Several trick elementary conclusions

In a Monoid, if an element is left and right invertible, then it is invertible.

Proof: By hypothesis we have
$$ a b = ca = 1 $$
hence we have
$$ b = cab = c$$
therefore $a$ is invertible.

Definition: In a Rng $A$, $z$ is called left(right) quasi-regular if there exists an element $z’$ such that $z’o z \equiv z’+z-z’z = 0\;(z o z’=0 )$.

If A is a Monoid, actually, $z$ is called left(right) quasi-regular if and only if $1-z$ is left invertible.

In a Rng, if $z’oz = 0$ and $z’’oz’ = 0$ then $z’’=z$, thus both $z$ and $z’$ are quasi-regular.

Proof: It is obvious right in a Monoid. Some trick occur while proving it.
$$ z’’ = z’’z’-z’ = z’’(z’z-z)-z’ = (z’’z’-z’’)z-z’=z’z-z = z $$
end the proof.

In a Ring,$1-az$ is invertible if and only if $1-za$ is invertible.

Proof: In fact, we have
$$ (1-az)^{-1} = 1+a(1-za)^{-1}z$$

In a Ring $M$, $z \in M$. If $\forall a \in M,\; 1-az$ is left invertible, then $1-az$ is invertible,so is $1-za$ why foregoing conclusion.

Proof: $\forall a \in M$,$1-az$ is left invertible.thus there exists $1-y$ such that
$$ (1-y)(1-az) = 1 $$
hence $y = (ya-a)z$ is left invertible.
Hence by forgoing conclusion $1-az$ is invertible,and so is $1-za$.

Note that:

A Monoid is a a set which is closed under an associative multiplication and contains an identity element.