A von Neumann $M$ is called prime if $M=M_1\bar{\otimes} M_2$ implies that $M_1$ or $M_2$ is a type $I$ von Neumann algebra.
A factor is prime if and inly if the factor cannot be factorized as the tensor product of two diffuse von Neumann algebras.
If a von Neumann algebra $M$ is prime, $N$ is a von Neumann subalgebra of $M$. Is $N$ also prime?
No. Every II$_1$ factor contains a hyperfinite II$_1$ subfactor $R$, which is not prime since $R = R \mathbin{\bar{\otimes}} R$. So just choose a prime II$_1$ factor $M$ and $N = R$.