Prove that a cyclic code of length $2^m-1$ is a binary $\left[2^m-1, 2^m-1-m, 3\right]$ code

Question

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.

Answer 1

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.

Answer 2

Look, Ma! no BCH bound.....

Every codeword $C(x)$ in the binary cyclic code of length $2^m-1$ with generator polynomial $g(x)$ is a multiple of $g(x)$ which is given to be the minimal polynomial of $\alpha$ which is a primitive element of $\mathbb F_{2^m}$. Since $g(\alpha) = 0$, we have that every codeword $C(x)$ has the property that $C(\alpha)= \sum_{i=0}^{2^m-2}C_i \alpha^i = 0$. Thus,

$$H = [1, \alpha, \alpha^2, \alpha^3, \cdots, \alpha^{2^m-2}]$$ is a parity-check matrix for the code. But the elements $1, \alpha, \alpha^2, \alpha^3, \cdots, \alpha^{2^m-2}$ are precisely the nonzero elements of $\mathbb F_{2^m}$. If we write each $\alpha^i$ as a column vector, we get the columns of $H$ are the $2^m-1$ binary nonzero $m$-tuples. Now, where have I seen that kind of parity-check matrix before? And what do I know about the code that is the null space of such a matrix??