TITLE: Is there a ring so that any two distinct non-zero elements do not commute?
QUESTION [9 upvotes]: I was wondering if there was a ring so that any two distinct non-zero elements do not commute.
Formally, is there a ring $R\not=\{0\}$ so that
$$\forall x,y\in R\setminus\{0\}, x\not= y\implies xy\not=yx$$

Obviously, $\{0\}$ and $\Bbb Z/2\Bbb Z$ work but I'm searching for a non-trivial example (or a proof that there is none).

If $R$ has a $1$, then if we have $x\in R$ so that $x\not=0$ and $x\not=1$ (that is, $R\not=\Bbb Z/2\Bbb Z$), $x$ and $1$ commute.
$$\boxed{R\text{ doesn't have }1}$$

Let $x\in R$
$(x+x)x=xx+xx=x(x+x)$
We have $x=0$ or $x+x=0$ or $x+x=x$
So $x+x=0$
$$\boxed{R\text{ has characteristic }2}$$

Suppose we have a nilpotent element $x\in R$
$\exists n\in\Bbb N^*,x^{n}=0$
If $n=1$, $x=0$
If $n=2$, $x^2=0$
If $n>2$, $x^{n-1}\not=x$ but they commute
$$\boxed{\text{The square of a nilpotent element is }0}$$

(Given by Arthur in the comments)
Let $x\in R$
$x$ and $x^2$ commute so $x=0$ or $x^2=0$ or $x=x^2$
$$\boxed{\text{Every element is either nilpotent or idempotent.}}$$

REPLY [2 votes]: Here is a slightly more general solution, along the same lines.
Consider an abelian (additive) group 
$$
G =\oplus_{\alpha\in A} {\mathbb Z}/2 {\mathbb Z}, 
$$
where $A$ is any set of cardinality $\ge 2$. The useful property that $G$ satisfies is that $x+x=0$ for every $x\in G$. Define the following product operation on $G$: For all $x, y\in G$, 
$$
xy=y
$$
unless $x=0$, in which case, of course, $xy=0$. It is then straightforward to verify that 
$G$ with these two binary operations is a ring where $xy=yx$ if and only if either $x=y$, or $x=0$, or $y=0$.