Checking if a codeword in a cyclic code generates it

Question

Let's have $g(x)$ be the generator polynomial of a cyclic code $C=<g(x)>$. For any $h(x)\in C$, what would be the condition so that $<h(x)> = <g(x)>$ - meaning h(x) generates the code (albeit not necessarily being THE generator polynimial)?

Answer 1

You would need $h(x)$ and $g(x)$ to generate the same ideal of $\mathbb{F}[x]/(x^n -1)$. You automatically have $\langle h(x) \rangle \subseteq \langle g(x) \rangle$. You need to have $g(x)$ be a multiple of $h(x)$ in order to guarantee equality. As $h(x)$ is already a multiple of $g(x)$, this means you need to have $h(x)$ and $g(x)$ differ by a unit of $\mathbb{F}[x]/(x^n -1)$.

For example, in $\mathbb{F}_{3}[x]/(x^3 -1)$, consider the code generated by $g(x) = x+1$. Then the polynomial $h(x) = -x^{2} -1 = (-x^2)(x+1)$, and $-x^{2}$ is a unit (since $(-x^{2})(-x)$$ = x^{3} = (x^{3}-1)+1 = 1$). Therefore $h(x)$ will also generate the code.