Is the algebra of bounded operators stable? Let $H$ be a separable Hilbert space. Is it true that there is an isomorphism of $C^*$-algebras $$B(H)\hat{\otimes} K(H)\cong B(H)$$ where $B(H)$ is the algebra of bounded operators, $K(H)$ is the ideal of compact operators?
 A: In response to Matt "calling my bluff" I realised that my claim takes some work to justify. The following is very much "thinking aloud through a head cold" so is not a polished exposition, and I have probably overlooked simplifications.

View $K(H)$ as generated by the matrix units $\{e_{ij} \colon i,j\geq 1\}$ viewed as rank-one operators on $H=\ell^2({\mathbb N})$.
For each $n\geq 1$ let $A_n$ be the closed subalgebra of $A\otimes K(H)$ generated by $\{a\otimes e_{ij} \colon a \in A, 1\leq i \leq n , 1\leq j\leq n\}$.
Let $\iota: A \to A$ be the identity map and define $\phi_n : K(H) \to {\mathbb M}_n$ by truncating to the top-left $n\times n$ corner. I claim that $E_n := \iota \otimes \phi_n : A\otimes K(H) \to A_n$ has the conditional expectation properties: $E_n(x)=x$ for all $x \in A_n$ and $E_n (xy) = E_n(x)E_n(y)$ if either $x$ or $y$ belongs to $A_n$. (I convinced myself of this with some $2\times 2$ block matrix heuristics, it shouldn't be hard to make the proof fully precise.)
Therefore, if $A\otimes K(H)$ has an identity element $u$, $E_n(u)$ is an identity element for $A_n$, for each $n$.
In particular, $A=A_1$ must have an identity element, which we denote by $1_A$. Then $1_A \otimes I_n$ must be the identity element for $A_n$ (because identity elements in an algebra are unique). We therefore get $E_n(u) = 1_A \otimes I_n$ for all $n$ which should lead to a contradiction with the assumption that $u \in A\otimes K(H)$, by some hacky approximation argument (e.g. find $u_0$ in $A\otimes {\mathbb M}_m= A_m$ for some $m$ such that $\Vert u-u_0\Vert \ll 1$, then $u_0 = E_{m+1}(u_0)$ has to be close to $1_A\otimes I_{m+1}$ which is a contradiction since elements of $A_m$ are annihilated by $E_{m+1}-E_m$ while $1_A\otimes I_{m+1}$ isn't).
