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) ?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.