description/proof of that for module with basis, components set of element w.r.t. basis is unique
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of basis of module.
Target Context
- The reader will have a description and a proof of the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(B\): \(= \{b_j \vert j \in J\}\), \(\in \{\text{ the bases for } M\}\)
\(m\): \(\in M\), \(= \sum_{j \in S} m^j b_j\) for a finite \(S \subseteq J\) and \(m^j \neq 0\)
//
Statements:
\(\exists S' \in \{\text{ the finite subsets of } J\} (m = \sum_{j \in S'} m'^j b_j \text{ where } m'^j \neq 0)\)
\(\implies\)
\(S' = S \land \forall j \in S (m^j = m'^j)\)
//
There is such an \(S\) by the definition of basis of module.
2: Note
The definition of basis of module does not directly require that the decomposition of \(m\) with respect to the basis is unique, but this proposition is claiming that the decomposition is inevitably unique.
3: Proof
Whole Strategy: Step 1: take \(S \cup S'\), and let \(m = \sum_{j \in S \cup S'} \widetilde{m}^j b_j = \sum_{j \in S \cup S'} \widetilde{m'}^j b_j\); Step 2: conclude that \(\widetilde{m}^j - \widetilde{m'}^j = 0\).
Step 1:
Let us take \(S \cup S'\).
For each \(j \in S \cup S'\), let \(\widetilde{m}^j = m^j\) for \(j \in S\) and \(\widetilde{m}^j = 0\) for \(j \notin S\).
Then, \(m = \sum_{j \in S \cup S'} \widetilde{m}^j b_j\).
For each \(j \in S \cup S'\), let \(\widetilde{m'}^j = m'^j\) for \(j \in S'\) and \(\widetilde{m'}^j = 0\) for \(j \notin S'\).
Then, \(m = \sum_{j \in S \cup S'} \widetilde{m'}^j b_j\).
Step 2:
As \(\sum_{j \in S \cup S'} \widetilde{m}^j b_j = \sum_{j \in S \cup S'} \widetilde{m'}^j b_j\), \(\sum_{j \in S \cup S'} \widetilde{m}^j b_j - \sum_{j \in S \cup S'} \widetilde{m'}^j b_j = 0\), but \(= \sum_{j \in S \cup S'} (\widetilde{m}^j - \widetilde{m'}^j) b_j\).
As \(B\) is linearly independent, \(\widetilde{m}^j - \widetilde{m'}^j\) for each \(j \in S \cup S'\), so, \(\widetilde{m}^j = \widetilde{m'}^j\).
Let us suppose that \(\widetilde{m}^j = \widetilde{m'}^j = 0\) for a \(j \in S \cup S'\). \(j \notin S\) and \(j \notin S'\), which would mean that \(j \notin S \cup S'\), a contradiction. So, \(\widetilde{m}^j = \widetilde{m'}^j \neq 0\), which implies that \(S = S'\).
For each \(j \in S\), \(m^j = \widetilde{m}^j = \widetilde{m'}^j = m'^j\).
