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Does anyone know an example of a unique factorization domain $R$ that is

(i) not a Dedekind domain (or equivalently, not a principal ideal domain) and

(ii) contains some irreducible element $r \in R$ such that the quotient $R/rR$ is finite?

I am grateful for any suggestions.

Algebrus
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    I suspect there is no such example, but I haven't been able to prove it. A UFD is noetherian and integrally closed, so if it is not a Dedekind domain then it must have Krull dimension $\geq 2$. By Krull's Hauptidealsatz $rR$ has codimension $\leq 1$ and since $R/rR$ is finite, then it has dimension $0$. In nice cases one has $\dim(R) = \dim(R/rR) + \mathrm{codim}(rR)$, which would give $2 \leq \dim(R) = \dim(R/rR) + \mathrm{codim}(rR) \leq 0 + 1 = 1$, a contradiction. However, in general we only have $\dim(R) \geq \dim(R/rR) + \mathrm{codim}(rR)$, which doesn't help. – Viktor Vaughn May 31 '18 at 05:42
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    @Quasicoherent: I do not think that UFDs are Noetherian, the polynomial ring over a field in countably many variables being a counterexample. So one also has to deal separately with non-Noetherian UFDs. However, thank you for sharing this interesting approach! So we have to find a ring which does not have one of these nice properties. – Algebrus May 31 '18 at 06:58
  • @Algebrus : maybe you could ask this on MathOverflow? – Watson May 31 '18 at 06:59
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    @Watson: Done. https://mathoverflow.net/questions/301617/ufd-containing-element-with-finite-quotient – Algebrus May 31 '18 at 07:08

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