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Recall the following relevant definitions. We say that

  • $b$ is divisible by $a$ in $R$, or $a\mid b$ in $R$, if $b = r a$ for some $r\in R$.

  • $a$ and $b$ are associates in $R$ if $a\mid b$ and $b\mid a$, (or, equivalently, if $aR = bR$).

  • $u\in R$ is a unit if it has a multiplicative inverse (a $v\in R$ such that $uv=vu=1$).

  • $a$ and $b$ are unit multiples in $R$ if $a = ub$ for some unit $u\in R$.

Given these definitions, my question is,

If $R$ is a commutative ring with unity and $a,b\in R$ are associates in $R$, are $a$ and $b$ unit multiples in $R$?

I was told that this not always true. But I encountered some difficulties in finding a counterexample.

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See the following paper,

When are Associates Unit Multiples?
D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.

It is mostly concerned with finding sufficient conditions on commutative rings that ensure that $Ra=Rb$ implies $a = bu$ for a unit $u$, but they do give some examples of $R$ where this fails. In particular this simple example of Kaplansky. Let $R=C[0,3]$, the set of continuous function from the interval $[0,3]$ to the reals. Let $f(t)$ and $g(t)$ equal $1-t$ on $[0,1]$, zero on $[1,2]$ but let $f(t)=t-2$ on $[2,3]$ and $g(t)=2-t$ on $[2,3]$. Then $f$ is not a unit multiple of $g$ in $R$ but each divides the other.

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  • $\begingroup$ Thank you very much for answering my question, but I think I am not quite clear about this answer. Here, if I want to find some $t(x)$ such that $f(x)=t(x)g(x)$, I should set $t(x)=1$ in the interval $[0,2]$, and $t(x)=-1$ in the interval $[2,3]$, but $t(2)=$?? Even if I determine the value of $t(2)$, $t(x)$ is not a continuous function, thus it does not belong to $R=C[0,3]$. Many thanks. $\endgroup$ – ShinyaSakai Dec 15 '10 at 1:49
  • $\begingroup$ I read the paper and find that $t(x)$ can be set to $1$ on $[0,1]$, $3-2t$ on $[1,2]$ and $-1$ on $[2,3]$. Now I understand. Many thanks. $\endgroup$ – ShinyaSakai Dec 15 '10 at 2:16
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This is true if $R$ is a domain. More generally this is true if $a$ or $b$ is not a zero divisor in $R$. Suppose, $a$ is not a zero divisor and there exist $r,s\in R$ such that $a=rb$ and $b=sa$, then $a=rsa$, so $a(1-rs)=0$. Since $a$ is not a zero divisor, $1-rs=0$, so $rs=1$. So, $r,s$ are units. So $a$ and $b$ are associates.

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  • $\begingroup$ Thanks. It is true that the counterexample can happen only when $a,b,1-rs$ are all zero divisors. $\endgroup$ – ShinyaSakai Dec 15 '10 at 1:56
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On earlier math forums I often cited the little-known article below on this topic (e,g. see sci.math Oct 15, 2008 google groups or mathforum, and Ask an Algebraist 2008, etc)

Beware that this equivalence, i.e. $\rm\ aR = bR \iff a/b\ $ is a unit in $\rm R$, generally fails when $\rm R $ has zero-divisors, so that there are at least a few different notions of "associate" that are of interest, e.g.

  • $\ a\sim b\ $ are $ $ associates $ $ if $\, a\mid b\,$ and $\,b\mid a$
  • $\ a\approx b\ $ are $ $ strong associates $ $ if $\, a = ub\,$ for some unit $\,u\ \,$ (a.k.a. unit multiples)
  • $\ a \cong b\ $ are $ $ very strong associates $ $ if $\,a\sim b\,$ and $\,a\ne 0,\ a = rb\,\Rightarrow\, r\,$ unit

See said paper below for much further discussion. See also the survey linked here for the effect that this bifurcation has on the notion of unique factorization ring and related matters.

When are Associates Unit Multiples?
D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.
http://math.la.asu.edu/~rmmc/rmj/vol34-3/andepage1.pdf
http://projecteuclid.org/handle/euclid.rmjm/1181069828

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