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

Question

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?

:)

Answer 1

I cannot speak about the cohomology thing because I would have to go back and look at the definitions.

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.