# Intersection of maximal ideals contains an ideal

Let $R$ be a commutative unitary ring and $M_{I}$ be the intesection of all maximal ideals contains $I$. Question: When for any two ideals $I$ and $J$ of $R$ there exists an ideal $K$ of $R$ such that $M_{I}+M_{J}=M_{K}$?

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## 1 Answer

A typical ring of dimension $2$, like $k[x,y]$, does not have that property. Indeed, if $I=(xy)$, $J=(x^2-y^2)$, then $M_I=I$, $M_J=J$, and $M_I+M_J=I+J$ contains $x^3$ and $y^3$ but not $x$ and $y$, so is not radical, so cannot be the intersection of any set of maximal ideals.

On the other hand, in a Dedekind domain, ideals of this form are exactly products of finitely many maximal ideals or the zero ideal. Scheme-theoretically, they correspond to finite sets of points on a curve or else the whole curve. This set is closed under addition, or, scheme-theoretically, intersection, so you're OK there.

Obviously, this also works in every local ring, where there is a unique maximal ideal.

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Ok, there are many examples which this is not true. But I think in J-semisimple ring (ring in which $J(R)=0$), has a topological equivalent in the space of maximal ideals with zariski topology. –  Ali Dec 10 '12 at 4:04