Let $A \subseteq B$ be two commutative integral domains which are $\mathbb{C}$-algebras.
Question 1: Is it possible to find mild conditions on $A, B, A \subseteq B$ such that the following property P is satisfied:
Property P: Every prime element in $A$ remains prime in $B$.
An empty example: $A=\mathbb{C}[x^2,x^3]$, $B=\mathbb{C}[x]$. There are no prime elements in $\mathbb{C}[x^2,x^3]$, so the property is satisfied. See this question.
A nice example: If $B$ is the integral closure of $A$ in the fraction field of $A$, then the property is satisfied, as was proved in an answer to this MO question.
Non-examples: (i) $A=\mathbb{C}[x^2]$, $B=\mathbb{C}[x^2,x^3]$. $x^2$ is prime (= irreducible in a UFD) in $\mathbb{C}[x^2]$, but $x^2$ is not prime in $\mathbb{C}[x^2,x^3]$ (it remains irreducible), since $x^2$ divides $x^6=x^2x^2x^2=x^3x^3$ but it does not divide $x^3$. (ii) $A=\mathbb{Z}$, $B=\mathbb{Z}[i]$. $2$ is prime in $\mathbb{Z}$ (= irreducible in a UFD), but $2$ is not prime in $\mathbb{Z}[i]$ (also, it is not irreducible), since $2=(1+i)(1-i)$ divides the product $(1+i)(1-i)$ but it does not divide $1+i$ or $1-i$; this example was presented in an answer to this question.
A plausible answer 1? Four mild conditions: (i) $Q(A)=Q(B)$. ($Q(D)$ denotes the fraction field of an integral domain $D$). (ii) $A^{\times}=B^{\times}$. ($S^{\times}$ denotes the invertible elements of a set $S$). (iii) $B$ is integral over $A$. (iv) $B$ is a finitely generated $A$-algebra.
Question 2: Is it possible to find mild conditions on $A, B, A \subseteq B$ such that the following property I is satisfied:
Property I: Every irreducible element in $A$ remains irreducible in $B$.
Notice that the first, empty example, is not valid anymore since $x^2,x^3$ are irreducibles in $\mathbb{C}[x^2,x^3]$, but reducibles in $\mathbb{C}[x]$.
Now I have asked the above question in MO also.
Thank you very much!