Apparently, every $R$-module $M$ is the quotient of a free module. The proof has been skipped in my notes. So if $F(S)$ denotes the free module generated by the set $S$, we define a map $f: F(M) \to M$ as $$v=(0, \dots, 0, 1, 0, \dots, 0) \mapsto m$$ where the vector $v$ is one at coordinate $m$ and zero otherwise and $$(0, 0, \dots, 0) \mapsto 0$$
Then by the first isomorphism theorem, $M \cong F(M) / ker(f)$ since $f$ is surjective. So far so good.
What disturbs me is that I think the kernel of $f$ is $0$. Something is wrong here.
How to simplify this best
$(a_1 + a_2 +a_3+... +a_n)^m$
for
$m=n, m<n, m>n$
I could only get
$\sum_{i=0}^{m}\binom{m}{i}a_i^i\sum_{j=0}^{m-i}\binom{m-i}{j}a_j ... $
