As discussed in this MO topic, every principal ideal domain has stable rank at most 2. The proof in the accepted answer uses the fact that PID is a unique factorization domain, but there can be no irreducibles in case of Bezout ring (every *finitely generated* ideal is principal), as shown by the example of the ring of all algebraic integers.

However, if one assumes the ring to be a Bezout *domain*, one can show that it is Hermitian, and thus of stable rank 2 (this paper, for example).

It looks like there is nothing about *infinitely generated* ideals in either PID or stable range conditions, so it is naturally to ask if there are Bezout rings of stable rank greater than 2. Surely they have to be non-UFD. The two basic examples I know of: algebraic integers and entire functions, but both are domains (and of stable rank 1, which makes the situation even more puzzling). Another example from the Wikipedia page is of no help either.

Question.Is the stable rank of Bezout rings bounded from above? Can it be strictly greater than 2?

PS. All rings are considered to be commutative.