I asked [this question at MSE](https://math.stackexchange.com/questions/3003201/extension-of-a-von-neumann-algebra-by-a-von-neumann-algebra) now I repeat it at MO:


Let $A,B,C$ be $3$ unital $C^\*$ algebras. Assume that we have the following short exact sequence of $C^\*$-algebras:


$$0\to A\to C\to B\to 0$$


Assume that $A,B$ are generated by their projections. Is $C$ necessarily generated by its projections, too?



> 
> Assume that $A,B$ are von Neumann algebras, is $C$ necessarily a von Neumann algebra, too?
> 
> 
> 


Does the last question has an obvious answer when $A,B$ (hence $C$) are commutative algebras?



Extension of a von Neumann algebra by a von Neumann algebra
Yes, it is. Let $C$ be a C\*-algebra and let $A \subseteq C$ be an ideal which is intrinsically a von Neumann algebra. Then the positive part of the unit ball of $A$ has a least upper bound in $A$ which must be a projection. (Its norm cannot be greater than $1$, so if it is not a projection then its square root also belongs to the unit ball and is larger.) It follows that $C \cong A \oplus B$, so if $B$ is also intrinsically a von Neumann algebra then so is $C$.


