Free Modules - Another problem regarding Bland Proposition 2.2.3

[Math Amateur]

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

I am trying to understand Section 2.2 on free modules and need help with the proof of Proposition 2.2.3 - that is some further help ... (for my first problem with the proof see my post "http://mathhelpboards.com/linear-abstract-algebra-14/free-modules-bland-proposition-2-2-3-a-13179.html")

Proposition 2.2.3 and its proof read as follows:View attachment 3531The first line of Bland's proof reads as follows:

"Proof: \(\displaystyle (1) \Longrightarrow (2)\). Since \(\displaystyle \{ x_\alpha \}_\Delta\) is a set of generators of \(\displaystyle M \), it is certainly the case that each \(\displaystyle x \in M\) can be expressed as \(\displaystyle x = \sum_\Delta x_\alpha a_\alpha\) where \(\displaystyle a_\alpha = 0\) for almost all \(\displaystyle \alpha \in \Delta\). ... ... "Now, by definition of a basis, \(\displaystyle \{ x_\alpha \}_\Delta\) being a basis for M certainly implies that each \(\displaystyle x \in M\) can be expressed as a sum \(\displaystyle x = \sum_\Delta x_\alpha a_\alpha\) ... ...

... ... BUT ... ... how do we know that each \(\displaystyle x \in M\) can be expressed as \(\displaystyle x = \sum_\Delta x_\alpha a_\alpha\) where \(\displaystyle a_\alpha = 0\) for almost all \(\displaystyle \alpha \in \Delta\) ... ...

... ... specifically ... ... what/where in the definition of a basis is the justification for adding the condition "where \(\displaystyle a_\alpha = 0\) for almost all \(\displaystyle \alpha \in \Delta\)"?

Hope someone can help,

Peter

 

[caffeinemachine]

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

I am trying to understand Section 2.2 on free modules and need help with the proof of Proposition 2.2.3 - that is some further help ... (for my first problem with the proof see my post "http://mathhelpboards.com/linear-abstract-algebra-14/free-modules-bland-proposition-2-2-3-a-13179.html")

Proposition 2.2.3 and its proof read as follows:View attachment 3531The first line of Bland's proof reads as follows:

"Proof: \(\displaystyle (1) \Longrightarrow (2)\). Since \(\displaystyle \{ x_\alpha \}_\Delta\) is a set of generators of \(\displaystyle M \), it is certainly the case that each \(\displaystyle x \in M\) can be expressed as \(\displaystyle x = \sum_\Delta x_\alpha a_\alpha\) where \(\displaystyle a_\alpha = 0\) for almost all \(\displaystyle \alpha \in \Delta\). ... ... "Now, by definition of a basis, \(\displaystyle \{ x_\alpha \}_\Delta\) being a basis for M certainly implies that each \(\displaystyle x \in M\) can be expressed as a sum \(\displaystyle x = \sum_\Delta x_\alpha a_\alpha\) ... ...

... ... BUT ... ... how do we know that each \(\displaystyle x \in M\) can be expressed as \(\displaystyle x = \sum_\Delta x_\alpha a_\alpha\) where \(\displaystyle a_\alpha = 0\) for almost all \(\displaystyle \alpha \in \Delta\) ... ...

... ... specifically ... ... what/where in the definition of a basis is the justification for adding the condition "where \(\displaystyle a_\alpha = 0\) for almost all \(\displaystyle \alpha \in \Delta\)"?

Hope someone can help,

Peter

Let $M$ be any $R$-module and $S$ be a subset of $M$.

Then the submodule of $M$ generated by $S$ is defined as

$$\{ \sum_{\text{finite}}a_is_i:a_i\in R, s_i\in S \}$$

Summing up infinitely many elements of a module doesn't make any sense.

Now coming to your problem.
If $X=\{x_\alpha\}_{\alpha\in J}$ is a basis of $M$, then $X$ generates $M$. Use the definition of "generation" as discussed above.

I hope this helped.