$R$ noetherian, then every finitely generated $R$-module A has a resolution

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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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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.