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Let \(\displaystyle{R}\) an associative ring ,where \(\displaystyle{1}\) is the unit. Prove that if \(\displaystyle{1-xy}\) has a multiplicative inverse then \(\displaystyle{1-yx}\) has a multiplicative inverse, where \(\displaystyle{x,y\in R}\).

Let \(\displaystyle{ 1-xy \in U(R)\Rightarrow \exists r \in U(R) : (1-xy)r=1=r(1-xy) ( \star\star )}\) .Relation \(\displaystyle{ (\star\star)}\) gives \(\displaystyle{ xyr=rxy (\star)}\).

\(\displaystyle { (1-xy)r = 1 \Rightarrow y(1-xy)rx=yx.}\) .Observing that \(\displaystyle{ (1-yx)+yx = 1}\) and using the previous relation, we get

\(\displaystyle{ 1-yx+y(1-xy)rx=1 \Rightarrow 1-yx + y(rx-xyrx)=1 (\star)\Rightarrow 1-yx + y(rx(1-yx))=1 \Rightarrow (1+yrx)(1-yx)=1} (1) \).

To show that \(\displaystyle { (1-yx)(1+yrx)=1 } \), we use \( \displaystyle (1) \) : \(\displaystyle{ 1-yx+yrx-yrxyx=1 (\star)\Rightarrow 1-yx +yrx -yxyrx =1 }\)

\(\displaystyle{ \Rightarrow (1-yx) + ( (1-yx)y)rx =1 \Rightarrow (1-yx)(1+yrx)=1 }\).So, \(\displaystyle{ 1+yrx }\) is both right and left multiplicative inverse of

\(\displaystyle {1-yx \Rightarrow 1+yrx = (1-yx)^{-1} }\)

A slightly more general result is the following:

\( \displaystyle 1_{R} - xy \) is left-invertible (resp. invertible) if and only if \( \displaystyle 1_{R} - yx \) is left-invertible (resp. invertible)