Is it true that quotient of a unique factorization domain by a prime ideal is a factorization domain?
Is it true at least for polynomial rings?
Could anyone give any reference of this fact?
Any help from anyone is welcome
2026-03-25 10:55:26.1774436126
On quotient of UFD
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If you're willing to assume that $R/p$ is noetherian, then every such ring is a factorization domain.
Statement: Every noetherian domain is a factorization domain.
Proof: Let $S$ be the set of ideals of the form $(x)$ for $x$ an element not expressible as a product of a unit and a finite number of irreducible elements. If it's nonempty, we may choose a maximal element, say $(a)$. As $a$ is not irreducible, $a=bc$ with $b,c$ not units nor associates of each other. So $(b)$ and $(c)$ properly contain the ideal $(a)$, and thus do not belong to $S$ by maximality of $(a)$ within $S$. So $b,c$ can be written as a product of a unit and a finite number of irreducibles, and therefore so can $(a)$. So $S$ is empty and we're done.
If you allow the non-Noetherian case, you will almost assuredly run in to counterexamples. You should be able to cook one up using a polynomial ring in infinitely many variables.