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All rings below are assumed to be commutative with unity.

It is well-known that in a Noetherian ring, the nilradical (the ideal of all nilpotent elements , or equivalently, the intersection of all prime ideals) is nilpotent (https://en.wikipedia.org/wiki/Nilpotent_ideal ) .

How much (or even at all) can we go in the converse direction ?

For a more explicit question, can we characterize those rings which satisfy a.c.c. on radical ideals and also the nilradical is nilpotent ?

(the only thing I am able to see is that in a ring satisfying a.c.c. on radical ideals, radical of every ideal is a finite intersection of prime ideals, hence the nilradical also is a finite intersection of prime ideals. )

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    $\begingroup$ $R$ satisfies the acc on radical ideals iff $R/nil(R)$ does, so it's sufficient to characterize reduced rings satsifying acc on radical ideals. At this point, it becomes questionable that the nilpotency of the nilradical matters at all. $\endgroup$ – rschwieb Jan 25 '18 at 17:45
  • $\begingroup$ @rschwieb $R/nil(R)$ need not be a finite product of domains. Consider $k[x,y]/(xy)$ $\endgroup$ – Lukas Heger Jan 29 '18 at 15:34
  • $\begingroup$ @MatheinBoulomenos Ah, I see now another thing that the $0$ dimensionality gave you. I'll delete my comment above and continue discussing it with you on your solution. Using the line of thought I had in mind, with what you wrote in your solution, i guess we only know it is a subdirect product of domains. $\endgroup$ – rschwieb Jan 29 '18 at 15:53
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Using the order-reversing correspondence between radical ideals of $R$ and closed subsets of $\operatorname{Spec}(R)$, acc on radical ideals is equivalent to dcc on closed subsets of $\operatorname{Spec}(R)$, which is by definition equivalent to $\operatorname{Spec}(R)$ being a Noetherian topological space.
Noetherian spaces have finitely many irreducible components, this translates to $R$ having only finitely many minimal prime ideals.

If we assume for a moment that $\operatorname{dim}(R) = 0$, then every prime ideal of $R$ is both maximal and minimal, so in particular, there are only finitely many and we have by the Chinese remainder theorem$$R/J(R) = R/\operatorname{nil}(R)\cong \displaystyle \prod_{\mathfrak{p} \in \operatorname{Spec}(R)}R/\mathfrak{p}$$ where each $R/\mathfrak{p}$ is a field.

This shows that $R/J(R)$ is a (commutative) semisimple ring, because it is a finite product of fields. Since we assume that $J(R)=\operatorname{nil}(R)$ is nilpotent, this means that $R$ is a semiprimary ring. (This notion is more prevalent in noncommutative algebra, it means that $J(R)$ is nilpotent and $R/J(R)$ is semisimple.)

Conversely, if we have any semiprimary commutative ring, then by the nilpotency of $J(R)$, we have $J(R)=\operatorname{nil}(R)$ (which is thus nilpotent) and $R/J(R)$ is a finite product of fields, thus $\operatorname{Spec}(R) \cong \operatorname{Spec}(R/J(R))$ (the isomorphism is as topological spaces) is finite and thus $R$ satisfies the acc on radical ideals. (And furthermore $R$ is $0$-dimensional.)

So we can say that the $0$-dimensional rings which satisfy your requirements are precisely the commutative semiprimary rings. These need not be Noetherian. A commutative counterexample is $k[x_1, x_2, \dots]/I$ where $I$ is generated by $x_ix_j$ for $i,j \in \Bbb N$ ($i=j$ is explicitly allowed) for a field $k$.
But there's an interesting theorem regarding Noetherian modules over semiprimary rings, see Hopkins-Levitzki theorem.

Edit: I'll call rings with the acc on radical ideals Spec-Noetherian for now, because that's the characterization I'll be using in the following a lot.

These are some (relatively) elementary properties of Noetherian spaces:

  • Noetherian spaces have finitely many irreducible components
  • Subspaces of Noetherian spaces are Noetherian
  • If $X=\bigcup_{i=1}^n X_i$ is a finite union of Noetherian spaces, then $X$ is Noetherian
  • $X$ is Noetherian iff every open subspace is quasi-compact

(In particular if $R=\prod_{i=1}^n R_i$ is a finite product of rings, then $R$ is Spec-Noetherian iff each $R_i$ is Spec-Noetherian.)
Using this, we can give another characterization of Spec-Noetherian rings. Open subsets of $X=\operatorname{Spec}(R)$ are of the form $X \setminus V(\mathfrak{a})$ where $\mathfrak{a}$ is a radical ideal. Since $D(f)=X\setminus V(f)$ where $f$ ranges over $R$ is a basis of $X$ consisting of compact open subspaces, the compact open subsets $U$ of $X$ are precisely the sets which may be written as a finite union $U=\bigcup_{i=1}^n D(f_i)$, where $f_i \in R$. (Cf. Atiyah-Macdonald exercise 1.7)

So $R$ is Spec-Noetherian iff for every radical ideal $\mathfrak{a} \subset R$, we can write $X \setminus V(\mathfrak{a}) = \bigcup_{i=1}^n D(f_i) = \bigcup_{i=1}^n (X \setminus V(f_i)) = X \setminus \bigcap_{i=1}^n V(f_i) = X \setminus V(f_1, \dots ,f_n)$, so $V(\mathfrak{a})=V(f_1, \dots f_n)$ which is equivalent to $\mathfrak{a}=\sqrt{(f_1, \dots, f_n)}$

So we get the following characterization of Spec-Noetherian rings, very similar to the characterizations of Noetherian rings.

Lemma A ring $R$ satisfies the acc on radical ideals iff every radical ideal is the radical of a finitely generated ideal.

For Noetherian rings, it is enough to know that every prime ideal is finitely generated. We can show the analogous result for Spec-Noetherian rings. For convenience, I'll use results about Oka families, to show this, since some part of this kind of proofs is always the same. (See here for an introduction))

Let $R$ be a commutative ring and let $\mathcal F$ be the family of ideals $I$ in $R$ such that $\sqrt{I}=\sqrt{(a_1, \dots, a_n)}$ for some $a_1, \dots, a_n \in R$. I claim that $\mathcal F$ is Oka family.
We will use the following fact: If $I,J$ are ideals and $a\in R$, then $\sqrt{IJ}=\sqrt{I} \cap \sqrt{J}$
Suppose $I$ is an ideal of $R$ such that $(I:(a)), I+(a) \in \mathcal F$.
We get $(I+(a)) \cdot (I:(a)) = I(I:(a))+(a)(I:(a)) \subset I$
Thus we get $\sqrt{(I+(a))\cdot (I:(a))} \subset \sqrt{I}$.
On the other hand $I \subset I+(a)$ and $I \subset (I:(a))$, so $I^2 \subset (I+(a))\cdot (I:(a))$, thus $\sqrt{I}=\sqrt{I^2} \subset \sqrt{(I+(a)) \cdot (I:(a))}$
So if we suppose $\sqrt{I+(a)}=\sqrt{(a_1, \dots, a_n)}$ and $\sqrt{(I:(a))}=\sqrt{(b_1, \dots, b_m)}$, we can conclude $\sqrt{I} = \sqrt{(I+(a))\cdot (I:(a)} = \sqrt{(a_1, \dots, a_n)} \cap \sqrt{(b_1, \dots, b_m)} ) = \sqrt{(a_1, \dots, a_n) \cdot (b_1, \dots, b_m)}=\sqrt{(a_1b_1, \dots , a_nb_m)}$
So $I \in \mathcal F$.
Thus $\mathcal F$ is an Oka family and we get the result.

Lemma An ideal maximal with the property that its radical is not the radical of a finitely generated ideal is prime.

Suppose that not every radical ideal of $R$ is the radical of a finitely generated ideal. Then we can use Zorn's lemma to show that the set of radicals which are not the radical ideal of a finitely generated ideal is prime. Indeed, if we have an ascending chain $I_1 \subset I_2 \subset \dots$ of radical ideals which are not the radical of a finitely generated ideal, then $I= \bigcup_{k=1}^\infty I_k$ is a radical ideal. If we had $I=\sqrt{(a_1, \dots a_n)}$, then every $a_i$ would be contained in some $I_k$, thus as we have an ascending chain, $I_k$ would contain all the $a_i$ for $k$ large enough. But then $I_k=I=\sqrt{(a_1, \dots , a_n)}$ which is impossible. Thus be conclude by Zorn's lemma. Combing this with the previous result, we get.

Lemma A ring satisfies the acc on radical ideals iff every prime ideal is the radical of a finitely generated ideal.

If we use the fact that $\operatorname{Spec}(R) = \bigcup_{\mathfrak{p} \in \operatorname{SpecMin}(R)} V(\mathfrak{p})$ is the decomposition into irreducible components, where $\operatorname{SpecMin}(R)$ is the set of minimal prime ideals, we get the following:

Lemma A ring $R$ is Spec-Noetherian, iff $R$ has finitely many minimal prime ideals and for every minimal prime ideal $\mathfrak{p}$ $R/\mathfrak{p}$ is Spec-Noetherian.

But $R/\mathfrak{p}$ is an integral domain so we have reduced the problem in some sense to integral domains.

There's a somewhat "dual" criterion that is only sufficient in general

Lemma If $R$ has only finitely many maximal ideals, then $R$ is Spec-Noetherian iff $R_{\mathfrak{m}}$ is Spec-Noetherian for each maximal ideal $\mathfrak{m}$

Proof: $\operatorname{Spec}(R) = \bigcup_{\mathfrak{m} \in \operatorname{SpecMax}(R)} \operatorname{Spec}(R_{\mathfrak{m}})$, so if this union is finite, we can apply the property of Noetherian spaces.

To see how this fails for rings with infinitely many maximal ideals, consider non-Noetherian von-Neumann-regular ring, e.g. an infinite product of fields.

For integral domains, I have no idea how to completely characterize those that are Spec-Noetherian, but here are some ideas for a special class of integral domains:

If $R$ is a valuation ring, then the ideals of $R$ are totally ordered by inclusion, so every radical ideal, being the intersection of a totally ordered chain of prime ideals is prime. Furthermore, since all the prime ideals form one chain, we get this easy lemma:

If $R$ is a valuation ring, then the following are equivalent:

  • $R$ is Spec-Noetherian
  • The Krull dimension of $R$ is finite
  • $\operatorname{Spec}(R)$ is finite
  • $\forall \mathfrak{p} \in \operatorname{Spec}(R)$ $\mathfrak{p}=\sqrt{(a)}$ for some $a \in R$

(The last point follows from the previous lemma on radical ideals, because every finitely generated ideal in a valuation ring is prinicpal.)

This makes it easy to construct examples of non-Noetherian Spec-Noetherian local integral domains of any given finite (non-zero) Krull dimension, because we can control the Krull dimension of $R$ by choosing the appropriate value group. (See for example here) And we can also construct local integral domains which are not Spec-Noetherian this way.

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    $\begingroup$ Instead of adopting that $\dim(R)=\{0\}$, how far can you get with the fact that $R/nil(R)$ is isomorphic to a finite product of domains? After all, the general problem is reduced to studying $R/nil(R)$. It seems each of the domains must satisfy the ACC on radical ideals as well. So it looks like the whole problem is reduced to domains. But I know next to nothing about the spectra of domains. Perhaps the problem is still not easy even with that assumption? $\endgroup$ – rschwieb Jan 29 '18 at 14:24
  • $\begingroup$ @rschwieb That might be something to try out. By Prop 2.20 in Eisenbud a Noetherian ring is a finite product of domains iff the localization at each maximal ideal is a domain, but his proof works for any ring with finitely many minimal primes. $\endgroup$ – Lukas Heger Jan 29 '18 at 15:31
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    $\begingroup$ I forgot that you'd need $0$ dimensionality to say that the map was onto :) So the best we have is that $R/nil(R)$ is a subdirect product of a finite number of domains. I haven't checked to see if this implies anything about the acc on radical ideals of the domains yet, but that would be the next logical thing to do. $\endgroup$ – rschwieb Jan 29 '18 at 15:54
  • $\begingroup$ @rschwieb Thanks for your suggestions. I have made some progess and reduced the problem to integral domains in another way. $\endgroup$ – Lukas Heger Jan 29 '18 at 19:44
  • $\begingroup$ Glad it helped... thanks for showing me something new $\endgroup$ – rschwieb Jan 29 '18 at 20:35

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