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
where
Let \(x=g^{\mathrm{dl}-q-1}\) and
A routine calculation shows
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
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
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
It is easy to verify that all the entries of B are in \(\mathbb {F}_{q}\). Suppose
Then, \(|S|=(q-1)^{\frac{h+1}{2}}\) since B is invertible. For \(0\le i\le h-1\), define
Then, the set of solutions to (6), denoted by V, is
where \({\overline{T}}=S-T\) for any subset \(T\subseteq S.\) By Inclusion–Exclusion principle,
for some \(r\ge 0.\) It follows that
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,
It suffices to show the identity
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
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
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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DOI: https://doi.org/10.1007/s11128-021-03010-6