description/proof of that for integral domain and nonzero element, multiplication map by element is injection
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of integral domain.
- The reader knows a definition of injection.
Target Context
- The reader will have a description and a proof of the proposition that for any integral domain and its any nonzero element, the multiplication map by the element is an injection.
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:
\(R\): \(\in \{\text{ the integral domains }\}\)
\(r\): \(\in R \setminus \{0\}\)
\(f_r\): \(: R \to R, r' \mapsto r r'\)
//
Statements:
\(f_r \in \{\text{ the injections }\}\)
//
2: Note
Replacing \(r r'\) with \(r' r\) does not make any difference, because any integral domain is commutative.
\(f_r\) is not any ring homomorphism unless \(r = 1\): \(f_r (1) = r 1 = r \neq 1\).
When \(r = 0\), \(f_r (r') = 0 r' = 0\), which is of course non-injective.
3: Proof
Whole Strategy: Step 1: let \(r', r'' \in R\) such that \(r' \neq r''\), suppose that \(f_r (r') = f_r (r'')\), and find a contradiction.
Step 1:
Let \(r', r'' \in R\) be any such that \(r' \neq r''\).
Let us suppose that \(f_r (r') = f_r (r'')\).
\(r r' = r r''\) and \(r (r' - r'') = 0\). As \(R\) is an integral domain, \(r = 0\) or \(r' - r'' = 0\), but \(r \neq 0\), so, \(r' - r'' = 0\), so, \(r' = r''\), a contradiction.
So, \(f_r (r') \neq f_r (r'')\).
