Free Modules - Another issue regarding Bland Proposition 2.2.4

[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 Corollary 2.2.4 - 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-corollary-2-2-4-issue-regarding-finite-generation-modules-13196.html" )Corollary 2.2.4 and its proof read as follows:View attachment 3538
View attachment 3539The last sentence in Bland's proof reads as follows:

" ... ... If \(\displaystyle f \ : \ R^{ ( \Delta ) } \ \rightarrow \ F \) is an isomorphism and \(\displaystyle f ( e_\alpha) = x_\alpha\) then it follows that \(\displaystyle \{ x_\alpha \}_\Delta\) is a basis for \(\displaystyle F\). ... ... "Although the above statement seems plausible I am unable to frame a formal and rigorous proof of the statement ...

Can someone please show me exactly why \(\displaystyle f \ : \ R^{ ( \Delta ) } \ \rightarrow \ F \) is an isomorphism and \(\displaystyle f ( e_\alpha) = x_\alpha\) implies that \(\displaystyle \{ x_\alpha \}_\Delta\) is a basis for \(\displaystyle F\)?

Help will be appreciated ...

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 Corollary 2.2.4 - 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-corollary-2-2-4-issue-regarding-finite-generation-modules-13196.html" )Corollary 2.2.4 and its proof read as follows:View attachment 3538
View attachment 3539The last sentence in Bland's proof reads as follows:

" ... ... If \(\displaystyle f \ : \ R^{ ( \Delta ) } \ \rightarrow \ F \) is an isomorphism and \(\displaystyle f ( e_\alpha) = x_\alpha\) then it follows that \(\displaystyle \{ x_\alpha \}_\Delta\) is a basis for \(\displaystyle F\). ... ... "Although the above statement seems plausible I am unable to frame a formal and rigorous proof of the statement ...

Can someone please show me exactly why \(\displaystyle f \ : \ R^{ ( \Delta ) } \ \rightarrow \ F \) is an isomorphism and \(\displaystyle f ( e_\alpha) = x_\alpha\) implies that \(\displaystyle \{ x_\alpha \}_\Delta\) is a basis for \(\displaystyle F\)?

Help will be appreciated ...

Peter

First we verify that $\{x_\alpha\}_{\Delta}$ generates $F$.
Let $m\in F$. Note that $f$ is surjective.
Therefore $f(r)=m$ for some $r\in R^{\delta}$.
Now $r=\sum_{\alpha\in \Delta} r_\alpha e_\alpha$ for some $r\alpha\in R$ (where almost all of $r_\alpha$'s are $0$).
Thus, $f(r)=\sum_{\alpha\in \Delta}r_\alpha f(e_\alpha)=\sum_{\alpha \in \Delta} r_\alpha x_\alpha$.

Thus $\{x_\alpha\}_{\alpha\in \Delta}$ generates $F$.

Now we need to verify that each element of $F$ can be expressed uniquely as an $R$-linear combination of $x_\alpha$'s.

Say $\sum_{\alpha\in \Delta} r_\alpha x_\alpha=0$.
Then $\sum_{\alpha\in \Delta}r_\alpha f(e_\alpha)=0$.
Therefore $f(\sum_{\alpha\in \Delta} r_\alpha e_\alpha) =0$.
Which gives $\sum_{\alpha\in \Delta} r_\alpha e_\alpha=0$.
This leads to $r_\alpha=0$ for all $\alpha\in \Delta$.

Hence we are done.