Artinian Modules - Bland - Proposition 4.24

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.2: Noetherian and Artinian Modules and need some help to fully understand the proof of part of Proposition 4.2.4 ... ...

Proposition 4.2.4 reads as follows:
View attachment 6126
View attachment 6127I need help to fully understand Part of the proof proving that \(\displaystyle (2) \Longrightarrow (3)\) ...In that part of the proof Bland seems to be assuming that

\(\displaystyle \bigcap_F M_\alpha = N \)

if and only if

\(\displaystyle \bigcap_F (M_\alpha / N ) = 0\)
In other words, if \(\displaystyle F = \{ 1, 2, 3 \}\) then

\(\displaystyle M_1 \cap M_2 \cap M_3\)

if and only if

\(\displaystyle M_1 / N \cap M_2 / N \cap M_3 / N\) But why exactly is this the case ... ...

... ... how do we formally and rigorously demonstrate that this is true ...Hope someone can help ...

Peter
====================================================

Proposition 4.2.4 refers to the (possibly not well known) concept of cogeneration so I am providing Section 4.1 Generating as Cogenerating Classes ... ... as follows ...
View attachment 6128
View attachment 6129
https://www.physicsforums.com/attachments/6130

 

[steenis]

To understand the proof of part $(2) \Rightarrow (3)$ op proposition 4.2.4, you must first understand and prove this, and let’s call this Lemma I:
Let $\{M_i\}_{i \in I}$ be a family of left-R modules and $N \leq M_i$ (submodule) for all $i \in I$, then:

$$\bigcap_{i \in I} (M_i/N) = 0 \Leftrightarrow \bigcap_{i \in I} M_i = N$$

The proof of the $(\Leftarrow)$ direction is as follows.
Let $\bar{x} \in \bigcap_{i \in I} (M_i/N)$, then for all $i \in I$ we have $\bar{x} = m_i + N$ for some $m_i \in M_i$.
Take $i \in I$ and $j \in I, j \neq i$, then $\bar{x} = m_i + N = m_j + N$,
thus $m_i - m_j \in N \subset M_j$, we also have $m_j \in M_j$ so $m_i = (m_i - m_j) + m_j \in M_j$.

Thus for all $j \in I$ we have $m_i \in M_j$, this means $m_i \in \bigcap_{i \in I} M_i = N$ thus $m_i \in N$ for all $i \in I$ and therefore $\bar{x} = 0$.

Can you do the $(\Rightarrow)$ direction ?
When you are ready, we will continue with the proof of part $(2) \Rightarrow (3)$ op proposition 4.2.4 of Bland.

 

[Math Amateur]

To understand the proof of part $(2) \Rightarrow (3)$ op proposition 4.2.4, you must first understand and prove this, and let’s call this Lemma I:
Let $\{M_i\}_{i \in I}$ be a family of left-R modules and $N \leq M_i$ (submodule) for all $i \in I$, then:

$$\bigcap_{i \in I} (M_i/N) = 0 \Leftrightarrow \bigcap_{i \in I} M_i = N$$

The proof of the $(\Leftarrow)$ direction is as follows.
Let $\bar{x} \in \bigcap_{i \in I} (M_i/N)$, then for all $i \in I$ we have $\bar{x} = m_i + N$ for some $m_i \in M_i$.
Take $i \in I$ and $j \in I, j \neq i$, then $\bar{x} = m_i + N = m_j + N$,
thus $m_i - m_j \in N \subset M_j$, we also have $m_j \in M_j$ so $m_i = (m_i - m_j) + m_j \in M_j$.

Thus for all $j \in I$ we have $m_i \in M_j$, this means $m_i \in \bigcap_{i \in I} M_i = N$ thus $m_i \in N$ for all $i \in I$ and therefore $\bar{x} = 0$.

Can you do the $(\Rightarrow)$ direction ?
When you are ready, we will continue with the proof of part $(2) \Rightarrow (3)$ op proposition 4.2.4 of Bland.

Hi Steenis ... thanks for the help ...Will try to show $$\bigcap_{i \in I} (M_i/N) = 0 \Longrightarrow \bigcap_{i \in I} M_i = N$$Assume that \(\displaystyle \bigcap_{i \in I} (M_i/N) = 0 \)Then ...

\(\displaystyle x \in \bigcap_{i \in I} M_i\)

\(\displaystyle \Longrightarrow x \in M_i\) for all \(\displaystyle i \in I\)

\(\displaystyle \Longrightarrow x + N \in M_i / N\) for all \(\displaystyle i \in I \)

\(\displaystyle \Longrightarrow x + N \in \bigcap_{i \in I} (M_i/N)\)

\(\displaystyle \Longrightarrow x + N \in \overline{0}\)

\(\displaystyle \Longrightarrow x \in N\) Now ... if we assume \(\displaystyle x \in N\) then argument above works essentially in reverse so that \(\displaystyle x \in \bigcap_{i \in I} M_i\)Is the above correct ... ?

Peter

 

[steenis]

Yes this is correct, we need this Lemma in the next proposition. I will give you now my proof of part $(2) \Rightarrow (3)$ op proposition 4.2.4., I hope it is correct.

M is a (left) R-module, let $N \leq M$ (submodule). We have to prove that $M/N$ is finitely cogenerated, given that every nonempty collection of submodules of $M$ has a minimal element.
To do this we have to prove that if $\{M_\alpha /N \}_{\alpha \in \Delta}$ is a family of submodules of $M/N$ such that $\bigcap_{\alpha \in \Delta} (M_\alpha/N) = 0$ then there is a finite subset $F \subset \Delta$ such that $\bigcap_{\alpha \in F} (M_\alpha/N) = 0$. (Can you see that we use the Correspondence Theorem for Modules here?)So let $\bigcap_{\alpha \in \Delta} (M_\alpha/N) = 0$ and

$$\mathscr{S} = \{ \bigcap_{\alpha \in \Gamma} M_\alpha \mbox{ } | \mbox{ } \Gamma \subset \Delta \mbox{ finite} \}$$.

By hypothesis $\mathscr{S}$ has a minimal element, say $\bigcap_{\alpha \in F} M_\alpha $ where $F$ is a finite subset of $\Delta$.

Suppose $\bigcap_{\alpha \in F} M_\alpha \neq N$. of course $N \subset \bigcap_{\alpha \in F} M_\alpha $, thus there is an $x \in \bigcap_{\alpha \in F} M_\alpha $ such that $x \notin N = \bigcap_{\alpha \in \Delta} M_\alpha $ (Lemma I).

This means that there is a $\beta \in \Delta$ such that $x \notin M_\beta$.

Then $x \notin \bigcap_{\alpha \in F \cup \{ \beta \}} M_\alpha $ but $x \in \bigcap_{\alpha \in F} M_\alpha $,

thus $\bigcap_{\alpha \in F \cup \{ \beta \}} M_\alpha \neq \bigcap_{\alpha \in F} M_\alpha $,

while $\bigcap_{\alpha \in F \cup \{ \beta \}} M_\alpha \subset \bigcap_{\alpha \in F} M_\alpha $.

This is a contradiction with the minimality of $\bigcap_{\alpha \in F} M_\alpha $ in $\mathscr{S}$ (notice: $F \cup \{ \beta \}$ is finite).

Therefore $\bigcap_{\alpha \in F} M_\alpha = N$ and $\bigcap_{\alpha \in F} (M_\alpha/N) = 0$, by Lemma I, ready.

 

[Math Amateur]

Yes this is correct, we need this Lemma in the next proposition. I will give you now my proof of part $(2) \Rightarrow (3)$ op proposition 4.2.4., I hope it is correct.

M is a (left) R-module, let $N \leq M$ (submodule). We have to prove that $M/N$ is finitely cogenerated, given that every nonempty collection of submodules of $M$ has a minimal element.
To do this we have to prove that if $\{M_\alpha /N \}_{\alpha \in \Delta}$ is a family of submodules of $M/N$ such that $\bigcap_{\alpha \in \Delta} (M_\alpha/N) = 0$ then there is a finite subset $F \subset \Delta$ such that $\bigcap_{\alpha \in F} (M_\alpha/N) = 0$. (Can you see that we use the Correspondence Theorem for Modules here?)So let $\bigcap_{\alpha \in \Delta} (M_\alpha/N) = 0$ and

$$\mathscr{S} = \{ \bigcap_{\alpha \in \Gamma} M_\alpha \mbox{ } | \mbox{ } \Gamma \subset \Delta \mbox{ finite} \}$$.

By hypothesis $\mathscr{S}$ has a minimal element, say $\bigcap_{\alpha \in F} M_\alpha $ where $F$ is a finite subset of $\Delta$.

Suppose $\bigcap_{\alpha \in F} M_\alpha \neq N$. of course $N \subset \bigcap_{\alpha \in F} M_\alpha $, thus there is an $x \in \bigcap_{\alpha \in F} M_\alpha $ such that $x \notin N = \bigcap_{\alpha \in \Delta} M_\alpha $ (Lemma I).

This means that there is a $\beta \in \Delta$ such that $x \notin M_\beta$.

Then $x \notin \bigcap_{\alpha \in F \cup \{ \beta \}} M_\alpha $ but $x \in \bigcap_{\alpha \in F} M_\alpha $,

thus $\bigcap_{\alpha \in F \cup \{ \beta \}} M_\alpha \neq \bigcap_{\alpha \in F} M_\alpha $,

while $\bigcap_{\alpha \in F \cup \{ \beta \}} M_\alpha \subset \bigcap_{\alpha \in F} M_\alpha $.

This is a contradiction with the minimality of $\bigcap_{\alpha \in F} M_\alpha $ in $\mathscr{S}$ (notice: $F \cup \{ \beta \}$ is finite).

Therefore $\bigcap_{\alpha \in F} M_\alpha = N$ and $\bigcap_{\alpha \in F} (M_\alpha/N) = 0$, by Lemma I, ready.

Hi Steenis ... thanks for the help ...

Working through your proof now ...

BUT ... just a minor point ...You write:

" ... ... Suppose $\bigcap_{\alpha \in F} M_\alpha \neq N$. of course $N \subset \bigcap_{\alpha \in F} M_\alpha $, ... ... "Can you explain exactly how/why we now that $N \subset \bigcap_{\alpha \in F} M_\alpha $ ... ?

Peter***EDIT***

Oh! ... maybe \(\displaystyle N \subset \bigcap_{\alpha \in F} M_\alpha\) ... because ...

... ...

... \(\displaystyle \bigcap_{\alpha \in \Delta} (M_\alpha/N) = 0 \Longrightarrow \bigcap_{\alpha \in \Delta} M_\alpha = N\) ... and we have \(\displaystyle F \subset \Delta\) so that \(\displaystyle \bigcap_{\alpha \in \Delta} M_\alpha \subset \bigcap_{\alpha \in F } M_\alpha\) ...... that is \(\displaystyle N \subset \bigcap_{\alpha \in F } M_\alpha\) ...Is that correct?

Peter

 

[Math Amateur]

Yes this is correct, we need this Lemma in the next proposition. I will give you now my proof of part $(2) \Rightarrow (3)$ op proposition 4.2.4., I hope it is correct.

M is a (left) R-module, let $N \leq M$ (submodule). We have to prove that $M/N$ is finitely cogenerated, given that every nonempty collection of submodules of $M$ has a minimal element.
To do this we have to prove that if $\{M_\alpha /N \}_{\alpha \in \Delta}$ is a family of submodules of $M/N$ such that $\bigcap_{\alpha \in \Delta} (M_\alpha/N) = 0$ then there is a finite subset $F \subset \Delta$ such that $\bigcap_{\alpha \in F} (M_\alpha/N) = 0$. (Can you see that we use the Correspondence Theorem for Modules here?)So let $\bigcap_{\alpha \in \Delta} (M_\alpha/N) = 0$ and

$$\mathscr{S} = \{ \bigcap_{\alpha \in \Gamma} M_\alpha \mbox{ } | \mbox{ } \Gamma \subset \Delta \mbox{ finite} \}$$.

By hypothesis $\mathscr{S}$ has a minimal element, say $\bigcap_{\alpha \in F} M_\alpha $ where $F$ is a finite subset of $\Delta$.

Suppose $\bigcap_{\alpha \in F} M_\alpha \neq N$. of course $N \subset \bigcap_{\alpha \in F} M_\alpha $, thus there is an $x \in \bigcap_{\alpha \in F} M_\alpha $ such that $x \notin N = \bigcap_{\alpha \in \Delta} M_\alpha $ (Lemma I).

This means that there is a $\beta \in \Delta$ such that $x \notin M_\beta$.

Then $x \notin \bigcap_{\alpha \in F \cup \{ \beta \}} M_\alpha $ but $x \in \bigcap_{\alpha \in F} M_\alpha $,

thus $\bigcap_{\alpha \in F \cup \{ \beta \}} M_\alpha \neq \bigcap_{\alpha \in F} M_\alpha $,

while $\bigcap_{\alpha \in F \cup \{ \beta \}} M_\alpha \subset \bigcap_{\alpha \in F} M_\alpha $.

This is a contradiction with the minimality of $\bigcap_{\alpha \in F} M_\alpha $ in $\mathscr{S}$ (notice: $F \cup \{ \beta \}$ is finite).

Therefore $\bigcap_{\alpha \in F} M_\alpha = N$ and $\bigcap_{\alpha \in F} (M_\alpha/N) = 0$, by Lemma I, ready.

Thanks again Steenis ... can now follow your proof ...

Only ... not sure where the Correspondence Theorem was used in the proof ... :( ...

Peter

 

[steenis]

Q1: You are working with a family $\{M_\alpha /N \}_{\alpha \in \Delta}$ of submodules of $M/N$. It is not mentioned explicitly, sorry, but implicitly this means that $N \leq M_\alpha$ (submodule) for each $\alpha \in \Delta$, otherwise $M_\alpha /N$ has no meaning.

Q2: To prove that $M/N$ is finitely cogenerated, you have to prove that if $\{K_\alpha \}_{\alpha \in \Delta}$ is a family of submodules of $M/N$ such that $\bigcap_{\alpha \in \Delta} K_\alpha = 0$ then there is a finite subset $F \subset \Delta$ such that $\bigcap_{\alpha \in F} K_\alpha = 0$.
By the Correspondenc Theorem of Modules, you know that a submodule $K_\alpha$ of $M/N$ has the form $M_\alpha /N$ where $M_\alpha$ is a submodule of $M$.
Therefore you have to prove that if $\{M_\alpha /N \}_{\alpha \in \Delta}$ is a family of submodules of $M/N$ such that $\bigcap_{\alpha \in \Delta} (M_\alpha/N) = 0$ then there is a finite subset $F \subset \Delta$ such that $\bigcap_{\alpha \in F} (M_\alpha/N) = 0$.

 

[Math Amateur]

Q1: You are working with a family $\{M_\alpha /N \}_{\alpha \in \Delta}$ of submodules of $M/N$. It is not mentioned explicitly, sorry, but implicitly this means that $N \leq M_\alpha$ (submodule) for each $\alpha \in \Delta$, otherwise $M_\alpha /N$ has no meaning.

Q2: To prove that $M/N$ is finitely cogenerated, you have to prove that if $\{K_\alpha \}_{\alpha \in \Delta}$ is a family of submodules of $M/N$ such that $\bigcap_{\alpha \in \Delta} K_\alpha = 0$ then there is a finite subset $F \subset \Delta$ such that $\bigcap_{\alpha \in F} K_\alpha = 0$.
By the Correspondenc Theorem of Modules, you know that a submodule $K_\alpha$ of $M/N$ has the form $M_\alpha /N$ where $M_\alpha$ is a submodule of $M$.
Therefore you have to prove that if $\{M_\alpha /N \}_{\alpha \in \Delta}$ is a family of submodules of $M/N$ such that $\bigcap_{\alpha \in \Delta} (M_\alpha/N) = 0$ then there is a finite subset $F \subset \Delta$ such that $\bigcap_{\alpha \in F} (M_\alpha/N) = 0$.

Hi Steenis ...

Just a note to say that I am revisiting Bland's section on Noetherian and Artinian Modules (I have only worked on two of the theorems anyway!) ...

I found your help in the above post essential to re-establish my understanding of the theorem ... so thanks ...

Possibly more questions coming on Artinian modules as I work further ...

Peter