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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.

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

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