Homomorphic Images of Free Modules .... Bland, Proposition 2.2.6 .... ....

[Math Amateur]

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.2 Free Modules ... ...

I need help with some aspects of the proof of Proposition 2.2.6 ...

Proposition 2.2.6 and its proof read as follows:
View attachment 8221
View attachment 8222
Near the end of Bland's proof we read the following:

" ... ... If $$x \in M$$, then $$x$$ can be written as $$x = \sum_{ \Delta } x_\alpha a_\alpha$$, where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$. It follows that $$( a_\alpha ) \in R_{ ( \Delta ) }$$, so $$f (( a_\alpha )) = \sum_{ \Delta } x_\alpha a_\alpha$$ and $$f$$ is an epimorphism. ... ... "My questions are as follows:Question 1

Why/how exactly does it follow from $$x \in M$$ and $$x = \sum_{ \Delta } x_\alpha a_\alpha$$ that $$( a_\alpha ) \in R_{ ( \Delta ) }$$ ... ?

Question 2

In the above quote Bland writes $$f (( a_\alpha )) = \sum_{ \Delta } x_\alpha a_\alpha$$ ... but what is going on ... ? At the start of the proof he defined $$f$$ this way ... but if he is not relying on this definition how does he calculate/formulate $$f$$ ...Question 3

Why/how exactly is $$f$$ an epimorphism ...Help will be appreciated ...

Peter

 

[steenis]

Read the proof this way:

$M$ is generated by $\{x_\alpha\}_\Delta$ (every module has a set of generators).

Define $f:R^{(\Delta)} = \bigoplus_\Delta R_\alpha \longrightarrow M$ (where $R_\alpha = R$) by

$f((a_\alpha)_\Delta) = \Sigma_\Delta x_\alpha a_\alpha $ for $(a_\alpha)_\Delta \in R^{(\Delta)}$ ($a_\alpha = 0$ for allmost all $\alpha \in \Delta$).

$f$ is $R$-linear (follow the proof in the book)

Now prove that $f$ is an epimorpism:

Take $x \in M$ then $x = \Sigma_\Delta x_\alpha, r_\alpha$ where $r_\alpha \in R$ (and $r_\alpha = 0$ for allmost all $\alpha \in \Delta$).

Obviously $(r_\alpha)_\Delta \in R^{(\Delta)} = \bigoplus_\Delta R_\alpha$ and it follows that

$f((r_\alpha)_\Delta) = \Sigma_\Delta x_\alpha r_\alpha = x$.

Therefore $f$ is an epimorpism.

That answers Q2 and Q3, you can anser Q1 yourself.