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.

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