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

Question

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) ?

Answer 1

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.