Idempotents in a local ring

Is it true that a local ring, i.e., a commutative ring with a unique maximal ideal, doesn't contain idempotent elements $\neq 0, 1$ ? Why ?

Any hint ?

Solution 1:

If $e$ is an idempotent which is not $0,1$, then $e(e-1)=e^2-e=0$ shows that both $e$ and $e-1$ are zero divisors and in particular not invertible. Hence they must be in the maximal ideal, but then $1=e-(e-1)$ is also in the maximal ideal - contradiction.

Solution 2:

Let $e$ be a nontrivial idempotent. Then $eR\oplus (1-e)R=R$ is a nontrivial splitting of $R$ into two proper ideals. But both $eR$ and $(1-e)R$ are contained in the maximal ideal: how could they add up to $R$? This shows no such $e$ can exist.

This works even for noncommutative local rings.