Free Modules: Solving Issue of Finite Generation Corollary 2.2.4

[Math Amateur]

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

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

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

\(\displaystyle x = \sum_\Delta x_\alpha a_\alpha \)

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

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

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 \(\displaystyle S\).

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

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

\(\displaystyle y = \sum_{\Delta} x_\alpha a_\alpha
\)

where \(\displaystyle x_\alpha \in S \) and \(\displaystyle a_\alpha \in R\)

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

\(\displaystyle x = \sum_{\Delta} x_\alpha a_\alpha
\)

for some \(\displaystyle x_\alpha, a_\alpha
\)

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

\(\displaystyle \sum_{\Delta} x_\alpha a_\alpha \in L \)

... ... BUT ... ... this is what is assumed in Bland's proof of Corollary \(\displaystyle 2.2.4\)?

Can someone please clarify this issue ...

Peter

 

[caffeinemachine]

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 \(\displaystyle (a_\alpha) \in R^{ ( \Delta ) }\), then \(\displaystyle \sum_\Delta x_\alpha a_\alpha \in F \) ... ... "My question is as follows:How, exactly, do we know that \(\displaystyle (a_\alpha) \in R^{ ( \Delta ) }\) implies that \(\displaystyle \sum_\Delta x_\alpha a_\alpha \in F\) ... ... that is, is it possible that for some \(\displaystyle (a_\alpha) \in R^{ ( \Delta ) }\) there is no element \(\displaystyle x \) such that \(\displaystyle x = \sum_\Delta x_\alpha a_\alpha \in F \)?To make sure my question is clear ... ...If F is a free R-module with basis \(\displaystyle \{ x_\alpha \}_\Delta \), then every element \(\displaystyle x \in F\) can be expressed (generated) as a sum of the form:\(\displaystyle x = \sum_\Delta x_\alpha a_\alpha \)... ... BUT ... ... does this mean that for any element \(\displaystyle (a_\alpha) \in R^{ ( \Delta ) }\) there is actually an element \(\displaystyle x \in F\) such that \(\displaystyle x = \sum_\Delta x_\alpha a_\alpha \)?... OR ... to put it another way ... could it be that for some element \(\displaystyle (a_\alpha) \in R^{ ( \Delta ) }\) there is actually NO element \(\displaystyle x \in F\) such that \(\displaystyle x = \sum_\Delta x_\alpha a_\alpha \)?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 \(\displaystyle S\).Bland defines the generation of a submodule of \(\displaystyle N\) of an \(\displaystyle R\)-module \(\displaystyle M\) as follows:View attachment 3535Now consider a submodule \(\displaystyle L\) of \(\displaystyle M\) such that \(\displaystyle L \subset N\).See Figure \(\displaystyle 1\) as follows:https://www.physicsforums.com/attachments/3536Now \(\displaystyle L\), like \(\displaystyle N\), will (according to Bland's definition) also be generated by \(\displaystyle S\), since every element \(\displaystyle y \in L\) will be able to be expressed as a sum\(\displaystyle y = \sum_{\Delta} x_\alpha a_\alpha\)where \(\displaystyle x_\alpha \in S \) and \(\displaystyle a_\alpha \in R\)This is possible since every element of \(\displaystyle N\) (and hence \(\displaystyle L\)) can be expressed this way.However ... ... if we consider \(\displaystyle x \in N\) such that \(\displaystyle x \notin L\) then \(\displaystyle x = \sum_{\Delta} x_\alpha a_\alpha\)for some \(\displaystyle x_\alpha, a_\alpha\)... ... BUT ... ... in this case, there is no \(\displaystyle (a_\alpha) \in R^{ ( \Delta ) } \) such that \(\displaystyle \sum_{\Delta} x_\alpha a_\alpha \in L \)... ... BUT ... ... this is what is assumed in Bland's proof of Corollary \(\displaystyle 2.2.4\)?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.

 

[Math Amateur]

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

 

[caffeinemachine]

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.

 

[blue_raver22]

Hi Peter,

Thank you for your question. I can understand why you are perplexed about the generation of a module or submodule by a set S. It is a common misconception that every element of a module can be expressed as a linear combination of elements from a generating set. However, this is not always the case.

In the case of a free module, we have the additional property that every element can be expressed as a linear combination of elements from a basis set, which is a special type of generating set. This is why we can say that every element of a free module can be generated by a set S.

In the proof of Corollary 2.2.4, Bland is using this property of free modules to show that if (a_\alpha) \in R^{ ( \Delta ) }, then \sum_\Delta x_\alpha a_\alpha \in F. This is because, by definition, F is a free module with basis \{x_\alpha\}_\Delta, so every element in F can be expressed as a linear combination of elements from this basis set. Therefore, the linear combination \sum_\Delta x_\alpha a_\alpha must also be an element of F.

To answer your question, it is possible for there to be some (a_\alpha) \in R^{ ( \Delta ) } such that there is no element x \in F satisfying x = \sum_\Delta x_\alpha a_\alpha. This is because not every element of a module can be expressed as a linear combination of elements from a generating set. However, in the case of a free module, every element can be expressed as a linear combination of elements from a basis set, which is why we can say that every element in F can be generated by a set S.

I hope this helps clarify the issue for you. If you have any further questions, please let me know.

Best,