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Does there exist any term (or, maybe, a "description"?) for commutative unital noetherian rings such that their Jacobson ideals are prime (and so, their maximal spectra are irreducible)? What is the relation of this condition to the primality of the nilradical?

Upd. Sorry; I was really stupid! Actually, the primality of the Jacobson radical is "independent from" that of the nilradical. Indeed, there exist local rings whose nil radical is not prime (see Konstantin Ardakov's example); on the other hand, a "semi-localization" of a indecomposable regular ring at any two closed points (take $R=k[X]$ and invert all polynomials whose values in $0$ and $1$ are non-zero) is regular indecomposable but its maximal spectrum consists of two closed points!

So, I am rather interested in the following question: for which $R$ its Jacobson radical contains a prime ideal? My collegues has told me that this is equivalent to the fact that the closed points of $Spec R$ lie in some single irreducible component of it, and there is a paper https://www.math.hmc.edu/~henriksen/publications/1977_Henriksen_Some_sufficient_conditions_for_the_Jacobson_radical_of_a_commutative_ring_with_identity_to_contian_a_prime_ideal.pdf on this question. Yet any further comments would be very welcome!

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  • $\begingroup$ Is (maximal) here optional in this question or not? The spectrum is connected precisely when the ring is not a non-trivial cartesian product. The nilradical is prime precisely when the spectrum is irreducible. So, you have two different topological properties here. $\endgroup$
    – M.G.
    Commented Nov 28, 2015 at 18:11
  • $\begingroup$ (cont.) For example, connectedness could imply irreducibility (i.e. when you have non-singularity and Noetherian). Not sure about more general than that. However, conversely, irreducible (=hyperconnected) implies connected. $\endgroup$
    – M.G.
    Commented Nov 28, 2015 at 18:19
  • $\begingroup$ I am terribly sorry! I am interested in irreducibility. $\endgroup$
    – Mikhail Bondarko
    Commented Nov 28, 2015 at 18:33
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    $\begingroup$ Yes; I was merely correcting your sentence "and so, their spectra are irreducible" by showing an example of a ring with prime Jacobson radical but reducible spectrum. $\endgroup$
    – user91132
    Commented Nov 29, 2015 at 13:35
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    $\begingroup$ It can happen that the maximal spectrum is disconnected but the prime spectrum is connected. Example: take the localisation of the integers at the ideal (6) (i.e. invert all integers that are coprime to 6). The resulting ring $A$ is still a domain so its prime spectrum is connected, but $A / J(A)$ is the direct product of two fields, so the maximal spectrum is disconnected. Explicitly, $3$ and $4$ map to non-trivial idempotents in $A / J(A)$. My earlier example ($k[[x,y]]/(xy)$) gives a Noetherian ring whose prime spectrum is disconnected and whose maximal spectrum is connected. $\endgroup$
    – user91132
    Commented Nov 30, 2015 at 9:07

1 Answer 1

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Let $A$ be a commutative ring with unity. Denote $X:=Spec(A)$.

  1. $X$ is connected if and only if $A$ cannot be written as a Cartesian product $A_1\times A_2$ for nontrivial $A_1$ and $A_2$. So, that gives you some description of $A$ (but I don't know if it is any useful to you).

Now, the nilradical of $A$ is prime if and only if $X$ is irreducible (aka. hyperconnected).

  1. Relation between irreducible and connected: hyperconnected always implies connected for set-theoretical topological reasons. Since you've imposed the Noetherian condition in your edit, for the inverse implication it suffices to additioinally assume that $X$ is smooth, that is, $A$ is a regular ring. I don't know if there is a more general condition that yields the inverse implication.
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  • $\begingroup$ Thank you! Yet it seems that connectedness and regular rings are of no interest for my purposes. $\endgroup$
    – Mikhail Bondarko
    Commented Nov 28, 2015 at 23:34
  • $\begingroup$ Oh sorry, I had "cached" in my mind the very first version of your question, where you mentioned connected spectra. $\endgroup$
    – M.G.
    Commented Nov 29, 2015 at 8:00
  • $\begingroup$ This version of the question also made some sense; you certainly do not have to apologize! $\endgroup$
    – Mikhail Bondarko
    Commented Nov 29, 2015 at 17:46
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    $\begingroup$ @MikhailBondarko The smoothness seems to be a necessary condition. Consider $A = k[x,y]/(xy)$. Then Spec A is connected but reducible. $\endgroup$
    – David Benjamin Lim
    Commented Nov 30, 2015 at 6:37
  • $\begingroup$ As for connectedness, algebraically it is just absence of nontrivial idempotents. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Nov 30, 2015 at 7:23

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