# 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?

• Welcome to Maths SX! These notions are independent. There exist non-noetherian integral domains and noetherian rings with zero-divisors. Jul 25, 2021 at 12:36
• 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. Jul 25, 2021 at 17:24

## 1 Answer

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.

• +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.) Jul 25, 2021 at 17:26