description/proof of that for module over division ring with finite basis, linearly independent subset can be supplemented with some elements of basis to be basis
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of basis of module.
- The reader knows a definition of linearly independent subset of module.
Target Context
- The reader will have a description and a proof of the proposition that for any module over any division ring with any finite basis, any linearly independent subset can be supplemented with some elements of the basis to be 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, \(B = \{b^1, ..., b^n\}\)
\(S\): \(\in \{\text{ the linearly independent subsets of } M\}\)
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Statements:
\(\exists B^` \subseteq B (S \cup B^` \in \{\text{ the bases of } M\})\)
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2: Note
In fact, inevitably \(\vert S \vert \le n\), by the proposition that for any module over any division ring with any finite basis, there is no basis that has any larger cardinality, because otherwise, we would have a basis whose cardinality was larger than \(n\), a contradiction.
3: Proof
Whole Strategy: Step 1: iteratively add an element of \(B\) to \(S\) to keep it still linearly independent; Step 3: see that eventually, the augmented \(S\) (called "S'") will be a basis.
Step 1:
If \(S\) does not span \(M\), let us add to \(S\) the 1st element of \(B\) such that the augmented subset is linearly independent, \(b^{j_1}\), calling the result \(S'\): such an \(b^{j_1}\) exists, because otherwise, for each \(j \in \{1, ..., n\}\), \(r_1 s^1 + ... + r_m s^m + r b^j = 0\) would have a not-all-are-\(0\) coefficients set, but in fact, \(r \neq 0\), because otherwise, \(r_1 s^1 + ... + r_m s^m = 0\) would have a not-all-are-\(0\) coefficients set, a contradiction against that \(S\) was linearly independent, then, \(b^j = r^{- 1} (- r_1 s^1 - ... - r_m s^m)\), so, \(S\) would span \(M\), because each element of \(M\) was a linear combination of \(B\), which would be a linear combination of \(S\).
If \(S'\) does not span \(M\), let us add to \(S'\) the next 1st element of \(B\) such that the augmented subset is linearly independent, \(b^{j_2}\), calling the result again \(S'\): such a \(b^{j_2}\) exists, because otherwise, for each \(j \in \{1, ..., n\} \setminus \{j_1\}\), \(r_1 s^1 + ... + r_m s^m + r'_1 b^{j_1} + r b^j = 0\) would have a not-all-are-\(0\) coefficients set, but in fact, \(r \neq 0\), because otherwise, \(r_1 s^1 + ... + r_m s^m + r'_1 b^{j_1} = 0\) would have a not-all-are-\(0\) coefficients set, a contradiction against that \(S'\) was linearly independent, then, \(b^j = r^{- 1} (- r_1 s^1 - ... - r_m s^m - r'_1 b^{j_1})\), so, \(S'\) would span \(M\), because each element of \(M\) was a linear combination of \(B\), which would be a linear combination of \(S'\).
And so on until all the possible element of \(B\) have been added.
Step 3:
After all, \(S'\) spans \(M\), because all the elements of \(B\) are some linear combinations of \(S'\), because while the elements added into \(S'\) are some linear combinations of \(S'\), all the other elements did not added into \(S'\) because they were already some linear combinations of \(S'\), and each element of \(M\) is a linear combination of \(B\), which is a linear combination of \(S'\).
\(S'\) is linearly independent and spans \(M\), so, is a basis of \(M\).
