Let $A$ and $B$ be $C^{\ast}-$ algebras and $A \otimes B$ denotes minimal(spatial) tensor product. Let $I$ and $J$ be be prime ideals of $A$ and $B$ respectively. The following fact should be easy but I am not able to see it.
$I \otimes J$ is prime ideal of $A \otimes B$.
It is clear that $I \otimes J$ is an ideal. Can someone please prove that it is infact prime ideal.
What you're asking fails trivially; we have the following easy exercise: if $A$ is a $C^*$-algebra, then $0$ is a prime ideal in $A$ if-f any pair of non-zero ideals in $A$ has non-zero intersection.
Take $A=\mathbb{C}$ and $B=C([0,1])$. Note that $0$ is a prime ideal in $A$, by the exercise above. It is also evident that any $C^*$-algebra is a prime ideal in itself. If what you were asking was true, we should have that $0=0\otimes B$ is a prime ideal in $A\otimes B=\mathbb{C}\otimes B=B$. Again by the exercise, this implies that any pair of non-zero ideals in $B$ has a non-zero intersection. But this is not true: take the ideals $I_1=\{f\in B: f(t)=0 \text{ for all }t\in[0,2/3]\}$ and $I_2=\{f\in B: f(t)=0\text{ for all }t\in[1/2,0]\}$. These are non-zero ideals, but their intersection is $0$.