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I certainly agree that if $R$ is an UFD then its ring of fractions $S^{-1}R$ is UFD as well. One may see the link.

What about the converse? It is stated in the link that 'The converse is also true for atomic domains'. So, it is clear that the converse of statement does not hold in general but I couldn't figure out example.

Fuat Ray
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  • $R = \prod_{i \in \mathbb{Z}} F_{2}$ is not UFD, but $R_{P} = F_{2}$ is UFD for all $P \in Spec(R)$. But for general localization, I'm still thinking. – jhzg Jan 07 '25 at 11:40
  • This is already an answer to the question. Notice that we fix $S$. If we quantify over $S$ it's trivial anyway (take $S=1$). – Martin Brandenburg Jan 07 '25 at 12:08
  • Could you please explain your example? I couldn't see why $R$ is not UFD, is there a quicker way to check? And lastly, what are the prime ideals of infinite product of rings that imply $R_P=F_2$? – Fuat Ray Jan 07 '25 at 12:51
  • The quotient field of any domain is a UFD. – Matthé van der Lee Jan 07 '25 at 13:05

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To try to clear up the loose ends in the post and in the comments:

Quantification of the statement

The post you are alluding to says that

If $S$ is a multiplicative set with certain properties in an atomic domain, we can say that if $S^{-1}R$ is a UFD, then so is $R$.

In other words it is not about "$S^{-1}R$ being a UFD for all $S$" implying $R$ is a UFD, which would certainly be true taking $S=\{1\}$ as Martin was saying in the comments.

The counterexample offered

The commutative rings known as von Neumann regular rings have the property that localizations at prime ideals are all fields. The given product ring is such a ring. The localizations of products of rings in general do not have this property. But obviously no product of two nonzero rings is a UFD (it cannot be even a domain.)

For the ring offered, though, it's easy to see the localizations at primes are all $F_2$: since $x^2=x$ in $R$, the same is true for $S^{-1}R$. Furthermore, $S^{-1}R$ is local. In a local ring, the only elements satisfying $x^2=x$ are $0$ and $1$. So that's all there is, $\{0,1\}$, the field of two elements.

I believe the ring was offered as an example of a ring such that

For every prime ideal $P$, taking $S=R\setminus P$ results in $S^{-1}R$ being a UFD, and yet $R$ is not a UFD. (It is not even a domain. In fact, all elements except the identity are zero divisors.)

which is not talking about the same quantification you seem to be interested in, but it interesting in its own right.

rschwieb
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