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A commutative ring $R$ is called Noetherian if any one of the following holds:

$1.$ Every ascending chain of ideals in $R$ stabilizes, that is, $$ I_1\subseteq I_2\subseteq I_3\subseteq\cdots $$ implies the existence of $n\in\mathbb{N}$ such that $I_n=I_{n+1}=I_{n+2}=\cdots$.

$2.$ Every ideal in $R$ is finitely generated.

But I. S. Cohen proved that for a ring to be Noetherian it suffices that every prime ideal is finitely generated. So for a ring to be Noetherian we need to check above condition $2$ only for prime ideals.

So the natural question is there any class of ideals for choosing ideals $I_i$ of Ascending Chain Condition for proving that $R$ is Noetherian?

Here are two cases which does not work:

$1.$ Every ascending chain of prime ideals terminates even $R$ is not noetherian. So we can not choose $I_i$ only from Spec($R$).

Example: $R=\dfrac{k[x_1,x_2,...]}{I^2}$, where $I=(x_1,x_2,...)$

$2$. Every chain of primary ideals terminates even $R$ is not Noetherian.

Example: $\prod_{i\in\Bbb N} F_i$ where the $F_i$ are fields.

This is clearly not Noetherian, and because it is commutative and von Neumann regular, all of its primary ideals are maximal.

Any references/ideas?

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1 Answer 1

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Claim: If every ascending chain of finitely generated ideals of $R$ stabilizes, then $R$ is Noetherian.

Proof: Let $I$ be an ideal of $R$. Setting $I_0 = (0)$, let us inductively construct ideals $I_n$ as follows. If $I_{n-1} \subsetneq I$, then choose $x_n \in I \setminus I_{n-1}$, and define $I_n = I_{n-1}+(x_n) = (x_1,\dots,x_n)$. We obtain a strictly ascending chain of finitely generated ideals $$0 \subsetneq (x_1) \subsetneq (x_1,x_2) \subsetneq \cdots.$$ By hypothesis this must terminate, so that some $(x_1,\dots,x_n) = I$. So $I$ is finitely generated, proving that $R$ is Noetherian.

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