It is well known that in $C^*$ algebra category, If $\mathcal{A}$ is a $C^*$ algebra and $\mathcal{I}$ is a norm closed ideal of $\mathcal{A}$, then $\mathcal{I}$ is also a $C^*$ algebra, and $\mathcal{A}/\mathcal{I}$ is also a $C^*$ algebra.
Similarly, we know that if $M\subset B(H)$ is a von Neumann algebra and $N$ is a $\sigma-$ WOT closed ideal of $M$, then $N$ is also a von Neumann algebra. However, it might not be a von Neumann subalgebra of $M$ since the unit element may not be the same, which means that $N$ is actually a von Neumann algebra when we consider $N\subset B(pH)$ where $p\in P(Z(M))$.
My question is, if we consider the quotient space $M/N$, we know that it will be a $C^*$ algebra, but will it still be a von Neumann algebra?
Here, someone may be confused since $M/N$ may not be a subspace of $B(H)$, just remember that we also have an abstract definition for a von Neumann algebra. $M/N$ is a von Neumann algebra if and only if it is a $C^*$ algebra and it will be isometrically isomorphic to the dual of a Banach space.
Any help will be truly grateful!
I'm not entirely clear what you are asking. A proper ideal in $M$ necessarily has a different unit than $M$.
When $N$ is a proper von Neumann ideal, it is not hard to show that there exists a central projection $p\in M$ with $N=pM$. Then $M/N$ is trivially isomorphic to $(1-p)M$, which is naturally a von Neumann subalgebra of $M$.