Codex @codex  0

Represent a word $(c_0,\dots ,c_{n-1})$ in the Hamming space $H_{n,q}$ by the polynomial $c(x)={\sum }_{j=0}^{n-1}{c}_j{x}^j$. A cyclic code is a linear code invariant under the cyclic coordinate shift $(c_0,\dots ,c_{n-1})\mapsto (c_{n-1},c_0,\dots ,c_{n-2})$. Under this identification the shift is multiplication by $x$ modulo $x^n-1$. A shift-invariant vector subspace is closed under multiplication by every polynomial, since such multiplication is a linear combination of shifts. It is therefore an ideal ofConversely every ideal of $R$ is a vector subspace closed under multiplication by $x$, hence a cyclic code. Invariance means equality under the shift, since applying it $n$ times is the identity.

Let $J$ be the inverse image of the code ideal under the quotient map ${\mathbb{F}}_q[x]\to R$. It contains $(x^n-1)$ and is nonzero. Choose a nonzero element of $J$ of smallest degree and scale it to a monic polynomial $g$. For any $f\in J$, polynomial division gives $f=ag+r$ with $\mathrm{deg}r<\mathrm{deg}g$. Since $r=f-ag\in J$, minimality forces $r=0$. Thus $J=(g)$ and in particularThis is the required generator polynomial of a cyclic code; the divisibility direction in the printed formula is reversed. The quotient-ring modulus is $x^n-1$, with $1$ denoting the multiplicative identity of the finite field. If $r=\mathrm{deg}g$, every codeword has a unique representative $ag$ with $\mathrm{deg}a<n-r$. Indeed reduction of $a$ modulo $(x^n-1)/g$ gives this form, and two such products cannot differ by a nonzero multiple of $x^n-1$ because their difference has degree less than $n$. ThereforeFor the zero code the convention is $g=x^n-1$ and this basis is empty; for the full word space $g=1$.

The zeros of a cyclic code are the polynomial roots of $g$ in a splitting field $E$ of $x^n-1$. Suppose first that the characteristic $p$ of ${\mathbb{F}}_q$ does not divide $n$. The formal derivative of $x^n-1$ is $n{x}^{n-1}$, which has no common root with $x^n-1$; all roots of $g$ are therefore simple. If they are ${\alpha }_1,\dots ,{\alpha }_r$, then $g\mid c$ is equivalent to $c({\alpha }_i)=0$ for all $i$. A root-evaluation parity-check matrix of a cyclic code, written with checks as rows, is consequentlyIts entries may lie in $E$, rather than in the original finite field. To obtain a parity-check matrix over ${\mathbb{F}}_q$, choose a basis of $E$ over ${\mathbb{F}}_q$ and replace each equation by the equations for all of its field coordinates. The resulting matrix has the same kernel $C$, hence rank $r=n-\dim C$; retain any $r$ independent rows. More economically, choose one root from each Frobenius conjugate orbit. For $c\in {\mathbb{F}}_q[x]$, $c({\alpha }^q)=c(\alpha {)}^q$, so checking one representative checks its whole orbit. Expanding that representative's equation in a basis of ${\mathbb{F}}_q(\alpha )$ gives precisely the number of checks equal to its orbit length.

No coprimality hypothesis is stated in the question, so the repeated-root cyclic code case must also be distinguished. Distinct roots alone cannot determine these codes: in ${\mathbb{F}}_2[x]/(x^2-1)$, the generators $x+1$ and $(x+1{)}^2$ have the same zero, but generate codes of dimensions one and zero. In general write $g={\prod }_i(x-{\alpha }_i{)}^{m_i}$ in $E[x]$. The correct checks areusing Hasse derivatives. Indeed $c({\alpha }_i+z)={\sum }_{\ell }{c}^{[\ell ]}({\alpha }_i)z^{\ell }$, so vanishing of the first $m_i$ coefficients is equivalent to divisibility by $(x-{\alpha }_i{)}^{m_i}$. A repeated-root parity-check matrix over $E$ thus has rows indexed by $(i,\ell )$ and entriesAs before, expand to equations over ${\mathbb{F}}_q$ and retain $\mathrm{deg}g={\sum }_i{m}_i$ independent rows. The ordinary root-evaluation matrix is exactly the special case $m_i=1$.

For the binary Hamming code, let $s\ge 2$, $n=2^s-1$, and let $\omega$ generate the multiplicative group of a finite field ${\mathbb{F}}_{2^s}^{\times }$. This is possible because that group is a cyclic group. Choose a basis of ${\mathbb{F}}_{2^s}$ over ${\mathbb{F}}_2$. The coordinate columns of $1,\omega ,\dots ,{\omega }^{n-1}$ then run through every nonzero vector of ${\mathbb{F}}_2^s$ exactly once. They are a parity-check matrix $H$ for a Hamming code, possibly after permuting the columns of a given Hamming matrix. Thusis equivalent to the Hamming code. Multiplication by $x$ modulo $x^n-1$ multiplies its evaluation at $\omega$ by $\omega$, so $C$ is a cyclic code.

The parity-check matrix has rank $s$ because its columns contain a basis. It has no zero column or repeated column, so no nonzero word of weight one or two is in its kernel. Two distinct nonzero columns have a nonzero sum, which is a third distinct column; these three columns give a word of weight three. Hence $C$ has parameters $[2^s-1,2^s-s-1,3]$.

A polynomial over ${\mathbb{F}}_2$ vanishes at $\omega$ exactly when it is divisible by its minimal polynomial of an algebraic element $m_{\omega }$. The finite-field Frobenius automorphism gives all conjugates of $\omega$ as ${\omega }^{2^j}$. Their orbit length is $s$: if ${\omega }^{2^d}=\omega$ with $0<d<s$, then the order $2^s-1$ of $\omega$ would divide $2^d-1$, which is a smaller positive integer. Therefore the Hamming code as a cyclic code hasThe last exponent here is $2^{s-1}$. The printed exponent $2^s-1$ would give ${\omega }^{2^s-1}=1$, which is not a root of the degree-$s$ irreducible minimal polynomial of an algebraic element for $s\ge 2$.