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I am curious exactly what are the differences between finitely generated and finitely presented? I understand that finitely generated means we have, for an $R$-module $M$ that there exists an epimorphism $$p:R^n\to M$$ and definitionally that finitely presented is when the kernel of $p$ is finitely generated that is $$h:R^m\to\ker p$$ is an epimorphism, so we get $$R^m\xrightarrow{h} R^n \xrightarrow{p} M\to 0$$ being exact.

What I don't get is what additional information does it supply? Wouldn't the kernel of any such epimorphism be finitely generated? If not got a good example of it not being the case?

Radulichnus
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Zelos Malum
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  • "Finitely generated" and "finitely presented" are certainly different for groups. The details are over my head (I am not a group theorist, hardly even a mathematician), but I have it on good hearsay that at one time the existence of a finitely generated infinite simple group was known, but the existence of a finitely presented infinite simple group was still an unsolved problem. – bof Apr 01 '18 at 06:42

1 Answers1

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For noetherian rings these properties are indeed equal, but not in general. Take your favorite non-noetherian ring $R$. It has an ideal $I$ which is not finitely generated. Then $R/I$ is finitely generated but not finitely presented.

Added later

Since it is not quite obvious, let me add a reason why $R/I$ being finitely presented would imply $I$ being finitely generated:

If $R/I$ is finitely presented, there is an exact sequence $0 \to K \to R^n \to R/I \to 0$ for some integer $n$ and some finitely generated module $K$. By applying Schanuel's lemma to that sequence and to the canonical exact sequence $0 \to I \to R \to R/I \to 0$, we obtain an isomorphism $R \oplus K \cong R^n \oplus I$. Now we see that $I$ is a quotient of a finitely generated module, and therefore is finitely generated as well.

Dune
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    Hmmm got a more concrete example of it? – Zelos Malum Sep 25 '15 at 12:46
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    You could take for $R$ the polynomial ring $\mathbb{k}[X_j : j \in J]$ in infinitely many indeterminates over some field $\mathbb{k}$ and $I = \langle X_j : j \in J \rangle$. Then $R/I \cong \mathbb{k}$ is a module you are looking for. – Dune Sep 25 '15 at 12:52
  • Ah yes now I see it, thank you – Zelos Malum Sep 25 '15 at 12:54
  • I think the concrete example is ok, but the statement in the answer should be changed. I think one should furthermore assume that $I$ should be a maximal ideal. Take for instance the polynomial ring with countably infinite indeterminate $R=K[x_i : i\in \mathbb N]$ and the (non maximal) ideal $I=<x_i : i\in 2\mathbb N>$ then $I$ is infinitely generated as an $R$-module and so is $R/I$. – quantum Jul 26 '16 at 06:33
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    @quantum $R$ is finitely generated as an $R$-module, and so is every quotient of $R$. There is no need for a further assumption. – Dune Jul 26 '16 at 09:58
  • Related: https://math.stackexchange.com/questions/3289357/finitely-generated-but-not-finitely-presented-module – Watson Jan 16 '22 at 11:39