Given an $R$-module $M$ arbitrary, show it is always possible to construct an exact sequence of $R$-modules $$0\longrightarrow K \longrightarrow L \longrightarrow M \longrightarrow 0,$$ with $L$ a free module.
I thought of considering $ K $ as being a free module and consider $ f: K \longrightarrow f (K) $, a homomorphism, which will be bijetor in the image, will soon take the basis of $ K $ in the base of $ f (K )$.
But I could not conclude anything with that.
You always may write an $R$-module $M$ as a quotient of a free $R$-module $L$, for instance, if $S$ is any generating set for $M$, the free $R$-module $L=R^{(S)}$, with basis $(e_s)_{s\in S}$, where each $e_s$ is defined as $$e_s[i]=\begin{cases}1&\text{if }i=s,\\ 0&\text{otherwise}.\end{cases}$$
The surjective homomorphism from $L$ to $M$ is defined as \begin{align} f:L&\longrightarrow M \\ \sum_{s\in S}\lambda_s e_s&\longmapsto \sum_{s\in S}\lambda_s\, s&&\text{(the sums are actually finite, by definition of $R^{(S)})$.} \end{align} If you denote $K$ the kernel of this linear map, you obtain the short exact sequence $$0\longrightarrow K\longrightarrow L\longrightarrow M\longrightarrow 0.$$ However, in general, $K$ (a.k.a. the module of linear relations between the generators in $S$) has no reason to be a
freemodule.