I'm reading up about the binary Golay code of length $23$. I know it's a cyclic code and I also know it's a quadratic residue code. I've read that we can consider the linear code over $\mathbb{F}_2$ generated by the $23$ vectors $c_i$ with coordinates $(c_i)_j = 1$ if $(j-i)$ is a quadratic residue mod $23$ and $0$ otherwise.
Why is this true? How does this relate to it being a quadratic residue code? I thought that $x^{23} + 1 = (x+1)(x^{11}+x^{10}+x^6+x^5+x^4+x^2+1)(x^{11} + x^9 + x^7 + x^6 + x^5 + x + 1)$, then we can chose either of the polynomials of degree $11$ to be a generator polynomial. Clearly the construction above would yield a completely different generator matrix than these generator polynomials would give. How is this possible?
The generator matrix of a linear code is not unique.
An $[n,k,d]_{q}$ code is a $k$-dimensional subspace of $\mathbb{F}_{q}^{n}$, and so there are $$\prod_{i=0}^{k-1} (q^{k}-q^{i})$$ different ordered bases; each one of these will give you a different generator matrix. And that is just the number of generator matrices for a given code, you have even more if you consider different but equivalent codes.
So choosing an irreducible $11$-degree factor of $x^{23}-1$ will give you a cyclic basis for your code, giving you a cyclic generator matrix. Using the quadratic residue construction gives you a different generator matrix, will most likely even give you a different code than you obtain from the cyclic construction (as in, a different set of codewords), but the codes are equivalent (ie can be obtained from each other by a permutation of coordinates).