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

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