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

[Math Amateur]

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 $$N$$ and $$M/N$$ are Noetherian, and let $$M_0$$ be a submodule of $$M$$. Then $$M_0 \cap N$$ and $$M_0 / ( M_0 \cap N ) \cong (M_0 + N) / N$$ are both finitely generated, so $$M_0$$ 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 $$M$$, namely $$M_0$$, is finitely generated ... ... and this means that M is Noetherian ... (Beachy, in his Proposition 2.4.3 has shown that every submodule of $$M$$ being finitely generated is equivalent to M being Noetherian ... ... )BUT ... I do not see how it follows in the above that ... ... $$M_0 \cap N$$ and $$M_0 / ( M_0 \cap N ) \cong (M_0 + N) / N$$ are both finitely generated ... ... AND ... exactly why it then follows that $$M_0$$ is finitely generated ... ...
Hope someone can help ...

Peter

 

[caffeinemachine]

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 $$N$$ and $$M/N$$ are Noetherian, and let $$M_0$$ be a submodule of $$M$$. Then $$M_0 \cap N$$ and $$M_0 / ( M_0 \cap N ) \cong (M_0 + N) / N$$ are both finitely generated, so $$M_0$$ 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 $$M$$, namely $$M_0$$, is finitely generated ... ... and this means that M is Noetherian ... (Beachy, in his Proposition 2.4.3 has shown that every submodule of $$M$$ being finitely generated is equivalent to M being Noetherian ... ... )BUT ... I do not see how it follows in the above that ... ... $$M_0 \cap N$$ and $$M_0 / ( M_0 \cap N ) \cong (M_0 + N) / N$$ are both finitely generated ... ... AND ... exactly why it then follows that $$M_0$$ 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?

 

[Math Amateur]

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

 

[caffeinemachine]

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?

 

[Math Amateur]

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