description/proof of that finite product module of modules with bases has this basis
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of product 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 any finite product module of modules with bases has this 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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(R\): \(\in \{\text{ the rings }\}\)
\(\{M_j \vert j \in J\}\): \(\subseteq \{\text{ the } R \text{ modules }\}\)
\(\times_{j \in J} M_j\): \(= \text{ the product module }\)
//
Statements:
\(\forall j \in J (B_j = \{b_{j, l_j} \vert l_j \in L_j\} \in \{\text{ the bases for } M_j\})\) where \(L_j\) is a possibly uncountable index set
\(\implies\)
\(B := \cup_{j \in J} \{\times_{j' \in J} \delta_{j, j'} b_{j, l_j} \vert l_j \in L_j\} \in \{\text{ the bases for } \times_{j \in J} M_j \}\)
//
2: Note
\(B\) is like \(\{(b_{1, l_1}, 0, ..., 0) \vert l_1 \in L_1\} \cup ... \cup \{(0, ..., 0, b_{n, l_n}) \vert l_n \in L_n\}\).
\(J\) needs to be finite, because otherwise, an \(m \in \times_{j \in J} M_j\) would not be any finite linear combination of \(B\).
3: Proof
Whole Strategy: Step 1: see that \(B\) is linearly independent; Step 2: see that each \(m \in \times_{j \in J} M_j\) is a finite linear combination of \(B\); Step 3: conclude the proposition.
Step 1:
Let us see that \(B\) is linearly independent.
Let \(S \subseteq B\) be any finite subset.
\(S = \cup_{j \in J} \{\times_{j' \in J} \delta_{j, j'} b_{j, l^`_j} \vert l^`_j \in L^`_j\}\), where \(L^`_j \subseteq L_j\) is a finite subset possibly empty: when \(L^`_j\) is empty, \(\{\times_{j' \in J} \delta_{j, j'} b_{j, l^`_j} \vert l^`_j \in L^`_j\}\) is empty, which means that no element of \(M_j\) has not been chosen.
Let \(\sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \times_{j' \in J} \delta_{j, j'} b_{j, l^`_j} = 0\).
Then, for each \(j'' \in J\), \((\sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \times_{j' \in J} \delta_{j, j'} b_{j, l^`_j}) (j'') = 0\), but the left hand side is \(\sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \delta_{j, j''} b_{j, l^`_j} = \sum_{l^`_{j''} \in L^`_{j''}} r_{{j''}, l^`_{j''}} b_{j'', l^`_{j''}}\), which implies that \(r_{{j''}, l^`_{j''}} = 0\) for each \(j'' \in J\) and each \(l^`_{j''} \in L^`_{j''}\), because \(B_{j''}\) is linearly independent.
That means that all the \(r_{j, l^`_j}\) s are \(0\).
So, \(B\) is linearly independent.
Step 2:
Let \(m \in \times_{j \in J} M_j\) be any.
For each \(j \in J\), \(m (j) = \sum_{l^`_j \in L^`_j} r_{j, l^`_j} b_{j, l^`_j}\), where \(L^`_j \subseteq L_j\) is a finite subset, because \(B_j\) is a basis for \(M_j\).
\(m = \sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \times_{j' \in J} \delta_{j, j'} b_{j, l_j}\), because for each \(j'' \in J\), \((\sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \times_{j' \in J} \delta_{j, j'} b_{j, l_j}) (j'') = \sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \delta_{j, j''} b_{j, l_j} = \sum_{l^`_{j''} \in L^`_{j''}} r_{j'', l^`_{j''}} b_{j'', l_{j''}} = m (j'')\).
So, \(m\) is a finite linear combination of \(B\).
Step 3:
So, \(B\) is a basis for \(\times_{j \in J} M_j\).
