# Example for Noetherian domain, non-UFD $R$ of dimension one such that every localization $R_P$ is UFD.

Theorem 179 in the book Commutative Rings by Kaplansky states that:

Let $R$ be an integral domain. The following conditions are necessary and sufficient for $R$ to be a UFD.

(1) $R$ satisfies the ascending condition on principal ideals

(2) In the polynomial ring $R[x]$ all minimal prime ideals are finitely generated.

(3) For any prime ideal $P$ of grade one in $R$, $R_P$ is a UFD.

(4) In any localization of $R[x]$ all invertible ideals are principal.

Now, I want to put this theorem in practice that is I'm looking for Noetherian domain $R$ , such that every localization $R_P$ at prime ideal $P$ is UFD but $R$ is not UFD.

As $R$ is Noetherian domain so the condition (1), (2) hold immediately.

If dim$R=1$ and because grade cannot exceeds dimension so every non zero prime ideal $P$ in $R$ has grade one then ht$P$=G$(P)$=1.

Now if $R$ is integrally closed then $R_P$ is DVR hence UFD then (3) holds.

At this point, I'm stuck because I have no idea to use condition (4).

I'm looking for concrete example such that (4) fails.

• The ring you are tailoring is called Dedekind domain: A Dedekind domain is an integrally closed Noetherian ring $R$ of dimension one. So basically your question becomes: "Does there exist a Dedekind domain which is not a UFD?'' the answer to which is YES. Example is $\mathbb{Z}[\sqrt{5}i]$. See example 13.6. of mathematik.uni-kl.de/~gathmann/class/commalg-2013/… – Hamed May 31 '17 at 23:59