The element that is an associate of everything. 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 $R$.
