Equivalent Conditions About Specific Element of a Ring

Groups, Rings, Domains, Modules, etc, Galois theory
Post Reply
Tsakanikas Nickos
Community Team
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Equivalent Conditions About Specific Element of a Ring

#1

Post by Tsakanikas Nickos »

Let \( \displaystyle R \) be an associative ring with unitary element \( \displaystyle 1_{R} \). Let \( \displaystyle r \in R \) such that \( \displaystyle r \cdot s = 1_{R} \) for some \( \displaystyle s \in R \). Show that the following conditions are equivalent:


(1) There exists \( \displaystyle s^{\prime} \in R \; , \; s^{\prime} \neq s \) such that \( \displaystyle r \cdot s^{\prime} = 1_{R} \).

(2) \( \displaystyle r \notin U(R) \)

(3) \( \displaystyle r \) is a left zero divisor; that is, there exists \( \displaystyle y \in R \; , \; y \neq 0_{R} \) such that \( \displaystyle r \cdot y = 0_{R} \).
Tsakanikas Nickos
Community Team
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Re: Equivalent Conditions About Specific Element of a Ring

#2

Post by Tsakanikas Nickos »

(i) \( \implies \) (ii) :
Suppose that \( \displaystyle r \in U(R) \). Then
\[ \displaystyle \exists t \in R : r \cdot t = 1_{R} = t \cdot r \]
But
\begin{align*}
r \cdot s = 1_{R} &\implies t \cdot ( r \cdot s ) = t \cdot 1_{R} \\
&\implies ( t \cdot r ) \cdot s = t \\
&\implies 1_{R} \cdot s = t \\
&\implies s = t
\end{align*}
Due to (i),
\[ \exists s^{\prime} \in R , s^{\prime} \neq s : \; r \cdot s^{\prime} = 1_{R} \]
With a similar way to the previous one, one easily show that \( s^{\prime} = t = s \), which is a contradiction.

(ii) \( \implies \) (iii) :
Since, by (ii), \( \displaystyle r \) isn't invertible, and because holds \( \displaystyle r \cdot s = 1_{R} \) for some \( \displaystyle s \in R \), for all \( \displaystyle x \in R \) we have that \( \displaystyle x \cdot r \neq 1_{R} \). Therefore
\[ \displaystyle s \cdot r \neq 1_{R} \implies s \cdot r - 1_{R} \neq 0_{R} \]
Set \( \displaystyle y = s \cdot r - 1_{R} \) and observe that \( \displaystyle y \neq 0_{R} \) and, additionally,
\begin{align*}
\displaystyle
r \cdot y
&= r \cdot ( s \cdot r - 1_{R} ) \\
&= r \cdot ( s \cdot r ) - r \cdot 1_{R} \\
&= (r \cdot s ) \cdot r - r \\
&= 1_{R} \cdot r - r \\
&= r - r \\
&= 0_{R}
\end{align*}
This means that \( \displaystyle r \) is a left zero divisor.


(iii) \( \implies \) (i) :
Since \( \displaystyle r \) is a left zero divisor,
\[ \displaystyle \exists s^{\prime} \in R \, , \, s^{\prime} \neq 0_{R} \; : \; r \cdot s^{\prime} = 0_{R} \]
If \( \displaystyle s = s^{\prime} \), then \( \displaystyle 1_{R} = r \cdot s = r \cdot s^{\prime} = 0_{R}, \) which is a contradiction. Hence \( \displaystyle s \neq s^{\prime} \) and therefore \( \displaystyle s - s^{\prime} \neq 0_{R}. \) Set \( \displaystyle y = s - s^{\prime} \) and notice that \( \displaystyle y \neq 0_{R} \; , \; y \neq s \) and, additionally,
\begin{align*}
\displaystyle
r \cdot y
&= r \cdot ( s - s^{\prime} ) \\
&= r \cdot s - r \cdot s^{\prime} \\
&= 1_{R} - 0_{R} \\
&= 1_{R}
\end{align*}
Hence \( \displaystyle r \) has more than one right inverses.
Post Reply

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

Register

Sign in

Who is online

Users browsing this forum: No registered users and 12 guests