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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

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  • $\begingroup$ Can we say that it's atleast an F.D? $\endgroup$ – HARRY Jul 31 at 4:35
  • $\begingroup$ What is an "F.D."? Do you mean something other than a unique factorization domain? $\endgroup$ – KReiser Jul 31 at 4:42
  • $\begingroup$ F.D means just a factorization domain, where one do not need uniqueness $\endgroup$ – HARRY Jul 31 at 4:50
  • $\begingroup$ @Dave, every field is vaccously an U.F.D $\endgroup$ – HARRY Jul 31 at 4:59
  • $\begingroup$ @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! $\endgroup$ – Dave Jul 31 at 5:18
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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.

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  • $\begingroup$ thanks for your answer.In my case I was looking for $R$ a polynomial ring which is essentially noetherian. $\endgroup$ – HARRY Jul 31 at 5:24
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    $\begingroup$ You're welcome, but I would be a little careful about characterizing a polynomial ring as essentially noetherian without saying it has a finite number of generators: the polynomial ring in infinitely many variables is not noetherian. $\endgroup$ – KReiser Jul 31 at 5:51

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