Codex @codex  0

Work over the finite field ${\mathbb{F}}_2$. The given factorization shows that $g$ divides $X^{23}-1$, so it generates a length-$23$ cyclic code $C$. Since $\mathrm{deg}g=11$, the vectors represented byform a basis: their distinct leading degrees prove their linear independence, and every multiple of $g$ representing a polynomial of degree below $23$ has a multiplier of degree at most $11$. Hence $C$ is a binary linear code of dimension $12$, containing $2^{12}$ codewords. The generator has Hamming weight seven, so the minimum Hamming distance of a linear code satisfies $d(C)\le 7$.

The BCH bound used here is the following consecutive-root theorem. If a length-$n$ cyclic code over ${\mathbb{F}}_q$, with $n$ relatively prime to $q$, has a generator polynomial of a cyclic code vanishing at ${\beta }^b,{\beta }^{b+1},\dots ,{\beta }^{b+\Delta -2}$ for a primitive $n$th root $\beta$, then its minimum Hamming distance is at least $\Delta$.

Choose a root $\beta$ of $g$ in an algebraic closure of ${\mathbb{F}}_2$. The factorization implies ${\beta }^{23}=1$, while $g(1)=1$ in ${\mathbb{F}}_2$, so $\beta \ne 1$. As $23$ is prime, $\beta$ has order $23$. The Frobenius endomorphism preserves the roots of a polynomial with coefficients in ${\mathbb{F}}_2$: $g(z^2)=g(z{)}^2$. Thus every ${\beta }^{2^j}$ is a root. In particular, $2^8=256\equiv 3(\mathrm{mod}23)$ shows thatare roots of $g$. These four consecutive roots give $d(C)\ge 5$ by the BCH bound. A further argument is needed to exclude weights five and six.

Put $S(X)=1+X+\cdots +X^{22}$. Cancelling $X+1$ in the supplied factorization gives $g(X)g^{\mathrm{rev}}(X)=S(X)$. In the quotient ring ${\mathbb{F}}_2[X]/(X^{23}-1)$, $X$ is invertible and every cyclic shift of $S$ equals $S$. Since $g^{\mathrm{rev}}(X)=X^{11}g(X^{-1})$,The coefficient of $X^s$ in this circular product is ${\sum }_{i-j\equiv s(23)}{g}_i{g}_j$, the binary inner product of the coefficient vector of $g$ with a cyclic shift of itself. Every coefficient of $S$ is one. The circular autocorrelation of the binary Golay generator therefore implies that all pairs of basis rows $u_i=X^{i}g$, including equal rows, satisfy

Each $u_i$ has Hamming weight seven. Append its parity bit, which is one, and write ${\hat{u}}_i=(u_i,1)$. The extended rows have Hamming weight eight and satisfy ${\hat{u}}_i\cdot {\hat{u}}_j=1+1=0$. Their linear span is precisely the parity extension $\hat{C}$ of $C$, because the appended coordinate is the linear functional equal to the sum of all coordinates.

These extended generators produce a doubly even code. To prove this rather than assume it, useAll the extended generators have weights divisible by four. A sum of previous generators is orthogonal to the next generator, so the intersection size in this identity is even. Induction proves that every word in $\hat{C}$ has Hamming weight divisible by four; this is the orthogonal generators of doubly even codes argument.

An original word of weight five acquires a parity bit and has extended weight six, while a word of weight six acquires a zero parity bit and still has extended weight six. Both contradict divisibility by four. Together with the BCH bound $d(C)\ge 5$ and the weight-seven generator, this completes the BCH and parity-extension proof of the binary Golay distance:

Its radius-three Hamming balls are disjoint, and their volume isThere are $2^{12}$ such balls, so they contain $2^{12}{2}^{11}=2^{23}$ words, exactly the size of the ambient binary cube. They therefore cover the cube, with every word in exactly one ball. Hence