MHB Bland - Proposition 4.1.1 - (4) => (1)

[Math Amateur]

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

I am trying to understand Chapter 4, Section 4.1 on generating and cogenerating classes and need help with the proof of $$(4) \Longrightarrow (1)$$ in Proposition 4.1.1.

Proposition 4.1.1 and its proof read as follows:

https://www.physicsforums.com/attachments/3656
View attachment 3657

In the proof of $$(4) \Longrightarrow (1)$$ in the text above, Bland writes:" ... ... Let $$\{ M_\alpha \}_\Delta$$ be a family of submodules of $$M$$ that spans $$M$$. If $$X = \{ x_1, x_2, \ ... \ ... \ , x_n \}$$ is a finite set of generators of $$M$$, then $$M = \sum_{ i = 1}^n x_i R = \sum_\Delta M_\alpha$$.

Thus, for each $$i$$, there is a finite set $$F_i \subseteq \Delta$$ such that $$x_i \in \sum_{F_i} M_\alpha$$. ... ... "My question is as follows:

Why, exactly, does the statement:

... ... for each $$i$$, there is a finite set $$F_i \subseteq \Delta$$ such that $$x_i \in \sum_{F_i} M_\alpha$$

follow from the two previous statements?Hope someone can help ... ...

Peter

 

[Fallen Angel]

Hi Peter,

The sum of any family of modules is a finitely nonzero sum.

 

[Math Amateur]

Hi Peter,

The sum of any family of modules is a finitely nonzero sum.

Hi Fallen Angel ... thanks for the help in this matter ... ... however, I am still finding this difficult to follow ... are you able to be more explicit and explain further ...

Peter

 

[Fallen Angel]

Hi Peter,

You got the equality $\displaystyle\sum_{i=1}^{n}x_{i}R=\displaystyle\sum_{\Delta}M_{\alpha}$.

So $x_{i}\in \displaystyle\sum_{\Delta}M_{\alpha}$.

Then, in principle $x_{i}=\displaystyle\sum_{\Delta}m_{\alpha}$ with $m_{\alpha}\in M_{\alpha}$.

But this sum is finitely non zero, i.e. $m_{\alpha}=0$ for almost every $\alpha \in \Delta$

So $x_{i}=\displaystyle\sum_{F_{i}\subset \Delta}m_{\alpha}$ where $F_{i}$ is finite.($F_{i}=\{\alpha \in \Delta \ : \ m_{\alpha}\neq 0\}$)

 

[Math Amateur]

Hi Peter,

You got the equality $\displaystyle\sum_{i=1}^{n}x_{i}R=\displaystyle\sum_{\Delta}M_{\alpha}$.

So $x_{i}\in \displaystyle\sum_{\Delta}M_{\alpha}$.

Then, in principle $x_{i}=\displaystyle\sum_{\Delta}m_{\alpha}$ with $m_{\alpha}\in M_{\alpha}$.

But this sum is finitely non zero, i.e. $m_{\alpha}=0$ for almost every $\alpha \in \Delta$

So $x_{i}=\displaystyle\sum_{F_{i}\subset \Delta}m_{\alpha}$ where $F_{i}$ is finite.($F_{i}=\{\alpha \in \Delta \ : \ m_{\alpha}\neq 0\}$)

Thanks for the help, Fallen Angel ... I can now understand the logic ...

Thanks again,

Peter