We will define the free $R$-modules.
Definition. Let $R$ be a ring with $1_R$ and $F$ an left $R$-module. We call $F$ free $R$-module, if $$F=\bigoplus_{i\in I} R_i$$ where $R_i:=\langle b_i \rangle \cong _RR,\ \forall i \in I $ and $I$ is a set of indexes (finite or infinite).
We will try to prove the following theorem.
Theorem. Let $R$ be a ring with $1_R$ and $M$ an left $R$-module. The following are equivalent.
- $M$ is a left, free $R$-module.
- $M$ has basis.
Proof. $1.\implies 2.$ According to our definition, $M=\bigoplus_{i\in I} R_i$, where $R_i:=\langle b_i \rangle \cong _RR,\ \forall i \in I $. and $I$ is a set of indexes (finite or infinite).
We define the set $$S:=\{e_i:=(\delta_{i,\lambda} )_{\lambda\in I}: i\in I\}\subseteq \bigoplus_{i\in I} R,$$ where $\delta_{i,\lambda} =1_R$, if $\lambda=i$ and $\delta_{i,\lambda} =0_R$ otherwise. Then, it's easy to observe that the set $S$ is $R$-basis for $\bigoplus_{i\in I} R$ has an $R$-basis and $\bigoplus_{i\in I} R_i \cong \bigoplus_{i\in I} R = M $, thus $M$ has an $R$-basis.
$2.\implies 1.$ We suppose that $S:=\{e_i \in M :i \in I \} \subseteq M$ is an $R$-basis for $M$. Obviously, the key is to define an $R$ module homomorphism $$\phi:M\longrightarrow \bigoplus_{i\in I} R.$$
And now my questions.
Questions. (1) Is the first part okey?
(2) The index set $I$ may be infinite, so does $S$. Then, does every element $m\in M=\langle S \rangle$ have a unique expression, as finite linear $R$-combination, in the form $m=\sum_{k=1}^{n} r_k e_k$, where $n\in \Bbb N , r_k \in R, e_k \in S$?
(3) Which the $R$-module isomorphism?
Thank you.