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 $D$?

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.

  • $\begingroup$ 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. $\endgroup$ – user237522 Nov 1 '16 at 23:12
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    $\begingroup$ 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 $3 = - \sqrt{-3} \cdot \sqrt{-3}$. $\endgroup$ – P. Koymans Nov 1 '16 at 23:15
  • $\begingroup$ 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 $U$). $\endgroup$ – user237522 Nov 1 '16 at 23:21
  • $\begingroup$ My $U$ and $D$ are of characteristic $0$. Recall that $\endgroup$ – P. Koymans Nov 1 '16 at 23:24
  • $\begingroup$ by definition $\mathbb{Z}_(2) = \{\frac{a}{b} : 2 \nmid b\} $\endgroup$ – P. Koymans Nov 1 '16 at 23:25

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