# Isn't the center of a von Neumann algebra on a separable Hilbert space a hyperfinite von Neumann subalgebra?

this is a very quick, probably dumb, question, I was reading this chapter from "Hochschild cohomology of von Neumann algebras" by Allan Sinclair and Roger M. Smith and I came across this theorem on page 78:

3.1.1 Theorem. If $\mathcal{N}$ is a hyperfinite von Neumann subalgebra of a von Neumann algebra $\mathcal{M}$ and if $\mathcal{V}$ is a dual normal $\mathcal{M}$-module, then

$H^{n}(\mathcal{M},\mathcal{V}) \cong H^{n}_{w}(\mathcal{M},\mathcal{V}) \cong H^{n}_{w}(\mathcal{M},\mathcal{V:}/\mathcal{N})$ and ......................................

That's not the complete text of the theorem, I was curious, I'm reading that abelian $C^*$-algebras are nuclear and nuclear $C^*$-algebras are amenable, amenable von Neumann algebras acting on separable Hilbert spaces are hyperfinite. Now let $M$ be a von Neumann algebra acting on a separable Hilbert space, the center of a $M$ is an abelian von Neumann subalgebra, but then that means that every von Neumann algebra on a separable Hilbert space has a hyperfinite von Neumann subalgebra which means that the theorem above is true for every von Neumann algebra on a separable Hilbert space no? what am I missing?

:)

But it is trivially true that every von Neumann algebra has hyperfinite subalgebras. In fact, every von Neumann algebra has finite-dimensional unital subalgebras. You can start with $\mathbb C\,1$, for instance. If the von Neumann algebra is nontrivial it will have a non trivial projection $p$, and then you can consider the two-dimensional subalgebra $\mathbb C\,p+\mathbb C\,(1-p)$. Etc.