Idempotents and Factorization

Question

In the ring $\mathbb{F}_2[x]$, let $f(x)$ be a reducible polynomial of degree $n$. If we happen to know that there is a non-trivial idempotent $g$ in $\mathbb{F}_2[x] / (f(x))$, then one of gcd($f,g$) and gcd($f,1-g$) is a proper factor of $f(x)$. Why is this true?

Answer 1

There are a lot of red herrings in this problem. If neither $\gcd(f,g)$ nor $\gcd(f,1-g)$ is a proper factor of $f$, then both are, without loss of generality, equal to $f$. Thus $$ g=hf\qquad\text{and}\qquad 1-g=kf $$ for some polynomials $h$ and $k$. Then $$ 1=g+(1-g)=\dots $$