Finitely Generated Modules and Their Submodules .... Berrick and Keating ....

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In summary, Berrick and Keating define a minimal subset of a finitely generated module $M$ as a smallest subset that generates the same submodule of $M$. This subset can be found by taking a finite set $S$ such that $L+\langle S\rangle =M$ and then selecting the smallest subset of $S$ that generates the same submodule as $S$. The inductiveness of a set $S$ that contains certain submodules of $M$ can also be shown to be true based on the definitions provided in section 1.2.18 of their book.
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I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with the proof of Lemma 1.2.21 ...

Lemma 1.2.21 and its proof reads as follows:
View attachment 6037Question 1In the above text by Berrick and Keating, we read the following:"... ... Since \(\displaystyle M\) is finitely generated, there is a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M. \ ... \ ... \ ... \)"
My problem is as follows:

I cannot see exactly why there exists a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M\). ... ... ... Can someone please demonstrate, rigorously and formally, that there exists a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M\)?

Question 2In the above text by Berrick and Keating, we read the following:"... ... Let \(\displaystyle S\) be the set of submodules \(\displaystyle X\) of \(\displaystyle M\) that contain \(\displaystyle x_1 R + \ ... \ ... \ , x_s R + L\) but do not contain \(\displaystyle x_0\). It is obvious that \(\displaystyle S\) is inductive ... ..." Can someone please explain exactly why \(\displaystyle S\) is inductive ... ... ?Hope someone can help ...

Peter========================================================================B&K's definition of "inductive" is contained in section 1.2.18 ... ... . which reads as follows:https://www.physicsforums.com/attachments/6038
View attachment 6039
 
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Peter said:
I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with the proof of Lemma 1.2.21 ...

Lemma 1.2.21 and its proof reads as follows:
Question 1In the above text by Berrick and Keating, we read the following:"... ... Since \(\displaystyle M\) is finitely generated, there is a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M. \ ... \ ... \ ... \)"
My problem is as follows:

I cannot see exactly why there exists a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M\). ... ... ... Can someone please demonstrate, rigorously and formally, that there exists a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M\)?

Question 2In the above text by Berrick and Keating, we read the following:"... ... Let \(\displaystyle S\) be the set of submodules \(\displaystyle X\) of \(\displaystyle M\) that contain \(\displaystyle x_1 R + \ ... \ ... \ , x_s R + L\) but do not contain \(\displaystyle x_0\). It is obvious that \(\displaystyle S\) is inductive ... ..." Can someone please explain exactly why \(\displaystyle S\) is inductive ... ... ?Hope someone can help ...

Peter========================================================================B&K's definition of "inductive" is contained in section 1.2.18 ... ... . which reads as follows:
The existence of a minimal $S$ is rather obvious. Since $M$ is finitely generated, one can find a finite set $S$ such that $L+\langle S\rangle =M$. Take a smallest subset of $S$ which generates the same submodule as $S$. This is a required minimial subset.

I will post about the inductiveness of $S$ when I have some more time. But at first look it semmed like this too follows immediately from the definitions.
 
  • #3
caffeinemachine said:
The existence of a minimal $S$ is rather obvious. Since $M$ is finitely generated, one can find a finite set $S$ such that $L+\langle S\rangle =M$. Take a smallest subset of $S$ which generates the same submodule as $S$. This is a required minimial subset.

I will post about the inductiveness of $S$ when I have some more time. But at first look it semmed like this too follows immediately from the definitions.
Thanks for the help, caffeinemachine BUT ... I do not follow ...

You write:"... ... Since $M$ is finitely generated, one can find a finite set $S$ such that $L+\langle S\rangle =M$ ... ... BUT ... why exactly is this true ... can you be more explicit ...

Peter
 

FAQ: Finitely Generated Modules and Their Submodules .... Berrick and Keating ....

1. What are finitely generated modules?

Finitely generated modules are modules over a ring where every element can be expressed as a finite linear combination of a fixed set of elements. In other words, the elements of a finitely generated module are generated by a finite number of elements.

2. What is the significance of submodules in finitely generated modules?

Submodules in finitely generated modules are important because they allow for the study of smaller, more manageable modules within a larger module. They also help to classify and understand the structure of a finitely generated module.

3. How are submodules related to generators in finitely generated modules?

Submodules are closely related to generators in finitely generated modules. In fact, a submodule is generated by a subset of the generators of the larger module. This means that the elements in the submodule can also be expressed as finite linear combinations of the generators of the larger module.

4. Can finitely generated modules have an infinite number of submodules?

No, finitely generated modules can only have a finite number of submodules. This is because a submodule is generated by a finite set of elements, and there are only a finite number of possible combinations of these elements within the larger module.

5. How are finitely generated modules used in practical applications?

Finitely generated modules have many practical applications, particularly in algebraic structures and linear algebra. They are used in areas such as coding theory, cryptography, and signal processing, where finite structures are commonly encountered. They also have applications in computer science and engineering, where they are used to model and analyze systems with finite resources.

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