My problem:
Let $\alpha$ be a primitive element of of $\textbf{F}_{2^{m}}$ and let $g\left(x\right) \in \textbf{F}_{2}\left[x\right]$ be the minimal polynomial of $\alpha$ with respect to $\mathbf{F}_{2}$. Show that a cyclic code of length $2^{m}-1$ with generator polynomial $g\left(x\right)$ is a binary $\left[2^m-1, 2^m-1-m, 3\right]$ code.
My work so far:
Let $n=2^m-1$ and let $\alpha$ be a primitive element in $\textbf{F}_{2^{m}}$. So, $\alpha$ will have order n.
Then, let the cyclic code of length $n$ of designed distance 3 be generated by the polynomial $g\left(x\right)$, of degree $m$.
Thus, our cyclic code generated by $g\left(x\right)$ will have size $n-m=2^m-1 – m$.
So, our code will be a binary $\left[2^m-1, 2^m-1-m, 3\right]$ code.
I'm not sure that I have approached this correctly, or if I should provide a better argument as to why the distance is 3 other than invoking that it has been designed that way. Any critiques or help would be appreciated.
Best Answer
So basically, the BCH bound says that if the generating polynomial of a cyclic code $C$ has $\delta-1$ consecutive roots $$\alpha^b,\alpha^{b+1},\ldots,\alpha^{b+\delta-2},$$ then the minimum distance is $d_{min}\geq \delta.$ Here $\alpha,\alpha^2$ are roots, so $\delta=3,$ so $d_{min}\geq 3.$
Now, it turns out that $$ |C| \times \textrm{Volume} =2^{2^m-m-1} \left\{1+\binom{2^m-1}{1}\right\}=2^{2^m-1} $$ so the (necessarily disjoint) volumes of all vectors at distance $\leq 1$ around the codewords fill out the whole space of binary vectors of lenth $2^m-1.$ This says that this cyclic code (Hamming Code) is perfect.