## An elementary result on noncommutative rings.

Let R be a (noncommutative) ring with identity, such that there are elements u and v with u v = 1.  Then the following are equivalent:

1. u w = 0 for some nonzero w.
2. u is not a unit.
3. u has other right inverses.

Proof. It’s easy to see that 1 and 3 are equivalent — if v is an inverse, then so is v + w, and vice-versa — and clearly they imply condition 2.  Showing that 2 implies the others isn’t as obvious, but it’s a nice one-liner formal trick.

In particular, note that u (1 – v u) = u – u v u = u – (u v) u = u – u = 0, so either v u = 1 and hence u is a unit, or w = 1 – vu is a nonzero element of R such that u w = 0.

In fact, by a theorem due to Kaplansky there are infinitely many such inverses if there are two, which we obtain simply by taking $w_n = (1 - v u) u^n$.  To see these are distinct, note that if

$(1 - v u) u^n = (1 - v u) u^m$

for some n < m, then multiplying through by $v^n$ on the right gives

$1 - vu = (1 - vu) u^{m-n},$

so

$1 = \left[v + (1-vu) u^{m-n-1}\right] u,$

i.e. u is actually a unit, contradicting the hypothesis.