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