description/proof of that for module, basis cannot be supplemented with element to keep linearly independent
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of basis of module.
- The reader admits the proposition that for any module, the inverse of each element is the element \(- 1\)-scalar multiplied.
Target Context
- The reader will have a description and a proof of the proposition that for any module, any basis cannot be supplemented with any element to keep linearly independent.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(B\): \(= \{b^j \vert j \in J\}\), \(\in \{\text{ the bases of } M\}\)
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Statements:
\(\forall m \in M \setminus B (B \cup \{m\} \notin \{\text{ the linearly independent subsets of } M\})\)
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2: Note
This property holds for any module basis: compare with the proposition that module linearly independent subsets or bases do not necessarily have some properties of vectors space linearly independent subsets or bases.
3: Proof
Whole Strategy: Step 1: see that \(m = \sum_{j^` \in J^`} m_{j^`} b^{j^`}\); Step 2: see that \(\sum_{j^` \in J^`} r_{j^`} b^{j^`} + r m = 0\) can be realized with some nonzero coefficients.
Step 1:
There is a finite subset, \(J^` \subseteq J\), such that \(m = \sum_{j^` \in J^`} m_{j^`} b^{j^`}\), where \(m_{j^`} \in R\).
So, \(0 = - m + m = - m + \sum_{j^` \in J^`} m_{j^`} b^{j^`}\).
Step 2:
Let us suppose that \(\sum_{j^` \in J^`} r_{j^`} b^{j^`} + r m = 0\), where \(r_{j^`} \in R\) and \(r \in R\).
Let us take \(r_{j^`} = m_{j^`}\) and \(r = - 1\).
\(\sum_{j^` \in J^`} m_{j^`} b^{j^`} + (- 1) m = \sum_{j^` \in J^`} m_{j^`} b^{j^`} + (- m)\), by the proposition that for any module, the inverse of each element is the element \(- 1\)-scalar multiplied, \(= 0\), by Step 1.
So, \(\sum_{j^` \in J^`} r_{j^`} b^{j^`} + r m = 0\) can be realized by some nonzero coefficients.
So, \(B \cup \{m\}\) is not linearly independent.
