The commutant of a tensor product In this whole post beginning with the second paragraph, by "tensor product" I will mean the operation done on Von Neumann algebras where one applies the homomorphism from here Tensor products of maps
to the algebraic tensor product of the Von Neumann algebras, and then takes the WO closure in the space $B(H \otimes H)$ where that "$\otimes$" is the tensor product of Hilbert spaces in the Hilbert sense.  
As part of the proof of some equalities relating the commutant of a tensor product to the tensor product of commutants, I am asked to prove "directly" that if $C(H)$ denotes the scalar operators on $H$, then $(B(H) \otimes C(H))'=C(H) \otimes B(H)$  I am having trouble proving that the left side is included in the right side.  Help please!
 A: Suppose that $\{ e_m\}$ is an orthonormal basis for $H$ and that $\{e_{mn}\}$ are the corresponding matrix units. Suppose that $T \in B(H \otimes H)$ with $$T (e_i \otimes e_j)=\sum_{k,l} T_{ij}^{kl} e_k \otimes e_l.$$ 
We now also assume that $T$ commutes with the operators $\{e_{mn} \otimes \iota\}$ (where $\iota:H \to H$ is the identity map). A direct calculation shows that $T$ must be of the form $\iota \otimes S$.
We have 
$$
T\circ (e_{mn} \otimes \iota) (e_i \otimes e_j) =\delta_{in} \sum_{kl} T_{mj}^{kl}e_k \otimes e_l,
$$
where $\delta_{in}$ is the Kronecker delta function.
On the other hand, we have
$$
(e_{mn} \otimes \iota)\circ T (e_i \otimes e_j)= \delta_{nk} \sum_{kl} T_{ij}^{kl} e_m \otimes e_l=\sum_l T^{nl}_{ij} e_m \otimes e_l.
$$
Comparing these two expressions, we see that $T_{mj}^{kl}$ must be $0$ unless $m=k$. We then have that $\delta_{in} T_{mj}^{ml}=T_{ij}^{nl}$ or $T_{mj}^{ml}=T_{ij}^{il}$. But this says that $T_{mj}^{ml}$ is the same for all $m$. Thus $T=\iota \otimes S$, where $Se_j =\sum_l T_{mj}^{ml}e_l$.  
As $\lambda \iota \otimes \frac{S}{\lambda}= \iota \otimes S$ for a nonzero constant $\lambda$, we have shown that $$(B(H) \otimes C(H))'\subseteq C(H) \otimes B(H).$$ The other inclusion is easy.
