It is known that Artinian (commutative) ring has finitely many maximal ideals.

(One proof is by considering finite intersections of maximal ideals, and a chain from it, then using Artinian property of ring and some lemma.)

I have one problem in following arguments given in Mastumuta's Ring theory.

Let $\mathfrak{p}_1,\mathfrak{p}_2\cdots$ be infinite family maximal ideals of Artinian ring $R$.

Then the chain $$\mathfrak{p}_1 \supset \mathfrak{p}_1\mathfrak{p}_2 \supset \cdots$$ is strictly decreasing (why it is strictly decreasing?)

I tried to use Nakayama lemma, but we do not know whether ideals are finitely generated - since at this argument of Matsumuta, it is not known that ideals are finitely generated. (At the end of proof, it is proved that ideals are finitely generated.)

In short, I want to prove that above chain of ideals is strictly decreasing, without assuming them to be finitely generated. How can we prove it?


If $\mathfrak p_1\cdots \mathfrak p_k=\mathfrak p_1\cdots \mathfrak p_{k-1}$, then $\mathfrak p_1\cap\cdots\cap \mathfrak p_k=\sqrt{\mathfrak p_1\cdots \mathfrak p_k}=\sqrt{\mathfrak p_1\cdots \mathfrak p_{k-1}}=\mathfrak p_1\cap\cdots\cap \mathfrak p_{k-1}$. In other words, $\mathfrak p_k$ is a prime ideal containing $\bigcap_{i=1}^{k-1} \mathfrak p_i$. But this is the case if and only if $\mathfrak p_k\supseteq \mathfrak p_j$ for some $j\le k-1$.

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    $\begingroup$ You can conclude more directly from $\mathfrak p_1\dotsm \mathfrak p_{k-1} = \mathfrak p_1\dotsm \mathfrak p_k \subseteq \mathfrak p_k$ that $\mathfrak p_j\subseteq \mathfrak p_k$ for some $1\le j\le k-1$. $\endgroup$ – Claudius Mar 15 '18 at 14:04
  • $\begingroup$ @Claudius - I suggest that you turn you comment into an answer. $\endgroup$ – Pierre-Yves Gaillard Mar 15 '18 at 22:21
  • $\begingroup$ @Claudius Yes, I guess I could have considered the radical of both members of that inclusion. It did not occur to me. $\endgroup$ – user228113 Mar 15 '18 at 23:07

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