by "quasi-local" ring did you mean a ring with only one maximal ideal or with finitely many maximal ideals?
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?
ok. the answer is yes. i'll prove that if is clean and the number of its idempotents is finite, then is a finite direct product of local rings. the proof is by induction on the number of idempotents. suppose i.e. the only idempotents of is and . then for any either or is a unit because is clean. now let be a maximal ideal of and . then and hence , for some and . thus . thus, since and hence cannot be a unit, is a unit and so is a unit. it follows that is a unit. so we just proved that every element outside is a unit and that means is a local ring. this completes the proof of our induction base. now suppose and the claim is true for any clean ring with the number of idempotents strictly less than n. choose an idempotent clearly and the number of idempotents of and is at most because and . so to finish the proof we only need to prove that if is clean, then and are clean too. this is quite easy and i'll prove it for the proof for is similar. first note that the identity element of is . now suppose that is any element of . since is clean, we have for some unit and an idempotent . then and it is clear that is a unit in and is an idempotent in . the proof is now complete.