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 $M$.

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    $\begingroup$ It is not true that $M$ is the sum of its finitely presented submodules, in general. Note that while $N_1\oplus N_2$ is finitely presented, $N_1+N_2$ may fail to be. You have to do an “external” direct limit. $\endgroup$ – egreg Sep 23 '16 at 8:49
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    $\begingroup$ Maybe you should look at the exact sequence $0 \to K \to R^I \to M \to 0$ and then look at the finitely generated submodules of $K$ whose images are contained in some $R^n \subset R^I$. Then the cokernels of those inclusions are finitely generated and they give a directed system. $\endgroup$ – user60589 Sep 23 '16 at 9:07

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 $$\displaystyle \varinjlim_{(N,S)\in\mathcal{S}}R^S/N\cong R^{(I)}/K.$$

  • $\begingroup$ Yes, sorry, i edit. $\endgroup$ – Anselm Sep 23 '16 at 10:15
  • $\begingroup$ Now there is a problem with the $S$, maybe $N \subset R^S \cap K$? $\endgroup$ – user60589 Sep 23 '16 at 10:22

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