description/proof of that for module over division ring with finite basis, there is no basis that has larger cardinality
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 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.
- The reader admits the proposition that for any module, any basis cannot be supplemented with any element to keep linearly independent.
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, there is no basis that has any larger cardinality.
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, \(B = \{b^1, ..., b^n\}\)
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Statements:
\(\lnot \exists B' \in \{\text{ the bases of } M\} (\vert B \vert \lt \vert B' \vert)\)
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2: Note
So, although the proposition that for any finite-dimensional vectors space, there is no basis that has more than the dimension number of elements is not guaranteed to hold for a module over a general ring, this proposition holds for any module over any division ring.
As there is no basis whose cardinality is larger than that of \(B\), also there is no basis whose cardinality is smaller than that of \(B\), because if there was a basis, \(B'\), such that \(\vert B' \vert \lt \vert B \vert\), \(B\) would be a basis whose cardinality was larger than that of \(B'\), a contradiction.
So, if there is a basis of cardinality, \(n \in \mathbb{N} \setminus \{0\}\), any basis will have the cardinality, \(n\).
3: Proof
Whole Strategy: Step 1: suppose that there was a \(B'\) such that \(\vert B \vert \lt \vert B' \vert\); Step 2: replace each element of \(B\) with an element of \(B'\) to keep being a basis; Step 3: find a contradiction.
Step 1:
Let us suppose there was a basis, \(B' = \{b'^l \vert l \in L\}\) where \(L\) is a possibly uncountable index set, such that \(\vert B \vert \lt \vert B' \vert\).
Step 2:
There would be a \(b'^{l_1} = r'_{l_1, j} b^j \in B'\) such that \(r'_{l_1, 1} \neq 0\), because while each \(b'^l\) could be expressed as \(r'_{l, j} b^j\), because \(B\) was a basis, if all the elements of \(B'\) were with \(r'_{l, 1} = 0\), while \(b^1\) was a linear combination of \(B'\), because \(B'\) was a basis, \(b^1\) would be a linear combination of \(\{b^2, ..., b^n\}\), a contradiction against that \(B\) was a basis: if \(b^1 = r_2 b^2 + ... + r_n b^n\), \(b^1 - r_2 b^2 - ... - r_n b^n = 0\), which would mean that \(B\) was not linearly independent.
Let us take \(B_1 := \{b'^{l_1}, b^2, ..., b^n\}\), which would be a basis, by 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.
Let us do that inductively to replace all the elements of \(B\).
Let us suppose that we had already a basis, \(B_{k - 1} := \{b'^{l_1}, ..., b'^{l_{k - 1}}, b^k, ..., b^n\}\).
There would be a \(b'^{l_k} = r'_{l_k, 1} b'^{l_1} + ... + r'_{l_k, k - 1} b'^{l_{k - 1}} + r'_{l_k, k} b^k + ... + r'_{l_k, n} b^n \in B'\) such that \(r'_{l_k, k} \neq 0\), because while each \(b'^l\) could be expressed as \(r'_{l, 1} b'^{l_1} + ... + r'_{l, k - 1} b'^{l_{k - 1}} + r'_{l, k} b^k + ... + r'_{l, n} b^n\), because \(B_{k - 1}\) was a basis, if all the elements of \(B'\) were with \(r'_{l, k} = 0\), while \(b^k\) was a linear combination of \(B'\), because \(B'\) was a basis, \(b^k\) would be a linear combination of \(\{b'^{l_1}, ..., b'^{l_{k - 1}}, b^{k + 1}, ..., b^n\}\), a contradiction against that \(B_{k - 1}\) was a basis: if \(b^k = r_1 b'^{l_1} + ... + r_{k - 1} b'^{l_{k - 1}} + r_{k + 1} b^{k + 1} + ... + r_n b^n\), \(b^k - r_1 b'^{l_1} - ... - r_{k - 1} b'^{l_{k - 1}} - r_{k + 1} b^{k + 1} - ... - r_n b^n = 0\), which would mean that \(B_{k - 1}\) was not linearly independent.
Let us take \(B_k := \{b'^{l_1}, ..., b'^{l_k}, b^{k + 1}, ..., b^n\}\), which would be a basis, by 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.
After all, let us take \(B_n := \{b'^{l_1}, ..., b'^{l_n}\}\), which would be a basis.
Step 3:
\(B_n \subset B'\), because \(\vert B_n \vert \le \vert B \vert \lt \vert B' \vert\).
\(B'\) would not be linearly independent, by the proposition that for any module, any basis cannot be supplemented with any element to keep linearly independent, a contradiction against that \(B'\) was a basis.
So, the supposition of Step 1 was wrong, so, there is no basis, \(B'\), such that \(\vert B \vert \lt \vert B' \vert\).
