ring satisfying a.c.c. on radical ideals, with nilpotent nilradical and every prime ideal maximal 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
 A: 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.
