Proof explanation: Every module is the quotient of free module

Question

Let $M$ be a left $R-$ module, it is said (here for example) that $M \approx F(M)/\sim$

The proof is apparently $F(M) \stackrel{\pi}\to M$ where $(0,\dots,1_m,\dots0) \stackrel{\pi}\to m$ and appeal to first isomorphism. theorem

Here is my gap in understanding. I thought $F(M) = \{\sum_{i = 1}^n r_im_i \}$, but there is also a canonical isomorphism (if $M$ is finite)

$$F(M) = \oplus_{i=1}^n Rm_i \stackrel{\omega}\approx \oplus_{m \in M} R = R^n$$

where $rm_i \stackrel{\omega}\to r$ is the isomorphism.

Are we saying taking $m_i = e_i$ the standard basis or something? How do we deal with the $r_i$ elements in front of the $e_is$?

Answer 1

The only "standard basis" available in this situation is $M$ itself (identified with a subset of $F(M)$).

The module $F(M)$ consists of all functions $f \colon M \to R$ such that $f(m) = 0$ for all but finitely many $m \in M$.

Addition of elements of $F(M)$, and multiplication by elements of $R$, are defined pointwise.

For each $m \in M$, let $e_m$ be the function $M \to R$ defined by: $$ e_m(m') = \begin{cases} 1 \text{ if } m' = m, \\ 0 \text{ if } m' \ne m. \end{cases} $$ Then for all $f \in F(M)$, $$ \label{eq:1}\tag{$*$} f = \sum_{m \in M}f(m)e_m. $$ The sum is well-defined, because of how $F(M)$ was defined.

If $(r_m)_{m \in M}$ is a family of elements of $R$, almost all of which are zero, then $r$ is precisely an element of $F(M)$, and \eqref{eq:1} therefore implies: $$ \sum_{m \in M} r_me_m = 0 \iff r = 0 \iff r_m = 0 \text{ for all } m \in M. $$ So, the family $(e_m)_{m \in M}$ is linearly independent, as well as generating $M$ - by \eqref{eq:1} again - i.e. it is a basis.

If, for all $m \in M$, we identify $m$ with $e_m$ (which is convenient, but can be rather confusing), \eqref{eq:1} becomes $$ f = \sum_{m \in M}f(m)m \quad (f \in F(M)), $$ and now $M$ itself is a basis for $F(M)$.

If $N$ is any module, and $j \colon M \to N$ is any function, then, by the defining property of free modules, there exists a homomorphism (necessarily unique), $\pi \colon F(M) \to N$, that extends $j$.

In particular, if we take $j \colon M \to M$ to be the identity function, it follows that there exists a (unique) homomorphism $\pi \colon F(M) \to M$ such that $\pi(m) = m$ for all $m \in M$, whence: $$ F(M)/\operatorname{Ker}\pi \cong \operatorname{Im}\pi = M. $$ The homomorphism $\pi$ is given explicitly by: $$ \pi(r) = \pi\left(\sum_{m \in M}r_mm\right) = \sum_{m \in M}r_m\pi(m) = \sum_{m \in M}r_mm \quad (r \in F(M)). $$ That is, $F(M)$ consists of "formal sums" of elements of $M$, and $\pi$ interprets these "sums" as actual sums in $M$.

$\operatorname{Ker}\pi$, of course, consists of all the "sums" that evaluate to zero in $M$.

It makes intuitive sense that quotienting out the "zero" sums from $F(M)$ leaves you essentially with $M$ itself.

The "$r_i$" you're worried about are, in the notation used here, the coefficients $r_m$ occurring in the formal "sums".