TITLE: indecomposable semi local Bezout ring
QUESTION [5 upvotes]: A commutative Bezout ring with $1$ is a ring in which every finitely generated  ideal is principal. Is there any charactrization for indecomposable semi local  Bezout ring?

For example , for a commutative   (indecomposable) local rign is Bezout ring if and only if it is a chain ring.

REPLY [2 votes]: Some partial ideas:
If semilocal is changed to semiperfect, then you can use the fact that commutative semiperfect rings are just products of local rings to extend the fact you mentioned in a different way:

A commutative, semiperfect ring is Bézout iff it is a serial ring

(Serial ring of course means it is a finite direct product of uniserial submodules, which in the case of commutative rings amounts to a finite product of chain rings.)
Things don't seem so nice for semilocal rings that aren't semiperfect. It's easy to create semilocal PIDs which have complex lattices of ideals. For example, you can take a finite collection of maximal primes in $\mathbb Z$ and semilocalize.
Commutative Bézout rings are distributive, though, and I don't immediately know if a semilocal, indecomposable, distributive ring has to be Bézout or not. That might be something to explore.