949: For Module with Basis, Components Set of Element w.r.t. Basis Is Unique

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description/proof of that for module with basis, components set of element w.r.t. basis is unique

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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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of basis of module.

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Target Context

• The reader will have a description and a proof of the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.

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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:
$R$: $\in \{\text{ the rings }\}$
$M$: $\in \{\text{ the }R\text{ modules }\}$
$J$: $\in \{\text{ the possibly uncountable index sets }\}$
$B$: $=\{b_j|j\in J\}$, $\in \{\text{ the bases for }M\}$
$m$: $\in M$, $=\sum _{j\in S}{m}^j{b}_j$ for a finite $S\subseteq J$ and $m^j\ne 0$
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Statements:
$\mathrm{\exists }{S}^{\prime }\in \{\text{ the finite subsets of }J\}(m=\sum _{j\in S^{\prime }}{m}^{\prime j}{b}_j\text{ where }{m}^{\prime j}\ne 0)$
$⟹$
$S^{\prime }=S\wedge \mathrm{\forall }j\in S(m^j=m^{\prime j})$
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There is such an $S$ by the definition of basis of module.

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2: Note

The definition of basis of module does not directly require that the decomposition of $m$ with respect to the basis is unique, but this proposition is claiming that the decomposition is inevitably unique.

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3: Proof

Whole Strategy: Step 1: take $S\cup S^{\prime }$, and let $m=\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^j{b}_j=\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^j{b}_j$; Step 2: conclude that ${\tilde{m}}^j-{\widetilde{m^{\prime }}}^j=0$.

Step 1:

Let us take $S\cup S^{\prime }$.

For each $j\in S\cup S^{\prime }$, let ${\tilde{m}}^j=m^j$ for $j\in S$ and ${\tilde{m}}^j=0$ for $j\notin S$.

Then, $m=\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^j{b}_j$.

For each $j\in S\cup S^{\prime }$, let ${\widetilde{m^{\prime }}}^j=m^{\prime j}$ for $j\in S^{\prime }$ and ${\widetilde{m^{\prime }}}^j=0$ for $j\notin S^{\prime }$.

Then, $m=\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^j{b}_j$.

Step 2:

As $\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^j{b}_j=\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^j{b}_j$, $\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^j{b}_j-\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^j{b}_j=0$, but $=\sum _{j\in S\cup S^{\prime }}({\tilde{m}}^j-{\widetilde{m^{\prime }}}^j)b_j$.

As $B$ is linearly independent, ${\tilde{m}}^j-{\widetilde{m^{\prime }}}^j$ for each $j\in S\cup S^{\prime }$, so, ${\tilde{m}}^j={\widetilde{m^{\prime }}}^j$.

Let us suppose that ${\tilde{m}}^j={\widetilde{m^{\prime }}}^j=0$ for a $j\in S\cup S^{\prime }$. $j\notin S$ and $j\notin S^{\prime }$, which would mean that $j\notin S\cup S^{\prime }$, a contradiction. So, ${\tilde{m}}^j={\widetilde{m^{\prime }}}^j\ne 0$, which implies that $S=S^{\prime }$.

For each $j\in S$, $m^j={\tilde{m}}^j={\widetilde{m^{\prime }}}^j=m^{\prime j}$.

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References

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