4

Let $x$ be an element of a ring and $d$ a divisor of $x$. Can we have $x \mid d$?

There's the trivial case where both $x$ and $d$ are units. Otherwise, we have $x=da$ and $d=xb$, thus $x=xba$, so the question reduces to: can an element divide itself?

kahen
  • 15,760
Jack M
  • 27,819
  • 7
  • 63
  • 129
  • 1
    So the trivial case is when $x$ and $d$ are associates, i.e. they differ by a unit: $x=da$ where $a$ is unit. Furthermore, if the ring is an integral domain, then this is a necessary condition. As you say, if $x=da$ and $d=xb$ gives $x=xba$ and in the case of integral domains, we get $1=ba$ and so $a$ is a unit. – Prism Apr 26 '14 at 19:02
  • 2
    Take a look at this thread. – Prism Apr 26 '14 at 19:11
  • @Prism So in the integral domain case, we could continue $x(1 - ba) = 0\implies ba=1$, and so either $b$ or $a$ is a unit? – Jack M Apr 26 '14 at 19:23
  • Actually, in an integral domain, the condition $ba=1$ implies that both $b$ and $a$ are units. – Prism Apr 26 '14 at 19:25
  • @prism think about composing an answer! – rschwieb Apr 26 '14 at 23:03
  • @rschwieb: I tried… :D – Prism Apr 26 '14 at 23:37

1 Answers1

2

This question has been previously considered in this thread. Here's a recap:

Robin Chapman gives a reference to the freely-available paper When are Associates Unit Multiples? by Anderson, Axtell, Forman and Stickles.

In the paper, the following notation is used: Let $R$ be a commutative ring. If two elements $a, b \in R$ satisfy $a\mid b$ and $b\mid a$, then they are called associates and we write $a\sim b$. If two elements $a, b\in R$ satisfy $a=u b$ for some unit $u$, then they are called strongly associates and we write $a\approx b$.

OP's question seems to be: Which elements $a, b\in R$ in the ring satisfy $a\sim b$?

It is clear that if $a\approx b$, then $a\sim b$ (see comments above). The paper above by Anderson et al. aims to describe (among other things) the commutative rings $R$ in which $a\sim b$ implies $a\approx b$ for all $a, b \in R$. If $R$ is an integral domain, then this is clearly true. (See Timothy Wagner's answer).

Here is a more general theorem due to Kaplansky:

If $R$ is principal ideal ring, or Artinian ring, or a ring satisfying $Z(R )\subset J(R )$ (here $Z(R )$ is the centre of $R$, and $J(R )$ is Jacobson radical of $R$), then $a\sim b$ implies $a\approx b$ for all $a, b \in R$.

Brian Tung
  • 34,160
Prism
  • 11,162