Show that the group $R^\times$ acts transitively on the set $\{s\in R| (s) = (r)\}$ for a fixed r.

Question

Maybe I'm having some issue with notation or something, but this is for an Integral Domain R. Fix some $r\in R$. Then we want to show the group $R^\times$ acts transitively on the set $\{s\in R | (s)=(r)\}$. Where (s) is the ideal generated by s.

Really not sure what $R^\times$ means...

Answer 1

For any (commutative) ring $R$, $R^\times$ denotes the group of units of $R$ ie. the group of all invertible elements $$R^\times = \{r \in R \mid \exists s\in R : rs=sr=1\}$$ with group operation given by the multiplication of the ring. This group is fundamental to the theory of rings. Just for convenience let me remark that elements $x,y\in R$ are called associate if there is a unit $u\in R^\times$ with $x=uy$ or equivalently $y=ux$.

The exercise reduces to the following two observations:

1. associate elements generate the same principle ideal, so for fixed $r\in R$ there is a welldefined group action $R^\times \curvearrowright \{s \in R\mid (s)=(r)\}$
2. in an integral domain $R$, elements, which generate the same principal ideal, are associate. Thus $R^\times \curvearrowright \{s \in R\mid (s)=(r)\}$ is transitive.

The first observation is immediate from plugging in the definition of being associate. The second observation requires the cancellation rule of integral domains.

If you have questions, feel free to ask andI will elaborate. For now I think it is good to leave the details as an exercise.