A commutative ring $R $ with 1 is called semiprimitive if its Jacobson radical is the zero ideal. Is there any characterization for zero-dimensional semiprimitive commutative rings?
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$\begingroup$ Since this will include all finitely generated algebras over a field which are integral domains (a class which this site is mostly concerned about), a non-trivial characterization would mean most of us can go home. $\endgroup$– MohanCommented Mar 1, 2017 at 16:57
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2 Answers
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Several nice characterizations for these rings are worked out as Exercise 4.15 in Lam's book "Exercises in Classical Ring Theory." These include (for commutative rings):
(A) $R$ is reduced and $K$-dim $R$=0.
(B) $R$ is von Neumann regular.
(C) The localizations of $R$ at its maximal ideals are all fields.
answered Mar 1, 2017 at 17:49
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These are exactly the zero-dimensional reduced commutative rings (a.k.a. "absolutely flat rings"). Clearly semiprimitive rings are reduced.
[EDIT: what follows is correct but much too complicated. See the comment by Luc Guyot.]
Conversely, assume $R$ is zero-dimensional and reduced. For every $\mathfrak{p}\in\mathrm{Spec}(R)$, $R_\mathfrak{p}$ is local, zero-dimensional and reduced, hence $\mathfrak{p}R_\mathfrak{p}=0$ (in other words, $R_\mathfrak{p}$ is a field). So if $x$ is in the Jacobson radical of $R$, then $x$ is zero in every $R_\mathfrak{p}$, hence $x=0$.
answered Mar 1, 2017 at 17:42
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$\begingroup$ If $R$ is zero-dimensional, then its primes are maximal so that the nilradical of $R$ coincides with the Jacobson radical of $R$. For this reason, I don't understand the need for "Conversely". Am I missing something? $\endgroup$ Commented Mar 1, 2017 at 21:54
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$\begingroup$ @LucGuyot: You are right! $\endgroup$ Commented Mar 2, 2017 at 6:59