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 \(\displaystyle (4) \Longrightarrow (1)\) in Proposition 4.1.1.

Proposition 4.1.1 and its proof read as follows:

https://www.physicsforums.com/attachments/3656
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In the proof of \(\displaystyle (4) \Longrightarrow (1)\) in the text above, Bland writes:" ... ... Let \(\displaystyle \{ M_\alpha \}_\Delta\) be a family of submodules of \(\displaystyle M\) that spans \(\displaystyle M\). If \(\displaystyle X = \{ x_1, x_2, \ ... \ ... \ , x_n \}\) is a finite set of generators of \(\displaystyle M\), then \(\displaystyle M = \sum_{ i = 1}^n x_i R = \sum_\Delta M_\alpha\).

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

Why, exactly, does the statement:

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

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