$0\to L\to R^{n}\to M \to 0$ 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.
 A: 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!
