All rings below are commutative with unity.

If $R$ has a.c.c. on radical ideals i.e. if $Spec R$ is Noetherian under Zariski topology, then so is $R[X]$, this is Theorem 2.5 in the following paper by Ohm and Pendleton

https://projecteuclid.org/euclid.dmj/1077377788 .

My question is : If $R$ satisfies a.c.c. on radical ideals, then is it true that so does $R[[X]]$ ? If this is not true in general, is there any additional hypothesis on $R$ that would force this ?

  • $\begingroup$ I have no time for a closer look at the moment, but the examples at the end of J. T. Arnold, Krull dimension in power series rings, Trans. Amer. Math. Soc. 177, 299-304 might be of interest. $\endgroup$ – Fred Rohrer Feb 19 at 21:22
  • $\begingroup$ In the aforementioned paper it is proved the following: If $V$ is a rank one non-discrete valuation ring, then $V$ has Noetherian spectrum, while $V[[X]]$ has an infinite chain of prime ideals, and therefore it does not have Noetherian spectrum. $\endgroup$ – user26857 Mar 4 at 19:04

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