Noetherian Modules and Submodules - J A Beachy, Proposition 2.4.5 .... ....

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In summary, in the conversation, Peter is seeking help with the proof of Proposition 2.4.5 in J A Beachy's book, Introductory Lectures on Rings and Modules. The proposition states that if a submodule and its quotient are both finitely generated, then the original module is also finitely generated. Peter is unsure about a particular step in the proof involving arbitrary submodules and is seeking clarification on how it shows that the submodule is finitely generated. Caffeinemachine provides a clear explanation of the step in question.
  • #1
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I am reading J A Beachy's Book, Introductory Lectures on Rings and Modules"... ...

I am currently focused on Chapter 2: Modules ... and in particular Section 2.4: Chain Conditions ...

I need help with the proof of Proposition 2.4.5 ...Proposition 2.4.5 reads as follows:
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In the above text by Beachy ... in the proof of part (a) ... we read the following:"... ... ... Conversely, assume that \(\displaystyle N\) and \(\displaystyle M/N\) are Noetherian, and let \(\displaystyle M_0\) be a submodule of \(\displaystyle M\). Then \(\displaystyle M_0 \cap N\) and \(\displaystyle M_0 / ( M_0 \cap N ) \cong (M_0 + N) / N\) are both finitely generated, so \(\displaystyle M_0\) is finitely generated ... ... ... "

I am very unsure of this part of the proof ... but overall Beachy seems to be trying to prove that an arbitrary submodule of \(\displaystyle M\), namely \(\displaystyle M_0\), is finitely generated ... ... and this means that M is Noetherian ... (Beachy, in his Proposition 2.4.3 has shown that every submodule of \(\displaystyle M\) being finitely generated is equivalent to M being Noetherian ... ... )BUT ... I do not see how it follows in the above that ... ... \(\displaystyle M_0 \cap N\) and \(\displaystyle M_0 / ( M_0 \cap N ) \cong (M_0 + N) / N\) are both finitely generated ... ... AND ... exactly why it then follows that \(\displaystyle M_0\) is finitely generated ... ...
Hope someone can help ...

Peter
 
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  • #2
Peter said:
I am reading J A Beachy's Book, Introductory Lectures on Rings and Modules"... ...

I am currently focused on Chapter 2: Modules ... and in particular Section 2.4: Chain Conditions ...

I need help with the proof of Proposition 2.4.5 ...Proposition 2.4.5 reads as follows:
In the above text by Beachy ... in the proof of part (a) ... we read the following:"... ... ... Conversely, assume that \(\displaystyle N\) and \(\displaystyle M/N\) are Noetherian, and let \(\displaystyle M_0\) be a submodule of \(\displaystyle M\). Then \(\displaystyle M_0 \cap N\) and \(\displaystyle M_0 / ( M_0 \cap N ) \cong (M_0 + N) / N\) are both finitely generated, so \(\displaystyle M_0\) is finitely generated ... ... ... "

I am very unsure of this part of the proof ... but overall Beachy seems to be trying to prove that an arbitrary submodule of \(\displaystyle M\), namely \(\displaystyle M_0\), is finitely generated ... ... and this means that M is Noetherian ... (Beachy, in his Proposition 2.4.3 has shown that every submodule of \(\displaystyle M\) being finitely generated is equivalent to M being Noetherian ... ... )BUT ... I do not see how it follows in the above that ... ... \(\displaystyle M_0 \cap N\) and \(\displaystyle M_0 / ( M_0 \cap N ) \cong (M_0 + N) / N\) are both finitely generated ... ... AND ... exactly why it then follows that \(\displaystyle M_0\) is finitely generated ... ...
Hope someone can help ...

Peter
the following is begin used:

Let $M$ be an $R$-module and $S$ be a submodule of $M$. Suppose $S$ and $M/S$ are both finitely generated. Then $M$ to is finitely generated.

This is easy to show. Let $\{s_1, \ldots, s_k\}$ be a generating set of $S$ and $\{\bar m_1, \ldots, \bar m_l\}$ be a generating set for $M/S$. Then $\{s_1, \ldots, s_k, m_1, \ldots, m_l\}$ is a generating set for $M$ (why?)

Can you see how this shows that $M_0$ is finitely generated?
 
  • #3
caffeinemachine said:
the following is begin used:

Let $M$ be an $R$-module and $S$ be a submodule of $M$. Suppose $S$ and $M/S$ are both finitely generated. Then $M$ to is finitely generated.

This is easy to show. Let $\{s_1, \ldots, s_k\}$ be a generating set of $S$ and $\{\bar m_1, \ldots, \bar m_l\}$ be a generating set for $M/S$. Then $\{s_1, \ldots, s_k, m_1, \ldots, m_l\}$ is a generating set for $M$ (why?)

Can you see how this shows that $M_0$ is finitely generated?
Hi caffeinemachine ... thanks fpr the reply ...

You write:

"... ... This is easy to show. Let $\{s_1, \ldots, s_k\}$ be a generating set of $S$ and $\{\bar m_1, \ldots, \bar m_l\}$ be a generating set for $M/S$. Then $\{s_1, \ldots, s_k, m_1, \ldots, m_l\}$ is a generating set for $M$ (why?) ... ... I cannot see why this follows ... can you help further ...

Peter
 
  • #4
Peter said:
Hi caffeinemachine ... thanks fpr the reply ...

You write:

"... ... This is easy to show. Let $\{s_1, \ldots, s_k\}$ be a generating set of $S$ and $\{\bar m_1, \ldots, \bar m_l\}$ be a generating set for $M/S$. Then $\{s_1, \ldots, s_k, m_1, \ldots, m_l\}$ is a generating set for $M$ (why?) ... ... I cannot see why this follows ... can you help further ...

Peter

Let $m$ be arbitrarily chosen in $M$. Then there exist $r_1, \dots, r_l\in R$ such that $\bar m= r_1\bar m_1+\cdots +r_l\bar m_l$. Thus $m-(r_1m_1+\cdots+r_lm_l)\in S$. So there are $a_1, \ldots, a_k\in S$ such that
$$m-(r_1m_1+\cdots+r_lm_l)= a_1s_1+\cdots +a_ks_k$$
This gives
$$m = r_1m_1+\cdots+r_lm_l + a_1s_1+\cdots +a_ks_k$$
Is this clear now?
 
  • #5
caffeinemachine said:
Let $m$ be arbitrarily chosen in $M$. Then there exist $r_1, \dots, r_l\in R$ such that $\bar m= r_1\bar m_1+\cdots +r_l\bar m_l$. Thus $m-(r_1m_1+\cdots+r_lm_l)\in S$. So there are $a_1, \ldots, a_k\in S$ such that
$$m-(r_1m_1+\cdots+r_lm_l)= a_1s_1+\cdots +a_ks_k$$
This gives
$$m = r_1m_1+\cdots+r_lm_l + a_1s_1+\cdots +a_ks_k$$
Is this clear now?

Yes, thanks caffeinemachine ... appreciate your help ...

Peter
 

FAQ: Noetherian Modules and Submodules - J A Beachy, Proposition 2.4.5 .... ....

What is a Noetherian module?

A Noetherian module is a module that satisfies the ascending chain condition. This means that for any sequence of submodules M1 ⊆ M2 ⊆ M3 ⊆ ... of the module M, there exists some integer N such that Mn = MN for all n ≥ N.

What is a Noetherian submodule?

A Noetherian submodule is a submodule that is also a Noetherian module. This means that it satisfies the ascending chain condition for its own submodules.

How is Proposition 2.4.5 used in Noetherian modules and submodules?

Proposition 2.4.5 states that if M is a Noetherian module and N is a submodule of M, then N is also a Noetherian module. This can be used to simplify proofs and arguments involving Noetherian modules and submodules.

Can a submodule of a Noetherian module fail to be Noetherian?

Yes, it is possible for a submodule of a Noetherian module to not be Noetherian. This can occur if the submodule does not satisfy the ascending chain condition.

In what fields of mathematics is the concept of Noetherian modules and submodules important?

The concept of Noetherian modules and submodules is important in abstract algebra, specifically in the study of modules over a ring. It is also used in commutative algebra and algebraic geometry.

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