Suppose I have an integral domain $R$ containing an element $a \in R$ with the following property:

$$(\forall r \in R)\, a \text{ is an associate of } r.$$

Is it true that the ring must be either the zero ring or the ring $\{0,1\}$?

"Associates" are defined as follows:

For $a,b \in R$, $a$ and $b$ are associates if $a \vert b$ and $b \vert a$.

I used the equivalence relation:

$$a \thicksim b \stackrel{def}\iff \text{$a$ and $b$ are associates}.$$

This leaves us with $( R\big/\!\sim) = \{ 0 \}$, but that's not the same as saying $R$ is the zero ring... is it?


$R$ must be the zero ring. Here's a proof.

By definition, if $a$ and $b$ are associates then each is a unit multiple of the other. In particular for this special $a$ we have that $b$ can be anything. Take $b=0$. Then $a=cb=0$ for some unit $c$, hence $a=0$. This uses the fact that $a$ is a multiple of $b$.

To finish the proof that $R$ is the zero ring, note that no matter what $b$ we choose, it must be a multiple of $a$. But $a=0$, hence $b=0$, so $0$ is the only element of $R$.

  • $\begingroup$ Okay, then what if we change the property to $\forall$ non-zero $r \in R$, then $a$ is associate to $r$? Do we still get the zero ring? $\endgroup$ – csasba Nov 11 '15 at 5:41
  • 3
    $\begingroup$ @csasba2 then we can conclude that $R$ is a field if it has a unit. $\endgroup$ – Matt Samuel Nov 11 '15 at 5:43
  • $\begingroup$ I believe that we do not need to use $1$ or units in a proof. We use the definition of associates by divisibility as in the question. $0$ and $a$ are associates. For every $b$, $b$ and $0$ are associates, $0\mid b$, $b=0$. $\endgroup$ – beroal Mar 16 '18 at 16:54

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