Entanglement-assisted quantum MDS codes from generalized Reed–Solomon codes

Abstract

In this paper, some new classes of entanglement-assisted quantum MDS codes (EAQMDS codes for short) are constructed via generalized Reed–Solomon codes over finite fields of odd characteristic. Among our constructions, there are many EAQMDS codes with new lengths. The lengths of these EAQMDS codes may not be divisors of \(q^2\pm 1\), which are new and have never been reported.

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Appendix

Appendix

Proof of lemma 3.1

Denote by \(\xi =(a_{0},a_{1},\cdots ,a_{h-1})\) and \(d=\frac{s-h}{2}+1 \). The system of Eq. 3.1 can be expressed in the matrix form

$$\begin{aligned} A\cdot {\mathbf {u}}^{\mathrm{T}} =(a_{0},a_{1},\cdots ,a_{h-1})^{\mathrm{T}}, \end{aligned}$$

where

$$\begin{aligned} A=\left( \begin{array}{cccc} 1 &{} 1 &{} \cdots &{} 1 \\ g^{\mathrm{dl}-q-1} &{} g^{3(\mathrm{dl}-q-1)} &{} \cdots &{} g^{(2h-1)(\mathrm{dl}-q-1)} \\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ g^{(d+h-2)l-q-1} &{} g^{3[(d+h-2)l-q-1]} &{} \cdots &{} g^{(2h-1)[(d+h-2)l-q-1]} \\ \end{array} \right) . \end{aligned}$$

Let \(x=g^{\mathrm{dl}-q-1}\) and

$$\begin{aligned} A^{\mathrm{T}}=\left( \begin{array}{cccc} 1 &{} x &{} \cdots &{} \mathrm{xg}^{(h-2)l} \\ 1 &{} x^{3} &{} \cdots &{} (\mathrm{xg}^{(h-2)l})^{3} \\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ 1 &{} x^{2h-1} &{} \cdots &{} (\mathrm{xg}^{(h-2)l})^{2h-1} \\ \end{array} \right) . \end{aligned}$$

A routine calculation shows

$$\begin{aligned} \text {det}(A^\mathrm{T})=\frac{1}{x\cdot (\mathrm{xg}^l)\cdot \cdots \cdot (\mathrm{xg}^{(h-2)l})}\left| \begin{array}{cccc} 1 &{} 1 &{} \cdots &{} 1 \\ 1 &{} x^{2} &{} \cdots &{} (\mathrm{xg}^{(h-2)l})^{2} \\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ 1 &{} x^{2h-2} &{} \cdots &{} (\mathrm{xg}^{(h-2)l})^{2h-2} \\ \end{array} \right| . \end{aligned}$$

It is obvious that \(\text {det}(A)\ne 0\) and \({\mathbf {u}}=(u_{0},u_{1},\cdots ,u_{h-1})\in (\mathbb {F}_{q^2})^h \).

Next we will show that (2) has a solution \({\mathbf {u}}=(u_{0},u_{1},\cdots ,u_{h-1})\in (\mathbb {F}^*_q)^h \). Let

$$\begin{aligned}&\mathbf {\beta _{0}}=(1,1,\cdots ,1), \\&\mathbf {\beta _{i}}=\left( g^{(d+i-1)l-q-1},g^{3[(d+i-1)l-q-1]},\cdots ,g^{(2h-1)[(d+i-1)l-q-1]}\right) \\&\text {for}\; i=1,2,\cdots ,h-1. \end{aligned}$$

It is easy to verify \(\beta _{i}^q=\beta _{h-i}\) for \(i=1,2,\cdots ,h-1.\) Assume \(\zeta =g^{\frac{q+1}{2}}\), then \(\zeta ^{q}=-\zeta .\) Raising all Eq. 3.1 to their qth powers leads to

$$\begin{aligned} \left\{ \begin{array}{lll} \beta _{0}{\mathbf {u}}=a_{0}\\ \\ (\beta _{1}+\beta _{h-1}){\mathbf {u}}=2a_{1}\\ \\ \vdots \\ \\ (\beta _{\frac{h-1}{2}}+\beta _{\frac{h+1}{2}}){\mathbf {u}}=2a_{\frac{h-1}{2}}\\ \\ \frac{\beta _{1}-\beta _{h-1}}{\zeta }{\mathbf {u}}=0\\ \\ \vdots \\ \\ \frac{\beta _{\frac{h-1}{2}}-\beta _{\frac{h+1}{2}}}{\zeta }{\mathbf {u}}=0.\end{array} \right. \end{aligned}$$
(6)

Denote by \(\alpha =\left( a_{0},2a_{1},\cdots ,2a_{\frac{h-1}{2}},0,\cdots ,0\right) ^\mathrm{T}\). The above Eq. (6) can be reformulated as \(B{\mathbf {u}}=\alpha \) with

$$\begin{aligned}B=\left( \beta _0^\mathrm{T}, \beta _1^\mathrm{T}+\beta ^\mathrm{T}_{h-1},\cdots , \frac{\beta ^\mathrm{T}_{1}-\beta ^\mathrm{T}_{h-1}}{\zeta },\cdots , \frac{\beta ^\mathrm{T}_{\frac{h-1}{2}} -\beta ^\mathrm{T}_{\frac{h+1}{2}}}{\zeta }\right) ^\mathrm{T}.\end{aligned}$$

It is easy to verify that all the entries of B are in \(\mathbb {F}_{q}\). Suppose

$$\begin{aligned} S=\left\{ {\mathbf {u}}\in \mathbb {F}_{q}^{n}|B{\mathbf {u}}=\alpha ,\;a_{i}\in \mathbb {F}^*_q, 1\le i\le h-1\right\} . \end{aligned}$$

Then, \(|S|=(q-1)^{\frac{h+1}{2}}\) since B is invertible. For \(0\le i\le h-1\), define

$$\begin{aligned} S_{i}=\left\{ {\mathbf {u}}\in \mathbb {F}_{q}^{n}|u_{i}=0,\;B{\mathbf {u}}=\alpha ,\;a_{j}\in \mathbb {F}^*_q\;\text {for all}\,0\le j\le \frac{h-1}{2}\; \right\} \;. \end{aligned}$$

Then, the set of solutions to (6), denoted by V, is

$$\begin{aligned} V=\overline{ S_{1}\bigcup S_{2}\bigcup \cdots \bigcup S_{\frac{h+1}{2}}}, \end{aligned}$$

where \({\overline{T}}=S-T\) for any subset \(T\subseteq S.\) By Inclusion–Exclusion principle,

$$\begin{aligned} |V|=|S|-\sum \nolimits ^{h-1}_{i=0}|S_{i}|+r \end{aligned}$$

for some \(r\ge 0.\) It follows that

$$\begin{aligned} |V|=(q-1)^{\frac{h+1}{2}}-\left( {\begin{array}{c}h\\ 1\end{array}}\right) (q-1)^{\frac{h-1}{2}}+r. \end{aligned}$$

Obviously \(|V|>0\). We only consider the case h is odd. The case h being even is similar and we omit the details. Therefore, there exists a solution satisfying \(u_{i}\in \mathbb {F}_{q}^{*}(i=0,1,\cdots ,h-1).\)

This completes the proof. \(\square \)

Proof of Lemma 3.3

We can calculate directly,

$$\begin{aligned} \langle {\mathbf {a}}_{2}^{qi+j},{\mathbf {v}}_{2}^{q+1}\rangle&=\sum \nolimits _{k=0}^{r-1}\sum \nolimits _{\nu =0}^{m-1}(g^{k}\theta ^{\nu })^{qi+j}\cdot \theta ^{\nu \cdot \frac{q+1}{2}}\\&=\sum _{k=0}^{r-1}g^{k(qi+j)}\sum _{\nu =0}^{m-1}\theta ^{\nu (qi+j+\frac{q+1}{2})}.\end{aligned}$$

It suffices to show the identity

$$\begin{aligned}\sum \nolimits _{v=0}^{m-1}\theta ^{v(qi+j+ \frac{q+1}{2})}=0\end{aligned}$$

holds for all \( 0\le i,j\le \frac{q+1}{2}+ \frac{q-1}{t}-2 ,\;t \ge 2.\) It is easy to check that the identity holds if and only if \(m\not \mid qi+j+\frac{q+1}{2}\). On the contrary, assume that \(m\mid \mathrm{qi}+j+\frac{q+1}{2}\). Let

$$\begin{aligned} qi+j+\frac{q+1}{2}=\mu \cdot m=q\cdot \frac{\mu (q-1)}{t}+\frac{\mu (q-1)}{t} \end{aligned}$$
(7)

where \(\mu \) is an integer. Since \(t\ge 2\), we have \(qi+j+\frac{q+1}{2}<q^{2}-1\), which implies \(0<\mu <t\). It suffices to discuss the following two cases.

Case 1::

If \(j+\frac{q+1}{2}\le q-1\), comparing remainder and quotient of module q on both sides of (7), we can deduce \(i=j+\frac{q+1}{2}=\mu \cdot \frac{q-1}{t}\). Since t is even, then \(\frac{q-1}{t}\mid \frac{q-1}{2}\). From \(\frac{q-1}{t}\mid j+1+\frac{q-1}{2}\), we can deduce that \(\frac{q-1}{t}\mid j+1\). Since \(j+1\ge 1\), then \(j+1\ge \frac{q-1}{t}\). Therefore, \(i=j+\frac{q+1}{2}\ge \frac{q+1}{2}+\frac{q-1}{t}-1\), which is a contradiction.

Case 2::

If \(j+\frac{q+1}{2}\ge q\), it takes

$$\begin{aligned} qi+j+\frac{q-1}{2}=q(i+1)+\left( j-\frac{q-1}{2}\right) =q\cdot \frac{\mu (q-1)}{t}+\frac{\mu (q-1)}{t}. \end{aligned}$$

In a similar way, \(j-\frac{q-1}{2}=i+1=\mu \cdot \frac{q-1}{t}\) which implies \(\frac{q-1}{t}\mid i+1\). Since \(i+1\ge 1\), then \(i+1\ge \frac{q-1}{t}\). Therefore, \(j=i+1+\frac{q+1}{2}\ge \frac{q+1}{2}+\frac{q-1}{t}-1\), which is a contradiction.

As a result, \(m\not \mid \mathrm{qi}+j+\frac{q+1}{2}\) which yields

$$\begin{aligned}\sum _{v=0}^{m-1}\theta ^{v(qi+j+ \frac{q+1}{2})}=0\end{aligned}$$

for all \( 0\le i,j\le \frac{q+1}{2}+ \frac{q-1}{t}-2 .\) This completes the proof. \(\square \)

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Jin, R., Cao, Y. & Luo, J. Entanglement-assisted quantum MDS codes from generalized Reed–Solomon codes. Quantum Inf Process 20, 73 (2021). https://doi.org/10.1007/s11128-021-03010-6

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Keywords

  • MDS code
  • EAQEC code
  • EAQMDS code
  • Generalized Reed–Solomon code