# Viewing $R/I$ as $R/I$-module

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!

• @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. – Prism Jul 9 '13 at 5:58

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.
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.
• 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? – Prism Jul 9 '13 at 5:54
• @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... – Prism Jul 9 '13 at 5:59