Let $A \subset B$ be Noetherian integral domains (not necessarily UFD's), and let $a \in A$ be irreducible (in $A$). Assume that $B$ is a finitely generated $A$-module and a free $A$-module.
Is it true that $a$ must be irreducible in $B$?
I do not mind to further assume that $A$ and $B$ are $\mathbb{C}$-algebras, if this helps.
My motivation: $A=k[x^2]$, $B=k[x^2][x^3]$, $a=x^2$; $a$ remains irreducible in $B$.
If there exists a counterexample, is there an additional mild assumption that will guarantee that $a$ remains irreducible?
Edit: What if we require, in addition, that $A,B$ are UFD's, and $A \subseteq B$ is a separable ring extension?
Thank you very much!
The answer is no. Take $A=\mathbb{R}[x]$ and $B=\mathbb{C}[x]$. Then $A \subseteq B$ and $x^2+1 \in A$ is irreducible. However, $x^2+1=(x+i)(x-i) \in B$.