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Thank you very much!

My problem is:

If $R$ is a commutative ring with identity, and $a, b$ are its elements that are divisible by each other, is it true that they must be associates?

Here, $a$ being divisible by $b$ means there exists an $r \in R$, such that $a=rb$; and $a$ and $b$ being associates means there exists an invertible element $u \in R$ such that $a=ub$.

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

Many thanks!

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3 Answers 3

up vote 13 down vote accepted

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.

This is freely available.

It is mostly concerned with finding sufficient conditions on commutative rings that ensure that $Ra=Rb$ imply $a$ and $b$ associates, 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 $f$ and $g$ are non-associates in $R$ but each divides the other.

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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. –  ShinyaSakai Dec 15 '10 at 1:49
    
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. –  ShinyaSakai Dec 15 '10 at 2:16

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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Thanks. It is true that the counterexample can happen only when $a,b,1-rs$ are all zero divisors. –  ShinyaSakai Dec 15 '10 at 1:56

Related questions arise frequently on sci.math, where I often point out the following standard reference on the topic.

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.

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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Thank you very much. This paper is very helpful. –  ShinyaSakai Dec 15 '10 at 2:01

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