Low-rank parity-check codes over finite commutative rings

Abstract

Low-Rank Parity-Check (LRPC) codes are a class of rank metric codes that have many applications specifically in network coding and cryptography. Recently, LRPC codes have been extended to Galois rings which are a specific case of finite rings. In this paper, we first define LRPC codes over finite commutative local rings, which are bricks of finite rings, with an efficient decoder. We improve the theoretical bound of the failure probability of the decoder. Then, we extend the work to arbitrary finite commutative rings. Certain conditions are generally used to ensure the success of the decoder. Over finite fields, one of these conditions is to choose a prime number as the extension degree of the Galois field. We have shown that one can construct LRPC codes without this condition on the degree of Galois extension.

Appendix: Example of intersection and product of submodules

Set $R= \mathbb {Z} _{4}$, $S=R\left[ \theta \right] =R\left[ X\right] /\left( X^{5}+X^{2}+1\right)$,

$$\begin{aligned} A=\left\langle 3\theta ^{3}+2\theta +3,2\theta ^{4}+2\theta ^{3}+3\theta +1\right\rangle , \end{aligned}$$

and

$$\begin{aligned} B=\left\langle \theta ^{4}+2\theta ^{3}+1,2\theta ^{4}+3\theta ^{3}+2\theta +3\right\rangle . \end{aligned}$$

• 1. The matrix whose rows are vector representations in the basis $\left( 1,\theta ,\theta ^{2},\theta ^{3},\theta ^{4}\right)$ of the generators of A is

  $$\begin{aligned} \varvec{M}_{A}=\left( \begin{array}{rrrrr} 3 &{} 2 &{} 0 &{} 3 &{} 0 \\ 1 &{} 3 &{} 0 &{} 2 &{} 2 \end{array} \right) . \end{aligned}$$

  Using elementary row operations, the matrix $\varvec{M}_{A}$ is equivalent to

  $$\begin{aligned} \widetilde{\varvec{M}_{A}}=\left( \begin{array}{rrrrr} 1 &{} 2 &{} 0 &{} 1 &{} 0 \\ 0 &{} 1 &{} 0 &{} 1 &{} 2 \end{array} \right) . \end{aligned}$$

  Thus, by Proposition 7, A is a free module of rank 2.

• 2. The matrix whose rows are vector representations in the basis $\left( 1,\theta ,\theta ^{2},\theta ^{3},\theta ^{4}\right)$ of the generators of B is

  $$\begin{aligned} \varvec{M}_{B}=\left( \begin{array}{rrrrr} 1 &{} 0 &{} 0 &{} 2 &{} 1 \\ 3 &{} 2 &{} 0 &{} 3 &{} 2 \end{array} \right) . \end{aligned}$$

  Using elementary row operations, the matrix $\varvec{M}_{B}$ is equivalent to

  $$\begin{aligned} \widetilde{\varvec{M}_{B}}=\left( \begin{array}{rrrrr} 1 &{} 0 &{} 0 &{} 2 &{} 1 \\ 0 &{} 2 &{} 0 &{} 1 &{} 3 \end{array} \right) . \end{aligned}$$

  Thus, by Proposition 7, B is a free module of rank 2.

• 3. We have

  $$\begin{aligned} A+B=\left\langle a, b, c, d \right\rangle \end{aligned}$$

  with $a = 3\theta ^{3}+2\theta +3, b = 2\theta ^{4}+2\theta ^{3}+3\theta +1, c= \theta ^{4}+2\theta ^{3}+1, d= 2\theta ^{4}+3\theta ^{3}+2\theta +3.$ The matrix whose rows are vector representations in the basis $\left( 1,\theta ,\theta ^{2},\theta ^{3},\theta ^{4}\right)$ of the generators of $A+B$ is

  $$\begin{aligned} \varvec{M}_{A+B}=\left( \begin{array}{rrrrr} 3 &{} 2 &{} 0 &{} 3 &{} 0 \\ 1 &{} 3 &{} 0 &{} 2 &{} 2 \\ 1 &{} 0 &{} 0 &{} 2 &{} 1 \\ 3 &{} 2 &{} 0 &{} 3 &{} 2 \end{array} \right) . \end{aligned}$$

  Using elementary row operations, the matrix $\varvec{M}_{A+B}$ is equivalent to

  $$\begin{aligned} \widetilde{\varvec{M}_{A+B}}=\left( \begin{array}{rrrrr} 1 &{} 2 &{} 0 &{} 1 &{} 0 \\ 0 &{} 1 &{} 0 &{} 1 &{} 2 \\ 0 &{} 0 &{} 0 &{} 1 &{} 3 \\ 0 &{} 0 &{} 0 &{} 0 &{} 2 \end{array} \right) . \end{aligned}$$

  Thus, by Proposition 7, ${{\,\textrm{frk}\,}}_{R}\left( A+B\right) =3$ and $A+B$ is not a free module.

• 4. We have

  $$\begin{aligned} A\cap B=\left\langle 2\theta ^{3}+2\right\rangle \end{aligned}$$

  and, by Proposition 7, $A\cap B$ is not a free module.

• 5. We have

  $$\begin{aligned} AB=\left\langle a, b, c, d \right\rangle \end{aligned}$$

  with $a=3\theta ^{3}+3\theta ^{2}+1$, $b=\theta ^{3}+2\theta ^{2}+3\theta +1$, $c = \ 3\theta ^{4}+2\theta ^{3}+\theta ^{2} + 3 \theta$, $d= 3\theta ^{4}+3\theta ^{3}+2\theta ^{2}+\theta +1$ and, by Proposition 7, AB is not a free module.