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"Let $A$ be a ring in which the zero ideal is the product of the maximal ideals $\mathfrak{m}_i, \, i=1, \cdots, r$. Then $A$ is Noetherian if and only if $A$ is Artinian."

This is Corollary 6.11, p. 78 from Atiyah and McDonald's "Introduction to Commutative Algebra". Their proof is quite elegant. I would like to share my approach for the proof, and some phenomenon that arises there.

Suppose that $A$ is Noetherian. I will show that $A$ is Artinian. So take a decreasing sequence of ideals $\alpha_0 \supset \alpha_1 \supset \cdots$. Adding to each term of the sequence the maximal ideal $\mathfrak{m}_i$, we obtain a new decreasing sequence of ideals of $A$ given by $\alpha_0 + \mathfrak{m}_i \supset \alpha_1 + \mathfrak{m}_i \supset \cdots$. Now take the image of this sequence under the canonical map $A \rightarrow A / \mathfrak{m}_i$ to obtain a decreasing sequence of $A$-submodules of the $A$-module $A /\mathfrak{m}_i$ given by $\frac{\alpha_0 + \mathfrak{m}_i}{\mathfrak{m}_i} \supset \frac{\alpha_1 + \mathfrak{m}_i} {\mathfrak{m}_i} \supset \cdots$. But $A /\mathfrak{m}_i$ is also a vector space over itself and since $A$ is Noetherian so will be $A /\mathfrak{m}_i$. But for vector spaces Noetherian and Artinian are equivalent notions, thus the decreasing sequence $\frac{\alpha_0 + \mathfrak{m}_i}{\mathfrak{m}_i} \supset \frac{\alpha_1 + \mathfrak{m}_i} {\mathfrak{m}_i} \supset \cdots$ must terminate, i.e. there exists some $N_i$ s.t. $\alpha_{N_i} - \alpha_{N_i + k} \subseteq \mathfrak{m}_i$ for any positive integer $k$. Repeating the same argument for all $i=1,\cdots,r$ and setting $N=\max_{i}\left\{N_i \right\}$ we have that $\alpha_{N} - \alpha_{N + k} \subseteq \prod_{i=1}^r \mathfrak{m}_i=0$ and so $\alpha_N=0$. Thus the initial sequence for $A$ must terminate and so $A$ is Artinian.

My question is twofold: 1) is the above argument technically correct? 2) If yes, how do we interpret the fact that the sequence terminates at the zero ideal?

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Is $\alpha_{N} = 0$ the only possible conclusion? –  Chris Leary Aug 21 '12 at 20:02
    
@ChrisLeary: I think so. –  Manos Aug 22 '12 at 1:06
    
@Manos-How about $\alpha_{N} = \alpha_{N+k}$ for all $k$? –  Chris Leary Aug 22 '12 at 3:20
    
@ChrisLeary: How can you prove that? –  Manos Aug 22 '12 at 13:26
    
@Manos-Their difference is $0$. –  Chris Leary Aug 22 '12 at 17:28
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