We consider a cyclic code $\mathcal{C}$ over a finite field $\mathbb{F}$, with generating idempotent $ε(x)$ and $\mathcal{C}_0$ a subcode, which contains all the (code)words of zero sum.
If $θ(x) = 1/n (x^{n-1}+...+x+1)$, I need to prove that:
If $\mathcal{C}_0$ is a proper subset of $\mathcal{C}$, then the polynomial $ε(x)-θ(x)$ is a generating idempotent of the code $\mathcal{C}_0$.
So far, I have noticed that $x^n-1=n(x-1)θ(x)$ and I have proven that $1-θ(x)$ is the generating idempotent of the cyclic code $\mathcal{E}_n$ over $\mathbb{F}$, which contains all zero-sum (code)words of length $n$. \
I don't know how to approach this at all. Thanks in advance for your help
$\newcommand{\c}{\mathcal C}$$\newcommand{\F}{\mathbb{F}}$All polynomials are viewed as elements in $\F(x)$ or $\F(x)/(x^n-1)$, depending on the context.
Note that a codeword $f(x)$ is zero sum means $f(1)=0$, an indentitity in $\F(x)$. Verify the $\c_0$ is also a cyclic code.
Since $\epsilon^2(x)=\epsilon(x)$, $\epsilon^2(1)=\epsilon(1)$. So $\epsilon(1)=0,1$. If $\epsilon(1)=0$, which means $\epsilon(x)\in C_0$, $\c$ will be the same as $C_0$, which is not true. Hence $\epsilon(1)=1$
$$(\epsilon(x)-\theta(x))(1)=\epsilon(1)-\theta(1)=1-1=0$$ which means $\epsilon(x)-\theta(x)\in\c_0$
Suppose $\alpha(x)=\sum_{i=0}^{n-1}\alpha_ix^i\in\c_0$. So, $\alpha(x)=\epsilon(x)\alpha_\epsilon(x)$ for some $\alpha_\epsilon(x)\in \F(x)$ and $\alpha(1)=0$.
The latter equality says $1$ is a root of $\alpha(x)$. Hence $\alpha(x)=\alpha_\epsilon(x)(x-1)$ for some $\alpha_\epsilon(x)\in \F(x)$. $\ $(In fact, $\alpha_\epsilon(x) = \sum_{i=0}^{n-2}\left(\sum_{j=0}^{i}\alpha_j\right)x^i$.) $$\alpha(x)\theta(x)=\alpha_1(x)(x-1)\theta(x)=\alpha_1(x)\frac1{n}(x^n-1)=0$$
So $$\alpha(x)=\alpha(x)\epsilon(x)=\alpha(x)(\epsilon(x)-\theta(x))$$
Hence $\epsilon(x)-\theta(x)$ is a generating idempotent of $\c_0$.