I want to solve the problem in the title.
A right $R$-module $M$ is finitely presented if there is an exact sequence
$$0\to K\to R^n\to M\to 0$$
with $K$ finitely generated, or equivalently if there is a sequence
$$R^m\to R^n\to M\to 0.$$
Let $M\in\text{Mod}-R$ and consider $\mathcal{S}=\{N\leq M :N\text{ is finitely presented}\}$.
$\mathcal{S}$ is a poset with $\subseteq$ relation. It's also directed: in fact
If $N_1, N_2$ are finitely presented, then we have exact sequences
$$0\to K_1\to R^{n_1}\to N_1\to 0$$
$$0\to K_2\to R^{n_2}\to N_2\to 0$$
So
$$0\to K_1\oplus K_2\to R^{n_1+n_2}\to N_1\oplus N_2\to 0$$
$K_1\oplus K_2$ is finitely generated, so $N_1\oplus N_2$ is finitely presented.
So i can consider $\displaystyle \varinjlim_{N\in \mathcal{}S}N=\bigoplus_{N\in \mathcal{S}}N=\bigcup_{N\in\mathcal{S}}N$.
I want that it is $M$. So i have to prove that each $m\in M$ belongs to some finitely presented submodule of $M$.
Best Answer
One has the exact sequence $$0\to K\to R^{(I)}\to M\to 0$$ We can assume w.l.o.g. that $K\subseteq R^{(I)}$, $M=R^{(I)}/K$. Let's consider now the set $$\mathcal{S}=\{(N,S):|S|<\infty,\;\;N\subseteq K\cap R^S, N\text{ is finitely generated}\}$$ with the order relation $(N,S)\leq (N',S') \iff N\subseteq N',S\subseteq S'$. This is a directed poset, in fact $N+N'$ is f.g. if $N,N'$ are, $S\cup S'$ is finite if $S,S'$ are. So by construction $\forall (N,S)\in \mathcal{S}$ the module $R^S/N$ is finitely presented. One verifies that $$\displaystyle \varinjlim_{(N,S)\in\mathcal{S}}R^S/N\cong R^{(I)}/K.$$