description/proof of that module linearly independent subsets or bases do not necessarily have some properties of vectors space linearly independent subsets or bases
Topics
About: module
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Note
- 3: Proof (imperfect)
Starting Context
- The reader knows a definition of basis of module.
- The reader knows a definition of generator of module.
- The reader admits the proposition that any ring is canonically a module with a \(1\)-element basis.
Target Context
- The reader will have a description and a proof of the proposition that module linearly independent subsets or bases do not necessarily have some properties of vectors space linearly independent subsets or bases.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
//
Statements:
not necessarily "\(\forall m \in M \setminus \{0\} (\{m\} \in \{\text{ the linearly independent subsets }\})\)"
\(\land\)
Not necessarily "\(\exists B \in \{\text{ the bases of } M\}\)"
\(\land\)
not necessarily "\(\forall S \in \{\text{ the linearly independent subsets of } M\} (\exists B \in \{\text{ the bases of } M\} (S \subseteq B))\)"
\(\land\)
not necessarily "\(\forall B = \{b^1, ..., b^d\} \in \{\text{ the bases of } M\}, \forall b'^k = r_j b^j \text{ such that } r_k \neq 0 (B \setminus \{b^k\} \cup \{b'^k\} \in \{\text{ the bases of } M\})\)"
\(\land\)
not necessarily "\(\forall S \in \{\text{ the finite generators of } M\} (\exists B \subseteq S (B \in \{\text{ the bases of } M\}))\)"
\(\land\)
not necessarily "\(\forall B_1, B_2 \in \{\text{ the bases of } M\} (\vert B_1 \vert = \vert B_2 \vert)\)"
//
2: Note
While a module may have a linearly independent subset or basis, some familiar properties of vectors space linearly independent subsets or bases are not guaranteed to the linearly independent subset or basis, so, this is a caveat against inadvertently assuming such properties.
A typical method of proving that something is not necessarily the case is to see a counterexample, but the succeeding "Proof (imperfect)" does not necessarily see a counterexample (just because the author could not immediately come up with a counterexample) but at least see why the corresponding proof for vectors spaces does not work for modules (which is the reason why it is qualified as "imperfect").
3: Proof (imperfect)
Whole Strategy: Step 1: see an example that a nonzero-\(1\)-element subset is not linearly independent; Step 2: see why the proof for the existence of basis for vectors spaces does not work for modules; Step 3: see an example that a linearly independent subset cannot be extended to be a basis; Step 4: see why the proof for replacing a basis element with a linear combination of the basis does not work for modules; Step 5: see an example that a finite generator cannot be reduced to be a basis; Step 6: see why the proof for bases cardinalities does not work for modules.
Step 1:
Let us see an example that an \(\{m\}\) where \(m \in M \setminus \{0\}\) is not linearly independent.
Let \(R = \mathbb{Z} / 6 = \{[0], [1], [2], [3], [4], [5]\}\) and \(M = R\), as the canonical module mentioned in the proposition that any ring is canonically a module with a \(1\)-element basis.
Let us take \(\{[2]\}\) where \([2] \in M \setminus \{0\}\).
\([3] [2] = [6] = [0]\), so, \(\{[2]\}\) is not linearly independent.
When \(M\) is a vectors space, \(r v = 0\) implies that \(r^{- 1} r v = r^{-1} 0 = 0\), which implies that \(v = 0\), but that does not work for a module, because \(r^{- 1}\) does not necessarily exist.
Step 2:
Let us see why the proof for the existence of basis for vectors spaces does not work for modules.
the proposition that any vectors space has a basis is based on the proposition that for any vectors space, any generator of the space, and any linearly independent subset contained in the generator, the generator can be reduced to be a basis with the linearly independent subset retained.
But while \(M\) is a generator, it is not proved that the generator has a linearly independent subset (because of Step 1), and even if it does, while \(\sum_{j \in \{1, ..., n\}} c^j b_j + c p = 0\) holds for each \(p \in M\) for a nonzero \(c\), \(p = c^{-1} \sum_{j \in \{1, ..., n\}} - c^j b_j\) may not be valid, because \(c^{-1}\) may not exist.
So, Proof of the proposition that for any vectors space, any generator of the space, and any linearly independent subset contained in the generator, the generator can be reduced to be a basis with the linearly independent subset retained does not work for modules.
Step 3:
Let us see an example that a linearly independent subset cannot be extended to be a basis.
Let \(R = \mathbb{Z}\) and \(M = R\), as the canonical module mentioned in the proposition that any ring is canonically a module with a \(1\)-element basis.
\(\{2\}\) is linearly independent, because \(r 2 = 0\) implies that \(r = 0\).
But \(\{2\}\) is not any basis, because \(1\) cannot be realized as any linear combination of \(\{2\}\).
\(\{2\}\) cannot be extended to be any basis, because any \(\{2, r\}\) is not linearly independent, because \(r 2 + - 2 r = 0\).
Step 4:
Let us see why the proof for replacing a basis element with a linear combination of the basis does not work for modules.
Proof of 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 does not work for modules, because \(d^k c^k = 0\) and \(c^k \neq 0\) does not necessarily imply \(d^k = 0\) (for example, for \(R = \mathbb{Z} / 6\), \([2] [3] = [6] = [0]\)), and even if it does, taking \(e'_k / c_k\) may not be valid, because \({c_k}^{- 1}\) may not exist.
Step 5:
Let us see an example that a finite generator cannot be reduced to be a basis.
Let \(R = \mathbb{Z}\) and \(M = R\), as the canonical module mentioned in the proposition that any ring is canonically a module with a \(1\)-element basis.
\(\{2, 3\}\) is a finite generator of \(M\), because for each \(m \in M\), \(m = m (3 + - 2) = - m 2 + m 3\).
But \(\{2, 3\}\) is not any basis, because it is not linearly independent, because \(3 2 + - 2 3 = 0\), and not \(\{2\}\) nor \(\{3\}\) is any basis, because \(1\) cannot be realized by each of them.
So, \(\{2, 3\}\) cannot be reduced to be any basis.
Step 6:
Let us see why the proof for bases cardinalities does not work for modules.
the proposition that for any finite-dimensional vectors space, there is no basis that has more than the dimension number of elements depends on the proposition that for any finite dimensional vectors space basis, replacing an element by any linear combination of the elements with any nonzero coefficient for the element forms a basis, which corresponds to Step 4.
