I have two questions, just the first one remains unsolved for me.
Is there an example of a commutative unital ring $B$, a unital subring $A \subsetneq B$ and a non-zero ideal $I$ of $B$ such that $I \subseteq A$?
Let $A \subset B$ be two commutative rings. Assume that any proper ideal of $A$ is also an ideal of $B$. Do we have $A=B$ ?
I noticed that $2)$ is wrong if we assume that any proper ideal of $B$ is included in (thus an ideal of) $A$; take $B$ to be a field. Actually, I've just noticed that $2)$ is wrong: take $A$ to be any field.
For $1)$ I thought to $B=\Bbb Z \times \Bbb Z \supsetneq A = \{(x,x),x\in \Bbb Z\}$. But it didn't work.
Thank you!
Let $R, S$ be commutative unital rings and let $S' \le S$ be a commutative unital subring. Let $A = R \times S'$ and $B = R \times S$. Clearly $A \le B$ and $(1,1) \in A$. If $I \subset R$ is a proper ideal, then $I \times \{0\} \subset A$ is a proper ideal too.
(Alternatively, following @user26857's excellent suggestion, one may start with $B$ and consider a unital subring $S$; then $A = S + I$ is a unital subring as desired.)