# R is a Noetherian ring, then every finitely generated R-module is finitely presented

Let $$M$$ be a finitely generated $$R$$-module. We need to show that there exists free R-modules $$F_1, F_2$$ of finite rank such that $$$$F_1 \rightarrow F_2 \rightarrow M \rightarrow 0$$$$ is an exact sequence.

There now exists a surjective homomorphism $$\varphi \colon R^n \to M$$ for some $$n \geq 1$$ such that $$R^n/\ker \varphi \cong M$$. Because $$R$$ is Noetherian, $$R^n$$ is also Noetherian and because $$\ker \varphi$$ is an ideal, we know that it is finitely generated. If I can now conlude that $$\ker \varphi$$ is free, then I have found a short exact sequence but I have no idea if this is even true.

• Notice that $F_1\to F_2$ does not need to be injective, so $F_1$ does not have to actually be isomorphic to $\ker\varphi$. Jan 7, 2020 at 14:13
• If you can find a surjective homomorphism $\alpha : R^m \to ker \varphi$, the sequence $R^m \to R^n \to M \to 0$ where the first arrow is obtained by composing $\alpha$ and the inclusion $\ker \varphi \hookrightarrow R^n$ might work. Jan 7, 2020 at 14:17
• In general, only for special finitely generated modules there exists an exact sequence of the form $0\to F_1\to F_2\to\dots\to F_n\to M\to0$ with $F_i$ finitely generated free. The problem doesn't require that the map $F_1\to F_2$ is injective. Jan 7, 2020 at 21:14

$$0 \rightarrow \ker(\phi) \rightarrow R^n \rightarrow M \rightarrow 0$$
and noted that $$\ker(\phi)$$ is finitely generated because it is an $$R$$-submodule of the Noetherian module $$R^n$$.
In general, if $$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$ is an exact sequence and $$A' \rightarrow A$$ is a surjection, then
$$A' \rightarrow B \rightarrow C \rightarrow 0$$ is exact too (where the first arrow is now the composition $$A' \rightarrow A \rightarrow B$$).