532: Linearly Independent Subset of Module

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definition of linearly independent subset of module

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Topics

About: module

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Note

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Starting Context

• The reader knows a definition of %ring name% module.

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Target Context

• The reader will have a definition of linearly independent subset of module.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the rings }\}$
$M$: $\in \{\text{ the }R\text{ modules }\}$
$\ast S$: $\subseteq M$, $\in \{\text{ the possibly uncountable sets }\}$
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Conditions:
$\mathrm{\forall }{S}^{\prime }\subseteq S,S^{\prime }\in \{\text{ the finite sets }\}$
(
$\sum _{p_j\in S^{\prime }}{r}^j{p}_j=0,r^j\in R$
$⟹$
$\mathrm{\forall }j(r^j=0)$
)
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2: Natural Language Description

For any module, $M$, over any ring, $R$, any (possibly uncountable) subset, $S\subseteq M$, such that for each finite subset, $S^{\prime }\subseteq S$, $\sum _{p_j\in S^{\prime }}{r}^j{p}_j=0$ implies that $r^j=0$ for each $j$, where $r^j$ is any element of $R$

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3: Note

As any vectors space is a module, 'linearly independent subset of vectors space' is nothing but 'linearly independent subset of module'.

We need to think of the linear combination of each finite subset instead of the linear combination of possibly infinite $S$, because the convergence of the linear combination of any infinite elements is not defined without any extra structure like topology or metric.

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References

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