If finite product of maximal ideals of ring $R$ is zero, then $R$ is Noetherian$\iff R$ is Artinian

I'm studying commutative algebra and now I am struggling to understand the following proof:

Proposition. Let $$R$$ be a commutative ring with $$1_R$$, $$t\in \Bbb{Z}^+$$ and $$P_1,\dots,P_t \in \mathrm{mSpec}(R)$$ maximal ideals of $$R$$ such that $$P_1\dotsb P_t=\{0_R\}$$. Then $$R$$ is Noetherian $$\iff$$ $$R$$ is Artinian.

My textbook contains the proof, but it's extremely bad-written.

So, could anybody write down clearly and step by step the proof of this proposition?

Notice that I have already studied short exact sequences, I know that if we have a $$K-$$vector space $$V$$ space ($$K$$ is a field), then $$\dim_K V<\infty \iff K-$$module $$V$$ is Noetherian $$\iff K-$$module $$V$$ is Artinian and at last if $$IM=0$$ then $$R/I-$$modules $$M$$ are equivalent to $$R-$$modules $$M$$.

PS: This Proposition is also in Atiyah and McDonald's "Introduction to Commutative Algebra", Corollary 6.11, p. 78 and in MSE there.

Thank you.

• Sorry, in your statement of the proposition, what are the $M_i$? – Santana Afton Mar 9 at 3:42
• Thank you for your comment. Where are you reffered to? – Chris Mar 11 at 22:42
• No worries! There was a typo I was confused about, and Eric Wofsey corrected it. – Santana Afton Mar 11 at 23:22

Suppose $$R$$ is Noetherian. For $$r=0,\dots,t$$, let $$I_r=P_1\dots P_r$$ (for $$r=0$$, this means $$I_r=R$$). We will prove by reverse induction on $$r$$ that $$I_r$$ is an Artinian $$R$$-module, so that in the last case $$r=0$$ we conclude $$I_0=R$$ is Artinian. In the base case $$r=t$$, $$I_t=0$$ by hypothesis is trivially Artinian.
Now suppose $$r and $$I_{r+1}$$ is Artinian; we will prove $$I_r$$ is Artinian. Note that $$I_r\supseteq I_rP_{r+1}=I_{r+1}$$ so we have a short exact sequence $$0\to I_{r+1}\to I_r\to I_r/I_{r+1}\to 0.$$ Since $$I_{r+1}$$ is Artinian, it suffices to show that $$I_r/I_{r+1}$$ is also Artinian. Now since $$I_{r+1}=I_rP_{r+1}$$, the module $$I_r/I_{r+1}$$ is annihilated by $$P_{r+1}$$, and so $$I_r/I_{r+1}$$ can be considered as a a module over the field $$K=R/P_{r+1}$$, with $$R$$-submodules of $$I_r/I_{r+1}$$ being the same thing as $$K$$-submodules of $$I_r/I_{r+1}$$. Since $$R$$ is Noetherian, $$I_r$$ and $$I_{r+1}$$ are Noetherian and hence $$I_r/I_{r+1}$$ is Noetherian as an $$R$$-module and thus also as a $$K$$-module. But since $$K$$ is a field, this implies $$I_r/I_{r+1}$$ is Artinian as a $$K$$-module, and hence also as an $$R$$-module. By the short exact sequence above we conclude $$I_r$$ is Artinian, completing the induction step.
That proves that if $$R$$ is Noetherian, then $$R$$ is Artinian. The proof of the converse is identical except you just swap the words "Noetherian" and "Artinian" everywhere.