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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?

57Jimmy
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  • This could be helpful, and shows that "is finitely presented" can replace "has a resolution by finitely generated free R-modules". – user26857 Apr 21 '21 at 09:35

1 Answers1

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Yes: consider free resolutions of the module $R/I$, where $I$ is an ideal of $R$.

Aphelli
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  • Let me spell this out. There is a resolution ending like $R^{n_1} \overset{d_1}{\to} R^{n_0} \overset{d_0}{\to} R/I \to 0$. We choose one of the standard generators of $R^{n_0}$ that maps to a generator of $R/I$ and consider the resulting, surjective map $R \to R/I$. Its kernel must be isomorphic to $I$ and coincides with the image of $d_1$ composed with the projection $R^{n_0} \to R$. So there is a surjection from a finitely generated free $R$-module to $I$, and thus $I$ is finitely generated. Right? – 57Jimmy Apr 21 '21 at 09:16
  • That’s what I thought – until I realized it’s actually slightly trickier than that. After all, there’s little reason why one element of the standard basis of $R^{n_0}$ must be sent to a generator of $R/I$. An actual argument (that actually uses a nontrivial result) works as follows: using the resolution, we see that $R/I$ is finitely presented. This implies that every “finite generation” of $R/I$ is a “finite presentation”. But $R/I$ is such a “finite generation”, so this implies $I$ finitely generated. – Aphelli Apr 21 '21 at 11:15
  • Well, ok, maybe not a standard generator. But there must be some element of $R^{n_0}$ that maps to a generator of $R/I$, as the map is surjective. And then we can extend this element to a basis of $R^{n_0}$ (not sure if this works over any ring though?). Would this work? – 57Jimmy Apr 22 '21 at 09:32
  • And I guess that the non-trivial result you mentioned is "if a module is finitely presented, then in whatever way one writes it as a quotient $M/I$, if $M$ is finitely generated then so is $I$", right? Which is the content of the link in the comment by user26857 – 57Jimmy Apr 22 '21 at 09:58