Show (1-a) is an indempotent.

Question

Question: If a is an idempotent in $\mathbb{Z}_{n}$, show that $\left ( 1-a \right ) $is also an idempotent.

Def: An element $\mathit{a}$ is an idempotent if $\forall a \in R: \space\ a^{2}=a$ where R is a commutative ring with unity.

Clearly, $\mathbb{Z}_{n}$ is a commutative ring with unity.

Suppose $\forall a \in \mathbb{Z}_{n}: a^{2}=a$

Indeed, R is a ring so $\forall a \in R: \left ( 1-a \right ) \in R$.

$\left ( 1-a \right )^{2}=1-a-a+a^{2}=1-2a+a^{2}$

At this point I'm slightly stuck. My intuition is this: If I can show $Char\left ( R \right )=2$,then, $2a=0$ or from observation $a^{2}=a$ so $a^{2}-a=0$ so $a\left ( a-1 \right )=0$. From this, either $a=0$ or $a=1$ and if $a=0$, we are done. If $a=1$, I cannot arrive at what is required.

Hint is appreciated.

Thanks in advance.

Answer 1

Well $a^2=a$ so $(1-a)^2=1-2a+a=1-a$. Done.

Answer 2

You have $1-2a+a^2=1-2a+a$ since $a^2=a$ henceforth $1-2a+a^2=1-a$.