# Finitely generated vs presented

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?

• "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 1, 2018 at 6:42

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.

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.

• Hmmm got a more concrete example of it? Sep 25, 2015 at 12:46
• 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, 2015 at 12:52
• Ah yes now I see it, thank you Sep 25, 2015 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 $R/I$. Jul 26, 2016 at 6:33
• @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, 2016 at 9:58