description/proof of that for 'modules - linear morphisms' isomorphism, linearly independent subset or basis of domain is mapped to linearly independent subset or basis
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of %ring name% module.
- The reader knows a definition of linear map.
- The reader knows a definition of %category name% isomorphism.
- The reader knows a definition of linearly independent subset of module.
- 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 'modules - linear morphisms' isomorphism, any linearly independent subset or basis of the domain is mapped to a linearly independent subset or 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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules }\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the 'modules - linear morphisms' isomorphisms }\}\)
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Statements:
\(\forall S_1 \in \{\text{ the linearly independent subsets of } M_1\} (f (S_1) \in \{\text{ the linearly independent subsets of } M_2\})\)
\(\land\)
\(\forall B_1 \in \{\text{ the bases of } M_1\} (f (B_1) \in \{\text{ the bases of } M_2\})\)
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2: Proof
Whole Strategy: Step 1: let \(S_1 = \{m_{1, j} \vert j \in J\}\) and take any finite \(J^` \subseteq J\), and see that \(\sum_{j^` \in J^`} r_{j^`} f (m_{1, j^`}) = 0\) implies that \(r_{j^`}\) s are \(0\); Step 2: let \(B_1 = \{{b_1}^j \vert j \in J\}\), and see that for each \(m_2 \in M_2\), there is a finite \(J^` \subseteq J\) such that \(m_2 = \sum_{j' \in J^`} r_{j'} f ({b_1}^{j'})\).
Step 1:
Let \(S_1\) be any linearly independent subset of \(M_1\).
\(S_1 = \{m_{1, j} \in M_1 \vert j \in J\}\) where \(J\) is a possibly uncountable index set.
Let \(J^` \subseteq J\) be any finite subset.
Let \(\sum_{j^` \in J^`} r_{j^`} f (m_{1, j^`}) = 0\).
\(\sum_{j^` \in J^`} r_{j^`} f (m_{1, j^`}) = f (\sum_{j^` \in J^`} r_{j^`} m_{1, j^`})\), because \(f\) is linear.
But \(\sum_{j^` \in J^`} r_{j^`} m_{1, j^`} = f^{- 1} \circ f (\sum_{j^` \in J^`} r_{j^`} m_{1, j^`}) = f^{- 1} (0) = 0\), because as \(f\) is a 'modules - linear morphisms' isomorphism, \(f^{- 1}\) is a linear morphism.
That implies that \(r_{j^`}\) s are \(0\), because \(S_1\) is linearly independent.
So, \(f (S_1)\) is linearly independent.
Step 2:
Let \(B_1 \subseteq M_1\) be any basis of \(M_1\).
\(B_1 = \{{b_1}^j \vert j \in J\}\) where \(J\) is a possibly uncountable index set.
While \(B_1\) is linearly independent, \(f (B_1)\) is linearly independent, by Step 1.
Let \(m_2 \in M_2\) be any.
There is a finite \(J^` \subseteq J\) such that \(f^{- 1} (m_2) = \sum_{j^` \in J^`} r_{j^`} {b_1}^{j^`}\), because \(B_1\) is a basis.
\(m_2 = f \circ f^{- 1} (m_2) = f (\sum_{j^` \in J^`} r_{j^`} {b_1}^{j^`}) = \sum_{j^` \in J^`} r_{j^`} f ({b_1}^{j^`})\), because \(f\) is linear, so, \(m_2\) is a linear combination of \(f (B_1)\).
So, \(f (B_1)\) is a basis of \(M_2\).
