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I have seen in a definition of the book An Introduction to Homological Algebra (Weibel) the following:

A ring $R$ is noetherian if every ideal is finitely generated. That is, every $R/I$ is finitely presented. It is well known that if $R$ is noetherian, then every finitely generated $R$-module is finitely presented. It follows that every finitely generated module $A$ has a resolution $F \longrightarrow A$ in which each $F_n$ is a finitely generated free $R$-module.

I do not understand when it says: that every finitely generated module $A$ has a resolution $F \longrightarrow A$ in which each $F_n$ is a finitely generated free $R$-module.

Could anyone help me?

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    $\begingroup$ It means that for $A$ any finitely generated $R$-module, there exists a sequence of finitely generated free $R$-modules $\left\{ F_{n} \right\}_{n \geq 0}$ along with a sequence of $R$-module homomorphisms $\left\{ d_{n} : F_{n} \rightarrow F_{n-1} \right\}_{n \geq 1}$, and a surjective $R$-module homomorphism $\epsilon : F_0 \rightarrow A$ such that $\operatorname{Img}d_{i} = \operatorname{ker}d_{i-1}$ for every $i \geq 1$, and $\operatorname{ker}\epsilon = \operatorname{img}d_1$. Are you asking why this is true? $\endgroup$ – Adam Higgins Dec 27 '18 at 14:38
  • $\begingroup$ Yes, I am asking why this is true. $\endgroup$ – idriskameni Dec 29 '18 at 11:06
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It's rather simple, though elliptic: a finitely generated $R$-module $A$ is the image of a finitely generated free $R$-module $F_0$ and the kernel $A_1$ of this morphism is itself a finitely generated $R$-module, so it is the image of (another) finitely generated free $R$-module $F_1$ and the kernel of this latter morphism is finitely generated, &c. The general assertion follows by a trivial induction.

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  • $\begingroup$ But how do you know that $A$ is the image of a finitely generated free $R$-module $F_0$ and the kernel $A_1$ of this morphism is itself a finitely generated $R$-module? I mean, how do you know you can do that? $\endgroup$ – idriskameni Dec 27 '18 at 14:47
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    $\begingroup$ If $G$ is a finite set of generators of $A$, you have a surjective map $\;R^G\longrightarrow A$ onto the corresponding generator of $R^G$ , and the kernels are finitely generated because $R$ is noetherian, so every submodule of a finitely generated $R$-module is finitely generated (that's why finitely generated modules over a noetherian ring are finitely presented). $\endgroup$ – Bernard Dec 27 '18 at 15:00
  • $\begingroup$ And what about being free? I can not see it. $\endgroup$ – idriskameni Dec 29 '18 at 13:17
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    $\begingroup$ What do you mean, precisely? $\endgroup$ – Bernard Dec 29 '18 at 13:25
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    $\begingroup$ They're all free by construction(and induction): each kernel being finitely generated, it is a uotient of a finitely generated free module. $\endgroup$ – Bernard Dec 29 '18 at 13:36

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