All rings below are assumed to be commutative with unity.

If a ring satisfies a.c.c. on radical ideals, every prime ideal is maximal , and the nilradical is nilpotent, then is it true that the ring is Artinian , or equivalently, is the ring Noetherian ?

Please help. Thanks in advance

  • 1
    $\begingroup$ I gave you an example of a non-Noetherian ring with a unique prime ideal and nilpotent nilradical yesterday. In fact it is exactly the same as the solution currently given below (I have no doubt it was conceived completely parallel to mine. It's a fairly natural choice.) Perhaps pay closer attention in the future? $\endgroup$ – rschwieb Jan 27 '18 at 12:57
  • $\begingroup$ @rschwieb sorry for giving the same example as you did, I didn't see your answer. $\endgroup$ – Lukas Heger Jan 27 '18 at 13:45
  • $\begingroup$ @MatheinBoulomenos of course, don’t worry about it! As I said it was completely parallel. I am more disappointed that the user does not apparently make use of solutions they receive. $\endgroup$ – rschwieb Jan 27 '18 at 16:28

The following example is taken from this answer:

Let $k$ be a field and $V$ be an infinite-dimensional vector space.

We can define a multiplication on $R=k \oplus V$ by $(a,v)(b,w)=(ab,aw+bv)$, this turns $R$ into a commutative ring. The nilradical of $R$ consists of all elements of the form $(0,v), v \in V$. Since we have $(0,v)(0,w)=(0,0)$, $\operatorname{nil}(R)^2=0$, so the nilradical is nilpotent.
Now $R/\operatorname{nil}(R)\cong k$, so $\operatorname{Spec}(R)$ has only one point. In particular $R$ is zero-dimensional and since the radical ideals of $R$ correspond bijectively to closed sets in $\operatorname{Spec}(R)$, $R$ satisfies the ACC on radical ideals.

Let $V_1 \subset V_2 \subset \dots$ be an infinite ascending chain of subspaces of $V$, then $0 \oplus V_1 \subset 0 \oplus V_2 \subset \dots$ is an infinite ascending chains of ideals, so $R$ is not Noetherian.

Edit: In the comments, you asked for an example such that $R/I$ is nilpotent for every ideal $I$. Since the nilradical of $R/I$ is $\sqrt{I}/I$, this is equivalent to asking that for every ideal $I$ there is an $\Bbb n \in \Bbb N$ such that $\sqrt{I}^n \subset I$. If we look at the example of $R$ above, we saw that $R$ has only one prime, which is the (nilpotent) nilradical. If we have any ideal $I$, then if $I=R$, we have $\sqrt{I}=I$, so the condition is satisfied. If $I$ is a proper ideal, then the radical of $I$ is the intersection of the primes containing it, so $\sqrt{I}=\operatorname{nil}(R)$, because $\operatorname{nil}(R)$ is the only prime. But then $\sqrt{I}^2=0 \subset I$.

It's worth pointing out that the rings satisfying the conditions of your question are precisely the commutative semiprimary rings, see my answer to your similar question here.

| cite | improve this answer | |
  • $\begingroup$ Do you think there is a counter example if we also impose the condition that the nilradical of $R/I$ is nilpotent for every ideal $I$ ? $\endgroup$ – user495643 Jan 27 '18 at 16:33
  • 1
    $\begingroup$ @misao the counterexample I gave already has this property. I edited in a justification. $\endgroup$ – Lukas Heger Jan 27 '18 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy