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Does there exist a commutative ring with unity such that the pairwise intersection of distinct maximal ideals is the Jacobson radical?

Ie. if $M_1, M_2$ are any pair of distinct maximal ideals then $M_1 \cap M_2 = J(R)$.

And if that is false, is there a ring, $R$ and $a$ such that any collection of maximal ideals $|M| \geq a$ satisfies $\bigcap M = J(R)$.

Or more weakly is there a subset of maximal ideals such that the above is true even if it is not true for every maximal ideal.

I know the case where $a$ is finite fails when $R$ is semiprimitive and is trivially true for rings with at most $a$ maximal ideals but I am unable to find an answer otherwise.

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  • $\begingroup$ Yes, that' what pairwise means. $\endgroup$ – user661541 Aug 31 '19 at 17:36
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If there exists a set of $n$ distinct maximal ideals intersecting to $J(R)$, then the Chinese remainder theorem says that $R/J(R)$ is isomorphic to $n$ fields, and such a ring has exactly $n$ maximal ideals.

So if just one set of $n$ exists, they are precisely the entire set of maximal ideals.

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  • $\begingroup$ So in a sense there are only trivial examples . $\endgroup$ – rschwieb Sep 1 '19 at 2:11

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