Using the snake lemma to prove that if $N$ and $M/N$ are finitely generated, then so is $M$.

Question

Let $M$ be a $R$-module and $N$ be a submodule. Suppose that both $N$ and $M/N$ are finitely generated. I want to prove that so is $M$, using the snake lemma.

(Following If $M$ and $N/M$ are finitely generated, then so is $N$.)

My idea is as follows:

We clearly two exact sequences: $$0\to R^{\oplus n}\to R^{\oplus n}\oplus R^{\oplus m}\to R^{\oplus m}\to 0$$ and $$0\to N \to M\to M/N\to 0.$$ We also have surjective morphisms $R^{\oplus n}\to N$ and $R^{\oplus m}\to M/N$. As soon as we have any $R$-module morphism $R^{\oplus n}\oplus R^{\oplus m}\to M$ that makes the desired diagram commute, the Snake lemma will imply that it is surjective and then we are done.

In order to create this morphism, I think we can sum $$R^{\oplus n}\to N\to M$$ and a lift of $$R^{\oplus m}\to M/N$$ to $M$. This certainly makes the left side of the diagram commute.

So, I have two doubts:

1) Does this make the right side of the diagram commute? That is, the square containing $R^{\oplus n}\oplus R^{\oplus m}$, $R^{\oplus m}$, $M$ and $M/N$.

2) How can I prove the existence of such a lift?

Since I'm in the beginning of P. Aluffi's Algebra Chapter 0, I would appreciate if the answer does not contain heavy category theory.

Answer 1

1) Yes it does, by definition of lift.

2) $R^{\oplus m}$ is freely generated by $e_1= (1,0,...0), e_2=(0,1,0,...),..., e_m$ so it suffices to find a lift for each $e_i$ : those exist by surjectivity.

Answer 2

Just to add, for (2), the existence of such a lift is the same as proving the “projectivity” of free modules. (Once we know what a projective module is.)

In other words, the existence of such a lift is just another interpretation of saying that a free $R$-module is projective. The proof of this fact is what @Maxime Ramzi pointed out in the wonderful answer.