Invertible elements of a ring
Invertible elements of a ring
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 \(\displaystyle{x,y\in R}\).
Re: Invertible elements of a ring
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
\(\displaystyle {1-yx \Rightarrow 1+yrx = (1-yx)^{-1} }\)
\(\displaystyle { (1-xy)r = 1 \Rightarrow y(1-xy)rx=yx.}\) .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
\(\displaystyle {1-yx \Rightarrow 1+yrx = (1-yx)^{-1} }\)
-
- Community Team
- Posts: 314
- Joined: Tue Nov 10, 2015 8:25 pm
Re: Invertible elements of a ring
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)
\( \displaystyle 1_{R} - xy \) is left-invertible (resp. invertible) if and only if \( \displaystyle 1_{R} - yx \) is left-invertible (resp. invertible)
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 19 guests