Note that this question was already posted on MSE: https://math.stackexchange.com/questions/4290741/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.
