Let $A, B$ be two simple C*-algebras, it is known that $A\otimes_{min}B$ is simple. Here, "simple" means there are no non-trivial closed two sided ideal, and the norm is using minimal norm(or spatial norm).
I want to verify this result, in Operator Algebras of Blackadar, this result is true because of the minimality of the spatial norm(P.187). But I can't figure out the proof. I believe the main idea of the proof is obtaining contradiction, then I get stuck. Can anyone give me some hints to proceed?
Let $\pi_A\otimes\pi_B$ be the representation of $A\odot B$ which induces the min norm i.e. $\|\pi_A\otimes \pi_B(x)\|=\|x\|_{min}$ (by Corollary II.6.4.9. in Blackadar) since $A$ and $B$ are simple this representation is faithful.
Suppose $A\otimes_{min} B$ is not simple, so there is a proper closed two-sided $*$-ideal $J\in A\otimes_{min} B$. Hence, there is a $*$-homomorphism $\phi:A\otimes_{min} B \rightarrow A\otimes_{min} B/J$ and $$\|\phi(\pi_A\otimes \pi_B(x))\|\leq \|\pi_A\otimes \pi_B(x)\|=\|x\|_{min}$$ (by Corollary II.1.6.6. in Blackadar), which contradicts the minimality of the min-norm.