Let $R$ be a ring with identity and $a,b \in R$ then prove that $1+ab$ is a unit if and only if $1+ba$ is a unit and find the inverse.
Then there exist an element say $s \in R$ such that $(1+ab)s =1$ then from here how will I approach to find the inverse of $1+ba$??
Since $1+ab$ is a unit in $R$, there exists an inverse $s$ such that
$$ (1+ab)s = s(1+ab) = 1. $$
Now let $t = 1 - bsa$. Then
\begin{align*} t(1+ba) &= 1 + ba - bsa - bsaba \\ &= 1 + ba - bs(1+ab)a \\ &= 1 + ba - ba \\ &= 1 \end{align*}
and likewise for $(1+ba)t$. So $1+ba$ is also a unit with the inverse $1-bsa$.
My intuition. Formally and heuristically, the inverse $s$ of $1+ab$ is
$$ s = (1+ab)^{-1} = 1 - ab + (ab)^2 - (ab)^3 + \cdots. $$
Likewise, if $1+ba$ has an inverse, it will have a formal expansion
\begin{align*} (1+ba)^{-1} &= 1 - ba + (ba)^2 - (ba)^3 + \cdots \\ &= 1 - b(1 - ab + (ab)^2 - \cdots)a \\ &= 1 - bsa. \end{align*}
Then we can begin the actual proof with this guess.
