When does a Noetherian ring becomes an Integral domain and vice versa?

Question

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?

Answer 1

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.