Let \(R\) be a (possibly non-commutative) ring with identity, and let \(u\) be an element of \(R\) with a right inverse. Prove that the following conditions on \(u\) are equivalent:
\(u\) has more than one right inverse;
\(u\) is a zero divisor;
\(u\) is not a unit.
Solution
Fix a right inverse \(v\) of \(u\), so \[ uv=1. \]
1. If \(wu=0\), then \(w=0\).
Using \(uv=1\), \[ w=w(uv)=(wu)v=0. \] Thus \(u\) has no nonzero left annihilator. In particular, under either standard convention for zero divisors in a noncommutative ring, condition 2 here is equivalent to the existence of some nonzero \(w\) with \(uw=0\).
2. Condition 1 implies condition 2.
Suppose \(v'\) is a second right inverse, with \(v'\ne v\). Then \[ u(v-v') =uv-uv' =1-1 =0, \] while \(v-v'\ne0\). Hence \(u\) is a zero divisor.
3. Condition 2 implies condition 3.
If \(u\) were a unit and \(uw=0\), then multiplying on the left by \(u^{-1}\) would give \(w=0\). Together with step 1, this shows that a unit cannot be a zero divisor. Therefore condition 2 forces \(u\) not to be a unit.
4. Condition 3 implies condition 1.
Assume \(u\) is not a unit. Since \(uv=1\), one must have \[ vu\ne1; \] otherwise \(v\) would be a two-sided inverse of \(u\). Set \[ w\coloneqq vu-1. \] Then \(w\ne0\), and \[ uw =uvu-u =(uv)u-u =0. \] Consequently \[ v'\coloneqq v+w \] is distinct from \(v\) and satisfies \[ uv' =uv+uw =1. \] Thus \(u\) has more than one right inverse.
5. The three conditions are equivalent.
Steps 2, 3, and 4 give the cycle \[ (1)\Longrightarrow(2)\Longrightarrow(3)\Longrightarrow(1). \]
6. Q.E.D.
Step 5 is the required equivalence.