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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.