description/proof of that for ring, if element has inverse and element multiplied by another element from left or right is \(1\), another element is inverse
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of ring.
- The reader admits the proposition that for any ring, if an element has an inverse, the inverse is unique.
Target Context
- The reader will have a description and a proof of the proposition that for any ring, if any element has an inverse and the element multiplied by any another element from left or right is \(1\), the another element is the inverse.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(r\): \(\in R\)
//
Statements:
(
\(\exists r^{- 1} \in R \land \exists r' \in R (r' r = 1)\)
\(\implies\)
\(r' = r^{- 1}\)
)
\(\land\)
(
\(\exists r^{- 1} \in R \land \exists r' \in R (r r' = 1)\)
\(\implies\)
\(r' = r^{- 1}\)
)
//
2: Note
Presupposing that \(r\) has an inverse is crucial for this proposition: any only \(1\) of \(r' r = 1\) and \(r r' = 1\) does not guarantee \(r\) having an inverse.
A special case is that if any ring square matrix, \(M\), is invertible and \(M' M = I\) or \(M M' = I\), \(M' = M^{- 1}\), because the \(n \times n\) ring matrices ring is a ring and the inverse of \(M\) is nothing but the inverse element of \(M\) in the matrices ring.
3: Proof
Whole Strategy: Step 0: see that \(r^{- 1}\) is unique; Step 1: suppose that \(r^{- 1}\) exists and \(r' r = 1\); Step 2: see that \(r' r r^{- 1} = 1 r^{- 1}\); Step 3: suppose that \(r^{- 1}\) exists and \(r r' = 1\); Step 4: see that \(r^{- 1} r r' = r^{- 1} 1\).
Step 0:
When \(r\) has an inverse, the inverse is the unique \(r^{- 1}\), by the proposition that for any ring, if an element has an inverse, the inverse is unique.
Step 1:
Let us suppose that \(r^{- 1}\) exists and \(r' r = 1\).
Step 2:
\(r' r r^{- 1} = 1 r^{- 1}\), but the left hand side is \(r' (r r^{- 1}) = r' 1 = r'\) and the right hand side is \(r^{- 1}\).
So, \(r' = r^{- 1}\).
Step 3:
Let us suppose that \(r^{- 1}\) exists and \(r r' = 1\).
Step 4:
\(r^{- 1} r r' = r^{- 1} 1\), but the left hand side is \((r^{- 1} r) r' = 1 r' = r'\) and the right hand side is \(r^{- 1}\).
\(r' = r^{- 1}\).
