Let $A \subseteq B$ be two $k$-algebras which are (commutative) integral domains. Assume that $A,B$ have the same field of fractions, $Q(A)=A(B)$.
Is it possible that there exists a prime element in $A$ which is not a prime element in $B$?
In other words, is there $a \in A$ such that $A/aA$ is an integral domain, but $B/aB$ is not an intrgral domain?
Remarks: Consider $A=k[x^2,x^3]$, $B=k[x]$.
(1) It is not an example, since there are no prime elements in $A$.
(2) Irreducibles in $A$ do not remain irreducibles (= prime, since $B$ is a UFD) in $B$; indeed, $x^2$ is irreducible in $A$, but reducible in $B$.
See this similar but not identical question.
Any hints and comments are welcome!
After the discussion in the comments, I wish to further assume that:
(i) $A^{\times}=B^{\times}=k^{\times}$. ($S^{\times}$ denotes the invertible elements in a set $S$).
(ii) $B$ is integral over $A$.
Is it true that every prime element in $A$ remains prime in $B$?
Observe that $A=\mathbb{Z}, B=\mathbb{Z}[i]$ is not a counterexample, since $Q(A)=\mathbb{Q} \subsetneq \mathbb{Q}(i)= Q(B)$.
Take $B$ the fraction field of $A$.