description/proof of that for linearly independent finite subset of module, induced subset of module with some linear combinations is linearly independent
Topics
About: module
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- 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 linearly independent finite subset of any module, the induced subset of the module with some linear combinations is linearly independent.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
\(S\): \(= \{p_1, ..., p_n\} \subseteq M\), \(\in \{\text{ the linearly independent subsets of } M\}\)
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Statements:
\(\forall \{d^1, ..., d^n\} \subset R \text{ such that } d^j \neq 0, \forall \{c^1_1, c^1_2, c^2_2, c^1_3, c^2_3, c^3_3, ..., c^1_n, ..., c^n_n\} \subset R \text{ such that } c^j_j \neq 0 (\{d^1 c^1_1 p_1, d^2 (c^1_2 p_1 + c^2_2 p_2), ..., d^n (c^1_n p_1 + ... + c^n_n p_n)\} \in \{\text{ the linearly independent subsets of } M\})\).
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2: Natural Language Description
For any ring, \(R\), any \(R\) module, \(M\), and any linearly independent subset of \(M\), \(S = \{p_1, ..., p_n\} \subseteq M\), for each \(\{d^1, ..., d^n\} \subset R\) such that \(d^j \neq 0\) and for each \(\{c^1_1, c^1_2, c^2_2, c^1_3, c^2_3, c^3_3, ..., c^1_n, ..., c^n_n\} \subset R\) such that \(c^j_j \neq 0\), \(\{d^1 c^1_1 p_1, d^2 (c^1_2 p_1 + c^2_2 p_2), ..., d^n (c^1_n p_1 + ... + c^n_n p_n)\} \subseteq M\) is a linearly independent subset of \(M\).
3: Proof
Let us think of \(t^1 d^1 c^1_1 p_1 + t^2 d^2 (c^1_2 p_1 + c^2_2 p_2) + ... + t^{n - 1} d^{n - 1} (c^1_{n - 1} p_1 + ... + c^{n - 1}_{n - 1} p_{n - 1}) + t^n d^n (c^1_n p_1 + ... + c^n_n p_n) = 0\) where \(t^j \in R\).
\((t^1 d^1 c^1_1 + t^2 d^2 c^1_2 + ... + t^n d^n c^1_n) p_1 + (t^2 d^2 c^2_2 + ... + t^n d^n c^2_n) p_2 + ... + (t^{n - 1} d^{n - 1} c^{n - 1}_{n - 1} + t^n d^n c^{n - 1}_n) p_{n - 1} + (t^n d^n c^n_n) p_n = 0\).
As \(\{p_1, ..., p_n\}\) is linearly independent, \(t^n d^n c^n_n = 0\), but as \(d^n \neq 0\) and \(c^n_n \neq 0\), \(t^n = 0\). \(t^{n - 1} d^{n - 1} c^{n - 1}_{n - 1} + t^n d^n c^{n - 1}_n = 0\), but as \(t^n = 0\), \(d^{n - 1} \neq 0\), and \(c^{n - 1}_{n - 1} \neq 0\), \(t^{n - 1} = 0\). And so on. After all, \(t^1 = ... = t^n = 0\).
So, \(\{d^1 c^1_1 p_1, d^2 (c^1_2 p_1 + c^2_2 p_2), ..., d^n (c^1_n p_1 + ... + c^n_n p_n)\}\) is linearly independent.
