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Let $R$ be (non-field) an Atomic (https://en.wikipedia.org/wiki/Atomic_domain) and a GCD domain (https://en.wikipedia.org/wiki/GCD_domain) of characteristic zero. Let $U(R)$ denote the multiplicative group of units in $R$. Let $G$ be the multiplicative group of the fraction field of $R$. If $G/U(R)$ is a free abelian group, then is it true that $R$ is a UFD (unique factorization domain) ?

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  • $\begingroup$ I'd be curious to know where this question came from. $\endgroup$ – Santana Afton Oct 9 '18 at 22:34
  • $\begingroup$ @SantanaAfton: really I just came up with it after noticing that the converse obviously holds i.e. if $R$ is a UFD of characteristic zero then the multiplicative group of the fraction field has all those properties ... now I asked myself whether the converse is true or not and I don't know any counterexamples even if I drop all those GCD domin and atomic domain assumptions ... but I wanted to play safe, so I added those extra assumptions in ... $\endgroup$ – user521337 Oct 9 '18 at 22:51
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    $\begingroup$ The ring $\mathbb{Z}[i]$ is a UFD, but $\mathbb{Q}(i)^\ast/\{-1,1\}$ ist not free abelian, since it $i\{-1,1\}$ is a torsion element in that group. The correct version is, that the multiplicative group modulo the units of the ring form a free abelian group. $\endgroup$ – Hagen Knaf Oct 9 '18 at 23:13
  • $\begingroup$ @HagenKnaf: thanks .. I was some how under the false impression that when $-1,1$ are the only torsion elements, then $U(R)=\{-1,1\}$ ... that is clearly wrong ... $\endgroup$ – user521337 Oct 9 '18 at 23:48
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Let $B$ be a basis of $G/U(R)$ and choose $P\subset R$ such that $P\rightarrow B, p\mapsto pU(R)$ is bijective. Then every $r\in R$ can be written as

$r=u\cdot p_1^{e_1}\cdot\ldots\cdot p_n^{p_n}$, $u\in U(R)$, $p_k\in P$ pairwise distinct.

This factorization is unique.

The elements $p\in P$ are prime: suppose $p$ divides a product $rs$ of elements of $R$. Then $rs=pq$ for some $q\in R$. Since one can obtain the factorization of $rs$ by substituting $q$ in this equation by its factorization, one sees that $p$ appears in the factorization of $rs$. On the other hand one can obtain the factorization through combining the factorizations of $r$ and $s$, which shows that $p$ must appear either in the factorization of $r$ or of $s$. Consequently $p$ divides $r$ or $s$.

Altogether this shows that $R$ is a UFD.

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  • $\begingroup$ how do you know we can choose elements from $R$ whose reduction image gives all the elements of $B$ ... ? Definitely we can choose elements from $Frac(R)$, but why $R$ ? $\endgroup$ – user521337 Oct 10 '18 at 14:18
  • $\begingroup$ hello ... are you there ...? $\endgroup$ – user521337 Oct 13 '18 at 16:01

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