Proof Check: Idempotents of a local commutative ring 
Let $R$ be a local commutative ring with 1. Prove that if $e\in R$ is idempotent, then $e=1$ or $e=0$.

My proof:
Let $m\subset R$ be the unique maximal ideal. We have that $R\cong R/m \oplus m$. Since $R/m$ is a field, then $e\in R/m$ is idempotent is trivially $0$ or $1$.
If $e\in m$, then $e\in J(R)$, the Jacobson radical. It follows that $1-e, 1-e^2$ are units. However, $1-e=1-e^2=(1-e)(1+e)$. By cancelling $1-e$, we have that $1=1+e$, and $e=0$.
My question is, can we always assume this decomposition of $R$, or is this only true under special requirements? And can I use the binomial theorem to solve it like I did?
 A: 
My question is, can we always assume this decomposition of $$, or is this only true under special requirements?

You only have $R\cong R/m\oplus m$ iff $m$ is generated by an idempotent, which you don't know.

If $e\in R/m$... if $e\in m$...

This whole business about treating $R/m\oplus m$ like $R/m \cup m$ and resolving two cases won't do. A direct sum is strictly bigger than the union of its pieces. (Are you possibly thinking of $(R\setminus{m})\cup m=R$?) There would actually be a third case where $e$ is in neither piece, but is a sum of two things.

By cancelling $1-e$ ...

Ok, you could do that, but $1-e$ is certainly idempotent and it's an easy exercise that an idempotent unit is $1$. Unfortunately, as I was getting at with the last point, it isn't fruitful to follow this case-by-case approach.  You would still be stuck with the case when $e$ isn't purely in one or the other.

As for solving the question, you should start with the fact that if $e\notin\{0,1\}$, then $R=eR\oplus (1-e)R$ is a nontrivial decomposition of $R$, and use the fact that every proper ideal has to be contained in a maximal ideal. Therefore...

There is also another way to approach it, if you believe that maximal ideals are prime ideals.  Just observe that $e(1-e)\in m$, and follow your nose for a different but similar solution.
A: The solution would be correct if you remove $R\cong R/m\oplus m$, which holds only in the special case when $m=\{0\}$. And your argument obviously doesn't need it.
You can also simplify the argument. Note that $e(1-e)=e-e^2=0$.
Suppose $e$ is not invertible; then $1-e$ is invertible, because of the properties of the Jacobson radical. Hence from $e(1-e)=0$ we obtain $e=0$.
If $e$ is invertible, then from $e(1-e)=0$ we obtain $1-e=0$.
