140
6 Asymmetric Quantum Codes
distance d and an RS code C
⊥
2 with minimum distance d − 1, so the inequality
p
m
≥ 2d − 2 must be satisfied.
Before proceeding further, we need the following result shown in [12, 56].
Lemma 6.5.1 (i) Let C be an [N , K , D 1 ] p m linear code and β be a basis of
F p m over F p . Then the p-ary expansion β(C) of C with respect to β is an
[m N, m K, D 2 ≥ D 1 ] p linear code.
(ii) Let C be an [N , K , D] p m linear code with Euclidean dual C
⊥ . Then the dual
code of the p-ary expansion β(C) of C with respect to β is the p-ary expansion
β
⊥
(C
⊥
) of the dual code C
⊥ with respect to β
⊥ .
We next construct AQECCs derived from RS codes.
Theorem 6.5.2 Let p be a prime number. Then there exists an
[[N = m( p
m
− 1), K = m( p
m
− 2d + 2), d z ≥ d/d x ≥ (d − 1)]] p
asymmetric quantum error-correcting code, where m ≥ 1 is an integer.
Proof Let C 1 = [n, k 1 , d 1 ] p m be the RS code generated by the polynomial
g 1 (x) = (x − 1)(x − α)(x − α
2
) · · · (x − α
(d−2)
),
where α is a primitive element in F p m and d ≥ 3 is a fixed integer. We know that C 1
is an [ p
m
− 1, p
m
− d, d] p m code.
Let U F p m be the set of units of F p m ordered in increasing order of the powers of
α:
U F p m = {α
0
= 1, α, α
2
, . . . , α
( p
m −2)
}.
The set of units of F p m , except the roots of g 1 (x), is denoted by
NR(g 1 (x)) = U F p m − R(g 1 (x)) =
= {α
d−1
, α
d
, α
d+1
, . . . , α
( p
m −2)
},
where NR(g 1 (x)) is also ordered in increasing order of the powers of α.
Let us next consider A be the set consisting of the last d − 2 elements of
NR(g 1 (x)), i.e.,
A = {α
[ p
m −(d−1)]
, α
[ p
m −(d−2)]
, . . . , α
( p
m −2)
}.
Let C 2 be the code generated by
g 2 (x) =
p
m −d
j=0
(x − α
j
).