Find minimum distance of the code

I want to show that the minimum distance of a narrow-sense BCH code over $\mathbb{F}_q$ with length $n$ and designed distance $\delta$ is equal to $\delta$ provided that it holds that $\delta \mid n$.

It holds that a BCH code with designed distance $\delta$ has minimum distance at least $\delta$.

So we need to show that there is a codeword with Hamming weight $\delta$.

How do we deduce the existence of such a codeword from the fact that $\delta \mid n$ ?

Can we consider that the length of the code is of the form $q^m-1$ ?

If so, then we could use the fact that a $q$-ary bch code of length $q^m-1$ with designed distance $\delta$ has dimension at least $q^m-1-m(\delta-1)$, couldn't we?

Then from the sphere packing bound and the above proposition we get that $\sum_{i=0}^{\lfloor \frac{d-1}{2} \rfloor} \binom{n}{i}(q-1)^i \leq q^{m(\delta-1)}$.

But does this help?

Or isn't the idea right?

• By the way, this and your previous questions contains in store.elsevier.com/The-Theory-of-Error-Correcting-Codes/… Apr 29 '16 at 0:20
• @Mr.Newman Is there maybe also an online version? Apr 30 '16 at 10:52
• Sorry, I can't find english edition ( Apr 30 '16 at 11:54

The Peterson theorem: if $n=\delta b$, then the distance of narrow-sense BCH with designed distance $\delta$ is equal to $\delta$
Proof: Let $\alpha$ has the order $n$ and $\alpha,\alpha^2,...,\alpha^r$ is a BCH-chain for $r=\delta-1$. Not hard to see $$x^n-1=(x^b-1)(1+x^b+...+x^{rb}).$$ Since $\alpha^{ib}\ne1$ for $i=1,...,r$ then all of chain elements are roots of $h(x)=1+x^b+...+x^{rb}$. So, $h(x)$ is a word of weight $\delta$.
• What do you mean with BCH-chain?  Do you mean the sequence of powers of $\alpha$ such that $c(\alpha^b)=c(\alpha^{b+1})=\dots+ c(\alpha^{b+\delta-2})=0$ if $c$ is a codeword? Apr 29 '16 at 8:13
• I mean by "BCH-chain" the chain of elements $\alpha,\alpha^2,...,\alpha^r$ from which we construct the code. Since $h(\alpha)=h(\alpha^2)=...=h(\alpha^r)=0$ then $h(x)$ is a codeword Apr 29 '16 at 8:30
• Is $h$ a word of weight $\delta$ since it has $\delta$ roots?  If so, we know that it holds that $$c(x) \in C \Leftrightarrow c(\alpha^b)=c(\alpha^{b+1})=\dots=c(\alpha^{b+\delta-2})=0$$ Wouldn't this mean that all the words of a BCH code have Hamming weight $\delta$ ? Or am I wrong? Apr 29 '16 at 9:31
• How to construct BCH code of the length $n$ over the field $GF(q)$ with designed distance $r$. We have to choose 1) $Q=GF(q^l)$ an extension of the field $GF(p)$ 2) $\alpha\in Q$ an element of order $n$ 3) a system $\alpha^s,\alpha^{s+1},...,\alpha^{s+r-1}$ (chain) Further, $g(x)$ be a minimal polynomial which annihilate any of $\alpha^i$. So, the cyclic code over $P$ with generating polynomial $g(x)$ is our BCH-code. Your case is narrow-sense, i.e. $s=1$. Any of codewords $c(x)$ can be represented in the form $$c(x)=g(x)t(x)(\mathrm{mod\ }(x^n-1))$$ Apr 29 '16 at 9:47
• So $h(x)$ is word of weight $\delta$ since weight of $(h_r,...,h_1,h_0)=(1,...,1,1)$ is equal to $\delta$. So any of codewords $c(x)$ should have at least $\delta$ roots but it's weight can be greater than $\delta$ Apr 29 '16 at 9:53