Is every Noetherian Ring an integral domain?

Question

And if this is so, could you please provide an example of an integral domain which is not a noetherian ring, and also a noetherian ring which is no unique factorization domain?

This would be really nice. Also I am asking this for self-study, currently I'm trying to understand how all the algebraic structures relate to one another.

Hint: The product of two Noetherian rings is again a Noetherian ring. — Tobias Kildetoft, Apr 14 '16 at 12:16
Second question: if $A$ is noetherian, then any quotient of $A[x]$ is noetherian. So the ring $\Bbb Z[x]/(x^2+5)$ is a noetherian integral domain which is not a UFD. — , Apr 15 '16 at 02:58

score 7 · Answer 1 · answered Apr 14 '16 at 12:23

7

No, a finite ring like $Z/n$ is always noetherian but not integral if $n$ is not a prime.

answered Apr 14 '16 at 12:23

Tsemo Aristide

87,475

score 3 · Answer 2 · answered Apr 14 '16 at 12:30

Integral domain, which is not noetherian:

$k[X_1, X_2, \dotsc]$ or a little bit more interesting: $R := \{f \in k[X,Y] ~|~ f(X,0) \in k \} \subset k[X,Y]$.

Any noetherian non-domain is in particular not an unique factorization domain. Here is an example for a noetherian domain, which is not UFD: $\mathbb Q[X,Y]/(X^2+Y^2-1)$. We have $X^2=(1+Y)(1-Y)$ in this ring, contradicting unique factorization.

Is every Noetherian Ring an integral domain?

2 Answers2