# Does sufficient information on the multiplicative group of the fraction field of a GCD and Factorization domain, captures unique factorization?

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) ?

• I'd be curious to know where this question came from. – Santana Afton Oct 9 '18 at 22:34
• @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 ... – user521337 Oct 9 '18 at 22:51
• 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. – Hagen Knaf Oct 9 '18 at 23:13
• @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 ... – user521337 Oct 9 '18 at 23:48

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.
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.
• 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 $R$ ? – user521337 Oct 10 '18 at 14:18