130
6 Asymmetric Quantum Codes
as follows: the smaller the cardinality of the defining set is, the greater is its dimension. According to this idea, we need to show the existence of distinct and specific
singleton cyclotomic cosets contained in the defining sets of codes C 1 and C, where
C is the code equivalent to the code C
⊥
2 . Additionally, we need to find the cardinality
of their defining sets as well as to show that they are mutually disjoint. These results
are presented in Lemma 6.3.2 to Lemma 6.3.7. They enable us to compute the exact
dimension of the corresponding AQECC, which is a hard task, since the dimension
of BCH codes is not known in general. In our construction, we use the code C 1 to
correct phase-shift errors and C
⊥
2 to correct qudit-flip errors.
We will utilize the following lemmas shown in [105].
Lemma 6.3.2 Let n = q
m
− 1, where q ≥ 3 is an odd prime power and m ≥ 3 is
an integer. Then
(i) the cyclotomic coset C [
q m −1
2 ] contains only one element;
(ii) the coset C [
q m −1
2 −1] contains the element
q
m −1
2
− q;
(iii) the coset C [
q m −1
2 +1] contains the element
q
m −1
2
+ q.
Proof See [105, Lemma 3.1] for a detailed proof.
Lemma 6.3.3 If n = q
m
− 1, where q ≥ 3 is an odd prime power and m ≥ 3 is
an integer, then the q-cosets C 1 , C 2 , . . . , C q−1 , C q+1 , . . . , C 2q−1 (modulo n) are
mutually disjoint and each of them has m elements.
Proof See [105, Lemma 3.2].
Lemma 6.3.4 If n = q
m
− 1, where q ≥ 3 is an odd prime power and m ≥ 3 is an
integer (if q = 3, m ≥ 4), then the q-cosets
C 0 , C 1 , C 2 , . . . , C q−1 , C q+1 , . . . , C 2q−1
are disjoint from the q-cosets C [
q m −1
2 +k] , where k = 0, 1, . . . , q − 1.
Proof See [105, Lemma 3.3].
Lemma 6.3.5 If n = q
m
− 1, where q ≥ 3 is an odd prime power and m ≥ 3 is an
integer (if q = 3, m ≥ 4), then the q-cosets
C 0 , C 1 , C 2 , . . . , C q−1 , C q+1 , . . . , C 2q−1
are disjoint from the q-cosets C [
q m −1
2 −k] , where k = 1, . . . , q − 1.
Proof See [105, Lemma 3.4].
Lemma 6.3.6 Let n = q
m
− 1, where q ≥ 3 is an odd prime power and m ≥ 3 is
an integer. Then we have
6 Asymmetric Quantum Codes
as follows: the smaller the cardinality of the defining set is, the greater is its dimension. According to this idea, we need to show the existence of distinct and specific
singleton cyclotomic cosets contained in the defining sets of codes C 1 and C, where
C is the code equivalent to the code C
⊥
2 . Additionally, we need to find the cardinality
of their defining sets as well as to show that they are mutually disjoint. These results
are presented in Lemma 6.3.2 to Lemma 6.3.7. They enable us to compute the exact
dimension of the corresponding AQECC, which is a hard task, since the dimension
of BCH codes is not known in general. In our construction, we use the code C 1 to
correct phase-shift errors and C
⊥
2 to correct qudit-flip errors.
We will utilize the following lemmas shown in [105].
Lemma 6.3.2 Let n = q
m
− 1, where q ≥ 3 is an odd prime power and m ≥ 3 is
an integer. Then
(i) the cyclotomic coset C [
q m −1
2 ] contains only one element;
(ii) the coset C [
q m −1
2 −1] contains the element
q
m −1
2
− q;
(iii) the coset C [
q m −1
2 +1] contains the element
q
m −1
2
+ q.
Proof See [105, Lemma 3.1] for a detailed proof.
Lemma 6.3.3 If n = q
m
− 1, where q ≥ 3 is an odd prime power and m ≥ 3 is
an integer, then the q-cosets C 1 , C 2 , . . . , C q−1 , C q+1 , . . . , C 2q−1 (modulo n) are
mutually disjoint and each of them has m elements.
Proof See [105, Lemma 3.2].
Lemma 6.3.4 If n = q
m
− 1, where q ≥ 3 is an odd prime power and m ≥ 3 is an
integer (if q = 3, m ≥ 4), then the q-cosets
C 0 , C 1 , C 2 , . . . , C q−1 , C q+1 , . . . , C 2q−1
are disjoint from the q-cosets C [
q m −1
2 +k] , where k = 0, 1, . . . , q − 1.
Proof See [105, Lemma 3.3].
Lemma 6.3.5 If n = q
m
− 1, where q ≥ 3 is an odd prime power and m ≥ 3 is an
integer (if q = 3, m ≥ 4), then the q-cosets
C 0 , C 1 , C 2 , . . . , C q−1 , C q+1 , . . . , C 2q−1
are disjoint from the q-cosets C [
q m −1
2 −k] , where k = 1, . . . , q − 1.
Proof See [105, Lemma 3.4].
Lemma 6.3.6 Let n = q
m
− 1, where q ≥ 3 is an odd prime power and m ≥ 3 is
an integer. Then we have