Idempotents and cyclic codes

Question

Let $C_1 = \langle e_1(x) \rangle$, $C_2 = \langle e_2(x) \rangle$ cyclic codes, where $e_1(x)$ and $e_2(x)$ are idempotents.

I know what cyclic codes and idempotents are, but why can one deduce the following: $C_1 \subset C_2 \Leftrightarrow e_1(x) e_2(x) = e_1(x)$?

Answer 1

If $C_1 \subset C_2$, then note that $e_1 \in C_1 \in C_2$. Hence $e_1 e_2 = e_2 e_1 = e_1$ because $e_1$ is an idempotent for $C_1$, and we can interpret $e_2$ as an element of $C_1$

For the other direction, if $e_1 e_2 = e_1$, then let $x \in C_1$. Now, $x = e_1 x = e_1 e_2 x = e_2 (e_1 x) = e_2 x$. Hence, $x = e_2 x$, and therefore $x \in C_2$.

Answer 2

I believe you are talking about an idempotent $e=e(x)\in F[x]/(x^n-1)$. The fact you are speaking of is actually true for idempotents in any commutative ring with identity.

Suppose $e,f$ are two idempotents in a commutative ring $R$.

Then it is elementary to show that $(e)\cap (1-e)=\{0\}$ and $(f)\cap (1-f)=\{0\}$.

Now if $(e)\subseteq (f)$, then $e-ef=e(1-f)\in (f)\cap (1-f)=\{0\}$. Therefore $e=ef$.

In the other direction, suppose $e=ef$: then clearly $e\in (f)$ and $(e)\subseteq (f)$.