# On quotient of UFD

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

• Can we say that it's atleast an F.D? – HARRY Jul 31 at 4:35
• What is an "F.D."? Do you mean something other than a unique factorization domain? – KReiser Jul 31 at 4:42
• F.D means just a factorization domain, where one do not need uniqueness – HARRY Jul 31 at 4:50
• @Dave, every field is vaccously an U.F.D – HARRY Jul 31 at 4:59
• @HARRY Yes I realised after posting my comment, apologies. For a counterexample in a non-Noetherian ring, I think the map sending $\mathbb{Z}[x_1,x_2,\ldots]$ to $\mathbb{Z}[\sqrt{2},\sqrt{\sqrt{2}},\ldots]$ via $x_i\mapsto 2^{1/(2^i)}$ is a homomorphism, so by the first isomorphism theorem we'll have a $\mathbb{Z}[x_1,x_2,\ldots]/P\cong\mathbb{Z}[\sqrt{2},\sqrt{\sqrt{2}},\ldots]$, but $\sqrt{2}=\sqrt{\sqrt{2}}\cdot\sqrt{\sqrt{2}}=\cdots$ I hope this is a bit more true than my previous comment! – Dave Jul 31 at 5:18

If you're willing to assume that $$R/p$$ is noetherian, then every such ring 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.
• thanks for your answer.In my case I was looking for $R$ a polynomial ring which is essentially noetherian. – HARRY Jul 31 at 5:24