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This is Exercise 2.9 of Reid's Undergraduate Commutative Algebra:

Let $0\longrightarrow L \longrightarrow M \longrightarrow N \longrightarrow 0$ be a short exact sequence $A$-modules. Prove that if $N$ and $L$ are finite over $A$, then so is $M$.

I am aware of an answer to this question here. However, when I first tried to prove this before finally giving up, I thought about considering the definition Reid gives of a finite $A$-module $M$, that is, being $\{s_1, \dots, s_r\}$ a generator set of $M$. Then $M$ is finite over $A$ if the map $\varphi :A^r \longrightarrow L$ given by $(f_1,\dots ,f_n) \mapsto \sum f_is_i$ is surjective. Also, there is an observation that if a short exact sequence is as the given one above, then $L \subset M$ and $N = M/L$, so I could consider the homomorphism $j$ and $\pi$, being the inclusion map and the quotient map respectively. My question then is:

Is there a way to prove this using only the existence of surjective maps $\varphi : A^r \longrightarrow L$ and $\psi: A^s \longrightarrow N$ (being $\{t_i,\dots, t_s\}$ a generator set of $N$), or maybe a more intuitive way to see what the candidate to generator set of $M$ might be?

P.D. I'm a newbie on this matters, so if the question doesn't make sense at all, or I'm just wrong about the approach I took, just tell me and I'll delete my answer. Thanks in advance!

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user313212
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    The basic fact you will need is that the map $\psi$ can be lifted to $M$, $A^s$ being a free module. That is, there exists a map $\theta:A^s\to M$ such that $\pi\circ\theta=\psi$. Once you have that, it is easy to check that the map $(j\circ\phi,\theta): A^r\oplus A^s=A^{r+s}\to M$ is onto. – Mohan Mar 28 '17 at 19:34
  • @Mohan Thank you! I'll try to follow the steps you wrote! – user313212 Mar 28 '17 at 20:01

2 Answers2

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$\newcommand{\coker}{\operatorname{coker}}$This can be done using the surjections $A^r\to L$, etc., but I don't think it'll make more clear what the explicit generating set is. Anyways, here's how to do it:

Suppose $L$ and $N$ are finite, so we have surjections $A^r\to L$ and $A^s\to N$ for some integers $r,s$. Recall that free modules are flat, which means precisely that since $M\to N$ is surjective, there is an induced map $A^s\to M$. Also, we get a map $A^r\to M$ by the composition $A^r\to L\to M$. Now, recall that for any $A$-modules $M_1,M_2,N$, we have $$\hom_A(M_1\oplus M_2,N)\cong\hom_A(M_1,N)\oplus\hom_A(M_2,N).$$

In particular, since we have maps $A^r\to M$ and $A^s\to M$, we get a corresponding map $A^{r+s}=A^r\oplus A^s\to M$, which in particular by the way we've constructed things will make the following diagram commute:

$$\begin{array}{cccccc} 0&\rightarrow&A^s&\rightarrow&A^{r+s}&\rightarrow&A^s&\rightarrow&0\\ & & \downarrow&&\downarrow&&\downarrow \\ 0&\rightarrow&L&\rightarrow&M&\rightarrow&N&\rightarrow&0\\ \end{array}$$

Now, by the Snake Lemma, there is an exact sequence, the second half of which is

$$\coker(A^r\to L)\to\coker(A^{r+s}\to M)\to\coker(A^s\to N)\to0$$

However, since $A^r\to L$ and $A^s\to N$ are surjective, the first and third terms are zero, and exactness implies that $\coker(A^{r+s}\to M)=0$, i.e. $A^{r+s}\to M$ is a surjective map. Alternatively, surjectivity is immediately implied by the Five Lemma.

So yeah, this argument uses some things you may not have seen before, and it may not be all that illuminating, but it certainly gives a clean argument using "fancy machinery".

You can do these things more explicitly without the lemmas and theorems I've mentioned, like the user Mohan suggested, but actually if you do that you'll basically just end up rewriting the proof you linked us to.

TY Mathers
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  • Thanks for your answer! And yes, most of the things you wrote are unknown to me yet. Hopefully in a couple of weeks I'll be able to understand it,. I'll take it as a personal challenge! Thank you very much again!! :) – user313212 Mar 28 '17 at 20:00
  • @user313212 glad I could help! – TY Mathers Mar 28 '17 at 20:06
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    What would be the induced map? – Santiago Estupiñán Sep 26 '18 at 02:53
  • @SantiagoEstupiñán the induced map $A^s\to M$? It's given by taking a basis (there's an obvious choice) $e_1,\dots,e_s$ for $A^s$, and taking any lifts $m_i\in M$ whose image in $M\to N$ is the same as the image of $e_i$ under $A^s\to N$. These $m_i$ exist since $M\to N$ is surjective. Now taking $e_i\mapsto m_i$ gives a map $A^s\to M$. – TY Mathers Sep 26 '18 at 05:13
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You can use the pull-back. Consider a pull back diagram $$\require{AMScd} \begin{CD} 0 @>>> L @>>> P @>>> A^n @>>> 0 \\ @. @| @VVV @VVV @. \\ 0 @>>> L @>>> M @>>> N @>>> 0 \\ \end{CD} $$ where $A^n\to N$ is surjective. Then also $P\to M$ is surjective.

Since $A^n$ is projective, the top row splits, so $P\cong L\oplus A^n$ is finitely generated; therefore also $M$ is finitely generated, being a homomorphic image of a finitely generated module.

This clearly tells you that in order to find a set of generators you put together a set of generators of $L$ (seen as a submodule of $M$) and elements of $M$ which map to a set of generators of $N$.

egreg
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