In the proof of every free R-module is projective, where do we use being free?

Question

In the proof of every free R-module is projective, where do we use being free? To prove this we define a homomorphism from the basis of F to N(For M->N->0) and extend it to a homomorphism from F to N, Why can't we defe if from the begening , from F?

Answer 1

The very replacement you propose to use in your proof is the definition of freeness. Free means, by definition, there is a one-to-one correspondence between functions from the basis and functions on the object generated by that basis. The fact that we could replace one by the other is the fact that the module is free.

This is often stated as an adjunction. Let $F$ be any set, and $M$ be the free module generated by it, and $N$ any other module. Then

$$\hom_{R\text{-Mod}}(M,N)\cong\hom_\text{Set}(F,\lvert N\rvert)$$

where $\lvert N\rvert$ is the set of elements of $N$. In words, this says there is a one-to-one correspondence between set functions on the basis, and $R$-linear homomorphisms on the free module generated by that basis. The map implementing this correspondence could be described as “extend by linearity”.

But perhaps you are familiar with another definition of freeness in a commutative algebra setting: an $R$-module $M$ is free if it is isomorphic to $R^n$ for some $n$ (if $n$ is not finite, we must understand this as $\bigoplus_n R$, rather than $\prod_n R.$) Then I should interpret your question as: how can I see that these two definitions of freeness the same?

On the one hand, $R$-linear morphisms from $R$ to $N$ are in one-to-one correspondence with elements of $N$. And $R$-linear morphisms from $R^n$ are just $n$-tuples of morphisms from $R.$

On the other hand, any free module with basis $F$ is easily seen to be isomorphic to $R^{\lvert F\rvert},$ where $\lvert F\rvert$ is the cardinality of $F$.