On my notes of the lesson I have a proof that the dual code of the Hamming code (7,4,3) is not cyclic. But a code is cyclic if and only if its dual is cyclic and I have proved by testing each word of the code that the Hamming code (7,4,3) is cyclic (because each shifting of a word of the code is a word of the code). Can someone help me finding the mistake in this proof please?
I consider the generating matrix of the dual code: $$G^t=H=\pmatrix{0& 0& 1\\ 0 &1& 0\\ 0&1& 1\\ 1 &0& 0\\ 1 &0 &1\\ 1 &1& 0 \\1& 1& 1}$$ $v_1$, $v_2$, $v_3$ are the polynomials associated to the rows of $G$. I consider $$R=\frac{F[x]}{(x^7-1)}$$ I observe that $v_1=x^3+x^4+x^5+x^6=x^3(1+x+x^2+x^3)$. So $1+x+x^2+x^3$ must be a multiple of $g(x)$.
$v_2=x+x^2+x^5+x^6=x(1+x+x^4+x^5)$. So $1+x+x^4+x^5$ must be a multiple of $g(x)$.`
I find $$MCD(g(x),1+x+x^2+x^3)=x-1$$
If there would be another generator of the dual it should divide $MCD(x^7-1,1+x+x^2+x^3)$, that means $g(x)\mid x-1$ and this is not true.
If I call $D$ the code associated to $x-1$, $\dim D = 6$, $\dim(D_{\rm dual})\ge 6$ and this is absurd because $\dim(D_{\rm dual})=3$.
I conclude that the Hamming code $(7,4,3)$ is not cyclic.
Thank you