I'm searching for a counterexample for $C^\*$-algebras $A$ and $B$ and essential ideals (I assume an ideal to be closed and only two-sided ideals) $I\subseteq A$, $J\subseteq B$ , such that the ideal $I\otimes\_{max} J$ is not essential in the (maximal tensor product) -$C^\*$-algebra $A\otimes\_{max} B$. I'm not sure if my idea works:


The algebra of compact operators $K(l^2(\mathbb{N}))$ is an essential ideal in the $C^\*$-algebra of bounded linear operators on $l^2(\mathbb{N})$, $B(l^2(\mathbb{N}))$. The reason is that $B(l^2(\mathbb{N}))$ has only the closed ideals $\{0\}$, $K(l^2(\mathbb{N}))$ and $B(l^2(\mathbb{N}))$, thus $K(l^2(\mathbb{N}))\cap M\neq 0$ for all closed nontrivial ideals $M\subseteq B(l^2(\mathbb{N}))$. 


But is $K(l^2(\mathbb{N}))\otimes\_{max} K(l^2(\mathbb{N}))$ essential in $B(l^2(\mathbb{N}))\otimes\_{max} B(l^2(\mathbb{N}))$? 


I don't think so, but I'm stuck to prove that there must be a nontrivial closed ideal $M$ in $B(l^2(\mathbb{N}))\otimes\_{max} B(l^2(\mathbb{N}))$ such that $K(l^2(\mathbb{N}))\otimes\_{max} K(l^2(\mathbb{N}))\cap M$ is trivial. For the proof it must be important that one takes the maximal norm-closure of $I\odot J\subseteq A\odot B$ ($\odot$ denotes the tensor product as $\*$-algebras), because particularly the spatial tensor product satisfies that if $I$ is essential in $A$, $J$ essential in $B$, then $I\otimes\_{min} J$ is essential in $A\otimes\_{min} B$ (and here is $B(l^2(\mathbb{N}))\otimes\_{min} B(l^2(\mathbb{N}))\neq B(l^2(\mathbb{N}))\otimes\_{max} B(l^2(\mathbb{N}))$. If my ideal don't work, what else can I do?
I appreciate your help.



is the maximal tensor product of compact operators an essential ideal?
My suggestion (I hope that nothing is wrong):


Let $$F:B(l^2(\mathbb{N}))\otimes\_{max} B(l^2(\mathbb{N}))\to B(l^2(\mathbb{N}))\otimes\_{min} B(l^2(\mathbb{N}))$$ be the canonical map and define $M:=\ker(F)$. Since $F$ is not injective, $M$ is nontrivial. 


Claim: $K(l^2(\mathbb{N}))\otimes\_{max} K(l^2(\mathbb{N})) \cap M = 0$.


First note that $K(l^2(\mathbb{N}))$ is nuclear (because it's a inductive limit of nuclear $C^\*$-algebras) and simple, therefore $$K(l^2(\mathbb{N}))\otimes\_{max} K(l^2(\mathbb{N}))\cong K(l^2(\mathbb{N}))\otimes\_{min} K(l^2(\mathbb{N}))$$ is simple.
It follows 
$K(l^2(\mathbb{N}))\otimes\_{max} K(l^2(\mathbb{N})) \cap M=\ker(F\_{|K(l^2(\mathbb{N}))\otimes\_{max} K(l^2(\mathbb{N}))})=0$, because $F\_{|K(l^2(\mathbb{N}))\otimes\_{max} K(l^2(\mathbb{N}))}$ is nonzero and $\ker(F\_{|K(l^2(\mathbb{N}))\otimes\_{max} K(l^2(\mathbb{N}))})$ is an ideal in $K(l^2(\mathbb{N}))\otimes\_{max} K(l^2(\mathbb{N}))$.


