# PIDs and associator rings

Let $R$ be a principal ideal domain and let $p \in R$ be prime. Show that the quotient $R/\langle p^k\rangle$ is an associator ring for any positive integer $k$ - that is, if $a, b \in R/\langle p^k\rangle$ generate the same ideal, then there exists a unit $u$ in $R/\langle p^k\rangle$ with $a=ub$ in this ring.

• So "associator ring" means every two generators of a given principal ideal are unit multiples of one another? (I only got 5 hits in google, and I couldn't get to a definition in any of them.) – rschwieb May 7 '18 at 20:11
• In my experience, they are usually called strongly associate rings. This gives some background: journals.math.tku.edu.tw/index.php/TKJM/article/viewFile/464/… – CPM May 19 '18 at 14:51

This property that $(a)=(b)$ implies $a=\lambda b$ for some unit $\lambda \in U(R)$ is called a strongly associate ring usually in my experience. Any ring which is presimplifiable has this property. I believe this kind of ring is called a Special Principal Ideal Ring. This kind of thing is studied here in a paper by Anderson, Axtell, Forman, and Stickles called When are associates unit multiples?: https://www.researchgate.net/publication/38371302_When_are_Associates_Unit_Multiples