564: For Linearly Independent Finite Subset of Module, Induced Subset of Module with Some Linear Combinations Is Linearly Independent

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description/proof of that for linearly independent finite subset of module, induced subset of module with some linear combinations is linearly independent

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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: Proof

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Starting Context

• The reader knows a definition of linearly independent subset of module.

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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.

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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:
$R$: $\in \{\text{ the rings }\}$
$M$: $\in \{\text{ the }R\text{ modules }\}$
$S$: $=\{p_1,...,p_n\}\subseteq M$, $\in \{\text{ the linearly independent subsets of }M\}$
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Statements:
$\mathrm{\forall }\{d^1,...,d^n\}\subset R\text{ such that }{d}^j\ne 0,\mathrm{\forall }\{c_1^1,c_2^1,c_2^2,c_3^1,c_3^2,c_3^3,...,c_n^1,...,c_n^n\}\subset R\text{ such that }{c}_j^j\ne 0(\{d^1{c}_1^1{p}_1,d^2(c_2^1{p}_1+c_2^2{p}_2),...,d^n(c_n^1{p}_1+...+c_n^n{p}_n)\}\in \{\text{ the linearly independent subsets of }M\})$.
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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 $d^j\ne 0$ and for each $\{c_1^1,c_2^1,c_2^2,c_3^1,c_3^2,c_3^3,...,c_n^1,...,c_n^n\}\subset R$ such that $c_j^j\ne 0$, $\{d^1{c}_1^1{p}_1,d^2(c_2^1{p}_1+c_2^2{p}_2),...,d^n(c_n^1{p}_1+...+c_n^n{p}_n)\}\subseteq M$ is a linearly independent subset of $M$.

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3: Proof

Let us think of $t^1{d}^1{c}_1^1{p}_1+t^2{d}^2(c_2^1{p}_1+c_2^2{p}_2)+...+t^{n-1}{d}^{n-1}(c_{n-1}^1{p}_1+...+c_{n-1}^{n-1}{p}_{n-1})+t^n{d}^n(c_n^1{p}_1+...+c_n^n{p}_n)=0$ where $t^j\in R$.

$(t^1{d}^1{c}_1^1+t^2{d}^2{c}_2^1+...+t^n{d}^n{c}_n^1)p_1+(t^2{d}^2{c}_2^2+...+t^n{d}^n{c}_n^2)p_2+...+(t^{n-1}{d}^{n-1}{c}_{n-1}^{n-1}+t^n{d}^n{c}_n^{n-1})p_{n-1}+(t^n{d}^n{c}_n^n)p_n=0$.

As $\{p_1,...,p_n\}$ is linearly independent, $t^n{d}^n{c}_n^n=0$, but as $d^n\ne 0$ and $c_n^n\ne 0$, $t^n=0$. $t^{n-1}{d}^{n-1}{c}_{n-1}^{n-1}+t^n{d}^n{c}_n^{n-1}=0$, but as $t^n=0$, $d^{n-1}\ne 0$, and $c_{n-1}^{n-1}\ne 0$, $t^{n-1}=0$. And so on. After all, $t^1=...=t^n=0$.

So, $\{d^1{c}_1^1{p}_1,d^2(c_2^1{p}_1+c_2^2{p}_2),...,d^n(c_n^1{p}_1+...+c_n^n{p}_n)\}$ is linearly independent.

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References

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