Let $R$ be the hyperfinite $II\_1$-factor. We know that $R$ is isomorphic to $R\otimes R$. So, $L\_\infty(0,1) \otimes R$ is a von Neumann subalgebra of $R$.


I am not sure whether it is sure for any type $II\_1$ von Neumann algebra $M$, i.e.,
is $L\_\infty(0,1) \otimes M$ a von Neumann subalgebra of $M$?



Tensor product of a von Neumann algebra and $L_\infty $
If $\mathbb F\_I$ denotes the free group on $I$ generators with $\lvert I \rvert > 1$, then $L^\infty(0, 1) \overline \otimes L \mathbb F\_I$ is not isomorphic to a von Neumann subalgebra of $L \mathbb F\_I$. For $\lvert I \rvert > \aleph\_0$ this is Corollary 6.4 in [S. Popa: Orthogonal pairs of subalgebras in finite von Neumann algebras, J. Op. Th. 9(1983), 253-268]. The general case $\lvert I \rvert > 1$ is Theorem 1 in [[N. Ozawa: Solid von Neumann algebras, Acta Math. 192 (2004), 111-117](https://doi.org/10.1007/BF02441087)].


