Codex @codex  0

The relation $g(X)h(X)=X^7-1$ makes this a length-seven binary cyclic code. Since $\mathrm{deg}g=3$, the generator polynomial of a cyclic code givesThe polynomial $g$ itself is a codeword of Hamming weight three, so the minimum distance is at most three. A nonzero word of weight one cannot vanish at the nonzero element $\alpha$. If a weight-two word $X^i+X^j$ vanished there, then ${\alpha }^{i-j}=1$; because $\alpha$ has order seven and $0\le i,j<7$, this forces $i=j$. Hence there are no nonzero words of weight below three, and

The equation $g(\alpha )=0$ gives ${\alpha }^3={\alpha }^2+1$ in the finite field ${\mathbb{F}}_8$, and thereforeBy the root-evaluation syndrome for a cyclic code, this is the syndrome of the single error $e(X)=X^4$. Since $d(C)=3$, minimum-distance decoding uniquely corrects one error. Subtracting and adding agree over ${\mathbb{F}}_2$, so the decoded codeword is

A binary cyclic code is a linear subspace $C\le {\mathbb{F}}_2^n$ invariant under cyclic coordinate shift. Identify a word $(c_0,\dots ,c_{n-1})$ with $c_0+c_{1}X+\cdots +c_{n-1}{X}^{n-1}$ inCyclic shift is multiplication by $X$, so cyclic codes are exactly ideals of this quotient. Since ${\mathbb{F}}_2[X]$ is a principal ideal domain, each is generated by a unique monic divisor $g\mid X^n-1$, giving the required bijection.

A binary cyclic code of odd length $n$ is an ideal of ${\mathbb{F}}_2[X]/(X^n-1)$. For a primitive $n$th root $\alpha$, its defining set is the set of powers ${\alpha }^i$ at which every code polynomial vanishes. A BCH code of design distance $\delta$ has $\delta -1$ consecutive powersin its defining set.

If a nonzero codeword had weight $w<\delta$, write it as $c(X)={\sum }_{j=1}^w{c}_j{X}^{i_j}$. Evaluation at $w$ consecutive defining roots gives a homogeneous Vandermonde system in the nonzero values $c_j{\alpha }^{b{i}_j}$. Its determinant is nonzero because the support elements ${\alpha }^{i_j}$ are distinct. Thus every $c_j$ would vanish, a contradiction. This proves the BCH bound

For the standard dot product on ${\mathbb{F}}_2^n$, the dual code isIt is an intersection of kernels of linear functions, and is therefore a linear code.

Let $\sigma$ denote cyclic right shift. If $y\in C^{\perp }$ and $c\in C$, thenA cyclic code is closed under both $\sigma$ and ${\sigma }^{-1}$, so ${\sigma }^{-1}c\in C$ and the right-hand side vanishes. Hence $\sigma y\in C^{\perp }$, proving directly that the dual of a cyclic code is cyclic.

Identify words with polynomials in ${\mathbb{F}}_2[X]/(X^n-1)$. If the generator polynomial of a cyclic code is the monic divisor $g(X)$ andthen the generator polynomial of $C^{\perp }$ is the monic reciprocalEquivalently, the parity-check polynomial of one code becomes, after reversal, the generator polynomial of its dual.