Quantum Cyclic Codes Over ${\mathbb{Z}}_m$

Abstract

Quantum codes over finite rings have the advantage of being able to adapt to quantum physical systems with arbitrary order. Furthermore, operations are much easier to execute in finite rings than they are in fields. This paper discusses quantum cyclic codes over the modulo m residue class ring ${\mathbb{Z}}_m$. A connection is established between the stabilizer codes over ${\mathbb{Z}}_m$ and the additive codes over an extension ring of ${\mathbb{Z}}_m$ that generalizes the well-known relationship between the stabilizer codes over ${\mathbb{F}}_q$ and the additive codes over ${\mathbb{F}}_{q^2}$. We prove that if the irreducible polynomial is selected according to a simple criterion, the additive codes which are self-orthogonal with respect to the conjugate inner product correspond to the stabilizer codes. The structure of cyclic stabilizer codes is developed, and some simple conditions for finding them are presented. We also define the quantum Bose-Chaudhuri-Hocquenghem (BCH) and quantum Reed-Solomon (RS) codes over ${\mathbb{Z}}_m$. Finally, new quantum cyclic codes over ${\mathbb{Z}}_m$ are given.

Appendix

1.1 Proof of Lemma 2

For two elements a + bx, c + dx in Q_i, we have

$$ {\sigma}_i\left(a+ bx+c+ dx\right)=a+b{x}^{p_i}+c+d{x}^{p_i}={\sigma}_i\left(a+ bx\right)+{\sigma}_i\left(c+ dx\right),\kern1.00em $$

and

$$ {\sigma}_i\left(\left(a+ bx\right)\left(c+ dx\right)\right)= ac+ ad{x}^{p_i}+ bc{x}^{p_i}+ bd{x}^{2{p}_i}={\sigma}_i\left(a+ bx\right){\sigma}_i\left(c+ dx\right).\kern1.00em $$

Thus, σ_i is compatible with addition and multiplication in Q_i. Moreover, one has σ_i(1) = 1. It follows that σ_i is a homomorphism from Q_i to Q_i. Since the kernel of σ_i is trivial, we conclude that Q_i is an automorphism of Q_i.

1.2 Proof of Lemma 3

For $\mathbf{u},\mathbf{w},{\mathbf{u}}^{\prime}\in {Q}_i^n$, it is easy to verify that $<\mathbf{u},\mathbf{w}{>}_{s_i}=-<\mathbf{w},\mathbf{u}{>}_{s_i}$. Then we know $<\mathbf{u},\mathbf{u}{>}_{s_i}=-<\mathbf{u},\mathbf{u}{>}_{s_i}$ and $<\mathbf{u},\mathbf{u}{>}_{s_i}=0$. Furthermore, we have

$$ <\mathbf{u}+{\mathbf{u}}^{\prime },\mathbf{w}{>}_{s_i}=\frac{\left(\mathbf{u}+{\mathbf{u}}^{\prime}\right)\cdot {\sigma}_i\left(\mathbf{w}\right)-{\sigma}_i\left(\mathbf{u}+{\mathbf{u}}^{\prime}\right)\cdot \mathbf{w}}{x-{x}^{p_i}} $$

$$ =\frac{\mathbf{u}\cdot {\sigma}_i\left(\mathbf{w}\right)-{\sigma}_i\left(\mathbf{u}\right)\cdot \mathbf{w}}{x-{x}^{p_i}}+\frac{{\mathbf{u}}^{\prime}\cdot {\sigma}_i\left(\mathbf{w}\right)-{\sigma}_i\left({\mathbf{u}}^{\prime}\right)\cdot \mathbf{w}}{x-{x}^{p_i}}=<\mathbf{u},\mathbf{w}{>}_{s_i}+<{\mathbf{u}}^{\prime },\mathbf{w}{>}_{s_i}, $$

which means the bilinearity holds. Hence, we can deduce that ${\left\langle \kern1em ,\kern1em \right\rangle}_{s_i}$ is a symplectic form. On the other hand, if we ignore the scalar, the matrix of the form with respect to the standard symplectic basis is equivalent to $\left(\begin{array}{ll}0& I\\ {}-I& 0\end{array}\right)$. It follows that the form ${\left\langle \kern1em ,\kern1em \right\rangle}_{s_i}$ is non-degenerate. The claim holds.

1.3 Proof of Lemma 4

For two vectors a = (a₁∣a₂), $\mathbf{b}=\left({\mathbf{b}}_1|{\mathbf{b}}_2\right)\in {R}_i^{2n}$, we have

$$ \psi \left(\mathbf{a}\right)\cdot {\sigma}_i\left(\psi \left(\mathbf{b}\right)\right)=\left({\mathbf{a}}_1+{\mathbf{a}}_2x\right)\cdot \left({\mathbf{b}}_1+{\mathbf{b}}_2{x}^{p_i}\right)={\mathbf{a}}_1\cdot {\mathbf{b}}_1+{\mathbf{a}}_1\cdot {\mathbf{b}}_2{x}^{p_i}+{\mathbf{a}}_2\cdot {\mathbf{b}}_1x+{\mathbf{a}}_2\cdot {\mathbf{b}}_2{x}^{p_i+1}, $$

and

$$ {\sigma}_i\left(\psi \left(\mathbf{a}\right)\right)\ast \psi \left(\mathbf{b}\right)=\left({\mathbf{a}}_1+{\mathbf{a}}_2{x}^{p_i}\right)\cdot \left({\mathbf{b}}_1+{\mathbf{b}}_2x\right)={\mathbf{a}}_1\cdot {\mathbf{b}}_1+{\mathbf{a}}_2\cdot {\mathbf{b}}_1{x}^{p_i}+{\mathbf{a}}_1\cdot {\mathbf{b}}_2x+{\mathbf{a}}_2\cdot {\mathbf{b}}_2{x}^{p_i+1}. $$

Thus, it follows that ${\left\langle \psi \left(\mathbf{a}\right)\kern1em ,\psi \left(\mathbf{b}\right)\right\rangle}_{s_i}=\left({\mathbf{a}}_2\cdot {\mathbf{b}}_1-{\mathbf{a}}_1\cdot {\mathbf{b}}_2\right)={\left\langle \mathbf{a}\kern1em ,\mathbf{b}\right\rangle}_a$.

1.4 Proof of Lemma 6

For a, b ∈ Q, we assume that $\mu (a)=\left({a}_1,\dots, {a}_l\right),\kern1em \mu (b)=\left({b}_1,\dots, {b}_l\right)$, where a_i, b_i ∈ Q_i for $i=1,\dots, l$. Then

$$ \sigma \left(a+b\right)={\mu}^{-1}\left({\sigma}_1\left({a}_1+{b}_1\right),\dots, {\sigma}_l\left({a}_l+{b}_l\right)\right) $$

$$ ={\mu}^{-1}\left({\sigma}_1\left({a}_1\right),\dots, {\sigma}_l\left({a}_l\right)\right)+{\mu}^{-1}\left({\sigma}_1\left({b}_1\right),\dots, {\sigma}_l\left({b}_l\right)\right)=\sigma (a)+\sigma (b), $$

and

$$ \sigma \left(\mathrm{ab}\right)={\mu}^{-1}\left({\sigma}_1\left({a}_1{b}_1\right),\dots, {\sigma}_l\left({a}_l{b}_l\right)\right)={\mu}^{-1}\left({\sigma}_1\left({a}_1\right){\sigma}_1\left({b}_1\right),\dots, {\sigma}_l\left({a}_l\right){\sigma}_l\left({b}_l\right)\right) $$

$$ ={\mu}^{-1}\left({\sigma}_1\left({a}_1\right),\dots, {\sigma}_l\left({a}_l\right)\right){\mu}^{-1}\left({\sigma}_1\left({b}_1\right),\dots, {\sigma}_l\left({b}_l\right)\right)=\sigma (a)\sigma (b). $$

Therefore, σ is compatible with addition and multiplication in Q. Moreover, we have σ(1) = 1. It follows that σ is a homomorphism from Q to Q. Because the kernel of σ is trivial, we conclude that σ is an automorphism of Q.

1.5 Proof of Lemma 9

Suppose that u = a + bx, w = c + dx, where $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\in {R}_i^n$. If p_i = 2, we have x² = −x − 1 in Q_i. There is

$$ {\displaystyle \begin{array}{rcl}\mathbf{u}\cdot \sigma \left(\mathbf{w}\right)& =& \left(\mathbf{a}+\mathbf{b}x\right)\cdot \left(\mathbf{c}+\mathbf{d}{x}^2\right)=\mathbf{a}\cdot \mathbf{c}+\mathbf{a}\cdot \mathbf{d}{x}^2+\mathbf{b}\cdot \mathbf{c}x+\mathbf{b}\cdot \mathbf{d}{x}^3\\ {}& =& \mathbf{a}\cdot \mathbf{c}-\mathbf{a}\cdot \mathbf{d}+\mathbf{b}\cdot \mathbf{d}+\left(\mathbf{b}\cdot \mathbf{c}-\mathbf{a}\cdot \mathbf{d}\right)x.\end{array}} $$

Therefore, if u ⋅ σ_i(w) equals 0, we must have b ⋅c −a ⋅d = 0, which is equivalent to ${\left\langle \mathbf{u}\kern1em ,\mathbf{w}\right\rangle}_{s_i}=0$ according to Lemma 4.

Conversely, if p_i ≠ 2, we know that x² = a in Q_i. There is

$$ \mathbf{u}\cdot {\sigma}_i\left(\mathbf{w}\right)=\left(\mathbf{a}+\mathbf{b}x\right)\cdot \left(\mathbf{c}+\mathbf{d}{x}^{p_i}\right)=\left(\mathbf{a}+\mathbf{b}x\right)\cdot \left(\mathbf{c}-\mathbf{d}x\right)=\mathbf{a}\cdot \mathbf{c}-\mathbf{a}\cdot \mathbf{d}x+\mathbf{b}\cdot \mathbf{c}x-\mathbf{b}\cdot \mathbf{d}{x}^2 $$

$$ =\mathbf{a}\cdot \mathbf{c}-a\mathbf{b}\cdot \mathbf{d}+\left(\mathbf{b}\cdot \mathbf{c}-\mathbf{a}\cdot \mathbf{d}\right)x. $$

Thus, if u ⋅ σ_i(w) equals 0, we must have b ⋅c −a ⋅d = 0, which is equivalent to ${\left\langle \mathbf{u}\kern1em ,\mathbf{w}\right\rangle}_{s_i}=0$ according to Lemma 4. This completes the proof.