Question

Show that, if R has an identity 1, the map ϕ : (R, ◦) → (R,×) sending a to 1 − a is a monoid isomorphism. In this case, an element a is left (right) quasi-regular iff 1 − a has a left (resp. right) inverse with respect to ring multiplication.

Answer 1

Question 1. Show that, if $R$ has an identity 1, the map $\phi:(R,\circ)\to(R,\cdot)$ sending $a$ to $1-a$ is a monoid isomorphism. In this case, an element $a$ is left (right) quasi-regular iff $1-a$ has a left (resp. right) inverse with respect to ring multiplication.

Solution. Indeed,

$\phi(a\circ b)=1-a\circ b=1-a-b+ab=(1-a)(1-b)=\phi(a)\phi(b),$

so, $\phi$ is a homomorphism of semigroups. Moreover, $\phi(0)=1$, so $\phi$ is a homomorphism of monoids ( is the identity of $(R,\circ)$). The bijectivity of $\phi$ is obvious: $1-a=1-b\Leftrightarrow a=b$, $a=1-(1-a)$. Thus, $\phi$ is an isomorphism. Note that

$a\circ b=0\Leftrightarrow 1=\phi(a\circ b)=\phi(a)\phi(b)=(1-a)(1-b).$

So, $a$ is left (right) quasi-regular iff $1-a$ is left (resp. right) invertible. $\square$