If a commutative ring $R$ is Noetherian, then every finitely generated $R$-module has a resolution by finitely generated free $R$-modules. It goes as follows:
- Start with a finitely generated $R$-module $M$, and choose a tuple of $n_0$ generators for $M$.
- Set $d_0:R^{n_0} \twoheadrightarrow M$ (the obvious map).
- Set $M_1:= \ker (d_0)$ and choose a tuple of $n_1$ generators for $M_1$.
- Set $d_1:R^{n_1} \to R^{n_0}$ with $\mathrm{im} (d_1) = M_1$ (the obvious map).
- Set $M_2:= \ker (d_1)$ and choose a tuple of $n_2$ generators for $M_2$.
- ...
In the end we have a free resolution
$$ \cdots \overset{d_2}{\to} R^{n_1} \overset{d_1}{\to} R^{n_0} \overset{d_0}{\to} M \to 0.$$
This procedure works because all the $M_i$ are finitely generated, by Noetherianity of $R$ (otherwise the free modules in the resolution would not be finitely generated, in general).
Question: Is the converse true, i.e., does the condition "every finitely generated $R$-module has a resolution by finitely generated free $R$-modules" imply that $R$ is Noetherian?
Yes: consider free resolutions of the module $R/I$, where $I$ is an ideal of $R$.