$\mathcal{K}(L^2(\mathbb{R}^m \times \mathbb{R}^n)) = \mathcal{K}(L^2(\mathbb{R}^m)) \otimes \mathcal{K}(L^2(\mathbb{R}^n))$?

QUESTION: Is it true that for the algebra of compact operators:

$\mathcal{K}(L^2(\mathbb{R}^m \times \mathbb{R}^n))$ is as a $C^{\ast}$-algebra isomorphic to $\mathcal{K}(L^2(\mathbb{R}^m)) \otimes \mathcal{K}(L^2(\mathbb{R}^n))$?

The latter tensor product is any $C^{\ast}$-tensor product (because the compact operators are nuclear it doesn't matter). On $L^2$ we use the ($\sigma$-finite) Lebesgue measure, but of course the algebra $\mathcal{K}(L^2)$ no longer depends on the measure. Clearly, $L^2(\mathbb{R}^{m} \times \mathbb{R}^n)$ can be identified with $L^2(\mathbb{R}^m) \otimes L^2(\mathbb{R}^n)$ by Fubini's theorem. This makes me think that $L^2$ has a better chance than other Hilbert spaces to make $\mathcal{K}(\mathcal{H}_1 \otimes \mathcal{H}_2) = \mathcal{K}(\mathcal{H}_1) \otimes \mathcal{K}(\mathcal{H}_2)$ hold.

EDIT: Of course $\mathcal{K}(\mathcal{H})$ denotes the compact operators on the Hilbert space $\mathcal{H}$.
 Obvious question: is $\mathcal K$ Kalkin algebra (the quotient of bounded operators by compact ones)? – Davide Giraudo Oct 13 '12 at 21:06 No.. the compact operators. – user44568 Oct 13 '12 at 23:27