Apologies if this is obvious since I am not an expert, but suppose that $\mathcal{M}$ and $\mathcal{N}$ are factors of arbitrary type. Is it true that the tensor product von Neumann algebra $\mathcal{M} \otimes \mathcal{N}$ is necessarily a factor as well?
An argument which I think points to the affirmative would be to simply note that the commutant of $(\mathcal{M}\otimes \mathcal{N})$ is $\mathcal{M}'\otimes \mathcal{N}'$ by the tensor product commutation theorem. Then I would like to say that \begin{equation} (\mathcal{M}\otimes \mathcal{N}) \,\cap\,(\mathcal{M}\otimes \mathcal{N})' = (\mathcal{M}\,\cap\,\mathcal{M}')\otimes (\mathcal{N}\,\cap\,\mathcal{N}') = \mathbb{C} I. \end{equation} However, I am unsure whether the first equality is valid, due to the closure involved in defining the tensor product algebra.
Also, I am aware of some specific cases, e.g. the tensor product of type II$_1$ factors is also a type II$_1$ factor, but cannot find a statement one way or the other for the general case.
Is the proof I gave valid? If not, are there any simple counterexamples? Thanks in advance for any help.
Your proof is valid. For the first equality, it actually follows from the fact that $(A \otimes B)’ = A’ \otimes B’$. Indeed,
$$(A \otimes B) \cap (C \otimes D) = [(A \otimes B)’ \vee (C \otimes D)’]’$$ $$= [(A’ \otimes B’) \vee (C’ \otimes D’)]’$$ $$= [(A’ \vee C’) \otimes (B’ \vee D’)]’$$ $$= (A’ \vee C’)’ \otimes (B’ \vee D’)’$$ $$= (A \cap C) \otimes (B \cap D)$$