= Solution
A cyclic binary code is a <linear subspace> of $\mathbb F_2^n$ invariant under cyclic shifts. Identify a word with its coefficient <polynomial> in $\mathbb F_2[X]/(X^n-1)$. Multiplication by $X$ is the shift, so the code is precisely an <ideal> of that <quotient ring>. Its inverse image in the <polynomial ring> is a <principal ideal> $(g)$ containing $(X^n-1)$; its unique monic generator divides $X^n-1$. Conversely each monic divisor generates such an <ideal>. Thus the bijection is with \b[monic divisors]; $g=1$ gives the full code and $g=X^n-1$ the <zero code>.
For odd $n$, some $m$ has $n\mid2^m-1$, so the cyclic multiplicative group of $\mathbb F_{2^m}$ contains a primitive $n$th root $\alpha$. If a nonzero word with the specified roots had weight $w<\delta$, write its nonzero positions as $r_1,\ldots,r_w$. The root equations give
$$
\sum_{j=1}^w c_j(\alpha^{r_j})^i=0\quad(1\le i\le w).
$$
Their matrix has <determinant> $\prod_j\alpha^{r_j}\prod_{j<l}(\alpha^{r_l}-\alpha^{r_j})\ne0$, by the Vandermonde formula and distinctness of the positions. Thus all coefficients would vanish, a contradiction. Hence \b[<minimum distance> is at least $\delta$].
For $n=7$, choose $\alpha$ with <minimal polynomial> $X^3+X+1$. The roots $\alpha,\alpha^2$ force also the conjugate $\alpha^4$, giving this cubic as generator. Its degree gives dimension four and its weight gives distance exactly three. Expressing $c(\alpha)=0$ in a basis of $\mathbb F_8$ gives a <parity-check matrix> with the seven distinct nonzero vectors of $\mathbb F_2^3$ as columns. This is \b[the binary $[7,4,3]$ <Hamming code>]. Its sixteen disjoint radius-one balls have $16(1+7)=128$ words, so one-error correction is perfect. The other primitive cubic produces an equivalent coordinate-permuted code.
Back to article page
