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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, 2018 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, 2018 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, 2018 at 15:53

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