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I have another homework to do, please give me some hints in order to solve this problem:

"Determine whether the dual of an arbitrary BCH code is a BCH code."

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Any suggestion is welcome – Math Geek Jan 12 '13 at 15:58
Suggestions: look up the definition of BCH code: look up the definition of dual of a code: look at some other examples where there are proofs that one code is the dual of another. – Gerry Myerson Jan 12 '13 at 19:43
@GerryMyerson The dual code contains all words $y$ such that $\langle x, y \rangle=0$ where $x \in C$. What can we get from that? – Evinda May 2 at 9:31
@Evinda, by itself, maybe not very much. But I also suggested looking up the definition of a BCH code. Maybe if you have that definition in front of you, it will be clear whether the dual of a BCH code is necessarily a BCH code. – Gerry Myerson May 2 at 10:48
According to my textbook: $$$$ The class of BCH codes is a generalization of the Hamming codes for multiple-error correction. $$$$ Let $\alpha$ be a primitive element of $\mathbb{F}_{q^m}$ and denote by $M^{(i)}(x)$ the minimal polynomial of $\alpha^i$ with respect to $\mathbb{F}_q$. A (primitive) BCH code over $\mathbb{F}_q$ of length $n=q^m-1$ with designed distance $\delta$ is a q-ary cyclic code generated by $g(x):=lcm(M^{(a)}(x),M^{(a+1)}(x), \dots, M^{a+\delta-2}(x))$ for some integer $a$ . $$$$ Does this help? – Evinda May 2 at 10:56

hint: RS codes are the special type of BCH codes,their dual are BCH too and all RS codes are MDS code and the dual of a MDS code is MDS too

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