Tensor product of operator subalgebras and properties of the trace Let $V$ be a vector space and let $\mathcal{A},\mathcal{B} \subset \text{End}(V)$ be two subalgebras of operators on $V$. Does the following statement hold?
If there exists an algebra isomorphism between $\mathcal{A} \otimes \mathcal{B}$ and $\mathcal{A} \vee \mathcal{B}$ (the algebra generated by the set $\mathcal{A} \cup \mathcal{B}$ by taking all possible linear combinations and products) then the following two properties hold:
i) $[\mathcal{A},\mathcal{B}]=0$
ii) $\operatorname{Tr}(AB)\propto \operatorname{Tr}(A) \operatorname{Tr}(B)$, for all $A\in\mathcal{A}$ and $B\in\mathcal{B}$
An example where this is true is given by: $V=V_A\otimes V_B$ and $\mathcal{A}=\{A\otimes I \mid A \in \text{End}(V_A)\}$ and $\mathcal{B}=\{I\otimes B \mid B \in \text{End}(V_B)\}$.
However, is this true in general? If yes, I would be glad if someone could provide a proof.
 A: There are natural inclusion maps from each of $\mathcal{A}$ and $\mathcal{B}$ to each of $\mathcal{A}\otimes\mathcal{B}$ and $A\vee B$.  Assuming you want the isomorphism $\mathcal{A}\otimes\mathcal{B}\cong A\vee B$ to preserve these inclusion maps, then (i) is immediate, since the images of $\mathcal{A}$ and $\mathcal{B}$ in $\mathcal{A}\otimes\mathcal{B}$ commute by definition of the ring structure on the tensor product.
However, (ii) can fail.  For instance, let $\mathcal{A}$ be the algebra of diagonal $5\times 5$ matrices whose diagonal entries have the form $(a,a,b,b,a)$ and let $\mathcal{B}$ be the algebra of diagonal $5\times 5$ matrices whose diagonal entries have the form $(c,d,c,d,c)$.  Then $\mathcal{A}\vee\mathcal{B}$ is the algebra of diagonal matrices whose diagonal entries have the form $(a,b,c,d,a)$, and is isomorphic to $\mathcal{A}\otimes\mathcal{B}$ (preserving the inclusions of $\mathcal{A}$ and $\mathcal{B}$).  However, given elements $A=(a,a,b,b,a)\in\mathcal{A}$ and $B=(c,d,c,d,c)\in\mathcal{B}$, we have $$Tr(AB)=2ac+ad+bc+bd$$ and $$Tr(A)Tr(B)=(3a+2b)(3c+2d)=9ac+6ad+6bc+4bd$$ and these are not proportional.
