# $0\to L\to R^{n}\to M \to 0$ is exact, prove $M$ is finitely presented if and only if $L$ is finitely generated.

Suppose $R$ is a ring, $0 \rightarrow L\rightarrow R^{n} \rightarrow M \rightarrow 0$ is a short exact sequence, prove $M$ is finitely presented if and only if $L$ is finitely generated.

• What is your definition of finitely presented? This can be taken to be the definition of finite presentation in fact. If you mean that $M$ is given by finitely many generators with finitely many relations between them, then that $L$ are the relations. Am I misunderstanding the question? – Dhruv Ranganathan May 11 '13 at 1:54
• the definition is: $M$ is finitely presented if $\ni$ exact sequence $R^{m}\to R^{n}\to M \to 0$ where $m,n \in \mathbb{N}$ – Alex May 11 '13 at 2:02
• @DhruvRanganathan: Dear Dhruv, I don't think this would ever be the definition of f.p. The point is that f.p. is about the existence of some finite presentation, while this question then aims at showing that if $M$ is f.p., then every finite generating set has a f.g. module of relations. Of course, they are equivalent; that is what this question is about. But the second condition is useless as a definition until you have proved this equivalence, since you can't check it (without proving the equivalence). Regards, – Matt E May 11 '13 at 14:46
• @DhruvRanganathan: P.S. Seeing your answer below, I guess that you are well aware of the issue in the preceding comment, so sorry about that. I just wanted it to be clear to anyone looking at this that the key points is to show that some finite generating set of a module having a f.g. module of relations implies that every finite generating set has the same property. – Matt E May 11 '13 at 14:48
• I learned of this results for the case of groups from the paper of B. H. Neumann, Some remarks on infinite groups, J. London Math. Soc. 12 (1937), 120-127. I needed it for some work of mine on (graded) Lie algebras, but it was really the same thing. I understand this is by now a standard result in universal algebra. – Andreas Caranti May 11 '13 at 15:07

Here is a sketch of a proof.

Suppose that $M$ is finitely presented (as defined in Alex's comment to my question). Choose a presentation $R^m\to R^n\to M\to0$.

Claim: For any epimorphism $\phi:R^l\to M$, $\ker \phi$ is finitely generated.

Proof: We have two sequences

$R^m\to R^n\to M\to 0$

$0\to K\to R^l\to M\to 0$.

Imagine an identity map between the $M$ on the top and bottom. There exists a map $\alpha:R^n\to R^l$ making the right square above. To see this, pick a basis and chase a diagram - this is not hard and a good exercise. This in turn furnishes a morphism $R^m\to K$ (top to bottom).

Now we're in good shape, we can use the snake lemma, to get an isomorphism $cok(R^m\to K)\cong cok(R^m\to R^l$). We can conclude that the cokernel is finite as an $R$-module. This tells us that $K$ is actually finitely generated (do you see why? Even though $R^m$ doesnt surject onto $K$, there's only "finite amount of stuff" left).

So thats one direction. The other one should be easy I believe: if you have a finitely generated kernel, you can pick a finite rank free module to map onto that, which produces the presentation you want.

• Dear Dhruv, There is a typo in your second $cok$; $R^m$ should be $R^n$. Regards, – Matt E May 11 '13 at 14:50