$M$ is free $R$-module $\iff$ $M$ has $R$-basis

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.

1. $$M$$ is a left, free $$R$$-module.
2. $$M$$ 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.

• I'm guessing that the definition is using the internal direct sum (which is a relation between submodules, despite the functional and equality notation) and not the external direct sum which has formal elements $\sum_{i \in I} \lambda_i e_i$. – Daniel Schepler Nov 29 '18 at 21:52
• @DanielSchepler Yes, indeed. But the two direct sums are in some way equivalent. So, it'doesn't really matter. Do you agree? Note that this definitions is taken form Rotman's book, Advanced Modern Algebra, AMS, 2012. – Chris Nov 29 '18 at 21:59
• Well, even when you have a canonical isomorphism $\alpha$, it's usual to make applications of that isomorphism explicit - so instead of $S := \{ e_i \ldots \}$ you would say something like $S := \{ \alpha(e_i) \ldots \}$. So, under the canonical isomorphism of the external direct sum $\bigoplus_{i\in I} R$ with $R$ under the hypothesis of $R$ being the internal direct sum of submodules $\langle b_i \rangle \simeq R$, what would $e_i$ be sent to? – Daniel Schepler Nov 29 '18 at 22:50
• Okay, let's take the external. It's easy then to do the transfromation. But my question is, shall we sent $m:=\sum_{i=1}^{n}r_i e_i \longmapsto \phi(m)= \phi (\sum_{i=1}^{n}r_i e_i) :=(r_1,...,r_n)$? And should these sums be finite? – Chris Nov 29 '18 at 22:58
• I'd say more specifically, my suggestion is to prove $M = \bigoplus_{i\in I} \langle b_i \rangle$ and $\langle b_i \rangle \simeq_R R$ for all $i \in I$ $\Leftrightarrow$ $\{ b_i \mid i \in i \}$ is a basis for $M$. (The part requiring $\langle b_i \rangle \simeq_R R$ is essential to rule out modules like $R / I$ where $I$ is a proper nontrivial ideal.) – Daniel Schepler Nov 30 '18 at 0:37