# Math Help - clean ring

1. ## clean ring

A commutative ring is said to be clean if each of its elements can be written as the sum of a unit and an idempotent. for example, every quasi-local ring is a clean ring.
If R is clean and it is not a finite product of quasi-local rings, can we conclude that the set of idempotent elements of R is infinite?

2. by "quasi-local" ring did you mean a ring with only one maximal ideal or with finitely many maximal ideals?

3. I mean a ring with unique maximal ideal(local ring).

4. Originally Posted by xixi
I mean a ring with unique maximal ideal(local ring).
ok. the answer is yes. i'll prove that if $R$ is clean and the number of its idempotents is finite, then $R$ is a finite direct product of local rings. the proof is by induction on $n,$ the number of idempotents. suppose $n = 2,$ i.e. the only idempotents of $R$ is $0$ and $1$. then for any $x \in R,$ either $x$ or $1 - x$ is a unit because $R$ is clean. now let $M$ be a maximal ideal of $R$ and $x \notin M$. then $Rx + M = R$ and hence $rx + y = 1 - x$, for some $y \in M$ and $r \in R$. thus $1-y=(r+1)x$. thus, since $y \in M$ and hence $y$ cannot be a unit, $1-y$ is a unit and so $(r+1)x$ is a unit. it follows that $x$ is a unit. so we just proved that every element outside $M$ is a unit and that means $R$ is a local ring. this completes the proof of our induction base. now suppose $n > 2$ and the claim is true for any clean ring with the number of idempotents strictly less than n. choose an idempotent $e \in R \setminus \{0,1\}.$ clearly $R = Re \oplus R(1-e)$ and the number of idempotents of $Re$ and $R(1-e)$ is at most $n-1$ because $1 \notin Re$ and $1 \notin R(1-e)$. so to finish the proof we only need to prove that if $R$ is clean, then $Re$ and $R(1-e)$ are clean too. this is quite easy and i'll prove it for $Re.$ the proof for $R(1-e)$ is similar. first note that the identity element of $Re$ is $e$. now suppose that $x = re$ is any element of $Re$. since $R$ is clean, we have $r = u + f$ for some unit $u \in R$ and an idempotent $f \in R$. then $x=re = ue + fe$ and it is clear that $ue$ is a unit in $Re$ and $fe$ is an idempotent in $Re$. the proof is now complete.