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Elements $a$ and $b$ of an integral domain are associates if $a\mathrel{\vdots}b$ and $b\mathrel{\vdots}a$

I have proved this fact. Then I tried to find out $a$ and $b$ which divide each other and does not associate. This may happen only in non-integral domains of course. But I couldn't find an example. Can someone suggest one?

Definition of associates.

$a$ and $b$ are associates if $a=b\epsilon$ where $\epsilon$ is invertible element of the ring.

$a\mathrel{\vdots}b$ means $a$ is divided by $b$. ($a=bc$). It is the same as $b|a$.

Ashot
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  • What is your definition of associates, if not in terms of each a divisor of the other? Also, what is the definition of a (vert dots) b or is that same as a | b? – coffeemath Dec 01 '18 at 09:03
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    @coffeemath I updated the question. – Ashot Dec 01 '18 at 09:06
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    Just deleted my "answer" and now I think the question is interesting! [already upvoted] – coffeemath Dec 01 '18 at 09:24
  • Just to be clear, do you need a commutative ring in an example? And I assume you require a ring having a unity (a 1 for multiplication a(1)=(1)a=a all a) so that a unit would be a divisor of 1? If not how is a unit defined? – coffeemath Dec 01 '18 at 09:54
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    I think if we found such elements in any ideal it would be counterexample of the theorem, but it would be nice to do it in a commutative ring with identity. – Ashot Dec 01 '18 at 10:23

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Here is an example due to Kaplansky.

Let $R=C([0,3])$ be the ring of continuous funtions $f:[0,3]\rightarrow\mathbb{R}$. I claim that $$R^{\times}=\left\lbrace f(t)\in R \hspace{2.5mm} | \hspace{2.5mm} f(t)\neq 0 \hspace{2.5mm} \forall t\in[0,3]\right\rbrace.$$ Indeed, if $f(t)g(t)=1$ for all $t\in[0,3]$ then $f(t)\neq 0$ because if there exists $s\in [0,3]$ such that $f(s)=0$ then $1=f(s)g(s)=0\cdot g(s)=0,$ a contradiction. Conversely, if $f(t)\neq 0$ then $\frac{1}{f(t)}\neq 0$ is a continuous function.

Now define the three elements $$a(t)=\begin{cases}1-t&t\in[0,1]\\0&t\in[1,2]\\t-2&t\in[2,3]\end{cases},$$ $$b(t)=\begin{cases}1-t&t\in[0,1]\\0&t\in[1,2]\\2-t&t\in[2,3]\end{cases},$$ and $$c(t)=\begin{cases}1&t\in[0,1]\\3-2t&t\in[1,2]\\-1&t\in[2,3]\end{cases}.$$ Then $a(t),b(t),c(t)\in R$ and it's clear that $c(t)a(t)=b(t)$ and $c(t)b(t)=a(t)$, so $a(t)\mid b(t)$ and $b(t)\mid a(t)$.

Let $u(t)\in R^{\times}$ be given. If $a(t)=b(t)u(t)$ then $$a(t)=\begin{cases}1-t&t\in[0,1]\\0&t\in[1,2]\\t-2&t\in[2,3]\end{cases}=\begin{cases}(1-t)u(t)&t\in[0,1]\\0&t\in[1,2]\\(2-t)u(t)&t\in[2,3]\end{cases}=b(t)u(t),$$ which implies that $$u(t)=\begin{cases}1&t\in[0,1]\\\star&t\in[1,2]\\-1&t\in[2,3]\end{cases}.$$ But due to the Intermediate Value Theorem we see that there exists $s\in[0,3]$ such that $u(s)=0$ because it attains all the values between $-1$ and $1$. Thus we conclude that $u(t)$ is not a unit and therefore we see that $a(t)$ and $b(t)$ are not associates.

YumekuiMath
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