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

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 $$D$$?

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 $$D$$?

Thank you very much!

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.
• Thanks! I also thought about $U=\mathbb{Z}$ and $D=\mathbb{Z}[\sqrt{-3}]$. Am I right? (What is the Krull dimension of $\mathbb{Z}[\sqrt{-3}]$?). This is why I deleted my additional question about $D$ being a UFD a few minutes ago. Nov 1, 2016 at 23:12
• The problem with $U = \mathbb{Z}$ and $D = \mathbb{Z}[\sqrt{-3}]$ is that your condition "every irreducible element of $U$ remains irreducible in $D$" is false in this case, since $3 = - \sqrt{-3} \cdot \sqrt{-3}$. Nov 1, 2016 at 23:15
• Oh, of course, thank you. Is it possible to find such $U$ and $D$ with characteristic zero? What if in the above answer we take $\mathbb{Z}$ instead of $\mathbb{Z}_{2}$? (probably this will not help, since there are infinitely many primes in $U$). Nov 1, 2016 at 23:21
• My $U$ and $D$ are of characteristic $0$. Recall that Nov 1, 2016 at 23:24