MHB Free Modules, Bases and Direct Sums/Products - Bland, Proposition 2.2.3

[Math Amateur]

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

Currently I am focused on Section 2.2 Free Modules ... ...

I need some help in order to fully understand the proof of the equivalence of (1) and (3) in Proposition 2.2.3 ...

Proposition 2.2.3 and its proof reads as follows:View attachment 5607Bland omits the proof of the equivalence of (1) and (3) ...

Can someone please help me to get started on a rigorous proof of the equivalence of (1) and (3) ... especially covering the case where $$\Delta$$ is an unaccountably infinite set ...

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

To help MHB members reading the above post with Bland's notation I am providing the following notes from Bland's text:View attachment 5608
https://www.physicsforums.com/attachments/5609

View attachment 5610View attachment 5611
View attachment 5612

 

[steenis]

$\Delta$ may be finite or infinite, whatever kind of infiniteness.

Using prop 2.1.10 of Bland, $M=\bigoplus _\Delta M_\alpha$ with $M _\alpha \leq M$, means that each $x\in M$ can be written uniquely as a finite sum $x=\Sigma _\Delta x _\alpha$, where $x _\alpha \in M$ for all $\alpha \in \Delta$ (i.e. only finitely many $x _\alpha$'s are nonzero).

(3) => (1) Given $\{x _\alpha \} _\Delta \subset M$
$M=\bigoplus _\Delta x _\alpha R$ means that each element $x\in M$ can be written uniquely as a finite sum $x=\Sigma _\Delta x _\alpha r _\alpha$ with $r _\alpha \in R$, where only finitely many $r _\alpha$'s are nonzero. This means that $\{x _\alpha \} _\Delta$ is a basis of $M$ by definition.

(1) => (3) Given $\{x _\alpha \} _\Delta$, with $x _\alpha \in M$, is a basis of $M$. So each element $x\in M$ can be written uniquely as a finite sum $x=\Sigma _\Delta x _\alpha r _\alpha$ with $r _\alpha \in R$, where only finitely many $r _\alpha$'s are nonzero. This means that $M=\bigoplus _\Delta x _\alpha R$ by prop 2.1.10.

 

[Math Amateur]

$\Delta$ may be finite or infinite, whatever kind of infiniteness.

Using prop 2.1.10 of Bland, $M=\bigoplus _\Delta M_\alpha$ with $M _\alpha \leq M$, means that each $x\in M$ can be written uniquely as a finite sum $x=\Sigma _\Delta x _\alpha$, where $x _\alpha \in M$ for all $\alpha \in \Delta$ (i.e. only finitely many $x _\alpha$'s are nonzero).

(3) => (1) Given $\{x _\alpha \} _\Delta \subset M$
$M=\bigoplus _\Delta x _\alpha R$ means that each element $x\in M$ can be written uniquely as a finite sum $x=\Sigma _\Delta x _\alpha r _\alpha$ with $r _\alpha \in R$, where only finitely many $r _\alpha$'s are nonzero. This means that $\{x _\alpha \} _\Delta$ is a basis of $M$ by definition.

(1) => (3) Given $\{x _\alpha \} _\Delta$, with $x _\alpha \in M$, is a basis of $M$. So each element $x\in M$ can be written uniquely as a finite sum $x=\Sigma _\Delta x _\alpha r _\alpha$ with $r _\alpha \in R$, where only finitely many $r _\alpha$'s are nonzero. This means that $M=\bigoplus _\Delta x _\alpha R$ by prop 2.1.10.

Hi Steenis,

Just reworking Section 2.2 of Bland ... and in particular Proposition 2.2.3 ...

In the above post you write:

"... ... (3) => (1) Given $\{x _\alpha \} _\Delta \subset M$
$M=\bigoplus _\Delta x _\alpha R$ means that each element $x\in M$ can be written uniquely as a finite sum $x=\Sigma _\Delta x _\alpha r _\alpha$ with $r _\alpha \in R$, where only finitely many $r _\alpha$'s are nonzero. This means that $\{x _\alpha \} _\Delta$ is a basis of $M$ by definition. ... ... "This argument seems to imply that the basis $\{x _\alpha \} _\Delta$ could well be infinite ... is that right?Hope you can help ...

Peter

 

[Math Amateur]

Hi Steenis,

Just reworking Section 2.2 of Bland ... and in particular Proposition 2.2.3 ...

In the above post you write:

"... ... (3) => (1) Given $\{x _\alpha \} _\Delta \subset M$
$M=\bigoplus _\Delta x _\alpha R$ means that each element $x\in M$ can be written uniquely as a finite sum $x=\Sigma _\Delta x _\alpha r _\alpha$ with $r _\alpha \in R$, where only finitely many $r _\alpha$'s are nonzero. This means that $\{x _\alpha \} _\Delta$ is a basis of $M$ by definition. ... ... "This argument seems to imply that the basis $\{x _\alpha \} _\Delta$ could well be infinite ... is that right?Hope you can help ...

Peter

My apologies steenis ... I did not read your previous post carefully enough ... and somehow (I don't quite know how ? ) missed your statement :

"... ... $\Delta$ may be finite or infinite, whatever kind of infiniteness. ... ... "

... careless of me ... sorry ...

But thanks again for your post ... your posts are enabling me to understand Bland's excellent book ...

Peter

 

[steenis]

No, I don't mind.

The basis may be infinite, but remember, the linear combinations $x=\Sigma _\Delta x _\alpha r _\alpha$
in which $x _\alpha$ are the basis elements, must be finite sums, thus only finitely many $r _\alpha \in R$ are nonzero.

 

[Math Amateur]

No, I don't mind.

The basis may be infinite, but remember, the linear combinations $x=\Sigma _\Delta x _\alpha r _\alpha$
in which $x _\alpha$ are the basis elements, must be finite sums, thus only finitely many $r _\alpha \in R$ are nonzero.

THanks for your generous help, steenis...

Peter