I was wondering whether the following statement is true: If a $C^*$algebra A is a primitive, and it contains a simple ideal F, then F is contained in every non-zero closed ideal of A.
We say a $C^*$ algebra is primitive if it has a faithful irreducible representation on a Hilbert space.
We say a (closed) ideal is simple if it doesn't contain any proper ideal.
Basically, we want to show that any closed ideal $I$ of $A$ must have a non-trivial intersection with $F$. If this holds, then since F is simple, it must be the cast that $I$ must contain $F$. But why is this true?
This is true. If $F = \{0\}$ there is nothing to prove. So assume $F$ is nontrivial. Recall that in a $C^\ast$-algebra, if $I$ and $J$ are closed ideals, then $I \cap J = IJ$ (see Murphy’s $C^\ast$-algebras and operator theory, page 82). Recall that a primitive ideal is always prime in a $C^\ast$-algebra, i.e., if $I$ is primitive, and $J_1J_2 \subset I$, then it must be the case that either $J_1 \subset I$ or $J_2 \subset I$ (see Theorem 5.4.5 of the same book). Combining the above, we see that in a primitive $C^\ast$-algebra, since $\{0\}$ is a primitive ideal, any two nonzero closed ideals must intersect nontrivially. But $F$, per your assumption, is simple. So for any nonzero $I$, its intersection with $F$ must be nontrivial, but its intersection with $F$ is an ideal contained in $F$, whence it must be $F$ itself, i.e., $F \subset I$.