When irreducible elements of a UFD remain irreducible in a ring extension

Question

Let $U$ be a Noetherian UFD and let $D$ be a Noetherian integral domain (not known to be a UFD) such that $U \subseteq D$. Further assume that $U$ and $D$ have the same finite Krull dimension.

Of course, generally, an irreducible (=prime) element of $U$ may become reducible in $D$.

What can be said about such pairs of domains with the additional property that every irreducible element of $U$ remains irreducible in $D$?

An example: $U=\mathbb{C}[x^2]$, $D=\mathbb{C}[x^2][x^3]$; if I am not wrong, every irreducible element of $\mathbb{C}[x^2]$ remains irreducible in $D=\mathbb{C}[x^2][x^3]$ (though not prime).

Edit: If my above question is too general, then I wish to ask the following question: Given an irreducible element $u \in U$, can one find a "nice" criterion which guarantees that $u$ remains irreducible in $D$?

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New edit: Another question:

If we further assume that $U \subseteq D$ is etale, then is it true that every irreducible element of $U$ remains irrdducible in $D$? or is it true that every prime element of $U$ remains prime in $D$?

Please see this recent question.

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Thank you very much!

Answer 1

This does not force $D$ to be a UFD as you originally asked, here is a counterexample. Take $U = \mathbb{Z}_{(2)}$ and $D = \mathbb{Z}_{(2)}[X]/(X^2 - 8)$. Then $U$ is a DVR and its only non-zero prime is $(2)$.

An easy computation shows that all units in $D$ are $a + bX$, where $a$ is a unit in $U$. Using this it is not hard to show that $2$ remains irreducible.

But clearly we have $$ 2^3 = 8 = X \cdot X $$ in $D$, showing that $D$ is not a UFD.