All rings are commutative and unital.

Suppose that $A$ is a ring in which the zero ideal can be written as a product of maximal ideals of $A$.

I try to prove that $A$ is Noetherian if and only if it is Artinian:

Proof. We begin by recalling the following fact:

Fact 1. Let $M$ be an $R$-module and $N$ is a submodule of $M$. Then $M$ is Noetherian/Artinian if and only if both $M/N$ and $N$ are Noetherian/Artinian.

Suppose $(0)=\mathfrak{m}_1\ldots\mathfrak{m}_k$ where the $\mathfrak{m}_i$ are maximal. Then consider the chain $$(0)=\mathfrak{m}_1\ldots\mathfrak{m}_k\subseteq\ldots\subseteq\mathfrak{m}_1\mathfrak{m}_2\subseteq\mathfrak{m}_1\subseteq A.$$

For each $1<i\leq k$, let $M_i = \mathfrak{m}_1\ldots\mathfrak{m}_{i-1}/\mathfrak{m}_1\ldots\mathfrak{m}_i$ as an $A$-module. Note that $\mathfrak{m}_i\subseteq\text{Ann}(M_i)$ and so, for each $1<i\leq k$, we can regard $M_i$ as an $A/\mathfrak{m}_i$-vector space with scalar multiplication defined by $$(a+\mathfrak{m}_i)(x+\mathfrak{m}_1\ldots\mathfrak{m}_i)=ax+\mathfrak{m}_1\ldots\mathfrak{m}_i.$$ It follows that $M_i$ is Artinian as an $A/\mathfrak{m}_i$-module if and only it is Noetherian as such. However, the $A/\mathfrak{m}_i$-submodules of $M_i$ are exactly the same as the $A$-submodules of $M_i$ and so $M_i$ is Artinian as an $A$-module if and only if it is Noetherian as an $A$-module. But each $M_i$ is Artinian/Noetherian (as an $A$-module) if and only if $A$ is a Artinian/Noetherian ring (by repeated application of Fact 1). Hence $A$ is an Artinian ring if and only if it is a Noetherian ring. //

I would be very grateful if someone could tell me whether this proof is correct or not.

Many thanks.

  • 2
    $\begingroup$ The same proof can be found in Atiyah-MacDonald, corollary 6.11. $\endgroup$ – Ayman Hourieh Dec 16 '14 at 21:17

Yes, your proof is correct; it's essentially the canonical proof.


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