description/proof of that for module over division ring basis, replacing element by linear combination of elements with nonzero coefficient for element forms basis
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 ring, any multiple of 0 is 0.
- The reader admits the proposition that for any module, each element \(0\)-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 and any basis, replacing any element of the basis by any linear combination of the basis with any nonzero coefficient for the element forms a basis.
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:
\(R\): \(\in \{\text{ the division rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\), with any basis, \(B = \{b^j \vert j \in J\}\) where \(J\) is any possibly uncountable index set
\(k\): \(\in J\)
\(b'^k\): \(= \sum_{j \in J^`} r_j b^j\) where \(J^` \subseteq J\) is any finite subset such that \(k \in J^`\) and \(r_k \neq 0\)
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Statements:
\(B' := (B \setminus \{b^k\}) \cup \{b'^k\} \in \{\text{ the bases of } M\}\)
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2: Note
So, although the proposition that for any finite dimensional vectors space basis, replacing any element by any linear combination of the elements with any nonzero coefficient for the element forms a basis may not hold for a module over a general ring, this proposition holds for any module over any division ring.
3: Proof
Whole Strategy: Step 1: see that \(B'\) is linearly independent; Step 2: see that \(B'\) spans \(M\); Step 3: conclude the proposition.
Step 1:
Let us see that \(B'\) is linearly independent.
Let \(L \subseteq J\) be any finite subset.
\(k \in L\) or \(k \notin L\).
Let us suppose that \(k \in L\).
Let us suppose that \(r'_k b'^k + \sum_{l \in L \setminus \{k\}} r'_l b^l = 0\).
\(r'_k \sum_{j \in J^`} r_j b^j + \sum_{l \in L \setminus \{k\}} r'_l b^l = 0\).
But the left hand side is \(r'_k r_k b^k + \sum_{j \in J^` \setminus \{k\}} r'_k r_j b^j + \sum_{l \in L \setminus \{k\}} r'_l b^l\), whose coefficient of \(b^k\) is \(r'_k r_k\), so, \(r'_k r_k = 0\), because \(B\) is linearly independent.
As \(R\) is a division ring, there is \({r_k}^{- 1} \in R\), and \(r'_k r_k {r_k}^{- 1} = 0 {r_k}^{- 1} = 0\), by the proposition that for any ring, any multiple of 0 is 0, but the left hand side is \(r'_k (r_k {r_k}^{- 1}) = r'_k 1 = r'_k\), so, \(r'_k = 0\).
So, \(0 b'^k + \sum_{l \in L \setminus \{k\}} r'_l b^l = 0 + \sum_{l \in L \setminus \{k\}} r'_l b^l\), by the proposition that for any module, each element \(0\)-scalar multiplied is \(0\), \(= \sum_{l \in L \setminus \{k\}} r'_l b^l = 0\), which implies that all the \(r'_l\) s are \(0\), because \(B\) is linearly independent.
So, all the coefficients of \(r'_k b'^k + \sum_{l \in L \setminus \{k\}} r'_l b^l\) are \(0\).
Let us suppose that \(k \notin L\).
Let us suppose that \(\sum_{l \in L} r'_l b^l = 0\).
All the \(r'_l\) s are \(0\), because \(B\) is linearly independent.
So, \(B'\) is linearly independent.
Step 2:
Let us see that each \(m \in M\) is a linear combination of \(B'\).
As \(r_k \neq 0\), there is \({r_k}^{- 1} \in R\).
\({r_k}^{- 1} b'^k = {r_k}^{- 1} \sum_{j \in J^`} r_j b^j = \sum_{j \in J^`} {r_k}^{- 1} r_j b^j = {r_k}^{- 1} r_k b^k + \sum_{j \in J^` \setminus \{k\}} {r_k}^{- 1} r_j b^j = 1 b^k + \sum_{j \in J^` \setminus \{k\}} {r_k}^{- 1} r_j b^j = b^k + \sum_{j \in J^` \setminus \{k\}} {r_k}^{- 1} r_j b^j\), so, \(b^k = {r_k}^{- 1} b'^k - \sum_{j \in J^` \setminus \{k\}} {r_k}^{- 1} r_j b^j\).
There is a finite subset, \(L \subseteq J\), such that \(m = \sum_{l \in L} r'_l b^l\), because \(B\) is a basis.
If \(k \in L\), \(m = r'_k b^k + \sum_{l \in L \setminus \{k\}} r'_l b^l = r'_k ({r_k}^{- 1} b'^k - \sum_{j \in J^` \setminus \{k\}} {r_k}^{- 1} r_j b^j) + \sum_{l \in L \setminus \{k\}} r'_l b^l\), which is a linear combination of \(B'\).
If \(k \notin L\), \(m = \sum_{l \in L} r'_l b^l\) is already a linear combination of \(B'\).
So, anyway, \(m\) is a linear combination of \(B'\).
Step 3:
So, \(B'\) is a basis of \(M\).
