Suppose that $R$ and $S$ are unital rings and that $S$ is a subring of $R$ in the weak sense where the multiplicative identities $1_R$ and $1_S$ are not assumed to be the same. In fact, assume $1_R \neq 1_S$. Then $S$ cannot contain any invertible element of $R$ (if $s \in S$ and there is an $r \in R$ with $sr=1_R$ then $0 = (1_Rs - 1_Ss)r = 1_R - 1_S$ so $1_R = 1_S$). This is sort of reminiscent of the statement a proper ideal in $R$ cannot contain any invertible element of $R$. My question is whether there must actually be some proper ideal $I$ of $R$ with $S \subset I \subset R$. Actually, I would like to ask the question with the rings replaced by C*-algebras - but a positive answer to the corresponding question about rings would clearly suffice.
Let $A$ and $B$ be unital C*-algebras with $B$ a sub-C*-algebra of $A$, but suppose $1_A \neq 1_B$. Does it necessarily follow that there is a proper closed ideal $I$ of $A$ satisfying $B \subset I \subset A$?
I'm also curious what happens if the assumption that $B$ is unital is dropped. In this case I'm not even certain whether $B$ can contain an invertible element of $A$.
Let $B$ be a non-unital sub-C*-algebra of a unital C*-algebra $A$. Can $B$ contain an invertible element of $A$? If not, does there have to exist a proper closed ideal $I$ of $A$ with $B \subset I \subset A$?