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 \begin{equation} F_1 \rightarrow F_2 \rightarrow M \rightarrow 0 \end{equation} 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.

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    $\begingroup$ Notice that $F_1\to F_2$ does not need to be injective, so $F_1$ does not have to actually be isomorphic to $\ker\varphi$. $\endgroup$ Jan 7, 2020 at 14:13
  • $\begingroup$ 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. $\endgroup$ Jan 7, 2020 at 14:17
  • $\begingroup$ 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. $\endgroup$
    – egreg
    Jan 7, 2020 at 21:14

1 Answer 1


In your post you produced an exact sequence

$$0 \rightarrow \ker(\phi) \rightarrow R^n \rightarrow M \rightarrow 0$$

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 $A' \rightarrow A \rightarrow B$).

Can you finish it from there?


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