4\in {\mathbb {F}}_{3^m}$ arbitrary.

• $(3^m)^2$ codes: $\langle (u-1)^{22}+3b(u),3(u-1)^5\rangle$, where

  • $\bullet$ $b(u)=2(u-1)^4+b^{[5,9)}(u)$,

  • $\bullet$ $b^{[5,9)}(u)=b_5(u-1)+b_6(u-1)^2+b_7(u-1)^3+b_8(u-1)^4$, where $\left( \begin{array}{c}b_5 \\ b_6 \\ b_7 \\ b_8\end{array}\right) = a_6\xi _6^{[5,9)}+a_8\xi _6^{[5,9)} =a_6\left( \begin{array}{c}{} \mathbf{1} \\ 0 \\ 0 \\ 2\end{array}\right) +a_8\left( \begin{array}{c}0 \\ 0 \\ \mathbf{1} \\ 2\end{array}\right) =\left( \begin{array}{c}a_6\\ 0\\ a_8\\ 2a_6+2a_8\end{array}\right)$ and $a_6,a_8\in {\mathbb {F}}_{3^m}$ arbitrary.

• $(3^m)^3$ codes: $\langle (u-1)^{20}+3b(u),3(u-1)^7\rangle$, where

  • $\bullet$ $b(u)=2(u-1)^2+b^{[7,13)}(u)$,

  • $\bullet$ $b^{[7,13)}(u)=\sum _{j=7}^{12}b_j(u-1)^{j-6}$, where $\left( \begin{array}{c} b_7 \\ b_8 \\ b_9 \\ b_{10} \\ b_{11} \\ b_{12}\end{array}\right) = a_8\xi _8^{[7,13)}+a_{10}\xi _{10}^{[7,13)}+a_{12}\xi _{12}^{[7,13)} = \left( \begin{array}{c|c|c} \mathbf{1} &{} 0 &{} 0 \\ 2 &{} 0 &{} 0 \\ 0 &{} \mathbf{1} &{} 0 \\ 0 &{} 1 &{} 0\\ 0 &{} 2 &{} \mathbf{1} \\ 0 &{} 1 &{} 0\end{array}\right) \left( \begin{array}{c} a_8 \\ a_{10} \\ a_{12}\end{array}\right)$ $=(a_8, 2a_{8}, a_{10}, a_{10}, 2a_{10}+a_{12}, a_{10})^{\mathrm{tr}}$ and $a_8,a_{10},a_{12}\in {\mathbb {F}}_{3^m}$ arbitrary.

• $(3^m)^4$ codes: $\langle (u-1)^{18}+3b(u),3(u-1)^9\rangle$, where

  • $\bullet$ $b(u)=2+b^{[9,17)}(u)$,

  • $\bullet$ $b^{[9,17)}(u)=\sum _{j=9}^{16}b_j(u-1)^{j-8}$, where

    $$\begin{aligned}&(b_9,b_{10},b_{11},b_{12},b_{13},b_{14},b_{15},b_{16})^{\mathrm{tr}} \\= & {} a_{10}\xi _{10}^{[9,17)}+a_{12}\xi _{12}^{[9,17)}+a_{14}\xi _{14}^{[9,17)}+a_{16}\xi _{16}^{[9,17)} \\= & {} \left( \begin{array}{c|c|c|c} \mathbf{1} &{} 0 &{} 0 &{} 0\\ 1 &{} 0 &{} 0 &{} 0 \\ 2 &{} \mathbf{1} &{} 0 &{} 0\\ 1 &{} 0 &{} 0 &{} 0 \\ 2 &{} 0 &{} \mathbf{1} &{} 0 \\ 1 &{} 1 &{} 2 &{} 0 \\ 2 &{} 0 &{} 0 &{} \mathbf{1} \\ 1 &{} 0 &{} 2 &{} 1\end{array}\right) \left( \begin{array}{c} a_{10} \\ a_{12} \\ a_{14}\\ a_{16}\end{array}\right) =\left( \begin{array}{c} a_{10} \\ a_{10} \\ 2a_{10}+a_{12} \\ a_{10} \\ 2a_{10}+a_{14} \\ a_{10}+a_{12}+2a_{14} \\ 2a_{10}+a_{16} \\ a_{10}+2a_{14}+a_{16}\end{array}\right) \end{aligned}$$

    and $a_{10},a_{12},a_{14},a_{16}\in {\mathbb {F}}_{3^m}$ arbitrary.

Remark

One can also use Theorem 3 to get the above results, by calculating binomial coefficients.