# Zero divisors vs right and left cancellations

Let $R$ be a ring (not necessarily commutative) with $1$.

Left zero divisor

$a\in R$ is said to be a left zero divisor (LZD) if there is a $b\neq 0$ in $R$ such that $ab=0$.

Right zero divisor

$a\in R$ is said to be a right zero divisor (RZD) if there is a $b\neq 0$ in $R$ such that $ba=0$.

Zero divisor

$a\in R$ is said to be a zero divisor (ZD) if it is both a left and a right zero divisor.

Left cancellation :     $a\neq 0, ab=ac\implies b=c$.

Right cancellation :     $a\neq 0, ba=ca\implies b=c$.

Prove that R has no non-trivial zero divisors if and only if both right and left cancellations hold.

Proof : Right cancellation holds will imply there are no nontrivial right zero divisors ($ba=0\implies ba=0a \implies b=0.$). Similarly left cancellation holds will imply no nontrivial left zero divisors. So clearly no element can be a zero divisor except 0 (for it can’t even be a LZD or an RZD)

The other direction is more complicated.

1. Assume $x$ is a nontrivial LZD and not a ZD (so x is not 0).
2. So there is a $y\neq 0$ such that $xy=0$.
3. If $yx=0$ also, then $x$ is an RZD and hence a ZD, a contradiction
4. So $yx\neq 0$.
5. $(yx)y=yxy=y(xy)=0$ and $y\neq 0$,  so yx is an LZD
6. $(x)(yx)=(xy)x=0$ and $x\neq 0$, so yx is an RZD
7. So $yx$ is a nontrivial ($yx\neq 0$ by Step 4) ZD.

So we have shown that $x,y\neq 0, xy=0, x\neq ZD\implies yx$ is a nontrivial ZD.

Let us show left cancellation holds assuming no nontrivial ZD in R. If $a\neq 0, ab=ac$, this implies $a(b-c)=0$. If $b-c=0$, we are done, else put $x=a, y=b-c$. Thus $yx=(b-c)a$ is a nontrivial ZD, a contradiction.

Let us show right cancellation holds assuming no nontrivial ZD in R. If $a\neq 0, ba=ca$, this implies $(b-c)a=0$. If $b-c=0$, we are done, else put $y=a, x=b-c$ and they are both not ZD as R has no nontrivial ZD. Thus $yx=a(b-c)$ is a nontrivial ZD, a contradiction.