Let $\omega$ be a primitive nth root of unity. Let $C=\langle g(x)\rangle=[e(x)]$ be a cyclic code in $R_n$. Show that $e(\omega ^i)=0$ or 1.
Proof:(i) Let $C=<g(x)>=<e(x)>$ be a cyclic group in $R_n$. Since $\omega$ is a primitive nth root of unity and since $e(x) \equiv a(x)g(x)$, we deduce that $g(\omega^i)=0$ which implies that $e(\omega^i)=0$. Now, suppose $e(\omega^i) \neq 0$. By a theorem and since $\omega$ is a primitive nth root of unity, we have $e(\omega^i)$ is a unity. Hence, $e(\omega^i)=1$.
I am no sure of this proof. May I know if this is correct? And from (i), how would I say that the dimension of C is just equal to the number of $\omega^i$ such that $e(\omega^i)=1$?
I assume the $e(x)$ you are talking about here is the generating idempotent of the cyclic code. What happens when $e(\omega^i)\neq0$? Can you recall the definition of an idempotent?
Also, if $e(\omega^i)=0$, then that just means $\omega^i$ is a zero of $e(x)$. How do these $n$th roots of unity relate to the zeros of $C$, considering that $e(x)$ generates $C$?