Quantum synchronizable codes from the Whiteman’s generalized cyclotomy

Abstract

Quantum synchronizable codes (QSCs) are special quantum error-correcting codes that can correct the effects of both quantum noise on qubits and misalignment in block synchronization. In this paper, using the factorizations of cyclotomic polynomials ${{\varPhi }}_{p_{1}p_{2}}(x)$, where p₁ and p₂ are distinct odd primes, we give a new method to construct QSCs whose synchronization capabilities can reach the best attainable tolerance against misalignment.

Appendix

Let p be an odd prime with p ≡ 1 (mod 4), $(\frac {\cdot }{p})_{4}$ denotes the quartic residue symbol. In fact, let a be an integer with $\gcd (a,p)=1$ and $(\frac qp)=1$, then $(\frac {a}{p})_{4}=1$ if and only if x⁴ ≡ a (mod p) has a solution over $\mathbb {Z}$; $(\frac {a}{p})_{4}=-1$ if and only if x⁴ ≡ a (mod p) has not any solution over $\mathbb {Z}$.

Proposition 4

For p₁ ≡ 1 (mod 4) and p₂ ≡ 3 (mod 4). Suppose that $(\frac {q}{p_{1}})=(\frac {q}{p_{2}})=1$.

• 1. Then ${{\varPhi }}_{p_{2}}(x)$ has not any self-reciprocal irreducible factor over $\mathbb {F}_{q}$.

• 2. If p₁ ≡ 5 (mod 8) and $(\frac {q}{p_{1}})_{4}=1$, then ${{\varPhi }}_{p_{1}}(x)$ has not any self-reciprocal irreducible factor over $\mathbb {F}_{q}$.

• 3. If p₁ ≡ 5 (mod 8) and $(\frac {q}{p_{1}})_{4}=-1$, then ${{\varPhi }}_{p_{1}}(x)$ has a self-reciprocal irreducible factor over $\mathbb {F}_{q}$.

Proof

Since $(\frac {q}{p_{1}})=(\frac {q}{p_{2}})$, q ∈ W₀ and q ≡ g^t (mod n), so q ≡ g^t (mod p₁) and q ≡ g^t (mod p₂). Hence t is even by $(\frac {q}{p_{1}})=(\frac {q}{p_{2}})=1$.

• 1. By p₂ ≡ 3 (mod 4), ord$_{p_{2}}(q)$ is odd. Then q^s ≡− 1 (mod p₂) does not hold. Hence ${{\varPhi }}_{p_{2}}(x)$ has not any self-reciprocal irreducible factor over $\mathbb {F}_{q}$ by Lemma 4.

• 2. Since $(\frac {q}{p_{1}})_{4}=1$, 4|t; by p₁ ≡ 5 (mod 8), ord$_{p_{1}}(q)$ is odd. By ord$_{p_{1}}(-1)=2$, q^s ≡− 1 (mod p₁) does not hold. Hence ${{\varPhi }}_{p_{1}}(x)$ has not any self-reciprocal irreducible factor over $\mathbb {F}_{q}$ by Lemma 4.

• 3. Since $(\frac {q}{p_{1}})_{4}=-1$, 2∥t; by p₁ ≡ 5 (mod 8), ord$_{p_{1}}(q)$ is even. Then there is an integer s such that q^s ≡− 1 (mod p₁). Hence ${{\varPhi }}_{p_{1}}(x)$ has a self-reciprocal irreducible factor over $\mathbb {F}_{q}$ by Lemma 4.

□

Proposition 5

For p₁ ≡ 1 (mod 4) and p₂ ≡ 3 (mod 4). Suppose that $(\frac {q}{p_{1}})=(\frac {q}{p_{2}})=-1$. Then ${{\varPhi }}_{p_{1}}(x)$ and ${{\varPhi }}_{p_{2}}(x)$ have self-reciprocal irreducible factor over $\mathbb {F}_{q}$.

Proof

Since $(\frac {q}{p_{1}})=(\frac {q}{p_{2}})$, q ∈ W₀ and q ≡ g^t (mod n), so q ≡ g^t (mod p₁). Hence t is odd due to $(\frac {q}{p_{1}})=-1$. Furthermore, ord$_{p_{1}}(q)=\frac {p_{1}-1}{\gcd (p_{1}-1,t)}$ is even. Then there is an integer s such that q^s ≡− 1 (mod p₁). Hence ${{\varPhi }}_{p_{1}}(x)$ has a self-reciprocal irreducible factor over $\mathbb {F}_{q}$ by Lemma 4

Similarly, ${{\varPhi }}_{p_{2}}(x)$ has self-reciprocal irreducible factor over $\mathbb {F}_{q}$. □

Proposition 6

For p₁ ≡ 3 (mod 4) and p₂ ≡ 3 (mod 4). Suppose that $(\frac {q}{p_{1}})=(\frac {q}{p_{2}})=-1$. Then ${{\varPhi }}_{p_{1}}(x)$, ${{\varPhi }}_{p_{2}}(x)$, and ${{\varPhi }}_{p_{1}p_{2}}(x)$ have self-reciprocal irreducible factor over $\mathbb {F}_{q}$.

Proof

Suppose that $(\frac {q}{p_{1}})=(\frac {q}{p_{2}})=-1$. Then ord$_{p_{1}}(q)$ and ord$_{p_{2}}(q)$ are even, thus ord$_{p_{1}p_{2}}(q)$ is also even. Then $q^{s_{i}}\equiv -1\pmod {p_{i}},i=1,2,$ and q^t ≡− 1 (mod p₁p₂) hold for some integer s₁,s₂,t. By Lemma 4, ${{\varPhi }}_{p_{1}}(x)$, ${{\varPhi }}_{p_{2}}(x)$, and ${{\varPhi }}_{p_{1}p_{2}}(x)$ have self-reciprocal irreducible factor over $\mathbb {F}_{q}$. □