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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!

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  • $\begingroup$ You probably need something like everything in sight being Noetherian and catenary. $\endgroup$
    – Neil Epstein
    Commented Feb 23, 2020 at 15:58
  • $\begingroup$ @NeilEpstein, thank you very much! You probably mean that the above propery is satisfied if $R,S$ are Noetherian and $S$ is catenary, as R. van Dobben de Bruyn has suggested (for example, if $R$ is a finitely generated $k$-algebra). So, for example, if $R$ is a finitely generated $k$-algebra, then if $R \subseteq S$ is integral and $Q(R)=Q(S)$, then every prime element in $R$ remains prime in $S$? $\endgroup$
    – user237522
    Commented Feb 23, 2020 at 20:42
  • $\begingroup$ @NeilEpstein, please what if we remove one of the following two conditions: (1) $R \subseteq S$ is integral. (2) $Q(R)=Q(S)$. For example, $R=k[x^2] \subsetneq k[x^2,x^3]=S$ with $Q(R)=k(x^2) \subsetneq k(x)=Q(S)$ and $x^2$ is prime in $k[x^2]$ but not prime in $k[x^2,x^3]$. I guess other conditions should be added in order to guarantee the above property (flatness instead of (2) is not enough since here the extension is free). What about being etale? math.stackexchange.com/questions/3548524/… $\endgroup$
    – user237522
    Commented Feb 23, 2020 at 20:57
  • $\begingroup$ Also, what if we replace "prime" by "irreducible"? $\endgroup$
    – user237522
    Commented Feb 23, 2020 at 22:13

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