5.2 BCH Codes—Part II
83
Let u
∗
= v
∗
− 1; it is clear that 0 ≤ u
∗
< n. Moreover, by the uniqueness of the
remainder, u
∗ is the remainder of u when considering the modulo n. This means
that the representatives v
∗ and u
∗ of v and u, respectively, are consecutive, i.e.,
v
∗
= u
∗
+ 1. The proof is complete.
5.2.2 Code Constructions
In this subsection we apply some results of Sect. 5.2.1 in order to construct families
of q-ary CSS quantum codes.
5.2.2.1 Construction I
In order to proceed further we need to prove Lemma 5.2.4.
Lemma 5.2.4 Let p be a prime number, p ≥ 5. Let n = p
2
− 1 and consider the
first 2 p − 2 p-ary cosets modulo n given by
C 0 = {0},
C 1 = {1, p},
C 2 = {2, 2 p},
C 3 = {3, 3 p},
. . .
C p−2 = {p − 2, (p − 2) p},
C p+1 = {p + 1},
C p+2 = {p + 2, 1 + 2 p},
. . .
C 2 p−1 = {2 p − 1, 1 + ( p − 1) p}.
Then, all these p-cosets modulo n are disjoint. Moreover, with exception of the cosets
C 0 and C p+1 , which contain only one element, all of them have exactly two elements.
Proof We know that p
2
− 1 > 1 + ( p − 1) p and p
2
− 1 > (p − 2) p are true. It is
clear that C 0 and C p+1 contain only one element. We next show that, except cosets
C 0 and C p+1 , all of them have exactly two elements.
Case 1: If l = lp, since l = lp < p
2
− 1 we obtain p = 1, a contradiction since
p is a prime.
Case 2: Assume that p + l = 1 + lp, where 2 ≤ l ≤ p − 1 is an integer. Then one
has l − 1 = p(l − 1). Since p + l = 1 + lp < p
2
− 1 and l − 1 = 0, one obtains
p = 1, which is a contradiction.