A question on $Z^{*}$ algebras A $Z^{*}$ algebra is  a  $C^{*}$ algebra which satisfies  each of the following equivalent conditions:


*

*All elements of $A$ are left zero divisor.

*All elements are right zero divisor.

*All elements are two sided zero divisors

*All positive elements are two sided zero divisor.
The commutative(topological) interpretation for this concept is the following:
For  a  locally compact Hausdorff space $X$, $C_{0}(X)$ is  NOT a  $Z^{*}$ algebra if  and only if $X$ is  an approximately $\sigma$- compact space (Briefly $A\sigma  C$ space). that is, there are a sequence of compact subsets $K_{n}$ of $X$ such that $\cup K_{n}$ is dense in $X$. Note that the product of two $A\sigma C$ spaces is  a $A\sigma C$ space. Equivalently if $X \times Y$ is  not a $A\sigma C$ space then $X$ or $Y$ is not a $A\sigma C$ space. This  is  a motivation to ask:
Assume that $A$ and $B$ are two $C^{*}$ algebras such that their minimal tensor product is a $Z^{*}$ algebra. Is it true to say that $A$ or $B$ is  a $Z^{*}$ algebra?
Note that if the answer to the following question were affirmative, then the answer to the above question would be affirmative, too:
positive element in C* tensor product
 A: Recall that a completely positive map $\phi\colon A\to B$ is said to be faithful if $\phi(a) \neq 0$ for all nonzero $a \in A^+$. 
Lemma: If $\phi_i\colon A_i\to B_i$ are faithful cp maps, then $\phi_1\otimes\phi_2\colon A_1\otimes_{\min}A_2\to B_1\otimes_{\min}B_2$ is faithful. 
Proof: Since $\phi_1\otimes\phi_2=(\mathrm{id}_{B_1}\otimes\phi_2)\circ(\phi_1\otimes\mathrm{id}_{A_2})$, we may assume one of $\phi_i$ (say $\phi_2$) is $\mathrm{id}$. Let a nonzero $a \in (A_1\otimes_{\min}A_2)^+$ be given. By definition of the minimal tensor product, there are states $f_i$ on $A_i$ such that $(f_1\otimes f_2)(a) > 0$. Observe that $(f_1\otimes f_2)(a) = f_1((\mathrm{id}_{A_1}\otimes f_2)(a))$, where $\mathrm{id}_{A_1}\otimes f_2\colon A_1\otimes A_2\to A_1\otimes{\mathbb C}\cong A_1$ is the slice map. Hence, $(\mathrm{id}_{A_1}\otimes f_2)(a) \in A^+\setminus\{0\}$ and 
$$(\mathrm{id}_{B_1}\otimes f_2)((\phi_1\otimes\mathrm{id}_{A_2})(a)) = \phi_1((\mathrm{id}_{A_1}\otimes f_2)(a)) \neq 0,$$ 
which implies $(\phi_1\otimes\mathrm{id}_{A_2})(a) \neq 0$. $\Box$
Now, if $A_i$ are non-$Z^*$ $\mathrm{C}^*$-algebras, then there are $a_i\in A_i$ which are not left zero divisors. This means that the completely positive maps $\phi_i\colon A_i\ni x\mapsto a_i^*xa_i \in A_i$ are faithful. By the above lemma, $\phi_i\otimes\phi_2$ is faithful, which means that $a_1\otimes a_2$ is not a left zero divisor. 
