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?
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 $\delta$.