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While doing exercise 4 of Chapter 6 in Atiyah & Macdonald's Introduction to Commutative Algebra, I got stuck at this step:

I have shown that $R/I$ is a Noetherian $R$-module. Here $R$ is a commutative ring with $1$ and $I$ is some ideal of $R$. How can I (no pun intended) conclude from here that $R/I$ is a Noetherian $R/I$-module?

Well, in the exercise $I$ is actually the annihilator of $R$-module $M$, but the argument above probably works for all ideals $I$.

Thanks!

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  • $\begingroup$ @BenjaLim: That's excellent! I think that is exactly what I was looking for. If you post that as answer, I will be happy to upvote and accept it. $\endgroup$
    – Prism
    Commented Jul 9, 2013 at 5:58

2 Answers 2

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If you already know that $R/I$ is a Noetherian $R$ - module this solves your problem. Namely because of the following. What is an $R$ - submodule of $R/I$? Well it's just an ideal of $R/I$! So to say that $R/I$ satisfies the ACC on $R$ - submodules is the same as saying it satisfies the ACC on ideals, or that $R/I$ is a Noetherian ring.

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    $\begingroup$ I think you deserve "Atiyah-Macdonald" badge :) $\endgroup$
    – Prism
    Commented Jul 9, 2013 at 6:01
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An ascending chain of $R/I$-submodules of $R/I$ can be viewed as an ascending chain of $R$-submodules of $R/I$. Then use the fact that $R/I$ is a Noetherian $R$-module.

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  • $\begingroup$ Dear Kevin, thanks for very fast answer! I just want to make this "viewing" precise. So, $R/I$ becomes $R/I$-module because we define $(r+I)(s+I)$ to be $r(s+I)$ where we use the old action of $R/I$ as an $R$-module. Is this correct? $\endgroup$
    – Prism
    Commented Jul 9, 2013 at 5:54
  • $\begingroup$ @Prism Any commutative ring is tautologically a module over itself. $\endgroup$
    – user38268
    Commented Jul 9, 2013 at 5:56
  • $\begingroup$ To complement: a ring, viewed as a module over itself has as the action, the ring multiplication. And I think your action is just the ring multiplication, right? $\endgroup$
    – awllower
    Commented Jul 9, 2013 at 5:58
  • $\begingroup$ @BenjaLim: You are right, of course. I just wanted to use the information I had already (which was $R/I$ is an Noetherian $R$-module), and I ended up confusing myself... $\endgroup$
    – Prism
    Commented Jul 9, 2013 at 5:59
  • $\begingroup$ @Kevin Lin: I have accepted BenjaLim's answer. I hope you don't mind :) I have given my (+1) to you as well :) $\endgroup$
    – Prism
    Commented Jul 9, 2013 at 6:01

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