[proofplan]
Let $P:=M\overline{\otimes}N\subseteq \mathcal{L}(H\otimes K)$. The proof uses the identity $Z(P)=P\cap P'$ and the commutant theorem for spatial tensor products, which gives $P'=M'\overline{\otimes}N'$. For an element in this intersection, normal slice maps show that all its right slices lie in $Z(M)$ and all its left slices lie in $Z(N)$; the slice-map characterization of spatial tensor products then forces the element to lie in $Z(M)\overline{\otimes}Z(N)$. The reverse inclusion follows by checking central elementary tensors and taking weak operator closures.
[/proofplan]
[step:Rewrite the center using the spatial tensor product commutant theorem]
Define
\begin{align*}
P:=M\overline{\otimes}N\subseteq \mathcal{L}(H\otimes K).
\end{align*}
By the definition of the center of a von Neumann algebra,
\begin{align*}
Z(P)=P\cap P'.
\end{align*}
The commutant theorem for spatial tensor products gives
\begin{align*}
P'=(M\overline{\otimes}N)'=M'\overline{\otimes}N'.
\end{align*}
Therefore
\begin{align*}
Z(P)=(M\overline{\otimes}N)\cap(M'\overline{\otimes}N').
\end{align*}
[/step]
[step:Use normal slice maps to force central slices]
Let $x\in Z(P)$. Let $\omega\in \mathcal{L}(K)_*$ be a normal linear functional on $\mathcal{L}(K)$, and let
\begin{align*}
\operatorname{id}_{\mathcal{L}(H)}\otimes\omega:\mathcal{L}(H\otimes K)\to\mathcal{L}(H)
\end{align*}
denote the corresponding normal right slice map. Since $x\in M\overline{\otimes}N$, the slice-map property gives
\begin{align*}
(\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M.
\end{align*}
Since $x\in M'\overline{\otimes}N'$, the same slice-map property gives
\begin{align*}
(\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M'.
\end{align*}
Thus
\begin{align*}
(\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M\cap M'=Z(M).
\end{align*}
Similarly, for every normal linear functional $\varphi\in \mathcal{L}(H)_*$, with left slice map
\begin{align*}
\varphi\otimes\operatorname{id}_{\mathcal{L}(K)}:\mathcal{L}(H\otimes K)\to\mathcal{L}(K),
\end{align*}
we obtain
\begin{align*}
(\varphi\otimes\operatorname{id}_{\mathcal{L}(K)})(x)\in N\cap N'=Z(N).
\end{align*}
[guided]
We want to prove that an operator $x\in Z(P)$ actually belongs to the smaller [tensor product](/page/Tensor%20Product) $Z(M)\overline{\otimes}Z(N)$. The available way to detect where an element of a spatial tensor product lives is to apply normal slice maps.
Fix a normal linear functional $\omega\in \mathcal{L}(K)_*$. The right slice map
\begin{align*}
\operatorname{id}_{\mathcal{L}(H)}\otimes\omega:\mathcal{L}(H\otimes K)\to\mathcal{L}(H)
\end{align*}
is the normal [linear map](/page/Linear%20Map) characterized on elementary tensors by
\begin{align*}
(\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(a\otimes b)=\omega(b)a
\end{align*}
for $a\in\mathcal{L}(H)$ and $b\in\mathcal{L}(K)$. Since $x\in Z(P)$, the previous step gives
\begin{align*}
x\in (M\overline{\otimes}N)\cap(M'\overline{\otimes}N').
\end{align*}
Applying the slice-map property to the inclusion $x\in M\overline{\otimes}N$ gives
\begin{align*}
(\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M.
\end{align*}
Applying the same property to the inclusion $x\in M'\overline{\otimes}N'$ gives
\begin{align*}
(\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M'.
\end{align*}
Hence the right slice commutes with every element of $M$ and belongs to $M$ itself, so
\begin{align*}
(\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M\cap M'=Z(M).
\end{align*}
The left slices are handled in the same way, but now we fix a normal linear functional $\varphi\in\mathcal{L}(H)_*$ and use the normal map
\begin{align*}
\varphi\otimes\operatorname{id}_{\mathcal{L}(K)}:\mathcal{L}(H\otimes K)\to\mathcal{L}(K).
\end{align*}
From $x\in M\overline{\otimes}N$ we get
\begin{align*}
(\varphi\otimes\operatorname{id}_{\mathcal{L}(K)})(x)\in N.
\end{align*}
From $x\in M'\overline{\otimes}N'$ we get
\begin{align*}
(\varphi\otimes\operatorname{id}_{\mathcal{L}(K)})(x)\in N'.
\end{align*}
Therefore
\begin{align*}
(\varphi\otimes\operatorname{id}_{\mathcal{L}(K)})(x)\in N\cap N'=Z(N).
\end{align*}
This is the point of using both descriptions of $x$: the first places each slice in the algebra, and the second places the same slice in the corresponding commutant.
[/guided]
[/step]
[step:Apply the slice-map characterization to obtain the forward inclusion]
The slice-map characterization of von Neumann subalgebra tensor products says the following: if $A\subseteq M$ and $B\subseteq N$ are von Neumann subalgebras and $y\in M\overline{\otimes}N$ satisfies
\begin{align*}
(\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(y)\in A
\end{align*}
for every $\omega\in\mathcal{L}(K)_*$ and
\begin{align*}
(\varphi\otimes\operatorname{id}_{\mathcal{L}(K)})(y)\in B
\end{align*}
for every $\varphi\in\mathcal{L}(H)_*$, then $y\in A\overline{\otimes}B$.
We apply this with $A:=Z(M)$, $B:=Z(N)$, and $y:=x$. The previous step verifies both slice hypotheses. Hence
\begin{align*}
x\in Z(M)\overline{\otimes}Z(N).
\end{align*}
Since $x\in Z(P)$ was arbitrary,
\begin{align*}
Z(P)\subseteq Z(M)\overline{\otimes}Z(N).
\end{align*}
[/step]
[step:Check central elementary tensors and close weakly]
Let $z\in Z(M)$ and $w\in Z(N)$. For arbitrary $m\in M$ and $n\in N$, the elementary tensor product multiplication rule gives
\begin{align*}
(z\otimes w)(m\otimes n)=zm\otimes wn.
\end{align*}
Since $z\in Z(M)$ and $w\in Z(N)$, we have $zm=mz$ and $wn=nw$, hence
\begin{align*}
zm\otimes wn=mz\otimes nw=(m\otimes n)(z\otimes w).
\end{align*}
Thus every elementary tensor $z\otimes w$ with $z\in Z(M)$ and $w\in Z(N)$ commutes with every elementary tensor $m\otimes n$ with $m\in M$ and $n\in N$.
The algebraic tensor product $M\odot N$ is weak-operator dense in $P=M\overline{\otimes}N$, and multiplication by a fixed bounded operator is weak-operator continuous on bounded sets against vector functionals. Since commutants are weak-operator closed, each $z\otimes w$ lies in $P'$. Also $z\otimes w\in P$. Therefore
\begin{align*}
z\otimes w\in P\cap P'=Z(P).
\end{align*}
Taking the weak operator closure of the linear span of such elementary tensors gives
\begin{align*}
Z(M)\overline{\otimes}Z(N)\subseteq Z(P).
\end{align*}
[/step]
[step:Conclude the center formula and the factor case]
Combining the two inclusions obtained above gives
\begin{align*}
Z(M\overline{\otimes}N)=Z(M)\overline{\otimes}Z(N).
\end{align*}
If $M$ and $N$ are factors, then
\begin{align*}
Z(M)=\mathbb{C}I_H
\end{align*}
and
\begin{align*}
Z(N)=\mathbb{C}I_K.
\end{align*}
Therefore
\begin{align*}
Z(M\overline{\otimes}N)=(\mathbb{C}I_H)\overline{\otimes}(\mathbb{C}I_K)=\mathbb{C}I_{H\otimes K}.
\end{align*}
Hence $M\overline{\otimes}N$ has scalar center, so $M\overline{\otimes}N$ is a factor.
[/step]
