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.
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2$\begingroup$ 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? $\endgroup$– Dhruv RanganathanMay 11, 2013 at 1:54
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$\begingroup$ the definition is: $M $ is finitely presented if $\ni$ exact sequence $R^{m}\to R^{n}\to M \to 0$ where $m,n \in \mathbb{N}$ $\endgroup$– AlexMay 11, 2013 at 2:02
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1$\begingroup$ @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, $\endgroup$– Matt EMay 11, 2013 at 14:46
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$\begingroup$ @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. $\endgroup$– Matt EMay 11, 2013 at 14:48
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$\begingroup$ 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. $\endgroup$– Andreas CarantiMay 11, 2013 at 15:07
1 Answer
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.
Also, see this interesting and nearly identical discussion on MO: https://mathoverflow.net/questions/1788/does-finitely-presented-mean-always-finitely-presented-answered-yes. Brian Conrad's answer is wonderfully instructive and worth reading!
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$\begingroup$ Dear Dhruv, There is a typo in your second $cok$; $R^m$ should be $R^n$. Regards, $\endgroup$– Matt EMay 11, 2013 at 14:50