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Thread: clean ring

  1. #1
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    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?
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  2. #2
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    by "quasi-local" ring did you mean a ring with only one maximal ideal or with finitely many maximal ideals?
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    I mean a ring with unique maximal ideal(local ring).
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  4. #4
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    Quote Originally Posted by xixi View Post
    I mean a ring with unique maximal ideal(local ring).
    ok. the answer is yes. i'll prove that if $\displaystyle R$ is clean and the number of its idempotents is finite, then $\displaystyle R$ is a finite direct product of local rings. the proof is by induction on $\displaystyle n,$ the number of idempotents. suppose $\displaystyle n = 2,$ i.e. the only idempotents of $\displaystyle R$ is $\displaystyle 0$ and $\displaystyle 1$. then for any $\displaystyle x \in R,$ either $\displaystyle x$ or $\displaystyle 1 - x$ is a unit because $\displaystyle R$ is clean. now let $\displaystyle M$ be a maximal ideal of $\displaystyle R$ and $\displaystyle x \notin M$. then $\displaystyle Rx + M = R$ and hence $\displaystyle rx + y = 1 - x$, for some $\displaystyle y \in M$ and $\displaystyle r \in R$. thus $\displaystyle 1-y=(r+1)x$. thus, since $\displaystyle y \in M$ and hence $\displaystyle y$ cannot be a unit, $\displaystyle 1-y$ is a unit and so $\displaystyle (r+1)x$ is a unit. it follows that $\displaystyle x$ is a unit. so we just proved that every element outside $\displaystyle M$ is a unit and that means $\displaystyle R$ is a local ring. this completes the proof of our induction base. now suppose $\displaystyle n > 2$ and the claim is true for any clean ring with the number of idempotents strictly less than n. choose an idempotent $\displaystyle e \in R \setminus \{0,1\}.$ clearly $\displaystyle R = Re \oplus R(1-e)$ and the number of idempotents of $\displaystyle Re$ and $\displaystyle R(1-e)$ is at most $\displaystyle n-1$ because $\displaystyle 1 \notin Re$ and $\displaystyle 1 \notin R(1-e)$. so to finish the proof we only need to prove that if $\displaystyle R$ is clean, then $\displaystyle Re$ and $\displaystyle R(1-e)$ are clean too. this is quite easy and i'll prove it for $\displaystyle Re.$ the proof for $\displaystyle R(1-e)$ is similar. first note that the identity element of $\displaystyle Re$ is $\displaystyle e$. now suppose that $\displaystyle x = re$ is any element of $\displaystyle Re$. since $\displaystyle R$ is clean, we have $\displaystyle r = u + f$ for some unit $\displaystyle u \in R$ and an idempotent $\displaystyle f \in R$. then $\displaystyle x=re = ue + fe$ and it is clear that $\displaystyle ue$ is a unit in $\displaystyle Re$ and $\displaystyle fe$ is an idempotent in $\displaystyle Re$. the proof is now complete.
    Last edited by NonCommAlg; Aug 22nd 2011 at 04:50 PM.
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