In a commutative principal ideal ring does $(a)=(b)\Rightarrow a = bu$ for a unit $u$, i.e. are associates unit multiples?

Question

Let $R$ be a commutative principal ideal ring with identity. Suppose that $a,b\in R$ and $(a)=(b)$. I'd like to know if there always exists $u\in R^\times$ such that $a=bu$.

I know several counterexamples if $R$ is not a principal ideal ring, so I'm a bit curious about whether it holds for principal ideal rings.

Usually people talk about $PID$, principal ideal domains. These are integral domains, so from $a=xb$ and $b=ya$ you can establish that $a(1-xy)=0$ so that indeed $x,y$ are units. Is there any reason to study non-domains? — Blah, Mar 30 '15 at 07:49
Yes, associates are unit multiples in principal ideal rings - see the paper we cite in this 2013 answer. $\ \ $ — Bill Dubuque, Mar 13 '25 at 12:54

score 4 · Accepted Answer · answered Mar 30 '15 at 10:21

4

A PIR is a direct product of PIDs and local PIRs. (This is well known and can be found in Zariski and Samuel, Commutative Algebra, or in Hungerford's paper "On the structure of principal ideal rings"). Now you can draw the conclusion.

answered Mar 30 '15 at 10:21

user26857

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In a commutative principal ideal ring does $(a)=(b)\Rightarrow a = bu$ for a unit $u$, i.e. are associates unit multiples?

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