MHB Free Modules: Solving Issue of Finite Generation Corollary 2.2.4

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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 Corollary 2.2.4.

Corollary 2.2.4 and its proof read as follows:View attachment 3533
View attachment 3534In the second last paragraph of Bland's proof above we read:

" ... ... If $$(a_\alpha) \in R^{ ( \Delta ) }$$, then $$ \sum_\Delta x_\alpha a_\alpha \in F $$ ... ... "My question is as follows:

How, exactly, do we know that $$(a_\alpha) \in R^{ ( \Delta ) }$$ implies that $$\sum_\Delta x_\alpha a_\alpha \in F$$ ... ... that is, is it possible that for some $$(a_\alpha) \in R^{ ( \Delta ) }$$ there is no element $$x $$ such that $$x = \sum_\Delta x_\alpha a_\alpha \in F $$?To make sure my question is clear ... ...

If F is a free R-module with basis $$\{ x_\alpha \}_\Delta $$, then every element $$x \in F$$ can be expressed (generated) as a sum of the form:

$$x = \sum_\Delta x_\alpha a_\alpha $$

... ... BUT ... ... does this mean that for any element $$(a_\alpha) \in R^{ ( \Delta ) }$$ there is actually an element $$x \in F$$ such that $$x = \sum_\Delta x_\alpha a_\alpha $$?

... OR ... to put it another way ... could it be that for some element $$(a_\alpha) \in R^{ ( \Delta ) }$$ there is actually NO element $$x \in F$$ such that $$x = \sum_\Delta x_\alpha a_\alpha $$?

Can someone please clarify this issue for me?

Peter
***EDIT***

I thought I would try to clarify just exactly why I am perplexed about the nature of the generation of a module or submodule by a set $$S$$.

Bland defines the generation of a submodule of $$N$$ of an $$R$$-module $$M$$ as follows:View attachment 3535Now consider a submodule $$L$$ of $$M$$ such that $$L \subset N$$.

See Figure $$1$$ as follows:https://www.physicsforums.com/attachments/3536Now $$L$$, like $$N$$, will (according to Bland's definition) also be generated by $$S$$, since every element $$y \in L$$ will be able to be expressed as a sum

$$y = \sum_{\Delta} x_\alpha a_\alpha
$$

where $$x_\alpha \in S $$ and $$a_\alpha \in R$$

This is possible since every element of $$N$$ (and hence $$L$$) can be expressed this way.However ... ... if we consider $$x \in N$$ such that $$x \notin L$$ then

$$x = \sum_{\Delta} x_\alpha a_\alpha
$$

for some $$x_\alpha, a_\alpha
$$

... ... BUT ... ... in this case, there is no $$(a_\alpha) \in R^{ ( \Delta ) } $$ such that

$$ \sum_{\Delta} x_\alpha a_\alpha \in L $$

... ... BUT ... ... this is what is assumed in Bland's proof of Corollary $$2.2.4$$?

Can someone please clarify this issue ...

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 Corollary 2.2.4.Corollary 2.2.4 and its proof read as follows:View attachment 3533View attachment 3534In the second last paragraph of Bland's proof above we read:" ... ... If $$(a_\alpha) \in R^{ ( \Delta ) }$$, then $$ \sum_\Delta x_\alpha a_\alpha \in F $$ ... ... "My question is as follows:How, exactly, do we know that $$(a_\alpha) \in R^{ ( \Delta ) }$$ implies that $$\sum_\Delta x_\alpha a_\alpha \in F$$ ... ... that is, is it possible that for some $$(a_\alpha) \in R^{ ( \Delta ) }$$ there is no element $$x $$ such that $$x = \sum_\Delta x_\alpha a_\alpha \in F $$?To make sure my question is clear ... ...If F is a free R-module with basis $$\{ x_\alpha \}_\Delta $$, then every element $$x \in F$$ can be expressed (generated) as a sum of the form:$$x = \sum_\Delta x_\alpha a_\alpha $$... ... BUT ... ... does this mean that for any element $$(a_\alpha) \in R^{ ( \Delta ) }$$ there is actually an element $$x \in F$$ such that $$x = \sum_\Delta x_\alpha a_\alpha $$?... OR ... to put it another way ... could it be that for some element $$(a_\alpha) \in R^{ ( \Delta ) }$$ there is actually NO element $$x \in F$$ such that $$x = \sum_\Delta x_\alpha a_\alpha $$?Can someone please clarify this issue for me?Peter***EDIT***I thought I would try to clarify just exactly why I am perplexed about the nature of the generation of a module or submodule by a set $$S$$.Bland defines the generation of a submodule of $$N$$ of an $$R$$-module $$M$$ as follows:View attachment 3535Now consider a submodule $$L$$ of $$M$$ such that $$L \subset N$$.See Figure $$1$$ as follows:https://www.physicsforums.com/attachments/3536Now $$L$$, like $$N$$, will (according to Bland's definition) also be generated by $$S$$, since every element $$y \in L$$ will be able to be expressed as a sum$$y = \sum_{\Delta} x_\alpha a_\alpha$$where $$x_\alpha \in S $$ and $$a_\alpha \in R$$This is possible since every element of $$N$$ (and hence $$L$$) can be expressed this way.However ... ... if we consider $$x \in N$$ such that $$x \notin L$$ then $$x = \sum_{\Delta} x_\alpha a_\alpha$$for some $$x_\alpha, a_\alpha$$... ... BUT ... ... in this case, there is no $$(a_\alpha) \in R^{ ( \Delta ) } $$ such that $$ \sum_{\Delta} x_\alpha a_\alpha \in L $$... ... BUT ... ... this is what is assumed in Bland's proof of Corollary $$2.2.4$$?Can someone please clarify this issue ...Peter
There is no problem here. What you have shown is, under your hypothesis, that $L=N$. This is expected, obvious actually.The submodule generated by a subset $S$, denoted $\langle S\rangle$, of a given module $M$ is the "smallest" submodule of $M$ which contains $S$. One way to describe it is $$\langle S\rangle =\bigcap_{K\text{ a submodule of } M, S\subseteq K} K$$Now if a subset $S$ of $M$ is contained in a submodule of $M$, $\langle S\rangle$ cannot be any larger than $N$. From your hypothesis, it is immediate that$$\langle S\rangle =L=N$$Considering an analogous situation for groups might be helpful.
 
caffeinemachine said:
There is no problem here. What you have shown is, under your hypothesis, that $L=N$. This is expected, obvious actually.The submodule generated by a subset $S$, denoted $\langle S\rangle$, of a given module $M$ is the "smallest" submodule of $M$ which contains $S$. One way to describe it is $$\langle S\rangle =\bigcap_{K\text{ a submodule of } M, S\subseteq K} K$$Now if a subset $S$ of $M$ is contained in a submodule of $M$, $\langle S\rangle$ cannot be any larger than $N$. From your hypothesis, it is immediate that$$\langle S\rangle =L=N$$Considering an analogous situation for groups might be helpful.
Thanks caffeinmachine ... Really appreciate your clarification of this matter ... most helpful ... thank you again ...

I guess in essence you are saying that since N is the smallest submodule containing S, that we must have L = N ...

Peter
 
Peter said:
Thanks caffeinmachine ... Really appreciate your clarification of this matter ... most helpful ... thank you again ...

I guess in essence you are saying that since N is the smallest submodule containing S, that we must have L = N ...

Peter
Yes. That's correct.
 
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