As the comment of Andreas Thom indicated [here](https://mathoverflow.net/questions/188344/simple-z-algebra), a separable $C^\star$ algebra $A$ can not contain a $Z^\star$ algebra.(A $Z^\star$ algebra is a $C^\star$ algebra which all elements are zero divisor). So separability is an obstruction for $A$ to **contain** a $Z^\star$ algebra.


Now we ask:



> 
> Is it the only obstruction? What type of other obstructions can be introduced? In particular is it true that every non separable algebra contains a $Z^\star$ algebra?
> 
> 
> 



Obstructions for $C^\star$ algebras to contain a $Z^\star$ algebra
$A$ contains a $Z^\*$ $C^\*$-subalgebra iff it contains an uncountable family $\{ a\_i \}$ of nonzero positive mutually orthogonal elements. If it has such a family, then $C^\*(\{ a\_i \})$ is $Z^\*$. If it does not have such a family and $B$ is a $C^\*$-subalgebra, then any maximal family $\{ a\_n \}$ of n.p.m.o. elements in $B$ is at most countable and hence one can define $a := \sum\_n 2^{-n}\| a\_n \|^{-1} a\_n \in B$, which is not a zero divisor. 
In particular, any $\mathrm{II}\_1$ factor does not contain a $Z^\*$ $C^\*$-subalgebra. 


