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A standard fact in commutative algebra:

Let $A$ be a ring in which there exists some finite number of maximal ideals whose product is $0$. Then $A$ is Artinian iff $A$ is Noetherian.

The usual proof involves noting that each $\frac{m_1...m_{i-1}}{m_1...m_i}$ is naturally an $\frac{A}{m_i}$-module. You look at a short exact sequence $$0 \rightarrow m_1...m_i \rightarrow m_1...m_{i-1} \rightarrow \frac{m_1...m_{i-1}}{m_1...m_i} \rightarrow 0 $$

and argue that the middle module is Artinian/Noetherian iff the other two are Artinian/Noetherian. Then by decreasing induction, each $m_1...m_i$ is Artinian iff it is Noetherian. In particular, for $i=0$, you get the desired result.

My question is, why is it OK to conflate $A$ being Noetherian (over itself), and $A$ being Noetherian as an $\frac{A}{m_i}$-module?

Surely there's a difference between the notion of being Artinian/Noetherian as $\frac{A}{m_i}$-modules, and being Artinian/Noetherian as $A$-modules. So for instance how can we justify using the assumption that the $\frac{A}{m_i}$-module $\frac{m_1...m_{i-1}}{m_1...m_i}$ is Noetherian to deduce anything about $A$ being Noetherian as a module over itself?

(For full details of the standard proof see Lemma 2.15 here: https://people.kth.se/~laksov/courses/algebradr01/notes/chains2.pdf)

SSF
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1 Answers1

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If $A$ is a commutative ring, $m$ an ideal of $A$, then the following hold:

  • if $M$ is an $A$-module with $mM = 0$, then $M$ is an $A/m$-module
  • if $M$ is an $A/m$-module, then $M$ is an $A$-module

With these in mind, if $M$ is an $A$-module with $mM = 0$, then the $A$-submodules of $M$ are precisely the $A/m$-submodules of $M$, and so $M$ is Artinian (resp. Noetherian) as an $A$-module if and only if it is Artinian (resp. Noetherian) as an $A/m$-module.

As explained in the linked proof, it is clear that $\frac{m_1...m_{i-1}}{m_1...m_i}$ is Artinian as an $A/m_i$-module iff it is Noetherian as an $A/m_i$-module (since $A/m_i$ is a field), and so by the previous part it is Artinian as an $A$-module iff it is Noetherian as an $A$-module.

Then consider the short exact sequence of $A$-modules $$0 \rightarrow m_1...m_i \rightarrow m_1...m_{i-1} \rightarrow \frac{m_1...m_{i-1}}{m_1...m_i} \rightarrow 0 $$ and apply induction as you describe.

Christopher
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