Proof Check: Idempotents of a local commutative ring

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.

Tags:

Abstract Algebra

Commutative Algebra

Ring Theory

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