An application of constacyclic codes to entanglement-assisted quantum MDS codes

Abstract

The constructions of entanglement-assisted quantum codes have been studied intensively by researchers. Nevertheless, it is hard to determine the number of shared pairs required for constructing entanglement-assisted quantum codes from linear codes. In this paper, by making use of the notion of decomposition for defining sets of constacyclic codes, we construct several new families of entanglement-assisted quantum MDS codes from constacyclic codes, some of which are of minimum distances greater than $q+1$. Moreover, we tabulate their parameters to illustrate what we find in this paper.

1 Introduction

Theoretical advantages of quantum mechanics to classical mechanics led scholars to study quantum computation and communication, which caused quantum bits (qubits) to be studied in place of classical bits. However, one principal difficulty for qubits was decoherence which overthrows the information in a superposition of qubits (Shor 1995). Shor realized that this key difficulty could be coped by introducing the first quantum code that encodes one qubit to a superposition of nine qubits and corrects at most one quantum error (Shor 1995). This pioneer paper of Shor encouraged researchers to develop quantum error correcting codes (QECCs). Called CSS construction, the first systematic construction for QECCs was explored by Calderbank et al. and Steane, independently in Calderbank and Shor (1996) and Steane (1996), respectively. According to CSS construction, one can construct a QECC from binary linear codes which are nested. Later, Gottesman developed a stabilizer formalism for QECCs that defines a QECC to be a subspace of ${C^{{2^{ \otimes n}}}}$ fixed by a commutative subgroup of Pauli matrix group on binary qubits, where ${C}^{{2^{ \otimes n}}}$ is an n-fold tensor product of the two-dimensional complex vector space $C^2$ (Gottesman 1997). Calderbank et al. (1998) turned constructing QECCs into finding self-orthogonal additive codes over the finite field of four elements with respect to trace inner product. As a generalization of the paper (Calderbank et al. 1998), Ketkar et al. (2006) defined Pauli matrices for highly states over $F_q$ and gave a way for constructing nonbinary quantum stabilizer codes from self-orthogonal additive codes over $F_{q^2}$ with respect to trace-alternating form, in particular, self-orthogonal linear codes over $F_{q^2}$ with respect to Hermitian inner product. Inspired by Ketkar et al. (2006), many scholars have focused on construction of new nonbinary stabilizer quantum codes (Aly et al. 2007; Chen et al. 2015; Grassl and Rötteler 2015; He et al. 2016; Hu et al. 2015; Jin et al. 2017; Kai and Zhu 2013; Kai et al. 2014; Liqin et al. 2016; Liu et al. 2017; Yuan et al. 2017; Zhang and Chen 2014; Zhang and Ge 2015).

Brun et al. (2006) developed a new systematic method for constructing QECCs. According to the construction given by Brun et al. (2006), the requirement of self-orthogonality for linear codes over $F_{q^2}$ was no longer needed, which allows us to quantize all linear codes over $F_{q^2}$. The QECCs obtained via the construction presented by Brun et al. (2006) are called entanglement-assisted quantum error correcting codes (EAQECCs). An EAQECC of length n and minimum distance d over $F_q$ is denoted by , and this EAQECC encodes k qubits to n-channel qubits via c pairs of maximally entanglement states and corrects up to $\lfloor {\frac{{d - 1}}{2}} \rfloor$ errors. An EAQECC with $n-k=c$ is called a maximal entanglement EAQECC. An EAQECC is called an entanglement-assisted quantum MDS code (EAQMDSC) if its parameters attain the entanglement-assisted singleton bound $k \le n - 2 ( {d - 1} ) + c$. For more details for EAQECCs, we refer the readers to (Brun et al. 2006, 2014; Lai and Brun 2013; Wilde and Brun 2008). There have been many studies on the constructions for EAQECCs which improve the parameters of existing ones (Chen et al. 2017; Fan et al. 2016; Guenda et al. 2018; Li et al. 2011; Lu et al. 2018; Lv et al. 2015; Qian and Zhang 2015, 2017). Fan et al. (2016) constructed several classes of EAQMDSCs from classical MDS codes with one or more shared entangled states. Guenda et al. (2018) proved that the number of shared pairs required is associated with the hull of linear codes and, using this connection, obtained methods for constructing EAQECCs with desired amount of entanglement. Qian and Zhang (2015) constructed EAQECCs from arbitrary [ n, k, d ] linear codes. Brun et al. (2006) also proved that there exists an EAQECC if there exits an ${[ {n,k,d} ]_q}$ linear code C with $c = \mathrm{{rank}} ( {H{H^\dag }} )$, where H is the parity check matrix of the code C and $H^\dag$ is the conjugate transpose of the matrix H. While this construction enables us to quantize all linear codes over $F_{q^2}$, determining the parameter c, which is the number of entanglement states required, is an open problem. Li et al. (2011) proposed a solution to this problem by making use of a decompose notion for the defining sets of cyclic codes and constructed EAQECCs from cyclic codes. Then, Chen et al. (2017) extended this decompose notion for the defining sets of cyclic codes to negacyclic codes over $F_{q^2}$ and obtained a few families of EAQMDSCs with minimum distances greater than $q+1$. They also constructed two classes of maximal entanglement EAQECCs. Motivated from these studies, in this paper, we consider a decomposition of defining sets of constacyclic codes over $F_{q^2}$ to determine the number of shared pairs required and, by taking advantage of this decomposition, we construct a few new families of EAQMDSs.

EAQMDSCs constructed in this paper are as follows:

• 1. For odd prime power $q \equiv 3 ( {\bmod 4} )$, and , where $q + 1 \le \lambda \le 2q - 2$.

• 2. For odd prime power $q \equiv 3 ( {\bmod 4} )$, , where $\frac{{q + 9}}{4} \le d \le q$.

• 3. For odd prime power $q \equiv 3 ( {\bmod 4} )$, , where $\frac{{3q + 7}}{4} \le d \le \frac{{5q + 1}}{4}$.

The contents of this paper are organized as follows: In Sect. 2, we present the fundamentals needed for following sections. In Sect. 3, we construct four new families of EAQMDSCs from constacyclic codes by making use of a decomposition for defining sets of constacyclic codes over $F_{q^2}$. Moreover, by tabulating the parameters of some of EAQMDSCs that we derive, we illustrate the findings in this paper. In Sect. 4, we conclude the paper.

2 Preliminaries

Let $F_{q^2}$ be a finite field of $q^2$ elements. A k-dimensional subspace of the vector space $F_{q^2}^n$ is a linear code of length n over $F_{q^2}$ and this linear code is denoted by ${[ {n,k} ]_{q^2}}$. An ${[ {n,k} ]_{q^2}}$ linear code C is an ${[ {n,k,d} ]_{q^2}}$ linear code if C detects $d-1$ errors but not d errors, where d is called the minimum distance of the code C. The Hermitian inner product ${\langle {x,y} \rangle _h}$ of the vectors $x = ( {{x_1},{x_2}, \ldots ,{x_n}} )$ and $y = ( {{y_1},{y_2}, \ldots ,{y_n}} )$ in $F_{q^2}^n$ is ${\langle {x,y} \rangle _h} = \sum \nolimits _{i = 1}^n {{x_i}y_i^q}$. The Hermitian dual ${C^{{ \bot _h}}}$ of a linear code C over $F_{q^2}$ of length n is the set of vectors in $F_{q^2}^n$ which are perpendicular to all vectors in C with respect to the Hermitian inner product. A parity check matrix H of an ${[ {n,k} ]_{q^2}}$ linear code C with respect to Hermitian inner product is an $( n - k ) \times n$ matrix whose rows constitute a basis for ${C^{{ \bot _h}}}$. Conjugate of a vector $x = ( {{x_1},{x_2}, \ldots ,{x_n}} )$ in $F_{q^2}^n$ is ${x^\dag } = ( {x_1^q,x_2^q, \ldots ,x_n^q} )$ and conjugate transpose of an $m \times n$ matrix $H = ( {{x_{i,j}}} )$ with entries in $F_{q^2}$ is an $n \times m$ matrix ${H^\dag } = ( {x_{j,i}^q} )$.

Let $\alpha$ be a nonzero element in $F_{q^2}$ with multiplicative order r. A linear code C over $F_{q^2}$ of length n is called an $\alpha$-constacyclic code if $\mu ( c ) \in C$ for all $c \in C$, where $\mu :F_{{q^2}}^n \rightarrow F_{{q^2}}^n$, $\mu ( {( {{c_1},{c_2}, \ldots ,{c_n}} )} ) = ( {\alpha {c_n},{c_1}, \ldots ,{c_{n - 1}}} )$. In the case that $\alpha =-1$, an $\alpha$-constacyclic code is called a negacyclic code. It is a well-known fact that an $\alpha$-constacyclic code C in $F_{q^2}^n$ can be viewed as an ideal in the quotient ring $\frac{{{F_{{q^2}}} [ x ]}}{{ \langle {{x^n} - \alpha } \rangle }}$ and so $C = \langle {g ( x )} \rangle$ for some g(x) dividing ${x^n} - \alpha$. Let $\left( {n,q} \right) = 1$. All roots of ${x^n} - \alpha$ over $F_{q^2}$ are $\gamma$, $\gamma ^{1+r}$,..., $\gamma ^{1+(n-1)r}$, where $\gamma$ is an rnth primitive root of unity in some extension field of $F_{q^2}$ and ${\gamma ^n} = \alpha$. Set ${O_{r,n}} = \{ {1,1 + r, \ldots ,1 + ( {n - 1} )r} \}$. The $q^2$-cyclotomic coset modulo rn containing i is the set ${C_i} = \{ {i{q^{2j}}\bmod rn:j \in N} \}$. The defining set $Z \subseteq {O_{r,n}}$ of an $\alpha$-constacyclic code $C = \langle {g ( x )} \rangle$ over $F_{q^2}$ is the set $Z = \{ {i \in {O_{r,n}}:g ( {{\gamma ^i}} ) = 0} \}$. The following is a lower bound for $\alpha$-constacyclic codes:

Theorem 1

(BCH bound for constacyclic codes) (Krishna and Sarwate 1990; Aydin et al. 2001) Let $( {n,q} ) = 1$. Let $\gamma$ be an rnth primitive root of unity, such that ${\gamma ^n} = \alpha$, where $\alpha$ is a nonzero element in $F_{{q^2}}$ with multiplicative order r. Then, the minimum distance of an $\alpha$-constacyclic code of length n over $F_{{q^2}}$ with the defining set including the set $\{ {1 + rj,\,l \le j \le l + d - 2} \}$ is at least d.

It is well known that the Hermitian dual of an $\alpha$-constacyclic code over $F_{q^2}$ is an $\alpha ^{-q}$-constacyclic code. Kai et al. (2014) give a necessary and sufficient condition for constacyclic codes over $F_{q^2}$ to contain their Hermitian duals.

Lemma 1

(Kai et al. 2014) Let C be a constacyclic code of length n over $F_{q^2}$ with defining set Z, where $( {n,q} ) = 1$. Then, ${C^{{ \bot _h}}}$ is included by C if and only if $Z \cap - qZ = \emptyset$.

Let c be a nonnegative integer. Via c pairs of maximally entanglement states, an EAQECC encodes k information qubits into n qubits, and can correct up to $\big \lfloor {\frac{{d - 1}}{2}} \big \rfloor$ errors which act on n qubits, where d is called minimum distance of the EA quantum code. From Wilde and Brun (2008), we have the following analog between linear codes over $F_{q^2}$ and EAQECCs.

Theorem 2

(Wilde and Brun 2008) If there exists an ${ [ {n,k,d} ]_{{q^2}}}$ linear code with parity check matrix H, then there exists an EAQECC having parameters , where $c=\mathrm{{rank}} ( {H{H^\dag }} )$.

Brun et al. (2006) establish a bound on the parameters of an EAQECC.

Proposition 1

(Brun et al. 2006) (Entanglement-Assisted (EA) Singleton bound) For an EAQECC, $k \le n - 2 ( {d - 1} ) + c$.

An EAQECC satisfying EA singleton bound is called an EAQMDSC. For an EAQECC, the number c of maximally entanglement states based on the linear codes is less than or equal to $n-k$, and if $c=n-k$, then this is called a maximal entanglement EAQECC.

3 Entanglement-assisted quantum MDS codes derived from constacyclic codes

In Chen et al. (2017), by taking advantage of a decomposition of the defining sets of negacyclic codes, Jianzhang Chen et al. determine the number of entanglement states of EAQECCs obtained from negacyclic codes over $F_{q^2}$. Then, this result is extended to constacyclic codes over $F_{q^2}$ in Liu et al. (2018) and Lu et al. (2018).

For a constacyclic code C over $F_{q^2}$ with defining set Z, define the sets ${Z_\beta } = \{ {i \in Z:-qi \in Z} \}$ and ${Z_\delta } = \{ {i \in Z:-qi \notin Z} \}$. Then, $Z = {Z_\beta } \cup {Z_\delta }$, ${Z_\beta } \cap {Z_\delta } = \emptyset$ and ${Z_\beta } = Z \cap - qZ$. We call $Z = {Z_\beta } \cup {Z_\delta }$ as a decomposition of Z. Note that constacyclic codes $C_\beta$ and $C_\delta$ with defining set $Z_\beta$ or $Z_\delta$, respectively, are codes over $F_{q^2}$, since $Z_\beta$ and $Z_\delta$ are union of some $q^2$-cyclotomic cosets, and in this case, $C = {C_\beta } \cap {C_\delta }$. Moreover, the definition of $Z_\delta$ implies that $- q{Z_\delta } \cap {Z_\delta } = \emptyset$, and by Lemma 1, $C_\delta ^{{ \bot _h}} \subseteq {C_\delta }$. We have the following result from Liu et al. (2018) and Lu et al. (2018).

Proposition 2

(Liu et al. 2018; Lu et al. 2018) Let C be a constacyclic code over $F_{q^2}$ with length n and defining set Z, where $\left( {n,q} \right) = 1$, and let $Z = {Z_\beta } \cup {Z_\delta }$ be a decomposition of Z. Then, the number c of entanglement states required for EAQECCs obtained from C is equal to $| {{Z_\beta }} |$.

3.1 Entanglement-assisted quantum MDS codes of length $n=q^2+1$

In Chen et al. (2017), authors study the construction of EAQMDSCs of length $q^2+1$ for odd prime power $q \equiv 1\left( {\bmod 4} \right)$. However, the case $q \equiv 3\left( {\bmod 4} \right)$ is not considered. In this section, using negacyclic codes over $F_{q^2}$, we construct EAQMDSCs codes of length $q^2+1$ for odd prime power q satisfying $q \equiv 3\left( {\bmod 4} \right)$ and $q > 3$.

Lemma 2

Let $q>3$ be an odd prime power of the form $q=4k+3$. Let $n=q^2+1$ and $s = {n / 2}$. For $0 \le i \le \frac{{{q^2} - 1}}{4}$, all $q^2$-cyclotomic cosets modulo 2n containing $s-2i$ are as follows:

• 1. For all $0 < i \le \frac{{{q^2} - 1}}{4}$, ${C_{s - 2i}} = \left\{ {s - 2i,s + 2i} \right\}$.

• 2. ${C_s} = \left\{ s \right\}$.

Proof

Since ${o_{2n}}\left( {{q^2}} \right) = 2$, the size of $q^2$-cyclotomic coset modulo 2n is less than or equal to 2. For $C_{s-2i}$, $0 \le i \le \frac{{{q^2} - 1}}{4}$, it follows from ${q^2}\left( {1 + 2j} \right) \equiv 1 + 2\left( {\frac{{{q^2} - 1}}{2} - j} \right) \bmod 2n$ that ${q^2}\left( {s - 2i} \right) = {q^2}\left( {1 + 2\left( {\frac{{{q^2} - 1}}{4} - i} \right) } \right) \equiv 1 + 2\left( {\frac{{{q^2} - 1}}{4} + i} \right) = s + 2i\bmod 2n$ and so ${C_{s - 2i}} = \left\{ {s - 2i,s + 2i} \right\}$. Moreover, if $i=0$, then $C_s=\left\{ {s} \right\}$. $\square$

Lemma 3

Let $q>3$ be an odd prime power of the form $q=4k+3$. Let $n=q^2+1$ and $s = {n / 2}$. Then

• 1. $- q{C_s} = {C_s}$,

• 2. $- q{C_{s - 2\left( {q - 1} \right) }} = {C_{s - 2\left( {q + 1} \right) }}$,

• 3. $- qZ \cap Z = \emptyset$, where $Z = \bigcup \nolimits _{i = 1}^\lambda {{C_{s - 2i}}}$, $1 \le \lambda \le q - 1$,

• 4. $- q{C_{s - 2}} = {C_{s - 2q}}$.

Proof

• 1. It follows from $\left( {q + 1} \right) \frac{{{q^2} + 1}}{2} \equiv 0\bmod 2n$ that $- q\frac{{{q^2} + 1}}{2} \equiv \frac{{{q^2} + 1}}{2}\bmod 2n$, and so, $- q{C_s} = {C_s}$.

• 2. Since ${q^2} \equiv - 1\bmod 2n$ and $- qs \equiv s\bmod 2n$, $- q\left( {s - 2\left( {q - 1} \right) } \right) \equiv s - 2\left( {q + 1} \right) \bmod 2n$. This implies that $- q{C_{s - 2\left( {q - 1} \right) }} = {C_{s - 2\left( {q + 1} \right) }}$.

• 3. It is enough to prove that $- qZ \cap Z = \emptyset$ for $Z = \bigcup \nolimits _{i = 1}^{q-1} {{C_{s - 2i}}}$. By Lemma 2, $Z = \left\{ {s - 2\left( {q - 1} \right) ,s - 2\left( {q - 2} \right) , \ldots ,s - 2,s + 2, \ldots ,s + 2\left( {q - 2} \right) ,s + 2\left( {q - 1} \right) } \right\}$. Suppose that $- qZ \cap Z \ne \emptyset$. Then, there exists $s \pm 2i$, $s \pm 2j$ for some $1 \le i,j \le q - 1$, such that $- q\left( {s \pm 2i} \right) \equiv s \pm 2j\bmod 2n$. This implies that $\pm qi \pm j \equiv 0\bmod n$. However, since $q + 1 \le qi + j \le {q^2} - 1 < n$ and $1 \le qi - j < n - q$, there does not exist $1 \le i,j \le q - 1$ satisfying the congruence $\pm qi \pm j \equiv 0\bmod n$. This is a contradiction.

• 4. It follows directly from $- qs \equiv s\bmod 2n$ and ${C_{s - 2q}} = \left\{ {s - 2q,s + 2q} \right\}$.

$\square$

Let $Z = {Z_1} \cup \left( {\bigcup \nolimits _{i = 2}^{q - 1} {{C_{s - 2i}}} } \right)$, where ${Z_1} = {C_s} \cup {C_{s - 2}} \cup {C_{s - 2q}}$. Then, by Lemma 3, it follows that $- q{Z_1} = {Z_1}$ and $( { - q\bigcup \nolimits _{i = 2}^{q - 1} {{C_{s - 2i}}} } ) \cap ( {\bigcup \nolimits _{i = 2}^{q - 1} {{C_{s - 2i}}} }) = \emptyset$. In this case, $\left| { - qZ \cap Z} \right| = 5$. Take C as a negacyclic code of length $q^2+1$ over $F_{q^2}$ with the defining set Z. Then, C is a ${\left[ {{q^2} + 1,{q^2} - 2q,2q + 2} \right] _{{q^2}}}$ negacyclic code. Applying Theorem 2 to the negacyclic code C and using Proposition 2, we have the following:

Theorem 3

Let $q>3$ be an odd prime power with $q \equiv 3\left( {\bmod 4} \right)$. Then, there exists EAQMDSC with parameters .

Proof

By the above argument, we have EAQECC with desired parameters. Since these parameters attain EA singleton bound, this is EAQMDSC. $\square$

We list parameters of some EAQMDSCs obtained via Theorem 3 in Table 1.

Table 1 Some parameters of entanglement-assisted quantum MDS codes obtained by Theorem 3

In addition to the EAQMDSCs obtained above, we have the following class of EAQMDSCs of length $q^2+1$ when $q \equiv 3\left( {\bmod 4} \right)$.

Theorem 4

Let $q>3$ be an odd prime power with $q \equiv 3\left( {\bmod 4} \right)$. Then, there exist EAQMDSCs with parameters , where $q + 1 \le \lambda \le 2q - 2$.

Proof

For each $q+1 \le \lambda \le 2q - 2$, define ${Z_\lambda } = \bigcup \nolimits _{i = 0}^\lambda {{C_{s - 2i}}}$ and $Z' = ( {\bigcup \nolimits _{i = 0}^1 {{C_{s - 2i}}} } ) \cup ( {\bigcup \nolimits _{j = q - 1}^{q + 1} {{C_{s - 2j}}} } )$. For $\lambda = q+1$, ${Z_{q + 1}} = Z' \cup ( {\bigcup \nolimits _{i = 2}^{q - 2} {{C_{s - 2i}}} })$. Then, by Lemma 3, $\left| {{Z_{q + 1}} \cap - q{Z_{q + 1}}} \right| = 9$. For $q+2 \le \lambda \le 2q - 2$, we get ${Z_\lambda } = Z' \cup ( {\bigcup \nolimits _{i = 2}^{q - 2} {{C_{s - 2i}}} } ) \cup ( {\bigcup \nolimits _{j = q + 2}^\lambda {{C_{s - 2j}}} } )$. Since $- qZ' = Z'$ by Lemma 3 (1), (2), and (4), it follows that $- qZ_\lambda = Z' \cup ( { - q\bigcup \nolimits _{i = 2}^{q - 2} {{C_{s - 2i}}} }) \cup ( { - q\bigcup \nolimits _{j = q + 2}^\lambda {{C_{s - 2j}}} } )$. Since $Z' \subseteq - q{Z_\lambda } \cap {Z_\lambda }$, $| { - q{Z_\lambda } \cap {Z_\lambda }} | \ge 9$. To get the result $| { - q{Z_\lambda } \cap {Z_\lambda }} | = 9$, we need to prove the following:

$$\begin{aligned} \left( { - q\bigcup \limits _{i = 2}^{q - 2} {{C_{s - 2i}}} } \right) \cap \left( {\bigcup \limits _{i = 2}^{q - 2} {{C_{s - 2i}}} } \right)= & {} \emptyset , \end{aligned}$$

(1)

$$\begin{aligned} \left( { - q\bigcup \limits _{i = 2}^{q - 2} {{C_{s - 2i}}} } \right) \cap \left( {\bigcup \limits _{j = q + 2}^\lambda {{C_{s - 2j}}} } \right)= & {} \emptyset , \end{aligned}$$

(2)

$$\begin{aligned} \left( {\bigcup \limits _{i = 2}^{q - 2} {{C_{s - 2i}}} } \right) \cap \left( { - q\bigcup \limits _{j = q + 2}^\lambda {{C_{s - 2j}}} } \right)= & {} \emptyset , \end{aligned}$$

(3)

$$\begin{aligned} \left( { - q\bigcup \limits _{i = q + 2}^\lambda {{C_{s - 2i}}} } \right) \cap \left( {\bigcup \limits _{i = q + 2}^\lambda {{C_{s - 2i}}} } \right)= & {} \emptyset . \end{aligned}$$

(4)

By Lemma 3 (3), the equality (1) holds. Since

$$\begin{aligned} \left( { - q\bigcup \limits _{i = 2}^{q - 2} {{C_{s - 2i}}} } \right) \cap \left( {\bigcup \limits _{j = q + 2}^\lambda {{C_{s - 2j}}} } \right) = \emptyset \end{aligned}$$

(5)

if and only if

$$\begin{aligned} \left( {\bigcup \limits _{i = 2}^{q - 2} {{C_{s - 2i}}} } \right) \cap \left( { - q\bigcup \limits _{j = q + 2}^\lambda {{C_{s - 2j}}} } \right) = \emptyset , \end{aligned}$$

(6)

it is enough to prove (2) instead of proving both (2) and (3). Suppose that $( { - q\bigcup \nolimits _{i = 2}^{q - 2} {{C_{s - 2i}}} } ) \cap \left( {\bigcup \nolimits _{j = q + 2}^\lambda {{C_{s - 2j}}} } \right) \ne \emptyset$. Then, there exist four cases for some integers $2 \le i \le q - 2$ and $q + 2 \le j \le \lambda$:

The case $- q\left( {s - 2i} \right) \equiv s - 2j\bmod 2n$: it follows that $qi + j \equiv 0\bmod n$. However, this contradicts with $0< 3q + 2 \le qi + j \le {q^2} - 2 < n$.

The case $- q\left( {s - 2i} \right) \equiv s + 2j\bmod 2n$: this implies that $qi - j \equiv 0\bmod n$. Since $0< 2 \le qi - j \le {q^2} - 3q - 2 < n$, this is a contradiction.

The proofs of the cases $- q\left( {s + 2i} \right) \equiv s - 2j\bmod 2n$ and $- q\left( {s + 2i} \right) \equiv s + 2j\bmod 2n$ are similar to the proofs of the above cases.

For (4), suppose that $\left( { - q\bigcup \nolimits _{i = q + 2}^\lambda {{C_{s - 2i}}} } \right) \cap \left( {\bigcup \nolimits _{i = q + 2}^\lambda {{C_{s - 2i}}} } \right) \ne \emptyset$. Then, for some integers $q + 2 \le i,j \le \lambda$, there are four cases.

The case $- q\left( {s - 2i} \right) \equiv s - 2j\bmod 2n$: this congruence is equivalent to $qi + j \equiv 0\bmod n$. However, this is a contradiction, since $n< q^2+3q+2 \le qi + j \le 2q^2-2 < 2n$.

The case $- q\left( {s - 2i} \right) \equiv s + 2j\bmod 2n$: then, $qi - j \equiv 0\bmod n$, which contradicts with $n< q^2+2 \le qi - j \le 2q^2-3q-2 < 2n$.

The proofs of the cases $- q\left( {s + 2i} \right) \equiv s - 2j\bmod 2n$ and $- q\left( {s + 2i} \right) \equiv s + 2j\bmod 2n$ are similar to the proofs of the cases $- q\left( {s - 2i} \right) \equiv s + 2j\bmod 2n$ and $- q\left( {s - 2i} \right) \equiv s - 2j\bmod 2n$, respectively.

Now, for each $q+1 \le \lambda \le 2q - 2$, let $C_{\lambda }$ be a negacyclic code of length $q^2+1$ over $F_{q^2}$ with the defining set $Z_{\lambda }$. Then, $C_{\lambda }$ is a ${\left[ {{q^2} + 1,{q^2} - 2\lambda ,2\lambda + 2} \right] _{{q^2}}}$ negacyclic code. Since $\left| { - q{Z_\lambda } \cap {Z_\lambda }} \right| = 9$, by Theorem 2 and Proposition 2, for each $q+1 \le \lambda \le 2q - 2$, we get EAQECC having desired parameters. Since EA singleton bound is attained, these EAQECCs are MDS. $\square$

We list parameters of some EAQECCs obtained via Theorem 4 in Table 2.

Table 2 Some parameters of entanglement-assisted quantum MDS codes obtained by Theorem 4

3.2 Entanglement-assisted quantum MDS codes of length $\frac{{{q^2} - 1}}{4}$

Let $q=4m+3$ be an odd prime power with $m\ge 1$ and $n=\frac{{{q^2} - 1}}{4}$. Let $\alpha$ be a $(r=4)$th primitive root of unity over $F_{q^2}$. Using $\alpha$-constacyclic codes over $F_{q^2}$ of length $\frac{{{q^2} - 1}}{4}$, we are going to construct EAQMDSCs of length $\frac{{{q^2} - 1}}{4}$. We have from Zhang and Chen (2014) that each $q^2$-cyclotomic coset modulo rn has exactly one element; that is, ${C_{1 + 4j}} = \left\{ {1 + 4j} \right\}$, $0 \le j \le n - 1$, since ${q^2} \equiv 1\bmod rn$.

We also have the following from Zhang and Chen (2014) with the notation ${C_{1 - 4j}} = {C_{1 + 4\left( {n - j} \right) }}$.

Lemma 4

(Zhang and Chen 2014) Let $n=\frac{{{q^2} - 1}}{4}$ and $r=4$, where $q>3$ is an odd prime power of the form $q=4m+3$. If $Z = \bigcup \nolimits _{j = - \frac{{q - 3}}{4}}^{\frac{{q - 3}}{2}} {{C_{1 + 4j}}}$, then $- q{Z} \cap {Z} = \emptyset$.

Since $1 + 4\left( {n - \left( \frac{{q + 1}}{4} \right) } \right) = 4n - q \equiv - q\bmod rn$, we have the following:

Lemma 5

Let $q=4m+3$ and $m \ge 1$ be an odd prime power. Let $n=\frac{{{q^2} - 1}}{4}$ and $r=4$. For $q^2$-cyclotomic cosets ${C_{1}}$ and ${C_{ - \frac{{q + 1}}{4}}}$ modulo rn, we have $- q{C_1} = {C_{ - \frac{{q + 1}}{4}}}$.

Theorem 5

Let q be an odd prime power of the form $4m+3$, $m \ge 1$. Then, for each integer $\frac{{q + 9}}{4} \le d \le q$, there exists an EAQMDSC having the parameters .

Proof

For each $0 \le \lambda \le \frac{{q - 3}}{4}$ and $0 \le \delta \le \frac{{q - 3}}{2}$, define the sets ${Z_{\lambda ,\delta }} = \bigcup \nolimits _{j = - \frac{{q + 1}}{4} - \lambda }^{\delta } {{C_{1 + 4j}}}$, ${Z_\lambda } = \bigcup \nolimits _{j = - \frac{{q + 1}}{4} - \lambda }^{ - \frac{{q + 1}}{4}} {{C_{1 + 4j}}}$ and ${Z_\delta } = \bigcup \nolimits _{j = - \frac{{q - 3}}{4}}^{ \delta } {{C_{1 + 4j}}}$. We are going to prove that $\left| {{Z_{\lambda ,\delta }} \cap - q{Z_{\lambda ,\delta }}} \right| = 2$. Since ${Z_{\lambda ,\delta }} = {Z_\lambda } \cup {Z_\delta }$, we get

$$\begin{aligned} {Z_{\lambda ,\delta }} \cap - q{Z_{\lambda ,\delta }}= & {} \left( {{Z_\lambda } \cup {Z_\delta }} \right) \cap \left( { - q{Z_\lambda } \cup - q{Z_\delta }} \right) \\= & {} \left( {{Z_\lambda } \cap - q{Z_\lambda }} \right) \cup \left( {{Z_\lambda } \cap - q{Z_\delta }} \right) \cup \left( {{Z_\delta } \cap - q{Z_\lambda }} \right) \cup \left( {{Z_\delta } \cap - q{Z_\delta }} \right) . \end{aligned}$$

Since ${C_1} \subseteq -q{Z_\lambda } \cap {Z_\delta }$ and ${C_{ - \frac{{q + 1}}{4}}} \subseteq {Z_\lambda } \cap - q{Z_\delta }$, to get the result $\left| {{Z_{\lambda ,\delta }} \cap - q{Z_{\lambda ,\delta }}} \right| = 2$, we are going to prove the following:

$$\begin{aligned} {Z_\delta } \cap - q{Z_\delta }= & {} \emptyset , \end{aligned}$$

(7)

$$\begin{aligned} {Z_\lambda } \cap - q{Z_\lambda }= & {} \emptyset , \end{aligned}$$

(8)

$$\begin{aligned} -q{Z_\lambda } \cap {Z_\delta }= & {} {C_1}, \end{aligned}$$

(9)

$$\begin{aligned} {Z_\lambda } \cap -q{Z_\delta }= & {} {C_{ - \frac{{q + 1}}{4}}}. \end{aligned}$$

(10)

It follows from Lemma 4 that equality (7) holds. For (8), suppose that ${Z_\lambda } \cap - q{Z_\lambda } \ne \emptyset$. Then, there exist some integers $- \frac{{q + 1}}{4} - \lambda \le i,j \le - \frac{{q + 1}}{4}$, such that $- q\left( {1 + 4i} \right) \equiv 1 + 4j\bmod rn$. This implies that $\frac{{q + 1}}{4} + qi + j \equiv 0\bmod n$. It follows from $- \frac{{q - 1}}{2} \le - \frac{{q + 1}}{4} - \lambda \le i,j \le - \frac{{q + 1}}{4}$ that $- \frac{{{q^2} - 1}}{2} + \frac{{q + 1}}{4} \le \frac{{q + 1}}{4} + qi + j \le - \frac{{{{\left( {q + 1} \right) }^2}}}{4} + \frac{{q + 1}}{4}$. This contradicts with $\frac{{q + 1}}{4} + qi + j \equiv 0\bmod n$, since $- 2n< - \frac{{{q^2} - 1}}{2} + \frac{{q + 1}}{4} \le \frac{{q + 1}}{4} + qi + j \le - \frac{{{{\left( {q + 1} \right) }^2}}}{4} + \frac{{q + 1}}{4} < - n$.

To prove the equality (9), for $- \frac{{q - 3}}{4} \le j \le \delta$ we count the integer(s) $- \frac{{q + 1}}{4} - \lambda \le i \le - \frac{{q + 1}}{4}$ satisfying $- q\left( {1 + 4i} \right) \equiv 1 + 4j\bmod rn$ or, equivalently, $\frac{{q + 1}}{4} + qi + j \equiv 0\bmod n$. It follows from $- \frac{{q - 1}}{2} \le - \frac{{q + 1}}{4} - \lambda \le i \le - \frac{{q + 1}}{4}$ and $- \frac{{q - 3}}{4} \le j \le \delta \le \frac{{q - 3}}{2}$ that $\frac{{ - {q^2} + q + 2}}{2} \le \frac{{q + 1}}{4} + qi + j \le \frac{{ - {q^2} + 2q - 5}}{4}$. Since $- 2n< \frac{{ - {q^2} + q + 2}}{2} \le \frac{{q + 1}}{4} + qi + j \le \frac{{ - {q^2} + 2q - 5}}{4} < 0$, the only possible value of $\frac{{q + 1}}{4} + qi + j$ is $-n$. The equality $\frac{{q + 1}}{4} + qi + j = - n$ implies that $j \equiv 0\bmod q$. Since $- \frac{{q - 3}}{4} \le j \le \delta \le \frac{{q - 3}}{2}$, j must be 0, and so, i must be $- \frac{{q + 1}}{4}$. This shows that $-q{Z_\lambda } \cap {Z_\delta } = {C_1}$. Therefore, the equality (9) holds. Then, since ${Z_\lambda } \cap - q{Z_\delta } = - q\left( { - q{Z_\lambda } \cap {Z_\delta }} \right) = - q{C_1} = {C_{ - \frac{{q + 1}}{4}}}$, the equality (10) holds.

Now, for each $0 \le \lambda \le \frac{{q - 3}}{4}$ and $0 \le \delta \le \frac{{q - 3}}{2}$, let $C_{\lambda ,\delta }$ be an $\alpha$-constacyclic code of length $\frac{q^2-1}{4}$ over $F_{q^2}$ with the defining set $Z_{\lambda ,\delta }$. Then, $C_{\lambda ,\delta }$ is an $\alpha$-constacyclic code with parameters ${\left[ {\frac{{{q^2} - 1}}{4},\frac{{{q^2} - 1}}{4} - d + 1,d} \right] _{{q^2}}}$, where $d = \frac{{q + 9}}{4} + \lambda + \delta$. In this case, since $\left| { - q{Z_{\lambda ,\delta } } \cap {Z_{\lambda ,\delta } }} \right| = 2$, by Theorem 2 and Proposition 2, for each $\frac{{q + 9}}{4} \le d \le q$, we get an EAQECC having the parameters . Since EA singleton bound is attained, EAQECCs that are constructed are MDS.

$\square$

We list parameters of some EAQMDSCs obtained via Theorem 5 in Table 3.

Table 3 Some parameters of entanglement-assisted quantum MDS codes obtained by Theorem 5

Since $- q\left( {1 + r\frac{{q - 3}}{4}} \right) \equiv 1 + r\frac{{q - 1}}{2}\bmod rn$, we have the following:

Lemma 6

Let $q=4m+3$, $m \ge 1$ be an odd prime power. Let $n=\frac{{{q^2} - 1}}{4}$ and $r=4$. For $q^2$-cyclotomic cosets ${C_{\frac{{q - 3}}{4}}}$ and ${C_{ \frac{{q - 1}}{2}}}$ modulo rn, we have $- q{C_{\frac{{q - 3}}{4}}} = {C_{ \frac{{q - 1}}{2}}}$.

We derive another class of EAQMDSCs with length $n=\frac{{{q^2} - 1}}{4}$, where the number of entanglement states required is $c=4$.

Theorem 6

Let q be an odd prime power of the form $4m+3$, $m \ge 1$. Then, for each integer $\frac{{3q + 7}}{4} \le d \le \frac{{5q + 1}}{4}$, there exists an EAQMDSC having the parameters .

Proof

For each $0 \le \lambda ,\delta \le \frac{{q - 3}}{4}$, define the sets ${Z_{\lambda ,\delta }} = \bigcup \nolimits _{j = - \frac{{q + 1}}{4} - \lambda }^{\frac{{q - 1}}{2} + \delta } {{C_{1 + 4j}}}$, ${Z_\lambda } = \bigcup \nolimits _{j = - \frac{{q + 1}}{4} - \lambda }^{-\frac{{q + 1}}{4}} {{C_{1 + 4j}}}$, and ${Z_\delta } = \bigcup \nolimits _{j = - \frac{{q - 3}}{4}}^{\frac{{q - 1}}{2} + \delta } {{C_{1 + 4j}}}$. We are going to prove that $\left| {{Z_{\lambda ,\delta }} \cap - q{Z_{\lambda ,\delta }}} \right| = 4$. Since ${Z_{\lambda ,\delta }} = {Z_\lambda } \cup {Z_\delta }$, we get

$$\begin{aligned} {Z_{\lambda ,\delta }} \cap - q{Z_{\lambda ,\delta }}= & {} \left( {{Z_\lambda } \cup {Z_\delta }} \right) \cap \left( { - q{Z_\lambda } \cup - q{Z_\delta }} \right) \\= & {} \left( {{Z_\lambda } \cap - q{Z_\lambda }} \right) \cup \left( {{Z_\lambda } \cap - q{Z_\delta }} \right) \cup \left( {{Z_\delta } \cap - q{Z_\lambda }} \right) \cup \left( {{Z_\delta } \cap - q{Z_\delta }} \right) . \end{aligned}$$

To prove that $\left| {{Z_{\lambda ,\delta }} \cap - q{Z_{\lambda ,\delta }}} \right| = 4$, it is enough to prove the following:

$$\begin{aligned} {Z_\lambda } \cap - q{Z_\lambda }= & {} \emptyset \end{aligned}$$

(11)

$$\begin{aligned} - q{Z_\lambda } \cap {Z_\delta }= & {} {C_1} \end{aligned}$$

(12)

$$\begin{aligned} {Z_\lambda } \cap - q{Z_\delta }= & {} {C_{ - \frac{{q + 1}}{4}}} \end{aligned}$$

(13)

$$\begin{aligned} {Z_\delta } \cap - q{Z_\delta }= & {} {C_{ \frac{{q - 3}}{4}}} \cup {C_{ \frac{{q - 1}}{2}}}. \end{aligned}$$

(14)

The equality (8) implies that the equality (11) holds. For (12), we count the integer(s) $- \frac{{q - 3}}{4} \le j \le \frac{{q - 1}}{2} + \delta$, such that $- q\left( {1 + 4i} \right) \equiv 1 + 4j\bmod rn$ or, equivalently, $\frac{{q + 1}}{4} + qi + j \equiv 0\bmod n$, where $- \frac{{q + 1}}{4} - \lambda \le i \le -\frac{{q + 1}}{4}$. It follows from $- \frac{{q - 1}}{2} \le - \frac{{q + 1}}{4} - \lambda \le i \le -\frac{{q + 1}}{4}$ and $- \frac{{q - 3}}{4} \le j \le \frac{{q - 1}}{2} + \delta \le \frac{{3q - 5}}{4}$ that $- 2n< - \frac{{{q^2} - q - 2}}{2} \le \frac{{q + 1}}{4} + qi + j \le -\frac{{{q^2} - 3q +4}}{4} < 0$. This implies that the only possible value of $\frac{{q + 1}}{4} + qi + j$ is $-n$. If $\frac{{q + 1}}{4} + qi + j=-n$, then $j \equiv 0 \bmod q$. Since $- \frac{{q - 3}}{4} \le j \le \frac{{3q - 5}}{4}$, j must be 0, and so, the equality (12) holds. Since ${Z_\lambda } \cap - q{Z_\delta } = - q\left( { - q{Z_\lambda } \cap {Z_\delta }} \right) = - q{C_1} = {C_{ - \frac{{q + 1}}{4}}}$, the equality (13) holds.

To prove the equality (14), for $- \frac{{q - 3}}{4} \le i \le \frac{{q - 1}}{2} + \delta$ we count the integer(s) $- \frac{{q - 3}}{4} \le j \le \frac{{q - 1}}{2} + \delta$ satisfying $- q\left( {1 + 4i} \right) \equiv 1 + 4j\bmod rn$ or, equivalently, $\frac{{q + 1}}{4} + qi + j \equiv 0\bmod n$. It follows from $- \frac{{q - 3}}{4} \le i,j \le \frac{{q - 1}}{2} + \delta \le \frac{{3q - 5}}{4}$ that $- n< - \frac{{{q^2} - 3q - 4}}{4} \le \frac{{q + 1}}{4} + qi + j \le \frac{{3{q^2} - q - 4}}{4} < 3n$. This implies that the possible values of $\frac{{q + 1}}{4} + qi + j$ are 0, n, and 2n.

The case $\frac{{q + 1}}{4} + qi + j=0$: Then, $4j+1 \equiv 0 \bmod q$. Since $- q< -q+4 \le 4j + 1 \le 3q - 4 < 3q$, $4j+1$ can be 0, q or 2q. These cases are impossible, since j is an integer and $q \equiv 3 \bmod 4$. Hence, there is no solution in this case.

The case $\frac{{q + 1}}{4} + qi + j=n$: Then, $4j+2 \equiv 0 \bmod q$. Since $- q< -q+5 \le 4j + 2 \le 3q - 3 < 3q$, $4j+2$ can be 0, q and 2q. The only possibility is $4j+2=2q$. If $4j+2=2q$, then $j = \frac{{q - 1}}{2}$ and $i = \frac{{q - 3}}{4}$.

The case $\frac{{q + 1}}{4} + qi + j=2n$: Then, $4j+3 \equiv 0 \bmod q$. Since $- q< -q+6 \le 4j + 3 \le 3q - 2 < 3q$, $4j+3$ can be 0, q, or 2q. The only possibility is $4j+3=q$. If $4j+3=q$, then $j = \frac{{q - 3}}{4}$ and $i = \frac{{q - 1}}{2}$.

All cases of $\frac{{q + 1}}{4} + qi + j$ imply that $- q{Z_\delta } \cap {Z_\delta } = {C_{ \frac{{q - 3}}{4}}} \cup {C_{ \frac{{q - 1}}{2}}}$. Hence, the equality (14) holds. Since the equalities (11), (12), (13), and (14) hold, we conclude that $\left| {{Z_{\lambda ,\delta }} \cap - q{Z_{\lambda ,\delta }}} \right| = 4$.

Now, for each $0 \le \lambda ,\delta \le \frac{{q - 3}}{4}$, let $C_{\lambda ,\delta }$ be an $\alpha$-constacyclic code of length $\frac{q^2-1}{4}$ over $F_{q^2}$ with the defining set $Z_{\lambda ,\delta }$. Then, $C_{\lambda ,\delta }$ is an $\alpha$-constacyclic code with parameters ${\left[ {\frac{{{q^2} - 1}}{4},\frac{{{q^2} - 1}}{4} - d + 1,d} \right] _{{q^2}}}$, where $d = \frac{{3q + 7}}{4} + \lambda + \delta$. In this case, since $\left| { - q{Z_{\lambda ,\delta } } \cap {Z_{\lambda ,\delta } }} \right| = 4$, by Theorem 2 and Proposition 2, for each $\frac{{3q + 7}}{4} \le d \le \frac{{5q + 1}}{4}$, we get an EAQECC having the parameters . Since EA singleton bound is attained, EAQECCs that are constructed are MDS. $\square$

We list parameters of some EAQMDSCs obtained via Theorem 6 in Table 4.

Table 4 Some parameters of entanglement-assisted quantum MDS codes obtained by Theorem 6

4 Conclusion

We derive four new families of EAQMDSCs from constacyclic codes over $F_{q^2}$ for lengths $q^2+1$ and $\frac{q^2-1}{4}$. The EAQMDSCs with length ${q^2+1}$ are of minimum distance more greater than $q+1$. When compared to quantum MDS codes existing in the literature, the EAQMDSCs with length $\frac{q^2-1}{4}$ have large minimum distances. For instance, while , , and quantum codes are obtained via the construction in Zhang and Chen (2014), for same length and dimensions, we get , , and EAQMDSCs of larger minimum distance via Theorem 5. We present Table 5 to indicate this comparison.

Table 5 List of comparisons between EAQMDSCs and quantum MDS codes

Recently, Lu et al. (2018) derived EAQMDSCs with length $\frac{q^2-1}{at}$, where $q=atm+1$, a is even or a is odd and t is even. We remark that this class of EAQMDSCs does not include our construction of EAQMDSCs with length $\frac{q^2-1}{4}$ since $q \equiv 3\left( {\bmod 4} \right)$ for our construction, but $q=4m+1$ for the construction given in Lu et al. (2018).

Chen et al. (2017) constructed EAQMDSc with $c=4$ and length $\lambda \left( q+1 \right)$, where $\lambda \ge 3$ is an odd integer dividing $q-1$. Since $q \equiv 3\left( {\bmod 4} \right)$ in our construction given in Theorem 6, their construction does not include ours. Moreover, they also constructed a class of EAQMDSCs of length $q^2+1$ for odd prime power $q \equiv 1\left( {\bmod 4} \right)$. However, the construction given in Theorems 3 and 4 is different from their construction, since $q \equiv 3\left( {\bmod 4} \right)$ for our construction.

Liu et al. (2018) derived two classes of EAQMDSCs with the parameters , $\frac{{3\left( {q + 1} \right) }}{4} \le d \le q$ and , $q + 1 \le d \le \frac{{5q + 1}}{4}$. However, our constructions in Theorems 5 and 6 give us larger classes than ones in Theorem 10 of Liu et al. (2018) since $\frac{{q + 9}}{4} \le \frac{{3 ( {q + 1} )}}{4}$ and $\frac{{3q + 7}}{4} \le q + 1$. For instance, letting $q=11$, we get EAQMDSCs having the parameters , $5 \le d \le 11$ and , $10 \le d \le 14$, while EAQMDSCs with parameters , $9 \le d \le 11$ and , $12 \le d \le 14$ are obtained via the construction in Liu et al. (2018). We give Table 6 to indicate this comparison.

Table 6 List of comparisons for EAQMDSCs between us and Liu et al. (2018)

For future studies, the construction of EAQECCs which are both maximal entanglement and MDS with respect to entanglement-assisted singleton bound is an open and attractive problem. While Guenda et al. (2018) obtained such EAQECCs from linear complementary dual (LCD) codes and Reed–Solomon codes based on the hull of classical linear codes, this problem is not solved completely.