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I'm solving this problem from "Introduction to Commutative Algebra" of Atiyah and Macdonald. Here is the problem:

Let $M$ be an $A$-module and let $N_1, N_2$ be submodules of $M$. If $M/N_1, M/N_2$ are Noetherian, so is $M/(N_1 \cap N_2)$.

I found a solution which states that we have the exact sequence $$0 \rightarrow M/N_1 \rightarrow M/(N_1 \cap N_2) \rightarrow M/N_2 \rightarrow 0$$

so $M/(N_1 \cap N_2)$ is Noetherian if and only if $M/N_1$ and $M/N_2$ are Noetherian.

I can't prove that this sequence is exact but can't find any counter-example for it. Can anyone help me? Thanks so much.

user26857
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le duc quang
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1 Answers1

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The sequence $0 \to M/N_1 \to M/(N_1 \cap N_2) \to M/N_2 \to 0$ is not exact, in general. A similar sequence that is exact is $0 \to N_2/(N_1 \cap N_2) \to M/(N_1 \cap N_2) \to M/N_2 \to 0$. The first sequence will be exact iff $N_1 + N_2 = M$, in which case the first terms of both sequences are isomorphic. However, the sequence that will solve your problem is the following:

$$0 \to M/(N_1 \cap N_2) \to M/N_1 \oplus M/N_2 \to M/(N_1 + N_2) \to 0$$

If $M/N_1, M/N_2$ are Noetherian, then so is $M/N_1 \oplus M/N_2$, so $M/(N_1 \cap N_2)$ is a submodule of a Noetherian module, hence is Noetherian. For why the sequence is exact, apply the reasoning in this answer.

Incidentally, I would advise you to stop using that particular set of solutions.

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zcn
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    So great. Thanks so much your help. I will throw that solutions immediately :-D – le duc quang Jun 29 '14 at 06:20
  • @leducquang: No problem. And actually, to amend my previous advice just a bit: perhaps you can simply be more on guard when you read those solutions. As you can see, even solution-writers make mistakes – zcn Jun 29 '14 at 06:23
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    We only need that there is a monomorphism $M/(N_1 \cap N_2) \to M/N_1 \times M/N_2$ (which is clear). It immediately implies the claim, since Noetherian modules are closed under products and submodules. We don't need exact sequences. – Martin Brandenburg Jun 29 '14 at 07:17
  • @MartinBrandenburg: That's true, the full short exact sequence is not needed (only exactness of $0 \to M/(N_1 \cap N_2) \to M/N_1 \oplus M/N_2$). However, I believed that the full answer would be of interest to the OP – zcn Jun 29 '14 at 07:33
  • @zcn Why need a new exact sequence if $0 \to N_2/(N_1 \cap N_2) \to M/(N_1 \cap N_2) \to M/N_2 \to 0$ solves the problem? – user26857 Jun 29 '14 at 11:12
  • @user26857: That's also true, as $N_2/(N_1 \cap N_2)$ is always a submodule of $M/N_1$, and extensions of Noetherian modules are Noetherian. The sequence with the direct sum was simply the first thing that occurred to me though – zcn Jun 29 '14 at 19:31
  • Why did you write that $M/N_1\cap N_2$ is a sub-module a neotherian module? It's a neotherian module since the short sequence $$0\to M/N_1\cap N_2\to M/N_1 \oplus M/N_2\to M/N_1+N_2\to 0$$ is exect and $M/N_1\oplus M/N_2$ is neotherian, isnt' it? – J. Doe Sep 11 '19 at 08:22