Let $A \subset C$ be two isomorphic unique factorization domains (UFD's).
Is it possible to find an integral domain $B$, $A \subset B \subset C$, such that $B$ is not a UFD?
I have tried to construct the following counterexample (but have not succeeded): $A=\mathbb{C}[p,q]$, $C=\mathbb{C}[x,y]$, where $p,q \in \mathbb{C}[x,y]$ are algebraically independent over $\mathbb{C}$ (this is equivalent to $\{p,q\}$ having a non-zero Jacobian: $p_xq_y-p_yq_x \neq 0$), and $B=\mathbb{C}[p,q][w]$, for some $w \in \mathbb{C}$, but the problem is to find concrete $p,q,w$ such that $B$ is not a UFD and not equals either $A$ or $C$.
Perhaps the following is a counterexample: $A=k[x_2,x_3,\ldots]$, $B=k[x_1^2,x_1^3,x_2,x_3,\ldots]$, $C=k[x_1,x_2,x_3,\ldots]$?
Related ideas (perhaps I should elaborate why they are relevent) are : Nagata's criterion for factoriality and this question.
Any ideas are welcome!
Yes, it's possible.
Let $K$ be a field, and let \begin{align*} C &= K[x]\qquad\qquad\qquad\;\,\\[4pt] B &= K[x^2,x^3]\\[4pt] A &= K[x^6]\\[4pt] \end{align*} Then \begin{align*} &{\small{\bullet}}\;\;A \subset B \subset C\\[4pt] &{\small{\bullet}}\;\;A,C\;\text{are isomorphic UFDs}\\[4pt] &{\small{\bullet}}\;\;B\;\text{is not a UFD}\\[4pt] \end{align*}
To see that $B$ is not a UFD, note that in $B$, the elements $x^2$ and $x^3$ are distinct irreducibles, so
$$(x^2)(x^2)(x^2) = x^6 = (x^3)(x^3)$$
gives two distinct factorizations of $x^6$ into irreducibles.
Hence, as claimed, $B$ is not a UFD.