6.6 Tensor Product Codes
145
Proof Let α be a primitive element of F q . Let C 1 be the Reed–Solomon (RS) code
over F q generated by the polynomial
g 1 (x) =
d 1 −2
i=0
(x − α
i
),
where d 1 ≥ 2.
Assume that C 2 is the RS code over F q generated by
g 2 (x) =
d 2 −2
i=0
(x − α
i
),
where d 1 ≤ d 2 < q − 1. We know that C 1 is an [q − 1, q − d 1 , d 1 ] q code, and C 2 is
an [q − 1, q − d 2 , d 2 ] q code. From construction, it follows that C 2 ⊂ C 1 .
Let C 3 be the RS code over F q generated by
g 3 (x) =
d 3 −2
i=0
(x − α
i
),
where d 3 ≥ 2, and suppose C 4 is the RS over F q generated by
g 4 (x) =
d 4 −2
i=0
(x − α
i
),
where d 3 ≤ d 4 < q − 1. Similarly, C 3 is an [q − 1, q − d 3 , d 3 ] q code, C 4 is an
[q − 1, q − d 4 , d 4 ] q code and C 4 ⊂ C 3 . We know that the product code C 1 ⊗ C 3
is an
[(q − 1)
2
, (q − d 1 )(q − d 3 ), d 1 d 3 ] q
code, and C 2 ⊗ C 4 is an
[(q − 1)
2
, (q − d 2 )(q − d 4 ), d 2 d 4 ] q
code. Applying Lemma 6.6.1, we have C 2 ⊗ C 4 ⊂ C 1 ⊗ C 3 . From Theorem 6.6.2,
the Euclidean dual (C 2 ⊗ C 4 )
⊥ of C 2 ⊗ C 4 is an
[(q − 1)
2
, (q − 1)
2
− (q − d 2 )(q − d 4 ), d
⊥
] q
code, where d
⊥
= min{q − d 2 , q − d 4 }. Applying the CSS construction to the
codes C 1 ⊗ C 3 , C 2 ⊗ C 4 and (C 2 ⊗ C 4 )
⊥ we obtain the desired code. The proof is
complete.