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Thank you very much! My problem is:
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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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 sufficent conditions on commutative ringsthat 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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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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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? |
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