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When does a Noetherian ring becomes an Integral domain and when does an Integral domain becomes Noetherian ring? That is what are the necessary conditions?

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    $\begingroup$ Welcome to Maths SX! These notions are independent. There exist non-noetherian integral domains and noetherian rings with zero-divisors. $\endgroup$
    – Bernard
    Jul 25, 2021 at 12:36
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    $\begingroup$ Her P and Q are pretty much unrelated, the answer to “when will a P ring be a Q ring?” will be “when it is already very close to a Q ring”. The solution below is a pretty good example, where most ideas are assumed to be finitely generated already. Honestly I know of no very interesting link. The most interesting theorems I can think of about when a ring is Noetherian are the Cohen-Kaplansky type theorems, and also the one in terms of when all sums of injectives and injective. $\endgroup$
    – rschwieb
    Jul 25, 2021 at 17:24

1 Answer 1

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Let $R$ be an integral domain. If every non-radical ideal of $R$ is finitely generated, then every ideal of $R$ is also finitely generated, i.e. the ring is Noetherian. So in particular, if every non-prime ideal in an integral domain is finitely generated, then the domain is Noetherian.

Proof :

Let $R$ be an integral domain. Let $0 \ne r ∈ R$ and $I$ be a proper ideal of $R$. Then r $\notin rI$. Because if $r ∈ rI$, then $r = ri$ for some $i ∈ I$ and then $0 \ne r$ and $R$ is a domain that implies $1 = i ∈ I$, contradicting $I$ is proper.

Also, since $R$ is a domain, via the natural map sending every $j ∈ I$ to $rj ∈ rI$, we have $rI ≈ I$ as $R$-modules, for every $0 \ne r ∈ R$.

Now let $0$ $\ne J$ be any proper ideal of $R$. Let $0 \ne x \in J$. Since $J$ is a proper ideal, so $x \notin xJ$.

But $x^2 \in xJ$. Thus $xJ$ is not a radical ideal of $R$, hence $xJ$ is finitely generated. Then due to $xJ≈ J$ as R-modules, $J$ is also finitely generated. Since $J$ was an arbitrary ideal, this proves the claim.

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    $\begingroup$ +1 because this is novel to me. I note also that “domain” is not necessary for this argument. I think all you need is for each nonzero proper ideal to contain a regular element (there are many such rings that aren’t domains.) $\endgroup$
    – rschwieb
    Jul 25, 2021 at 17:26

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