Direct proof of minimum distance of the binary Golay code

Question

Let $\alpha$ be a primitive $23$-rd root of unity in $GF(2^{11})$. Consider the binary cyclic code generated by $g(x)=\prod_{i \in I}(x-\alpha^{i})$, where $I=\left \{1, 2, 4, 8, 16, 9, 18, 13, 3, 6, 12 \right \}$. It can easily be proven that, if $w$ is the weight of an even weight code word, then $4$ divides $w(w-1)$. Is there a simple way of knowing that $A_{i}= A_{23-i}$, where $A_{i}$ is the number of words of weight $i$? Because then, one would have $0=A_{18}=A_{5}$ and also $A_{6}=0$ and by the BCH-bound, the minimum distance is at least $5$, so, since it cannot be more than $7$, it must be $7$.

Answer 1

Actually not hard at all. Since, over $GF(2)$, $$x^{23}-1=(x-1)(x^{22}+x^{21}+ \ldots+ 1)=(x-1)q(x)g(x)$$ with $g$ and $q$ irreducible, we have that $$x^{22}+x^{21}+ \ldots+ 1=q(x)g(x)$$ hence $c=(1,1, \ldots, 1)$ is a codeword. Thus, if $d$ is a codeword of weight $i$, then $c+d$ is a codeword of weight $23-i$. We have established a bijection between weight $i$ and weight $23 -i$ codewords.