In any ring $R$:product of non-unit and any element is non unit.

Question

My question may seem duplicate for :

Product of a non-unit with any other element

and

Can non-units be multiplied together to form units?

But these above don't give complete answer:

Because both of them , proved it in the case of commutative ring with unity , and for the case of non commutative ring with unity , they just gave example that there exists left inverse and right inverse.

Definition of unit: $a\in R$ is unit iff it has both left and right inverse.

So: $a\in R$ is non unit iff it doesn't have left or right inverse.

Now my approach to problem:

Let $a\in R$ be non unit and let $b \in R$

case $1$: $b$ is unit:

For the sake of contradiction assume $ab$ is unit.

then $\exists m \in R$ such that $(ab)m=1_{R}$ and $m(ab)=1_{R}$

which implies $a=(bm)^{-1}$ so $a$ is unit which is contradiction since $a$ is nonunit.

note:$(bm)^{-1}$ is unit because $b,m$ are units.

Case $2$: b is non unit:

For the sake of contradiction assume $ab$ is unit.

then $\exists m \in R$ such that $(ab)m=1_{R}$ and $m(ab)=1_{R}$

Now by associativity $a(bm)=1_{R}$ and $(ma)b=1_{R}$ which means $a$ has right inverse and b has left inverse.

But there is no contradiction until I also show $a$ has left inverse and $b$ has right inverse.

But I don't how to continue.

Can you , please , help me to continue solution?

Answer 1

Consider the abelian group $A=\Bbb Z^{\omega}$ of sequences of integers and let $R=\operatorname{End}(A)$ be its ring of endomorphisms.

For $f\in A$, define $D(f)$ per $$ D(f)(n)=f(n+1)$$ and $I(f)$ per $$ I(f)=\begin{cases}f(n-1)&n>0\\0&n=0\end{cases}$$ As $D$ is not injective and $I$ is not onto, $D$ cannot have a left inverse and $I$ cannot have a right inverse. So neither $D$ nor $I$ is a unit. But their product is a unit, in fact $D\circ I=\operatorname{id}_A$.