MHB Noetherian Modules - Bland - Proposition 4.2.3 - (3) => (1)

[Math Amateur]

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

I am trying to understand Chapter 4, Section 4.2 on Noetherian and Artinian modules and need help with the proof of $$ (3) \Longrightarrow (1) $$ in Proposition 4.2.3.

Proposition 4.2.3 and its proof read as follows:

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The first line of the proof of $$ (3) \Longrightarrow (1) $$ reads as follows:

"If $$ M_1 \subseteq M_2 \subseteq M_3 \subseteq \ ... \ $$ is an ascending chain of submodules of M then $$\bigcup_{ i = 1 }^{ \infty } M_i $$ is a finitely generated module of $$M$$. ... ... "

My question is as follows:

How do we know that if

... ... $$ M_1 \subseteq M_2 \subseteq M_3 \subseteq \ ... \ $$ is an ascending chain of submodules of $$M$$

then

... ... $$\bigcup_{ i = 1 }^{ \infty }$$ is a finitely generated module of $$M$$ ...

That is ... why exactly does this follow?

Peter

 

[Fallen Angel]

Hi Peter,

$\cup_{i=0}^{\infty}M_{i}$ is a submodule of $M$.

The point 3) states that every submodule of $M$ is finitely generated.

 

[Math Amateur]

Hi Peter,

$\cup_{i=0}^{\infty}M_{i}$ is a submodule of $M$.

The point 3) states that every submodule of $M$ is finitely generated.

Thanks Fallen Angel ... ... so the union of submodules is a submodule ... ... including the union of an infinite set of submodules?

Thanks for the help!

Peter***EDIT***

PROBLEM! I now believe that the union of a set of modules is not necessarily a module!

 

[Euge]

PROBLEM! I now believe that the union of a set of modules is not necessarily a module!

You're right Peter, but the union of an ascending chain of submodules of an $R$-module $M$ is a submodule of $M$.

 

[Math Amateur]

You're right Peter, but the union of an ascending chain of submodules of an $R$-module $M$ is a submodule of $M$.

Thanks Euge ... ...

But ... why/how does the fact that the submodules are in an ascending chain, make a difference?

Peter

 

[Euge]

Thanks Euge ... ...

But ... why/how does the fact that the submodules are in an ascending chain, make a difference?

Peter

Sorry, but I don't understand the question. You cannot guarantee closure under addition holds for $\cup M_i$ if the sequence $M_i$ is not increasing. For example, the subset $A := 2\Bbb Z \cup 3\Bbb Z$ of $\Bbb Z$ is not closed under addition since $2, 3 \in A$ but $2 + 3 = 5 \notin A$. Therefore, $2\Bbb Z \cup 3\Bbb Z$ is not a submodule of $\Bbb Z$. For an infinite sequence counterexample, let $p_i$ represent the $i$-th prime, and let $B = \bigcup_{i = 2}^\infty p_i\Bbb Z$. Then $3, 5 \in B$, but $5 + 3 = 8 \notin B$ since $8$ is not divisible by any odd prime. So $B$ is not a submodule of $\Bbb Z$.

 

[Math Amateur]

Sorry, but I don't understand the question. You cannot guarantee closure under addition holds for $\cup M_i$ if the sequence $M_i$ is not increasing. For example, the subset $A := 2\Bbb Z \cup 3\Bbb Z$ of $\Bbb Z$ is not closed under addition since $2, 3 \in A$ but $2 + 3 = 5 \notin A$. Therefore, $2\Bbb Z \cup 3\Bbb Z$ is not a submodule of $\Bbb Z$. For an infinite sequence counterexample, let $p_i$ represent the $i$-th prime, and let $B = \bigcup_{i = 2}^\infty p_i\Bbb Z$. Then $3, 5 \in B$, but $5 + 3 = 8 \notin B$ since $8$ is not divisible by any odd prime. So $B$ is not a submodule of $\Bbb Z$.

Thanks Euge ... I think you have answered my question ...

I was asking if you could explain further exactly why the union of an ascending chain of submodules of an R-module M is a submodule of M ... but I think you have done this ... actually, still reflecting on what you have written ...

It seems that you are saying that the union of a set of submodules is often not a submodule because addition is not closed in the union ... ... BUT ... when we are dealing with an ascending chain, then closure under addition is assured ...

Thanks again for your help ... ..

Peter

 

[Fallen Angel]

Yes it is, the key of the question is as follows, given $a,b \in \displaystyle\cup_{i=0}^{\infty}M_{i}$ we got that there exists $n,m$ with $a\in M_{n}$, $b\in M_{m}$, without loss of generality we can assume $n\geq m$.

Then, the chain condition assures you that $a,b\in M_{n}$ since $M_{m}\subseteq M_{n}$ and $M_{n}$ is a module :D