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


1. All elements of $A$ are left zero divisor.
2. All elements are right zero divisor.
3. All elements are two sided zero divisors
4. 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](https://mathoverflow.net/questions/58186/positive-element-in-c-tensor-product)



A question on $Z^{*}$ algebras
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. 


