MHB Free Modules - Another problem regarding Bland Proposition 2.2.3

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The discussion centers on understanding Proposition 2.2.3 from Paul E. Bland's book on free modules, specifically the proof's assertion that any element in module M can be expressed as a finite sum of basis elements, where almost all coefficients are zero. The key point raised is the need for clarification on how the definition of a basis justifies this condition. Participants emphasize that the submodule generated by a set only involves finite sums, making the condition necessary. The conversation highlights the importance of grasping the definitions of basis and generation in module theory for a clearer understanding of the proof. Understanding these concepts is crucial for comprehending the structure of free modules.
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I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Section 2.2 on free modules and need help with the proof of Proposition 2.2.3 - that is some further help ... (for my first problem with the proof see my post "http://mathhelpboards.com/linear-abstract-algebra-14/free-modules-bland-proposition-2-2-3-a-13179.html")

Proposition 2.2.3 and its proof read as follows:View attachment 3531The first line of Bland's proof reads as follows:

"Proof: $$(1) \Longrightarrow (2)$$. Since $$ \{ x_\alpha \}_\Delta$$ is a set of generators of $$M $$, it is certainly the case that each $$x \in M$$ can be expressed as $$x = \sum_\Delta x_\alpha a_\alpha$$ where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$. ... ... "Now, by definition of a basis, $$ \{ x_\alpha \}_\Delta$$ being a basis for M certainly implies that each $$x \in M$$ can be expressed as a sum $$x = \sum_\Delta x_\alpha a_\alpha$$ ... ...

... ... BUT ... ... how do we know that each $$x \in M$$ can be expressed as $$x = \sum_\Delta x_\alpha a_\alpha$$ where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$ ... ...

... ... specifically ... ... what/where in the definition of a basis is the justification for adding the condition "where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$"?

Hope someone can help,

Peter
 
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Peter said:
I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Section 2.2 on free modules and need help with the proof of Proposition 2.2.3 - that is some further help ... (for my first problem with the proof see my post "http://mathhelpboards.com/linear-abstract-algebra-14/free-modules-bland-proposition-2-2-3-a-13179.html")

Proposition 2.2.3 and its proof read as follows:View attachment 3531The first line of Bland's proof reads as follows:

"Proof: $$(1) \Longrightarrow (2)$$. Since $$ \{ x_\alpha \}_\Delta$$ is a set of generators of $$M $$, it is certainly the case that each $$x \in M$$ can be expressed as $$x = \sum_\Delta x_\alpha a_\alpha$$ where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$. ... ... "Now, by definition of a basis, $$ \{ x_\alpha \}_\Delta$$ being a basis for M certainly implies that each $$x \in M$$ can be expressed as a sum $$x = \sum_\Delta x_\alpha a_\alpha$$ ... ...

... ... BUT ... ... how do we know that each $$x \in M$$ can be expressed as $$x = \sum_\Delta x_\alpha a_\alpha$$ where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$ ... ...

... ... specifically ... ... what/where in the definition of a basis is the justification for adding the condition "where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$"?

Hope someone can help,

Peter

Let $M$ be any $R$-module and $S$ be a subset of $M$.

Then the submodule of $M$ generated by $S$ is defined as

$$\{ \sum_{\text{finite}}a_is_i:a_i\in R, s_i\in S \}$$

Summing up infinitely many elements of a module doesn't make any sense.

Now coming to your problem.
If $X=\{x_\alpha\}_{\alpha\in J}$ is a basis of $M$, then $X$ generates $M$. Use the definition of "generation" as discussed above.

I hope this helped.
 
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