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Which result is closest to the classical

General Hilbert's Nullstellensatz: Finite type algebras over Jacobson rings are Jacobson.

and constructively true at the same time? And where can I find a proof of it?

I skimmed through the books by Lombardi/Quitté and Mines/Richman/Ruitenberg but there seems to be no discussion of Jacobson ring-related methods.

The usual definition of Jacobson rings relies heavily on prime ideals so perhaps one should find a suitable constructive version of the definition of a Jacobson ring first. I'm curious for your ideas.

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  • $\begingroup$ Are these rings commutative? $\endgroup$
    – darij grinberg
    Commented Jul 18, 2017 at 16:03
  • $\begingroup$ Yes, they are. dummy characters $\endgroup$
    – Jakob Werner
    Commented Jul 18, 2017 at 16:05
  • $\begingroup$ I'd be curious to know the meaning of "being constructively true at the same time". In particular, do you want something uniform in the ground Jacobson ring? $\endgroup$
    – YCor
    Commented Jul 18, 2017 at 16:13
  • $\begingroup$ "one should find a suitable constructive version of the definition of a Jacobson ring first": Actually, there is one on the Wikipedia page: "In every quotient ring, the nilradical is equal to the Jacobson radical". The Jacobson radical has several equivalent constructive definitions (see mathoverflow.net/questions/57877/… ). $\endgroup$
    – darij grinberg
    Commented Jun 8, 2018 at 20:27
  • $\begingroup$ @darijgrinberg You are right. However »every quotient ring« seems to be very strong, constructively. For example consider the statement that $ \mathbb{Z} $ is Jacobson. If $ I = \langle n \rangle$ is principal, then it is rather easy to prove that $ \operatorname{rad}(I) = \operatorname{Jac}(I) $. (Reduce to the case that $ n $ is prime and use that $\mathbb{Z}/n$ is a field. I think this should work intuitionistically.) If however $I$ is not assumed to be principal (or equivalently, finitely generated), I don't even know how to start. $\endgroup$
    – Jakob Werner
    Commented Jun 10, 2018 at 13:49

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If you define a Jacobson ring to be a ring $A$ such that $\sqrt{I}=\mathrm{Jac}(I)$ for every ideal $I$ of $A$ as suggested by Darij Grinberg, then the general Nullstellensatz holds constructively. It is also provable that $\mathbb{Z}$ is Jacobson. The proof is in this preprint of mine: A constructive proof of general Nullstellensatz for Jacobson rings, arXiv:2406.06078 [math.AC].

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