= Solution
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 $\deg g=11$, the vectors represented by
$$
g,\ Xg,\ \ldots,\ X^{11}g
$$
form 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 $d(C)\leq7$.
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 $\beta^b,\beta^{b+1},\ldots,\beta^{b+\Delta-2}$ 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 $\beta\neq1$. 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, $2^8=256\equiv3\pmod{23}$ shows that
$$
\beta,\ \beta^2,\ \beta^3,\ \beta^4
$$
are roots of $g$. These four consecutive roots give $d(C)\geq5$ 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 $g^{\mathrm{rev}}(X)=X^{11}g(X^{-1})$,
$$
g(X)g(X^{-1})=X^{-11}S(X)\equiv S(X)\pmod{X^{23}-1}.
$$
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 $u_i=X^ig$, including equal rows, satisfy
$$
u_i\cdot u_j=1\quad\text{in }\mathbb F_2.
$$
Each $u_i$ has <Hamming weight> seven. Append its parity bit, which is one, and write $\widehat u_i=(u_i,1)$. The extended rows have <Hamming weight> eight and satisfy $\widehat u_i\cdot\widehat u_j=1+1=0$. Their <linear span> is precisely the <parity extension> $\widehat C$ 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, use
$$
\operatorname{wt}(v+w)=\operatorname{wt}(v)+\operatorname{wt}(w)
-2|\operatorname{supp}(v)\cap\operatorname{supp}(w)|.
$$
All 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 $\widehat C$ 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> $d(C)\geq5$ and the weight-seven generator, this completes the <BCH and parity-extension proof of the binary Golay distance>:
$$
\boxed{d(C)=7,\qquad C\text{ has parameters }[23,12,7].}
$$
Its radius-three <Hamming balls> are disjoint, and their volume is
$$
v_{23}(3)=1+23+\binom{23}{2}+\binom{23}{3}
=1+23+253+1771=2048=2^{11}.
$$
There 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
$$
\boxed{C\text{ is the perfect binary Golay code, correcting every pattern of at most three errors}.}
$$
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