## Clean rings

Posted: September 15, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Definition. A ring $R$ with 1 is called clean if for every $a \in R$ there exist a unit $b \in R$ and an idempotent $c \in R$ such that $a = b+c.$

Examples. Every commutative local ring $R$ is clean. The reason is that for every $a \in R,$ either $a$ or $a-1$ is a unit. It is also obvious that a direct sum or product of clean rings is clean.

Problem. Let $R$ be a commutative clean ring. Prove that if the number of idempotent elements of $R$ is finite, then $R$ is a finite direct product of local rings.

Solution. The proof is by induction on $n,$ the number of idempotent elements of $R.$ Suppose that $n = 2,$ i.e. the only idempotent elements of $R$ are $0$ 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 $y \in M, \ r \in R$. Thus $1-y=(r+1)x$ and so, since $y \in M$ and hence $y$ cannot be a unit, $1-y$ is a unit. Therefore $x$ is a unit and so $R$ is local. Now suppose $n > 2.$ Choose an idempotent $e \in R \setminus \{0,1\}.$ Clearly $R = Re \oplus R(1-e)$ and the number of idempotent elements in both $Re$ and $R(1-e)$ is at most $n-1$ because $1 \notin Re$ and $1 \notin R(1-e)$. So to apply induction hypotheis on $Re$ and $R(1-e)$ and finish the proof, we only need to prove that both $Re$ and $R(1-e)$ are clean. First note that the identity element of $Re$ is $e$. Now let $x = re \in Re.$ Since $R$ is clean, we have $r=u+f$ for some unit $u \in R$ and an idempotent $f \in R.$ Then  $x=re=ue + fe.$ Clearly  $ue$ is a unit in $Re$ and $fe$ is an idempotent in $Re.$ So $Re$ is clean. A similar argument shows that $R(1-e)$ is clean too. $\Box$