description/proof of that for module over division ring, nonzero element nonzero-scalar multiplied is nonzero
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of %ring name% module.
- The reader admits the proposition that for any module, \(0\) each-scalar multiplied is \(0\).
Target Context
- The reader will have a description and a proof of the proposition that for any module over any division ring, each nonzero element each-nonzero-scalar multiplied is nonzero.
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:
\(R\): \(\in \{\text{ the division rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
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Statements:
\(\forall m \in M \setminus \{0\}, \forall r \in R \setminus \{0\} (r m \neq 0)\)
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2: Note
Compare with the proposition that for a module, a nonzero element a-nonzero-scalar multiplied is not necessarily nonzero.
3: Proof
Whole Strategy: Step 1: suppose that \(r m = 0\), and find a contradiction.
Step 1:
Let \(m \in M \setminus \{0\}\) and \(r \in R \setminus \{0\}\) be any.
Let us suppose that \(r m = 0\).
As \(R\) is a division ring and \(r \neq 0\), there is \(r^{- 1} \in R\).
\(r^{- 1} (r m) = r^{- 1} 0 = 0\), by the proposition that for any module, \(0\) each-scalar multiplied is \(0\), but the left hand side is \((r^{- 1} r) m = 1 m = m\), so, \(m = 0\), a contradiction against that \(m \neq 0\).
So, \(r m \neq 0\).
