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 $$(2) \Longrightarrow (3)$$ ...In that part of the proof Bland seems to be assuming that

$$\bigcap_F M_\alpha = N $$

if and only if

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

$$M_1 \cap M_2 \cap M_3$$

if and only if

$$M_1 / N \cap M_2 / N \cap M_3 / N$$ 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 ...
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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 $$\bigcap_{i \in I} (M_i/N) = 0 $$Then ...

$$x \in \bigcap_{i \in I} M_i$$

$$\Longrightarrow x \in M_i$$ for all $$i \in I$$

$$\Longrightarrow x + N \in M_i / N$$ for all $$i \in I $$

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

$$\Longrightarrow x + N \in \overline{0}$$

$$\Longrightarrow x \in N$$ Now ... if we assume $$x \in N$$ then argument above works essentially in reverse so that $$x \in \bigcap_{i \in I} M_i$$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 $$N \subset \bigcap_{\alpha \in F} M_\alpha$$ ... because ...

... ...

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