The answer to this question says the following: Let $R$ be a finitely generated $k$-algebra, where $k$ is a field. If $p \in R$ is a prime element, then $p$ is a prime element in $\tilde{R}$, the integral closure of $R$ in its fraction field $Q(R)$.

Now, let $S$ be a $k$-algebra such that $R \subseteq S \subseteq \tilde{R} \subset Q(R)$. Clearly, $R \subseteq S$ is integral.

When the following property is satisfied? If $p \in R$ is a prime element, then $p$ is a prime element in $S$.

A condition that may be relevant: $R^{\times}=S^{\times}$ (= same invertible elements).

More generally (see the following questions: i and ii):

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.

A nice example: If $B$ is the integral closure of $A$ in the fraction field of $A$, then the property is satisfied, as was quoted above.

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.

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 $\mathbb{C}[x]$.

Thank you very much!

The answer to this question says the following: Let $R$ be a finitely generated $k$-algebra, where $k$ is a field. If $p \in R$ is a prime element, then $p$ is a prime element in $\tilde{R}$, the integral closure of $R$ in its fraction field $Q(R)$.

Now, let $S$ be a $k$-algebra such that $R \subseteq S \subseteq \tilde{R} \subset Q(R)$. Clearly, $R \subseteq S$ is integral.

When the following property is satisfied? If $p \in R$ is a prime element, then $p$ is a prime element in $S$.

A condition that may be relevant: $R^{\times}=S^{\times}$ (= same invertible elements).

More generally (see the following questions: i and ii):

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.

A nice example: If $B$ is the integral closure of $A$ in the fraction field of $A$, then the property is satisfied, as was quoted above.

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.

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 $\mathbb{C}[x]$.

Remark: I do not mind to further assume that $B$ is a finitely generated $A$-algebra, or even that $B=A[b_1,b_2]$ or $B=A[b]$, for some $b_1,b_2,b \in B$.

Thank you very much!

The answer to this question says the following: Let $R$ be a finitely generated $k$-algebra, where $k$ is a field. If $p \in R$ is a prime element, then $p$ is a prime element in $\tilde{R}$, the integral closure of $R$ in its fraction field $Q(R)$.

Now, let $S$ be a $k$-algebra such that $R \subseteq S \subseteq \tilde{R} \subset Q(R)$. Clearly, $R \subseteq S$ is integral.

When the following property is satisfied? If $p \in R$ is a prime element, then $p$ is a prime element in $S$.

A condition that may be relevant: $R^{\times}=S^{\times}$ (= same invertible elements). See the following similar questions: i, ii.

Thank you very much!

The answer to this question says the following: Let $R$ be a finitely generated $k$-algebra, where $k$ is a field. If $p \in R$ is a prime element, then $p$ is a prime element in $\tilde{R}$, the integral closure of $R$ in its fraction field $Q(R)$.

Now, let $S$ be a $k$-algebra such that $R \subseteq S \subseteq \tilde{R} \subset Q(R)$. Clearly, $R \subseteq S$ is integral.

When the following property is satisfied? If $p \in R$ is a prime element, then $p$ is a prime element in $S$.

A condition that may be relevant: $R^{\times}=S^{\times}$ (= same invertible elements).

More generally (see the following questions: i and ii):

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.

A nice example: If $B$ is the integral closure of $A$ in the fraction field of $A$, then the property is satisfied, as was quoted above.

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.

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 $\mathbb{C}[x]$.

Thank you very much!